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# Quantitative Aptitude Questions (Boats and Streams) for IBPS Clerk Mains Day-15
## Quantitative Aptitude Questions (Boats and Streams) for IBPS Clerk Mains Day-15:
Dear Readers, IBPS is conducting Online Mains Examination for the recruitment of Clerical Cadre. Main Examination of IBPS Clerk was scheduled from 21st Jan 2018. To enrich your preparation here we have providing new series of Boats and Streams Questions – Quantitative Aptitude Questions. Candidates those who are appearing in IBPS Clerk Main Exam can practice these Quantitative Aptitude Average questions daily and make your preparation effective.
Practice Aptitude Questions (Boats and Streams) Day-15
maximum of 10 points
### Click “Start Quiz” to attend these Questions and view Solutions
1. A boat covers 12 km upstream and 18 km downstream in 6 hours, while it covers 18 km upstream and 12 km downstream in 7 hours. Find the speed of the current.
1. 3.285 km/hr
2. 2.528 km/hr
3. 3.723 km/hr
4. 2.083 km/hr
5. 4.625 km/hr
1. If a man can row 14.6 km downstream in 12 minutes and his rowing speed in still water is 60 km/hr, how much distance can he cover upstream in 20 minutes(in km)?
1. 20
2. 14.33
3. 12.66
4. 14.66
5. 15.66
1. A boat can cover 9.6 km upstream in 24 minutes. If the speed of the current is 1/4th of the boat in still water, then how much distance( in km) can the boat cover downstream in 36 minutes?
1. 36 km
2. 24 km
3. 48 km
4. 32 km
5. None of these
1. A man rows to a place 48 km apart and back in 16 hours. He finds that he can row 6 km downstream and 4 km upstream in the same time. The speed of the stream is?
1. 1.25 km/hr
2. 3 km/hr
3. 2 km/hr
4. 1.8 km/hr
5. None of these
1. Raji can swim at 10 km/hr in still water. The river flows at 3 km/hr and it takes 12 hours more upstream than downstream for the same distance. How far is the place?
1. 182 km
2. 164 km
3. 176 km
4. 160 km
5. 175 km
1. The speed of a boat in still water is 18 km/hr and the speed of the stream is 3 km/hr. It takes a total 11 hours to row upstream from point X to Point Y and downstream from Point Y to Point Z. If the distance from X to Y is one third of the distance between Y and Z. What is the total distance travelled by the boat(both upstream and downstream)?
1. 190 km
2. 180 km
3. 210 km
4. 170 km
5. 105 km
1. A man can row a boat at a speed of 10 km/hr in still water. He goes to a certain point upstream and back to the starting point in a river. The speed of the flowing water is 4 km/hr. What is the average speed of the boat for that journey?
1. 9.2 km/hr
2. 8.4 km/hr
3. 9.6 km/hr
4. 7.2 km/hr
5. None of these
1. A boat covers 24 km in 6 hours along the stream and 10 km in 5 hours again the stream. What is the speed of the stream and speed of still water?
1. 3 Km/hr,1 Km/hr
2. 4 Km/hr,1.5 Km/hr
3. 2.5 Km/hr, 1 Km/hr
4. 4 Km/hr,1.5 Km/hr
5. None of these
1. A boat running upstream takes 8 hours 48 minutes to cover a certain distance. While it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water respectively?
1. 2.1
2. 3:2
3. 8:3
4. Can’t be determined
5. None of these
1. A boat covers a certain distance downstream in 1 hour, while it comes back in 1 and half hours. If the speed of the stream be 3 km/hr, What is the speed of the boat in still water?
1. 14 km/hr
2. 15 km/hr
3. 13 km/hr
4. 12 km/hr
5. None of these
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https://thepronotes.com/c-program-to-generate-random-number-using-middle-square-method/
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C Program to Generate Random Number Using Middle-Square Method
The pseudo-random numbers can be generated from Middle-Square Method. In this post, we will look at the way the middle square method works and will also look at its Code in C.
Implementation of Mid-Square Method:
The mid-square method works by taking the middle term of the squared number of the seed. Leading 0 can be added if necessary.
Example of the random number generation using Mid-Square Method:
Let’s take a seed n=62 Then, n2=3844
Now, taking the middle two-term the new seed is n=84. (i.e. We removed the first and last digit to get the middle term.
Now, new seed n=84, n2=7056,
so the new seed is n=5
seed(n)=25,
n2=625
Then to get the 2-digit seed from the square, we can add leading 0 as below:
n2=0625,
So, seed(n) =62, n2=3844,
Then the new seed becomes 84 and so on.
Mid Square Method Doesn’t Work for Odd Length Seed.
The mid-square method for random number generation only works if the number of digit of the seed is even. It is because if the seed is of odd length, then from its square, not always, the exact middle term of the same length of seed cannot be extracted.
For example: when seed of odd length is taken
For example, let’s take a number of odd length n=674
It has digit length 3, now n2=454,276
As you can see the middle term having 3 digits cannot be extracted from here.
Explanation of Code:
Before I head into code, I just want to explain what I have done in the following Code.
1) A seed is taken and the number of digits in the seed is calculated as if it has odd length the mid-square method cannot be applied. The digit is copied to count as well. So, count has the original length of the entered seed.
2) If the seed has an odd length then a message “mid-square method cannot be applied in odd length is applied.” is shown.
However, if the seed has even length, then the square of the seed is calculated and again the digit present in the square is calculated.
3)Then it is divided by 10, ceil(digit/4.0) times to remove the least significant values from the seed.
4) Then, the seed after dividing(seed from step 3) mod (10 ^count) (count: original length of seed) is calculated to get the final seed.
Example:
1) let’s take a number seed=5865, here the number of digits in the seed=4, which is even. So digit=4, count=digit=4
2) So mid-square method can be applied here. So the square of seed is calculated. (5856)²=34398225, Here digit=8
3) Now, 34398225 is to be divided by 10, ceil(digit/4.0) times i.e. ceil(8/4.0)=2 times.
So, 34398225/10/10= 343982
4) Seed from step 3 mod (10^count) is to be done now where count is the original length of the entered seed.
So, seed from step 3=343982,
10^count=10^4=10000
Then, 343982 mod 10000=3982
Code in C:
``````#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
void intermediate_odd(long long,int);
int count_digit(long long, int);
int main()
{
int a[20]; /*to store the random numbers */
long long seed; /*seed is the initaial number */
int digit; /*to know the number of digit in the square obtained */
int n; /*how many random numbers to generate */
int i,j; /*to control the loop*/
long long p;/* to copy something */
int count; /*to count the original length of the seed which will be used as remainder */
int divisor; /* use this to divide and find remainder*/
printf("Enter a number to generate the random number: ");
scanf("%lld",&seed);
p=seed;
digit=0;
digit=count_digit(p,digit);
count=digit; /*the same digit is to be used with 10 power to get reaminder */
if(digit%2!=0) /*if the number of digits is odd then the mid square multiplicate method cannot be applied */
{
printf("\nThe mid square multipicate method cannot be applied as the number of digit of seed is %d(odd)",digit);
}
else /*if even */
{
printf("\n How many random number you want to generate: ");
scanf("%d",&n);
printf("\nThe random numbers are:\n");
for(i=0;i<n;i++)
{
seed=seed*seed;
p=seed;
digit=0;
digit=count_digit(p,digit);
digit=ceil(digit/4.0);
for(j=0;j<digit;j++)
{
seed=seed/10;
}
divisor=pow(10,count);
seed=seed%divisor;
a[i]=seed;
printf("%lld\t",seed);
digit=0;/* checking if the intermediate seed has odd length */
p=seed;
digit=count_digit(p,digit);
if(digit%2!=0)
{
intermediate_odd(seed,digit);
exit(0);
}
}
}
getch();
return(0);
}
int count_digit(long long seed,int digit) //to count the number of digit
{
long long p;
p=seed;
while(p!=0) /*digit count*/
{
p=p/10;
digit++;
}
return digit;
}
void intermediate_odd(long long seed, int digit)
{
printf("\nThe intermediate seed %d has odd length %d and therefore mid square method cannot be applied from here",seed,digit);
}``````
Outputs:
Also Read: C Program to Implement Selection Sort
Do comment if you have any queries or if you are confused in any steps.
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## Puzzle : Count possible combinations of coins
Q. Given a sum of value S and infinite supply of coins each of value C = { C1, C2, .. , Cn}, how many ways can you make the change? The order of coins doesn’t matter.
For example, for S = 5 and S = {1,2}, there are three solutions: (1, 1, 1, 1, 1), (1, 1, 1, 2), (1, 2, 2) . So output should be 3. For S = 10 and C = {2, 5, 3, 6}, there are five solutions: (2,2,2,2,2), (2,2,3,3), (2,2,6), (2,3,5) and (5,5). So the output should be 5.
Yes, I agree; this question is hard. But can we simplify it to come up with a solution that will solve this tough one!
What if C was { 1 } and S was 10? Number of combinations possible is 1; Ten times 1 is the only way to form the sum 10. That was quick and easy, right? 😀
Lets take it to next level. This time, lets try with more coins: 1 and 2 and same sum: 10. Yes, many combinations flashed my mind too. Lets cross this bridge one step at a time. Lets first list out all the combinations
How do I convert this to a program!!! Lets walk through the process that happened in my brain when I wrote down those combinations.
1. I took zero 2 and tried to form the remaining sum with all 1. I could use ten 1 to form the sum.
2. I took one 2 and tried to form the rest of the sum with all 1.
3. Took three 2 and got six 1
4. Took four 2 and got four 1
5. Took five 2 and got two 1
6. Took six 2 and got zero 1
Good question (if this already came up in your mind)! what if I tried to find the count of 2 by giving values to number of 1. Lets work it out.
Both cases, I did nothing but iterate from 0 till a number which when multiplied with the chosen coin (2 in first case and 1 in second case) is larger than sum. In every step, I tried to form the remaining “sum” with rest of the denominations.
We encountered more steps than in the previous one but at the end it turned out to be the same number of combinations. Obvious! Trying out both ways helped us discover an optimization: Larger Denominations First. Lets shorten it to LDF 😉
So lets transform our knowledge to algorithm now. For that, lets set some terminologies.
Terminologies:
1. D : All denominations(coins) in ascending order. Why sorting? Sorting is just to implement LDF (Larger Denominations First) and is optional.
2. D[i] : value of i-th coin . Assume that index starts with 1.
3. S : sum to be formed.
4. F(i, S) : Returns the maximum combinations possible with first ‘i’ denominations to form sum ‘S’
5. F(i, n, S) : Returns the maximum combinations possible with first ‘i’ denominations using i-th denomination ‘n’ times. This function is to find the maximum combinations possible with remaining denominations while keeping the count of i-th denomination to a particular value.
For our case where D = { 1, 2 } and S = 10:
• F(2, 10) should give us 6 since there are 6 different ways to form a sum of 10 with 1 and 2
• F(1, 10) should give 1.
• F(2, 0, 10) should give 1
• F(2, 1, 10) should give 1
• F(2, 2, 10) should give 1
• F(2, 3, 10) should give 1
• F(2, 4, 10) should give 1
• F(2, 5, 10) should give 1
• F(2, 6, 10) should give 0
• F(2, 7, 10) should give 0
Lets try to define F(i, S) in terms of F(i, n, S).
This is nothing but mathematical representation of a for-loop with variable x iterating from 0 to (S/D[i]), beyond which the x*D[i] will be greater than S.
Now we should define F(i, n, S). As stated before, F(i, n, S) is the maximum number of combinations possible with D[i] used ‘n’ times. With the number of D[i] fixed to n, we just need to iterate through the rest of the coins. So it can be defined as
It is a recursive function which reduces the sum by D[i]*n and finds the maximum combinations with rest of the coins. But what should it do when it reaches the last denomination? It should be able to divide the sum without any remainder; if not, there is no possible combination with the chosen numbers. Simple!
Now, lets try to apply our new algorithm.
F(2, 10) = F(2, 0, 10) +
F(2, 1, 10) +
F(2, 2, 10) +
F(2, 3, 10) +
F(2, 4, 10) +
F(2, 5, 10)
= F(1, 0, 10) + F(1, 2, 10) + F(1, 3, 10) + F(1, 4, 10) + F(1, 5, 10) + F(1, 6, 10) + F(1, 7, 10) ++ F(1, 8, 10) + F(1, 9, 10) + F(1, 10, 10) + F(1, 0, 8) + F(1, 1, 8) + F(1, 2, 8) + F(1, 3, 8) + F(1, 4, 8) + F(1, 5, 8) + F(1, 6, 8) + F(1, 7, 8) ++ F(1, 8, 8) +
F(1, 0, 8) + F(1, 1, 8) + F(1, 2, 8) + F(1, 3, 8) + F(1, 4, 8) + F(1, 5, 8) + F(1, 6, 8) + F(1, 7, 8) ++ F(1, 8, 8) +
F(1, 0, 6) + F(1, 1, 6) + F(1, 2, 6) + F(1, 3, 6) + F(1, 4, 6) + F(1, 5, 6) + F(1, 6, 6) +
F(1, 0, 4) + F(1, 1, 4) + F(1, 2, 4) + F(1, 3, 4) + F(1, 4, 4) +
F(1, 0, 2) + F(1, 1, 2) + F(1, 2, 2) +
F(1, 0, 0)
= 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 +
0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 +
0 + 0 + 0 + 0 + 0 + 1 +
0 + 0 + 0 + 1 +
0 + 1 +
1
= 6
That worked!
Lets write some Java now..
package fun.puzzles;
import java.util.Arrays;
import java.util.List;
public class CoinCombinationsPuzzle {
private final List<Integer> coins;
public CoinCombinationsPuzzle(List<Integer> coins) {
this.coins = coins;
}
public int maximumPossibleCombinations(int numOfCoins, Integer sum) {
coins.sort(null);
int maxIteration = sum/coins.get(numOfCoins);
int maxPossibleCombinations = 0;
for (int i = 0; i <= maxIteration; i++) {
maxPossibleCombinations += maximumPossibleCombinations(numOfCoins, i, sum);
}
return maxPossibleCombinations;
}
public int maximumPossibleCombinations(int numOfCoins, int count, Integer sum) {
// safety check
if (numOfCoins < 0) { return 0; }
if (numOfCoins < 1) {
if (count * coins.get(numOfCoins) == sum) {
return 1;
}
else {
return 0;
}
}
// Reduce the fixed denomination from sum
sum -= (count * coins.get(numOfCoins));
int maxIteration = sum/coins.get(numOfCoins-1);
int maxPossibleCombinations = 0;
for (int i = 0; i <= maxIteration; i++) {
maxPossibleCombinations += maximumPossibleCombinations(numOfCoins-1, i, sum);
}
return maxPossibleCombinations;
}
public int maximumPossibleCombinations(Integer sum) {
return maximumPossibleCombinations(coins.size()-1, sum);
}
public void printMaximumPossibleCombinations(Integer sum) {
System.out.printf("Coins: %-50s Sum: %-5d MaximumPossibleCombinations: %-5d\n",
coins, sum, maximumPossibleCombinations(coins.size()-1, sum));
}
public static void main(String[] args) {
List<Integer> coins;
CoinCombinationsPuzzle ccp;
coins = Arrays.asList(1, 2);
ccp = new CoinCombinationsPuzzle(coins);
ccp.printMaximumPossibleCombinations(5);
coins = Arrays.asList(1, 2);
ccp = new CoinCombinationsPuzzle(coins);
ccp.printMaximumPossibleCombinations(10);
coins = Arrays.asList(1, 2, 3);
ccp = new CoinCombinationsPuzzle(coins);
ccp.printMaximumPossibleCombinations(100);
coins = Arrays.asList(1, 2, 5, 10);
ccp = new CoinCombinationsPuzzle(coins);
ccp.printMaximumPossibleCombinations(1074);
}
}
Output:
This solution can still be optimized by caching the results of recursive method: maximumPossibleCombinations(int numOfCoins, int count, Integer sum)
## Transform A,B,…AA,AB,.. to 1,2,..27,28,..
NOTE: Below code is Python 3.
base = 26
def transform(row):
res = []
for field in [tmp.strip() for tmp in row.split(',')]:
ival = 0
power = 0
for c in field[::-1]:
ival += pow(base,power)*(ord(c)-ord('A')+1)
power += 1
res.append(ival)
return res
print(transform("A, B, Z, AA, AB, AAA"))
Output:
[1, 2, 26, 27, 28, 703]
## Google Code jam Solutions: Problem A. Store Credit
Problem
You receive a credit C at a local store and would like to buy two items. You first walk through the store and create a list L of all available items. From this list you would like to buy two items that add up to the entire value of the credit. The solution you provide will consist of the two integers indicating the positions of the items in your list (smaller number first).
Input
The first line of input gives the number of cases, N. N test cases follow. For each test case there will be:
• One line containing the value C, the amount of credit you have at the store.
• One line containing the value I, the number of items in the store.
• One line containing a space separated list of I integers. Each integer P indicates the price of an item in the store.
• Each test case will have exactly one solution.
Output
For each test case, output one line containing “Case #x: ” followed by the indices of the two items whose price adds up to the store credit. The lower index should be output first.
Limits
5 ≤ C ≤ 1000
1 ≤ P ≤ 1000
Small dataset
N = 10
3 ≤ I ≤ 100
Large dataset
N = 50
3 ≤ I ≤ 2000
Sample
Input Output 3 100 3 5 75 25 200 7 150 24 79 50 88 345 3 8 8 2 1 9 4 4 56 90 3 Case #1: 2 3 Case #2: 1 4 Case #3: 4 5
C++ Solution:
/*
Author : Sreejith Sreekantan
*/
#include
#include
#include
#include
using namespace std;
int main(int argc, char const *argv[])
{
int numOfTestInstances;
cin >> numOfTestInstances;
std::vector itemWeight;
for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
{
int limit;
cin >> limit;
int numOfItems;
cin >> numOfItems;
itemWeight.clear();
itemWeight.reserve(numOfItems);
for (int itemNum = 0; itemNum < numOfItems; ++itemNum)
{
cin >> itemWeight[itemNum];
}
cout << "Case #" << testInstanceNum + 1 << ": ";
for (int i = 0; i < numOfItems; ++i)
{
for (int j = i + 1; j < numOfItems; ++j)
{
if (itemWeight[i] > limit)
{
break;
}
if (itemWeight[j] > limit)
{
continue;
}
if (itemWeight[i] + itemWeight[j] == limit)
{
if (i < j)
{
cout << i + 1 << " " << j + 1 << endl;
}
else
{
cout << j + 1 << " " << i + 1 << endl;
}
}
}
}
}
return 0;
}
Python Solution:
#! /usr/bin/python
numOfInstances = int(raw_input())
for x in xrange(1,numOfInstances+1):
limit = int(raw_input())
numOfItems = int(raw_input())
d = dict()
items = [int(x) for x in raw_input().split()]
for y in xrange(1,numOfItems+1):
if items[y-1] not in d:
d[items[y-1]] = set()
print d
for x in d.keys():
if x <= limit:
index = d[x].pop()
if (limit-x) in d and len(d[limit-x])>0:
index2=d[limit-x].pop()
print min(index, index2), max(index, index2)
## Google Code jam Solutions: Problem B. Reverse Words
Problem
Given a list of space separated words, reverse the order of the words. Each line of text contains L letters and W words. A line will only consist of letters and space characters. There will be exactly one space character between each pair of consecutive words.
Input
The first line of input gives the number of cases, N.
N test cases follow. For each test case there will a line of letters and space characters indicating a list of space separated words. Spaces will not appear at the start or end of a line.
Output
For each test case, output one line containing “Case #x: ” followed by the list of words in reverse order.
Limits
Small dataset
N = 5
1 ≤ L ≤ 25
Large dataset
N = 100
1 ≤ L ≤ 1000
Sample
Input Output 3 this is a test foobar all your base Case #1: test a is this Case #2: foobar Case #3: base your all
C++ Solution:
/*
Author : Sreejith Sreekantan
Description : Problem B. Reverse Words https://code.google.com/codejam/contest/351101/dashboard#s=p1
*/
#include
#include
#include
#include
#include
#include
using namespace std;
int main(int argc, char const *argv[])
{
int numOfTestInstances;
cin >> numOfTestInstances;
for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
{
istringstream in;
string s;
cin >> ws;
getline(cin ,s);
replace(s.begin(), s.end(), ' ', '\n');
in.str(s);
stack stack_s_rev;
while (in >> s)
{
stack_s_rev.push(s);
}
cout << "case #" << testInstanceNum+1 << ": ";
while(!stack_s_rev.empty())
{
cout << stack_s_rev.top() << " ";
stack_s_rev.pop();
}
cout << endl;
}
return 0;
}
## Google Code jam Solutions: Problem A. Alien Language
Problem
After years of study, scientists at Google Labs have discovered an alien language transmitted from a faraway planet. The alien language is very unique in that every word consists of exactly L lowercase letters. Also, there are exactly D words in this language.
Once the dictionary of all the words in the alien language was built, the next breakthrough was to discover that the aliens have been transmitting messages to Earth for the past decade. Unfortunately, these signals are weakened due to the distance between our two planets and some of the words may be misinterpreted. In order to help them decipher these messages, the scientists have asked you to devise an algorithm that will determine the number of possible interpretations for a given pattern.
A pattern consists of exactly L tokens. Each token is either a single lowercase letter (the scientists are very sure that this is the letter) or a group of unique lowercase letters surrounded by parenthesis ( and ). For example: (ab)d(dc) means the first letter is either a or b, the second letter is definitely d and the last letter is either d or c. Therefore, the pattern (ab)d(dc) can stand for either one of these 4 possibilities: add, adc, bdd, bdc.
Input
The first line of input contains 3 integers, L, D and N separated by a space. D lines follow, each containing one word of length L. These are the words that are known to exist in the alien language. N test cases then follow, each on its own line and each consisting of a pattern as described above. You may assume that all known words provided are unique.
Output
For each test case, output
Case #X: K
where X is the test case number, starting from 1, and K indicates how many words in the alien language match the pattern.
Limits
Small dataset
1 ≤ L ≤ 10
1 ≤ D ≤ 25
1 ≤ N ≤ 10
Large dataset
1 ≤ L ≤ 15
1 ≤ D ≤ 5000
1 ≤ N ≤ 500
Sample
Input Output 3 5 4 abc bca dac dbc cba (ab)(bc)(ca) abc (abc)(abc)(abc) (zyx)bc Case #1: 2 Case #2: 1 Case #3: 3 Case #4: 0
C++ Solution:
/*
Author : Sreejith Sreekantan
Description : Problem A. Alien Language (https://code.google.com/codejam/contest/90101/dashboard#s=p0)
*/
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define rep(v,N) for(int v=0; v<n; v++)="" #define="" <span="" class="hiddenSpellError" pre="define " data-mce-bogus="1">rep2(v,M,N) for(int v=M; v<n; v++)="" #define="" for(v,c)="" for(__typeof(c.begin())="" v="C.begin();" v!="C.end();" ++v)="" vi="" std::vector<int="">
#define vll std::vector
#define pb push_back
#define Sort(C) std::sort(C.begin(), C.end())
#define RSort(C) std::sort(C.rbegin(), C.rend())
#define Copy(ans,out) copy(ans.begin(),ans.end(), ostream_iterator<__typeof(ans[0])>(out, " "))
// #define SMALL
#define LARGE
int main(int argc, char const *argv[])
{
freopen("a.in", "rt", stdin);
// freopen("a.out", "wt", stdout);
#ifdef SMALL
freopen("as.in", "rt", stdin);
freopen("as.out", "wt", stdout);
#endif
#ifdef LARGE
freopen("al.in", "rt", stdin);
freopen("al.out", "wt", stdout);
#endif
unsigned int L, D, N;
cin >> L >> D >> N;
std::vector dict(D);
std::vector inp(D);
rep(i, D)
{
cin >> dict[i];
}
rep(i, N)
{
cin >> inp[i];
replace(inp[i].begin(), inp[i].end(), '(', '[');
replace(inp[i].begin(), inp[i].end(), ')', ']');
// cout << inp[i]<< endl;
int c = 0;
rep(j, D)
{
if (regex_match(dict[j], regex(inp[i])) ) c++;
}
cout << "Case #" << i+1 << ": " << c << endl;
}
return 0;
}
## Brain Teaser
T : Number of test cases
M: value of M
N: Number of strings
For each string, ascii value of each character of the string is raised to M and multiplied together. For each test case the sum of above values is ODD or EVEN
Input:
1
10 2
ac ab
Output:
ODD
TIP: Challenge lies in finding the solution without doing all the mathematical operations mentioned in the question. (yes, also called optimizing..!)
Solution:
#!/usr/bin/python
T = int(raw_input())
while T>0:
T = T - 1
tmp = raw_input()
M = int(tmp.split()[0])
K = int(tmp.split()[1])
tmp = raw_input()
final_odd = False
for s in tmp.split():
odd = None
for i in s:
if ord(i)%2==0: # the power is even
odd = False
else: # the power is odd
if odd==None: # if odd is True, odd*odd=odd
odd = True
# odd + odd = even
# odd + even = odd
# even + even = even
if final_odd != odd:
final_odd = True
else:
final_odd = False
if final_odd:
print "ODD"
else:
print "EVEN"
## Find the biggest black square from an N*N matrix of black or white squares
Question:
Find the biggest black square from an N*N matrix of squares. 1 means it’s black, 0 means it’s white. Output should be the list of squares which forms the largest black square.
Input : ‘{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}’
Output : {(3#0,3#1,3#2,3#3),(4#0,4#1,4#2,4#3),(5#0,5#1,5#2,5#3),(6#0,6#1,6#2,6#3)}
Solution(Python):
# Author : Sreejith Sreekantan
# Description :
# Find the biggest black square from an N*N matrix of squares. 1 means it's black, 0 means it's white
# Input : '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
# Output : {(3#0,3#1,3#2,3#3),(4#0,4#1,4#2,4#3),(5#0,5#1,5#2,5#3),(6#0,6#1,6#2,6#3)}
#!/usr/bin/python
def largestSquareAt(sq, i, j, k=1):
if not ( i+k<len(sq) and="" j+k<len(sq[i])="" ):="" ="" ="" return="" 0="" for="" x="" in="" range(0,k):="" if="" sq[i+x][j+k-1]="='0':" y="" sq[i+k-1][j+y]="='0':" 1="" +="" largestsquareat(sq,="" i,="" j,="" k+1) ="" ="" def="" biggestsquare(input1):="" len(input1)="=0" or="" not="" (="" input1.count('{')="=input1.count('}')" )="" :="" ''="" i="str(input1)" j="i" for="" j]="" resx="-1" rexy="-1" ressize="-1" res="[]" range(0,len(j)):="" res.append([])="" range(0,len(j[x])):="" res[x].append(largestsquareat(j,="" x,="" y))="" ressize<res[x][y]:="" resy="y" resstring="" ressize="">0:
resstring = "{"
for x in range(resx,resx+ressize):
if x>resx:
resstring += ","
resstring += "("
for y in range(resy,resy+ressize):
if y>resy:
resstring += ","
resstring += (str(x)+"#"+str(y))
resstring += ")"
resstring += "}"
return resstring
input = '{0#1#1#0,0#1#1#1,0#1#1#1,1#1#1#1}'
input1 = '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input2 = '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#0#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input3= '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,0#1#0#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input4 = '{0#0#0#0,0#0#0#0,0#0#0#0,0#0#0#0}'
print biggestSquare('')
print biggestSquare(input1)
print biggestSquare(input2)
print biggestSquare(input3)
print biggestSquare(input4)
## Given a string of length N, output “Correct” if brackets close in correct order else output “Incorrect”
Question:
Given a string of length N, output “Correct” if brackets close in correct order else output “Incorrect”
Input : ({}[((({{}})[{()}]))])
Solution(Python):
# Author : Sreejith Sreekantan
# Description :
# Given a string of length N, output "Correct" if brackets close in correct order else output "Incorrect"
# Input : ({}[((({{}})[{()}]))])
#
#!/usr/bin/python
def validString(input1):
stk = []
flag = True
for x in input1:
if x in ['{', '(', '[']:
stk.append(x)
else:
if len(stk)==0:
flag = False
elif x == '}' and not stk[len(stk)-1]=='{' :
flag = False
elif x == ')' and not stk[len(stk)-1]=='(' :
flag = False
elif x == '}' and not stk[len(stk)-1]=='{' :
flag = False
if not flag:
return "Incorrect"
else:
stk.pop()
return "Correct"
input1 = '({}[((({{}})[{()}]))])'
input2 = '({}((({{}})[{()}]))])'
print validString(input1)
print validString(input2)
## Count Inversions in an array
This file contains all of the 100,000 integers between 1 and 100,000 (inclusive) in some order, with no integer repeated.
Your task is to compute the number of inversions in the file given, where the ith row of the file indicates the ith entry of an array.
Tips:
If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j
Eg: The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3).
You can get working program from here
Input files : Input1 Input2
Answer for input1 :
Answer for input 2: 25
## Algorithm Puzzles
Q:
You are a given a unimodal array of n distinct elements, meaning that its entries are in increasing order up until its maximum element, after which its elements are in decreasing order. Give an algorithm to compute the maximum element that runs in O(log n) time.
Ans. Divide and Conquer
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CC-MAIN-2023-14
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latest
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en
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https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-3-parallel-and-perpendicular-lines-common-core-cumulative-standards-review-selected-response-page-212/1
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## Geometry: Common Core (15th Edition)
Area= length$\times$ width The side length (and width) of the square is 7n$^{3}$. To find the area take $7n^{3}\times7n^{3}$ $=49n^6$ *Remember when you multiply the powers of 3, you add them together instead of multiplying.
| 80 | 265 |
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crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00676.warc.gz
| 12,823,278 | 9,118 |
Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dgrub Structured version Visualization version GIF version
Theorem dgrub 24210
Description: If the 𝑀-th coefficient of 𝐹 is nonzero, then the degree of 𝐹 is at least 𝑀. (Contributed by Mario Carneiro, 22-Jul-2014.)
Hypotheses
Ref Expression
dgrub.1 𝐴 = (coeff‘𝐹)
dgrub.2 𝑁 = (deg‘𝐹)
Assertion
Ref Expression
dgrub ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑀𝑁)
Proof of Theorem dgrub
Dummy variables 𝑛 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 simp1 1130 . . . 4 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝐹 ∈ (Poly‘𝑆))
2 simp2 1131 . . . . 5 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑀 ∈ ℕ0)
3 dgrub.1 . . . . . . . . 9 𝐴 = (coeff‘𝐹)
43coef3 24208 . . . . . . . 8 (𝐹 ∈ (Poly‘𝑆) → 𝐴:ℕ0⟶ℂ)
51, 4syl 17 . . . . . . 7 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝐴:ℕ0⟶ℂ)
65, 2ffvelrnd 6503 . . . . . 6 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝐴𝑀) ∈ ℂ)
7 simp3 1132 . . . . . 6 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝐴𝑀) ≠ 0)
8 eldifsn 4453 . . . . . 6 ((𝐴𝑀) ∈ (ℂ ∖ {0}) ↔ ((𝐴𝑀) ∈ ℂ ∧ (𝐴𝑀) ≠ 0))
96, 7, 8sylanbrc 572 . . . . 5 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝐴𝑀) ∈ (ℂ ∖ {0}))
103coef 24206 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → 𝐴:ℕ0⟶(𝑆 ∪ {0}))
11 ffn 6185 . . . . . 6 (𝐴:ℕ0⟶(𝑆 ∪ {0}) → 𝐴 Fn ℕ0)
12 elpreima 6480 . . . . . 6 (𝐴 Fn ℕ0 → (𝑀 ∈ (𝐴 “ (ℂ ∖ {0})) ↔ (𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ∈ (ℂ ∖ {0}))))
131, 10, 11, 124syl 19 . . . . 5 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝑀 ∈ (𝐴 “ (ℂ ∖ {0})) ↔ (𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ∈ (ℂ ∖ {0}))))
142, 9, 13mpbir2and 692 . . . 4 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑀 ∈ (𝐴 “ (ℂ ∖ {0})))
15 nn0ssre 11498 . . . . . . 7 0 ⊆ ℝ
16 ltso 10320 . . . . . . 7 < Or ℝ
17 soss 5188 . . . . . . 7 (ℕ0 ⊆ ℝ → ( < Or ℝ → < Or ℕ0))
1815, 16, 17mp2 9 . . . . . 6 < Or ℕ0
1918a1i 11 . . . . 5 (𝐹 ∈ (Poly‘𝑆) → < Or ℕ0)
20 0zd 11591 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → 0 ∈ ℤ)
21 cnvimass 5626 . . . . . . 7 (𝐴 “ (ℂ ∖ {0})) ⊆ dom 𝐴
22 fdm 6191 . . . . . . . 8 (𝐴:ℕ0⟶(𝑆 ∪ {0}) → dom 𝐴 = ℕ0)
2310, 22syl 17 . . . . . . 7 (𝐹 ∈ (Poly‘𝑆) → dom 𝐴 = ℕ0)
2421, 23syl5sseq 3802 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → (𝐴 “ (ℂ ∖ {0})) ⊆ ℕ0)
253dgrlem 24205 . . . . . . 7 (𝐹 ∈ (Poly‘𝑆) → (𝐴:ℕ0⟶(𝑆 ∪ {0}) ∧ ∃𝑛 ∈ ℤ ∀𝑥 ∈ (𝐴 “ (ℂ ∖ {0}))𝑥𝑛))
2625simprd 483 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → ∃𝑛 ∈ ℤ ∀𝑥 ∈ (𝐴 “ (ℂ ∖ {0}))𝑥𝑛)
27 nn0uz 11924 . . . . . . 7 0 = (ℤ‘0)
2827uzsupss 11983 . . . . . 6 ((0 ∈ ℤ ∧ (𝐴 “ (ℂ ∖ {0})) ⊆ ℕ0 ∧ ∃𝑛 ∈ ℤ ∀𝑥 ∈ (𝐴 “ (ℂ ∖ {0}))𝑥𝑛) → ∃𝑛 ∈ ℕ0 (∀𝑥 ∈ (𝐴 “ (ℂ ∖ {0})) ¬ 𝑛 < 𝑥 ∧ ∀𝑥 ∈ ℕ0 (𝑥 < 𝑛 → ∃𝑦 ∈ (𝐴 “ (ℂ ∖ {0}))𝑥 < 𝑦)))
2920, 24, 26, 28syl3anc 1476 . . . . 5 (𝐹 ∈ (Poly‘𝑆) → ∃𝑛 ∈ ℕ0 (∀𝑥 ∈ (𝐴 “ (ℂ ∖ {0})) ¬ 𝑛 < 𝑥 ∧ ∀𝑥 ∈ ℕ0 (𝑥 < 𝑛 → ∃𝑦 ∈ (𝐴 “ (ℂ ∖ {0}))𝑥 < 𝑦)))
3019, 29supub 8521 . . . 4 (𝐹 ∈ (Poly‘𝑆) → (𝑀 ∈ (𝐴 “ (ℂ ∖ {0})) → ¬ sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ) < 𝑀))
311, 14, 30sylc 65 . . 3 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → ¬ sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ) < 𝑀)
32 dgrub.2 . . . . . 6 𝑁 = (deg‘𝐹)
333dgrval 24204 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → (deg‘𝐹) = sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ))
3432, 33syl5eq 2817 . . . . 5 (𝐹 ∈ (Poly‘𝑆) → 𝑁 = sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ))
351, 34syl 17 . . . 4 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑁 = sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ))
3635breq1d 4796 . . 3 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝑁 < 𝑀 ↔ sup((𝐴 “ (ℂ ∖ {0})), ℕ0, < ) < 𝑀))
3731, 36mtbird 314 . 2 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → ¬ 𝑁 < 𝑀)
382nn0red 11554 . . 3 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑀 ∈ ℝ)
39 dgrcl 24209 . . . . . 6 (𝐹 ∈ (Poly‘𝑆) → (deg‘𝐹) ∈ ℕ0)
4032, 39syl5eqel 2854 . . . . 5 (𝐹 ∈ (Poly‘𝑆) → 𝑁 ∈ ℕ0)
411, 40syl 17 . . . 4 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑁 ∈ ℕ0)
4241nn0red 11554 . . 3 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑁 ∈ ℝ)
4338, 42lenltd 10385 . 2 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → (𝑀𝑁 ↔ ¬ 𝑁 < 𝑀))
4437, 43mpbird 247 1 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝑀 ∈ ℕ0 ∧ (𝐴𝑀) ≠ 0) → 𝑀𝑁)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 382 ∧ w3a 1071 = wceq 1631 ∈ wcel 2145 ≠ wne 2943 ∀wral 3061 ∃wrex 3062 ∖ cdif 3720 ∪ cun 3721 ⊆ wss 3723 {csn 4316 class class class wbr 4786 Or wor 5169 ◡ccnv 5248 dom cdm 5249 “ cima 5252 Fn wfn 6026 ⟶wf 6027 ‘cfv 6031 supcsup 8502 ℂcc 10136 ℝcr 10137 0cc0 10138 < clt 10276 ≤ cle 10277 ℕ0cn0 11494 ℤcz 11579 Polycply 24160 coeffccoe 24162 degcdgr 24163 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-8 2147 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-rep 4904 ax-sep 4915 ax-nul 4923 ax-pow 4974 ax-pr 5034 ax-un 7096 ax-inf2 8702 ax-cnex 10194 ax-resscn 10195 ax-1cn 10196 ax-icn 10197 ax-addcl 10198 ax-addrcl 10199 ax-mulcl 10200 ax-mulrcl 10201 ax-mulcom 10202 ax-addass 10203 ax-mulass 10204 ax-distr 10205 ax-i2m1 10206 ax-1ne0 10207 ax-1rid 10208 ax-rnegex 10209 ax-rrecex 10210 ax-cnre 10211 ax-pre-lttri 10212 ax-pre-lttrn 10213 ax-pre-ltadd 10214 ax-pre-mulgt0 10215 ax-pre-sup 10216 ax-addf 10217 This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-3or 1072 df-3an 1073 df-tru 1634 df-fal 1637 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-mo 2623 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ne 2944 df-nel 3047 df-ral 3066 df-rex 3067 df-reu 3068 df-rmo 3069 df-rab 3070 df-v 3353 df-sbc 3588 df-csb 3683 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-pss 3739 df-nul 4064 df-if 4226 df-pw 4299 df-sn 4317 df-pr 4319 df-tp 4321 df-op 4323 df-uni 4575 df-int 4612 df-iun 4656 df-br 4787 df-opab 4847 df-mpt 4864 df-tr 4887 df-id 5157 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-se 5209 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-pred 5823 df-ord 5869 df-on 5870 df-lim 5871 df-suc 5872 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-isom 6040 df-riota 6754 df-ov 6796 df-oprab 6797 df-mpt2 6798 df-of 7044 df-om 7213 df-1st 7315 df-2nd 7316 df-wrecs 7559 df-recs 7621 df-rdg 7659 df-1o 7713 df-oadd 7717 df-er 7896 df-map 8011 df-pm 8012 df-en 8110 df-dom 8111 df-sdom 8112 df-fin 8113 df-sup 8504 df-inf 8505 df-oi 8571 df-card 8965 df-pnf 10278 df-mnf 10279 df-xr 10280 df-ltxr 10281 df-le 10282 df-sub 10470 df-neg 10471 df-div 10887 df-nn 11223 df-2 11281 df-3 11282 df-n0 11495 df-z 11580 df-uz 11889 df-rp 12036 df-fz 12534 df-fzo 12674 df-fl 12801 df-seq 13009 df-exp 13068 df-hash 13322 df-cj 14047 df-re 14048 df-im 14049 df-sqrt 14183 df-abs 14184 df-clim 14427 df-rlim 14428 df-sum 14625 df-0p 23657 df-ply 24164 df-coe 24166 df-dgr 24167 This theorem is referenced by: dgrub2 24211 coeidlem 24213 coeid3 24216 dgreq 24220 coemullem 24226 coemulhi 24230 coemulc 24231 dgreq0 24241 dgrlt 24242 dgradd2 24244 dgrmul 24246 vieta1lem2 24286 aannenlem2 24304
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https://socratic.org/questions/how-many-boxes-of-envelopes-can-you-buy-with-12-if-one-box-costs-3
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# How many boxes of envelopes can you buy with $12 if one box costs$3?
Nov 15, 2014
To solve this problem one would need to create an algebra expression to represent the cost of the box with the number of boxes in order to determine the total cost of the transaction.
Cost of a box = $3 Number of boxes = n (this is what we are looking for) Total cost =$12
(Cost of the box) x (Number of boxes) = (Total Cost)
($3) x (n) = ($12)
$3 n = 12$
$\frac{3 n}{3} = \frac{12}{3}$ to isolate the $n$ divide by 3 on both sides
$n = 4$
You can buy 4 boxes with $12 if the boxes cost$3 each.
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https://scicomp.stackexchange.com/questions/24561/strang-splitting
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Strang splitting
I have recently come across the Strang splitting and have some questions. For the differential equation of the form
$$dy/dt = (L_1 + L_2)y$$
Strang splitting implement the time splitting as
$$\begin{eqnarray*} \tilde y_1 & = & e^{L_1 \Delta t/2} y_0, & & \bar y_1 = e^{L_2 \Delta t} \tilde y_1, & & y_1 = e^{L_1 \Delta t/2} \bar y_1 \\ \tilde y_2 & = & e^{L_1 \Delta t/2} y_1, & & \bar y_2 = e^{L_2 \Delta t} \tilde y_2, & & y_2 = e^{L_1 \Delta t/2} \bar y_2 \\ ... \\ \tilde y_n & = & e^{L_1 \Delta t/2} y_{n-1}, & & \bar y_n = e^{L_2 \Delta t} \tilde y_n, & & y_n = e^{L_1 \Delta t/2} \bar y_n \\ \end{eqnarray*}$$
However, from the equations above, it is obviously that the half time step with $y_1$ and $\tilde y_2$ can be combined into a single time step, so it is equivalent to:
$$\begin{eqnarray*} \tilde y_1 & = & e^{L_1 \Delta t/2} y_0, & & \bar y_1 = e^{L_2 \Delta t} \tilde y_1 \\ \tilde y_2 & = & e^{L_1 \Delta t} \bar y_1, & & \bar y_2 = e^{L_2 \Delta t} \tilde y_2 \\ ... \\ \tilde y_n & = & e^{L_1 \Delta t} \bar y_{n-1}, & & \bar y_n = e^{L_2 \Delta t} \tilde y_n \\ y_n & = & e^{L_1 \Delta t/2} \bar y_n \end{eqnarray*}$$
This look like that it is exactly the same as the first order time splitting scheme except the first and last half time step, and the computation is faster with the reduction. Am I missing something here?
Also, what is the time splitting method with four order? How does it look like explicitly? I am looking for some numerical algorithms for solving nonlinear Schroedinger equation.
• You're not missing anything, I think: the first-order splitting scheme shadows the true solutions to second-order accuracy—it gives a second-order accurate solution to a modified problem where initial values are modified slightly (by $O(\Delta t)$), and your analysis proves as much. It even says so in epubs.siam.org/doi/abs/10.1137/0705041 on p.511. Commented Jul 27, 2016 at 23:25
This look like that it is exactly the same as the first order time splitting scheme except the first and last half time step, and the computation is faster with the reduction. Am I missing something here?
You're exactly correct, as others have already said.
Also, what is the time splitting method with four order?
There are many higher-order splitting schemes, some developed very recently. This is the most comprehensive list I know of and includes various methods of order 4.
• Thanks. It seems that a good choice is a splitting with real coefficient and least number of step $s$. Can you explain a little bit why there are many difference method? I expect there is a unique BCH expansion for a given order. What are the advantage of complex coefficient over the real coefficient? Commented Jul 29, 2016 at 0:30
• @hwlau Please go ahead and post any questions you have as new questions. It's not recommended to post additional questions as comments. Commented Jul 29, 2016 at 10:38
You can combine those two steps since both of them involve the same operator ($L_1$).
For the fourth-order method I recommend that you check out my answer to a similar question on Stack Overflow and also Section 4 in the following paper (see also the discussion below):
Yoshida, H. (1990). Construction of higher order symplectic integrators. Physics Letters A, 150(5–7), 262–268. http://doi.org/10.1016/0375-9601(90)90092-3
Let $a = L_1$ and $b = L_2$. The analysis of this type of schemes is done using the Baker-Campbell-Hausdorff (BCH) formula, which tells us that the expression for $c$ in $$\exp\{t a\} \exp\{t b\} = \exp\{t c\},$$ is given in terms of Lie brackets involving $a$ and $b$ (recall that $[a, b] = ab - ba$): $$c = a + b + \frac{t}{2} \left[a, b\right] + \frac{t^2}{12} \left( \left[a, \left[a, b\right]\right] + \left[b, \left[b, a\right]\right] \right) + \frac{t^3}{24} \left[a, \left[b, \left[b, a\right]\right]\right] + \dotsb$$ The BCH formula for the particular case of the Strang splitting is given by $$\exp\left\{\tfrac{t}{2} a\right\} \exp\left\{t b\right\} \exp\left\{\tfrac{t}{2} a\right\} = \exp\left\{t c_1 + t^3 c_3 + t^5 c_5 + \dotsb \right\},$$ where \begin{align*} c_1 &= a + b, \\ c_3 &= -\frac{1}{24} \left[a, \left[a, b\right]\right] + \frac{1}{12} \left[b, \left[b, a\right]\right], \\ &\dotsb \end{align*} In general, $c_3$ does not vanish and this implies that the scheme is second order. In order to construct higher order methods, one can choose a splitting such that the appropriate terms in the BCH formula vanish (the paper by Yoshida cited above discusses this in depth).
Particular choices of $a$ and $b$ will yield symplectic methods but from the discussion above we see that the order of the scheme does not necessarily have to do with it being symplectic.
• It is still quite vague. In my understading, we can only have symplectic integrator when the operator splitting takes the form $L_1(q)$ and $L_2(p)$ which is not always the case, right? Commented Jul 29, 2016 at 0:40
• That is one way of obtaining a symplectic integrator. The reason why that is the case is because the systems $dq/dt = L_1(q), dp/dt = 0$ and $dq/dt = 0$, $dp/dt = L_2(p)$ can be solved exactly, and the solutions are of course a symplectic mappings. Therefore, the composition of $\exp{t L_1}$ and $\exp{t L_2}$ will be symplectic.One could devise other symplectic integrators, though (e.g.: exploiting particular features of the vector fields under consideration). Commented Jul 29, 2016 at 9:50
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# What Is The First Decimal Place?
## What is 1/3 as a decimal rounded to 2 decimal places?
Fraction to Decimal Conversion Tablesfraction = decimal1/3 = 0.32/3 = 0.61/4 = 0.253/4 = 0.751/5 = 0.22/5 = 0.43/5 = 0.61/6 = 0.165/6 = 0.8319 more rows.
## Is 1.5 a natural number?
A natural number is an integer greater than 0. Natural numbers begin at 1 and increment to infinity: 1, 2, 3, 4, 5, etc. Natural numbers are also called “counting numbers” because they are used for counting.
## What does round to 2 decimal places mean?
“Two decimal places” is the same as “the nearest hundredth”. … So, for example, if you are asked to round 3.264 to two decimal places it means the same as if your are asked to round 3.264 to the nearest hundredth. Some questions, like the example below, will ask you to “show your answer correct to two decimal places.”
## What’s a decimal place?
noun. the position of a digit to the right of a decimal point; a specific number of digits to the right of the decimal point in a line of numbers.
## What is 1/3 as a decimal rounded to 3 decimal places?
0.333Example: Convert 1 3 to a Decimal Answer = 0.333 (accurate to only 3 decimal places !!)
## What is the decimal form of 1 by 3?
Common Fractions with Decimal and Percent EquivalentsFractionDecimalPercent1/30.333…33.333…%2/30.666…66.666…%1/40.2525%3/40.7575%21 more rows•Feb 21, 2017
## What is 1/3 as a decimal?
0.333333333333331/3 as a decimal is 0.33333333333333.
## How do you read decimal places?
The first column to the right of the decimal point has place value 1/10 (tenths). The second column to the right of the decimal point has place value 1/100 (hundredths). The third column to the right of the decimal point has place value 1/1000 (thousandths).
## What is the value of 5?
5 is in hundreds place and its place value is 500, 4 is in tens place and its place value is 40, 8 is in ones place and its place value is 8.
## Is 1.0 a decimal number?
In much the same way that whole numbers pivot between negative and positive on zero, decimals extend out infinitely from zero, since a whole number is expressed as 1.0, and a decimal of that is 1.947; everything to the right of the decimal is less than 1 and is, in a sense, encapsulated within the zero.
## What is 2.738 to 2 decimal places?
Since it is larger than, you can round the 38 up to 40 . So now the number you have is 2.740 , but since the 0 does not need to be included, you have 2.74 , which is 2 decimal places.
## Is 1 a decimal number?
Decimal notation For writing numbers, the decimal system uses ten decimal digits, a decimal mark, and, for negative numbers, a minus sign “−”. The decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; the decimal separator is the dot “.” in many countries, but also a comma “,” in other countries.
## How do you count decimal places?
If a number has a decimal point , then the first digit to the right of the decimal point indicates the number of tenths. For example, the decimal 0.3 is the same as the fraction 310 . The second digit to the right of the decimal point indicates the number of hundredths.
## What is a single decimal number?
The decimal number system consists of ten single- digit numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The number after 9 is 10. The number after 19 is 20 and so forth. … All these schemes have a number of digits that is a power of 2.
## How do you round to the first decimal place?
To round to a decimal place:look at the first digit after the decimal point if rounding to one decimal place or the second digit for two decimal places.draw a vertical line to the right of the place value digit that is required.look at the next digit.if it’s 5 or more, increase the previous digit by one.More items…
## What does it mean by give your answer to 1 decimal place?
To write the number 0.76 correct to one decimal place: If the digit in the second decimal place is 5 or more then add 1 to the digit in the first decimal place. If the digit in the second decimal place is less than 5, then leave the digit in the first decimal place as it is.
## What is 0.2 rounded to a whole number?
We cannot write 0.2 as a whole number, but it can be rounded to a whole number. As tenth digit is less than 5, so upon rounding 0.2 to nearest whole number we will get 0. As 0 is also a whole number, therefore, upon rounding 0.2 to nearest whole number we will get 0.
## What does 2 decimal places look like?
Rounding to a certain number of decimal places 4.732 rounded to 2 decimal places would be 4.73 (because it is the nearest number to 2 decimal places). … 4.735 is halfway between 4.73 and 4.74, so it is rounded up: 4.735 rounded to 2 decimal places is 4.74. Examples. (a) What is 5.79 rounded to 1 decimal place.
## What does 5 decimal places mean?
“To five places” means “to five decimal places”. First, I count out the five decimal places, and then I look at the sixth place: 3.14159 | 265… I’ve drawn a little line separating the fifth place from the sixth place. This can be a handy way of “keeping your place”, especially if you are dealing with lots of digits.
## Whats does 13 mean?
The number 13 brings the test, the suffering and the death. It symbolises the death to the matter or to oneself and the birth to the spirit: the passage on a higher level of existence. (In Tarot, no. 13 card is named as Death, but it mostly means death of a struggling period and new beginning s.
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Several types of triangles exist, including scalene, isosceles, equilateral, right, obtuse and acute. Triangles are categorized according to their sides, angles or a combination of both. More »
www.reference.com Math Geometry Shapes
The main kinds of triangles are: equilateral, isosceles, scalene, acute, right and obtuse. These names can be combined to further describe the type of triangle. More »
In geometry, the triangle is a polygon that has three sides. There can be different types of triangles, but the sum of the three angles in this polygon is equal to 180 degrees. In the context of religion, alchemy and occ... More »
www.reference.com Math Geometry Shapes
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An isosceles triangle is a specific type of triangle where only two of the side lengths and two of the internal angles are equal. There are also other types of triangles like equilateral triangles and scalene triangles. More »
The main kinds of triangles are: equilateral, isosceles, scalene, acute, right and obtuse. These names can be combined to further describe the type of triangle. More »
An isosceles trapezoid contains a single line of symmetry. A trapezoid is referred to as an isosceles trapezoid when its two non-parallel edges are equal in length. More »
The perimeter of a triangle is the sum of all the sides of the triangle, and you determine the perimeter of any triangle by adding the lengths of all three sides. Since an isosceles triangle has two sides of equal length... More »
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# Mean vs. Median
Suppose we have a set of values , defined as:
if we were to compute the arithmetic mean, we would obtain:
However, notice that this mean does not really look entirely "right" and that is due to the exceptional value which drags the mean towards a very high value compared to the rest of the values in the set.
This is why, it is useful sometimes to compute the median which gives a more correct overview of the average of the set. In order to compute the median, we sort the set in increasing order, thereby obtaining :
and calculate the arithmetic mean between the two values in the middle of the set:
which gives us a median that is a better presentation of the middle of the original set .
# Standard Deviation
The standard deviation can be thought of measuring how far the data values lie from the mean.
Let there be a random variable with mean value
where the operator denotes the average or expected value of , then the standard deviation of is the quantity:
## Example
Given a set of measurements with the values:
Calculate the arithmetic average of the elements of the set:
Calculate the difference squared between each element of the set and the mean to obtain the set :
By summing up the elements of () and then dividing by the number of elements minus one we obtain the variance :
and the standard deviation is the square root of the variance:
By subtracting and adding the standard deviation from the mean we found out the bounds between which most values are placed.
For the lower bound, we have:
For the upper bound, we have:
which means that most values are placed between and .
# Chebyshev's Theorem
The fraction of any set of numbers lying within k-standard deviation of those numbers of the mean of those numbers is at least:
where:
and
## Example
Taking values for (), we can observe the following:
• for , we obtain , meaning that at least of the values must be within two standard deviations from the mean.
• for , we obtain , meaning that at least of the values must be within three standard deviations from the mean.
• for , we obtain , meaning that at least of the values must be within four standard deviations from the mean.
Given the previous example, we know that the mean is and the standard deviation is . Now we take values: for , we have and, subtracting that value from the mean, we obtain the lower bound: ; then we add the value to the mean, obtaining the upper bound: meaning that at least of the values must be between the lower bound and the upper bound . The same can be performed for other values of .
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https://byjus.com/question-answer/in-the-garbar-jhala-aminabad-a-shopkeeper-first-raises-the-price-of-a-jewellery-by/
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Question
# In the Garbar Jhala, Aminabad a shopkeeper first raises the price of a Jewellery by x% then he decreases the new price by x%. After one such up-down cycle, the price of a Jewellery decreased by Rs. 21025. After a second up-down cycle, the jewellery was sold for Rs. 484416. What was the original price of the jewellery?
A
Rs. 5,00,000
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B
Rs. 6,00,625
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C
Rs. 525625
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D
Rs. 5,26,000
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Solution
## The correct option is A Rs. 525625Let the original price be p, then the decrease in value of p after one cycle. =p(x100)2=21025……(i) Again the final value after second cycle ⇒p(1+x100)(1−x100)(1+x100)(1−x100)=484416⇒p[1−(x100)2]2=484416……(ii) Dividing equation (ii) by equation (i) [1−(x100)2]2(x100)2=48441621025=2304100⇒ 1−(x100)2(x100)=√2304100=4810 Let x100=k, then 1−k2k=4810⇒10k2+48k−10=0⇒5k2−24x−5=0⇒k=5 or k=−15(inadmissible value)So,x=20%Hence,p(x100)2=21025⇒p=525625
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Discover a lot of information on the number 230: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
Mathematical properties of 230
Is 230 a prime number? No
Is 230 a perfect number? No
Number of divisors 8
List of dividers 1, 2, 5, 10, 23, 46, 115, 230
Sum of divisors 432
Prime factorization 2 x 5 x 23
Prime factors 2, 5, 23
How to write / spell 230 in letters?
In letters, the number 230 is written as: Two hundred and thirty. And in other languages? how does it spell?
230 in other languages
Write 230 in english Two hundred and thirty
Write 230 in french Deux cent trente
Write 230 in spanish Doscientos treinta
Write 230 in portuguese Duzentos trinta
Decomposition of the number 230
The number 230 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 0 : ... Find out more about the number 0
Other ways to write 230
In letter Two hundred and thirty
In roman numeral CCXXX
In binary 11100110
In octal 346
In US dollars USD 230.00 (\$)
In euros 230,00 EUR (€)
Some related numbers
Previous number 229
Next number 231
Next prime number 233
Mathematical operations
Operations and solutions
230*2 = 460 The double of 230 is 460
230*3 = 690 The triple of 230 is 690
230/2 = 115 The half of 230 is 115.000000
230/3 = 76.666666666667 The third of 230 is 76.666667
2302 = 52900 The square of 230 is 52900.000000
2303 = 12167000 The cube of 230 is 12167000.000000
√230 = 15.165750888103 The square root of 230 is 15.165751
log(230) = 5.4380793089232 The natural (Neperian) logarithm of 230 is 5.438079
log10(230) = 2.3617278360176 The decimal logarithm (base 10) of 230 is 2.361728
sin(230) = -0.61606420405336 The sine of 230 is -0.616064
cos(230) = -0.78769594164506 The cosine of 230 is -0.787696
tan(230) = 0.7821091508568 The tangent of 230 is 0.782109
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The K5 Learning Blog urges parents to be pro-active in helping their children reach their full academic potential.
# Comparing Fractions and Mixed Numbers with Same Denominator
In grades 3, 4 and 5 students spend quite some time covering fractions. One of the areas that causes a bit of confusion is comparing a fraction and a mixed number. It takes careful consideration to work out which of the fraction or mixed number is greater than, less than, or if they are equal to each other.
Let’s take you through how to compare fractions and mixed numbers with the same denominator. We’ll use examples to demonstrate.
## Comparing an improper fraction and a mixed number with the same denominator
When comparing an improper fraction (a fraction in which the top number is greater than or equal to the bottom number) and a mixed number with the same denominator (bottom number), you’ll first need to convert the mixed number to an improper fraction.
For example, let’s compare the improper mixed number 4 1/3 with the improper fraction 15/3.
## Converting mixed numbers into improper fractions
1. First, we need to convert the mixed number to an improper fraction.
2. Multiply the denominator (bottom number) by the whole number:
4. The answer is the numerator (top number):
The denominator stays the same:
Now we can compare the numbers:
## Converting an improper fraction into a mixed number
You can also work on the improper fraction to work out which mixed number is greater than, less than or if they are equal.
Here’s an example:
We’re looking to convert the improper fraction into a mixed number:
1. Divide the denominator (bottom number) into the numerator (top number). This answer becomes the whole number:
2. The remainder becomes the numerator:
3. The denominator remains the same. So the mixed number is:
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# What s the numerical value of 1/a + 1/b + 1/c ? 1) a+b+c =
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What s the numerical value of 1/a + 1/b + 1/c ?
1) a+b+c = 1;
2) abc = 1.
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05 Aug 2003, 10:48
I might be falling for the Joe Bloggs answer here but:
C
given (2) gives us a common denominator if we were to require adding 1/each = 1
and (1) gives us the derived numerator = 1
Thus we have 1/1 as the answer.
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05 Aug 2003, 11:16
how did u get the derived numerator as a+b+c? Shouldn't it be ab+bc+ca?
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05 Aug 2003, 21:03
I will vote with Prashant.
E. None of the conditions are sufficient.
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05 Aug 2003, 21:19
E. I tryed different numbers and got different answers.
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06 Aug 2003, 02:44
Konstantin Lynov wrote:
What s the numerical value of 1/a + 1/b + 1/c ?
1) a+b+c = 1;
2) abc = 1.
Solution: Let's simplify the equation first. This leads us to (bc+ac+ab)/abc;
Neither statement one, not two gives us sufficient info to find values for the given equation. Combining two statements together, we get three variables and only two equations. Thus, E is the correct answer.
_________________
Respect,
KL
Re: DS practice #4 [#permalink] 06 Aug 2003, 02:44
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## Sunday, August 5, 2012
### YeeHaw! Wild About Numbers!
Megan and I joined together to make this new number packet! Here’s what it has! We just listed it on tpt.
There are a variety of games for dot plates, rekenreks, fives frames, tens frames and dominoes. For each of the tools there are games that address these common core standards:
· Solve addition and subtraction word problems with 10/20.
· Identify whether the number is greater than or less than another number.
· Relate counting to addition and subtraction.
· Decompose numbers less than or equal to 10 into pairs in more than one way.
· Represent addition and subtraction with objects, fingers, mental images, drawings, sounds, acting out situations, verbal explanations, expressions, or equations.
· Fluently add and subtraction within 5/10.
· Add and subtraction within 10/20.
· Determine the unknown whole number in an addition or subtraction equation.
· Determine if equations involving addition and subtraction are true or false.
· Solves addition and subtraction problems within 10/20.
· For any number from 1 to 9, find the number that makes 10 when added to the given number.
· Understand subtraction as an unknown addend problem.
The same games are played with each of the math tools allowing you to spend time practicing the standards instead of teaching another new game. There are 8-9 games for each tool.
For each tool (rekenreks, fives frames, etc.) the following games and items are included:
· Mini Lesson Questions for “Math Talks”
· Teacher Set
· Student Set
· My Number Books
· Add it Up! Cowboy!
· Number Riddles
· Find Your Partner
· Boot Stompin’ Game
· Memory Dominoes
· Domino Dice
· Make 5, Make 10
The large sets of rekenreks, fives frames, tens frames, dot plates and dominoes are used in large group. Questions are provided for each tool that can be used in mini lessons or during your “math talk” or mental math time. For example the questions might be:
1. Flash a domino card. Say, “My card has 6 dots. Make your card have enough dots to make 10 dots together.” Repeat.
2. Flash a domino. Ask, “Was that closer to 5 or closer to 10?” Repeat.
Your kiddos will say YeeHaw!....and so will you!
We Teach Like A Rockstar said...
You were born to be a kindergarten teacher and create activities that engage the students and teachers! Another great unit that I cannot wait to buy from you! I am going to be presenting in INDY at the SDE conference. I was hoping to see you there. I guess we will have to cross paths another time. Enjoy your school year with your daughters! Amber ~ Kindergarten Rocks Blog
Kim and Megan said...
Welcome to the SDE world! It is great how many bloggers are now coming on board! It is a great company. I have worked with them for about 20 years!Hope to see you "on the road"!
Deedee Wills said...
So cute!!! Smart! Smart! Smart!
Just Wild About Teaching said...
this is so cute =) thanks for sharing
Just Wild About Teaching
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If one root is common between two quadratic e...
If one root is common between two quadratic equations, x2 + 4x + 6 and ax2 + bx+ c, then find the product of the roots of ax2 + bx+ c. (Given: a, b and c are natural numbers.)
• a)
4
• b)
3
• c)
-2
• d)
6
FREE This question is part of
If one root is common between two quadratic equations, x2 + 4x + 6 and...
Solution: Discriminant of (x2 + 4x + 6) < 0 The equation will not have real roots. i.e., x2 + 4x + 6 has complex roots and they are conjugates of each other. One of the roots is also a root of ax2 + bx + c. Conjugate roots always occur in pairs.
So, the product of the roots of (ax2 + bx + c) is same as that of (x2 + 4x + 6). v a = 1 , b = 4 and c = 6 The product = c/a = 6 Hence, option 4.
View all questions of this test
Related Free Doc
If one root is common between two quadratic equations, x2 + 4x + 6 and...
To find the product of the roots of the quadratic equation ax^2 + bx + c, we need to first find the roots of the equation x^2 + 4x + 6. Let's solve this equation first.
Solving the equation x^2 + 4x + 6 = 0
Using the quadratic formula, the roots of this equation can be found using the formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For the given equation, a = 1, b = 4, and c = 6.
x = (-4 ± √(4^2 - 4(1)(6)))/(2(1))
x = (-4 ± √(16 - 24))/2
x = (-4 ± √(-8))/2
x = (-4 ± 2√2i)/2
x = -2 ± √2i
So, the roots of the equation x^2 + 4x + 6 = 0 are -2 + √2i and -2 - √2i.
Now, let's find the product of the roots of the given quadratic equation ax^2 + bx + c.
Product of the roots = (root1) * (root2)
= (-2 + √2i) * (-2 - √2i)
= (-2)^2 - (√2i)^2
= 4 - 2i^2
= 4 - 2(-1)
= 4 + 2
= 6
Therefore, the product of the roots of the quadratic equation ax^2 + bx + c is 6.
Hence, the correct answer is option D) 6.
Explore Courses for CAT exam
If one root is common between two quadratic equations, x2 + 4x + 6 and ax2 + bx+ c, then find the product of the roots of ax2 + bx+ c. (Given: a, b and c are natural numbers.)a)4b)3c)-2d)6Correct answer is option 'D'. Can you explain this answer?
Question Description
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MTEL Test Information Guide
# Test Information Guide
## Overview and Test ObjectivesField 65: Middle School Mathematics
### Test Overview
Format Computer-based test (CBT); 100 multiple-choice questions, 2 open-response items 4 hours (does not include 15-minute CBT tutorial) 240
The Massachusetts Tests for Educator Licensure (MTEL) are designed to measure a candidate's knowledge of the subject matter contained in the test objectives for each field. The MTEL are aligned with the Massachusetts educator licensure regulations and, as applicable, with the standards in the Massachusetts curriculum frameworks.
The test objectives specify the content to be covered on the test and are organized by major content subareas. The chart below shows the approximate percentage of the total test score derived from each of the subareas.
The test assesses a candidate's proficiency and depth of understanding of the subject at the level required for a baccalaureate major according to Massachusetts standards. Candidates are typically nearing completion of or have completed their undergraduate work when they take the test.
Sub area I 20%, Sub area II 30%, Sub area III 20%, Sub area IV 10%, and Sub area V 20%.
### Test Objectives
Table outlining test content and subject weighting by sub area and objective.
Subareas Range of Objectives Approximate Test Weighting
Multiple-Choice
I Number System and Quantity 01–02 20%
II Algebra, Functions, and Modeling 03–07 30%
III Geometry and Measurement 08–11 20%
IV Statistics and Probability 12–13 10%
80%
Open-Response*
V Integration of Knowledge and Understanding
Mathematics Curriculum Framework: Concepts and Skills 14 10%
Statistics, Probability, and Algebra 15 10%
20%
*The open-response items may relate to topics covered in any of the subareas.
#### Subarea I–Number System and Quantity
##### 0001: Apply the structure and properties of number systems.
For example:
• Apply and extend understanding of the place value system to represent, estimate, and perform operations on the full system of real numbers in a variety of ways (e.g., graphic, numerical, physical, symbolic).
• Reason about the order and absolute value of rational numbers.
• Apply and extend understanding of number systems (e.g., complex numbers, irrational numbers).
• Analyze the relationships between operations (e.g., multiplication as repeated addition).
• Apply and extend understanding of prime and composite numbers, divisibility, least common multiples, and greatest common factors to model and solve real-world and mathematical problems.
• Analyze standard algorithms for operations on real numbers (e.g., decimals, fractions, integers).
• Justify and apply order of operations and the use of inverse and identity elements to solve problems.
• Model and solve problems using the properties of integer exponents (e.g., scientific notation).
• Apply and extend understanding of number properties (e.g., associative, commutative, distributive) to model and solve problems.
##### 0002: Use rational numbers, ratios, and proportional relationships.
For example:
• Represent fractions, arithmetic operations on fractions, and problems involving fractions using a variety of visual models (e.g., area models, diagrams, tiles) and equations.
• Solve real-world and mathematical problems with integers and other rational numbers (e.g., decimals, fractions).
• Apply ratios, rates, unit rates, and proportionality to solve a variety of problems, including percent problems (e.g., discounts, interest, percent increase and decrease, taxes, tips).
• Solve problems involving conversions between decimals (e.g., finite, repeating), percents, and fractions using visual models and strategies based on place value and properties of operations.
• Compare and interpret rational numbers (e.g., equivalent fractions, multiplication as scaling with fractions).
• Use benchmark numbers, rounding, and number sense to estimate mentally and assess the reasonableness of solutions to problems.
#### Subarea II–Algebra, Functions, and Modeling
##### 0003: Use patterns to model and solve problems.
For example:
• Make conjectures about patterns presented in numerical, geometric, and tabular forms.
• Represent patterns and relations using symbolic notation.
• Identify patterns of change created by linear, quadratic, and exponential functions.
• Model and solve problems using patterns, relations, sequences, and series (e.g., arithmetic, Fibonacci, geometric).
• Identify, express, and apply patterns of change in proportional, linear, and inversely proportional situations.
##### 0004: Apply algebraic techniques to expressions and equations.
For example:
• Translate between verbal descriptions and algebraic sentences that represent mathematical situations in various forms (e.g., graphic, numerical, symbolic, tabular).
• Model situations with algebraic expressions, equations, and inequalities, including those with fractional and decimal coefficients and those with infinitely many or no solutions.
• Evaluate algebraic expressions for a given value of a variable and express one variable in terms of another variable.
• Apply properties of real numbers in algebraic contexts to manipulate and simplify algebraic expressions (e.g., polynomials, rational expressions) and solve equations and inequalities, including those with fractional and decimal coefficients and integer exponents.
##### 0005: Demonstrate knowledge of relations and functions.
For example:
• Distinguish between relations and functions using a variety of representations (e.g., graphic, symbolic, tabular, verbal) and use relations and functions to describe relationships between quantities.
• Analyze various representations (e.g., graphic, symbolic, tabular, verbal) of functions and relations with respect to their characteristics (e.g., continuity, domain, intercepts, inverses).
• Generate, interpret, and translate between various representations (e.g., algebraic, graphic, tabular) of real-world situations.
• Identify and analyze piecewise-defined functions and addition, subtraction, and composition of functions from real-world and mathematical situations.
• Identify the effects of transformations such as f of open parens x plus k close parens, f of open parens x close parens plus k, and k times f of open parens x close parens on the graph of a function.
##### 0006: Apply the properties of linear relations and functions.
For example:
• Analyze connections between proportional relationships, direct variation, rates of change, and linear models and use these connections to build linear functions.
• Analyze the relationship between the equation of a line and its graph and interpret slope and intercepts in real-world and mathematical contexts.
• Determine the equation of a line from different types of information (e.g., graph, one point and slope, two points).
• Apply a variety of methods for solving systems of linear equations and inequalities (e.g., elimination, graphing, substitution).
• Apply knowledge of linear equations and inequalities, systems of equations, linear functions, and slope of a line to analyze situations and solve problems.
##### 0007: Apply the principles and properties of nonlinear relations and functions.
For example:
• Identify and express patterns of change in quadratic and exponential functions and the types of real-world relationships that these functions can model.
• Translate between different representations (e.g., algebraic, graphic, tabular, verbal) of quadratic and exponential functions.
• Analyze properties and features of quadratic relations, functions, and systems (e.g., graphs, maxima/minima, real roots).
• Model and solve problems involving quadratic relations, functions, and systems using a variety of techniques (e.g., completing the square, factoring, graphing, quadratic formula).
• Model and solve problems involving exponential growth (e.g., compound interest, population growth) and decay (e.g., half-life).
• Analyze properties and graphs of linear, quadratic, exponential, and absolute value functions.
#### Subarea III–Geometry and Measurement
##### 0008: Apply principles, concepts, and procedures related to measurement.
For example:
• Apply and extend understanding of quantities and units to convert within measurement systems and use these conversions in solving multistep, real-world problems.
• Apply formulas to find measures (e.g., area, length, volume) involving two- and three-dimensional figures (e.g., composite shapes) including those with fractional measures.
• Analyze the effect of changing linear dimensions on measures of length, area, or volume.
• Calculate and analyze the effect of measurement error and rounding on computed quantities.
• Use degrees to calculate, estimate, and analyze angle measures (e.g., coterminal angles).
• Solve real-world and mathematical problems using right triangle trigonometry (e.g., cosine, sine, tangent).
##### 0009: Apply the principles of Euclidean geometry and proof.
For example:
• Identify, use, and understand the relationships between the building blocks (e.g., postulates, undefined terms) of Euclidean geometry.
• Classify geometric relationships and solve problems using the properties of lines (e.g., parallel, perpendicular) and angles (e.g., supplementary, vertical).
• Apply the principles of congruence, similarity, and proportional and spatial reasoning (e.g., indirect measurement, informal geometric constructions, scale drawings) to solve real-world and mathematical problems.
• Analyze and prove theorems within the axiomatic structure of Euclidean geometry.
##### 0010: Apply properties of two- and three-dimensional figures.
For example:
• Classify plane figures in a hierarchy based on their properties (e.g., angles, diagonals, sides).
• Apply deductive reasoning to justify properties of and relationships between triangles, quadrilaterals, and other polygons.
• Apply the Pythagorean theorem and its converse to solve real-world and mathematical problems and to derive special right triangle relationships.
• Apply properties of arcs, angles, and segments associated with circles to solve real-world and mathematical problems.
• Analyze the properties and compare the measures (e.g., surface area, volume) of three-dimensional figures (e.g., cones, cylinders, prisms, pyramids, spheres).
• Translate between two- and three-dimensional representations of geometric figures (e.g., conic sections, cross sections, nets, perspective and isometric drawings).
• Derive properties of three-dimensional figures from two-dimensional figures.
##### 0011: Apply the principles and properties of coordinate and transformational geometries.
For example:
• Classify, represent, and analyze geometric figures (e.g., circles, polygons) in the coordinate plane.
• Apply concepts of distance, midpoint, slope, and parallel and perpendicular lines to classify and analyze figures in the coordinate plane.
• Apply transformations (e.g., dilations, reflections, rotations, translations) to figures in the coordinate plane and analyze their effects on congruence, similarity, and symmetry.
• Use coordinate and transformational geometry to prove theorems and solve problems.
#### Subarea IV–Statistics and Probability
##### 0012: Understand the principles, techniques, and applications of statistics.
For example:
• Construct and interpret frequency distributions, tables, charts, and graphs (e.g., box plots, dot plots, histograms, stem-and-leaf plots).
• Describe and summarize numerical data sets by identifying clusters, modes (e.g., peaks), gaps, and symmetry and by considering the context in which the data were collected.
• Use measures of center (e.g., mean, median) and measures of variability (e.g., interquartile range, standard deviation) for numerical data from random samples to draw informal comparative inferences about two populations, and determine the effects of transformations on these measures.
• Demonstrate knowledge of normal probability distributions and use percentile scores to solve problems.
• Evaluate real-world situations to determine appropriate sampling techniques and methods for gathering and organizing data.
• Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities, interpret correlation coefficients, and solve problems involving linear regression models.
##### 0013: Understand the principles of probability.
For example:
• Use and interpret a variety of representations for situations involving probability (e.g., organized lists, tables, tree diagrams, Venn diagrams).
• Compute theoretical probabilities for simple and compound events using a variety of approaches (e.g., addition and multiplication rules).
• Select simulations to generate frequencies for compound events.
• Make connections between probability and geometry.
• Use probability models to explore and understand real-world phenomena.
#### Subarea V–Integration of Knowledge and Understanding
##### 0014: Prepare an organized, developed analysis on a topic cited from the Massachusetts Mathematics Curriculum Framework grades 5–8.
For example:
• Identify related prerequisite skills and explain their relevance to the provided standard.
• Create appropriate representations to model and describe the standard.
• Critique whether a given situation aligns to the standard.
##### 0015: Prepare an organized, developed analysis on a topic related to one or more of the following: statistics, probability, and algebra.
For example:
• Create appropriate graphs and/or diagrams, including all proper labels, to model and describe a given real-world situation.
• Apply appropriate mathematical techniques to make a prediction or comparison regarding the situation.
• Make a recommendation or argument based on the prediction or comparison.
• Discuss factors that could influence the accuracy of the prediction/comparison and recommendation/argument.
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# GSEB Class 9 Maths Notes Chapter 12 Surface Areas and Volumes
This GSEB Class 9 Maths Notes Chapter 12 Surface Areas and Volumes covers all the important topics and concepts as mentioned in the chapter.
## Surface Areas and Volumes Class 9 GSEB Notes
Area of a triangle:
We are familiar with the following formula used to find the area of a triangle which is applicable to all the types of triangle:
Area of a triangle = $$\frac{1}{2}$$ × base × height
For some particular types of triangle, the same formula can be represented in other forms.
For a right-angled triangle, the formula can be taken as below:
Area of a right triangle = $$\frac{1}{2}$$ × product of the sides forming the right angle
• In case of an equilateral triangle, its height can be found using Pythagoras’ theorem. In that case the formula will be changed as below:
Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}$$ × (side)2
• In case of an isosceles triangle, if all the sides are known, the height corresponding to the base can be found using Pythagoras’ theorem and the original formula can be , used.
• But in case of a scalene triangle, the height cannot be found even if all the three sides are known.
Heron’s formula:
For any type of triangle, if all the sides are known, its area can be calculated by the following formula given by mathematician Heron:
Area of a triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
Here, a, b and c denote the length of three sides of the triangle and s = $$\frac{a+b+c}{2}$$ is the semi perimeter of the triangle.
Units of area:
Area of small planar figures are usually denoted using units square metre (m2) or square centimetre (cm2), etc. The area of plots of land are usually denoted in hectares. 1 hectare = 10,000 m2
An important result: if the sides of a triangle measure 2n2 + 1, 2n2 + 2 and 4n2 + 1, where n is a natural number, its semiperimeter s will be 4n2 + 2 and its area will be 2n(2n2 + 1) = n × s. Here, area will be given by a natural number always, not an irrational number. There can be other forms also.
Example 1:
Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm (see the given figure).
Here, the perimeter of the triangle = 32 cm. a = 8 cm and b = 11 cm.
∴ Third side c = 32 cm – (8 + 11) cm = 13 cm
Here, 2s = 32, i.e., s = 16cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11)cm = 5cm and
s – c = (16 – 13) cm = 3 cm.
Now, area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{16 \times 8 \times 5 \times 3}$$ cm2
= 8√30 cm2
Example 2:
A triangular park ABC has sides 120 m, 80 m and 50 m (see the given figure). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ 20 per metre leaving a space 3 m wide for a gate on one side.
For finding area of the park, we have 2s = 50m + 80m + 120 m = 250 m s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m and
s – c = (125 – 50) m = 75 m.
Area of the park = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{125 \times 5 \times 45 \times 75}$$ m2
= 375√15 m2
Also. perimeter of the park = AB + BC + CA
= 250m
Then, length of the wire needed for fencing
= 250 m – 3 m (to be left for gate)
= 247 m
So, the cost of fencing = ₹ 20 × 247 = ₹ 4940
Example 3:
The sides of a triangular plot are in the ratio of 3:5:7 and its perimeter is 300 m. Find its area.
Suppose that the sides, in metres, are 3x, 5x and 7x (see the given figure).
Then, 3x + 5x + 7x = 300 (perimeter of the triangle)
∴ 15x = 300
∴ x = 20.
So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m, i.e., 60 m, 100 m and 140 m.
Here, s = $$\frac{60+100+140}{2}$$m = 150 m
∴ Area of the plot
= $$\sqrt{150(150-60)(150-100)(150-140)}$$m2
= $$\sqrt{150 \times 90 \times 50 \times 10}$$ m2
= 1500√3 m2
Application of Heron’s formula in finding areas of quadrilaterals :
In a quadrilateral, if the measures of four sides and one diagonal / is given, its area can be found by finding the areas of two triangles formed by the diagonal with the help of Heron’s formula.
In quadrilateral ABCD, if h1 and h2 are the lenghts of altitudes drawn from B and D respectively to AC, then
Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD
= ($$\frac{1}{2}$$ × AC × h1) + ($$\frac{1}{2}$$ × AC × h2
= $$\frac{1}{2}$$ × AC × (h1 + h2)
We do have the formule for the area of some special quadrilaterals as:
Area of a square = (length)2 = $$\frac{(\text { diagonal })^{2}}{2}$$
Area of a rectangle = length × breadth
Area of a parallelogram = base × corresponding altitude
Area of a rhombus = $$\frac{1}{2}$$ × product of two diagonals
Area of a trapezium = $$\frac{1}{2}$$ × stun of parallel sides × distance between parallel sides
Example 1:
Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions (see the given figure). She divided the field in two parts by joining the midpoint of the longest side to the opposite vertex and grew potatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m )
Let, ABC be the field where wheat is grown.
Also let ACD be the field which has been divided in two parts by joining C to the midpoint E of AD.
For the area of triangle ABC, we have
a = 200 m, b = 240 m, c = 360 m
∴ s = $$\frac{200+240+360}{2}$$ m = 400 m
So, area for growing wheat
= $$\sqrt{400(400-200)(400-240)(400-360)}$$m2
= $$\sqrt{400 \times 200 \times 160 \times 40}$$ m2
= 16000 √2 m2 = 1.6 × √2 hectares
= 2.26 hectares (nearly)
Let us now calculate the area of triangle ACD.
Here, we have s = $$\frac{240+320+400}{2}$$m = 480 m
So, area of ∆ACD
= $$\sqrt{480(480-240)(480-320)(480-400)}$$m2
= $$\sqrt{480 \times 240 \times 160 \times 80}$$m2
= 38400 m2 = 3.84 hectares
We notice that the line segment joining the b midpoint E of AD to C divides the triangle AGD in two parts equal in area.
Therefore, area for growing potatoes
= area for growing onions
= (3.84 + 2) hectares = 1.92 hectares
Example 2:
Students of a school staged a rally for \ cleanliness campaign. They walked through ; the lanes in two groups. One group walked
through the lanes AB, BC and CA; while the other through AC, CD and DA (see the given figure). Then they cleaned the area \ enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠B = 90°, which group cleaned more area and by how much ? Find the total area cleaned by the students (neglecting the width of the lanes).
In ∆ABC, ∠B = 90°; AB = 9 m and BC = 40 m.
Then, AC = $$\sqrt{9^{2}+40^{2}}$$ m
= $$\sqrt{81+1600}$$ m
= √1681 m = 41 m
Therefore, the first group has to clean the area of triangle ABC, which is right angled. Area of ∆ ABC = $$\frac{1}{2}$$ × base × height
= $$\frac{1}{2}$$ × 40 × 9 m2 = 180 m2
The second group has to clean the area of triangle ACD, which is scalene having sides 41m, 15 m and 28 m.
Here, s = $$\frac{41+15+28}{2}$$m = 42 m
Then, area of ∆ACD
= $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{42(42-41)(42-15)(42-28)}$$ m2
= $$\sqrt{42 \times 1 \times 27 \times 14}$$ m2 = 126 m2
So, first group cleaned 180 m2 which is (180 – 126) m2, i.e., 54 m2 more than the area cleaned by the second group.
Total area cleaned by all the students = (180 + 126) m2 = 306 m2.
Example 3:
Sanya has a piece of land which is in the shape of a rhombus (see the given figure). She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?
Let, ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m
i.e., AB = AD = 100 m
Let, diagonal BD=160m
Then, semiperimeter s of ∆ABD is given by
s = $$\frac{100+100+160}{2}$$ m = 180 m
Hence, area of ∆ABD
= $$\sqrt{180(180-100)(180-100)(180-160)}$$ m2
= $$\sqrt{180 \times 80 \times 80 \times 20}$$ m2
= 4800 m2
Therefore, each of them will get an area of 4800 m2.
Alternative method:
Draw CE ⊥ BD (see the given figure).
As BD = 160 m, we have
DE = 160 m2 = 80 m
Now, DE2 + CE2 = DC2
CE = $$\sqrt{\mathrm{DC}^{2}-\mathrm{DE}^{2}}$$
CE = $$\sqrt{100^{2}-80^{2}}$$ m = 60 m
Therefore, area of ∆BCD = $$\frac{1}{2}$$ × 160 × 60 m2
= 4800 m2
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Скачать презентацию The Mass of a One Cent Coin T
3866f83b26175a89c04dec103d656eb8.ppt
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The Mass of a One Cent Coin T. H. The Professional Development Service for Teachers is funded by the Department of Education and Skills under the National Development Plan
What is the Mass of a One Cent Coin? PDST Resources for Leaving Certificate Physics 2
The mass of a 1 cent coin § A student placed a 1 cent coin on this balance. § The scale gave the mass as zero. Why? PDST Resources for Leaving Certificate Physics 3
Sensitivity § The mechanical balance is not sensitive enough to measure the mass of a 1 cent coin. § How can we solve this problem? PDST Resources for Leaving Certificate Physics 4
More coins please! § We found the mass of forty coins to be 90 g. § What is the average mass of each coin? PDST Resources for Leaving Certificate Physics 5
Answer § The average mass of each coin is 90 ÷ 40 = 2. 5 g PDST Resources for Leaving Certificate Physics 6
Question § Would you say that each coin has the same mass? Say why. PDST Resources for Leaving Certificate Physics 7
Answer § To see if each coin has the same mass one could try finding the mass of different groups of, say, 20 coins. If the groups have different masses, then the coins are not all the same. PDST Resources for Leaving Certificate Physics 8
A worn coin § If we had an extremely accurate balance, we could find the individual mass of different coins. § It would be no surprise if their masses were slightly different, due to different amounts of wear. PDST Resources for Leaving Certificate Physics 9
What do you notice in this picture? PDST Resources for Leaving Certificate Physics 10
ZERO ERROR § The electronic balance showed a reading of 10 g when nothing was being weighed on it. § This zero error must be subtracted from subsequent readings. PDST Resources for Leaving Certificate Physics 11
Average mass § There are 29 coins on this balance. Find the average mass of each coin. (Don’t forget the 10 g zero error. ) PDST Resources for Leaving Certificate Physics 12
Answer § 77. 5 – 10 = 67. 5 ÷ 29 = 2. 33 g § The last time we got an average of 2. 5 g. How can we tell which scale is more accurate? PDST Resources for Leaving Certificate Physics 13
Calibration § We can test a balance with a known mass, e. g. a 100 g mass. PDST Resources for Leaving Certificate Physics 14
Average mass § There are 35 coins on this balance. Find the average mass of a one cent coin. PDST Resources for Leaving Certificate Physics 15
Answer § 80÷ 35 = 2. 29 g § We now have three results for the average mass of a one cent coin: 2. 5 g, 2. 33 g and 2. 29 g. § Are one cent coins made with small difference in mass? PDST Resources for Leaving Certificate Physics 16
The Standard Mass of a One Cent Coin. § In fact, one cent coins are all minted to exact specifications. The standard mass of a one cent coin is 2. 27 g. § Find the percentage error for each of our results: 2. 5 g, 2. 33 g and 2. 29 g. PDST Resources for Leaving Certificate Physics 17
Percentage Errors PDST Resources for Leaving Certificate Physics 18
Specifications of a One Cent Coin Diameter (mm): 16. 25 Thickness (mm): 1. 67 Weight (g): 2. 27 Shape: Round Colour: Red Composition: Copper covered steel Edge: Smooth (See: http: //www. fleur-decoin. com/eurocoins/coins. asp) PDST Resources for Leaving Certificate Physics 19
Question § How would find out how many coins are in this bag. § You’re not allowed to open the bag or count the coins through the plastic. PDST Resources for Leaving Certificate Physics 20
A Suggestion § You could find the mass of the bag of coins. § The mass of the plastic is negligible. § If the standard mass of one cent is 2. 27 g, and the bag of coins has mass 337 g, how many coins are in the bag? PDST Resources for Leaving Certificate Physics 21
Answer This suggests that there are 148 coins in total. The mass of the plastic bag accounts for the 0. 49 g. PDST Resources for Leaving Certificate Physics 22
Further Questions § The mintage of 1 c coins from 1999 to 2006 was 8, 400, 000 coins. § Find the total mass of this mintage in tonnes. PDST Resources for Leaving Certificate Physics 23
ANSWERS PDST Resources for Leaving Certificate Physics 24
Question § How could you prove that a 1 c coin is not made from solid copper? PDST Resources for Leaving Certificate Physics 25
One Cent Coins Are Not Made From Solid Copper § You can pick up a 1 c coin with a magnet. This couldn’t be done if they were made from solid copper. (Try picking up a copper calorimeter with a magnet. ) § You could find the density of a 1 c coin and compare it with the density of copper. PDST Resources for Leaving Certificate Physics 26
Question § How much money would you have if you had your mass in 1 c coins? PDST Resources for Leaving Certificate Physics 27
Your Mass in One Cent Coins § Suppose you have a mass of 65 kg. Then: PDST Resources for Leaving Certificate Physics 28
Trivia § The Greek one cent coin is called a “lepto”. § The elementary particles – electrons, muons, tauons and neutrinos – are called “leptons. ” PDST Resources for Leaving Certificate Physics 29
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Math Central - mathcentral.uregina.ca
Quandaries & Queries
Q & Q
Topic: inverses
start over
8 items are filed under this topic.
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The inverse of y = 5^x 2014-03-07 From Melody:how do you find the inverse of y = 5^xAnswered by Penny Nom. What is f^-1(3) when f(x)=2x-1? 2009-01-03 From Peter:how do you do functions like f^-1(3) when f(x)=2x-1?Answered by Penny Nom. Inverses of Quadratic Functions 2007-10-11 From Elliot:I'm a special education tchr trying to demonstrate how to find the inverse (using algebra) of quadratic functions. Any help would be greatly appreciated.Answered by Steve La Rocque and Harley Weston. y = log(x) + x. Solve for x. 2005-08-26 From Alain:I have the following equation: y = ln(x) + x How do I solve for x? Answered by Penny Nom. An inverse 2004-09-09 From Hillary:I cannot figure out the inverse of this function. f(x)= 1/2x -1Answered by Penny Nom. Differentiating inverses 2002-11-20 From Amy:f(x)= x3+x+1, a=1 find g'(a) (g = f -1). I am having trouble finding g(a).Answered by Penny Nom. Range of a function 2000-11-21 From David Bell:Given a rational function such as f(x) = (8x-3)/(4x-1). How can the range be found.Answered by Penny Nom. Inverses of functions 1999-11-01 From Leanne Hickey:Let f(x) = 2x2 -3x + 2. Find f-1(4) given the fact that f(2) = 4. So the question is finding the inverse of 4, he said it's easier than it looks.Answered by Penny Nom.
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How to find centroid of a hemisphere using Pappus's centroid theorem?
• kshitij
In summary, Pappus's centroid theorem can be used to calculate the centroid of a semi-circular ring and disc of radius r by finding the distance d from the center of mass to the base. This method can also be extended to calculate the centroid of hemispheres and cones by using the concept of wedge elements and considering the distance from a flat surface instead of a diameter. The volume of a 4D ball, or 3-sphere, can be calculated as ##\frac 12\pi^2r^4##, which is used in the formula for finding the distance to the centroid of a hemisphere. Visualization and understanding of these concepts may require practice and the use of figures.
kshitij
Homework Statement
To find the centroid of a hollow and solid hemisphere.
Relevant Equations
Pappus's centroid theorem
I recently learned how to calculate the centroid of a semi-circular ring of radius ##r## using Pappus's centroid theorem as
##\begin{align}
&4 \pi r^2=(2 \pi d)(\pi r)\nonumber\\
&d=\frac {2r}{\pi}\nonumber
\end{align}##
Where ##d## is the distance of center of mass of the ring from its base.
Similarly its second theorem can also be used to calculate C.O.M of semi-circular disc of radius ##r##,
##\begin{align}
&\frac{4}{3} \pi r^3=(2 \pi d)(\dfrac {\pi r^2}{2})\nonumber\\
&d=\frac {4r}{3 \pi}\nonumber
\end{align}##
Now this is great because now I don't have to do the lengthy calculations to find the C.O.M using ##d=\dfrac{\int ydm}{M}##
So I wanted to extend this to find the centroid of hemispheres and cones but then how do I find the area or volume traced by rotating a hemisphere?
How do I extend this to 3D object or can I extend this to 3D objects?
One way to understand Pappus' centroid theorems is in terms of wedge elements.
If we take a semicircular disc and rotate it only through a small angle ##d\theta## we produce a wedge element. Now copy this, not rotated around the axis but into parallel planes. Like laying out the segments of an orange, core piece uppermost on a flat table.
We can think of the thickness of a wedge at any point as a weighting. This is proportional to the distance from the core. E.g. compress each segment into a semicircular lamina of some uniform thickness but maintaining the mass, so the further from the core the greater the density. The flattening is arranged so that the thickness is preserved at the centroid.
This weighting pattern is how the centroid of the original semicircle is defined. It is the "torque arm" of each bit of the semicircle about the axis.
The purpose of writing all that is to find a way to think about it in higher dimensions.
For the centroid of a hemisphere, the 'axis' is the flat surface. The weighting is the distance from that surface, not from a diameter of it.
Starting with a hemisphere, we can create a row of hemispheric elements. In each element, the density distribution is weighted to represent the distance from the flat surface.
Each hemisphere has now a thickness in thefourth dimension, and these total ##2\pi d##.
On this basis, the distance from the flat surface of a hemisphere to its centroid should be (vol of 4D ball)/((2##\pi##)(vol of hemisphere))##=\frac{\frac 12\pi^2r^4}{2\pi.\frac 23\pi r^3}=\frac 38r##, which I believe is correct.
Note that a 4D ball is termed a 3-sphere, not a 4-sphere.
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haruspex said:
One way to understand Pappus' centroid theorems is in terms of wedge elements.
If we take a semicircular disc and rotate it only through a small angle ##d\theta## we produce a wedge element. Now copy this, not rotated around the axis but into parallel planes. Like laying out the segments of an orange, core piece uppermost on a flat table.
We can think of the thickness of a wedge at any point as a weighting. This is proportional to the distance from the core. E.g. compress each segment into a semicircular lamina of some uniform thickness but maintaining the mass, so the further from the core the greater the density. The flattening is arranged so that the thickness is preserved at the centroid.
This weighting pattern is how the centroid of the original semicircle is defined. It is the "torque arm" of each bit of the semicircle about the axis.
The purpose of writing all that is to find a way to think about it in higher dimensions.
For the centroid of a hemisphere, the 'axis' is the flat surface. The weighting is the distance from that surface, not from a diameter of it.
Starting with a hemisphere, we can create a row of hemispheric elements. In each element, the density distribution is weighted to represent the distance from the flat surface.
Each hemisphere has now a thickness in thefourth dimension, and these total ##2\pi d##.
On this basis, the distance from the flat surface of a hemisphere to its centroid should be (vol of 4D ball)/((2##\pi##)(vol of hemisphere))##=\frac{\frac 12\pi^2r^4}{2\pi.\frac 23\pi r^3}=\frac 38r##, which I believe is correct.
Note that a 4D ball is termed a 3-sphere, not a 4-sphere.
I don't understand what you said there about understanding Pappus theorem, but I just wanted to find the area/volume enclosed by a hollow hemisphere, solid hemisphere, hollow cone and a solid cone on rotating these about their base, i.e., how did you get
Volume of 4D ball = ##\frac 12\pi^2r^4## ?
How do you manage to imagine all these in your head without a figure or anything? I cannot understand even the simplest things without a diagram!
Delta2
kshitij said:
I don't understand what you said there about understanding Pappus theorem, but I just wanted to find the area/volume enclosed by a hollow hemisphere, solid hemisphere, hollow cone and a solid cone on rotating these about their base, i.e., how did you get
Volume of 4D ball = ##\frac 12\pi^2r^4## ?
How do you manage to imagine all these in your head without a figure or anything? I cannot understand even the simplest things without a diagram!
For the volumes and surface volumes of n-spheres, see https://en.m.wikipedia.org/wiki/N-sphere.
It would be difficult to illustrate what I wrote with diagrams. Visual imagination takes practice. Here's one you could try:
Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus. Now I paint a red line on the outside, the short way around. These two coloured circles are linked.
Now I make a small hole in the wall of the torus and pull it inside out through the hole. So the blue line is now outside and the red line inside. Are they still linked?
kshitij and Delta2
haruspex said:
For the volumes and surface volumes of n-spheres, see https://en.m.wikipedia.org/wiki/N-sphere.
It would be difficult to illustrate what I wrote with diagrams. Visual imagination takes practice. Here's one you could try:
Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus. Now I paint a red line on the outside, the short way around. These two coloured circles are linked.
Now I make a small hole in the wall of the torus and pull it inside out through the hole. So the blue line is now outside and the red line inside. Are they still linked?
By those two circles being linked do you mean that they are concentric??
If so, then to answer your problem I think maybe that first we extend that small hole we made so that we can slit the tube (torus) along its length (along its circumference) such that now instead of a tube we are left with a strip of rubber and then after we turn the slitted tube (strip) inside out we stitch it back together along its length such that now we are only left with the small hole which we started with and so our entire arrangement is the same just the only difference is that the blue line is now outside and the red one is inside and they are still concentric or linked.
I assumed that changing the size of the hole will not change our problem so we can increase the hole such that its long enough to cover the entire length of the tube along the circumference such that we can slit the whole tube with that small hole.
kshitij said:
I assumed that changing the size of the hole will not change our problem
I don't have a proof but I have heard that changing the size of a hole in any shape doesn't change it homeomorphically (idk if that's the word)
that is why a ring is homomorphic to a cylinder or a disc is homomorphic to a cone or a cup
kshitij said:
do you mean that they are concentric??
No. Lnked like links in a chain. Each passes through the other.
haruspex said:
No. Linked like links in a chain. Each passes through the other.
What?? How are they linked like a chain? I thought that those two circles are like rings one surrounded by the other and there is a layer of rubber in between them separating each other
haruspex said:
Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus
That means that you painted a circle along the outer circumference (because its the long way round) of the tube but on the inside surface of the tube
haruspex said:
Now I paint a red line on the outside, the short way around.
And this means that you painted the circle on the inside circumference of the tube but this time on the outside surface
kshitij said:
And this means that you painted the circle on the inside circumference of the tube but this time on the outside surface
No, think of a bicycle tube. The short way around is the way you would pick it up with one hand, fingers wrapped around it.
kshitij
haruspex said:
No, think of a bicycle tube. The short way around is the way you would pick it up with one hand, fingers wrapped around it.
Oh I get it now! (hopefully)
Okay so here is my second attempt
I use the same process as earlier that is I increase the size of the hole such that the tube is now a strip, and the hole is increased such that blue circle remains intact but the red circle is cut into a line, now if we flip this strip and stitch it back together we can see that red circle which earlier was wrapped around the blue circle is now wrapped outside the blue circle.
And the blue circle which was earlier on the inside surface after flipping comes on the outside surface and the red circle which was out is now on the inside surface thus they are untangled! i.e., they are no longer linked
kshitij said:
Okay so here is my second attempt
I use the same process as earlier that is I increase the size of the hole such that the tube is now a strip, and the hole is increased such that blue circle remains intact but the red circle is cut into a line, now if we flip this strip and stitch it back together we can see that red circle which earlier was wrapped around the blue circle is now wrapped outside the blue circle.
And the blue circle which was earlier on the inside surface after flipping comes on the outside surface and the red circle which was out is now on the inside surface thus they are untangled! i.e., they are no longer linked
So it should look somewhat like this,
sysprog
kshitij said:
Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
kshitij and Delta2
haruspex said:
Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
So that means that we cannot increase the length of the hole such that the tube is converted to a strip?
Because by this method it does completely makes sense why the red circle passed through the blue one (because we literally cut and stitched the red circle in the process)
But yes if that is not allowed then it is way too hard for me to imagine how a tube is inverted inside out through a small hole, infact I cannot believe how can people do this in their head!
kshitij said:
So that means that we cannot increase the length of the hole such that the tube is converted to a strip?
Yes and no.
Let's take the bicycle tube model and say the two radii are r<R.
If you cut it right around the direction of rotation of the wheel you get a short fat cylinder, length ## 2 \pi r## and radius R. If you cut it instead the other way you get a long narrow tube, length ## 2 \pi R## radius r.
If you do both you get a rectangle ## 2 \pi R## by ## 2 \pi r##
Sticking it back together, you have a choice of which to do first. Does the order matter?
haruspex said:
Yes and no.
Let's take the bicycle tube model and say the two radii are r<R.
If you cut it right around the direction of rotation of the wheel you get a short fat cylinder, length ## 2 \pi r## and radius R. If you cut it instead the other way you get a long narrow tube, length ## 2 \pi R## radius r.
If you do both you get a rectangle ## 2 \pi R## by ## 2 \pi r##
Sticking it back together, you have a choice of which to do first. Does the order matter?
Wait, can we cut it such that we get a rectangle? doesn't that break any rules? is a toroid with single hole isomorphic to rectangle?
I agree that with can cut it in both ways that you mentioned, i.e., cutting it such that we get a short fat cylinder and cutting it such that we get a long narrow tube (as it makes sense that a short fat cylinder when stretched along its length gives a long narrow pipe) but how can we do both of these together to get a rectangle?
Consider a short fat cylinder, which can be thought of as a rectangular strip of paper whose shorter sides are joined together, now if we want to get a rectangle from this strip, we will have to cut this strip and cutting is not equivalent to increasing a size of hole. As I understand increasing or decreasing the size of hole is allowed but cutting or stitching two surfaces aren't allowed.
I only increased the size of the hole in my attempt, and then flipped the strip (short fat cylinder) inside out and decreased the size of hole in reverse such that we got back the original shape with which we started with.
haruspex said:
Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
I was thinking about this a lot and even though I don't know how it will happen but I think that it should happen, i.e., the red and blue circles shouldn't be linked after inverting the tube inside out.
This is because, the red circle which was initially outside the tube, after flipping has to be inside that tube, and similarly blue circle which was initially inside, has to be on the outside after flipping. Now it doesn't matter where the blue circle is on the outside, as long as the red circle is inside the tube, both of these circles cannot be linked!
And this makes me think that those two circles weren't linked in the first place itself! And that must have something to do with the fact that our tube had a hole in it. If there was no hole then it makes sense that both circles are linked and cannot pass through each other.
Now what I think is happening is that, as we are pulling the inside surface of the tube out it passes through the inside of the red circle, now again I am not clear about how is that happening, I was finding toroids in my home and the only similar thing I found is my bicycle tube but to access that I'll have to rip off my tyres! Why don't we produce more things that are toroids? The other alternative is that I have to cut my shirt's sleeves and then stitch it end to end but I don't think even this is a good idea.
Given that we are allowing cutting the strip and laying it out as a rectangle we can dispense with the hole; except that when we reassemble it as a torus it must be equivalent to having pulled it through a hole.
As I mentioned, there are two ways we can stitch the rectangle back to be a torus. Obviously one way will get exactly back to where we started. What happens if we do it the other way?
kshitij said:
it doesn't matter where the blue circle is on the outside, as long as the red circle is inside the tube, both of these circles cannot be linked!
and yet, they must be.
haruspex said:
Given that we are allowing cutting the strip and laying it out as a rectangle we can dispense with the hole; except that when we reassemble it as a torus it must be equivalent to having pulled it through a hole.
As I mentioned, there are two ways we can stitch the rectangle back to be a torus. Obviously one way will get exactly back to where we started. What happens if we do it the other way?
But why is cutting the strip allowed? wouldn't that change the shape we are dealing with?
haruspex said:
Obviously one way will get exactly back to where we started. What happens if we do it the other way?
As I said before that now it makes sense why the circles won't be linked after we stitch it back the other way because we are literally cutting both the circle such that they changed from a circle to a line, but the cutting isn't allowed
After stitching it back I cannot believe how the red circle will still be linked with the blue one?
kshitij said:
View attachment 287813
After stitching it back I cannot believe how the red circle will still be linked with the blue one?
Try it with a sheet of paper. Draw a line across the width on one side and along the length on the other.
You can turn it into a cylinder by bringing together two opposite edges and gluing them together, then into a torus by doing the same with the opposite ends of the cylinder.
You can do it either so that the first fold leaves the line inside as a circle and the outside line straight, or so that the outside line is a circle around the cylinder and the inside line is straight.
Check linkage of the lines in the finished torus in each case.
haruspex said:
Try it with a sheet of paper. Draw a line across the width on one side and along the length on the other.
You can turn it into a cylinder by bringing together two opposite edges and gluing them together, then into a torus by doing the same with the opposite ends of the cylinder.
You can do it either so that the first fold leaves the line inside as a circle and the outside line straight, or so that the outside line is a circle around the cylinder and the inside line is straight.
Check linkage of the lines in the finished torus in each case.
I think that the linkage should be different in both cases but I should check it myself
haruspex said:
Try it with a sheet of paper. Draw a line across the width on one side and along the length on the other.
You can turn it into a cylinder by bringing together two opposite edges and gluing them together, then into a torus by doing the same with the opposite ends of the cylinder.
You can do it either so that the first fold leaves the line inside as a circle and the outside line straight, or so that the outside line is a circle around the cylinder and the inside line is straight.
Check linkage of the lines in the finished torus in each case.
Well first I learned that making a cylinder is relatively easier but making a toroid is way too difficult, the paper doesn't want to change its surface's curvature and IIRC there is a theorem like that by Gauss (??).
Anyway, to the conclusion, the blue circle are red circle are not linked like a chain in one but they are linked in folding the other way around as I expected. 100% their linkages are different, they have to be different!
kshitij said:
Well first I learned that making a cylinder is relatively easier but making a toroid is way too difficult, the paper doesn't want to change its surface's curvature and IIRC there is a theorem like that by Gauss (??).
Anyway, to the conclusion, the blue circle are red circle are not linked like a chain in one but they are linked in folding the other way around as I expected. 100% their linkages are different, they have to be different!
Right, but for each of those two ways of folding the paper there were two ways you could have done it. If the lines are red and blue, your first fold could have have formed a circle on the inside with either the red line or the blue, or formed a circle on the outside with either the red or the blue.
So the reconstructed torus has any of four configurations:
Linked rings, red inside, blue outside
Linked rings, blue inside, red outside
Unlinked rings, red inside, blue outside
Etc.
The question becomes, which of the possible transitions could pulling the torus inside out through a hole produce? If it started with red outside, blue inside then obviously it ends with red in blue out, but they could be linked or unlinked.. and it should be obvious the linkage status cannot change.
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kshitij
haruspex said:
Right, but for each of those two ways of folding the paper there were two ways you could have done it. If the lines are red and blue, your first fold could have have formed a circle on the inside with either the red line or the blue, or formed a circle on the outside with either the red or the blue.
So the reconstructed torus has any of four configurations:
Linked rings, red inside, blue outside
Linked rings, blue inside, red outside
Unlinked rings, red inside, blue outside
Etc.
The question becomes, which of the possible transitions could pulling the torus inside out through a hole produce? If it started with red outside, blue inside then obviously it ends with red in blue out, but they could be linked or unlinked.. and it should be obvious says the linkage status cannot change.
Oh yes, I see it now,
If we did the first fold horizontally (and the horizontal fold should be in opposite direction to the unfolding otherwise we will end up with what we started with) instead of vertically then the circles won't be linked and thus that is not what is happening in inverting a tube with linked circles!
Also I noticed that the tube was narrower and longer before inverting it inside out.
1. What is Pappus's centroid theorem?
Pappus's centroid theorem is a mathematical theorem that states that the centroid of a plane figure can be found by taking the average of the centroids of its cross-sections along a line parallel to the plane figure's axis of symmetry.
2. How does Pappus's centroid theorem apply to finding the centroid of a hemisphere?
In the case of a hemisphere, the cross-sections along a line parallel to its axis of symmetry are circles. By taking the average of the centroids of these circles, we can find the centroid of the hemisphere.
3. What is the formula for finding the centroid of a hemisphere using Pappus's centroid theorem?
The formula is: C = (4R/3π) * (3R/4) = R/π, where R is the radius of the hemisphere.
4. Can Pappus's centroid theorem be used for any other shapes besides a hemisphere?
Yes, Pappus's centroid theorem can be applied to any shape with an axis of symmetry, such as a cone, cylinder, or sphere.
5. Are there any limitations to using Pappus's centroid theorem?
One limitation is that the shape must have an axis of symmetry in order for the theorem to be applicable. Additionally, the shape must have a constant cross-sectional area along the axis of symmetry.
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# MSU ME 221 - Lecture 06 (20 pages)
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ME221 Statics LECTURE 6 Sections 3 6 3 9 ME221 Lecture 6 1 Homework 2 Due today Quiz 3 Today ME221 Lecture 6 2 Homework 3 Chapter 3 problems 48 55 57 61 62 65 72 Chapter 4 problems 2 4 10 11 18 24 39 43 Due Monday June 7 ME221 Lecture 6 3 Moment of a Couple Let F1 F2 B y Mo rA x F2 rB x F1 rB rA x F1 rAB x F1 C C F1 d ME221 rB O F1 rAB rB A F2 d A rA x The Moment of two equal and opposite forces is called a Couple z Lecture 6 4 Moment of a Couple continued The two equal and opposite forces form a couple no net force pure moment The moment depends only on the relative positions of the two forces and not on their position with respect to the origin of coordinates ME221 Lecture 6 5 Moment of a Couple continued Since the moment is independent of the origin it can be treated as a free vector meaning that it is the same at any point in space The two parallel forces define a plane and the moment of the couple is perpendicular to that plane ME221 Lecture 6 6 Equivalent Force Systems Replacing a given set of forces with an equivalent force moment system ME221 Lecture 6 7 Equivalent Force Systems The action of a force tends to translate the body along the direction of the force and rotate it about an axis not located along the line of action of the force The External Effects remain the same if the force is moved from one point to another from A to B if ME221 Lecture 6 8 1 Point B is on the line of action of the force A force can be replaced by an equal magnitude force provided it has the same line of action and does not disturb equilibrium B A ME221 Lecture 6 9 2 Point B is not on the line of action of the force Start with a rigid body having force F applied to it Replace this force with a force and couple at A Add zero to the body in the form of adding F and F ME221 B F A F Lecture 6 F 10 Next group two forces to create couple C Let rB A be the relative position of B with respect to A Define couple with cross product C C rB A x F In essence the force F has F been moved from point B to A with the addition of couple C ME221 Lecture 6 A F B rB A F F 11 For More Than One Force A similar procedure is used when there are many forces on the body R Fi C ci ri x F i ME221 Lecture 6 12 Example Problems ME221 Lecture 6 13 ME 221 Statics Exam 1 Review ME221 Lecture 6 14 Exam 1 Wednesday June 2 No class on Monday May 31 ME221 Lecture 6 15 Exam Format Similar to homework problems 5 or 6 maybe 7 problems No new concepts Closed books and closed notes 25 of final grade Will need a calculator ME221 Lecture 6 16 Exam 1 Helpful Hints Study by working homework problems Review the examples in the book Read the exam questions carefully Work the problems you know first Use a logical flow to show your knowledge Be complete check your answers ME221 Lecture 6 17 Exam 1 Pitfalls Vectors have both magnitude and direction Most answers will have units Calculated answers should be in decimal form Use 3 significant figures where appropriate Answers are generally worth 1 3 of points Steps leading to answers are worth 2 3 ME221 Lecture 6 18 Exam 1 Topics All of Chapters 1 2 except 2 11 Springs Up to 3 4 in Chapter 3 Newton s Laws Units Scalars vectors defining adding etc Laws of sine and cosine Resolution of a vector into components ME221 Lecture 6 19 Exam 1 Topics 3D coordinates unit base vectors Directional cosines Non orthogonal vector components Scalar cross products of vectors 2D 3D equilibrium problems Moments couples Drawing good free body diagrams ME221 Lecture 6 20
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Grade 8 Mathematics Module 4, Topic B, Overview | EngageNY
## Grade 8 Mathematics Module 4, Topic B, Overview
In Topic B, students work with constant speed, a concept learned in Grade 6 (6.RP.A.3), but this time with proportional relationships related to average speed and constant speed. These relationships are expressed as linear equations in two variables. Students find solutions to linear equations in two variables, organize them in a table, and plot the solutions on a coordinate plane (8.EE.C.8a). It is in Topic B that students begin to investigate the shape of a graph of a linear equation. Students predict that the graph of a linear equation is a line and select points on and off the line to verify their claim. Also in this topic is the standard form of a linear equation, ax + by = c, and when a, b ≠ 0, a non-vertical line is produced. Further, when a or b = 0, then a vertical or horizontal line is produced.
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Integration Questions
Question Number 102183 by bemath last updated on 07/Jul/20
$$\int\:\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta}\:{d}\theta\:? \\$$
Answered by Dwaipayan Shikari last updated on 07/Jul/20
$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left({sin}\theta−{cos}\theta\right)−\frac{\theta}{\mathrm{2}}+{C} \\$$$${I}=\int\frac{{cos}\theta}{{sin}\theta−{cos}\theta}{d}\theta \\$$$${I}=\int\frac{{sin}\theta}{{sin}\theta−{cos}\theta}−\mathrm{1}{d}\theta \\$$$$\mathrm{2}{I}=\int\frac{{sin}\theta+{cos}\theta}{{sin}\theta−{cos}\theta}−\mathrm{1}{d}\theta \\$$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({sin}\theta−{cos}\theta\right)−\frac{\theta}{\mathrm{2}}+{Constant} \\$$
Answered by bobhans last updated on 07/Jul/20
$$\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta}\:×\:\frac{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta+\mathrm{cos}\:^{\mathrm{2}} \theta}{−\mathrm{cos}\:\mathrm{2}\theta} \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left(\mathrm{cos}\:\mathrm{2}\theta\right)}{\mathrm{cos}\:\mathrm{2}\theta}−\int\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\mathrm{2}\theta}\:{d}\theta\:=\: \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{ln}\mid\mathrm{cos}\:\mathrm{2}\theta\mid\:−\int\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{2}\theta}\:{d}\theta\:= \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{cos}\:\mathrm{2}\theta\mid−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sec}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{2}}\theta\:+\:{c}\:= \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{cos}\:\mathrm{2}\theta\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{sec}\:\mathrm{2}\theta+\mathrm{tan}\:\mathrm{2}\theta\mid−\frac{\mathrm{1}}{\mathrm{2}}\theta\:+\:{c} \\$$$$\\$$$$\\$$
Answered by 1549442205 last updated on 07/Jul/20
$$\mathrm{Putting}\:\mathrm{t}=\mathrm{tan}\frac{\theta}{\mathrm{2}}\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{d}\theta \\$$$$\mathrm{cos}\theta=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} },\:\mathrm{sin}\theta=\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} },\:\mathrm{d}\theta=\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\$$$$\mathrm{F}=\mathrm{2}\int\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt}==−\int\frac{\mathrm{t}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\mathrm{dt}+\int\frac{\mathrm{t}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}}\mathrm{dt} \\$$$$=−\int\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}} \\$$$$=−\mathrm{arctan}\:\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{1}\mid \\$$$$\frac{−\boldsymbol{\theta}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \frac{\boldsymbol{\theta}}{\mathrm{2}}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \frac{\boldsymbol{\theta}}{\mathrm{2}}+\mathrm{2}\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\theta}}{\mathrm{2}}−\mathrm{1}\mid+\mathrm{C} \\$$
Answered by mathmax by abdo last updated on 08/Jul/20
$$\mathrm{I}\:=\int\:\frac{\mathrm{cos}\theta}{\mathrm{sn}\theta−\mathrm{cos}\theta}\mathrm{d}\theta\:\Rightarrow\:\mathrm{I}\:=\int\:\:\frac{\mathrm{d}\theta}{\mathrm{tan}\theta−\mathrm{1}}\:\mathrm{changement}\:\mathrm{tan}\theta\:=\mathrm{x}\:\mathrm{give} \\$$$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{x}−\mathrm{1}\right)}\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\$$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}}{\mathrm{x}−\mathrm{1}}\:+\frac{\mathrm{bx}+\mathrm{c}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\$$$$\mathrm{wehave}\:\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\mathrm{xF}\left(\mathrm{x}\right)\:=\mathrm{0}\:=\mathrm{a}+\mathrm{b}\:\Rightarrow\mathrm{b}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\mathrm{F}\left(\mathrm{0}\right)\:=−\mathrm{1}\:=−\mathrm{a}\:+\mathrm{c}\:\Rightarrow\mathrm{c}\:=\mathrm{a}−\mathrm{1}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arctanx}\:+\mathrm{c} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{tan}\theta\:−\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)−\frac{\theta}{\mathrm{2}}\:+\mathrm{c}\:. \\$$
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#### What is 525 percent of 180?
How much is 525 percent of 180? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 525% of 180 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 525% of 180 = 945
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Calculating five hundred and twenty-five of one hundred and eighty How to calculate 525% of 180? Simply divide the percent by 100 and multiply by the number. For example, 525 /100 x 180 = 945 or 5.25 x 180 = 945
#### How much is 525 percent of the following numbers?
525% of 180.01 = 94505.25 525% of 180.02 = 94510.5 525% of 180.03 = 94515.75 525% of 180.04 = 94521 525% of 180.05 = 94526.25 525% of 180.06 = 94531.5 525% of 180.07 = 94536.75 525% of 180.08 = 94542 525% of 180.09 = 94547.25 525% of 180.1 = 94552.5 525% of 180.11 = 94557.75 525% of 180.12 = 94563 525% of 180.13 = 94568.25 525% of 180.14 = 94573.5 525% of 180.15 = 94578.75 525% of 180.16 = 94584 525% of 180.17 = 94589.25 525% of 180.18 = 94594.5 525% of 180.19 = 94599.75 525% of 180.2 = 94605 525% of 180.21 = 94610.25 525% of 180.22 = 94615.5 525% of 180.23 = 94620.75 525% of 180.24 = 94626 525% of 180.25 = 94631.25
525% of 180.26 = 94636.5 525% of 180.27 = 94641.75 525% of 180.28 = 94647 525% of 180.29 = 94652.25 525% of 180.3 = 94657.5 525% of 180.31 = 94662.75 525% of 180.32 = 94668 525% of 180.33 = 94673.25 525% of 180.34 = 94678.5 525% of 180.35 = 94683.75 525% of 180.36 = 94689 525% of 180.37 = 94694.25 525% of 180.38 = 94699.5 525% of 180.39 = 94704.75 525% of 180.4 = 94710 525% of 180.41 = 94715.25 525% of 180.42 = 94720.5 525% of 180.43 = 94725.75 525% of 180.44 = 94731 525% of 180.45 = 94736.25 525% of 180.46 = 94741.5 525% of 180.47 = 94746.75 525% of 180.48 = 94752 525% of 180.49 = 94757.25 525% of 180.5 = 94762.5
525% of 180.51 = 94767.75 525% of 180.52 = 94773 525% of 180.53 = 94778.25 525% of 180.54 = 94783.5 525% of 180.55 = 94788.75 525% of 180.56 = 94794 525% of 180.57 = 94799.25 525% of 180.58 = 94804.5 525% of 180.59 = 94809.75 525% of 180.6 = 94815 525% of 180.61 = 94820.25 525% of 180.62 = 94825.5 525% of 180.63 = 94830.75 525% of 180.64 = 94836 525% of 180.65 = 94841.25 525% of 180.66 = 94846.5 525% of 180.67 = 94851.75 525% of 180.68 = 94857 525% of 180.69 = 94862.25 525% of 180.7 = 94867.5 525% of 180.71 = 94872.75 525% of 180.72 = 94878 525% of 180.73 = 94883.25 525% of 180.74 = 94888.5 525% of 180.75 = 94893.75
525% of 180.76 = 94899 525% of 180.77 = 94904.25 525% of 180.78 = 94909.5 525% of 180.79 = 94914.75 525% of 180.8 = 94920 525% of 180.81 = 94925.25 525% of 180.82 = 94930.5 525% of 180.83 = 94935.75 525% of 180.84 = 94941 525% of 180.85 = 94946.25 525% of 180.86 = 94951.5 525% of 180.87 = 94956.75 525% of 180.88 = 94962 525% of 180.89 = 94967.25 525% of 180.9 = 94972.5 525% of 180.91 = 94977.75 525% of 180.92 = 94983 525% of 180.93 = 94988.25 525% of 180.94 = 94993.5 525% of 180.95 = 94998.75 525% of 180.96 = 95004 525% of 180.97 = 95009.25 525% of 180.98 = 95014.5 525% of 180.99 = 95019.75 525% of 181 = 95025
1% of 180 = 1.8 2% of 180 = 3.6 3% of 180 = 5.4 4% of 180 = 7.2 5% of 180 = 9 6% of 180 = 10.8 7% of 180 = 12.6 8% of 180 = 14.4 9% of 180 = 16.2 10% of 180 = 18 11% of 180 = 19.8 12% of 180 = 21.6 13% of 180 = 23.4 14% of 180 = 25.2 15% of 180 = 27 16% of 180 = 28.8 17% of 180 = 30.6 18% of 180 = 32.4 19% of 180 = 34.2 20% of 180 = 36 21% of 180 = 37.8 22% of 180 = 39.6 23% of 180 = 41.4 24% of 180 = 43.2 25% of 180 = 45
26% of 180 = 46.8 27% of 180 = 48.6 28% of 180 = 50.4 29% of 180 = 52.2 30% of 180 = 54 31% of 180 = 55.8 32% of 180 = 57.6 33% of 180 = 59.4 34% of 180 = 61.2 35% of 180 = 63 36% of 180 = 64.8 37% of 180 = 66.6 38% of 180 = 68.4 39% of 180 = 70.2 40% of 180 = 72 41% of 180 = 73.8 42% of 180 = 75.6 43% of 180 = 77.4 44% of 180 = 79.2 45% of 180 = 81 46% of 180 = 82.8 47% of 180 = 84.6 48% of 180 = 86.4 49% of 180 = 88.2 50% of 180 = 90
51% of 180 = 91.8 52% of 180 = 93.6 53% of 180 = 95.4 54% of 180 = 97.2 55% of 180 = 99 56% of 180 = 100.8 57% of 180 = 102.6 58% of 180 = 104.4 59% of 180 = 106.2 60% of 180 = 108 61% of 180 = 109.8 62% of 180 = 111.6 63% of 180 = 113.4 64% of 180 = 115.2 65% of 180 = 117 66% of 180 = 118.8 67% of 180 = 120.6 68% of 180 = 122.4 69% of 180 = 124.2 70% of 180 = 126 71% of 180 = 127.8 72% of 180 = 129.6 73% of 180 = 131.4 74% of 180 = 133.2 75% of 180 = 135
76% of 180 = 136.8 77% of 180 = 138.6 78% of 180 = 140.4 79% of 180 = 142.2 80% of 180 = 144 81% of 180 = 145.8 82% of 180 = 147.6 83% of 180 = 149.4 84% of 180 = 151.2 85% of 180 = 153 86% of 180 = 154.8 87% of 180 = 156.6 88% of 180 = 158.4 89% of 180 = 160.2 90% of 180 = 162 91% of 180 = 163.8 92% of 180 = 165.6 93% of 180 = 167.4 94% of 180 = 169.2 95% of 180 = 171 96% of 180 = 172.8 97% of 180 = 174.6 98% of 180 = 176.4 99% of 180 = 178.2 100% of 180 = 180
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# Which point is collinear with the points (3,-5),(2,-3),and (-1,3)
mr-mayonnaise | Certified Educator
Points are said to be collinear if they lie on the same straight lines as shown on a graph. These points also are connected by the same relationship or ratio. This can be found using the slope equation m = (y2-y2/x2-x2)
`m = (y_2 - y_1)/(x_2 - x_1)`
`m=(-3-(-5))/(2-3)`
`m= 2/-1 = -2`
Putting this into slope-intercept form (y=mx+b) gives us the line that all collinear points can be found on. By graphing the line using the above points, we find that the y-intercept is 1.
`y=mx+b`
`y= -2*x+1`
If you plot these points on a graph you can find a line that descends at the required ratio to show all collinear points.
Collinear points includes: (0,1), (1,-1), (-2, 5) and so on. As long as the coordinates fulfill the above slope-intercept equation the point can be considered collinear!
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# How can I interpret a regression statistics table in Excel?
Oct 20, 2015
I assume you mean this:
The "Coefficients" are the slope or y-intercept in this case. "HH SIZE" refers to the Slope, and of course, Intercept is the y-intercept.
If you multiply the Standard Error by $1.96$, you get the Associated Error for either the Intercept or the Slope. The Associated Error is basically the uncertainty you have.
For example, in a standard physics lab course, bare minimum, here's what you would need to know:
• Slope
• Intercept
• Slope Standard Error ($S {E}_{\text{slope}}$)
• Slope Associated Error ($A {E}_{\text{slope}}$)
• Intercept Standard Error ($S {E}_{\text{int}}$)
• Intercept Associated Error ($A {E}_{\text{int}}$)
The sample standard deviation is:
$s = \sqrt{\frac{1}{N - 1} {\sum}_{i = 1}^{N} {\left({x}_{i} - \overline{x}\right)}^{2}}$
where $N$ is the number of trials, ${x}_{i}$ is each individual value, and $\overline{x}$ is the average of said values.
The Standard Error is:
$S E = \frac{s}{\sqrt{N}}$
where $s$ is the standard deviation above, and:
$A E = 1.96 \cdot S E$
Here is an example of an Ohm's law analysis I did using a similar regression statistics table:
Oftentimes, even in a quantitative analysis course, you only need to further know the coefficient of determination ${R}^{2}$. The closer it is to $1$, the better it is, but it is only for a linear fit line.
Other than that, I have not had to use any other quantity on the regression statistics table in my 7 University semesters.
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# Poundal/square foot to Millipascal converter| pdl/ft^2 to mPa conversion
Are you struggling with converting Poundal/square foot to Millipascal? Don’t worry! Our online “Poundal/square foot to Millipascal Converter” is here to simplify the conversion process for you.
Here’s how it works: simply input the value in Poundal/square foot. The converter instantly gives you the value in Millipascal. No more manual calculations or headaches – it’s all about smooth and effortless conversions!
Think of this Poundal/square foot (pdl/ft^2) to Millipascal (mPa) converter as your best friend who helps you to do the conversion between these pressure units. Say goodbye to calculating manually over how many Millipascal are in a certain number of Poundal/square foot – this converter does it all for you automatically!
## What are Poundal/square foot and Millipascal?
In simple words, Poundal/square foot and Millipascal are units of pressure used to measure how much force is applied over a certain area. It’s like measuring how tightly the air is pushing on something.
The short form for Poundal/square foot is “pdl/ft^2” and the short form for Millipascal is “mPa”.
In everyday life, we use pressure units like Poundal/square foot and Millipascal to measure how much things are getting squeezed or pushed. It helps us with tasks like checking tire pressure or understanding the force in different situations.
## How to convert from Poundal/square foot to Millipascal?
If you want to convert between these two units, you can do it manually too. To convert from Poundal/square foot to Millipascal just use the given formula:
``mPa = Value in pdl/ft^2 * 1488.1639436``
here are some examples of conversion,
• 2 pdl/ft^2 = 2 * 1488.1639436 = 2976.3278872 mPa
• 5 pdl/ft^2 = 5 * 1488.1639436 = 7440.819718 mPa
• 10 pdl/ft^2 = 10 * 1488.1639436 = 14881.639436 mPa
### Poundal/square foot to Millipascal converter: conclusion
Here we have learn what are the pressure units Poundal/square foot (pdl/ft^2) and Millipascal (mPa)? How to convert from Poundal/square foot to Millipascal manually and also we have created an online tool for conversion between these units.
Poundal/square foot to Millipascal converter” or simply pdl/ft^2 to mPa converter is a valuable tool for simplifying pressure unit conversions. By using this tool you don’t have to do manual calculations for conversion which saves you time.
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# 11.67 kg to lbs - 11.67 kilograms to pounds
Before we move on to the more practical part - it means 11.67 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 11.67 kg to lbs? 11.67 kilograms it is equal 25.7279459754 pounds, so 11.67 kg is equal 25.7279459754 lbs.
## 11.67 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).
From time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but hard to use.
Then, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was prepared of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was switched by a new definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 11.67 kilogram to pounds
You know a little about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is in use also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 11.67 kg?
11.67 kilogram is equal to 25.7279459754 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 11.67 kg in lbs
Theoretical section is already behind us. In this part we want to tell you how much is 11.67 kg to lbs. Now you learned that 11.67 kg = x lbs. So it is time to know the answer. Let’s see:
11.67 kilogram = 25.7279459754 pounds.
It is an accurate result of how much 11.67 kg to pound. You may also round it off. After it your outcome is exactly: 11.67 kg = 25.674 lbs.
You know 11.67 kg is how many lbs, so have a look how many kg 11.67 lbs: 11.67 pound = 0.45359237 kilograms.
Obviously, in this case you can also round it off. After rounding off your result will be as following: 11.67 lb = 0.45 kgs.
We are also going to show you 11.67 kg to how many pounds and 11.67 pound how many kg results in tables. See:
We are going to start with a table for how much is 11.67 kg equal to pound.
### 11.67 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
11.67 25.7279459754 25.6740
Now see a table for how many kilograms 11.67 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
11.67 0.45359237 0.45
Now you know how many 11.67 kg to lbs and how many kilograms 11.67 pound, so we can go to the 11.67 kg to lbs formula.
### 11.67 kg to pounds
To convert 11.67 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 25.7279459754 result in pounds
The first version of a formula give you the most exact result. In some situations even the smallest difference can be considerable. So if you want to get an accurate outcome - first formula will be the best for you/option to calculate how many pounds are equivalent to 11.67 kilogram.
So let’s move on to the another formula, which also enables conversions to know how much 11.67 kilogram in pounds.
The shorter formula is down below, see:
Amount of kilograms * 2.2 = the outcome in pounds
As you see, this formula is simpler. It can be better option if you want to make a conversion of 11.67 kilogram to pounds in quick way, for example, during shopping. You only need to remember that your result will be not so exact.
Now we are going to show you these two formulas in practice. But before we will make a conversion of 11.67 kg to lbs we want to show you easier way to know 11.67 kg to how many lbs without any effort.
### 11.67 kg to lbs converter
Another way to learn what is 11.67 kilogram equal to in pounds is to use 11.67 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on first formula which we showed you in the previous part of this article. Thanks to 11.67 kg pound calculator you can quickly convert 11.67 kg to lbs. You only need to enter amount of kilograms which you need to calculate and click ‘convert’ button. You will get the result in a second.
So let’s try to calculate 11.67 kg into lbs using 11.67 kg vs pound calculator. We entered 11.67 as a number of kilograms. This is the result: 11.67 kilogram = 25.7279459754 pounds.
As you see, our 11.67 kg vs lbs converter is user friendly.
Now we are going to our main issue - how to convert 11.67 kilograms to pounds on your own.
#### 11.67 kg to lbs conversion
We are going to start 11.67 kilogram equals to how many pounds calculation with the first version of a formula to get the most exact outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 25.7279459754 the outcome in pounds
So what need you do to know how many pounds equal to 11.67 kilogram? Just multiply number of kilograms, this time 11.67, by 2.20462262. It is exactly 25.7279459754. So 11.67 kilogram is 25.7279459754.
It is also possible to round it off, for example, to two decimal places. It is equal 2.20. So 11.67 kilogram = 25.6740 pounds.
It is high time for an example from everyday life. Let’s convert 11.67 kg gold in pounds. So 11.67 kg equal to how many lbs? And again - multiply 11.67 by 2.20462262. It is exactly 25.7279459754. So equivalent of 11.67 kilograms to pounds, if it comes to gold, is 25.7279459754.
In this example it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 11.67 kilogram 25.674 pounds.
Now we can move on to examples converted using short formula.
#### How many 11.67 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 25.674 the result in pounds
So 11.67 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, in this case 11.67, by 2.2. Let’s see: 11.67 * 2.2 = 25.674. So 11.67 kilogram is exactly 2.2 pounds.
Do another conversion using this version of a formula. Now calculate something from everyday life, for instance, 11.67 kg to lbs weight of strawberries.
So calculate - 11.67 kilogram of strawberries * 2.2 = 25.674 pounds of strawberries. So 11.67 kg to pound mass is equal 25.674.
If you know how much is 11.67 kilogram weight in pounds and can calculate it using two different formulas, we can move on. Now we want to show you all results in charts.
#### Convert 11.67 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Thanks to this you can easily compare 11.67 kg equivalent to lbs results.
Let’s start with a 11.67 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
11.67 25.7279459754 25.6740
And now see 11.67 kg equal pound chart for the second formula:
Kilograms Pounds
11.67 25.674
As you can see, after rounding off, if it comes to how much 11.67 kilogram equals pounds, the outcomes are not different. The bigger amount the more considerable difference. Please note it when you need to do bigger number than 11.67 kilograms pounds conversion.
#### How many kilograms 11.67 pound
Now you learned how to calculate 11.67 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 11.67 kilogram to pounds and ounces calculation?
We are going to show you how you can convert it little by little. Let’s begin. How much is 11.67 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 11.67, by 2.20462262. So 11.67 * 2.20462262 = 25.7279459754. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 11.67 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So your outcome is 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.
As you see, conversion 11.67 kilogram in pounds and ounces quite simply.
The last calculation which we will show you is conversion of 11.67 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it you need another formula. Before we give you it, see:
• 11.67 kilograms meters = 7.23301385 foot pounds,
• 11.67 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 11.67 foot pounds to kilograms meters you have to multiply 11.67 by 0.13825495. It gives 0.13825495. So 11.67 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 11.67 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 11.67 kilogram into pounds conversions.
We showed you not only how to do a calculation 11.67 kilogram to metric pounds but also two other conversions - to know how many 11.67 kg in pounds and ounces and how many 11.67 foot pounds to kilograms meters.
We showed you also other way to do 11.67 kilogram how many pounds calculations, this is with use of 11.67 kg en pound calculator. This is the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you can do 11.67 kilogram equal to how many pounds calculation - on your own or with use of our 11.67 kgs to pounds calculator.
So what are you waiting for? Let’s calculate 11.67 kilogram mass to pounds in the best way for you.
Do you need to make other than 11.67 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 11.67 kilogram equal many pounds.
### How much is 11.67 kg in pounds
To quickly sum up this topic, that is how much is 11.67 kg in pounds , we prepared for you an additional section. Here we have for you all you need to know about how much is 11.67 kg equal to lbs and how to convert 11.67 kg to lbs . Have a look.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 11.67 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 11.67 kilogram to pounds? The correct answer is 25.7279459754 lbs.
You can also calculate how much 11.67 kilogram is equal to pounds with another, easier type of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So in this case, 11.67 kg equal to how much lbs ? The result is 25.7279459754 pounds.
How to convert 11.67 kg to lbs in an easier way? It is possible to use the 11.67 kg to lbs converter , which will do all calculations for you and you will get a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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×
# 2-EDGE CONNECTED GRAPH
2 2-edge connected graph means the graph is always connected if we remove any edge of that graph. For checking a graph is a two-edge connected graph , we just need to check whether for every sub-tree of the graph there should be an edge going from that sub-tree to outside vertex and that vertex is already visited.This can be solved by using DFS and in linear time. PSUEDOCODE : The below code returns the minimum value of the arrival time of the back-edge that the sub-tree is connected with. time=0; 2edgeconnected(v) { visited[v]=1; arr[v]=time++; int xyz=arr[v]; for all w adjecent to v do { if(!visited[w]) xyz=min(xyz,2edgeconnected(w)); else xyz=min(xyz,arr[w]); } if(xyz==arr[v]) abort //graph is not two-edge connected. return xyz; } asked 03 Jul '14, 12:06 2★trek 327●4●12●18 accept rate: 0%
0 nice one!! answered 03 Jul '14, 16:04 4★suryanit 11●1●3●4 accept rate: 0%
0 Equality will hold for the root node, u need to take care of that case answered 06 Dec '17, 23:21 1 accept rate: 0%
0 Equality will hold for the root node, u need to take care of that case answered 06 Dec '17, 23:22 1 accept rate: 0%
0 Useful one... But same problem can be solved by following two checks: Graph is connected (Simple DFS) Graph does not contains a bridge. (Another DFS) Read more about bride finding here. If both of above two conditions hold true, graph is 2-edge connected. Time Complexity of this Solution: O(N+M) Above two DFS can also be combined into a single DFS to improve time complexity. Anyways, Nice editorial. answered 06 Dec '17, 23:52 3.6k●13●51 accept rate: 24%
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question asked: 03 Jul '14, 12:06
question was seen: 3,594 times
last updated: 06 Dec '17, 23:52
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# How can I reconcile cardinality and and subsets in Set Theory?
I have a problem with subsets in Set Theory.
a set A is a subset of set B if all the elements of A are also elements of B.
{1,2} is a subset of {1,2,3}
Simple enough.
But, as I understand it, the word "subset" means a set within another set. So, intuitively if A is in B It would mean that B is {{1,2},3}
This would mean that {1,2,3} and {{1,2},3} and {1,{2,3}} are the same set as brackets would be arbitrary ways of grouping elements in subsets within the set. This is generally the way we understand and draw sets as circles within circles. Circles are drawn depending on how we want to group the content but they have no substance in and on themselves. This is also backed up by the fact that two identical elements are the same element, and by that token, subset A is entirely inside the set B.
But when looking at cardinalities, it seems those 3 sets are very different as {1,2,3} has a cardinality of 3 while {{1,2},3} and {1,{2,3}} have a cardinality of 2. If they were the same set, their cardinalities would be ambiguous. Therefore, they aren't the same set.
So, I conclude that I am wrong in the way I see subsets as subsets are NOT sets within sets, they are just sets which content is also included in another set.
Right?
In a nutshell, it's hard to tell if a set is only its elements or if a set is the collection of the elements AND the brackets around it.
It can't be more basic than that but I'm confused since both interpretations could make sense but they are incompatible.
If brackets are included a subset is not a set within a set (since it would make cardinality inconsistent), and if brackets are not included how can cardinality depend on sets within sets? It's not just a problem of semantics since I get actual inconsistencies depending on how I understand it
No matter how I see it, I can't seem to reconcile subsets and cardinality.
EDIT: a user completely edited the title but the new formulated question was not the question I asked (and some things I considered important in my text had been erased too). So I decided to rollback to what it was.
Thank you very much though for editing the format. If needed I'm happy to edit anything myself, if I'm asked to. My username can only be associated with my own words. It seems fair.
Brackets do make a huge difference in Math!
There is a fundamental difference between $$A\subset B$$ ($$A$$ is a subset of $$B$$) and $$A\in B$$ ($$A$$ is a member of $$B$$).
The former means that any member of $$A$$ is a member of $$B$$, while the latter means that $$A$$ itself is a member of $$B$$. This said $$\{1,2\} \subset \{1,2,3\}$$ but $$\{1,2\} \not\in\{1,2,3\}$$ while $$\{1,2\} \not\subset \{\{1,2\},3\}$$ but $$\{1,2\} \in\{\{1,2\},3\}.$$
• This clarifies everything, thank you. I see I intuitively considered subsets as members while they are not. I haven't read the part of the textbook where they talk about members yet, which explains the confusion. For whatever reasons, it feels to me like members should be subsets and subsets should be members etymologically but as long as the ambiguity is lifted, it's alright. Thank you very much. May 24 at 1:19
• The buzzword I'm used to is "elements" instead of "members", which also explains the ∈ symbol May 24 at 12:41
a set $$A$$ is a subset of set $$B$$ if all the elements of $$A$$ are also elements of $$B$$.
is correct. Simple enough, as you say.
You tie yourself in vocabulary knots when you change the definition to
the word "subset" means a set within another set.
and then struggle with what "within" means.
Stop overthinking and redefining. Your paragraph beginning "But" is right. Those are three different sets, one of which has cardinality 3 and the other two cardinality 2.
The brackets are not part of a set, they are the punctuation you use to specify the set.
This may help you: What is the difference between $x$ and $\{x\}$ when $x$ itself is a set?
• Thank you. I don't think I was overthinking it. There was something missing for me to fully grasp the concept. And as Ruy said, the missing part of the puzzle was the "members". Now that I know what members are, I realize I mixed up both concepts and now the ambiguity is totally lifted. I needed to know that to carry on. Thank both of you. Your help is very appreciated. May 24 at 1:23
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Thursday
May 5, 2016
Homework Help: math
Posted by katie on Friday, September 2, 2011 at 4:48pm.
You deposit \$2200 in an account that pays 3% annual interest. After 15 years, you withdraw the money, what is the balance if the interest is compounded quarterly?
so I figure you would get 2650.00 help please
• math - MathMate, Friday, September 2, 2011 at 6:26pm
What you probably did was calculated simple interest for 15 years on \$1000 and added to \$2200 to give \$2650.
Compound interest formula are based on the number of periods, n, the interest was compounded.
The interest being compounded 4 times a year, so there are 15*4=60 periods of 3 months each. The corresponding interest rate for each period is therefore r = 3%/4=0.0075.
The formula for the future value using compound interest is:
FV = Principal * (1+r)^n
=2200*1.0075^60
=2200*1.565681
=\$3444.50
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# PSAT Math : How to find the area of a right triangle
## Example Questions
### Example Question #4 : How To Find The Area Of A Right Triangle
The perimeter of a right triangle is 40 units. If the lengths of the sides are , , and units, then what is the area of the triangle?
Explanation:
Because the perimeter is equal to the sum of the lengths of the three sides of a triangle, we can add the three expressions for the lengths and set them equal to 40.
Perimeter:
Simplify the x terms.
Simplify the constants.
Subtract 8 from both sides.
Divide by 4
One side is 8.
The second side is
.
The third side is
.
Thus, the sides of the triangle are 8, 15, and 17.
The question asks us for the area of the triangle, which is given by the formula (1/2)bh. We are told it is a right triangle, so we can use one of the legs as the base, and the other leg as the height, since the legs will intersect at right angles. The legs of the right triangle must be the smallest sides (the longest must be the hypotenuse), which in this case are 8 and 15. So, let's assume that 8 is the base and 15 is the height.
The area of a triangle is (1/2)bh. We can substitute 8 and 15 for b and h.
.
The answer is 60 units squared.
### Example Question #4 : Right Triangles
Figure not drawn to scale.
In the figure above, the line segment BC is a diameter of the circle and has a length of 12 units. If the measure of the arc AC is equal to 120 degrees, then what is the area, in square units, of the region outside of the triangle and inside the circle?
36π – 18√3
36π – 36
144π – 36√3
144π – 18√3
36π – 36√3
36π – 18√3
Explanation:
Since segment BC is a diameter, it divides the circle into two arcs with measures of 180 degrees each. In general, if an angle is inscribed in a circle, its measure is half of the measure of the arc which it intercepts. As a result, angle BAC, which intercepts an arc of 180 degrees, must have a 90 degree measure. In other words, angle BAC is a right angle, and triangle BAC is a right triangle.
We are also told that the measure of arc AC is 120 degrees. This means that inscribed angle ABC, which intercepts arc AC, has a measure of one-half of 120 degrees, or 60 degrees.
In triangle BAC, angle A is 90 degrees and angle B is 60 degrees. The measure of angle C must be 30 degrees, because the sum of all three angles must equal 180, as they would in any triangle.
Therefore, triangle BAC is a special 30-60-90 right triangle. Segment BC is the hypotenus of the triangle. According to the properties of 30-60-90 triangles, the smallest side is equal to one-half the length of the hypotenuse. This means that the length of AB must be 6. Also, the length of the side across from the 60 degree angle is equal to the shortest side of the triangle multiplied by √3. Thus, the length of AC is 6√3.
Now that we have the lengths of the sides of the triangle, we can find its area. The area of a triangle is equal to one-half of the product of the base and the height. In the case of any right triangle, the base and height are the legs of the triangle, which intersect at a right angle.
area of triangle = (1/2)bh
= (1/2)(6)(6√3)
=18√3
Next, we must find the area of the circle so that we can find the area of the region outside of the triangle and inside the circle.
Because the diameter of the circle is 12, the radius must be one-half of the diameter, or 6.
The area of a circle is given by the formula below:
area of circle = πr2
= π(62)
= 36π
Therefore, the area of the circle is 36π, and the area of the triangle is 18√3. The area that is outside the triangle and inside the circle is equal to the difference between the area of the circle and the triangle.
area of region = 36π – 18√3
The answer is 36π – 18√3.
### Example Question #3 : Right Triangles
Figure not drawn to scale.
In the figure above, rays PA and PB are tangent to circle O at points A and B, respectively. If the diameter of circle O is 16 units and the length of line segment PO is 17 units, what is the area, in square units, of the quadrilateral PAOB?
240
136
60
120
68
120
Explanation:
Because PA and PB are tangent to circle O, angles PAO and PBO must be right angles; therefore, triangles PAO and PBO are both right triangles.
Since AO and OB are both radii of circle O, they are congruent. Furthermore, because PA and PB are external tangents originating from the same point, they must also be congruent.
So, in triangles PAO and PBO, we have two sides that are congruent, and we have a congruent angle (all right angles are congruent) between them. Therefore, by the Side-Angle-Side (SAS) Theorem of congruency, triangles PAO and PBO are congruent.
Notice that quadrilateral PAOB can be broken up into triangles PAO and PBO. Since those triangles are congruent, each must comprise one half of the area of quadrilateral PAOB. As a result, if we find the area of one of the triangles, we can double it in order to find the area of the quadrilateral.
Let's determine the area of triangle PAO. We have already established that it is a right triangle. We are told that PO, which is the hypotenuse of the triangle, is equal to 17. We are also told that the diameter of circle O is 16, which means that every radius of the circle is 8, because a radius is half the size of a diameter. Since segment AO is a radius, its length must be 8.
So, triangle PAO is a right triangle with a hypotenuse of 17 and a leg of 8. We can use the Pythagorean Theorem in order to find the other leg. According to the Pythagorean Theorem, if a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse, then:
a2 + b2 = c2
Let us let b represent the length of PA.
82 + b2 = 172
64 + b2 = 289
Subtract 64 from both sides.
b2 = 225
Take the square root of both sides.
b = 15
This means that the length of PA is 15.
Now let's apply the formula for the area of a right triangle. Because the legs of a right triangle are perpendicular, one can be considered the base, and the other can be considered the height of the triangle.
area of triangle PAO = (1/2)bh
= (1/2)(8)(15) = 60
Ultimately, we must find the area of quadrilateral PAOB; however, we previously determined that triangles PAO and PBO each comprise half of the quadrilateral. Thus, if we double the area of PAO, we would get the area of quadrilateral PAOB.
Area of PAOB = 2(area of PAO)
= 2(60) = 120 square units
### Example Question #4 : How To Find The Area Of A Right Triangle
If the hypotenuse of a triangle is 5 meters, which of the following is the closest value to the area of the triangle?
45
5
26
12
54
12
Explanation:
The answer is 12. In this circumstance, the area of the triangle cannot be smaller than its hypotenuse length, and cannot be bigger than its hypotenuse squared (that would be the area of a square).
### Example Question #2 : Triangles
Triangle ABC is drawn between the points A(4, 3), B(4, 8), and C(7, 3). What is the area of ABC?
Explanation:
Drawing a quick sketch of this triangle will reveal that it is a right triangle. The lines AB and AC form the height and base of this triangle interchangeably, depending on how you look at it.
Either way the formula for the area of the triangle is the distance from A to B multiplied by the distance from A to C, divided by 2.
This is
### Example Question #6 : How To Find The Area Of A Right Triangle
A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?
15
6
10
3
12
6
Explanation:
The area of a triangle is denoted by the equation 1/2 b x h.
b stands for the length of the base, and h stands for the height.
Here we are told that the perimeter (total length of all three sides) is 12, and the hypotenuse (the side that is neither the height nor the base) is 5 units long.
So, 12-5 = 7 for the total perimeter of the base and height.
7 does not divide cleanly by two, but it does break down into 3 and 4,
and 1/2 (3x4) yields 6.
Another way to solve this would be if you recall your rules for right triangles, one of the very basic ones is the 3,4,5 triangle, which is exactly what we have here
### Example Question #1 : How To Find The Area Of A Right Triangle
The ratio for the side lengths of a right triangle is 3:4:5. If the perimeter is 48, what is the area of the triangle?
108
240
96
48
50
96
Explanation:
We can model the side lengths of the triangle as 3x, 4x, and 5x. We know that perimeter is 3x+4x+5x=48, which implies that x=4. This tells us that the legs of the right triangle are 3x=12 and 4x=16, therefore the area is A=1/2 bh=(1/2)(12)(16)=96.
### Example Question #15 : How To Find The Area Of A Right Triangle
The length of one leg of an equilateral triangle is 6. What is the area of the triangle?
Explanation:
The base is equal to 6.
The height of an quilateral triangle is equal to , where is the length of the base.
### Example Question #1 : How To Find The Area Of A Right Triangle
Note: Figure NOT drawn to scale.
Refer to the above diagram. In terms of area, is what fraction of ?
Insufficient information is given to answer the question.
Explanation:
The area of a triangle is half the product of its base and its height.
The area of is
The area of is
Therefore, is of .
Note that actually finding the measure of is not necessary.
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# What times what equals 72?
Relevance
8x9=72...hold your ten fingers up...(only when multiplying by 9) if your multiplying 9 and 4...hold down your ring finger on your left hand...then look at how many fingers are on the left side of your ring finger (the 4th finger because your mulitplying it by 4) ...there are 3 fingers up...then on the right...6 fingers...36 is the answer...
hold down the 8th finger. (the middle finger on your left hand) on the left side of the downed finger are 7 fingers up. On the right...2...the answer is 72...8x9=72
• 3 years ago
72 Times 3
• Anonymous
9 times 8 = 72
72 times 1=72
2 times 36=72
3 times 24=72
4 times 18=72
6 times 12=72
1 and 72
2 and 36
8 and 9
• Anonymous
3 years ago
9€8
9 x 8 = 72.
• 6 years ago
The factors of 72 is:
1x72=72
2x36=72
3x24=72
4x18=18
6x12= 72
8x9=72
These are the factors of 72
1,2,3,4,6,8,9,12,18,24,36,72
• achain
Lv 5
You can write an endless list of "what times what" = 72
However, if you get prime factors of 72 and combine them the way you want, you can write some 2 factors combinations.
72 = 1*2*2*2*3*3 (prime factors)
8 and 9
1x72
2x36
3x24
4x 18
6x 12
8x9
9x8
12x 6
18x 4
24 x 3
36X2
72x 1
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Lecture 5 Post
# Lecture 5 Post - Lecture 5 Wednesday 31 August 2011 Chapter...
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Lecture 5 Wednesday 31 August 2011 Chapter 2: Motion along a straight line (cont’d) Constant acceleration along the x -axis Vertical free-fall motion along the y -axis
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How long does it take after the brakes are applied before the car comes to a stop? 1. 2.0 s 2. 4.0 s 3. 6.0 s 4. 8.0 s A car is moving in a straight line towards the positive x -axis with initial speed 30 m/s. The brakes are applied and the acceleration of the car is a x = ! 5.0 m/s 2 . Question Solution Here we use the constant acceleration expression for the velocity v x = v 0 x + a x t . Solve for t to get t = v x ! v 0 x a x = 0 ! (30 m/s) ! 5.0 m/s 2 = 6.0 s . !
How far does the car go after the brakes are applied before the car comes to a stop? 1. 90 m 2. 120 m 3. 150 m 4. 180 m A car is moving in a straight line towards the positive x -axis with initial speed 30 m/s. The brakes are applied and the acceleration of the car is then a x = ! 5.0 m/s 2 . Example
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Lecture 5 Post - Lecture 5 Wednesday 31 August 2011 Chapter...
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Ask a homework question - tutors are online
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# 2017 AMC 10A Problems/Problem 14
## Problem
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%$
## Solution
Let $m$ = cost of movie ticket
Let $s$ = cost of soda
We can create two equations:
$m = \frac{1}{5}(A - s)$
$s = \frac{1}{20}(A - m)$
Substituting we get:
$m = \frac{1}{5}(A - \frac{1}{20}(A - m))$
which yields:
$m = \frac{19}{99}A$
Now we can find s and we get:
$s = \frac{4}{99}A$
Since we want to find what fraction $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
$\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23\%\qquad \mathrm}$ (Error compiling LaTeX. ! Missing } inserted.)
2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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## Convert terajoule to kilojoule
terajoule kilojoule
How many terajoule in 1 kilojoule? The answer is 1.0E-9.
We assume you are converting between terajoule and kilojoule.
You can view more details on each measurement unit:
terajoule or kilojoule
The SI derived unit for energy is the joule.
1 joule is equal to 1.0E-12 terajoule, or 0.001 kilojoule.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between terajoules and kilojoules.
Type in your own numbers in the form to convert the units!
## Quick conversion chart of terajoule to kilojoule
1 terajoule to kilojoule = 1000000000 kilojoule
2 terajoule to kilojoule = 2000000000 kilojoule
3 terajoule to kilojoule = 3000000000 kilojoule
4 terajoule to kilojoule = 4000000000 kilojoule
5 terajoule to kilojoule = 5000000000 kilojoule
6 terajoule to kilojoule = 6000000000 kilojoule
7 terajoule to kilojoule = 7000000000 kilojoule
8 terajoule to kilojoule = 8000000000 kilojoule
9 terajoule to kilojoule = 9000000000 kilojoule
10 terajoule to kilojoule = 10000000000 kilojoule
## Want other units?
You can do the reverse unit conversion from kilojoule to terajoule, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Terajoule
The SI prefix "tera" represents a factor of 1012, or in exponential notation, 1E12.
So 1 terajoule = 1012 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## Definition: Kilojoule
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilojoule = 103 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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What's between an irrational and rational number? [closed]
There is a rational number between two irrational numbers, and an irrational number between two rational numbers. So what's between an irrational and rational number?
I know about rational numbers being found between 2 real numbers but I don't know how it applies to this.
• between distinct REAL numbers there are infinitely many rational and irrational numbers. – sranthrop Jul 18 '17 at 22:16
Rational numbers and irrational numbers are real numbers.
For any $x,y\in{\bf R}$ with $x<y$, there exists rational $a$ and irrational $b$ such that $a,b\in(x,y)$ by density:
What does it mean for rational numbers to be "dense in the reals?"
Proof that the set of irrational numbers is dense in reals
Both rational and irrational numbers are dense in the real line. This means for any two real numbers (rational or irrational), there is at least one rational number and at least one irrational number between them. It follows from this fact that there are infinitely many rational and irrational numbers between each pair of real numbers.
This can be a difficult concept to grasp, but as mentioned in the answers above, if you pick any two real numbers, there will be a rational number between them. Moreover, there will also be an irrational number between them.
This also implies that you can find a rational number as close to an irrational number as you want (without being equal). For examples, the number $$3.14159265358979323846264338327950288$$ is a rational number which is less than $1 \times 10^{-35}$ bigger than $\pi$. Using an approximation of $\pi$ using even more decimal, you would also find even more rational numbers that are between that number and $\pi$. In fact, you could also imagine more irrational numbers between them. Here's a process how to find one. You take the number $\pi$ which as an infinite decimal expansion that never repeats, and you change $1$ digit really far in the expansion, but fixing everything else. In that way, you would obtain an other irrational number between the number above and $\pi$.
As you might have realized, there is nothing special about the numbers that I picked above, any two numbers will have that property. This is due to the fact that both rationals and irrationals are dense in the real number.
However, on result which is even more interesting in this area, and I hope you'll soon see and understand the result is that even though both sets are infinite, there is immensely more irrational numbers than there are rational numbers.
You can actually see a longer discussion on that subject on the following post:
Cardinality of the Irrationals
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# JavaScript: Find the armstrong numbers of 3 digits
## JavaScript Conditional Statement and loops: Exercise-9 with Solution
Write a JavaScript program to find the armstrong numbers of 3 digits.
Note : An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3**3 + 7**3 + 1**3 = 371.
Pictorial Presentation:
Sample Solution:-
HTML Code:
``````<!DOCTYPE html>
<html>
<meta charset=utf-8 />
<title>Three digit armstrong numbers</title>
<body>
</body>
</html>
```
```
JavaScript Code:
``````function three_digit_armstrong_number()
{
for (var i = 1; i < 10; ++i)
{
for (var j = 0; j < 10; ++j)
{
for (var k = 0; k < 10; ++k)
{
var pow = (Math.pow(i,3) + Math.pow(j,3) + Math.pow(k,3));
var plus = (i * 100 + j * 10 + k);
if (pow == plus)
{
console.log(pow);
}
}
}
}
}
three_digit_armstrong_number();
```
```
Sample Output:
```153
370
371
407
```
Flowchart:
Live Demo:
See the Pen javascript-conditional-statements-and-loops-exercise-9 by w3resource (@w3resource) on CodePen.
Improve this sample solution and post your code through Disqus
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## JavaScript: Tips of the Day
Creates an array of elements, ungrouping the elements in an array produced by zip and applying the provided function
Example:
```const tips_unzip = (arr, fn) =>
arr
.reduce(
(acc, val) => (val.forEach((v, i) => acc[i].push(v)), acc),
Array.from({
length: Math.max(...arr.map(x => x.length))
}).map(x => [])
)
.map(val => fn(...val));
console.log(tips_unzip([[2, 15, 200], [3, 25, 300]], (...args) => args.reduce((acc, v) => acc + v, 0)));
```
Output:
```[5, 40, 500]
```
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# How to prove this sum of integrals
How to prove this ? $$\displaystyle \sum\limits_{k=0}^{\infty }{\int_{2k\pi }^{\left( 2k+1 \right)\pi }{{{\text{e}}^{-\frac{x}{2}}}\frac{\left| \sin x-\cos x \right|}{\sqrt{\sin x}}}}\text{d}x=\frac{2\cdot \sqrt[4]{8}{{\text{e}}^{-\frac{\pi }{8}}}}{1-{{\text{e}}^{-\pi }}}$$
-
Any attempts from your side? – Eckhard Dec 29 '12 at 15:42
In the summand integral, perform a change of variables, $x= 2 \pi k + t$: $$\int_{2 \pi k}^{2 \pi k + \pi} \exp\left(-\frac{x}{2}\right) \frac{| \sin x - \cos x|}{\sqrt{ \sin x}} \mathrm{d}x = \exp(-\pi k) \int_0^\pi \exp(-t/2) \frac{| \sin t - \cos t|}{\sqrt{ \sin t}} \mathrm{d}t$$ Therefore: $$\sum_{k=1}^\infty \int_{2 \pi k}^{2 \pi k + \pi} \exp\left(-\frac{x}{2}\right) \frac{| \sin x - \cos x|}{\sqrt{ \sin x}} \mathrm{d}x = \sum_{k=0}^\infty \exp(-\pi k) \cdot \int_0^\pi \exp(-t/2) \frac{| \sin t - \cos t|}{\sqrt{ \sin t}} \mathrm{d}t$$ The problem is now split into two simpler problems: a trivial sum, and less trivial integral. The integral admits an anti-derivative: $$\frac{\mathrm{d}}{\mathrm{d}t} \left( 2 \sqrt{\sin t} \exp(-t/2) \right) = \exp(-t/2) \frac{\cos t - \sin t}{\sqrt{\sin t}}$$ You should be able to finish it off now.
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https://www.fmaths.com/square-root/readers-ask-what-is-square-root-of-5.html
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| 811,796,975 | 10,506 |
# Readers ask: What Is Square Root Of 5?
## What is the square root of 5 simplified?
The numerical value of the square root of 5, which has been shortened to 50 decimal places is as follows: 2.23606797749978969640917366873127623544061835961152… This is the simplified value of square root of 5.
## What is the square root of 5 in fraction form?
The square root of 5 is not a rational number. This is because 5 cannot be expressed as a fraction where both the numerator and denominator have even exponents. This means that a rational number cannot have been squared to get 5.
## How do you find the square root of 5?
Therefore, the value of root 5 is, √ 5 = 2.2360… You can find the value of the square root of all the non-perfect square number with the help of the long division method.
## Is the square root of 5 A whole number?
This is definitely a whole number, an integer, and a rational number. This is not a rational number because it is not possible to write it as a fraction. If we evaluate it, the square root of 5 will have a decimal value that is non-terminating and non-repeating. This makes it an irrational number.
You might be interested: Often asked: How To Get Square Root Of Decimal Number?
## What is the perfect square of 5?
List of Perfect Squares
NUMBER SQUARE SQUARE ROOT
4 16 2.000
5 25 2.236
6 36 2.449
7 49 2.646
## What is the value of root 6?
√ 6 = 2.449 Thus, we have found the value of root 6.
## Which is the square root of 144?
The value of the square root of 144 is equal to 12. In radical form, it is denoted as √ 144 = 12.
## What are the roots of 4?
Square Root From 1 to 50
Number Square Root Value
3 1.732
4 2
5 2.236
6 2.449
## How do you find the root value?
The product property of square roots states that for any given numbers a and b, Sqrt (a × b) = Sqrt (a) × Sqrt (b). Because of this property, we can now take the square roots of our perfect square factors and multiply them together to get our answer. In our example, we would take the square roots of 25 and 16.
## Is 5 a real number?
Rational and irrational numbers form real numbers set. 5 consists of digits only so it is natural, but as mentioned above it is also integer, rational and real.
## Is √ 5 is an irrational number?
√5 is an irrational number.
## Is 5 a whole number?
In mathematics, whole numbers are the basic counting numbers 0, 1, 2, 3, 4, 5, 6, … and so on. 17, 99, 267, 8107 and 999999999 are examples of whole numbers. Whole numbers include natural numbers that begin from 1 onwards.
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# Show that l2 space separable
1. May 12, 2010
### complexnumber
1. The problem statement, all variables and given/known data
1. Prove that if a metric space $$(X,d)$$ is separable, then
$$(X,d)$$ is second countable.
2. Prove that $$\ell^2$$ is separable.
2. Relevant equations
3. The attempt at a solution
1. $$\{ x_1,\ldots,x_k,\ldots \}$$ is countable dense subset. Index the
basis with rational numbers, $$\{ B(x,r) | x \in A, r \in \mathbb{Q} \}$$ is countable (countable $$\times$$ countable).
2. What set is a countable dense subset of $$\ell^2$$?
2. May 17, 2010
### ninty
2. Let A = the set of sequences with only finitely many non-zero components(N of them), where each term is a member of the rationals.
We can show that the we can approximate every element of $$\ell^2$$ by sequences in A, hence the closure is $$\ell^2$$. (The set $$\ell^2$$ \ A are the limit points)
If you think about it, between any reals there's a rational number
So for each term, we can get a rational that is of distance $$\frac{\epsilon}{N}$$ of it.
Then the distance is $$N*\frac{\epsilon}{N}$$.
Take limit as N goes to infinity.
It's late here so I'm not really capable of putting all this into nice sentences.
3. May 17, 2010
### Landau
1. correct
2. this comes down to the fact that R (or C) is separable; just restrict to rationals and finite sequences (see ninty's reply).
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# According to some analysts, whatever its merits, the
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According to some analysts, whatever its merits, the proposal to tax away all capital gains on short-term investments would, if enacted, have a disastrous effect on Wall Street trading and employment.
(A) its merits, the proposal to tax
(B) its merits may be, the proposal of taxing
(C) its merits as a proposal, taxing
(D) the proposal’s merits, to tax
(E) the proposal’s merits are, taxing
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03 Sep 2005, 18:48
A. D is not correct guys. D doesnot have pronoun for noun and even its noun is not clear too.
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04 Sep 2005, 19:43
"whatever the proposal’s merits are" sounds better than "whatever the proposal’s merits' that sounds like it is not a complete sentence.
Even if I don't like the +ing form in choice E, I would still go for E.
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04 Sep 2005, 20:33
I think D). Sounds like "its" is refering to "some analysts" when it is suppose to refer to "the proposal". That gets rid of A and B, C sucks, and E uses "taxing" which doesn't sit right with me.
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05 Sep 2005, 06:42
E for me.
D sounds incomplete to me. and so do A and C.
The use of 'may be' and the pronoun 'its' in B makes it unacceptable.
OA/OE Plz.
krishna
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05 Sep 2005, 08:26
Only relevent answer choices are B and E.
rest of the choices are missing a verb, which is needed to complete a appositive or absolute clause...
E is definetly wrong, because the sentence is unclear in its meaning. There is no clause or phrase that is stating the proposal,
B states the proposal correctly and its refer to the possessive form of proposal. Moreover, B establishes the contrast indicating that whatever the merits or benifits (MAY BE), it is not a good proposal.
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05 Sep 2005, 08:34
The OA is A.
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05 Sep 2005, 17:38
whats the source of this question?
I am not sure if OA is correct...
I mean what is "its" referring to?
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05 Sep 2005, 18:07
Tough one ! I am still trying to understand why A and not the others...
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06 Sep 2005, 01:03
Can't explain but i did chose A. "the proposal to tax" sounded best to me.
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06 Sep 2005, 01:03
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# Physics
does a baseball thrown at 100 miles per hour have a lot of force?
1. It has zero force, but it does have momentum. If something gets in the path of the ball, that something will exert a force.
Force*time= change mometnum= massball*change in velocityball
posted by bobpursley
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# Gas Station Problem - Dijkstra's Algorithm variation
## I am trying to find an algorithm which finds the least expensive route from one town to another.
This is the general setup.
There are a series of one-way roads from some towns to other towns. Not all towns have roads between them. Let R be the set of all the roads and let T be the set of all towns. For two towns u, v ∈ T, let (u→v)∈ R be the one-way road from u to v and let w(u, v) be the amount of gasoline required to travel along this road.
Let S ⊂ T be the set of towns which have gas stations. At a town u ∈ S, one can choose to pay c(u) dollars to fill up the gas tank. Note that the cost does not change depending on how much gas must be bought; if you need to buy gas, fill up the whole tank. The cost is only affected by which town u is.
The car’s tank can hold k units of gas and it is full at the start.
## My ideas:
I believe this uses a shortest path graph algorithm, but I am unsure how to implement it. I believe it uses Dijkstra's algorithm. I also recognize that the run-time depends on |T|, |R| and |S|.
Any advice will help! thanks :)
• Please do not vandalize your post if you cannot delete it. Even though you may have a solution now, we want to preserve the question and answers so that they can help others as well. Apr 14, 2020 at 6:25
We note that we can assume that the source node has a gas station with refilling cost as $$0$$, even if it doesn't. It just makes the algorithm cleaner, as we start out with a full tank, and we note that no optimal solution is ever going to come back to the source again and refill. Similarly, we also assume that the $$\text{target}$$ also has a gas station with some arbitrary cost of refilling - the cost here wouldn't matter. So in effect, we set $$S = S \cup \{\text{source, target}\}$$.
In the first phase of the algorithm, for every $$s \in S$$, we want to find all the other towns in $$S$$ that we can reach without refueling, if we start out with a full tank from $$s$$. For this, we can take two approaches:
1. Run Floyd–Warshall using $$w(u, v)$$ as the weight of every edge, to find the shortest distance from any town to any other. Now, for every $$s \in S$$, we iterate through every other $$t \in S$$ and if the shortest distance from $$s$$ to $$t$$ is $$\leq k$$, then we can reach $$t$$ from $$s$$ without refueling. So we create a new graph $$G'$$, whose vertex set is $$S$$, and add a directed edge $$(s, t)$$ with a weight of $$c(s)$$. This approach takes $$\mathcal{O}(|T|^3)$$ time and $$\mathcal{O}(|T|^2)$$ space.
2. Another way to construct the same $$G'$$, would be to start a Dijkstra's from every $$s \in S$$, and stop when the distance is more than $$k$$, which in the worst case wouldn't help our complexity. This approach would require $$\mathcal{O}(|S| \times (|R| + |T|\log|T|)$$ time and $$\mathcal{O}(|T|^2)$$ space.
Now on to the second phase of the algorithm, where we have the graph $$G'$$ with us, which again is a directed weight graph, but now the weights are costs of refilling. So all we have to do is run a Dijkstra's on this graph starting from $$\text{source}$$. The answer is the shortest distance to $$\text{target}$$.
So the total algorithm needs $$\mathcal{O}(|S| \times (|R| + |T|\log|T|)$$ time and $$\mathcal{O}(|T|^2)$$ space.
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# Help with Fibonacci Identity
## Main Question or Discussion Point
Can someone guide me on how to prove that
$$F_{4n+3} + F_{4n+6} = F_{2n+1}^2 + F_{2n+4}^2$$
either side of the above is the difference
$$(F_{2n+2}*F_{2n+3} + F_{2n+4}^2) - (F_{2n}*F_{2n+1} + F_{2n+2}^2)$$
I intend to post this sequence $$F_{2n}*F_{2n+1} + F_{2n+2}^2$$, with a comment re a few properties thereof, on Sloane's online encyclopedia of integer sequences but would like to verify the above identity first.
Related Linear and Abstract Algebra News on Phys.org
First thing springs to mind is to try use general formula for nth
Fibbonacci number:
$$F_{n}=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$$
In order to proceed with math induction.But I'm unsure will it work or not.
I'm sure there are better methods ,though.
Last edited:
First thing springs to mind is to try use general formula for nth
Fibbonacci number:
$$F_{n}=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$$
In order to proceed with math induction.But I'm unsure will it work or not.
I'm sure there are better methods ,though.
Thanks
I think there is an identity for the following that works:
$$F_{i}*F_{j} + F_{i+1}*F_{j+1} = F_{?}$$
Let j = i = 2n+1 then
$$F_{2n+1}^{2} + F_{2n+2}^{2} = F_{4n+3}$$
[Tex]F_{2n+2}^{2} + F_{2n+3}^{2} = F_{4n+5}[/tex]
[Tex]F_{2n+2}^{3} + F_{2n+4)^{2} = F_{4n+7}[/tex]
\\
[Tex]F_{4n+3} +F_{4n+6} = F_{4n+3} + F_{4n+7} - F_{4n+5}[/tex]
[Tex] =F_{2n+1}^{2} + F_{2n+4}^{2}[/tex]
Last edited:
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Transcendental functions
Exponential and logarithmic functions, inverse functions
Trigonometric (cyclometric) functions and inverse trigonometric functions (arc-functions)
Transcendental functions:
· Exponential and logarithmic functions are mutually inverse functions
- Exponential function y = ex <=> x = ln y, e = 2.718281828...the base of the natural logarithm,
exponential function is inverse of the natural logarithm function, so that eln x = x.
- Logarithmic function y = ln x = log e x <=> x = e y, where x > 0
the natural logarithm function is inverse of the exponential function, so that ln(ex) = x.
- Exponential function y = ax <=> x = loga y, where a > 0 and a is not 1
exponential function with base a is inverse of the logarithmic function, so that
- Logarithmic function y = log a x <=> x = a y, where a > 0a is not 1 and x > 0
the logarithmic function with base a is inverse of the exponential function, so that loga (ax) = x.
· Trigonometric (cyclometric) functions and inverse trigonometric functions (arc functions)
Trigonometric functions are defined as the ratios of the sides of a right triangle containing the angle equal to the argument of the function in radians.
Or more generally for real arguments, trigonometric functions are defined in terms of the coordinates of the terminal point Q of the arc (or angle) of the unit circle with the initial point at P(1, 0).
sin2x + cos2x = 1
- The sine function y = sin x is the y-coordinate of the terminal point of the arc x of the unit circle. The graph of the sine function is the sine curve or sinusoid.
In a right-angled triangle the sine function is equal to the ratio of the length of the side opposite the given angle to the length of the hypotenuse.
- The arc-sine function y = sin-1x or y = arcsin x is the inverse of the sine function, so that its value for any argument is an arc (angle) whose sine equals the given argument.
That is, y = sin-1x if and only if x = sin y. For example,
Thus, the arc-sine function is defined for arguments between -1 and 1, and its principal values are by convention taken to be those between -p/2 and p/2.
- - - - - - -
- The cosine function y = cos x is the x-coordinate of the terminal point of the arc x of the unit circle. The graph of the cosine function is the cosine curve or cosinusoid.
In a right-angled triangle the cosine function is equal to the ratio of the length of the side adjacent the given angle to the length of the hypotenuse.
- The arc-cosine function y = cos-1x or y = arccos x is the inverse of the cosine function, so that its value for any argument is an arc (angle) whose cosine equals the given argument.
That is, y = cos-1x if and only if x = cos y. For example,
Thus, the arc-cosine function is defined for arguments between -1 and 1, and its principal values are by convention taken to be those between 0 and p.
- - - - - - -
- The tangent function y = tan x is the ratio of the y-coordinate to the x-coordinate of the terminal point of the arc x of the unit circle, or it is the ratio of the sine function to the cosine function.
In a right-angled triangle the tangent function is equal to the ratio of the length of the side opposite the given angle to that of the adjacent side.
- The arc-tangent function y = tan-1x or y = arctan x is the inverse of the tangent function, so that its value for any argument is an arc (angle) whose tangent equals the given argument.
That is, y = tan-1x if and only if x = tan y. For example,
Thus, the arc-tangent function is defined for all real arguments, and its principal values are by convention taken to be those strictly between -p/2 and p/2.
- - - - - - -
- The cosecant function y = csc x is the reciprocal of the sine function.
In a right-angled triangle the cosecant function is equal to the ratio of the length of the hypotenuse to that of the side opposite to the given angle.
- The arc-cosecant function y = csc-1x or y = arccsc x is the inverse of the cosecant function, so that its value for any argument is an arc (angle) whose cosecant equals the given argument.
That is, y = csc-1x if and only if x = csc y. For example,
Thus, the arc-cosecant function is defined for arguments less than -1 or greater than 1, and its principal values are by convention taken to be those between -p/2 and p/2.
- - - - - - -
- The secant function y = sec x is the reciprocal of the cosine function.
In a right-angled triangle the secant function is equal to the ratio of the length of the hypotenuse to that of the side adjacent to the given angle.
- The arc-secant function y = sec-1x or y = arcsec x is the inverse of the secant function, so that its value for any argument is an arc (angle) whose secant equals the given argument.
That is, y = sec-1x if and only if x = sec y. For example,
Thus, the arc-secant function is defined for arguments less than -1 or greater than 1, and its principal values are by convention taken to be those between 0 and p.
- - - - - - -
- The cotangent function y = cot x is the reciprocal of the tangent function, or it is the ratio of the cosine function to the sine function.
In a right-angled triangle the cotangent function is equal to the ratio of the length of the side adjacent to the given angle to that of the side opposite it.
- The arc-cotangent function y = cot-1x or y = arccot x is the inverse of the cotangent function, so that its value for any argument is an arc (angle) whose cotangent equals the given argument.
That is, y = cot-1x if and only if x = cot y. For example,
Thus, the arc-cotangent function is defined for all real arguments, and its principal values are by convention taken to be those strictly between 0 and p.
Calculus contents A
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# Is total variation continuous?
Given a sequence of signed measures $<\nu_j>$, if it happens that $\nu=\sum\limits_{j = 1}^\infty \nu_j$ is still a valid signed measure (then it can be proved that each partial sum $\nu_n=\sum\limits_{j = 1}^n \nu_j$ is valid signed measure), do we have $\lim\limits_{n\to \infty}|\sum\limits_{j = 1}^n \nu_j|=|\sum\limits_{j = 1}^\infty \nu_j|$ ($=|\lim\limits_{n\to \infty} \sum\limits_{j = 1}^n \nu_j|$)? Thanks!
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Which topology do employ on the space of measures? The one induced by total variation norm? – shuhalo Feb 19 '11 at 8:46
And measures on which space, by the way ? – BS. Feb 19 '11 at 12:19
Sorry for the late response. I'm concerned with the equality in the post which looks like a continuity property, so I used the term "continuous" in the title. Of course it would be better if we can establish a setting in which topology is equipped and then continuity of total variation of signed measures has the usual meaning. But currently I have no idea about this. – zzzhhh Feb 21 '11 at 10:59
Assuming I understood the question correctly, the answer is no. Consider measures on $\{0,1\}^\omega$ with the product topology and Borel $\sigma$-algebra. Let $\mu_i$ be the uniform measure on the set with $i$-th coordinate equal to 0. This sequence converges by your definition to the uniform measure, but all $\mu_i$ are far (in total variation) from the uniform measure. (To fit your description we can take $\nu_i=\mu_i-\mu_{i-1}$).
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Sorry but what is the uniform measure? – zzzhhh Feb 21 '11 at 11:00
The uniform measure is the probability measure of total measure 1, and for each $i$, $\mathbb{P}(x_i=0)=\frac12$ independently. – Ori Gurel-Gurevich Feb 21 '11 at 18:57
I refrained from asking what is total measure. Could you please refer me to a textbook containing conceptions like uniform measure, total measure so that I can understand your reply after reading it? Thanks! – zzzhhh Feb 22 '11 at 10:40
I'm really nor very good with books recommendations, but any rigorous book on probability theory should be helpful. Perhaps the following reformulation will help: the space is just $I=[0,1]$. $\mu$ is the Lebesgue measure restricted to $I$. $I_n$ is the set of all reals in $I$ whose $n$-th binary digit is 0. $\mu_n$ is twice the Lebesgue restricted to $I_n$, i.e. $\mu_n(A)=2\mu(A \cap I_n)$. Then for every measurable $A$ we have $\mu_n(A) \to \mu(A)$, but the total variation distance between any $\mu_n$ and $\mu$ is 1. – Ori Gurel-Gurevich Feb 22 '11 at 18:22
_ Thank you! – zzzhhh Feb 23 '11 at 7:04
Am I misunderstanding your question? Because it seems to me that if you consider the $\nu_n = \frac{(-1)^n}{n} \delta$ where $\delta$ is the Dirac mass, then we've got the usual situation with the alternating harmonic series.
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Yes, this is an example that the equality holds. But do we have counterexamples? – zzzhhh Feb 21 '11 at 11:00
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# Mechanics help !
1. Nov 24, 2011
### awesome4444
Mechanics help :(!
I REALLY NEED HELP :(
Two blocks are connected by a cord which passes over a frictionless pulley at D.
Block A has a mass of 40 kg and block B a
mass of 120 kg. A rests on a horizontal
surface, while B rests on a ramp with the slope
shown. B is initially in a position 3 m from the
stop C as shown. The coefficient of kinetic
friction between the blocks and the surface
beneath is μK = 0.20 for both A and B.
(a) Find the acceleration of the two blocks and
the tension T in the cord, assuming that the
blocks just begin to move at time t = 0.
2. Nov 24, 2011
### grzz
Re: Mechanics help :(!
Diagram?
3. Nov 24, 2011
### awesome4444
4. Nov 24, 2011
### grzz
Re: Mechanics help :(!
First we draw the FBD for mass A and for mass B.
How many forces act on each?
5. Nov 24, 2011
### awesome4444
Re: Mechanics help :(!
on mass 1 we have forcetension - forcefriction + m*g = ma
mass2 -frocetension-Forcefriction2+(m*g)sin theta= m2a
the thing thats confusing me is if we could have assumed that the aceleration of B is same as acceleration of a before it hits the barrier in the diagram
6. Nov 25, 2011
### grzz
Re: Mechanics help :(!
To avoid confusion, instead of mass 1 let us call it mass A as in diagram.
Re 1st equation: I do not know why you have m*g.
Until mass B hits the barrier at C we HAVE TO assume that the acceleration of B is the same (in magnitude) as that of A because otherwise the string either snaps or becomes slack and so there would not be any tension at all!
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# If ${\sec ^2}\theta + {\tan ^2}\theta + 1 = 2$, then find the value of $\sec \left( { - \theta } \right)$:A. $- 2$B. $- \dfrac{1}{2}$C. $1$D. $\pm 1$
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Hint : Solve using trigonometric identities.
Given that: ${\sec ^2}\theta + {\tan ^2}\theta + 1 = 2$
Converting the above equation in the terms of Sin and Cos, we get
$\Rightarrow \dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + 1 = 2{\text{ }}\left( {\because \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}{\text{ and }}\sec \theta = \dfrac{1}{{\cos \theta }}} \right) \\ \Rightarrow \dfrac{{1 + {{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = 2 \\ \Rightarrow \dfrac{{1 + 1}}{{{{\cos }^2}\theta }} = 2{\text{ }}\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right) \\ \Rightarrow \dfrac{2}{{{{\cos }^2}\theta }} = 2 \\ \Rightarrow {\cos ^2}\theta = 1 \\ \Rightarrow \cos \theta = \pm 1{\text{ }} \ldots \, \ldots \left( 1 \right) \\$
We know that, $\cos \left( { - \theta } \right) = \cos \left( \theta \right)$
$\therefore \sec \left( { - \theta } \right) = \sec \left( \theta \right) = \dfrac{1}{{\cos \theta }}$
Put the value of $\cos \theta$ from equation $\left( 1 \right)$, we get
$\sec \left( { - \theta } \right) = \pm 1$
Note: In these types of problems, where there is no direct formula for the given trigonometric terms, one should always try to convert them to some trigonometric terms which have some relation using trigonometric relations and identities so as to make the problem easier to calculate.
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# Thread: Distance from a point to a set
1. ## Distance from a point to a set
I need to find $\displaystyle (in R^2) \rho((4,2), A_2)$ where $\displaystyle A_2=\{(x,y)|x^2+y^2=1 \}$.
I know that the definition says that this distance is the greatest lower bound of $\displaystyle \{\rho(x,a)|a \in A\}$. So the distance between $\displaystyle \rho((4,2), (x,y))= \sqrt{(x-4)^2+(y-2)^2}=\sqrt{x^2+8x+16+y^2-4y+4}= \sqrt{21-8x-4y}$ and then I am not sure what to do.
2. The easy way is to write the equation of the line through $\displaystyle (0,0)~\&~(4,2)$.
Find the its intersection with the circle.
One of those two points is the closest.
3. Thanks so much! I had made this problem quite hard by trying to use polar coordinates, but this method is much simpler. Thank you so much!
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Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up? New to KöMaL?
# KöMaL Problems in Physics, April 2017
Show/hide problems of signs:
## Problems with sign 'M'
Deadline expired on May 10, 2017.
M. 368. Let us experiment with a spinning top, or with any other type of gyroscope. Measure how the number of revolution of the top changes as a function of time, once it was set in motion and left to spin.
(6 pont)
statistics
## Problems with sign 'G'
Deadline expired on May 10, 2017.
G. 599. The acceleration of an object starting from rest at $\displaystyle t_0=0$, and moving along a straight path, is $\displaystyle 4~{\rm m/s}^2$. The magnitude of the acceleration of another object, undergoing uniform circular motion of radius 9 m, is also $\displaystyle 4~{\rm m/s}^2$.
$\displaystyle a)$ What is the instantaneous speed of each object at $\displaystyle t_1=5$ s?
$\displaystyle b)$ How much distance do they cover during the 5 s?
(3 pont)
solution (in Hungarian), statistics
G. 600. A helicopter flies between two cities daily. Its normal speed in each direction is 120 km/h. On a windy day in one direction its speed was 140 km/h, whilst in the other direction it was 100 km/h, thus the total time of the whole journey increased by 15 minutes.
What is the distance between the two cities?
(3 pont)
solution (in Hungarian), statistics
G. 601. A 100 gram and a 50 gram weights were glued together, and hung to the end of a vertical spring. When the glue dried out the 50 gram weight came unstuck, and fell down.
At what acceleration did the other object start to move?
(3 pont)
solution (in Hungarian), statistics
## Problems with sign 'P'
Deadline expired on May 10, 2017.
P. 4927. When a battery of unknown internal resistance and electromotive force is loaded with a resistor of resistance 10 ohm the terminal voltage across the battery is 6 V and the energy stored in the battery decreases by 7.2 J in each second.
$\displaystyle a)$ What is the internal resistance of the battery?
$\displaystyle b)$ What is its electromotive force?
$\displaystyle c)$ What is the terminal voltage when the resistance of the load is increased to 20 ohms?
(4 pont)
solution (in Hungarian), statistics
P. 4928. $\displaystyle a)$ At what accelerations do the trolleys shown in the figure start to move if the wheels can roll easily, and the mass of the pulley and air drag are negligible? Data: $\displaystyle m_1=1$ kg, $\displaystyle m_2=2$ kg, $\displaystyle M=5$ kg.
$\displaystyle b)$ For other mass values, in what interval can the value of the initial acceleration of the trolley of mass $\displaystyle M$ be?
(5 pont)
solution (in Hungarian), statistics
P. 4929. A rigid rod of length $\displaystyle 3L$, and of negligible mass, shown in the figure can rotate in the vertical plane about a fixed horizontal axle, which is at a distance of $\displaystyle L$ from the left end of the rod. To the ends of the rod small objects of masses $\displaystyle m$ and $\displaystyle 2m$ are attached, and then at a certain moment the rod is released from its horizontal position.
$\displaystyle a)$ Determine the speeds of the objects when the rod is vertical.
$\displaystyle b)$ At this moment what is the force exerted by the rod on the axle?
$\displaystyle c)$ What is the acceleration of the objects at the moment right after the rod was released?
(4 pont)
solution (in Hungarian), statistics
P. 4930. There is a uniform density right-triangular cross-section prism of mass $\displaystyle M$ on the horizontal ground. The prism lies on its smallest face, and the surface of the ground is rough enough. The acute angle of the triangle at its vertex which is on the ground is $\displaystyle \alpha=60^\circ$. A thread attached to the midpoint of the top edge of the prism, and a small object of mass $\displaystyle m$ is fixed to the other end of the thread. The bob of the simple pendulum is released from the horizontal position as shown in the figure. The prism is observed to tip when the thread makes an angle of $\displaystyle 20^\circ$ with the horizontal.
$\displaystyle a)$ What is the mass ratio $\displaystyle M/m$?
$\displaystyle b)$ To what value should this ratio $\displaystyle M/m$ be increased in order that the moving pendulum should not be able to make the prism tip at all?
(5 pont)
solution (in Hungarian), statistics
P. 4931. In a thermally insulated horizontal cylinder an easily moveable heat conducting piston separates two samples of gas of the same type, and of different temperature and different pressure. Initially the piston is fixed, but later the fixing ceased. What initial conditions may result in that after the equilibrium is reached, the piston is not on that side where initially the gas had smaller pressure?
(4 pont)
solution (in Hungarian), statistics
P. 4932. The figure shows the cyclical processes of a heat engine which is operated with a certain amount of ideal, noble gas. Both of the cyclical processes $\displaystyle ABCA$ and $\displaystyle CDEC$ consist of isothermal isobaric and isochoric processes. What is the ratio of the two efficiencies?
(5 pont)
solution (in Hungarian), statistics
P. 4933. Two unfixed spheres of mass 1 g, and of radius 4 mm are charged to $\displaystyle Q_1=10^{-10}$ C, and $\displaystyle Q_2=-2\cdot 10^{-10}$ C and are placed at a distance of 1 m. The spheres are left to move freely.
How long does it take for them to be at a distance of 25 cm?
(5 pont)
solution (in Hungarian), statistics
P. 4934. A 2-metre long straight current carrying wire radiates energy to its environment at a rate of 10 J per second, whilst its temperature remains the same.
$\displaystyle a)$ What is the electric field strength in the wire, if the diameter of the wire is 4 mm, and the magnitude of the magnetic flux density at the surface of the wire is 0.002 Vs/m$\displaystyle {}^2$?
$\displaystyle b)$ What type of material may the wire be made of?
(4 pont)
solution (in Hungarian), statistics
P. 4935. The wavelengths of a photon and an electron are the same. Which of them has greater kinetic energy?
(5 pont)
solution (in Hungarian), statistics
P. 4936. During the decay of a stationary, free neutron what may the greatest value of the kinetic energy of the electron be?
(5 pont)
solution (in Hungarian), statistics
P. 4937. In the future men will be able to build spaceships which are able to take us to distant star systems. Suppose that one of such a spaceship leaves the Earth at the escape speed, and due to its special engine every day its kinetic energy is doubled. (Its rest mass remains the same.)
Estimate how much does the captain of the spaceship get older when he travels to Alpha Centauri, which is at a distance of 4.3 light years from the Earth.
(6 pont)
solution (in Hungarian), statistics
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## 1. Sections of a cone: parabola, hyperbola, ellipse
Chapter 11
Conic Sections
Sections of a cone: circle, parabola, hyperbola, ellipse:
When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations:
(a) When b = 90o, the section is a circle .
(b) When a < b < 90o, the section is an ellipse .
(c) When b = a; the section is a parabola .
(In each of the above three situations, the plane cuts entirely across one nappe of the cone).
(d) When 0 £ b < a; the plane cuts through both the nappes and the curves of intersection is a hyperbola .
Circle: Set of points in a plane equidistant from a fixed point. A circle with radius r and centre (h, k) can be represented as (x – h)+ (y – k)= r2
Parabola: Set of points in a plane that are equidistant from a fixed-line and point. A parabola with a > 0, focus at (a, 0), and directrix x = – a can be represented as y= 4ax
In parabola y= 4ax, the length of the latus rectum is given by 4a.
Ellipse: The sum of distances of a set of points in a plane from two fixed points is constant. An ellipse with foci on the x-axis can be represented as:
Hyperbola: The difference of distances of set of points in a plane from two fixed points is constant. The hyperbola with foci on the x-axis can be represented as:
## 2. Standard equations, Properties and Application of a circle
Standard equations , Properties and Application of a circle
Circle:
Definition: A circle is the set of all points in a plane that are equidistant from a fixed
point in the plane.
The fixed point is called the centre of the circle and the distance from the centre
to a point on the circle is called the radius of the circle.
Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on
the circle . Then, by the definition, | CP | = r . By the distance formula,
we have
(x-h)2 + (y-k)2= r2
This is the required equation of the circle with centre at (h,k) and radius r .
General form of Equation of a Circle
The general equation of any type of circle is represented by:
x2 + y2 + 2gx + 2fy + c = 0, for all values of g, f and c.
Adding g2 + f2 on both sides of the equation gives,
x2 + 2gx + g2+ y2 + 2fy + f2= g2 + f2 − c ………………(1)
Since, (x+g)= x2+ 2gx + g2 and (y+f)=y+ 2fy + f2 substituting the values in equation (1), we have
(x+g)2+ (y+f)= g+ f2−c …………….(2)
Comparing (2) with (x−h)+ (y−k)= a2, where (h, k) is the center and ‘a’ is the radius of the circle.
h=−g, k=−f
a2 = g2+ f2−c
Therefore,
x+ y+ 2gx + 2fy + c = 0, represents the circle with centre (−g,−f) and radius equal to a2 = g+ f2− c.
• If g+ f> c, then the radius of the circle is real.
• If g+ f= c, then the radius of the circle is zero which tells us that the circle is a point that coincides with the center. Such a type of circle is called a point circle
• g+ f<c, then the radius of the circle become imaginary. Therefore, it is a circle having a real center and imaginary radius.
N.B.: Standard Equation of Circle:
centre (0, 0) and Radius (r)
Equation of circle in centre radius form:
Centre (h, k), Radius = r
Equation of circle in General form:
Where (–g, –f ) centre
r2 = g2 + f2 – c .
Equation of circle with points P(x1, y1) and Q(x2, y2) as extremities of diameter is
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
Equation of circle through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3) is
Area of circle = πr2
Perimeter = 2πr, where r is the radius.
Example: Find an equation of the circle with centre at (0,0) and radius r.
Solution :Here h = k = 0. Therefore, the equation of the circle is x2 + y2 = r2
Example: Find the equation of the circle with centre (–3, 2) and radius 4.
Solution: Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is
(x + 3) 2 + (y –2)2 = 16
Example : Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0
Solution : The given equation is
(x2 + 8x) + (y22+ 10y) = 8
Now, completing the squares within the parenthesis, we get
(x2+ 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)2 + (y + 5)2 = 49
i.e. {x – (– 4)} 2+ {y – (–5)} 2 = 72
Therefore, the given circle has centre at (– 4, –5) and radius 7.
Example : Find the equation of the circle which passes through the points (2, – 2), and (3, 4) and whose centre lies on the line x + y = 2.
Solution: Let the equation of the circle be (x – h)2 + (y – k)2 = r2.
Given that the circle passes through the points (2, –2) and (3, 4).
Thus,
(2 – h)2 + (–2 – k)2 = r2….(1)
and (3 – h)2 + (4 – k)2 = r2….(2)
Also, given that the centre lies on the line x + y = 2.
h + k = 2 ….(3)
Solving the equations (1), (2) and (3), we get
h = 0.7, k = 1.3 and r2 = 12.58
Hence, the equation of the required circle is
(x – 0.7)2 + (y – 1.3)2 = 12.58
## 3. Standard equations, Properties and Application of a parabola
Standard equations , Properties and Application of a parabola
Parabola:
Definition: A parabola is the set of all points in a plane that are equidistant from a fixed line
and a fixed point (not on the line) in the plane.
The fixed line is called the directrix of the parabola and the fixed point F is called the focus 11. (‘Para’ means ‘for’ and ‘bola’ means ‘throwing’, i.e., the shape described when you throw a ball in the air).
A line through the focus and perpendicular to the directrix is called the axis of the parabola.
The point of intersection of parabola with the axis is called the vertex of the parabola.
Standard equations of parabola
The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola.
Let P(x, y) be any point on the parabola such that
PF = PB, ... (1)
where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance
formula, we have
(PF) 2 = (x – a)2 + y2 and (PB) 2 = (x + a) 2
Since PF = PB, we have
(x – a) 2 + y2 = (x + a) 2
or x2 – 2ax + a2 + y2 = x2 + 2ax + a2
or y2 = 4ax ( a > 0). is Standard equation of Parabola.
Latus rectum
Definition : Latus rectum of a parabola is a line segment perpendicular to the axis of
the parabola, through the focus and whose end points lie on the parabola .
To find the Length of the latus rectum of the parabola y2 = 4ax .
By the definition of the parabola, AF = AC.
But AC = FM = 2a
Hence AF = 2a.
And since the parabola is symmetric with respect to x-axis AF = FB and so
AB = Length of the latus rectum = 4a.
Example : Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 8x.
Solution The given equation involves y2, so the axis of symmetry is along the x-axis.
The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 2.
Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is
x = – 2 .
Length of the latus rectum is 4a = 4 × 2 = 8.
Properties:
1. Parabola is symmetric with respect to the axis of the parabola.If the equation
has a y2 term, then the axis of symmetry is along the x-axis and if the
equation has an x2 term, then the axis of symmetry is along the y-axis.
2. When the axis of symmetry is along the x-axis the parabola opens to the
(a) right if the coefficient of x is positive,
(b) left if the coefficient of x is negative.
3. When the axis of symmetry is along the y-axis the parabola opens
(c) upwards if the coefficient of y is positive.
(d) downwards if the coefficient of y is negative
Example: Find the equation of the parabola with focus (2,0) and directrix x = – 2.
Solution :Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the
parabola. Hence the equation of the parabola is of the form either
y2 = 4ax
• y2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola
is to be of the form y2 = 4ax with a = 2.
Hence the required equation is
y2 = 4(2)x = 8x
Example: Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).
Solution: Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the
y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form
x2 = 4ay. thus, we have
x2 = 4(2)y, i.e., x2= 8y.
## 4. Standard equations, Properties and Application of a ellipse
Standard equations , Properties and Application of a ellipse
Ellipse
Definition: An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse.
The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse
We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b
Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse:
Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F2P
(Since, F1P = F1O + OP)
= c + a + a – c = 2a
a2 = b2 + c2
Special cases of an ellipse In the equation c2 = a2 – b2 obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape.
Case (i) When c = 0, both foci merge together with the centre of the ellipse and a2 = b2, i.e., a = b, and so the ellipse becomes circle . Thus, circle is a special case of an ellipse.
Case (ii) When c = a, then b = 0. The ellipse reduces to the line segment F1F2 joining the two foci.
Eccentricity
Definition: The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., e= c/a
Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.
Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin and the line from O through F2 be the positive x-axis and that through F1as the negative x-axis.
Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) .
Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a
so given PF1 + PF2 = 2a.
Hence any point on the ellipse satisfies satisfies
the geometric condition and so P(x, y) lies on the ellipse.
Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines.
Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines.
Similarly, we can derive the equation of the ellipse.
These two equations are known as standard equations of the ellipses.
Properties:
1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse.
2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator.
Latus rectum
Definition: Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse.
Let the length of AF2 be l. Then the coordinates of A are (c, l ),
i.e., (ae, l )
Since A lies on the ellipse
we have
• l2 = b2 (1 – e2) but e2=c2/a2 = 1- (b2/a2)
Therefore, l2 = b4/a2
• l = b2/a
Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 2l =2b2/a
Example: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse
Solution Since denominator of x2/25 is larger than the denominator of y2/9 , the major
axis is along the x-axis.
a = 5 and b = 3. Also
c =√(a2 – b2)
=√(25 – 9)
=4
Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and
(5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the
eccentricity is 4/5. and latus rectum is 2b2/a = 18 /5
## 5. Standard equations, Properties and Application of a hyperbola
Standard equations , Properties and Application of a hyperbola
Hyperbola
Definition : A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.
The term “difference” that is used in the definition means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.
The distance between the two foci by 2c, the distance between two vertices (the length of the transverse axis) by 2a.
b= √(c2- a2)
2b is the length of the conjugate axis
we have
BF1 – BF2 = AF2 – AF1 (by the definition of the hyperbola)
BA +AF1– BF2 = AB + BF2– AF1
i.e., AF1 = BF2
So that, BF1 – BF2 = BA + AF1– BF2 = BA = 2a
Eccentricity
Definition : similarly, an ellipse, the ratio e = c / a is called the eccentricity of the
hyperbola. Since c ³a, the eccentricity is never less than one.
In terms of the eccentricity, the foci are at a distance of ae from the centre.
Standard equation of Hyperbola: The equation of a hyperbola is simplest if
the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The
two such possible orientations.
Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O
be the origin and the line through O through F2 be the positive x-axis and
that through F1 as the negative x-axis.
Let P(x, y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a.
So given, PF1 – PF2 = 2a
On squaring again and further simplifying, we get
(Since c2 a2 = b2)
Or
Note.: A hyperbola in which a = b is called an equilateral hyperbola.
Therefore, no portion of the curve lies between the
lines x = + a and x = – a, (i.e. no real intercept on the conjugate axis).
Similarly, we can derive the equation of the hyperbola.
These two equations are known as the standard equations of hyperbolas.
Properties:
1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola.
2. The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.
has transverse axis along x-axis of length 6.
has transverse axis along y-axis of length 10.
Latus rectum:
Definition Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.
The length of the latus rectum in hyperbola is 2b2/a
Example Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus rectum is 36.
Solution Since foci are (0, ±12), it follows that c = 12.
Length of the latus rectum = 2b2/a = 36
or b2 = 18a
Therefore c2 = a2 + b2; gives
144 = a2 + 18a
i.e., a2 + 18a – 144 = 0,
So a = – 24, 6.
Since a cannot be negative, we take a = 6 and so b2 = 108.
Therefore, the equation of the required hyperbola is
Or,
i.e., 3y2 – x2 = 108
Question :1 An ellipse passes through the foci of the hyperbola, 9x2 −4y2 =36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 1/2. Find the equation of ellipse.
Solution:
Equation of hyperbola is 9x2 −4y2 =36 or x2/4 − y2/9 = 1
(Here a < b)
Focus = (0, ± be)
Eccentricity = e = √(1+4/9) = √13/3
So, Foci of hyperbola: (0, ±√13)
Standard equation of the ellipse, x2/a2 + y2/b2 = 1 …(i)
Eccentricity = e’ = √(1-a2/b2) …(ii)
ee’ = 1/2 (given)
Using eccentricity value of hyperbola, e’ = 1/2 x 3/√13 = 3/2√13
(ii) e’2 = (1-a2/b2)
9/52 = (1-a2/b2)
Find the value of b2 form (i) using focii 13/b2 = 1 => b2 = 13
9/52 = (1-a2/13)
9/4 = 13 – a2
a2 = 43/4
Now equation of ellipse is 4x2/43 + y2/13 = 1
Question 2: The locus of a point P(α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola x2/a2 – y2/b2 = 1 is
(a) an ellipse
(b) a circle
(c) a hyperbola
(d) a parabola
Solution:
Tangent to the hyperbola x2/a2 – y2/b2 = 1 is y = mx ± √(a2m2 – b2)
Given that y = αx + β is the tangent of hyperbola.
m = α and a2m2 – b2 = β2
Therefore, a2 α 2 – b2 = β2
Locus is a2 x 2 – y2 = b, which is parabola.
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# college Trig
posted by .
Solve this equation on the interval 0 θ < 2π. Round your answer(s) to two decimal places.
2 sin θ + 3 = 2
(smaller value)
(larger value)
## Respond to this Question
First Name School Subject Your Answer
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# good trignometric inequality
• Apr 7th 2013, 10:16 PM
earthboy
good trignometric inequality
prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
• Apr 7th 2013, 10:30 PM
earthboy
Re: good trignometric inequality
I started by trying to show the expression as the sum of squares:
$\displaystyle x^2\sin x + x \cos x + x^2 + \frac{1}{2}$
$\displaystyle = x^2\sin x + x \cos x + x^{2}(\sin^{2} x + \cos^{2} x)+\frac{1}{4}+ \frac{1}{4}$
$\displaystyle =(x^{2} \cos^{2} x + x \cos x + \frac{1}{4})+x^{2} \sin^{2} x + x^{2} \sin x + \frac{1}{4}$
$\displaystyle =(x \cos x + \frac{1}{2})^2 + x^2 \sin^2 x + x^2 \sin x + \frac{1}{4}$
any suggestion on how to go about from here......???
Advanced thanks for any help or suggestions....(Happy)
• Apr 7th 2013, 10:48 PM
Gusbob
Re: good trignometric inequality
Quote:
Originally Posted by earthboy
prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
Wow this is a fun inequality. This is what I did, it is similar to yours.
$\displaystyle 0\leq\frac{(x\sin(x)+x)^2}{2}=\frac{1}{2}(x^2\sin^ 2(x) + 2x^2\sin(x)+x^2)$
$\displaystyle 0\leq\frac{(x\cos(x)+1)^2}{2}=\frac{1}{2}(x^2\cos^ 2(x) + 2x\cos(x)+1)$
$\displaystyle 0\leq \frac{1}{2}(x^2(\sin^2(x)+\cos^2(x))+2(x^2\sin(x)+ x\cos(x))+x^2+1)$
$\displaystyle 0\leq x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}$
I'm not sure about if the strict inequality case is true. If it is, the adjustment would be minor (just show that if one of the expressions is zero if the other one is strictly positive).
• Apr 8th 2013, 04:42 AM
topsquark
Re: good trignometric inequality
Quote:
Originally Posted by earthboy
prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
I did it pretty much term by term.
Note that
$\displaystyle x^2~sin(x) + x^2 = x^2(sin(x) + 1)$
As sin(x) has a minimum of -1, this expression is always 0 or positive.
Note that
$\displaystyle x~cos(x) \geq 0$ since both x and cos(x) are odd functions.
And of course 1/2 is always positive.
-Dan
• Apr 8th 2013, 10:42 PM
earthboy
Re: good trignometric inequality
(Ninja)great proof! Gusbob!
Quote:
Originally Posted by topsquark
Note that
$\displaystyle x~cos(x) \geq 0$ since both x and cos(x) are odd functions.
-Dan
Thanks Dan!(Nod), but can you or anybody please explain how this part is true ???
As $\displaystyle x$ is any real number, what about if $\displaystyle x=120$ , then $\displaystyle x \cos x = -60$ which is $\displaystyle \leq 0$.
• Apr 15th 2013, 12:23 AM
ohiosuba
Re: good trignometric inequality
great proof! Gusbob!
___________________
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• Apr 15th 2013, 07:15 AM
topsquark
Re: good trignometric inequality
Quote:
Originally Posted by earthboy
(Ninja)great proof! Gusbob!
Thanks Dan!(Nod), but can you or anybody please explain how this part is true ???
As $\displaystyle x$ is any real number, what about if $\displaystyle x=120$ , then $\displaystyle x \cos x = -60$ which is $\displaystyle \leq 0$.
| 1,119 | 3,173 |
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# What is the derivative of arctan(1/x)?
Jul 31, 2015
The derivative is: $\frac{- 1}{{x}^{2} + 1}$
#### Explanation:
$\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$
So
$\frac{d}{\mathrm{dx}} \arctan \left(u\right) = \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$
And
$\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) = \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
$= \frac{1}{1 + \frac{1}{x} ^ 2} \cdot \frac{- 1}{x} ^ 2$
$= {x}^{2} / \left({x}^{2} + 1\right) \cdot \frac{- 1}{x} ^ 2$
$= \frac{- 1}{{x}^{2} + 1}$
Faster Method?
Use the fact that $\arctan \left(\frac{1}{x}\right) = a r c \cot \left(x\right)$
and
$\frac{d}{\mathrm{dx}} a r c \cot \left(x\right) = - \frac{1}{1 + {x}^{2}}$
to go straight to the answer.
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Algebra Tutorials!
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## Completing the Square
The Vertex Form of a Quadratic Function
The format for a quadratic equation given above,
y = a · x 2 + b · x + c, where the letter x represents the input, the letter y represents the value of the output and the letters a, b and c are all numbers, is called standard form.
Other ways of writing the equations for quadratic functions include vertex form,
y = a · (x - h) 2 + k,
where the letter x represents the value of the input, the letter y represents the value of the output and the letters a, h and k all represent numbers. Just as in standard form, in vertex form the number a cannot be equal to zero. Converting a quadratic equation to vertex form is often quite helpful as it allows you to determine exactly where the graph of the quadratic equation reaches its “low point†or “high point†very easily. Every single quadratic formula can be converted to vertex form. The process for doing this conversion is called completing the square.
What the Vertex Form of a Quadratic can tell you about the graph
The vertex form of a quadratic function:
y = a · (x - h) 2 + k,
also tells you whether the graph of the quadratic is smiling or frowning. To check, simply look at the value of a, as you would if the equation had been written in standard form. If the value of a is positive then the quadratic is smiling and if the value of a is negative then the quadratic will be frowning.
The vertex form of a quadratic equation can also tell you about the location of the highest point (on a frowning quadratic) or the lowest point (on a smiling quadratic – see Figure 1 on the next page). This point (the highest point on a frowning quadratic or the lowest point on a smiling quadratic) is called the vertex.
The x-coordinate of the vertex is the number h that appears inside the parentheses of the vertex form and the y-coordinate of the vertex is the number k that appears outside the parentheses in the vertex form.
Figure 1: (a) In this quadratic, a = -1 and the shape of the graph is a “frown.†The vertex in this case is the highest point on the graph. (b) In this quadratic a = 0.5 and the shape of the graph is a “smile.†The vertex in this case is the lowest point on the graph.
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# Boyle's LawPage 2
#### WATCH ALL SLIDES
Curved lines are hard to recognise, so we plot the volume against the reciprocal of pressure (ie. 1/p)
This time the points lie close to a straight line through the origin.
This means volume is directly proportional to 1/pressure or
volume is inversely proportional to pressure
Slide 11
## This leads us back to Boyle’s Law
Boyle’s Law: for a fixed mass of gas kept at constant temperature the volume of the gas is inversely proportional to its pressure.
Slide 12
## Problem
A deep sea diver is working at a depth where the pressure is 3.0 atmospheres. He is breathing out air bubbles. The volume of each air bubble is 2 cm2. At the surface the pressure is 1 atmosphere. What is the volume of each bubble when it reaches the surface?
Slide 13
## How we work this out
We assume that the temperature is constant, so Boyle’s Law applies:
Formula first: P1 x V1 = P2 x V2
Then numbers:= 1.0 x 2 = 3.0 x V2
Now rearrange the numbers so that you have V2 on one side, and the rest of the numbers on the other side of the ‘equals’ symbol.
Slide 14
Here’s what you should have calculated
V2 = 3.0 x 2
1.0
therefore volume of bubbles = 6 cm3
Note that P1 and P2 have the same unit, as will V1 and V2
Go to page:
1 2
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https://leetcode.ca/2019-09-21-1391-Check-if-There-is-a-Valid-Path-in-a-Grid/
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##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1391.html
# 1391. Check if There is a Valid Path in a Grid (Medium)
Given a m x n grid. Each cell of the grid represents a street. The street of grid[i][j] can be:
• 1 which means a street connecting the left cell and the right cell.
• 2 which means a street connecting the upper cell and the lower cell.
• 3 which means a street connecting the left cell and the lower cell.
• 4 which means a street connecting the right cell and the lower cell.
• 5 which means a street connecting the left cell and the upper cell.
• 6 which means a street connecting the right cell and the upper cell.
You will initially start at the street of the upper-left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets.
Notice that you are not allowed to change any street.
Return true if there is a valid path in the grid or false otherwise.
Example 1:
Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).
Example 2:
Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)
Example 3:
Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).
Example 4:
Input: grid = [[1,1,1,1,1,1,3]]
Output: true
Example 5:
Input: grid = [[2],[2],[2],[2],[2],[2],[6]]
Output: true
Constraints:
• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• 1 <= grid[i][j] <= 6
Related Topics: Depth-first Search, Breadth-first Search
## Solution 1. Union Find
// OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
// Time: O(MN)
// Space: O(MN)
class UnionFind {
private:
vector<int> id, rank;
int find (int i) {
if (id[i] == i) return i;
return id[i] = find(id[i]);
}
public:
UnionFind(int n) : id(n), rank(n, 0) {
for (int i = 0; i < n; ++i) id[i] = i;
}
void connect(int i, int j) {
int p = find(i), q = find(j);
if (p == q) return;
if (rank[p] > rank[q]) id[p] = q;
else {
id[q] = p;
if (rank[p] == rank[q]) rank[p]++;
}
}
bool connected(int i, int j) { return find(i) == find(j); }
};
class Solution {
int M, N;
int h(int x, int y) { return x * N + y; }
const int dirs[4][2] = { {0,-1},{0,1},{-1,0},{1,0} };
const int neighbor[6][2] = { {0,1}, {2,3}, {0,3}, {1,3}, {0,2}, {1,2} };
public:
bool hasValidPath(vector<vector<int>>& A) {
M = A.size(), N = A[0].size();
UnionFind uf(M * N);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int n : neighbor[A[i][j] - 1]) {
int x = i + dirs[n][0], y = j + dirs[n][1];
if (x < 0 || x >= M || y < 0 || y >= N) continue;
int r = n ^ 1;
auto &rn = neighbor[A[x][y] - 1];
if (rn[0] != r && rn[1] != r) continue;
uf.connect(h(x, y), h(i, j));
}
}
}
return uf.connected(h(0,0), h(M-1,N-1));
}
};
## Solution 2. DFS
// OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
// Time: O(MN)
// Space: O(MN)
class Solution {
int M, N;
vector<vector<int>> A;
vector<vector<bool>> vis;
int h(int x, int y) { return x * N + y; }
int dx[4] = {1,-1,0,0}, dy[4] = {0,0,1,-1}, t[6] = {4|8, 1|2, 8|1, 4|1, 8|2, 4|2};
bool dfs(int i, int j) {
if (i == M - 1 && j == N - 1) return 1;
vis[i][j] = 1;
for (int k = 0; k < 4; ++k) {
if (t[A[i][j] - 1] >> k & 1 ^ 1) continue; // If A[i][j] can't extend to this direction, skip
int x = i + dx[k], y = j + dy[k];
if (x < 0 || x >= M || y < 0 || y >= N || vis[x][y]) continue;
int rk = k ^ 1;
if (t[A[x][y] - 1] >> rk & 1 ^ 1) continue; // If A[x][y] can't extend back, skip
if (dfs(x, y)) return 1;
}
return 0;
}
public:
bool hasValidPath(vector<vector<int>>& A) {
M = A.size(), N = A[0].size();
this->A = A;
vis.assign(M, vector<bool>(N));
return dfs(0, 0);
}
};
• class Solution {
public boolean hasValidPath(int[][] grid) {
int rows = grid.length, columns = grid[0].length;
boolean[][] visited = new boolean[rows][columns];
visited[0][0] = true;
queue.offer(new int[]{0, 0});
while (!queue.isEmpty()) {
int[] cell = queue.poll();
if (cell[0] == rows - 1 && cell[1] == columns - 1)
return true;
List<int[]> reachableCells = reachableCells(grid[cell[0]][cell[1]], grid, cell, rows, columns);
for (int[] reachableCell : reachableCells) {
int row = reachableCell[0], column = reachableCell[1];
if (!visited[row][column]) {
visited[row][column] = true;
queue.offer(new int[]{row, column});
}
}
}
return false;
}
public List<int[]> reachableCells(int street, int[][] grid, int[] cell, int rows, int columns) {
List<int[]> reachableCells = new ArrayList<int[]>();
int row = cell[0], column = cell[1];
if (street == 1) {
int newColumn1 = column - 1, newColumn2 = column + 1;
if (newColumn1 >= 0) {
int newStreet = grid[row][newColumn1];
if (newStreet == 1 || newStreet == 4 || newStreet == 6)
}
if (newColumn2 < columns) {
int newStreet = grid[row][newColumn2];
if (newStreet == 1 || newStreet == 3 || newStreet == 5)
}
} else if (street == 2) {
int newRow1 = row - 1, newRow2 = row + 1;
if (newRow1 >= 0) {
int newStreet = grid[newRow1][column];
if (newStreet == 2 || newStreet == 3 || newStreet == 4)
}
if (newRow2 < rows) {
int newStreet = grid[newRow2][column];
if (newStreet == 2 || newStreet == 5 || newStreet == 6)
}
} else if (street == 3) {
int newRow = row + 1, newColumn = column - 1;
if (newRow < rows) {
int newStreet = grid[newRow][column];
if (newStreet == 2 || newStreet == 5 || newStreet == 6)
}
if (newColumn >= 0) {
int newStreet = grid[row][newColumn];
if (newStreet == 1 || newStreet == 4 || newStreet == 6)
}
} else if (street == 4) {
int newRow = row + 1, newColumn = column + 1;
if (newRow < rows) {
int newStreet = grid[newRow][column];
if (newStreet == 2 || newStreet == 5 || newStreet == 6)
}
if (newColumn < columns) {
int newStreet = grid[row][newColumn];
if (newStreet == 1 || newStreet == 3 || newStreet == 5)
}
} else if (street == 5) {
int newRow = row - 1, newColumn = column - 1;
if (newRow >= 0) {
int newStreet = grid[newRow][column];
if (newStreet == 2 || newStreet == 3 || newStreet == 4)
}
if (newColumn >= 0) {
int newStreet = grid[row][newColumn];
if (newStreet == 1 || newStreet == 4 || newStreet == 6)
}
} else if (street == 6) {
int newRow = row - 1, newColumn = column + 1;
if (newRow >= 0) {
int newStreet = grid[newRow][column];
if (newStreet == 2 || newStreet == 3 || newStreet == 4)
}
if (newColumn < columns) {
int newStreet = grid[row][newColumn];
if (newStreet == 1 || newStreet == 3 || newStreet == 5)
}
}
return reachableCells;
}
}
############
class Solution {
private int[] p;
private int[][] grid;
private int m;
private int n;
public boolean hasValidPath(int[][] grid) {
this.grid = grid;
m = grid.length;
n = grid[0].length;
p = new int[m * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int e = grid[i][j];
if (e == 1) {
left(i, j);
right(i, j);
} else if (e == 2) {
up(i, j);
down(i, j);
} else if (e == 3) {
left(i, j);
down(i, j);
} else if (e == 4) {
right(i, j);
down(i, j);
} else if (e == 5) {
left(i, j);
up(i, j);
} else {
right(i, j);
up(i, j);
}
}
}
return find(0) == find(m * n - 1);
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private void left(int i, int j) {
if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
p[find(i * n + j)] = find(i * n + j - 1);
}
}
private void right(int i, int j) {
if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
p[find(i * n + j)] = find(i * n + j + 1);
}
}
private void up(int i, int j) {
if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
p[find(i * n + j)] = find((i - 1) * n + j);
}
}
private void down(int i, int j) {
if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
p[find(i * n + j)] = find((i + 1) * n + j);
}
}
}
• // OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
// Time: O(MN)
// Space: O(MN)
class UnionFind {
private:
vector<int> id, rank;
int find (int i) {
if (id[i] == i) return i;
return id[i] = find(id[i]);
}
public:
UnionFind(int n) : id(n), rank(n, 0) {
for (int i = 0; i < n; ++i) id[i] = i;
}
void connect(int i, int j) {
int p = find(i), q = find(j);
if (p == q) return;
if (rank[p] > rank[q]) id[p] = q;
else {
id[q] = p;
if (rank[p] == rank[q]) rank[p]++;
}
}
bool connected(int i, int j) { return find(i) == find(j); }
};
class Solution {
int M, N;
int h(int x, int y) { return x * N + y; }
const int dirs[4][2] = { {0,-1},{0,1},{-1,0},{1,0} };
const int neighbor[6][2] = { {0,1}, {2,3}, {0,3}, {1,3}, {0,2}, {1,2} };
public:
bool hasValidPath(vector<vector<int>>& A) {
M = A.size(), N = A[0].size();
UnionFind uf(M * N);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int n : neighbor[A[i][j] - 1]) {
int x = i + dirs[n][0], y = j + dirs[n][1];
if (x < 0 || x >= M || y < 0 || y >= N) continue;
int r = n ^ 1;
auto &rn = neighbor[A[x][y] - 1];
if (rn[0] != r && rn[1] != r) continue;
uf.connect(h(x, y), h(i, j));
}
}
}
return uf.connected(h(0,0), h(M-1,N-1));
}
};
• class Solution:
def hasValidPath(self, grid: List[List[int]]) -> bool:
m, n = len(grid), len(grid[0])
p = list(range(m * n))
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def left(i, j):
if j > 0 and grid[i][j - 1] in (1, 4, 6):
p[find(i * n + j)] = find(i * n + j - 1)
def right(i, j):
if j < n - 1 and grid[i][j + 1] in (1, 3, 5):
p[find(i * n + j)] = find(i * n + j + 1)
def up(i, j):
if i > 0 and grid[i - 1][j] in (2, 3, 4):
p[find(i * n + j)] = find((i - 1) * n + j)
def down(i, j):
if i < m - 1 and grid[i + 1][j] in (2, 5, 6):
p[find(i * n + j)] = find((i + 1) * n + j)
for i in range(m):
for j in range(n):
e = grid[i][j]
if e == 1:
left(i, j)
right(i, j)
elif e == 2:
up(i, j)
down(i, j)
elif e == 3:
left(i, j)
down(i, j)
elif e == 4:
right(i, j)
down(i, j)
elif e == 5:
left(i, j)
up(i, j)
else:
right(i, j)
up(i, j)
return find(0) == find(m * n - 1)
• func hasValidPath(grid [][]int) bool {
m, n := len(grid), len(grid[0])
p := make([]int, m*n)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
left := func(i, j int) {
if j > 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) {
p[find(i*n+j)] = find(i*n + j - 1)
}
}
right := func(i, j int) {
if j < n-1 && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) {
p[find(i*n+j)] = find(i*n + j + 1)
}
}
up := func(i, j int) {
if i > 0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4) {
p[find(i*n+j)] = find((i-1)*n + j)
}
}
down := func(i, j int) {
if i < m-1 && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) {
p[find(i*n+j)] = find((i+1)*n + j)
}
}
for i, row := range grid {
for j, e := range row {
if e == 1 {
left(i, j)
right(i, j)
} else if e == 2 {
up(i, j)
down(i, j)
} else if e == 3 {
left(i, j)
down(i, j)
} else if e == 4 {
right(i, j)
down(i, j)
} else if e == 5 {
left(i, j)
up(i, j)
} else {
right(i, j)
up(i, j)
}
}
}
return find(0) == find(m*n-1)
}
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https://mathematica.stackexchange.com/questions/85523/sum-over-multiple-indices-that-take-specific-values
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# Sum over multiple indices that take specific values
Pretty simple question: how do I sum over multiple indices that can only take specific values all together?
To clarify: let's say I have the following sum:
$\sum p_{a,b,c} f(a,b,c)$
where $f$ is a function and $p$ is a coefficient (number) still in symbolic form, both dependent on indices $a,b,c$. Here {$a$, $b$, $c$} can take values specified by a list of three numbers, like {1,2,3}, {3,6,7} and so on. My idea is that I have a list of triplets, and I would like the sum to run on these triplets by assigning $a$ to the first number, $b$ to the second and $c$ to the third. So let's say that for a list of triplets
list={{1,2,3},{3,6,7},{2,6,9}}
The outcome should be
$p_{1,2,3}f[1,2,3]+p_{2,6,9}f[2,6,9]+p_{3,6,7}f[3,6,7]$
I was able to do it with Apply if there was only a function, but the coefficients $p$ need to carry the indices so it becomes tricky.
• I'm not understanding what you mean by p_{1,2,3} as that translates to {p_,2 p_, 3 p_}. But if p_ is really a function that takes a list as an argument, then you might want to try Sum[p[list[[i]]] f[list[[i, 1]], list[[i, 2]], list[[i, 3]]], {i, Length[list]}] which gets you f[1, 2, 3] p[{1, 2, 3}] + f[2, 6, 9] p[{2, 6, 9}] + f[3, 6, 7] p[{3, 6, 7}].
– JimB
Jun 9, 2015 at 13:47
• So $p$ is actually also a function of $a,b,c$? Jun 9, 2015 at 13:51
• no, $p$ carries indices $a,b,c$ but they are just labels. i edited the question. Jun 9, 2015 at 13:55
• So $p$ is symbolic and not numerical? Jun 9, 2015 at 14:00
• If p is subscripted (and using subscripts can have definite but sometimes unexpected consequences), then the following might be what you want: Sum[Subscript[p, list[[i, 1]], list[[i, 2]], list[[i, 3]]] f[list[[i, 1]], list[[i, 2]], list[[i, 3]]], {i, Length[list]}] with output f[1, 2, 3] Subscript[p, 1, 2, 3] + f[2, 6, 9] Subscript[p, 2, 6, 9] + f[3, 6, 7] Subscript[p, 3, 6, 7].
– JimB
Jun 9, 2015 at 14:10
Here's one way:
Total[Subscript[p, #] & /@ lst f @@@ lst]
or
Total[Subscript[p, #] & /@ lst f /@ lst]
But you're probably making a mistake defining your p's this way, especially if you later need to do something with them. If you are willing to define p as a function, then
Total[p @@@ lst f @@@ lst]
would work.
• (p @@@ lst).(f @@@ lst) will also work. Jun 9, 2015 at 22:41
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Weighting Factor: Definition, Uses
Statistics Definitions > Weighting Factor
Contents:
1. What is a Weighting Factor?
2. Use in Sampling.
3. Use in Nuclear Medicine.
1. What is a Weighting Factor?
A weighting factor is a weight given to a data point to assign it a lighter, or heavier, importance in a group. It is usually used for calculating a weighted mean, to give less (or more) importance to group members. It is also used in statistical sampling for adjusting samples and in nuclear medicine for calculating effective doses.
Simple Example
For example, let’s say you take three tests in class. The last test is much harder than the first two tests, so your professor gives it less weight. The weights for the three tests are:
The weighted mean formula uses a weighting factor, wi.
The percents listed after the tests are the weighting factors. For example, test 1 has a weighting factor of 40% while test 3 has a weighting factor of 20%. Let’s say you score 80, 80, and 85 points. The weighted mean for the three tests is found by:
1. Multiplying your scores by the percentage weights:
.4(80) = 32
.4(80) = 32
.2(95) = 19
2. Adding the numbers up. 32 + 32 + 19 = 83.
See more examples in Weighted Mean.
2. Use In Sampling
Weighting factors are used in sampling to make samples match the population. For example, let’s say you took a sample of the population and had 41% female and 59% male. You know from census data that females should make up 51% of the population and males 49%. In order to make sure that you have a representative sample, you could add a little more “weight” to data from females. To calculate how much weight you need, divide the known population percentage by the percent in the sample. For this example:
Known population females (51) / Sample Females (41) = 51/41 = 1.24.
Known population males (49) / Sample males (59) = 49/59 = .83.
3. Use In Nuclear Medicine
Weighting factors are used extensively in radiologic and nuclear medicine to calculative effective doses for procedures. The calculations for Tissue Weighting Factors (sometimes called Radiologic Weighting Factors) account for the fact that different parts of the body absorb radiation at different rates.
A tissue weighting factor(WT) is assigned to body parts, with more radiosensitive parts given higher weighting factors.
Effective dose = individual organ dose values * WT.
Tissue weighting factors (ICRP) are:
• WT = 0.12: stomach, colon, lung, red bone marrow, breast, remainder tissues,
• WT = 0.04: urinary bladder, oesophagus, liver, thyroid,
• WT = 0.01: bone surface, skin, brain, salivary glands.
Reference:
European Nuclear Society. Tissue Weighting Factor. Retrieved 9/20/2006 from: https://www.euronuclear.org/info/encyclopedia/t/tissue-weight-factor.htm
International Commission on Radiological Protection. The 2007 Recommendations of the International Commission on Radiological Protection. ICRP publication 103. Ann ICRP. 2007;37 (2-4): 1-33. Available from PubMed.
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If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.
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# Distributing and multiplying polynomials pdf
Foiling, and pushing through multiplying binomials also sometimes called foiling is done frequently in your algebra classes. In the model, the length and width of a rectangle represent the two binomials. Every term in one polynomial must be multiplied by every term in the other polynomial. When multiplying binomials, use foil firstouterinnerlast example 3. Note that the product rule is used when distributing through the parentheses. It is so tempting to ignore this rule, but you are sure to have a wrong answer if you do. In this problem the exponent is 2, so it is multiplied two times. The volume v is equal to the area of the base b times the height h. It looks the same as the distributive property that we learned in. A w m m what methods can you use to add polynomials.
G j2 w0p1e2 r ik nuatja n 5saofbtjw 3a trke m hlplgcx. Rather than only factoring or only multiplying, the puzzles give students the opportunity to employ both operations. Introduction to multiplying polynomials she loves math. When working a subtraction problem, we will distribute the negative first and. How do you distribute with whole numbers and fractions. Polynomials are those expressions that have variables raised to all sorts of powers and multiplied by all types of numbers.
The foundation for multiplying any pair of polynomials is distribution and monomial multiplication. Multiplying binomials by polynomials old video khan. Just as we can multiply numbers, so also we can multiply polynomials. Lets look at the original area model and think about multiplying a different way. Add, subtract, multiply, divide and factor polynomials stepbystep. Change the division into addition by distributing negative one through the 2nd set of parentheses. Teachers pay teachers is an online marketplace where teachers buy and sell original educational materials. This video branches off of the how to foil video by explaining how to multiply polynomials with more terms. Type of multiplication monomial by a monomial monomial by other polynomials binomial by a binomial how follow exponent laws distributive property and exponent laws distributive property twice foil example 4a2. The exponents are always monomial operations with standard form 6 perimeter 2 laws of exponents anything to the first power is itself 20120 anything to the zero power is one 2001 when multiplying exponents that. To multiply any two polynomials, use the distributive property and multiply each term of one polynomial by each term of the other polynomial. Lesson 5 multiplying two binomials la19 example 3 use foil to multiply binomials geometry a shipping box is shaped like a rectangular prism. Here i have a binomial, meaning 2 terms being multiplied by a trinomial which has 3 terms. If we are adding or subtracting the exponnets will stay the same, but when we multiply or divide the exponents will be changing.
This website uses cookies to ensure you get the best experience. Multiply negative 4x squared by the whole expression 3x squared plus 25x minus 7. Next, we consider multiplying a monomial by a polynomial. Multiplying monomials by polynomials video khan academy. Distribute algebra tiles and copies of the multiplying polynomials using algebra tiles activity sheet. These unique features make virtual nerd a viable alternative to private tutoring. Step 1 distribute each term of the first polynomial to every term of the second polynomial. To raise a product to a power, raise each factor in the product to that power.
Multiplying polynomials and monomials when finding the product of a monomial and a polynomial, we multiply the monomial by each term of the polynomial. To group like terms, you can line them up vertically. A term is a number, variable or the product of a number and variables. Im going to do this multiplication using distributing, but i have to do the distributing kind of twice. I start by having students work on the entry ticket as soon as they enter the class as the year has progressed it has become more and more automatic that students take out their binders and get to work on the entry ticket rather than milling around or socializing. Distribute each term of the first polynomial to every term of the second polynomial. The work of solving the puzzles in polynomial puzzles 2 practice will help students to become more fluent with multiplying and factoring polynomials. Factoring polynomials metropolitan community college. Equations inequalities system of equations system of inequalities basic operations algebraic properties partial fractions polynomials rational expressions sequences power sums. A polynomial with just two terms is called a binomial. Remember that when you multiply two terms together you must multiply the coefficient numbers and add the exponents. Multiplying binomials with radicals old next lesson. Matho is played like bingo, however students work out the math problems before marking them off their board. Polynomials and factoring adding polynomials means combining like terms ex.
The simplest multiplication involving polynomials is where were taking a number through a set of parentheses. A video introduction of multiplying polynomials using distribution. Basically, this is the same as multiplying binomials except you cannot use the shortcut foil. Multiplying larger degree polynomials using distributing. Multiplying trinomials and polynomials when multiplying trinomials or polynomials, you just distribute all of the terms in the first polynomial. When you work with polynomials you need to know a bit of vocabulary, and one of the words you need to feel comfortable with is term. In this nonlinear system, users are free to take whatever path through the material best serves their needs. T r2c0s1 u29 nkxuqtbae 2s vodfktawzazrpe n plelhcz. We have seen this operation before when distributing through parentheses. Distributive property and factoring worksheet author. Multiplying binomials with radicals old our mission is to provide a free, worldclass education to anyone, anywhere.
Next we consider multiplying a monomial by a polynomial. T 3 3mqaadne u 2w iztrh6 8i pntf tilnji btee 0 pafllgpexbdrla e y1n. Multiplying polynomials using algebra tiles virginia department of. One method is to use an area model, but another way to multiply polynomials without having to draw diagrams, is to multiply polynomials using distribution. The factors and equivalent product for this model are. So were going to have to distribute this negative 4x squared over every term in the expression. By using this website, you agree to our cookie policy. To multiply monomials with the same base, keep the base and add the powers. These differences are the coefficients of the resulting polynomial.
So if you multiply anything times a whole expression, you really just use the distributive property to multiply each term of the expression by the negative 4x squared. We have seen this operation before with distributing through parenthesis. Multiplying polynomials is to activate students prior knowledge about working with polynomials. Multiplying polynomials examples, solutions, videos. This matho activity covers multiplying polynomials including distributing and simplifying, multiplying binomials, finding the square of a binomial, and finding the area of a shaded region. Discover on your own how to simplify by distributing the binomial with the trinomial. If we are adding or subtracting polynomials, the exponents will stay the same, but when we multiply or divide the exponents will be changing. The following are rules regarding the multiplying of variable expressions. Adding and subtracting polynomials date period kuta. And just as some numerical multiplication is easier than others, so it is with polynomials.
When youre multiplying two binomials together, you can use an easy to remember method called foil. Free practice questions for ged math simplifying, distributing, and factoring. Whether we are working with binomials, trinomials, or larger polynomials, the process is fundamentally the same. Techniques for multiplying polynomials examples, solutions. Multiplying binomials by polynomials video khan academy. Distributing monomials by brandi earl teachers pay teachers. J r vmhatd 4e u dwfi itqh 8 bi 5n4f wixnzi lt gem bprrze j1a 0ldg zeqbhrsap.
When multiplying polynomials, it is critical that we correctly implement the distributive property of multiplication. In these lessons, we will learn how to multiply polynomials. Multiplying polynomials using distribution is an important mathematical skill that is used often in and after algebra. Be sure to change the subtraction to addition before combining like terms. This 3x has to get distributed onto each of these 3 terms and then this positive 2 gets distributed onto all 3 of those terms. Nov 25, 20 this video branches off of the how to foil video by explaining how to multiply polynomials with more terms. The simplest type of polynomial multiplication involves a monomial multiplier. Multiplying binomials by polynomials old video khan academy. R q2j0 u192l gk xu ltga9 1saoyf atjw va urueg blcl7ce. Multiply binomials as with multiplying a monomial by a binomial, you can use algebra tiles to multiply two binomials. Unit 5 polynomials study guide term an expression made up of variables and constants. Here are the steps required for multiplying polynomials. When youre multiplying polynomials you could use the area model or the table where youre drawing rectangles and finding the area that way or you can just be really careful using the distributing property, just make sure if you choose to use distributing, that youre being specially careful with the minus signs the negative signs. Multiplying polynomials using distributing there are different methods for multiplying polynomials.
Instruct students to model each expression with the tiles. Virtual nerds patentpending tutorial system provides incontext information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. All three methods will be used to solve the same two multipli cation problems. Draw arrows to indicate that all terms have been distributed.
165 463 177 1190 1314 1543 703 119 561 578 1256 263 1412 438 49 584 1395 1361 785 1261 685 1066 7 862 1323 113 700 571 857 633 397 818
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# How Measurement Mishaps Can Increase Roof Replacement Cost
Measuring for a new roof can be complex, and miscalculations can lead to a higher roof replacement cost. If you want to roof your house by yourself, you will need to understand all the intricacies of measuring your roof. Math calculations must be done to account for special features of the roof after the measurements are done. A professional home remodeling company can make short work of measuring accurately for a roof replacement. Here is what you need to know if you are doing it yourself.
1. Basic Square Footage
You will need a measurement of the basic square footage of the outline of your roof. Remember that this includes every part of the house that is covered by a roof. Include every ell. To do this, you have to measure the length and the width of each rectangle that makes up the base of your house. Figure the area of each rectangle by multiplying the length times the width. Then, add all these figures together.
2. Overhang
The basic area is not good enough for the calculation of overall roof coverage area. You will also need to add on an extra measure for the overhang of the roof that goes beyond the edge of the house. If you do not include this figure in your reckoning, you will be short on materials when you go to do the roof. It is best to include this roof replacement cost in the beginning rather than pay more for it later.
3. Pitch
The basic square footage and overhang are measurements of the roof if it were level. Assuming you are not replacing a flat roof, you will need to consider the slope or pitch of the roof in your calculations for roof replacement cost.
Pitch is described as the number of feet the roof rises for every 12 feet of distance. For example, a low pitch might be 3:12, a medium pitch might be 6:12, and a steep pitch might be higher than 9:12. These categories are important in determining the number of shingles needed.
4. Conversion
A special number is used to convert the flat, level measurements into 3 dimensional figures using the pitch of the roof as a guide. This number is called an approximate average roofing multiplier. For a low pitched roof, it could be from 1.15 to 1.25. For a medium pitched roof, the multiplier might be between 1.25 and 1.4. A steep roof would have a multiplier of anywhere from 1.41 to 1.7.
The basic measurement of the roof is then multiplied by the approximate average roofing multiplier for the given roof, and you arrive at a new, more accurate figure. If you use the wrong multiplier or do not do the calculations correctly, you can throw off your roof replacement cost budget.
5. Special Features
The measurements of a roof can be altered by special features such as dormers, end-walls, valleys, and chimneys. These must all be accounted for and there is no one neat, simple formula for figuring them in.
If you want to get all your shingles at the same time to get a price break, it is important to figure correctly. If you want to make sure you get enough of the same kind of shingles, you need the proper number. Finally, if you do not want to overbuy, you need an accurate estimate of your roofing material needs. A home remodeling company can help you measure and calculate to reduce the roof replacement cost.
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2009 Final
# 2009 Final - MATH 203 Final Examination Student Name...
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MATH 203 Final Examination April 27, 2009 Student Name: Student Number: McGill University Faculty of Science FINAL EXAMINATION MATH 203 Principles of Statistics I April 27th, 2009 9 a.m. - 12 Noon Answer directly on the test (use front and back if necessary). Calculators are allowed. One 8.5” × 11” two-sided sheet of notes is allowed. Language dictionaries are allowed. There are 17 pages to this exam and 2 pages of tables. The total number of marks for the exam is 100. Examiner: Professor Russell Steele Associate Examiner: Professor David Stephens 1
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MATH 203 Final Examination April 27, 2009 Question 1: (10 points) The Canadian government has decided that there was some suspicious, potentially illegal activity regarding a company’s revenue stream. As part of their investigation, they wanted to establish that there was a significant increase in the mean transaction amount before and after a particular event. The government statistician took a random sample of transactions with 50 customers who were billed both before and after the date of interest (i.e. each customer was billed twice). The data are summarized in the table below: Mean Std Dev 25%ile Median 75%ile Sample size Before event 98.82 20.49 83.99 101.2 109.3 50 After event 108.3 17.94 96.25 107.8 120.1 50 Difference [After - Before] 9.47 19.81 5.188 9.01 24.72 50 Test for a significant increase in the mean transaction amount before and after the event at α = 0 . 05. 2
MATH 203 Final Examination April 27, 2009 Question 2: (10 points) Professor Steele gave out a surprise quiz during one of his classes. The quiz had 3 questions. Each question was worth 2 points , however no partial credit was given. Assume that the probability distribution (i.e. probability mass function) for the number of correctly answered questions by a randomly selected student is: Number of correctly answered questions 0 1 2 3 Probability 0.05 0.20 0.40 0.35 (a) What is the distribution (i.e. the probability mass function) for the total score on the quiz for a randomly selected student? [2 points] (b) What is the expectation and variance of the total quiz score for a randomly selected student? [2 points] 3
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MATH 203 Final Examination April 27, 2009 (c) What is the distribution (i.e. the probability mass function) for the average (or mean ) of two randomly selected student quiz scores? [3 points] (d) What is the approximate distribution for the average (or mean ) of 100 randomly selected student quiz scores? [3 points] 4
MATH 203 Final Examination April 27, 2009 Question 3: (6 points) In a 1995 paper in Astrophysics and Space Science, Higgins and Henrikson discussed the distribution of gamma-ray bursts in halo neutron star-comet models. They based most of their conclusions on a simulation model where the velocities of comets ejected from globular cluster stellar systems were assumed to be Normally distributed with mean velocity (in km/s) equal to 200 and standard deviation 110.
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A mixture of sand and cement contains, 3 parts of sand and 5 : GMAT Problem Solving (PS)
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# A mixture of sand and cement contains, 3 parts of sand and 5
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A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
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01 Jun 2014, 14:05
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ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
Source: Indian CAT
We have total of 8 parts: 3 parts of sand and 5 parts of cement.
In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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02 Jun 2014, 12:32
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ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
This is just a weighted average question, so we can apply the formula for that: with $$C$$ as the concentrations and $$V$$ as the volumes...
$$C_1*\frac{V_1}{V_1 + V_2} + C_2*\frac{V_2}{V_1 + V_2} = C_{final}$$
If sand:cement=3:5, then the concentration of sand in the initial mixture is $$\frac{3}{8}$$. Since we are asked for the proportion of the mixture that should be replaced, we can assume a total volume of $$1$$ and let $$x$$ be the "amount" of the mixture to be replaced by pure sand (i.e., concentration of 1). We can then write the following equation:
$$(\frac{3}{8})*(1-x) + (1)*(x)=\frac{1}{2}$$
Thus, $$x=\frac{1}{5}$$
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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02 Jun 2014, 20:15
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ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
Source: Indian CAT
You can use the scale method here too.
A mix has 3/8 of sand. Another is all sand so fraction of sand is 1. You have to mix them to get 1/2 sand.
w1/w2 = (1 - 1/2)/(1/2 - 3/8) = 4/1
So the mix should be 4 parts and only sand should be 1 part. Hence 1/5 of the mix must have been replaced by sand.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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04 Jun 2014, 01:08
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Initial
Sand ............... Cement ............... Total
$$\frac{3}{8}$$ ..................... $$\frac{5}{8}$$ ...................... 1
Requirement
$$\frac{4}{8} ..................... \frac{4}{8} ..................... 1$$
Just focus on cement:
To have $$\frac{4}{8}$$ cement, we require to remove
$$\frac{5}{8} - \frac{4}{8} = \frac{1}{8}$$cement
In Mixture, the portion of cement is $$\frac{5}{8};$$so to remove $$\frac{1}{8}$$ cement, mixture required to be removed
$$= \frac{\frac{1}{8} * 1}{\frac{5}{8}}$$
$$= \frac{1}{5}$$
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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25 Jun 2014, 15:42
Bunuel wrote:
ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
Source: Indian CAT
We have total of 8 parts: 3 parts of sand and 5 parts of cement.
In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.
Bunuel: I do not quite understand the highlighted section above. 1 part of the mixture will contain 3/8 sand and 5/8 cement. How did you come up with "1 part of cement contains 3/5 parts of sand"?
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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25 Jun 2014, 20:01
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One more method:
Sand ............. Cement .............. Total
$$\frac{3}{8}$$ .................. $$\frac{5}{8}$$ ........................ 1
Let "x" quantity of mixture be removed;
$$\frac{3}{8} - \frac{3x}{8}$$ ........... $$\frac{5}{8} - \frac{5x}{8}$$ ............... 1 - x
$$\frac{3}{8} - \frac{3x}{8} + x$$ .......... $$\frac{5}{8} - \frac{5x}{8}$$ .............. 1-x+x
Resultant should be half sand & half cement
Two options to set up the equation
Option I
$$\frac{3}{8} - \frac{3x}{8} + x = \frac{1}{2}$$
$$x = \frac{1}{5}$$
Option II
$$\frac{5}{8} - \frac{5x}{8} = \frac{1}{2}$$
$$x = \frac{1}{5}$$
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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25 Jun 2014, 20:09
1
KUDOS
Game wrote:
Bunuel wrote:
ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
Source: Indian CAT
We have total of 8 parts: 3 parts of sand and 5 parts of cement.
In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.
Bunuel: I do not quite understand the highlighted section above. 1 part of the mixture will contain 3/8 sand and 5/8 cement. How did you come up with "1 part of cement contains 3/5 parts of sand"?
Not Bunuel, but seems I can explain it
Mixture .................. Sand ............... Cement
1 ............................ $$\frac{3}{8}$$ ................... $$\frac{5}{8}$$
Multiply by $$\frac{8}{5}$$ to all above
$$\frac{8}{5}$$ ........................ $$\frac{3}{8} * \frac{8}{5}$$ ............. $$\frac{5}{8} * \frac{8}{5}$$
$$\frac{8}{5}$$ ........................ $$\frac{3}{5}$$ ....................... 1
From the above, we can say that 1 part of cement comes with $$\frac{3}{5}$$ parts of sand in $$\frac{8}{5}$$ quantity of mixture.
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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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25 Jun 2014, 20:19
S1 - 3 parts element 1 , 5 parts element 2
S2 - x parts of mixture removed i.e, -(3/8 * x) parts element 1, -(5/8 * x) parts element 2
S3 - x parts of element 1 added
We thus have 3 - (3*x)/8 +x / 5 - (5*x)/8 = 1/1
x=1.6 . i.e, 1.6/8 = 1/5 of the mixture removed.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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18 Aug 2015, 22:18
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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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18 Aug 2015, 22:53
ConnectTheDots wrote:
A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/8
Source: Indian CAT
It is easier to answer this type of question using allegation rule.
3/8 parts are sand in the original mixture.
we are adding with only sand to make the new mixture 1:1
therefore, 1/1 part is sand that we add.
3/8 1/1
1/2
1/2 : 1/8
so, the ratio is 1/2 : 1/8 or 4:1
That is, if you remove 4 parts from the original mixture and add 1 part sand, the resultant mixture is 1:1
To elaborate further, if the original mixture is 8 kg (3 kg sand and 5 kg cement), you remove 4 kg of the mixture which contains 1.5 kg sand and 2.5 kg of cement.
Now, add 1 kg of sand. The new mixture becomes 2.5 kg of sand and 2.5 kg of cement.
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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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06 Sep 2015, 14:53
let x=fraction of mixture to be substituted
3/8-3x/8+x=1/2
x=1/5
Last edited by gracie on 11 Nov 2015, 19:39, edited 1 time in total.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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07 Sep 2015, 12:36
How can we use the alligation method/shortcut to solve this?
My attempt:
Sand...................Cement
3/8......................5/8
.......\.................../
.........\............../
..............4/8
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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13 Nov 2016, 14:21
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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24 Dec 2016, 12:40
Lets say we have 3 Kg of Sand and 5 Kg of Cement in the total 8 Kg mixture. We want to make it 4 Kg Sand and 4 Kg Cement.
To do so, we need to remove 1 Kg of Cement.
Each 800 grams of mixture will have 500 grams Cement and 300 grams of Sand. So we need to remove 1.6 Kg ( i.e. 1/5th) of mixture to remove 1 Kg of cement.
In the process we have also removed 600 grams of Sand (i.e. remaining sand is 2.4 Kg)
Now we replace the mixture with 1.6 Kg of Sand making it 4 Kg Cement and 4 Kg Sand.
Hence the answer is replace 1/5th of the mixture with Sand.
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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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26 Dec 2016, 04:14
miva0601 wrote:
How can we use the alligation method/shortcut to solve this?
My attempt:
Sand...................Cement
3/8......................5/8
.......\.................../
.........\............../
..............4/8
Old Mix...................Sand-only mix
3/8......................1/1
.......\.................../
.........\............../
..............4/8
1/2......................1/8
Therefore the resulting mix will have a ratio Old Mix to Sand-only mix of (1/2)/(1/8) = 4 to 1.
From here we know that every 4 part of the old mix we need to have one part of sand only mix. So in the final mix the old mix will be 4/5 and the sand-only mix will be 1/5. Answer is C we need to substitute 1/5 of the old mix and replace with san.
Anyone could suggest if I have applied the method correctly? (Bunuel )
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
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15 Jan 2017, 18:47
Think of "x" as the original amount of mixture & "y" as the amount we are replacing of the original mixture
(3/8)x - (3/8)y + y = (1/2)x
5y = x --> therefore, y=(1/5)x
C.
Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink] 15 Jan 2017, 18:47
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## Quick Answer: What Does Postulate Mean In Geometry?
A statement, also known as an axiom, which is taken to be true without proof. Postulates are the basic structure from which lemmas and theorems are derived. The whole of Euclidean geometry, for example, is based on five postulates known as Euclid’s postulates.
## What is a postulate in geometry examples?
A postulate is a statement that is accepted without proof. Axiom is another name for a postulate. For example, if you know that Pam is five feet tall and all her siblings are taller than her, you would believe her if she said that all of her siblings are at least five foot one.
## What is meant by postulate in mathematics?
Postulate. A postulate is an assumption, that is, a proposition or statement, that is assumed to be true without any proof. Once a theorem has been proven it is may be used in the proof of other theorems. In this way, an entire branch of mathematics can be built up from a few postulates.
## What is a postulate simple definition?
1: demand, claim 2 a: to assume or claim as true, existent, or necessary b: to assume as an axiom or as a hypothesis advanced as an essential presupposition, condition, or premise of a train of reasoning (as in logic or mathematics)
You might be interested: FAQ: What Is Theorem 10-A In Geometry?
## What are the 5 postulates in geometry?
Euclid’s Postulates
• A straight line segment can be drawn joining any two points.
• Any straight line segment can be extended indefinitely in a straight line.
• Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
• All right angles are congruent.
## What is a postulate in geometry with triangles?
Postulate. If two angles of one triangle are equal in measure. to two angles of another triangle, then the two. triangles are similar. Side-side-side (SSS) Similarity Theorem If the three sides of one triangle are proportional to the three corresponding sides of another triangle, then the triangles are similar.
## What is a postulate and theorem in geometry?
In geometry, a postulate is a statement that is assumed to be true based on basic geometric principles. An example of a postulate is the statement “through any two points is exactly one line”. A theorem is a mathematical statement that can and must be proven to be true.
## What is a corollary in math?
Mathematics. a proposition that is incidentally proved in proving another proposition. an immediate consequence or easily drawn conclusion.
## What best defines a postulate?
A postulate is a statement accepted to be true without proof so the correct answer is choice.
## What is postulates in science?
A postulate (also sometimes called an axiom) is a statement that is agreed by everyone to be correct. This is useful for creating proofs in mathematics and science, (also seen in social science)Along with definitions, postulates are often the basic truth of a much larger theory or law.
You might be interested: Quick Answer: 4) Why Is It Important To Study Geometry?
## What does Hypothisize mean?
: to suggest (an idea or theory): to make or suggest (a hypothesis) See the full definition for hypothesize in the English Language Learners Dictionary.
## What does postulate 3 mean?
Postulate 3: Through any two points, there is exactly one line.
## What are the 5 famous postulates?
Geometry/Five Postulates of Euclidean Geometry
• A straight line segment may be drawn from any given point to any other.
• A straight line may be extended to any finite length.
• A circle may be described with any given point as its center and any distance as its radius.
• All right angles are congruent.
## What is fifth postulate?
Euclid’s fifth postulate: If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
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# Explain Hash Table in simple terms
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Hi Experts,
Can you give me a step by step explanation of how a hash table works and how to implement it with a single linked list.
Please be as specific and explain with simple terms.
Thanks,
Dennis
p.s. In addition, if you want to put a link to supplement your explanation that would be ok. But please explain first.
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Graphics Expert
CERTIFIED EXPERT
Commented:
Before explain hash table, lets explain hash.
Hash is a number or a small set of numbers, calculated from a large set of numbers.
Supose you have a phonebook, composed by names and phone numbers and for each name you have a hash number. For example, if we add the code of the characters of each name, probably we have different results for all the names.
A=1, B=2, ..., Z=26
JOHN --> 10 + 15 + 8 + 14 = 47
ANNE --> 1 + 14 + 14 + 5 = 34
There are special algorithms to calculate hash numbers such that they are unique values or with few chances of equal hash numbers for different names.
Now suppose we have a phonebook where we have hash numbers as keys, something like:
01 0
02 0
:
:
33 0
34 2337=3431
35 0
:
:
47 2344-7839
In a common phonebook we have the names in alphabetic order and respectives numbers associated to them. If we want a phone number, the searching algorithm will scan the list in a character by character basis, until finding the name, then returning the phone number associated to such name.
In our hash oriented phonebook the algorithm gets the name, say JOHN, calculates its hash number, 34 and returns the phone number at register 34. No searching.
Easy to understand that the second algorithm is faster than the first one. There is a price to that: a lot of cells will be empty, and the database size is larger that the one of the traditional phonebook. Hash tables waste more memory and disk space, because the table is sparse. Optimized hashing can reduce the database size, but it will be larger than the traditional one.
Probably you will ask if different names can have the same hash number. Yes, it happens, but there are techniques to solve the collisions.
Now suppose we have much more data than a single phone number, say, for each name we have a street addres, security number, credit card number, e-mail, father's name, etc. We can't waste disk space with so large empty records, so we use a trick: we have a hash table as below:
01 0
02 0
:
:
33 0
34 1
35 0
:
:
47 2
such that if we want JONH's data, the algorithm returns "2", not the data. Then we use such number, 2, as a key number in another table, as below:
1 JONH 2344-7839 4532 Rodeo Street VISA 12345678
2 MARY 2337=3431 1234 Durango Rd MC 98765432
3 etc.
and find the data we are looking for.
That way, the hash table, which wastes space with empty registers, is small because stores just small key numbers. The real data is in a dense table, so no wasting room in the large data records.
By using hash tables, we find data very quickly and with the described trick, by wasting very few disk room.
Jose
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Commented:
Thanks Jose,
Explanation was very concise.
I am going to pose another question about the same topic, can you please look for it?
Thanks,
Dennis
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# Monotonic Function
(redirected from Monotone operator)
## monotonic function
[¦män·ə¦tän·ik ′fəŋk·shən]
(mathematics)
## Monotonic Function
(or monotone function), a function whose increments Δf(x) = f(x′) − f(x) do not change sign when Δx = x′ − x > 0; that is, the increments are either always nonnegative or always nonpositive. Somewhat inaccurately, a monotonic function can be defined as a function that always varies in the same direction. Different types of monotonic functions are represented in Figure 1. For example, the function y = x3 is an increasing function. If a function f(x) has a derivative f′(x) that is nonnegative at every point and that vanishes only at a finite number of individual points, then f(x) is an increasing function. Similarly, if f′(x) ≤ 0 and vanishes only at a finite number of points, then f(x) is a decreasing function.
Figure 1
A monotonicity condition can hold either for all x or for x on a given interval. In the latter case, the function is said to be monotonic on this interval. For example, the function y = increases on the interval [−1,0] and decreases on the interval [0, +1]. A monotonic function is one of the simplest classes of functions and is continually encountered in mathematical analysis and the theory of functions. If f(x) is a monotonic function, then the following limits exist for any X0:
and
References in periodicals archive ?
Convex analysis and monotone operator theory in Hilbert Spaces.
It is well-known that a monotone operator T is maximal monotone if and only if R(T + [lambda]J) = [X.sup.*] for every [lambda] > 0 (cf.
(A) If H(q) is a diagonally monotone operator family in H with compact resolvent, then all eigenvalues [E.sup.H(q).sub.i] of H(q) are also diagonally monotone.
Kumam, "A new hybrid iterative method for solution of equilibrium problems and fixed point problems for an inverse strongly monotone operator and a nonexpansive mapping," Journal of Applied Mathematics and Computing, vol.
where A is a monotone operator and the set of zero point of A is denoted by [A.sup.-1] (0).
This problem has been investigated by use of a variety of nonlinear analyses such as fixed point theorem for mixed monotone operator [7, 15, 22-25], maximal principle [6], Banach's contraction mapping principle [26-29], and the linear operator theory [27, 30, 31].
It is a well known result that if [phi] is proper, convex and lower semicontinuous, then [partial derivative][phi] is a maximal monotone operator. We refer the reader to the books by Barbu [4], Brezis [5] and Morosanu [17] for further details on the properties of monotone operators and subdifferentials of convex functions in Hilbert spaces.
for all v, w [member of] X and that a monotone operator on a reflexive Banach space is hemicontinuous if and only if it is demicontinuous (see e.g.
By the assumption L < 1/[K.sup.2]m([OMEGA]), it turns out that [PHI]' is a strongly monotone operator. So, by applying Minty-Browder theorem [29, Theorem 26.A], [PHI]': X [right arrow] [X.sup.*] admits a Lipschitz continuous inverse.
where B is k-Lipschitzian and [eta]-strongly monotone operator. Then he proved that if the sequence {[[alpha].sub.n]} satisfies appropriate conditions, the sequence {[x.sub.n]} generated by (6) converges strongly to the unique solution [x.sup.*] [member of] [F.sub.ix] (T) of the variational inequality
Let A : D(A) [subset] H [right arrow] H be a (possibly multivalued) maximal monotone operator. Consider the difference equation
To study solvability of the nonlinear problem (2.1), we shall use the variational approach and monotone operator theory (see [4,19-22]).
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### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
### A Mixed-up Clock
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There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
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I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
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48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Multiplication Squares
##### Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Making Pathways
##### Age 7 to 11 Challenge Level:
Can you find different ways of creating paths using these paving slabs?
### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Two Primes Make One Square
##### Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Mystery Matrix
##### Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Give Me Four Clues
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Fractions in a Box
##### Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
### What Two ...?
##### Age 7 to 11 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Divide it Out
##### Age 7 to 11 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Path to the Stars
##### Age 7 to 11 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### In the Money
##### Age 7 to 11 Challenge Level:
There are a number of coins on a table. One quarter of the coins show heads. If I turn over 2 coins, then one third show heads. How many coins are there altogether?
### What's in the Box?
##### Age 7 to 11 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### Which Is Quicker?
##### Age 7 to 11 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Factor-multiple Chains
##### Age 7 to 11 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### What Do You Need?
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Flashing Lights
##### Age 7 to 11 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
### Tom's Number
##### Age 7 to 11 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Fitted
##### Age 7 to 11 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Odds and Threes
##### Age 7 to 11 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### What Is Ziffle?
##### Age 7 to 11 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### A Square Deal
##### Age 7 to 11 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
### Becky's Number Plumber
##### Age 7 to 11 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Three Dice
##### Age 7 to 11 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Ip Dip
##### Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
### Crossings
##### Age 7 to 11 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Number Tracks
##### Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Money Measure
##### Age 7 to 11 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Multiplication Series: Number Arrays
##### Age 5 to 11
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Sets of Four Numbers
##### Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### A Dotty Problem
##### Age 7 to 11 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
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Select subject
Equations
# Equations
A mathematical statement that indicates that the value of the LHS is equal to the value of the RHS is called an equation.
Lesson Demo
• A mathematical statement that indicates that the value of the LHS is equal to the value of the RHS is called an equation.
• An equation puts a condition on the variable.
• The value for which the equation is satisfied is the solution of the equation.
• The value of the variable in an equation that satisfies the equation, or makes its LHS equal to its RHS, is the solution.
• An equation can contain numbers and variables.
• An equation is said to be algebraic equation if it consists of a variable.
• A single variable equation will have a unique solution.
• An equation that does not have any variable is called a numerical or an arithmetic equation.
• Different numerical values for the variable are substituted in an algebraic equation, and the solution is obtained by using a method called the trial and error method.
If there is no sign of equality between the LHS and the RHS, then it is not an equation.
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Fringed Loomwork: the Mathematical Approach
By Jane Tyson
(Jane is an Australian. She has not misspelled things and is not wrong using "maths" where Americans would use "math.")
They say you learn something new every day. Well, perhaps it's old, but new to me/you anyway. I must have been away from school on the day they taught arithmetic progression.1 Lucky for me, my daughter wasn't!
I was struggling with trying to guesstimate how much thread I'd need for the weft of a fringed choker, given that I wanted to make it in one piece.
Adding in thread is a pet peeve of mine, so I much prefer to brave knots and tangles than add to an existing piece of thread. Rachel (my daughter) said, "that's easy," and whipped out her maths exercise book with its formulae for Arithmetic Progression.
["Adding in thread": you prefer to brave knots and tangles? Wouldn't you get knots if you had to add thread? Do you mean you would prefer tangles to knots?]
It almost worked like a charm. It certainly was much closer than I would have guessed by using a measuring tape to measure the thread I needed. I felt that "near enough" would be "good enough," but I was about a 30cm (a foot) and several curses short. I think I have identified the error and corrected it.
The choker is made up of two 'arms' and a V-shaped fringe.
If you work out the amount of weft thread needed for one "arm" and half the fringe (see below), all you need to do is double the total, then double the total again to account for the weft thread on the return pass.
O.K. here goes:
1. The "arm" of the choker is a rectangle, so all you need to do is measure how wide it is (in this case, 4/10 of an inches or 1 cm). Multiply that by the number of rows of beads (50), giving you 20 inches or 50 cm. This number is quadrupled, as I just explained. Now we have 80 inches (6' 8") or 2 metres.
2. That is pretty straightforward. It is the fringe that is tricky. Here we use arithmetic progression. The formula for it is: s = n ¸ 2(2a + (n-1)d)
Where S = the sum of the length of the fringes. (This is the answer we want).
n = the number of fringes (38 in this case)
a = the length of the shortest fringe (1 inch or 2.5 cm)
d = the difference in length between the first and second fringe (1/8 inch or 3 mm).
Using metrics this would be:
S = 38 ¸ 2 (2 x 25mm + [38 ¸ 1] x 3)
S = 19 x (50 +111)
S = 19 x 161
S = 3.059 m
Or in the English system:
S = 38 ¸ 2 (2 x 1 + [38 ¸ 1] X 1/8)
S = 19 x (2 + 4 5/8)
S = 19 x 6 5/8
S = 125 7/8 inches
S = 3 yards, 1 1/2 feet
Now we multiply this answer by 2 because we only measured half the fringe.
S = 6.118 m or 7 yards.
Add the "arms" we calculated above:
+ 2 m or 80 inches (6' 8")
S = 8.118 m or 9 yards, 8 inches
This gives you an exact measurement of the distance travelled, but it still needs some refining. You will need to add some thread for the start thread, the end tail and 'turns' between rows. You should allow 3mm (1/8 inch) for each turn = 174 x 3mm = 0.522mm (1' 9 3/4") and 10cm (4") each for the start thread and end tail.
The grand total comes to 8.84 metres or 9 yards, 2 feet and 5 3/4 inches).*
After all of this, you probably think that it is easier to make a rough guess and be done with it!
* In reality, this is slightly long because the longest, center fringe should not be doubled, but it is better to have too much thread than too little and the difference is rather small.
P.S. I (pfjr) did the conversions between English and metric. Guess which one was much easier to work with.
__________________________________________________
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Friday, June 24, 2011
Class X, PHYSICS, "Scalar and Vectors"
SCALAR
"Scalar quantity are those physical quantity which are completely specified by their magnitude express with suitable unit. They do not require any mention of the direction for complete their specification is called scalar quantity."
OR
" Scalar quantity are those physical quantity which require magnitude , express with suitable unit only is called scalar quantity."
CHARACTERISTICS OF SCALAR QUANTITY
1, Scalar quantity can be added,subtracted,multiplied,divided according to the ordinary algebraic rule.
2, Two scalars are equal if they have same unit.
REPRESENTATION
It can be represented by the numbers with decimals. (positive negative)
EXAMPLE
Mass,Distance,Temperature,volume,speed e.t.c
VECTOR
"VECTOR quantity are those physical quantity which do not require only their magnitude express with suitable unit. But they also require a particular direction for complete their specificaton is called vector quantity."
OR
" vector quantity are those physical quantity which require magnitude , express with suitable unit as well as proper direction is called vector quantity."
CHARACTERISTICS OF VECTOR QUANTITY
1, vector quantity can not be added,subtracted,multiplied, divided according to the ordinary algebraic rule.
2, It can be added,subtracted,multiplied,divided according to the some special rules like head and tail rule,Graphical method e.t.c.
3, vector always treats as positive.
REPRESENTATION
It can be represented by an arrow with headline. The length of an arrow represents its magnitude and the headline represents the direction of the vector(figure 1.1)
------------------------------------->
(figure 1.1)
EXAMPLE
Weight,Displacement,Velocity,Acceleraton,Torque,Momentum e.t.c
"The process of combining of two or more vector to produce a signal vector having the combinig effect of all the vector is called the resultant of the vector and this process is known as the addition of a vector".
Suppose we have two vector A and B having the different magnitude and direction.
1, First of all chose a suitable scale and representation of all the vector have been drawn on the paper.
2, Put all the vector for finding the resultant of given vector such that the head of the first vector join the tail of the second vector.
3, Now join the tail of the first vector with tail of the second vector such that it join the two vector with head to head and tail to tail by another.
4, The new vector R will be the resultant of the given vector.
5, It can be measured by the Dee or any suitable mean.This method is called the head and tail or tip to tail rule.
RESOLUTION OF A VECTOR
"The process of splitting up of a signal vector into two or more vector is called the resolution of a vector"
OR
"The process of splitting up of a signal vector into its components is called the resolution of a vector"
RECTANGULAR COMPONENTS
A vector which is not along x-axis or y-axis it can be resolved into infinite number, but generally a vector can be resolved into its components at a right angle to each other
MATHEMATICALLY PROVED
suppose a vector F is denoted by a line AB which makes an angle @ with horizontal surface OX. From a point A draw perpendicular to the horizontal surface OX.
The line AB represents its vertical component and it is denoted by Fy.The line OB represents its horizontal component and it is denoted by Fx. Now in the triangle AOB
Sin@= AB/OA {sin@= Perpendicular/Hypotenuse}
or sin@= Vy/V
or Vy= V sin@
Similarly
Cos@= OB/OA {sin@= Base/Hypotenuse}
or Cos@= Fx/F
or Vx= V Cos@
For the triangle
Tan@= AB/OB {Tan@= per/hyp)
or Tan@= Vy/Vx
or @=Tan-1 =Vy/Vx
SUBTRACTION OF A VECTOR
"It is defined as the Addition of A to the negative of a B is called the subtraction of a vector (A-B)"
1 comment:
1. thnks bro good notes and good work
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$\newcommand{\Cat}{{\rm Cat}}$ $\newcommand{\A}{\mathcal A}$ $\newcommand{\freestar}{ \framebox[7pt]{\star} }$
## 3. Generating functions
1. Suppose $R(x,y,z)=\frac{P(x,y,z)}{Q(x,y,z)}$ is a rational generating function, and $C(z)$ is a non-analytic generating function (for example $C(z)=\sum_n n! z^n$). Let $F(x,y)={R(x,y,z) \odot_z C(z)}{|_{z=1}}$, where $\odot$ denotes the Hadamard product.
#### Problem 3.1.
[Jay Pantone] Can ACSV do anything with this?
• ### Embedding algebraic generating functions
Suppose $F(x)$ is an algebraic generating function. It is known that it can be written as the diagonal of a rational generating function $R(x,y)$.
#### Problem 3.2.
[Mark Wilson] If $F(x)$ is combinatorial (that is $F(x)=\sum_n a_n x^n, a_n \in \mathbb Q_{\geq 0}$), does there exist $R(x,y)$ that is also combinatorial?
For example if $F(x)=\frac{x}{\sqrt{1-x}}$, then we can take $R(x,y)=\frac{2xy}{2-x-y}$, but none of the general methods find this. A subprogram of ACSV is to reduce the case of algebraic generating functions to that of rational generating functions, and combinatorial ones are much nicer.
• #### Problem 3.3.
[Mark Wilson] If the answer to Problem 3.2 is yes, then does there exist an efficient way to find these?
• ### $\mathbb N$-D-finiteness
Suppose $F(x)=\sum_n a_n x^n, a_n \in \mathbb Z_{\geq 0}$. $F$ is said to be $\mathbb N$-algebraic if \begin{align*} F_0&=P_o(x,F_0,F_1,\dots,F_n)\\ F_1&=P_1(x,F_0,F_1,\dots,F_n)\\ &\vdots\\ F_k&=P_k(x,F_0,F_1,\dots,F_n) \end{align*} where the coefficient of $P_0,P_1,\dots,P_k$ are nonnegative integers.
#### Problem 3.4.
[Igor Pak] What is the $D$-finite analog of that? Replace $P_i$’s with ODE’s. What would be the asymptotics of this?
• #### Problem 3.5.
[Igor Pak] It is known that the diagonal of a rational generating function is still $D$-finite. We want to define $\mathbb N$-D-finite. Is it the diagonal of a rational generating function $R(z)=\sum_{\omega} b_\omega z^\omega, b_\omega \in \mathbb Z_{\geq 0}$?
• Suppose we have $F_{2n}(t,x,y)$ defined recursively as follows: \begin{align*} F_2(t,x,y)&=t^2 x+ty\\ q_{2n}(x,y)&=\frac 1 2 (F_{2n}(t,x,y)+F_{2n}(t,x,-u))\\ r_{2n}(x,y)&=[t^{2n}]q_{2n}(x,y)\\ F_{2n+2}(t,x,y)&=t^{2n} r_{2n}(x(t^2 x+ty+1),tx+y)-F_{2n}(t,x,y). \end{align*}
#### Problem 3.6.
[Sheila Sundaram] Conjecturally, the coefficients of $q_{2n}(x,y)$ are nonnegative. Can ACSV help with this?
1. Remark. [Sheila Sundaram] Thank you to Stefan Trandafir for verifying the conjecture up to degree 200!
The reference for this problem is
Sundaram, Sheila The homology of partitions with an even number of blocks. J. Algebraic Combin. 4 (1995), no. 1, 69–92.
The polynomials in t,x,y are generated according to Algorithm 2.11. They originate in a plethystic recurrence involving the homogeneous symmetric functions.
The coefficient of t^{2n} in q_{2n} is r_{2n}(x,y), for whose coefficients I can give an explicit recurrence, see Conjecture 3.1. Those particular coefficients were later proved to be nonnegative in Benjamin Joseph’s MIT thesis.
But the main conjecture, that ALL the coefficients in q_{2n}(t,x,y) are nonnegative, is still open.
Cite this as: AimPL: Analytic combinatorics in several variables, available at http://aimpl.org/combinseveral.
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# Difference between revisions of "2011 AIME II Problems/Problem 4"
## Problem 4
In triangle $ABC$, $AB=20$ and $AC=11$. The angle bisector of $\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of the intersection of $AC$ and $BM$. The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solutions
### Solution 1
$[asy] pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); [/asy]$ Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$. It follows that $\triangle BPC \sim \triangle DD'C$, so $$\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}$$ by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that $AP = PD'$. Thus, $$\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},$$ and $m+n = \boxed{51}$.
### Solution 2 (mass points)
Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$, so we assign $m(B) = 11, m(C) = 20, m(D) = 31$. Since $AM = MD$, then $m(A) = 31$, and $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$, so $m+n = \boxed{51}$.
### Solution 3
By Menelaus' Theorem on $\triangle ACD$ with transversal $PB$, $$1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.$$ So $m+n = \boxed{051}$.
### Solution 4
We will use barycentric coordinates. Let $A = (1, 0, 0)$, $B = (0, 1, 0)$, $C = (0, 0, 1)$. By the Angle Bisector Theorem, $D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)$. Since $M$ is the midpoint of $AD$, $M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)$. Therefore, the equation for line BM is $20x = 31z$. Let $P = (x, 0, 1-x)$. Using the equation for $BM$, we get $$20x = 31(1-x)$$ $$x = \frac{31}{51}$$ Therefore, $\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}$ so the answer is $\boxed{051}$.
### Solution 5
Let $DC=x$. Then by the Angle Bisector Theorem, $BD=\frac{20}{11}x$. By the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.$ Notice that $[\triangle BAM]=[\triangle BMD]$ since their bases have the same length and they share a height. By the sin area formula, we have that $$\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.$$ Simplifying, we get that $\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.$ Plugging this into what we got from the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}$
### Solution 6 (quick Menelaus)
First, we will find $\frac{MP}{BP}$. By Menelaus on $\triangle BDM$ and the line $AC$, we have $$\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.$$ This implies that $\frac{MB}{BP}=1-\frac{MP}{BP}=\frac{51}{62}$. Then, by Menelaus on $\triangle AMP$ and line $BC$, we have $$\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.$$ Therefore, $\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.$ The answer is $\boxed{051}$. -brainiacmaniac31
2011 AIME II (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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## Electrostatics – One mark and two marks questions for practice
A collection of one mark and two marks questions from ELECTROSTATICS. Practise these for scoring a better and winning the edge over your coscholars!
## Electrostatics : A Practice Test Paper – 1
Time: 45 min
Max. Marks: 25
1. How does the force between two point charges change, if the dielectric constant of the medium in which they are kept, increases? [1]
2. Name the
## Numericals from Kinematics
PHYSICS TEST (CLASS XI) Kinematics
1. A particle is projected at 60º to the horizontal with a kinetic energy K. What is the kinetic energy at the highest point
2. Which of the following statements is false for a particle moving in a circle with a constant angular speed?
(a) the velocity vector is tangent to the circle
(b) the acceleration vector is tangent to the circle
(c) the acceleration vector points to the centre of the circle
(d) the velocity and acceleration vectors are perpendicular to each other
3. A car is moving in a circular horizontal track of radius 10.0 m with a constant speed of 10.0m/s.A plumb bomb is suspended from the roof of the car by a light rigid rod of length1.00m. What is the angle made by the rod with the track is (g= 10 m/s²)
4. The distance traveled by an object along the axes are given by x=2t², y = t² – 4, z = 3 t – 5. What is the initial velocity of the particle
5. A boy playing o the roof of 10m high building throws a ball with a speed of 10 m/s at an angle of 30º the horizontal. How far from the throwing point will the ball be at the height of 10m from the ground?
6. If a body travels half of its path in the last second of its fall from rest; find the time and height of its fall?
7. An object A is kept fixed at the point x = 3m and y=1.25m on a moving along the +x direction with an acceleration 1.5 m/s. At the same instant a stone hitting the object during its downward motion at an angle pf 45º to the horizontal .all the motions are in X-Y plane .Find u and time after which stone hits the object. Take g = 10m/s²
8. Three particles a, b and c are situated at the vertices of an equilateral triangle ABC of side d at t=0. Each of the particles moves with constant speed v. A always has its velocity along BC and C along CA. At what time will the particles meet each other?
9. A man standing on road has to hold his umbrella at 30º with the vertical to keep the rain away he throws the umbrella and starts running at 10 km/h. he finds that raindrops with respect to [a]the road, [b] the moving man.
10. A man can swim at speed of 3 km/h in still water. He wants to cross a 500 m wide river flowing at 2 km/h. he keeps himself always at an angle of 120º with the
## A Random Test from Magnetism
Each question carries one mark
• What is a magnetic dipole?
• What is the difference between electric field lines and magnetic field lines?
• Write an expression for the magnetic
## ELECTROSTATICS TEST PAPER FOR CLASS XII
1. What orientation of an electric dipole in a uniform electric field corresponds to its stable equilibrium?
2. What is the area of the plates of a parallel plate capacitor of capacitance
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# GMAT Book 3: Profits
#### GMAT Syllabus for Profit, Loss, Discounts
You could get one or two questions from Profit and Loss in the GMAT quant section - in both variants viz., problem solving and data sufficiency.
## Syllabus Covered in Wizako's GMAT Book | Profits
Profit and Loss is one of the easy concepts in GMAT Math. Basic concepts, formulae are covered and are followed by examples and exercise problems. Wizako's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The Chapter begins with the explanation of the meaning of cost price and selling price.
2. The meaning of the terms profit and loss is explained and the relationship between profit, loss, cost price and selling price is given.
3. Expressing profit and loss as a percentage of cost price.
4. Converting percentages of profit or loss expressed in terms of selling price to percentages of cost price.
5. Explanation of the meaning of the terms marked price (retail price) and discount and the relationship between selling price, marked price, and discount.
6. 4 Illustrative examples to grasp the concept with ease.
7. 25 solved examples with detailed explanations, a few highlighting applicable shortcuts.
8. 37 exercise problems with answer key and explanatory answers.
9. An objective type test with 35 GMAT level multiple choice questions in the work book. Answer key and explanatory answers are provided for all questions.
Here is a typical solved example in Wizako's GMAT Book from this chapter
### Sample Question
The cost price of 40 articles is equal to the selling price of 35 articles. What is the profit /loss percentage made by the trader?
Let the cost price of one article be $1. So, the cost price of 40 articles is$40
So, the selling price of 35 articles = $40 Let us compute the cost price for 35 articles. If the cost of 1 article =$1, the cost price of 35 articles = $35 Therefore, the profit/loss made on 35 articles = S.P of 35 articles - C.P of 35 articles i.e.,$40 - $35=$5, or a profit of \$5
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Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up? New to KöMaL?
# Problem C. 1329. (January 2016)
C. 1329. King Sanislaus sends an envoy to a distant empire. In three days, a courier is sent back to the King's court with a message. It takes the courier two days to cover the same distance as the envoy takes three days to cover. With the answer, the courier returns to the envoy, and catches up with them exactly as they arrive at the destination. However, the emperor, who is a great friend of king Sanislaus, has been exiled in the meantime, so the entire envoy returns immediately, and this time they all travel at the speed of the courier. How many days elapse between the envoy starting their journey and returning to the court?
(5 pont)
Deadline expired on February 10, 2016.
### Statistics:
148 students sent a solution. 5 points: 120 students. 4 points: 2 students. 3 points: 1 student. 2 points: 5 students. 1 point: 6 students. 0 point: 14 students.
Problems in Mathematics of KöMaL, January 2016
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# Determine the inverse for f(x)= 2(x-4)^2+5
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
A function f(x) and its inverse f^-1(x) are related by: f(f^-1(x)) = x
For f(x) = 2*(x - 4)^2 + 5, the value of f(x) is the same for any x = a and x = -a. As a result it is not possible to determine the inverse function.
A function y = f(x) is a relation where for any value of x, the value of y is unique. The inverse of the given function would be one where each value of x gives 2 possible values of y, this is not permitted for any function.
The function f(x) = 2*(x - 4)^2 + 5 does not have an inverse.
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search
# You owe \$200,000 on a mortgage loan You wish to repay the loan with
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You owe \$200,000 on a mortgage loan. You wish to repay
the loan with 10 equal payments, one at the end of each
year for the next 10 years, and a separate final \$80,000
balloon payment at the end of 11 years. What is the
appropriate amount of each of the 10 equal annual
payments? The interest rate is 8% compounded annually.
Instrument BA II Plus calculator.
Solution
This problem can be solved in two steps using the TI BA II
Plus.
1) We need to find the PV of the \$80,000 balloon payment
made at the end of the year 11 with interest rate 8%.
FV = \$80,000
N = 11
I/Y = 8
Solving for PV we get = \$34,310.63
We will subtract the PV of the balloon payment from our
mortgage loan of \$200,000 to get:
\$200,000 - \$34,310.63 = \$165,689.37 This is the amount
we need to pay in 10 equal end of year instalments.
2) Solving for the annual payments:
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You owe \$200,000 on a mortgage loan. You wish to repay the loan with 10 equal payments, one at the end of each year for the next 10 years, and a separate final \$80,000 balloon payment at the end of 11 years. What is the appropriate amount of each of the 10 equal annual payments? The interest rate is 8% compounded annually. If possible, please explain how to find answer using Texas Instrument BA II Plus calculator. Solution This problem can be solved in two steps using the TI BA II Plus. 1) We need to find the PV of the \$80,000 balloon payment made at the end of the year 11 with interest rate ...
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# How to solve formulas in algebra
The solver will provide step-by-step instructions on How to solve formulas in algebra. So let's get started!
## How can we solve formulas in algebra
One of the most important skills that students need to learn is How to solve formulas in algebra. If you're struggling to solve a trig identity, there's a tool that can help. A trig identity solver is a online tool that can simplify an equation containing trigonometric functions. It can be very helpful to plug in an equation and have the tool generate the steps to solve it.
There are a few different ways to solve rate of change problems. The most common way is to find the equation of the line that represents the rate of change, and then use that equation to find the desired value. Another way is to use a graph to find the rate of change. This can be done by finding the slope of the line on the graph, or by using the average rate of change formula.
The known variables are usually called y 1 , y 2 , ..., y n . A system of two linear equations can always be solved by arranging the equations so that the unknowns are on one side and the knowns are on the other side. Therefore, a system of two linear equations has six possible arrangements: If there are three or more unknowns, then it may be necessary to use more than one arrangement. For example, if there are five unknowns, they could be arranged in two parallel rows such as (0, 0), (1, 1), (2, 3), (3, 5), and (4, 6). Alternatively, they could be arranged in a column such as (0, 0), (1, 1), (2, 3), (3, 4), (4, 5), and (5, 6). To solve a system of equations you must solve each equation for its corresponding unknown variable. Once you have solved all of the equations to determine all of the unknown variables you can use these values to solve for any remaining variables.
This can be simplified to x=log32/log8. By using the Powers Rule, you can quickly and easily solve for exponents. However, it is important to note that this rule only works if the base of the exponent is 10. If the base is not 10, you will need to use a different method to solve for the exponent. Nevertheless, the Powers Rule is a useful tool that can save you time and effort when solving for exponents.
The best log problem solver can be described as the person who can take care of log problems in the most efficient way possible. This person should always have a good understanding of how to deal with log problems and be able to apply various techniques and strategies when needed. They should also be able to interpret the information from logs, which are often presented in a complex manner. This ability is vital for anyone who wants to become a successful log problem solver. The best log problem solver will know how to find the root cause of any given log problem, which is crucial in order to get rid of it as soon as possible. It is also important for them to be able to resolve any given log problem quickly and efficiently. If you are interested in learning more about this topic, read this article about why you should become a better log problem solver:
## We cover all types of math problems
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Xandria King
The app is greater. It shows the answers very quickly. The only thing lacking is the draw/write function. Sometimes it doesn't get the problem unless you write it clearly again if the draw/write function was there it would have been much easier. Thank you, developers, for creating such a useful app for understanding math easily.
Irene Campbell
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## Binomial Theorem Quiz 3
Binomial Theorem is one of the most important chapters of Algebra in the JEE syllabus and other engineering exams. For JEE Mains, it has 4% weightage and for JEE Advanced, it has 2.42% weightage..
Q1.
•
•
•
•
Solution
Q2. The coefficient of x28 in the expansion of (1 + x3 − x6)30 is
• 1
• 0
•
•
Solution
Q3.
• 255
• 61
• 127
• None of these
Solution
Q4.
•
•
•
•
Solution
Q5.
• x
• (1 + x)1/3
• (1 - x)1/3
• (1 - x)-1/3
Solution
Q6.
• 210
• 105
• 70
• 112
Solution
Q7.
• 2n + 1
• 2n
• n
• n + 1
Solution
Q8.
•
•
•
•
Solution
Q9. If C0,C1,C2, ...,Cn are the binomial coefficients, then 2 X C1 + 23 X C3 + 25 X C5 + ... equals
•
•
•
•
Solution
Q9. If the coefficients of three consecutive terms in the expansion of (1 + x)n are in the ratio 1:7:42, then the value of n is
• 60
• 70
• 55
• None of these
Solution
#### Written by: AUTHORNAME
AUTHORDESCRIPTION
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Binomial Theorem-Quiz-3
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# Question
You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.
Define a pair (u,v) which consists of one element from the first array and one element from the second array.
Find the k pairs (u1,v1),(u2,v2) …(uk,vk) with the smallest sums.
Example 1:
Given nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Return: [1,2],[1,4],[1,6]
The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
Example 2:
Given nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Return: [1,1],[1,1]
The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
Example 3:
Given nums1 = [1,2], nums2 = [3], k = 3
Return: [1,3],[2,3]
All possible pairs are returned from the sequence:
[1,3],[2,3]
# Solution
We can use a heap to store k candidate pairs. For every numbers in nums1, its best partner (yields min sum) always starts from nums2[0] since arrays are all sorted; And for a specific number in nums1, its next candidate should be [this specific number] + nums2[current_partner_index + 1], unless out of boundary for nums2.
class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
k = Math.min(nums1.length * nums2.length, k);
List<int[]> res = new ArrayList<>(k);
PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> (a[0] + a[1]) - (b[0] + b[1]));
int l = 0, r = 0;
for(int i = 0; i < Math.min(k, nums1.length); i++)
q.offer(new int[]{nums1[i], nums2[0], 0});
for(int i = 0; i < k; i++) {
int[] arr = q.poll();
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Question about mg on oscilations course
Tags:
1. Oct 3, 2015
Remixex
On a test our teacher asked about a system composed of (string -> mass -> string -> mass) hanging, that began to oscillate up and down.
We all considered weight (mg) when applying Newton's second law to find the associated differential equation.
When we met our teacher again he said that we shouldn't have, the weight has no dynamic effect on the system, because the weight force is not variable over time (as opposed to a simple or double pendulum, there it depends of the angle).
Should i NOT include weight on the sum of forces if said weight doesn't depend on the angle?
This was supposed to be a normal mode exercise but since it was hanging, we all encountered a forced oscillation (forced by mg)
2. Oct 3, 2015
nasu
What makes the system to oscillate? Are these strings elastic?
3. Oct 3, 2015
Remixex
Damn, i actually meant spring, yes they are elastic, of course, didn't mean string :D
4. Oct 4, 2015
Chandra Prayaga
If you first do the simpler problem of one spring + mass hanging from a ceiling, you will find that the weight only changes the equilibrium position of the mass. You can stretch or compress the spring from that equilibrium position, and it will oscillate with the same frequency depending on the spring constant and the mass. The differential equation for oscillations about equilibrium is still the same. This is what your teacher meant when he said the weight has no dynamic effect. The same thing happens if you have a two spring - two mass system
5. Oct 4, 2015
sophiecentaur
Why would including the weight force in your equation of motion affect the period of the oscillation? But would there also not be a mean force from the spring, balancing the weight?
6. Oct 4, 2015
Chandra Prayaga
At this stage, instead of wondering why something would and something else would not happen, we should actually write down the equation of motion, and see how the different features happen. This is done, in every introductory book on calculus based physics, for the single spring + mass program. Once you work that out, you will find that extending it to the two-spring + two-mass problem is not difficult. To answer your last questions in anticipation of your working it out, the weight indeed does not effect the period of oscillation, and yes, there would be a stretch in the spring to balance the weight. That stretch would be the equilibrium position, and oscillations occur about that equilibrium position.
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# An Elementary Geometry and Trigonometry
Thompson, Bigelow, and Brown, 1872 - Geometry - 238 pages
### Contents
GEOMETRY 1 RATIO AND PROPORTION 25 RELATIONS OF POLYGONS 31 BOOK III 49 BOOK IV 64 BOOK VI 89
LOGARITHMS 1 CHAPTER II 11 CHAPTER III 17 CHAPTER IV 29 CHAPTER V 41 CHAPTER VI 53
### Popular passages
Page 30 - Hence -,- = -76" dn that is a" : b" = c" : dn THEOREM IX. 23 1 If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a : b = c : d...
Page 49 - A circle is a plane figure bounded by a curved line, called the circumference, every point of which is equally distant from a point within called the center.
Page 96 - Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.
Page 12 - In an isosceles triangle the angles opposite the equal sides are equal.
Page 11 - If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.
Page 23 - If two triangles have two sides of one respectively equal to two sides of the other, and the angles contained by those sides supplementary, the triangles are equal in area.
Page 20 - ... polygon, is equal to twice as many right angles as the polygon has sides minus two.
Page 71 - The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude.
Page 40 - At a point 200 feet from, and on a level with the base of a tower, the angle of elevation of the top of the tower is observed to be 60° : what is the height of the tower?
Page 23 - In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference.
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# Counting Sort
Till now, all the sorting algorithms we have learned were comparison sort i.e., they compare the values of the elements to sort them but the counting sort is a non-comparison sort. It means that it sorts the input without comparing the values of the elements.
Comparison sorts have a lower bound of $\Omega(n\lg{n})$ i.e., they can't perform better than this. And this is where non-comparison sorts get interesting because they can beat this lower bound and can perform faster like in linear time ($\Theta(n)$). This is because non-comparison sorts are generally implemented with few restrictions like counting sort has a restriction on its input which we are going to study further.
So, the restriction of the counting sort is that the input should only contain integers and they should lie in the range of 0 to k, for some integer k. For example, if the value of k is 10, then all the inputs must be between 0 to 10.
Also, we don't want the value of k to be too high because that will increase the running time of the counting sort. And we will see that the counting sort takes $\Theta(n)$ time when k is $O(n)$.
So, let's see the working of the counting sort.
## Idea Behind Counting Sort
Suppose, there are 10 boxes and 10 stones. Now, you have to keep all these 10 stones in 10 boxes but in a sorted order i.e., the smallest stone will be in the first box and the largest stone will be in the 10th. Of course, there are many different ways to do this but let's talk about one particular way which is related to the counting sort.
Suppose, you picked up a stone and you know that there are exactly 5 stones which are smaller than this stone, then without giving a single thought, you would place it in the 6th box. This is the exact idea behind the counting sort.
Thus in the counting sort, we need an extra array to store the output like the boxes in the previous example. This array is going to store all the numbers which are in the input, so the size of this array should be n.
As the name suggests, we start by counting the number of times a number is in the input array. For example, if 2 is appearing 4 times in the input array then we count this number and store it in a different temporary array. And the way we are going to store this number in this temporary array is by changing the value of that element of the array whose index is equal to the number itself to a value number of times it is appearing. For example, if 2 is appearing 4 times, then the element with index 2 of the temporary array will have a value of 4. This is also illustrated in the picture given below.
So, if the range of the number in the input is large i.e., if the value of k is large, then the size of this temporary array will also be large. For example, if there are numbers from 0 to 10,000 in the input array, then we are going to need a temporary array at least of size 10,001 (range is starting from 0) to store how many times 10,000 is appearing in the input array. So, the size of this input array is $O(k)$.
This is also one of the reasons why we want the value of k to be small because a large value of k would require a temporary array of large size.
With the help of this temporary array, we know how many elements are smaller than a particular element and then we can use the logic of the stones and the boxes to fill up the numbers in sorted order to the new output array.
We start by making the temporary array.
Then, we fill the values of the temporary array with the help of the input array i.e., by counting the number of times an element is appearing in the input array.
Lastly, we fill our output array using the temporary array.
Now, let's see the proper working of the counting sort and how to properly implement it.
## Working of Counting Sort
So, we start with a temporary of size k+1 and then we iterate over the input array and store the number of times an element is appearing in the input array to the temporary array.
As stated above, we need the count of the numbers smaller than a particular number to do the counting sort, so now we modify the temporary array to store the number of elements which are smaller than a particular element. This is shown in the picture given below.
Now, we know the number of elements smaller than a particular element, thus the last task is to fill up the output array.
Thus, we require arrays of sizes n (output array) and k (temporary array) for the process of counting sort. So, the counting sort has a space complexity of $O(n+k)$. From here, you can also see that reducing the value of k is beneficial to us.
After understanding the logic and the working of counting sort, let's write a code for it.
## Code for Counting Sort
Our first task is to make a temporary array of size k+1 and make all its elements 0 i.e.,
temp_array[k+1] → Initializing a temporary array
for i in 0 to k → Iterating over the array
temp_array[i] = 0 → Making each element 0
Now, our next task is to store the number of occurrence of an element to the temporary array. So if an element is 2, then we have to store its count of occurrence to temp_array[2]
Thus, for any element A[i], we are going to store its occurrence to temp_array[A[i]]. So every time we find an element A[i], we will increase the count of temp_array[A[i]] by 1. For example, every time we find 2, we will increase temp_array[2] by 1.
for i in 1 to A.length
temp[A[i]] = temp[A[i]] + 1
Now, the next task is to modify the elements of the temporary array and store the number of elements smaller than or equal to a particular element. This is simple, we just need to add all the elements of the temporary array before that particular element.
So, we can start iterating over the temporary array from the index 1 and store the sum as shown in the above picture.
for i in 1 to k
temp_array[i] = temp_array[i]+ temp_array[i-1]
Now, we are left with the final task of making the output array. For this, we need to iterate over the input array and for each element of the input array, we need to find the correct position where it should be in the output array.
So, let's say we are on an element A[i] of the input array, then temp_array[A[i]] stores the number of elements less than or equal to the number A[i].
Now in the output array, A[i] will go to the index temp_array[A[i]] i.e., output_array[temp[A[i]]] = A[i].
Since A[i] is in the correct place in the output array, we need to decrease the count of it in the temporary array i.e., temp_array[A[i]] = temp_array[A[i]] - 1, so the places don't coincide in case of the duplicate numbers.
You can look at the pictures given below for the pictorial representaion of the counting sort.
Stability is also a property of an algorithm. Stability is when the numbers with same values appear in the same sequence in both input and the output. For example, if there are 4 twos in the input then the 2 which appeared first in the input will also appear first in the output and the 2 which appeared second in the input will also appear second in the output and so on.
Counting sort is a stable algorithm
As stated above, we are going to iterate to each element of the input array to find its correct position in the output array. Now, we have the option to make this iteration forward or backward. But take a case of iterating backward on the input array to fill up the output array. Here, among the common elements, the element which is at the last in the input array will be filled at last in the output array. This is shown in the picture given below. In this case, the counting sort will be a stable algorithm but this won't be possible if we iterate forward.
for i in A.length downto 1
output_array[temp[A[i]]] = A[i]
temp_array[A[i]] = temp_array[A[i]] - 1
So, let's sum up the above points and write the entire code for the counting sort.
COUNTING-SORT(A, output_array, k)
temp_array[k+1]
for i in 0 to k
temp_array[i] = 0
for i in 1 to A.length
temp[A[i]] = temp[A[i]] + 1
for i in 1 to k
temp_array[i] = temp_array[i]+ temp_array[i-1]
for i in A.length downto 1
output_array[temp[A[i]]] = A[i]
temp_array[A[i]] = temp_array[A[i]] - 1
• C
• Python
• Java
#include <stdio.h>
void counting_sort(int a[], int output_array[], int k, int size) {
int temp_array[k+1];
int i;
for(i=1; i<=k; i++) {
temp_array[i] = 0;
}
for(i=1; i<=size; i++) {
temp_array[a[i]] = temp_array[a[i]] + 1;
}
for(i=1; i<=k; i++) {
temp_array[i] = temp_array[i]+ temp_array[i-1];
}
for(i=size; i>=1; i--) {
output_array[temp_array[a[i]]] = a[i];
temp_array[a[i]] = temp_array[a[i]] - 1;
}
}
int main() {
//array is starting from 1. So, fake element -100 at index 0.
int a[] = {-100, 4, 8, 1, 3, 10, 9, 2, 7, 5, 6};
int output_array[11];
counting_sort(a, output_array, 10, 10); //k is 10. size for function is 10, array from 1 to 10.
//printing array
int i;
for(i=1; i<=10; i++) {
printf("%d ",output_array[i]);
}
printf("\n");
return 0;
}
def counting_sort(a, output_array, k, size):
temp_array = [0]*(k+1)
for i in range(1, k+1):
temp_array[i] = 0
for i in range(1, size+1):
temp_array[a[i]] = temp_array[a[i]] + 1
for i in range(1, k+1):
temp_array[i] = temp_array[i]+ temp_array[i-1];
for i in range(size, 0, -1):
output_array[temp_array[a[i]]] = a[i]
temp_array[a[i]] = temp_array[a[i]] - 1
if __name__ == '__main__':
#list is starting from 1. So, fake element -100 at index 0.
a = [-100, 4, 8, 1, 3, 10, 9, 2, 7, 5, 6]
output_array = [0]*11
counting_sort(a, output_array, 10, 10) #k is 10. size for function is 10, list from 1 to 10.
for i in range(1, 11):
print(output_array[i])
class Count {
public static void countingSort(int a[], int outputArray[], int k, int size) {
int[] tempArray = new int[k+1];
for(int i=1; i<=k; i++) {
tempArray[i] = 0;
}
for(int i=1; i<=size; i++) {
tempArray[a[i]] = tempArray[a[i]] + 1;
}
for(int i=1; i<=k; i++) {
tempArray[i] = tempArray[i]+ tempArray[i-1];
}
for(int i=size; i>=1; i--) {
outputArray[tempArray[a[i]]] = a[i];
tempArray[a[i]] = tempArray[a[i]] - 1;
}
}
public static void main(String[] args) {
//array is starting from 1. So, fake element -100 at index 0.
int a[] = {-100, 4, 8, 1, 3, 10, 9, 2, 7, 5, 6};
int[] outputArray = new int[11];
countingSort(a, outputArray, 10, 10); //k is 10. size for function is 10, array from 1 to 10.
//printing array
for(int i=1; i<=10; i++) {
System.out.print(outputArray[i] + " ");
}
System.out.println("");
}
}
## Analysis of Counting Sort
The analysis of the counting sort is simple. For the first for loop i.e., to initialize the temporary array, we are iterating from 0 to k, so its running time is $\Theta(k)$. The next loop is running from 1 to A.length and thus has a running time of $\Theta(n)$. The next for loop is again iterating over the temporary array from 1 to k and thus has a running time of $\Theta(k)$. The last loop is again $\Theta(n)$.
So, the overall running time of the counting sort is $\Theta(k+n)$.
As stated earlier in this chapter, when the value of k will be around the value of n i.e., $k=O(n)$, then the running time of the counting sort will reduce to $O(n)$ and thus the counting sort will run in linear time. And that's why we were saying that we want the value of k to be of $O(n)$.
Software is a gas; it expands to fill its container.
- Nathan Myhrvold
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# challenge question -- Factor the polynomial completely
#### lookagain
##### Elite Member
Edit:
Demonstrate at least two methods for factoring the following polynomial
completely over the integers.
$$\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$
Last edited:
#### MarkFL
##### Super Moderator
Staff member
Method 1:
Factor first by grouping:
$$\displaystyle (x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)$$
Now, for the quartic factor assume it may be factored as follows:
$$\displaystyle x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$$
Equating coefficients, we find:
$$\displaystyle a+b=0$$
$$\displaystyle ab+2=1$$
and so one solution is $$\displaystyle (a,b)=(1,-1)$$ and we have:
$$\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ which means:
$$\displaystyle x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)$$
Method 2:
Let:
$$\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:
$$\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:
$$\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:
$$\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$
#### Subhotosh Khan
##### Super Moderator
Staff member
A) Give the completely factored form over the integers of the following polynomial, and
B) demonstrate at least two methods for doing so.
$$\displaystyle x^5 + x^4 + x^3 + x^2 + x + 1$$
(x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)
or
(x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)
#### lookagain
##### Elite Member
Here is another:
(x^5 + 1) + (x^4 + x^3 + x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =
(x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =
(x + 1)(x^4 + x^2 + 1) =
(You can continue simplifying this with one of the above steps in
any of the appropriate above posts.)
Last edited:
#### daon2
##### Senior Member
Here's another way.
Since $$\displaystyle x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}$$ the roots of this polynomial are exactly the set $$\displaystyle \{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}$$, i.e. the roots of unity, ignoring the positive real root. They are $$\displaystyle e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1$$.
We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:
$$\displaystyle (x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1$$
$$\displaystyle (x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1$$
$$\displaystyle x-(-1) = x+1$$
There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.
#### soroban
##### Elite Member
Hello, lookagain
This is a variation of daon's solution.
Demonstrate at least two methods for factoring the following polynomial:
. . . $$\displaystyle P(x) \;=\;x^5 + x^4 + x^3 + x^2 + x + 1$$
$$\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$$
. . . . .$$\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$$
. . . . .$$\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$$
. . . . .$$\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)$$
#### lookagain
##### Elite Member
Hello, lookagain
This is a variation of daon's solution.
$$\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$$
. . . . .$$\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$$
. . . . .$$\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$$
. . . . .$$\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)$$
MarkFL said:
Method 2:
Let:
$$\displaystyle S=x^5+x^4+x^3+x^2+x+1$$ and so:
$$\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$$ hence:
$$\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$$ thus:
$$\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)$$
These two (MarkFL's and soroban's versions) look essentially the same to me.
- - - - - - - - - - - - - - - - -
Others:
(x^5 + x^2) + (x^4 + x) + (x^3 + 1) =
x^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =
(x^3 + 1)(x^2 + x + 1) =
(x + 1)(x^2 - x + 1)(x^2 + x + 1)
. . . . . . . . . . . . . . . . . . . . .
(x^5 + x^3 + x) + (x^4 + x^2 + 1) =
x(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =
(x^4 + x^2 + 1)(x + 1) =
(x^2 - x + 1)(x^2 + x + 1)(x + 1)
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## Section7.7Power Series
###### Motivating Questions
• What is a power series?
• What are some important uses of power series?
• What is the interval of convergence? For what values does a power series converge?
So far, each infinite series we have discussed has been a series of real numbers, such as
$$1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^k} + \cdots = \sum_{k=0}^{\infty} \frac{1}{2^k}\text{.}\label{E-geom12}\tag{7.18}$$
In the remainder of this chapter, we will include series that involve a variable. For instance, if in the geometric series in Equation(7.18) we replace the ratio $r = \frac{1}{2}$ with the variable $x\text{,}$ we have the infinite (still geometric) series
$$1 + x + x^2 + \cdots + x^k + \cdots = \sum_{k=0}^{\infty} x^k\text{.}\label{E-geomx}\tag{7.19}$$
In this section we will focus on more general series of powers. As a motivation, consider the following example.
###### Example7.46
Consider the power series defined by
\begin{equation*} f(x) = \sum_{k=0}^{\infty} \frac{x^k}{2^k}\text{.} \end{equation*}
What are $f(1)$ and $f\left(\frac{3}{2}\right)\text{?}$ Find a general formula for $f(x)$ and determine the values for which this power series converges.
Solution
If we evaluate $f$ at $x=1$ we obtain the series
\begin{equation*} \sum_{k=0}^{\infty} \frac{1}{2^k} \end{equation*}
which is a geometric series with ratio $\frac{1}{2}\text{.}$ So we can sum this series and find that
\begin{equation*} f(1) = \frac{1}{1-\frac{1}{2}} = 2\text{.} \end{equation*}
Similarly,
\begin{equation*} f(3/2) = \sum_{k=0}^{\infty} \left(\frac{3}{4}\right)^k = \frac{1}{1-\frac{3}{4}} = 4\text{.} \end{equation*}
In general, $f(x)$ is a geometric series with ratio $\frac{x}{2}\text{,}$ so
\begin{equation*} f(x) = \sum_{k=0}^{\infty} \left(\frac{x}{2}\right)^k = \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x} \end{equation*}
provided that $-1 \lt \frac{x}{2} \lt 1$ (which ensures that the ratio is less than 1 in absolute value). Thus, the power series that defines $f(x)=\frac{2}{2-x}$ converges for $-2 \lt x \lt 2\text{.}$
As Example7.46 illustrates, a power series may converge for some values of $x$ and not for others. In this section, we will learn how to determine the interval of $x$-values where a power series converges. In the following sections we will show how power series may be used to obtain polynomial approximations of functions.
### SubsectionPower Series
###### Power Series
A power series centered at $x = a$ is a function of the form
$$\sum_{k=0}^{\infty} c_k(x-a)^k\label{eq-8-6-power-series}\tag{7.20}$$
where $\{c_k\}$ is a sequence of real numbers and $x$ is an independent variable.
For example, the series
\begin{equation*} 1 + (x-2) + (x-2)^2 + (x-2)^3 + \cdots \end{equation*}
is a power series centered at $x = 2$ with $c_i = 1$ for $i \ge 1 \text{,}$ and a geometric series
\begin{equation*} \sum_{n=1}^{\infty} bx^n = b + bx + bx^2 + bx^3 + \cdots = \sum_{n=0}^{\infty} b(x-0)^n \end{equation*}
is a power series centered at $x=0$ with $c_i = b$ for $i \ge 1 \text{.}$
Convergence of power series is similar to convergence of series. Namely, a power series will converge if its sequence of partial sums converges. In general, a power series may converge for some values of $x \text{,}$ and diverge for others.
###### Convergence of Power Series
For a fixed value of $x \text{,}$ the power series
\begin{equation*} \sum_{k=0}^{\infty} c_k(x-a)^k \end{equation*}
converges to $L$ if the sequence of partial sums $S_0(x), S_1(x), S_2(x), \ldots$ converges to $L \text{.}$ That is,
\begin{equation*} \lim_{n \rightarrow \infty} S_n(x) = L \text{.} \end{equation*}
The set of $x$ values at which a power series $\sum_{k=0}^{\infty} c_k(x-a)^k$ converges is always an interval centered at $x=a\text{.}$ For this reason, the set on which a power series converges is called the interval of convergence. Half the length of the interval of convergence is called the radius of convergence.
###### Example7.47
Consider the series $\sum_{n=0}^{\infty} \frac{ (x-1)^n}{2^n} \text{.}$ This is a power series about $x = 1 \text{.}$
1. Does the series converge or diverge when $x =2 \text{?}$
2. Does it converge or diverge when $x = 3 \text{?}$
3. For what values of $x$ will the series converge?
1. The series converges at $x=2\text{.}$
2. It diverges at $x=3\text{.}$
3. The series converges for values of $x$ satisfying $-1 \lt x \lt 3 \text{.}$
Solution
1. To test, we can substitute in $x=2$ into the series, giving:
\begin{equation*} \sum_{n=0}^\infty \frac{(2-1)^n}{2^n} = \sum_{n=0}^\infty \frac{1}{2^n}. \end{equation*}
This series is geometric with ratio $\frac{1}{2}\text{,}$ so it converges (in fact, to 2).
2. We make the substitution $x=3$ into the series, and get
\begin{equation*} \sum_{n=0}^\infty \frac{(3-1)^n}{2^n} = \sum_{n=0}^\infty \frac{2^n}{2^n} = \sum_{n=0}^\infty 1 = 1 + 1 + 1 + \cdots\text{.} \end{equation*}
Since this series is unbounded, it diverges.
3. To figure out exactly when the series converges, we'll start by using the Ratio Test. Here $a_n= \frac{(x-1)^n}{2^n}\text{.}$ So, notice
\begin{equation*} \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-1|^{n+1}}{2^{n+1}} \cdot \frac{2^n}{|x-1|^n} = \frac{|x-1|}{2}. \end{equation*}
Thus, the series converges whenever $\frac{|x-1|}{2}\lt 1$ and diverges whenever $\frac{|x-1|}{2} \gt 1 \text{.}$ Examining that first inequality yields:
\begin{align*} \frac{|x-1|}{2} \amp \lt 1 \\ |x-1| \amp \lt 2 \\ -2 \amp\lt x-1 \lt 2 \\ -1 \amp\lt x \lt 3 \end{align*}
So now we know that the series converges when $-1 \lt x \lt 3$ and diverges when $x \lt -1$ and when $x \gt 3 \text{.}$ Unfortunately, the ratio test can't tell us what happens exactly at $x=-1$ or at $x=3\text{,}$ because that's when $\left|\frac{a_{n+1}}{a_n}\right|=1\text{.}$ However, we can just check those directly. We determined in part b. that the series diverges at $x=3\text{,}$ so we just need to check $x=-1\text{:}$
\begin{equation*} \sum_{n=0}^\infty \frac{(-1-1)^n}{2^n}= \sum_{n=0}^\infty \frac{(-2)^n}{2^n} = \sum_{n=0}^\infty -1 = -1 -1 -1 \cdots\text{.} \end{equation*}
This series is unbounded below, so it diverges.
Thus, the series converges exactly when $-1 \lt x \lt 3$ and diverges everywhere else.
### SubsectionFinding the Interval of Convergence
A power series defines a function $f$ whose domain is the set of $x$ values for which the power series converges. We therefore write
\begin{equation*} f(x) = \sum_{k=0}^{\infty} c_k(x-a)^k\text{.} \end{equation*}
To determine the values of $x$ for which a power series
\begin{equation*} \sum_{k=0}^{\infty} c_k (x-a)^k\text{,} \end{equation*}
centered at $x = a$ will converge, we apply the Ratio Test with $a_k = | c_k (x-a)^k |\text{.}$ The series converges if $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} \lt 1\text{.}$
Observe that
\begin{equation*} \frac{a_{k+1}}{{a_k}} = | x-a | \frac{| c_{k+1} |}{| c_{k} |}\text{,} \end{equation*}
so when we apply the Ratio Test, we get
\begin{equation*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} |x-a| \frac{| c_{k+1} |}{| c_{k} |}\text{.} \end{equation*}
Note suppose that
\begin{equation*} \lim_{k \to \infty} \frac{| c_{k+1} |}{| c_{k} |} = L\text{,} \end{equation*}
so that
\begin{equation*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = |x-a| \cdot L\text{.} \end{equation*}
There are three possibilities for $L\text{:}$ $L$ can be $0\text{,}$ it can be a finite positive value, or it can be infinite. Based on this value of $L\text{,}$ we can determine for which values of $x$ the original power series converges.
• If $L = 0\text{,}$ then the power series converges on $(-\infty, \infty)\text{.}$
• If $L$ is infinite, then the power series converges only at $x = a\text{.}$
• If $L$ is finite and nonzero, then the power series converges absolutely for all $x$ that satisfy
\begin{equation*} |x-a| \cdot L \lt 1 \end{equation*}
or for all $x$ such that
\begin{equation*} |x-a| \lt \frac{1}{L}\text{,} \end{equation*}
which is the interval
\begin{equation*} \left(a-\frac{1}{L}, a+\frac{1}{L}\right)\text{.} \end{equation*}
This interval is centered at $a$ and has radius $R = \frac{1}{L} \text{.}$ Because the Ratio Test is inconclusive when the $|x-a| \cdot L = 1\text{,}$ the endpoints $a \pm \frac{1}{L}$ have to be checked separately.
###### Finding the Interval of Convergence
To find the interval of convergence of the power series $\sum_{k=0}^{\infty} c_k(x-a)^k \text{,}$ apply the ratio test to obtain
\begin{equation*} \lim_{k \rightarrow \infty} |x - a|\frac{|c_{k+1}|}{|c_k|} = |x - a| \cdot L \end{equation*}
There are three cases.
• If $L = 0 \text{,}$ then the power series converges for all values of $x \text{.}$ Thus, the the interval of convergence is $(-\infty, +\infty)$ and the radius of convergence is $R = \infty \text{.}$
• If $L = \infty \text{,}$ then the power series converges only at $x = a \text{,}$ and the radius of convergence is $R = 0 \text{.}$
• If $L$ is finite and nonzero, then the power series converges for all $x \in (a - \frac{1}{L}, a + \frac{1}{L}) \text{,}$ so the radius of convergence is $R = \frac{1}{L} \text{.}$ The power series may or may not converge at each endpoint, so testing for convergence at $x = a - R$ and $x = a + R$ is necessary.
###### Example7.48
Let $f(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}\text{.}$ Determine the interval of convergence of this power series.
Solution
First we will plot some of the partial sums of this power series to get an idea of the interval of convergence. Let
\begin{equation*} S_n(x) = \sum_{k=1}^{n} \frac{x^k}{k^2} \end{equation*}
for each $n \geq 1\text{.}$ Figure7.49 shows plots of $S_{10}(x)$ (in red), $S_{25}(x)$ (in blue), and $S_{50}(x)$ (in green).
The behavior of $S_{50}$ in particular suggests that $f(x)$ appears to be converging to a particular curve on the interval $(-1,1)\text{,}$ while growing without bound outside of that interval. Thus, the interval of convergence might be $-1 \lt x \lt 1\text{.}$ To verify our conjecture, we apply the Ratio Test. Now,
\begin{equation*} a_k = \frac{x^k}{k^2}\text{,} \end{equation*}
so
\begin{align*} \lim_{k \to \infty} \frac{\left| a_{k+1} \right|}{ \left| a_k \right|} \amp = \lim_{k \to \infty} \frac{ \frac{|x|^{k+1}}{(k+1)^2} }{ \frac{| x|^{k}}{k^2} }\\ \amp = \lim_{k \to \infty} |x| \left(\frac{k}{k+1}\right)^2\\ \amp = |x| \lim_{k \to \infty} \left(\frac{k}{k+1}\right)^2\\ \amp = |x|\text{.} \end{align*}
Therefore, the Ratio Test tells us that $f(x)$ converges absolutely when $| x | \lt 1$ and diverges when $| x | \gt 1\text{.}$ Because the Ratio Test is inconclusive when $|x| = 1\text{,}$ we need to check $x = 1$ and $x = -1$ individually.
When $x = 1\text{,}$ observe that
\begin{equation*} f(1) = \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} \end{equation*}
This is a $p$-series with $p \gt 1\text{,}$ which we know converges. When $x = -1\text{,}$ we have
\begin{equation*} f(-1) = \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}\text{.} \end{equation*}
This is an alternating series, and since the sequence $\left\{ \frac{1}{n^2} \right\}$ decreases to 0, the power series converges by the Alternating Series Test. Thus, the interval of convergence of this power series is $-1 \le x \le 1\text{.}$
###### Example7.50
Determine the interval of convergence of each power series.
1. $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$
2. $\sum_{k=1}^{\infty} kx^k$
3. $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$
4. $\sum_{k=1}^{\infty} \frac{x^k}{(2k)!}$
5. $\sum_{k=1}^{\infty} k!x^k$
1. $[0,2)\text{.}$
2. $(-1,1)\text{.}$
3. $(-5,3)\text{.}$
4. $(-\infty, \infty)\text{.}$
5. $\{0\}\text{.}$
Solution
1. We use the Ratio Test with $a_k = \frac{|x-1|^k}{3k}\text{:}$
\begin{align*} \lim_{k \to \infty} \frac{ \frac{|x-1|^{k+1}}{3(k+1)} }{ \frac{|x-1|^k}{3k} } \amp = \lim_{k \to \infty} \frac{3k|x-1|^{k+1}}{3( k+1)|x-1|^{k}}\\ \amp = |x-1| \lim_{k \to \infty} \frac{k}{k+1}\\ \amp = |x-1|\text{.} \end{align*}
So the power series $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$ converges absolutely when $|x-1| \lt 1$ or when $0 \lt x \lt 2$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.
• When $x=0$ our power series is $\sum_{k=1}^{\infty} \frac{(-1)^k}{3k}$ which is just a scalar multiple of the alternating harmonic series and so converges.
• When $x=2$ our power series is $\sum_{k=1}^{\infty} \frac{1}{3k}$ which is just a scalar multiple of the harmonic series and so diverges.
Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{(x-1)^k}{3k}$ is $[0,2)\text{.}$ Note that the interval is centered at $x = 1$ and has radius R = 1.
2. We use the Ratio Test with $a_k = k|x|^k\text{:}$
\begin{align*} \lim_{k \to \infty} \frac{ (k+1)|x|^{k+1} }{ k|x|^k } \amp = |x|\lim_{k \to \infty} \frac{k+1}{k}\\ \amp = |x|\text{.} \end{align*}
So the power series $\sum_{k=1}^{\infty} kx^k$ converges absolutely when $|x| \lt 1$ or when $-1 \lt x \lt 1$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.
• When $x=-1$ our power series is $\sum_{k=1}^{\infty} (-1)^k k\text{.}$ Since $k \to \infty$ as $k \to \infty\text{,}$ this series diverges by the Divergence Test.
• When $x=1$ our power series is $\sum_{k=1}^{\infty} k$ which again diverges by the Divergence Test.
Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} kx^k$ is $(-1,1)\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = 1\text{.}$
3. We use the Ratio Test with $a_k = \frac{k^2|x+1|^k}{4^k}\text{:}$
\begin{align*} \lim_{k \to \infty} \frac{ \frac{(k+1)^2|x+1|^{k+1}}{4^{k+1}} }{ \frac{k^2|x+1|^k}{4^k} } \amp = \lim_{k \to \infty} \frac{4^k(k+1)^2|x+1|^{k+1}}{4^{k+1}k^2|x+1|^k}\\ \amp = \frac{1}{4}|x+1| \lim_{k \to \infty} \left(\frac{k+1}{k}\right)^2\\ \amp = \frac{1}{4}|x+1|\text{.} \end{align*}
So the power series $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$ converges absolutely when $\frac{1}{4}|x+1| \lt 1$ or when $-5 \lt x \lt 3$ and diverges outside this interval. To completely determine the interval of convergence, we need to check what happens at the endpoints of this interval.
• When $x=-5$ our power series is $\sum_{k=1}^{\infty} (-1)^k k^2\text{.}$ Since $k^2 \to \infty$ as $k \to \infty\text{,}$ this series diverges by the Divergence Test.
• When $x=3$ our power series is $\sum_{k=1}^{\infty} k^2\text{,}$ which again diverges by the Divergence Test.
Therefore, the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{k^2(x+1)^k}{4^k}$ is $(-5,3)\text{.}$ Note that the interval is centered at $x = -1$ and has radius $R = 4\text{.}$
4. We use the Ratio Test with $a_k = \frac{|x|^k}{(2k)!}\text{:}$
\begin{align*} \lim_{k \to \infty} \frac{ \frac{|x|^{k+1}}{(2(k+1))!} }{ \frac{|x|^k}{(2k)!} } \amp = \lim_{k \to \infty} |x|\frac{(2k)!}{(2(k+1))!}\\ \amp = |x| \lim_{k \to \infty} \frac{1}{(2k+2)(2k+1)}\\ \amp = 0\text{.} \end{align*}
So the power series $\sum_{k=1}^{\infty} \frac{x^k}{(2k)!}$ converges absolutely on the interval $(-\infty, \infty)\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = \infty\text{.}$
5. We use the Ratio Test with $a_k = k!|x|^k\text{:}$
\begin{align*} \lim_{k \to \infty} \frac{ (k+1)!|x|^{k+1} }{ k!|x|^k} \amp = \lim_{k \to \infty} |x|(k+1)\\ \amp = \infty \end{align*}
unless $x=0\text{.}$ So the interval of convergence of the power series $\sum_{k=1}^{\infty} \frac{x^k}{k!}$ is $\{0\}\text{.}$ Note that the interval is centered at $x = 0$ and has radius $R = 0\text{.}$
###### Example7.51
Suppose that the power series
\begin{equation*} \sum_{n=1}^\infty C_n (x+2)^n \end{equation*}
converges at $x=2$ and diverges at $x=-8\text{.}$
1. Does the series converge or diverge at $x=-5\text{?}$
2. Does the series converge or diverge at $x=5\text{?}$
3. What are all the possibilities for the radius of convergence of the power series?
1. The series converges at $x=-5.$
2. The series diverges at $x=5.$
3. The radius of converges is at least 4 and at most 6; that is, $4 \leq R \leq 6\text{.}$
Solution
Since the power series is centered at $x=-2\text{,}$ the fact that the series converges at $x=2$ tells us that the radius of convergence is greater than or equal to 4, since 4 is the distance between 2 and -2. Similarly, since the series diverges at $x=-8\text{,}$ this tells us that the radius of convergence is less than or equal to 6. Thus, we have $4 \leq R \leq 6\text{,}$ and we are guaranteed that if an $x$ is less than 4 away from $-2\text{,}$ then the series will converge there. Similarly, if an $x$ is more than 6 away from -2, the series must diverge.
Thus, since -5 is 3 away from -2, and 5 is 7 away from -2, the series will converge at $x=-5$ and diverge at $x=5\text{.}$
### SubsectionSummary
• A power series is a series of the form
\begin{equation*} \sum_{k=0}^{\infty} a_kx^k\text{.} \end{equation*}
• A power series always converges at at least one point. If the power series is centered at $x = a \text{,}$ the power series either converges only at $x = a \text{,}$ or it converges for all $x \in (-\infty, +\infty) \text{,}$ or it converges for all $x$ in a finite interval $(a - R, a + R)$ where $R$ is the radius of convergence. In the latter case, the power series may or may not converge at the endpoints $x = a - R$ and $x = a+R \text{,}$ so these points have to be checked separately.
### SubsectionExercises
Find the radius of convergence of the following series:
\begin{equation*} \sum_{n=1}^\infty \frac{(2n)!x^n}{(n!)^2} \end{equation*}
Consider the power series
\begin{equation*} \sum_{n=1}^\infty C_n(x-1)^n\text{,} \end{equation*}
with radius of convergence $R\text{.}$
1. If the series converges at $x=3$ and diverges at $x=5\text{,}$ what are the possible values for $R\text{?}$
2. If the series converges at $x=4$ and diverges at $x=-2\text{,}$ what are the possible values for $R\text{?}$
3. Is $-3 \lt x \lt 3$ a possible interval of convergence for the series?
4. Is $-2 \leq x \lt 4$ a possible interval of convergence for the series?
In this exercise we will begin with a strange power series and then find its sum. The Fibonacci sequence $\{f_n\}$ is a famous sequence whose first few terms are
\begin{equation*} f_0 = 0, f_1 = 1, f_2 = 1, f_3 = 2, f_4 = 3, f_5 = 5, f_6 = 8, f_7 = 13, \cdots\text{,} \end{equation*}
where each term in the sequence after the first two is the sum of the preceding two terms. That is, $f_0 = 0\text{,}$ $f_1 = 1$ and for $n \geq 2$ we have
\begin{equation*} f_n = f_{n-1} + f_{n-2}\text{.} \end{equation*}
Now consider the power series
\begin{equation*} F(x) = \sum_{k=0}^{\infty} f_kx^k\text{.} \end{equation*}
We will determine the sum of this power series in this exercise.
1. Explain why each of the following is true.
1. $xF(x) = \sum_{k=1}^{\infty} f_{k-1}x^k$
2. $x^2F(x) = \sum_{k=2}^{\infty} f_{k-2}x^k$
2. Show that
\begin{equation*} F(x) - xF(x) - x^2F(x) = x\text{.} \end{equation*}
3. Now use the equation
\begin{equation*} F(x) - xF(x) - x^2F(x) = x \end{equation*}
to find a simple form for $F(x)$ that doesn't involve a sum.
4. Use a computer algebra system or some other method to calculate the first 8 derivatives of $\frac{x}{1-x-x^2}$ evaluated at 0. Why shouldn't the results surprise you?
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition D 5 Copyright © Orchard Publications Matrix Operations For matrix multiplication, the operation is row by column. Thus, to obtain the product , we multiply each element of a row of by the corresponding element of a column of ; then, we add these products. Example D.3 Matrices and are defined as and Compute the products and Solution: The dimensions of matrices and are respectively ; therefore the product is feasible, and will result in a , that is, The dimensions for and are respectively and therefore, the product is also feasible. Multiplication of these will produce a matrix as follows: Check with MATLAB: C=[2 3 4]; D=[1 1 2]’; % Define matrices C and D. Observe that D is a column vector C*D, D*C % Multiply C by D, then multiply D by C ans = 7 Here, B and A are not conformable for multiplication B A p × n m × p A B A B C D C 2 3 4 = D 1 1 2 = C D D C C D 1 3 3 1 × × C D 1 1 × C D 2 3 4 1 1 2 2 ( ) 1 ( ) 3 ( ) 1 ( ) 4 ( ) 2 ( ) + + 7 = = = D C 3 1 1 3 × × D C 3 3 × D C 1 1 2 2 3 4 1 ( ) 2 ( ) 1 ( ) 3 ( ) 1 ( ) 4 ( ) 1 ( ) 2 ( ) 1 ( ) 3 ( ) 1 ( ) 4 ( ) 2 ( ) 2 ( ) 2 ( ) 3 ( ) 2 ( ) 4 ( ) 2 3 4 2 3 4 4 6 8 = = =
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Appendix D Matrices and Determinants D 6 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications ans = 2 3 4 -2 -3 -4 4 6 8 Division of one matrix by another, is not defined. However, an analogous operation exists, and it will become apparent later in this chapter when we discuss the inverse of a matrix. D.3 Special Forms of Matrices A square matrix is said to be upper triangular when all the elements below the diagonal are zero. The matrix of (D.4) is an upper triangular matrix. In an upper triangular matrix, not all elements above the diagonal need to be non zero. (D.4) A square matrix is said to be lower triangular , when all the elements above the diagonal are zero. The matrix of (D.5) is a lower triangular matrix. In a lower triangular matrix, not all elements below the diagonal need to be non zero.
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# Symmetry of real and imaginary parts in FFT [duplicate]
INPUT
x[0] = (0.00 + j 0.00)
x[1] = (1.00 + j 0.00)
x[2] = (2.00 + j 0.00)
x[3] = (3.00 + j 0.00)
x[4] = (4.00 + j 0.00)
x[5] = (5.00 + j 0.00)
x[6] = (6.00 + j 0.00)
x[7] = (7.00 + j 0.00)
x[8] = (8.00 + j 0.00)
x[9] = (9.00 + j 0.00)
x[10] = (0.00 + j 0.00)
x[11] = (0.00 + j 0.00)
x[12] = (0.00 + j 0.00)
x[13] = (0.00 + j 0.00)
x[14] = (0.00 + j 0.00)
x[15] = (0.00 + j 0.00)
FFT:
X[0] = (45.00 + j 0.00)
X[1] = (-25.45 + j 16.67)
X[2] = (10.36 + j -3.29)
X[3] = (-9.06 + j -2.33)
X[4] = (4.00 + j 5.00)
X[5] = (-1.28 + j -5.64)
X[6] = (-2.36 + j 4.71)
X[7] = (3.80 + j -2.65)
X[8] = (-5.00 + j 0.00)
X[9] = (3.80 + j 2.65)
X[10] = (-2.36 + j -4.71)
X[11] = (-1.28 + j 5.64)
X[12] = (4.00 + j -5.00)
X[13] = (-9.06 + j 2.33)
X[14] = (10.36 + j 3.29)
X[15] = (-25.45 + j -16.67)
From the above FFT out put I noticed the following:
Re(x[1])=Re(x[15]), Im(x[1])=-Im(x[15])
Re(x[2])=Re(x[14]), Im(x[2])=-Im(x[14])
Re(x[3])=Re(x[13]), Im(x[3])=-Im(x[13])
Re(x[4])=Re(x[12]), Im(x[4])=-Im(x[12])
and so on
Is this a proven result that
Re(X[n])=Re(x[N-n]), and Im(X[n])=-Im(x[N-n]) for 0<n<N-1, where N is no. of DFT points?
If yes then is there any particular condition under which this is true ? What is the gerenralised result?
If this is a general rule then I can save alot in memory and arithmatics as I am concerned only in magnitudes of the DFT output, and not in phase.
## 1 Answer
Yes, this is always true if the input to the DFT is real valued. It's called the "conjugate complex symmetry", because $$X_{N-n} = {X_n}^*$$ where $X_n$ is the DFT output and $()^*$ denotes the conjugate. It can be proven by inserting the property into the transformation formula of time domain sequence $x_k$: $$X_n = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\ X_{N-n} = \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\ =\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}}$$ Using $exp(-j2\pi k) = 1 \:\: \forall \: k$ we find $$X_{N-n} = \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}$$ Now we exploit that $x_k$ is real for all $k$ to rewrite the above: $$X_{N-n} = \sum_{k=0}^{N-1}\left(x_k e^{-j\frac{2\pi k n}{N}}\right)^* = X_n^*$$ It's now straightforward to show that the same is valid for the IDFT. (real spectrum -> complex conjugate symmetric time domain sequence). Moreover, the opposite statement is also true: if the input to the DFT (IDFT) is complex conjugate symmetric its output is real valued.
• How you create these boxes of formula ? Is it an images ? like this one XN−n=Xn∗ please suggest. Commented Jun 1, 2017 at 11:25
• You can use Latex markdown in posts. It is rendered by MathJax.
– Deve
Commented Jun 5, 2017 at 8:35
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# What's the simplest rational not expressible as a sum of a given number of unit fractions?
This is essentially the same as the closed question Representation of rational numbers as the sum of 1/k but I hope I can make a case for it as an MO-worthy question.
Ed Pegg, Jr., in his Math Games column for 19 July 2004 at the MAA website, http://www.maa.org/editorial/mathgames/mathgames_07_19_04.html writes, "Here is an interesting sequence of fractions that would likely would [sic] have fascinated Ahmes: $$1/2, 2/3, 4/5, 8/11, 14/17, 19/23, 24/29, 49/59, 65/71, 76/83, 61/157, 183/191, 260/269, 289/299.$$ $8/11 = 1/2 + 1/6 + 1/21 + 1/77$. This is the simplest Egyptian fraction that requires 4 parts. $14/17 = 1/2 + 1/4 + 1/20 + 1/55 + 1/187$ requires 5 parts. 289/299 is the simplest fraction that requires 14 parts. One might think that this sort of thing was well known, but it isn't.... What is the simplest fraction that requires 15 parts, 16 parts, and beyond?"
Pegg never defines "simplest," but presumably it means smallest (positive) denominator and, among fractions with the same denominator, smallest (positive) numerator. So the general question would be, given $s$, what's the simplest rational that can be expressed as a sum of $s$ unit fractions, but not fewer?
In this form, it's probably an open, and maybe impossible, problem (that is, I don't think anyone will find a simple formula for the rational as a function of $s$), so let me ask a bit less. Has there been any advance beyond 14 since 2004? Are there any bounds in the literature (that is, bounds on the "complexity" of the rational as a function of $s$)?
I note that Pegg gives no source for his list of 14. The Online Encyclopedia of Integer Sequences does not recognize the sequence of numerators, nor the sequence of denominators. Before anyone suggests typing "Egyptian fractions" into Google, or looking at the Wikipedia article on that subject, I hope he or she will verify that the particular question I'm asking is in fact answerable by such means.
EDIT: As per the comments, it appears that only the first four terms in Pegg's list are correct, and that the current state of knowledge is $${1\over2},{2\over3},{4\over5},{8\over11},{16\over17},{77\over79},{732\over733}.$$
Also as per the comments, if we are after $f(s)=\min\lbrace b:N(a,b)=s{\rm\ for\ some\ }a,1\le a\lt b\rbrace$ then $f(s)\ge e^{Cn^2}$ for some $C>0$, and, conjecturally, $f(s)\ge e^{e^{Cn}}$ for some $C>0$.
At this point I will gladly settle for a calculation of $s(8)$.
• 14/17 = 1/2 + 1/4 + 1/14 + 1/476 appears to be an error; I believe 16/17 = 1/2 + 1/3 + 1/10 + 1/128 + 1/32640 is the simplest requiring 5 terms. Jul 28, 2010 at 10:52
• "He conjectures that the true result is $N(a,b)=O(\log\text{}\log b)$, and he in fact establishes the inequalities $$\sum_{a=1}^{b-2}N(a,b)>{\textstyle\frac 1{2}}(b-2)(\log\text{}\log b-1),\quad N(b-1,b)>\log\text{}\log b-1.$$ " Jul 28, 2010 at 12:23
• I haven't checked the intervening numbers, but by hand I found 289/299 = 1/2 + 1/3 + 1/8 + 1/156 + 1/552. I checked also just using the greedy algorithm, and that gives 1/2 + 1/3 + 1/8 + 1/122 + 1/39795 + 1/1935522680. So I have no idea what Pegg's numbers are supposed to represent, but I can't see any relation between them and the stated problem. Time permitting I'll try calculating the sequence, but I anticipate the correct denominators will grow much more rapidly, so it'll be hard to calculate more than 7-8 terms. Jul 28, 2010 at 18:44
• My apologies to all for taking Pegg's numbers on faith. Guy, Unsolved Problems In Number Theory, D11, writes, "Victor Meally...noted that 2/3, 4/5 and 8/11 are the [simplest rationals] that need 2, 3 and 4 [unit fractions].... Stephane Vandemergel, in a 93-04-28 letter, states that 16/17 requires 5 [unit] fractions, and 77/79 needs 6." With these numbers, I found oeis.org/A097048 which gives 732/733 as the next, and last known, term. So unless there has been some advance since then, exact values are only known to $s=7$. As for bounds... (see next comment) Jul 29, 2010 at 0:47
• Lasse: re: "Is it obvious / known that for any n there is a rational number that can be written as a sum of n unit fraction, but not of n-1?" Maybe not obvious, but it is known. If s_i is the i'th number in Sylvester's sequence then (s_i-1)/s_i requires i terms. E.g. 1/2 requires 1 term, 2/3 requires 2, 6/7 requires 3, 42/43 requires 4, etc. Jul 29, 2010 at 3:57
$$s(8) = \frac{27538}{27539}$$.
Update: those links no longer available, code is available via github at https://github.com/hvds/seq/tree/master/least_eg.
The package includes both PARI/GP code and C code using the GNU GMP library to calculate the results, as well as a synopsis of the results for each denominator from 2 to 27539 which may be of use for further analysis.
I estimate the PARI code would have taken about a CPU-year to find the result; the C code runs over 20 times faster on my machine, and I don't understand why the difference is so great. (I'd appreciate email if someone can explain.)
• Bravo. Bravissimo. Well done, and many thanks. Sep 14, 2010 at 23:10
• The links are broken (2019). Can the files kindly be made available again? Nov 26, 2019 at 10:59
• Sorry I no longer have the original tarball; the C code available via github under github.com/hvds/seq/tree/master/least_eg should be the same or close to what was there. Nov 27, 2019 at 11:20
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# The speed of a projectile at its highest point is ${{v}_{1}}$ and at the point half the maximum height is ${{v}_{2}}$. If $\dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{2}{5}}$ then find the angle of projection. A. $45{}^\circ$ B. $30{}^\circ$ C. $37{}^\circ$ D.$60{}^\circ$
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Hint: As a first step, you could make a rough diagram of the projectile motion under discussion. Then, you could consider the motion part by part. You could apply Newton’s equation of motion for the case of motion from 0 to H and H to$\dfrac{H}{2}$ separately. Now using those relations along with the given conditions we will get the angle of the projectile.
Formula used:
Equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$
In the question, we are given the speed of a projectile at the highest point as ${{v}_{1}}$ and at the point that is half the maximum height be${{v}_{2}}$. We are given that,
$\dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{2}{5}}$…………………………………………. (1)
We are supposed to find the angle of projection.
We know that at the highest point the vertical component of velocity is zero. Also, for a projectile motion, the horizontal component of velocity remains the same, that is,
$v\cos \theta ={{v}_{2}}\cos \alpha ={{v}_{1}}$
$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\cos \alpha =\sqrt{\dfrac{2}{5}}$ ………………………………………. (2)
Now, we could consider the motion from$\dfrac{H}{2}$ to $H$,
Let us recall the Newton’s equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$
$\Rightarrow {{v}_{1y}}^{2}-{{v}_{2y}}^{2}=2{{a}_{y}}s$
$\Rightarrow 0-\left( {{v}_{2}}^{2}{{\sin }^{2}}\alpha \right)=2\left( -g \right)\dfrac{H}{2}$
$\Rightarrow {{v}_{2}}^{2}\left( 1-{{\cos }^{2}}\alpha \right)=gH$
From (2),
$\Rightarrow {{v}_{2}}^{2}\left( 1-\dfrac{2}{5} \right)=gH$
$\therefore {{v}_{2}}=\sqrt{\dfrac{5}{3}}gH$ ……………………………………….. (3)
Now from (1),
$\dfrac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\dfrac{2}{5}}$
$\Rightarrow {{v}_{1}}={{v}_{2}}\times \sqrt{\dfrac{2}{5}}$
$\Rightarrow {{v}_{1}}=\sqrt{\dfrac{5}{3}}gH\times \sqrt{\dfrac{2}{5}}$
$\therefore {{v}_{1}}=\sqrt{\dfrac{2}{3}}gH$
Similarly, for the motion from 0 to H, we have,
${{v}_{{{1}_{y}}}}^{2}-{{\left( v\sin \theta \right)}^{2}}=2\left( -g \right)H$
$\Rightarrow {{v}^{2}}{{\sin }^{2}}\theta =2gH$
$\Rightarrow \dfrac{{{v}_{1}}^{2}}{{{\cos }^{2}}\theta }{{\sin }^{2}}\theta =2gH$
$\dfrac{2}{3}gH{{\tan }^{2}}\theta =2gH$
$\Rightarrow {{\tan }^{2}}\theta =3$
$\Rightarrow \tan \theta =\sqrt{3}$
$\therefore \theta =60{}^\circ$
Therefore, we found that the angle of the projectile in the given question is found to be $60{}^\circ$ .
Hence, option D is found to be the correct answer.
Note:
We have assigned negative signs for the acceleration of the projectile, because we have chosen the convention in such a way. For questions like this you could consider the motion related to each point mentioned in the question. Doing so, you will easily solve the problem and also avoid confusions.
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# Feynman screw jack device
Feynman mentions in his book, The Feynman Lectures on physics
Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4-5. A handle 20 inches long is used to turn the screw, which has 10 threads to the inch. We would like to know how much force would be needed at the handle to lift one ton (2000 pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately 126 inches. The handle must thus travel 1260 inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight W applied to the end of the handle. So we find out that W is about 1.6 pounds. This is a result of the conservation of energy.
I dont understand the meaning of the bold part of the text. How it's connected with the example he's carrying on ?
• Conservation of energy often is a useful shortcut that you can use to get answers to questions that could have been solved in other, more complicated ways. You could draw diagrams, and do vector math to figure out the relationship between force on the top of the jack and force on the end of the handle, but Feynman blew past all that by noticing that if the jack was perfectly frictionless, then any value other than 1.6 pounds would lead to a violation of conservation of energy. Commented Apr 8, 2018 at 20:51
Energy (work done = force $\times$ distance) is conserved so $1.6 \times 1260 \approx 2000 \times 1$
Which is the applied force on handle $\times$ the distance moved by the applied force is equal to the load $\times$ distance moved by the load
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# Variance of Sample Variance
Given $$X_1,...,X_n$$ iid to a certain distribution (not necessarily normal), with $$\mathbb{E}(X_i)=\mu$$ and $$\mathbb{V}(X_i)=\sigma^2$$, I'm trying to deduce the standard and mean squared error of the estimator $$\widehat{\sigma}^2=S_n^2$$, where $$S_n^2$$ is the sample variance, given by$$S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X}_n)^2.$$ To do so I need its variance, $$\mathbb{V}(S_n^2)$$. Since I know the expectation $$\mathbb{E}(S_n^2)=\sigma^2$$, I started by expanding$$\mathbb{V}(S_n^2)=\mathbb{E}(S_n^4)-(\mathbb{E}(S_n^2))^2=\mathbb{E}(S_n^4)-\sigma^4$$ but I'm stuck with the expansion of the term $$\mathbb{E}(S_n^4)$$. Any ideas?
P.S. - I saw user940's answer on this question, but I was looking for a different approach, also not assuming normal distributed random variables.
• Emm how different approach you want? I have typed a 2-3 pages of full derivation for this starting from the Casella Berger Ex hints also. So basically it was just an expansion. If you start from this definition of $S^2$, I believe you will end up facing with the same expansion problem, but slightly longer.
– BGM
Commented Oct 17, 2017 at 11:02
• You're right, I guess there is no way around it. I ended up doing the expansion and left my solution below. Commented Oct 17, 2017 at 12:00
Let $\mu_k$ denote the $k$th central momentum of $X_i$, i.e, $\mu_k=\mathbb{E}((X_i-\mu)^k)$, and $Z_i\equiv X_i-\mu$ for all $i$. Thus $\mathbb{E}(Z_i)=0$. Since$$\mathbb{V}(S_n^2)=\mathbb{E}(S_n^4)-(\mathbb{E}(S_n^2))^2=\mathbb{E}(S_n^4)-\sigma^4,$$we derive an expression of $\mathbb{E}(S_n^4)$ in terms of $n$ and the moments. We can rewrite $S_n^2$ as\begin{align*} S_n^2=\frac{n\sum_{i=1}^n Z_i^2-(\sum_{i=1}^nZ_i)^2}{n(n-1)}. \end{align*} Squaring leads to\begin{align*} S_n^4=\frac{n^2(\sum_{i=1}^nZ_i^2)^2-2n(\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2+(\sum_{i=1}^nZ_i)^4}{n^2(n-1)^2} \end{align*} and so\begin{align*} \mathbb{E}(S_n^4)=\frac{n^2\mathbb{E}((\sum_{i=1}^nZ_i^2)^2)-2n\mathbb{E}\left((\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2 \right)+\mathbb{E}((\sum_{i=1}^n Z_i)^4)}{n^2(n-1)^2}. \end{align*} Since $Z_1,...,Z_n$ are independent, we have that, for distinct $i,j,k$,\begin{align*} \mathbb{E}(Z_iZ_j)=0,\hspace{5mm}\mathbb{E}(Z_i^3Z_j)=0,\hspace{5mm}\mathbb{E}(Z_i^2Z_jZ_k)=0 \end{align*} and\begin{align*} E(Z_i^2Z_j^2)=\mu_2^2=\sigma^4,\hspace{5mm}\mathbb{E}(Z_i^4)=\mu_4. \end{align*} Then, with simple algebraic simplifications, it can be shown that the following holds\begin{align*} \mathbb{E}((\sum_{i=1}^nZ_i^2)^2)&=n\mu_4+n(n-1)\sigma^4,\\ \mathbb{E}\left((\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2 \right)&=n\mu_4+n(n-1)\sigma^4,\\ \mathbb{E}((\sum_{i=1}^n Z_i)^4)&=n\mu_4+3n(n-1)\sigma^4. \end{align*} Substituting these into the expansion of $\mathbb{E}(S_n^4)$ and simplifying leads to\begin{align*} \mathbb{E}(S_n^4)=\frac{(n-1)\mu_4+(n^2-2n+3)\sigma^4}{n(n-1)} \end{align*} and so\begin{align}\label{var} \mathbb{V}(S_n^2)=\mathbb{E}(S_n^4)-\sigma^4=\frac{(n-1)\mu_4+(n^2-2n+3)\sigma^4}{n(n-1)}-\sigma^4=\frac{\mu_4}{n}-\frac{\sigma^4(n-3)}{n(n-1)}. \end{align}
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# SBP-Program
## How to determine degree of monomial?
How do you determine the degree of a monomial?
To determine the degree of a monomial add the exponents of all letters (variables) in this monomial.
### Determining degree of monomial
Example 1.
Determine the degree of the monomial
5n
Solution.
5n = 5n1
The literal part of the monomial is
n1
The answer: the degree of the monomial 5n is 1.
Example 2.
Determine the degree of the monomial
6n3
Solution.
The literal part of the monomial is
n3
The answer: the degree of the monomial 6n3 is 3.
Example 3.
Determine the degree of the monomial
47a3b2
Solution.
The literal part of the monomial is
a3b2
Add the exponents of the variables:
3 + 2 = 5
The answer: the degree of the monomial 47a3b2 is 5.
Example 4.
Determine the degree of the monomial
5a2h3s4
Solution.
Add the exponents of the variables:
2 + 3 + 4 = 9
The answer: the degree of the monomial 5a2h3s4 is 9.
More examples are available on What is the degree of the monomial?
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# Liverpoololympia.com
Just clear tips for every day
# Is 1m 100cm?
## Is 1m 100cm?
There are 100 centimeters in 1 meter.
How do you convert from 1 cm to m?
How to Convert Centimeters to Meters. To convert a centimeter measurement to a meter measurement, divide the length by the conversion ratio. The length in meters is equal to the centimeters divided by 100.
Which is bigger cm or m?
A centimeter is 100 times smaller than one meter (so 1 meter = 100 centimeters).
### What is a 1 cm?
1 centimeter is equal to 0.3937 inches, or 1 inch is equal to 2.54 centimeters. In other words, 1 centimeter is less than half as big as an inch, so you need about two-and-a-half centimeters to make one inch.
What is 1cm 1km?
Centimeters to kilometers conversion table
Centimeters (cm) Kilometers (km)
1 cm 0.00001 km
2 cm 0.00002 km
3 cm 0.00003 km
4 cm 0.00004 km
What is 10 cm called?
A decimeter is a unit of length in the metric system. The term “Deci” means one-tenth, and therefore decimetre means one-tenth of a meter. Since a meter is made up of 100 cm, one-tenth of 100 cm is 10 cm. Thus one decimeter measures 10 cm. The symbol used to write decimeter is dm.
#### What is 100 km called?
The Metric System
nanometer(nm) 11,000,000,000 of a meter
hectometer(hm) 100 meters
kilometer(km) 1000 meters
Megameter(Mm) 1,000,000 meters
Gigameter(Gm) 1,000,000,000 meters
What is 1 km equal to in meters?
1,000 meters
1 kilometre is equal to 1,000 meters, which is the conversion factor from kilometres to meters.
What is 100m called?
1 hectometer (hm) = 100 meters. 10 hectometers = 1 kilometer (km)
## How thick is 1mm?
1/32 inch
1mm = just over 1/32 inch. 2mm = just over 1/16 inch. 3mm = almost 1/8 inch.
What size is a cm?
A measure of length in the metric system. There are 100 centimeters in a meter and 2½ centimeters in an inch. Tumor sizes are often measured in centimeters (cm) or inches.
How do u convert cm to km?
As we know, 1 centimeter is equal to 1/100000 kilometer. i.e. 1 cm = 0.00001 km. Thus, to convert cm to km, multiply the number of cm by 0.00001. In other words, divide the number of cm by 100000 to get the value in km.
### What is 1cm to 5km?
We know that on a scale of 1cm, the actual distance is 5km. At An actual distance of 5km, the scale on the map is 1cm. cm. So, we get that the distance on the map will be 2.8cm if the actual distance is 14km.
How wide is 1 m?
39.37 in
Meters to Inches table
Meters Inches
1 m 39.37 in
2 m 78.74 in
3 m 118.11 in
4 m 157.48 in
What is after km?
In the Metric System, the units of measurement that comes after kilometers are megameters. One megameter is equivalent to one million meters.
#### Where does the earth end?
“The Kármán line is an approximate region that denotes the altitude above which satellites will be able to orbit the Earth without burning up or falling out of orbit before circling Earth at least once,” Bossert said. “It is typically defined as 100 kilometers [62 miles] above Earth,” Igel added.
How high is space?
62 miles
A common definition of space is known as the Kármán Line, an imaginary boundary 100 kilometers (62 miles) above mean sea level. In theory, once this 100 km line is crossed, the atmosphere becomes too thin to provide enough lift for conventional aircraft to maintain flight.
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# Proving that $n\mid A_n$ with integers that are relatively prime
Problem: Let $1 \leq b_1 < b_2 < \dots < b_{\phi(n)} < n$ be integers relatively prime with $n$, and $B_n = b_1 b_2 \cdots b_{\phi(n)}$. Consider the sum
$$1/b_1 + 1/b_2 + \dots + 1/b_{\phi(n)} = A_n/B_n$$
Prove that $n\mid A_n$. Is it true that $n^2\mid A_n$
I am thinking of rewriting the fractions on the left side to have the same denominator,$B_n$, but I am not sure if that helps me or not?
-
You may need some restrictions, as $2$ does not divide $A_2=1$. If $n>2$ is the only necessary restriction, look at $n=6$. – Aaron Jul 28 '13 at 1:18
Sorry I am a mistake in typing the problem. Do you still think I need the restrictions? – Username Unknown Jul 28 '13 at 1:20
In the calculations below, we assume that $n\gt 2$.
Imagine bringing the terms to a common denominator $B_n$. Then the numerator is a sum of products $b_1b_2 \cdots \hat{b_i} \cdots b_{\varphi(n)}$. Here $\hat{b_i}$ indicates that the term $b_i$ is missing. Let $c_i$ be the number in the interval $1$ to $n$ such that $c_ib_i \equiv 1\pmod{n}$. Then $$b_1b_2 \cdots \hat{b_i} \cdots b_{\varphi(n)}\equiv c_iB_n\pmod{n}.$$ It follows that $$A_n \equiv (c_1+c_2+\cdots +c_{\varphi(n)})B_n\pmod{n}.$$ Now the $c_i$ travel, in some order, through the $b_i$, so their sum is congruent to $b_1+b_2+\cdots+b_{\varphi(n)}$.
The $b_i$ come in pairs with sum $n$, so their sum is congruent to $0$ modulo $n$. This completes the proof.
As to divisibility by $n^2$, there is the easy counterexample $n=4$.
I have this for the case of n=4. $\phi(4)=2$. $1/b_1 + 1/b_2 = A_2/B_2$, which is the same as $b_2 +b_1=A_2$. How am I suppose to the contradiction from here? – Username Unknown Jul 28 '13 at 2:33
We have in this case $b_1=1$ and $b_2=3$. Binging to the common denominator $1\cdot 3$, we find that $A_4=4$ and $B_4=3$. Thus $n$, that is, $4$, divides $A_4$. But $4^2$ certainly does not divide $A_4$. So $n=4$ is a counterexample to the assertion that $n^2$ divides $A_n$. By the way, if $n$ is a prime greater than $3$, then $n^2$ does divide $A_n$. – André Nicolas Jul 28 '13 at 2:42
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Explore BrainMass
# Optimization problem dealing with a fence and area.
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
A farmer has 600 feet of fencing with which to enclose a rectangular plot. What is the maximum area he can enclose?
Hint: Find a model for the area of the rectangular plot and maximize by completing the square
https://brainmass.com/math/optimization/optimization-problem-determining-maximum-area-6168
#### Solution Preview
First let's find the function we want to maximize.
Le x and y be sides of a rectangular plot. We know that its perimeter has to be 600 feet.
So 2x+2y=600
=> x+y=600
=> ...
#### Solution Summary
The maximum area of a fenced enclosure is found by completing the square.
\$2.49
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## What is a scale definition?
(Entry 1 of 7) 1: an instrument or machine for weighing. 2a: a beam that is supported freely in the center and has two pans of equal weight suspended from its ends —usually used in plural. b: either pan or tray of a balance.
## What is scale short answer?
A scale is a set of levels or numbers which are used in a particular system of measuring things or are used when comparing things.
## What is a scale in math?
In math, a scale in graphs can be defined as the system of marks at fixed intervals, which define the relation between the units being used and their representation on the graph. Each small interval or division measures 100 ml. Fun Facts. A ruler is often called a scale.
## What is scale and its types?
Introduction: There are 4 types of scales, based on the extent to which scale values have the arithmetic properties of true numbers. The arithmetic proper- ties are order, equal intervals, and a true zero point. From the least to the most mathematical, the scale types are nominal, ordinal, interval, and ratio.
## What is scale example?
more The ratio of the length in a drawing (or model) to the length on the real thing. Example: in the drawing anything with the size of “1” would have a size of “10” in the real world, so a measurement of 150mm on the drawing would be 1500mm on the real horse.
## Does scale mean size?
If you refer to the scale of something, you are referring to its size or extent, especially when it is very big.
## What is on a scale?
—used with a range of numbers to show the size, strength, or quality of something I liked it.
## What does size and scale mean?
Size is the physical dimensions of an object. Scale is the relative size of different objects in relation to each other or a common standard. In design when we talk about scale we’re usually talking about size, however scale is simply a relative comparison of some measurable quality.
## What do we use scale for?
When we say something is to scale it means it is a standardized measurement using actual dimensions that are represented by a smaller unit of measurement. These smaller increments of measurement allow us to represent a larger object in a proportionate way but at a reduced size, not full size.
## How do you read scales?
A scale is shown as a ratio, for example 1:100. A drawing at a scale of 1:100 means that the object is 100 times smaller than in real life scale 1:1. You could also say, 1 unit in the drawing is equal to 100 units in real life.
You might be interested: Question: What to plant in the fall?
## How is a scale written?
Scale tells the size relationship between the map or model and the real earth. It can be written in three ways, the verbal scale, the graphic scale, and the representative fraction (RF) or fractional scale. The verbal scale is a simple statement of the scale such as one inch equals one mile or 1″ = 1 mile.
## What is scale diagram?
Scale Diagram – a drawing in which measurements are proportionally reduced or enlarged. from actual measurements.
## What are 3 types of scales?
Three Types of Scale:
• Fractional or Ratio Scale: A fractional scale map shows the fraction of an object or land feature on the map.
• Linear Scale: A linear scale shows the distance between two or more prominent landmarks.
• Verbal Scale: This type of scale use simple words to describe a prominent surface feature.
## What are the 5 types of measurements?
Types of data measurement scales: nominal, ordinal, interval, and ratio.
## What are the 4 types of measurement scales?
Each of the four scales (i.e., nominal, ordinal, interval, and ratio) provides a different type of information.
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Thanks for whoever can help. I'm a dope when it comes to this.
Attachments:
problems.JPG
X=-24 for number 1
I don't know a single thing about maths(that's a lie, I just suck at it), but there's a general rule that applies here:
"If you don't know what to say, answer A."
Originally Posted by iforgotmyname
I'd love to get quoted once...
Being dutch is ubercool.
That's why I used the smiley, and not stuff like: .
Number 2 is 4Xsquared
And number 3 is C.
Here's one possible solution:
Just say you couldn't do it because THERE'S AN ELEPHANT IN THE WAY!
Quote by p o e
lmfao man thats so sick and depraved and yet funny all at once
my hats off to you IbanezSA160, you have embodied the Pit into one little poem
the third answer is clearly "C"...however i have no idea about the other ones
This water's dark and coldGod's not where you hopedThis moment come and goneIt's time we all moved on
Number 1: Multiply 2x-7 by the other denominator (2) and multiply x+2 by the other denominator (5) so you get
4x-14=5x+10
Then you subtract 10 and subtract 4x and you get
-24=x
Done with #1
With that knowledge you can figure out #2.
As for three, set it up like this:
60/220 = 30/x
60x = 6600
x = 110
C
EDIT: nevermind, JamieB is right, it's A. I need to read questions more carefully.
God I wish I was still in algebra
EDIT:
Actually, forget all that crap. Go with the elephant! It's bulletproof!
Last edited by rush5757 at Dec 5, 2007,
The last question is A.
The stopping distance is proportional to the square of the speed, so when speed halves, the stopping distance quarters.
Co-President of UG's Tubgirl Virgins Club
I still need a legitimate answer for number 2. The possible answers are: A. 8x^3 + 12x^3 +18x B. 8x^3 - 18x C. 8x^3 +18 D. 8x^3 - 12x^2 +18x
Thanks for your help so far though.
The answer to 2 is 8x^3
Co-President of UG's Tubgirl Virgins Club
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How to write a many-one reduction proof
Writing a proof by contradiction is fairly formulaic--first you assume the opposite, then derive a contradiction. I would like to know the steps and conventions for writing a many-one reduction proof. This answer was somewhat helpful, but like other explanations I've heard, it sounds too similar to a proof by contradiction for me to understand the difference.
My specific points of confusion:
• The exact meaning of $A \leq _{m} B$
• Do we reduce an unknown problem to a known problem, or the opposite
• Does the reduction function transform a known problem into an unknown problem, or the opposite
• How a many-one reduction is different from a proof by contradiction. It seems like a simpler version of the same logic; if so, how is the simplification justified
I am only interested in many-one reductions of language/Turing machine problems (in case the term has meaning in other contexts as well). Any help is appreciated, including links and illustrative examples. Thanks.
A many-one reduction is not a "proof-by-contradiction" per se, it is an application of a theorem. The theorem is:
Let $L,L'\subseteq \{0,1\}^*$, and $L \leq_m L'$. Then, it holds that
1. If $L$ is undecidable, then $L'$ is undecidable.
2. if $L'$ is decidable, then $L$ is decidable.
I will omit the proof, there are plenty of resources where you can read up on it. You can view the statement $L \leq_m L'$ intuitively as "The language $L$ is easier to decide than the language $L'$". And that's what a reduction proof does. You show that one language (the hard, known one, like the halting problem $H_0$) is easier to decide than a new, unknown language $L$. By showing $H_0 \leq_m L$ you've essentially shown (by application of the above theorem) that the halting problem is easier than $L$. Thus, $L$ cannot be decided. So, to answer your questions:
1. $A \leq_m B$ means that the language $A$ is easier to decide than the language $B$.
2. To show hardness, you reduce "YOUR_HARD_LANGUAGE $\leq_m$ SOME_NEW_LANGUAGE". To show computability, you can reduce in the other direction, but usually it's simpler to just construct a TM that solves the given problem.
3. A reduction $L \leq_m L'$ takes an instance $x \in \{ 0,1\}^*$ and does the following with it: If $x \in L$, you make sure that the reduction $f(x) \in L'$. If $x \not \in L$, then the reduction maps $f(x) \not \in L'$. So - if you prove hardness, you transform a known hard problem into the unknown problem.
4. It is a contradiction in the sense of "If I could solve the new problem, then I can solve the hard problem, which cannot be, since it is proven to be hard". This contradiction is hidden in the Theorem above and you just slap it (implicitly) on every reduction you do.
I hope that helped. As I said, my main intuitive approach to reductions is "How could I use a new language $L'$ to solve an old, hard language $L$?".
A many-one reduction is not a contradiction and should not be thought of one, even informally. A many-one reduction from problem A to problem B is an efficient translation of instances of problem A into instances of problem B such that "yes" instances are mapped to "yes" instances and "noes" to "noes".
Even when you use many-one reductions to prove NP-completeness, you're not using contradiction. All you're saying is that instances of every problem in NP can be transformed efficiently into instances of some particular problem. It follows that a deterministic polynomial-time algorithm for that problem would give deterministic polynomial-time algorithms for all problems in NP. But, even here, there is no contradiction: we do not know that P $\neq$ NP so there is nothing to contradict.
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