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### Course: Grade 7 math (FL B.E.S.T.)>Unit 8
Lesson 3: Measures of center
# Statistics intro: Mean, median, & mode
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set. Created by Sal Khan.
## Want to join the conversation?
• I've heard of both the arithmetic mean and the geometric mean. What's the difference?
• Think about it this way. The arithmetic mean of a bunch of numbers is the number a that satisfies
x₁ + x₂ + x₃ + .... = a + a + a + ... + a
The geometric mean is the number b that satisfies
x₁ * x₂ * x₃ * ... = b * b * b * ... * b
There is also a harmonic mean, which is the number h that satisfies
1/x₁ + 1/x₂ + 1/x₃ + ... = 1/h + 1/h + ... + 1/h.
If the set of numbers were (2 , 4 , 6 , 8 , 10) , how would you find the mode?
(There are no numbers repeated in the above question.)
• The mode is 'No Mode' or 'None'.
Mode is used to find the number of times a number appears for statistics.
• is centeral tendancy the same thing as mean?? What is the difference??
• The arithmetic mean is one example of a statistic that describes the central tendency of a dataset. But any other formula or process that takes a dataset and generates a single number that represents a "typical" value is also a measure of central tendency. That includes the median and mode as well as more exotic things like the midrange or the arithmetic mean when you ignore the largest and smallest value. All of these numbers attempt to capture the spirit of a dataset by giving you a sense of a single "usual" value, and that is what makes them measures of central tendency..
• if there is a question such as:
what is the mode of 2,2,3,5,6,5?
would it be 2 or 5?
• It's always possible that there are two modes, and sometimes there is no mode at all. So since 2 and 5 are both repeated the same time, they are both modes of your data set.
• Here are some quick definitions.
Statistics: the study and manipulation of data (basically it's just data)
Descriptive statistics: Showing data while summarizing or using a smaller set of numbers.
Average: (these are just words to describe average) In general, mean, typical, middle, central tendency.
Arithmetic mean (or just mean, I just like arithmetic mean more): The process of adding a set of numbers, then dividing by the amount of numbers. (Example: 4+3+1+6+1+7= 22 divided by 6. The answer to that is 3 2/3.)
Median: The "middle" number in a set. Rule: When faced with a set of numbers where the amount of numbers in it are all even, you take the two middle numbers and add them. Then you divide them by two to get the median. (In the set of numbers we had earlier [1, 1, 3, 4, 6, 7], 3 and 4 were the middle numbers. We add them together to get 7, then divide it by 2 to get a median of 3.5.) However, this DOES NOT apply to a set of numbers with an odd set of numbers in it.
Mode: The most common number in a set (the number that repeats itself the most). Rule: If all the numbers are represented equally (basically if there are the same amount of each number, if that makes sense,) there is no mode. For example, in the set of numbers from earlier (1, 1, 3, 4, 6, 7), the mode would be 1. But if we remove one of the ones, since every number would be represented equally, there would be no mode.
Hope this helped.
• That really helps
(1 vote)
• If two numbers are the most common in a set ( example: 1,2,3,3,4,5,6,6,7), what would be the mode?
• A data set can have more than one mode. Unlike the mean, the mode is not necessarily unique. Your example is "bimodal" - it has two modes: 3 and 6.
• How would you use average in real life?
• There are countless applications. I'll give some examples. The normal body temperature is 98.6 degrees Fahrenheit. How was this exact temperature chosen?This number was given by a German doctor Carl Reinhold August Wunderlich, after examining millions of readings taken from 25,000 German patients and taking their average. The mileage of automobiles is calculated by finding the average volume of fuel consumed by the automobile. Each and every science experiment done in the lab involves calculation of the average reading after repeating the experiment many times, so that error is minimized. In fact, calculating the average is one of the most essential mathematical skills. One would need this knowledge regardless of which field he/she works in.
• What if the numbers are 1,3,5,6,7,8,23,42,76,83,93 how do you find the median
• You put the numbers in order (as you've done) and count how many numbers there are. If there's an odd number of numbers (as in this case), you pick the number in the middle of the list, and that's the median. If there's an even number of numbers, you take the two numbers in the middle, add them together, and divide them by two.
• Does anyone know an easy way (such as a song or rhyme) to memorize what mean, median, and mode are?
Or you can try this one I made up:
The mode is the first one to be seen,
Occurring the most, there is nowhere it can hide,
While the median lies in between,
With the same number of numbers to either side,
Finally the mean. The mean is mean!
You have to add up all the numbers, then divide. | HuggingFaceTB/finemath | |
# Tag Info
## Hot answers tagged angular-momentum
6
Basically the rings don't fall into Saturn for the same reason the Moon doesn't fall into the earth. The rings are billions of little moons, each in it's own stable, or largely stable orbit. The rings are also likely resupplied with new ring material from Enseladus, Saturn's 2nd closest moon. (ice volcanoes due to strong tidal forces that can shoot ice ...
4
The general question is quite hard to tackle I think, because a rigorous motivation of Hilbert space would end up in the theory of operator algebras (see e.g. this answer) and the OP is probably not interested in these aspects at the moment. As for the example of spin, the Hilbert space in this case is still an $L^2$ space, but the functions are no longer ...
4
Yes. This commutation relation is that of the Lie algebra $\mathfrak{so}(3)$ corresponding to the rotation group in three dimensions. Thus the commutation relation states that the Pauli matrices generate rotations. To understand why this is the commutation relation of $\mathfrak{so}(3)$, one can draw a diagram showing that the commutator of two ...
4
That you can only ever know one of the three components of angular momentum is best seen not through the uncertainty principle, but on the states themselves. Since $[L_i,L_j] = \epsilon_{ijk} L_k$, the three momentum operators are pairwise not simultaneously diagonalizable (since simultaneous diagonalizability implies that the operators commute), meaning ...
3
The book where the derivation is described sufficiently pedagogically is Ballentine's Quantum mechanics - A Modern development, chapter 3. I am going to give a sketch of the 30-page chapter. (Beware, I suppress vector notation) Transformations of the quantum state are expressible as unitary transformations. The first order expansion of a unitary ...
3
For $d=3$ the group theoretic meaning of total angular momentum is that it is the Casimir operator of $SO(3)$. For $SO(d)$ where $d>3$ you have more than one Casimir operator, so it's not clear what you mean by "total angular momentum" In particular the number of Casimir operators is $[d]/2$, where $[d]=d$ or $d-1$ depending whether $d$ is even or odd.
3
Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required to ...
2
The angular momentum operators obey the commutation relations $$[L_i,L_j]=i\sum_k\epsilon_{ijk}L_k$$ A nonzero commutation relation means we can't have a state vector which is an eigenstate of more than one angular momentum operator at the same time. This also leads to an uncertainty principle.
2
Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$ The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and ...
2
Looking at your video, it appears that the boomerang is not turning fast enough to return. This typically means that it is too heavy. I wrote a couple of answers earlier [here](http://physics.stackexchange.com/a/156122/26969_ and here to explain some of the physics of boomerangs; perhaps you will find the physics there hard to understand, but then these are ...
2
Just to clarify to Robin Ekman's answer, superpositions of the Pauli matrices exponentiate to $SU(2)$, not $SO(3)$, but both these Lie groups have $\mathfrak{so}(3)$ as their Lie algebra - but I am sure you already know this. Also, there is another way to look at the problem that you might find helpful, even though it is a mathematical insight rather than a ...
1
What you have to assume a defined motion, where $\theta(t)$ is known. If the mass was spinning with a constant rate there would be no torque needed. The basic equation of motion is $$T = I_{zz} \ddot{\theta}$$ where $I_{zz}=\int r^2 {\rm d}m$ is the mass moment of inertia about z. That internal torque is the translated into shear stress with $$\tau = ... 1 The wave function only contains all the information about the system im so far as you consider it. Meaning each qualitatively different physical system needs its a modified Hilbert space to fit what can happen with the system. In case you have something like spin on its own in H_{Spin} and you want to look at a freely moving particle in H_{free} that ... 1 For a full shell, the addition of the expectation values of any angular momentum L_i is zero, and similarly for the spin operators \sigma_i. This is not hard to see - for l(l+1) as the expectation value of L^2 for a s,p,d,f subshell, the basis of that subshell is spanned by states indexed by integers between -l and l, and since that l is also the ... 1 Let us put an hat on operators, which acts non-trivially on kets:$$ [\hat{J}_a,\hat{O}^s_{\ell}] ~=~ \hat{O}^s_m~[J_a^s]_{m\ell}, \tag{4.1}$$The matrix element$$[J_a^s]_{\ell^{\prime}\ell}~\in~\mathbb{C}$$is just a complex number and hence commutes with a ket. Hence$$[\hat{J}_a,\hat{O}^s_{\ell}]|j,m,\alpha\rangle ~\stackrel{(4.1)}{=}~ ...
1
The point is that the spin operator is defined to be (1/2) times SU(2) generator while the orbital angular momentum is defined to be only SU(2)(or SO(3), is the same) generator. The proof is the same, and is " representation independent", in the sense that the structure of identity multiplied by something plus a linear combination of sigma matrices ...
1
The identity you used, $$\exp(i\theta \, \hat s)=\cos(\theta)+i\sin(\theta)\,\hat s, \tag{\ast}$$ is crucially dependent on the operator $\hat s$ being idempotent, and particularly on the fact that $\hat{s}^2=\mathbb1$. This is generally not the case for angular momenta other than spin-1/2. In general, the total angular momentum is a scalar within the ...
1
First, angular momentum isn't measured about an axis. It's measured about a point. Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in. Now, about ...
1
Let me assume that the object has spherical symmetry, however, for solving the present problem it is painted on its surface with different colors. So, imagining a plane section that contains the orbit of the object around the earth. The section of the plane through the object is a circle and we will see different points of the circumference painted in ...
1
"By the uncertainty principle" is the answer. In more detail, let's say we're talking about x and y axes. The first measurement puts the electron into an eigenstate of the spin X observable (the question of how it does this is the quantum measurement problem). Whichever of the two X eigenstates this "collapse" ends up in, it not an eigenstate of the spin Y ...
1
Nothing prevents the electron's spin from being measured along a particular axis, and then subsequently measured along an axis perpendicular to the first. In this situation, however, the spins along the perpendicular axes would not be known simultaneously, so the uncertainty principle would not be violated. As an example, say that we perform our own ...
1
The answer to your question is that to a first approximation the direction of the spaceship will not change, so the upper diagram is the correct one. However the direction of the spaceship will change very slightly due to a phenomenon called the geodetic effect. The easiest way to see this is to replace the spaceship by a gyroscope, and make the gyroscope ...
Only top voted, non community-wiki answers of a minimum length are eligible | open-web-math/open-web-math | |
# [LeetCode] #1# Two Sum : 数组/哈希表/二分查找/双指针
1. Two Sum
Total Accepted: 241484 Total Submissions: 1005339 Difficulty: Easy
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
• 给定一个数组和一个目标值
• 找出数组中两个成员，两者之和为目标值
• 假设一定存在一个解
• 算法核心:
• 三种方法:
• 暴力搜索: O(n^2): 超时
• 哈希表: O(n)
• 先快排, 后二分查找: O(nlogn) + O(nlogn)
• 先快排, 后使用双指针分别指向数组头和尾，同时双向遍历数组: O(nlogn) + O(n)
• 实现细节:
• 算法逻辑相对简单
• 实现细节相对容易
• 哈希表
• 实现
1 class Solution {
2 public:
3 vector<int> twoSum(vector<int>& nums, int target) {
4 vector<int> res;
5 unordered_map<int, int> m;
6
7 for (int i = 0; i < nums.size(); i++) m[nums[i]] = i;
8
9 for (int i = 0; i < nums.size(); i++) {
10 if (m.count(target - nums[i]) && m[target - nums[i]] != i) {
11 res.push_back(i);
12 res.push_back(m[target - nums[i]]);
13 return res;
14 }
15 }
16
17 return res;
18 }
19 };
• 结果
• 快排-二分查找
• 实现
1 class Solution {
2 public:
3 vector<int> twoSum(vector<int>& B, int target) {
4 vector<int> res;
5 vector<pair<int, int>> A;
6
7 for (int i = 0; i < B.size(); i++) {
8 A.push_back(make_pair(B[i], i));
9 }
10
11 my_qsort(A, 0, A.size() - 1);
12
13 for (int i = 0; i <= A.size(); i++) {
14 int left = i + 1, right = A.size() - 1;
15 while (left <= right) {
16 int mid = left + (right - left) / 2;
17 if (A[mid].first == target - A[i].first) {
18 res.push_back(A[i].second);
19 res.push_back(A[mid].second);
20 return res;
21 }
22 if (A[mid].first < target - A[i].first) left = mid + 1;
23 else right = mid - 1;
24 }
25 }
26
27 return res;
28 }
29 private:
30 void my_qsort(vector<pair<int, int>>& A, int l, int r) {
31 if (l > r) return;
32
33 pair<int, int> key = A[l];
34 int nl= l, nr = r;
35 while (l < r) {
36 pair<int, int> tmp;
37 while (A[r].first >= key.first && l < r) r--;
38 while (A[l].first <= key.first && l < r) l++;
39
40
41 tmp = A[l];
42 A[l] = A[r];
43 A[r] = tmp;
44 }
45 A[nl] = A[l];
46 A[l] = key;
47
48 my_qsort(A, nl, l - 1);
49 my_qsort(A, l + 1, nr);
50 }
51 };
• 结果
• 快排-双指针
• 实现
1 class Solution {
2 public:
3 vector<int> twoSum(vector<int>& B, int target) {
4 vector<int> res;
5 vector<pair<int, int>> A;
6
7 for (int i = 0; i < B.size(); i++) {
8 A.push_back(make_pair(B[i], i));
9 }
10
11 int left = 0, right = A.size() - 1;
12 my_qsort(A, 0, A.size() - 1);
13
14 while (left < right) {
15 if (A[left].first + A[right].first < target) left++;
16 else if (A[left].first + A[right].first > target) right--;
17 else {res.push_back(A[left].second), res.push_back(A[right].second); return res;};
18 }
19
20 return res;
21 }
22 private:
23 void my_qsort(vector<pair<int, int>>& A, int l, int r) {
24 if (l > r) return;
25
26 pair<int, int> key = A[l];
27 int nl= l, nr = r;
28 while (l < r) {
29 pair<int, int> tmp;
30 while (A[r].first >= key.first && l < r) r--;
31 while (A[l].first <= key.first && l < r) l++;
32
33
34 tmp = A[l];
35 A[l] = A[r];
36 A[r] = tmp;
37 }
38 A[nl] = A[l];
39 A[l] = key;
40
41 my_qsort(A, nl, l - 1);
42 my_qsort(A, l + 1, nr);
43 }
44 };
• 结果
©️2019 CSDN 皮肤主题: 大白 设计师: CSDN官方博客 | HuggingFaceTB/finemath | |
### Home > CC3 > Chapter 8 > Lesson 8.2.2 > Problem8-71
8-71.
Use $<$, $>$, or $=$ to compare the number pairs below.
1. $0.183 \text{____} 0.18$
Notice that $0.18$ can also be looked at as $0.180$. This may make the numbers easier to compare.
$>$
1. $−13\text{____} −17$
Although $17$ is a larger number, since it is negative, it is farther left on the number line than $-13$. This means that $-13$ is larger than $−17$.
$>$
1. $0.125\text{____}\frac{1}{8}$
Converting the fraction into a decimal will make this relationship easier to compare.
1. $−6_\text{________}4$
See (b).
1. $72\%\text{____}\frac{35}{30}$
Converting the fraction into a percent will make this relationship easier to compare.
$<$
1. $−0.25\text{____}-0.05$
See (b). | HuggingFaceTB/finemath | |
# Superposition of two plane waves
• ballzac
#### ballzac
Hi,
I'm not after much help here. I already have an answer, but I want to make sure that I haven't made any stupid mistakes and that I understand the question. I also have a query about the second part of the question.
## Homework Statement
What is the probability density of the superposition of two plane waves of differing angular frequencies, whose wave vectors point in different directions, and what is the frequency at which this probability density oscillates.
## Homework Equations
The two plane waves are
$$\Psi_1(\textbf{r},t)=exp[i(\textbf{k}\cdot \textbf{r}-\omega t)]$$
and
$$\Psi_2(\textbf{r},t)=exp[i(\textbf{k}'\cdot \textbf{r}-\omega' t)]$$
## The Attempt at a Solution
$$\rho=|\Psi_1+\Psi_2|^2$$
$$=|exp[i(\textbf{k}\cdot \textbf{r}-\omega t)]+exp[i(\textbf{k}'\cdot \textbf{r}-\omega' t)]|^2$$
$$=2+exp[i[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]]+exp[i[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]]$$
$$=2+cos[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]+isin[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)]+cos[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]+isin[(\textbf{k}'\cdot \textbf{r}-\omega' t)-(\textbf{k}\cdot \textbf{r}-\omega t)]$$
Due to the properties of even and odd functions, this simplifies to
$$2(1+cos[(\textbf{k}\cdot \textbf{r}-\omega t)-(\textbf{k}'\cdot \textbf{r}-\omega' t)])$$
$$=2(1+cos[(\textbf{k}-\textbf{k}')\cdot\textbf{r}-(\omega+\omega')t])$$
This seems to be simplified as much as possible, and answers the first part of the question. As for the second part, it seems that the angular frequency is just $$\omega +\omega'$$
making the frequency
$$\nu=\frac{\omega +\omega'}{2\pi}$$
My question is, if I have the right answer here, then can't this be violated? For instance, if one angular frequency is an integer multiple of the other, won't the resultant frequency just be the lower frequency?
P.S It is also required that I find the time-average of the probability density as a function of position. I would assume that for a cosine function with
$$\omega \neq 0$$
the average would be zero, which would mean in this case the answer is simply
$$\rho_{av} (\textbf{r}) = 2$$
Does this sound right? It sounds plausible to me that the time-average probability is constant over all space, and of course this is not a normalised wave function, so I don't see any problem with having a probability over 1. Do I understand this correctly?
Last edited:
You made a small error in your derivation. The resultant angular frequency should be the difference of the two frequencies, which is the beat frequency you learned about in lower level physics. The rest of your conclusions sound accurate. The probability density is greater than one because you're only using two plane waves. You would need to superpose an infinite number of plane waves with different wavelengths to obtain a normalizable function. This is done in standard QM texts.
ah, okay. It did seem kind of odd that it was a sum and not difference of angular frequencies. I'll have a look through it to see where I went wrong. Thanks heaps.
I've fixed up my arithmetic and arrived at the result you suggested. Just another quick question. The value of $$\rho$$ is the same whether $$\omega$$ is larger or $$\omega'$$ is larger, so does that mean that the resultant frequency is the absolute value of the difference? i.e.
$$\nu_\rho=\frac{|\omega-\omega'|}{2\pi}$$
I guess it has to be, otherwise I wouldn't know which way around to put the two omegas in the difference. Thanks again for your help.
Last edited: | HuggingFaceTB/finemath | |
# math
posted by .
The width of a rectangular prism is w centimeters. The height is 2 centimers less than the width. The length is 4 centimeters more than the width. If the volume of the prism is 8 times the measure of the length, find the dimensioon of the prism.
• math -
V = Lwh
w = width
w - 2 = height
w + 4 = length
8(w + 4) = V
8(w + 4) = w(w - 2)(w + 4)
Solve for w
• math -
-2
• math -
w=2
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# Ideal of a Polynomial ring $R$ which is not principal.
Let $R$ be the ring given by $R=\mathbb Z+x\mathbb Q[x]$. Then show that:
1) $R$ is an integral domain and its units are $+1$ and $-1$.
2) $x$ is not prime in $R$ and describe the quotient ring $R/(x)$.
3) Compute an ideal of $R$ which is not principal.
$R$ being a subset of $\mathbb Q[x]$ inherits non-zero divisor, unity, and commutativity. So $1$st question is done. For the third one I think the ideal $(2,{1/2}x)$ may be but can't give proper proof. Please help me for $2$nd and $3rd$ ones. Thanks in advance.
• $R$ doesn't inherit its units from $Q[x]$. For example, $2$ is a unit in $Q[x]$. Is it in $R$?
– user14972
Commented Sep 12, 2017 at 1:32
• No no its correct. I am said that only unity, commutativity and non-zero divisors can be inherits. Units can't be an inheritable property. Sir please help me for other ones. @Hurkyl Commented Sep 12, 2017 at 1:36
• You also said that the first question is done, but the first question says something about units.
– user14972
Commented Sep 12, 2017 at 9:24
## 2 Answers
This is going to sound a little strange at first, but it ends up at your solution. There is quite a bit of checking for you to do.
The set $S=\mathbb Z\times \mathbb Q/\mathbb Z$ is an Abelian group under coordinatewise addition, and it becomes a ring if you use the multiplication $(y, q+\mathbb Z)(z, p+\mathbb Z)=(yz, yp+\mathbb Z+zq+\mathbb Z)$. It's called the trivial extension of $\mathbb Q/\mathbb Z$ by $\mathbb Z$.
Now, it is not hard to prove that $S\cong R/(x)$ for the $R$ in your problem. Just look at $(x)=xR=x\mathbb Z+x^2\mathbb Q[x]$ and consider the quotient carefully.
You can check that $I=\{0\}\times \mathbb Q/\mathbb Z$ is an ideal of $S$, and furthermore $ab=0$ for any $a,b\in I$. Therefore $S$ isn't an integral domain... and that should be enough information for you to conclude that $(x)$ is not prime in $R$.
Finally, $x\mathbb Q[x]$ is your candidate for a non-finitely-generated ideal. You'll see that you have to generate everything of the form $x\mathbb Q$ using integer multiples of your generators (which won't be possible with finitely many generators.)
• What is the quotient will? Is it $ax+b+(x)$, where $0<a<1$ and $b$ in $Z$. Also $a$ must be in $Q$. Am I correct ? @rschwieb Commented Sep 12, 2017 at 2:56
• @abcdmath Right, you can find a representative of each coset with the properties you mentioned. Then you can see what the logical map is to $S$ ... Commented Sep 12, 2017 at 20:26
Let me give some remarks about the ring $R=\mathbb Z+x\mathbb Q[x]$. This happens to be a particular construction of the more general construction $D+xD_S[x]$, where $D$ is an integral domain and $S$ is a multiplicatively closed subset of $D$ such that $0\notin S$.
In general, there is no a "good" relationship between $D$ and $D+xD_S[x]$, except when $S=D\setminus \{0\}$. In this case there are many interesting properties related to both $D$ and $D+xD_S[x]=D+K[x]$ (here $K$ is the fraction field of $D$).
For example, in this paper by Costa and Zafrullah it's proved the following remarkable result:
Theorem 1: $D+xK[x]$ is a bézout domain iff $D$ is a bézout domain.
Proof: This is corollary 4.13 in the paper cited above.
Now, in your case, $D=\Bbb Z$ is a bézout domain (it's even a PID), so by the above theorem $\mathbb Z+x\mathbb Q[x]$ is also a bézout domain.
This explains why in order to produce a non principal ideal you need to take a non finitely generated ideal, e.g., the ideal given in rschwieb's answer: $x\Bbb Q[x]$. This also justifies why your example, namely $(2,1/2x)$ won't work, since it's a finitely generated ideal.
On the other hand, rschwieb already answered what is the form of $R/(x)$ and from there he proved that $x$ is not prime, but there is another way to proof the later result.
It's enough to show that $x$ is not irreducible and indeed, in the ring $\mathbb Z+x\mathbb Q[x]$ we have the non-trivial factorization of $x$: $$x=\Bigl(\frac{1}{2}x\Bigr)2.$$ This shows that $x$ is not irreducible in $\mathbb Z+x\mathbb Q[x]$.
More generally, Costa and Zafrullah proved in their paper above a nice result about the structure of the prime ideals of $D+xK[x]$. We have the following:
Theorem 2: The nonzero prime ideals of $D+xK[x]$ are the ideals $P+xK[x]$ where $P$ is a prime ideal of $D$, and the principal ideals $f(x)(D+xK[x])$, where $f(X)$ is irreducible in $K[x]$ and $f(0)=1$.
Proof: This is theorem 4.21 in the paper cited above.
So by the theorem above we can conclude again that $x$ is not prime in $\mathbb Z+x\mathbb Q[x]$ by noticing that $x$ is irreducible in $\Bbb Q[x]$, but $x(0)=0$.
There has been many researching in commutative algebra about the ring $D+xD_S[x]$ and similar constructions. If you want to explore and deepen these constructions I recommend you the following articles:
1. Costa, D.; Mott, J.L.; Zafrullah, M. The Construction $D+XD_S[X]$. Journal of Algebra. 53 (1978), p. 423-439.
2. Zafrullah M. The $D+XD_S[X]$ construction from GCD domains. Journal of Pure and Applied Algebra. 50 (1988), p. 93-107.
3. Jackson, T.; Zafrullah, M. Examples in modern algebra with which students can play. Problems, Resources, and Issues in Mathematics Undergraduate Studies. 6.4 (1996), p. 351-354. | HuggingFaceTB/finemath | |
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# If d is a positive integer and f is the product of the first
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If d is a positive integer and f is the product of the first [#permalink]
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If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?
(1) 10^d is a factor of f
(2) d>6
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23 Sep 2005, 02:14
I think we have had this one before.
But I can't find it in the search function.
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23 Sep 2005, 04:36
I think it is C
1 is insufficient, 10, 100, 1000... are factors of f
2 is insufficient dcan be 7, 8, ...
If we take 1 and 2 together:
From 1 we have that 2^d * 5^d is a factor of f
In f there are 7 fives, so d cannot be bigger than 7
From 2 we have that d is bigger than 6, therefore, d = 7
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23 Sep 2005, 05:10
This question uses similar logic.
http://www.gmatclub.com/phpbb/viewtopic ... ht=#124646
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23 Sep 2005, 08:40
how did you figure out that 30! has 7 5s in it?
30!=2^a*3^b*5^c...29
now, 30/5=6 5s
then you have the over 30/10=3 5s
30/15, gives you 2 5s
so IMHO there should be 6+3+2= 11 5s in 30!
jdtomatito wrote:
I think it is C
1 is insufficient, 10, 100, 1000... are factors of f
2 is insufficient dcan be 7, 8, ...
If we take 1 and 2 together:
From 1 we have that 2^d * 5^d is a factor of f
In f there are 7 fives, so d cannot be bigger than 7
From 2 we have that d is bigger than 6, therefore, d = 7
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Joined: 28 Dec 2004
Posts: 3357
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Schools: Wharton'11 HBS'12
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23 Sep 2005, 08:43
Oops I take it back...
I realized there are 30/5=6 5s...
then we need to look at 25 which is another factor of 30!, so you get one more 5 there...TOtal is 7...
23 Sep 2005, 08:43
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Discrete Mathematics with Graph Theory (3rd Edition) 89
# Discrete Mathematics with Graph Theory (3rd Edition) 89 -...
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Section 4.3 87 (a) [BB] We have a 2 = p201. r2 with gcd(p, r2) = 1 and b = rf3 s. Also gcd(r2, s) = 1 since gcd(r, s) = 1. We conclude that there are two possibilities: If f3 = 1, then gcd(a 2 , b) = p, while if f3 > 1, then gcd(a 2 , b) = p2. (b) Here, a 3 = p301. r3 and b = rf3 s. Now there are three possibilities. If f3 = 1, then gcd(a 3 , b) = p. If f3 = 2, then gcd(a 3 , b) = p2, and if f3 ~ 3, then gcd(a 3 , b) = p3. (c) Here we have a 2 = p201. r2 and b 3 = p3 f3 s 3. There are two possibilities. If a = 1, then gcd(a 2 , b 3 ) = p2. If a> 1, then gcd(a 2 ,b 3 ) = p3. 29. [BB] Let the prime decompositions of a and b be a = pflp~2 ... p~r and b = qflqg2 ... q~ •. Since gcd(a, b) = 1, we know that Pi =1= qj for any i and j. Thus, ab = pfl ... p~rqfl ... q~. with no simplification possible. Since ab = x 2 , it follows that each ai and each f3i must be even. But this means that a and b are perfect squares. 30. (a) We proceed exactly as in the solution to Problem 17. We write a and b in the form for certain primes Pl, P2, .. . , Pr and integers alo a2, .. . , ar, f31. f32, . .. , f3r.
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# Verify whether the following sequences are G.P. If so, write tn : 5,15,155,1255, ... - Mathematics and Statistics
Sum
Verify whether the following sequence is G.P. If so, write tn:
sqrt(5), 1/sqrt(5), 1/(5sqrt(5)), 1/(25sqrt(5)), ...
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#### Solution
sqrt(5), 1/sqrt(5), 1/(5sqrt(5)), 1/(25sqrt(5)), ...
t1 = sqrt(5), "t"_2 = 1/sqrt(5), "t"_3 = 1/(5sqrt(5)), "t"_4 = 1/(25sqrt(5), ...
Here, "t"_2/"t"_1 = "t"_3/"t"_2 = "t"_4/"t"_3 = 1/5
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = sqrt(5), "r" = 1/5
tn = arn–1
∴ tn = sqrt(5) (1/5)^("n" - 1)
= (5)^(1/2) (5)^(1 - "n")
= (5)^(3/2 - "n").
Concept: Sequence and Series - Geometric Progression (G.P.)
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#### APPEARS IN
Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Exercise 4.1 | Q 1. (iii) | Page 50
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# RD Sharma Class 10 and Chapter Circles Maths Solutions
## Maths Class 10 and Chapter Circles RD Sharma Solutions Filter
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### RD SHARMA Class 10 | CIRCLES
If radii of the two concentric circles are 15cm and 17cm, then the length of each chord of one circle which is tangent to other is: 8cm (b) 16cm (c) 30cm (d) 17cm ### RD SHARMA Class 10 | CIRCLES
If from a point `P ,` tangents `P Qa n dP R` are drawn to the ellipse `(x^2)/2+y^2=1` so that the equation of `Q R` is `x+3y=1,` then find the coordinates of `Pdot` ### RD SHARMA Class 10 | CIRCLES
In Figure, if tangents `P Aa n dP B` are drawn to a circle such that `/_A P B=30^0` and chord `A C` is drawn parallel to the tangent `P B ,` then `/_A B C=` `60^0` (b) `90^0` (c) `30^0` (d) None of these ### RD SHARMA Class 10 | CIRCLES
In two concentric circles, a chord of length 24cm of larger circle becomes a tangent to the smaller circle whose radius is 5cm. Find the radius of the larger circle. In this chapter, we will learn about Circles and the details on Tangents of Circle. Several theorems and solved problems based on the properties of the circle are provided in this chapter. Doubtnut has made it much easier for the students to understand the subject with the help of the video tutorials. The theorems and solved problems are presented in a simple way that students can understand easily. Since following a theorem is crucial for scoring good marks in examination, so the video solutions are prepared in such a way so that students won’t find any difficulties while looking at the examples and will be able to solve all problems with ease.
It is observed that students following the NCERT textbooks find it quite perplexing in getting an in-depth understanding of various concepts. Math is such a subject that demands rigorous practise for achieving success. RD Sharma books help students to a certain extent in getting some good idea on how to clear their queries. But this does not lead to getting all clear ideas where they fail to attempt the tricky questions that are set in the CBSE board exams. This leads the students to get low marks in the subject, which leads to finding themselves low confident as well.
Owing to the various complications or problems faced by the students, a team of experts of Doubtnut has introduced online study materials. This makes it much easier for them to find all the essential and useful idea on how to solve even the trickiest questions without fail. Therefore, it can be the right choice for the students to opt for it if they genuinely wish to gain good marks in the examination. So, it becomes possible to practice in the best way with the help of various examples and exercises. It is possible to get the best solution with the support of the RD Sharma Solutions class 10 maths. The concepts explained in the video tutorials are helpful in preparation of engineering entrance examination like JEE Main and Advanced.
## Topics and subtopics of RD Sharma class 10 Chapter 10
Introduction: Since you have studied already about Circles, so we will mainly focus on Tangents and practice on the same with several examples and problems. Here in this chapter, various concepts like tangents, tangents to a circle, the number of tangents from a point on a circle, etc. are introduced. For the convenience of the students, several solved examples are provided here to clear these concepts and to solve the problems quickly.
Some important point to remember:
1) A circle can have infinite tangents (Tangent is a line that intersects the circle at one point. There is infinite number of points on a circle).
2) Secant is a line that intersects a circle in two points.
3) The Point of contact is a common point of a tangent to a circle and the circle.
Theorems are given below:
### Theorem 1
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
A circle is given with its center point “C”. “AB” is a tangent at the point of contact“X”. Prove that CX is perpendicular to AB. Proof:
Let Y be a point on AB.
Connect CY. Suppose it touches the circle at Z
Hence,
CY > CZ
CY > CX (since, CX = CZ = radius)
Therefore, it will be same with any other points on the circle.
Let us try an exercise:
“AB” is a tangent of a circle at a point “X”. AB meets a line CY at point Y. The radius CX = 6 cm. The line CY = 14 cm. Find out the length of XY. Solution:
Given, CX = radius = 6 cm
CY = 14 cm.
Since, XY is a tangent,
Hence, CX is perpendicular to XY(since tangent at any point of the circle is perpendicular to the radius through point of contact)
So, CXY = 90o
Hence, Δ CXY is a right-angle triangle.
In right angle triangle, using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
• (CY)2 = (CX)2 + (XY)2
• (14)2 = (6)2 + (XY)2
• 196 = 36 + (XY)2
• 196 – 36 = (XY)2
• (XY)2 = 160
• XY = √160 = 12.6 cm
### Theorem 2
The lengths of tangents drawn from an external point to a circle are equal.
A circle is drawn with its centre point C and two tangents are drawn from a point P outside the circle connecting at A and B. Prove PA = PB
Solution: Let us connect AC, PC, and BC
As PA is tangent,
PA | CA (Tangent at any point of the circle is
The point of contact
So, PAC = 90o
Hence, Δ PCA is a right-angle triangle.
Similarly,
PB is a tangent.
PB _|_ CB (Tangent at any point of the circle is
The point of contact
So, PBC = 90o
Hence, Δ PCB is a right-angle triangle.
In Δ PCA and Δ PCB
PC is a common line.
Since PA and PB are tangents, PAC and PBC are right angles
Therefore,
In Δ PCA and Δ PCB PAC = PBC (both are 90o) PC = PC (common line between the triangles)
CA = CB (radius of the circle)
Δ PCA Δ PCB
PA = PB
Therefore, tangents drawn from an external point to a circle are equal in lengths.
Let’s do an exercise to understand the above theorem:
• If PA and PB are tangents from a point P to the circle
with centre Cand the angle between PA and PB is
600, then what will be PCA =? Solution: Given BPA = 600
Connect P, A, B, and C first.
Since PA is a tangent to the circle. CA PA( Tangent at any point of circle is perpendicular to the radius through
the point of contact
so, CAP = 90o
In Δ PCA and Δ PCB600 PAC = PBC (both are 90o)
PC = PC (common line between the triangles)
CA = CB (radius of the circle)
Δ PCA Δ PCB
So, CPA = CPB
= ½ BPA
= ½ x 60o
= 30o
In triangle Δ PCA, PCA + PAC + CPA = 180o
• PCA+ 90o + 30o = 180o
• PCA = 180o- 90o- 30o= 60o Hence proved.
Some exercises are given below:
Exercise: In two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact. Prove it. Solution: Let us take two concentric circles.
Where AB is the chord of the larger circle which touches the
smaller circle at P. Connect the centre C and P. Now to
Prove chord AB is bisected at the point P i.e. AP = PB
Since AB is tangent to the smaller circle
So, CP _|_ AB (Tangent at any point of circle is perpendicular to the radius through
the point of contact)
Now, connect CA and CB.
In Δ APC and Δ BPC, CPA = CPB (both are 90o)
PC = PC (common line between the triangles)
CA = CB (radius of the circle)
Δ APC ΔBPC
Therefore, AP = PB Hence proved.
Exercise: Two concentric circles are of radii 7 cm and 5 cm. Find the length of the chord of the larger circle which touches the smaller circle. Solution:
Solution: Let the centre of the two concentric circles be P and chord XY
of the larger circle touches the smaller perplexing at Y. Since XZ is
tangent to the smaller circle, hence PY XZ.
Given,
PY = 3 cm
PX = 5 cm
Applying Pythagoras theorem in PXY,
PY2 + XY2 = PX2
9 + XY2 = 25
XY2 = 25 - 9
XY2 = 16 cm = 4 cm
Since,
XY = YZ (Perpendicular from centre of circle bisects the chord)
XZ = 2XY = 2 × 4 cm = 8 cm
Hence, length of chord of larger circle is 8 cm.
Exercise: The tangents drawn at the ends of a diameter of a circle are parallel. Prove this. Solution: Let us take a circle with centre C. XY is its diameter. AB and DE are two tangents of the circle connect at the endpoints X and Y. Now we have to prove AB is parallel to DE.
As we learned already that the radius is perpendicular to the tangent at the point of contact. Hence CYA = 90o and CXD = 90o
Again,CYB = CXE= 90o
Since the opposite angles are equal, hence the tangents must be parallel to each other.
Since alternate interior angles are equal, hence lines PQ and RS must be parallel.
## Importance of Doubtnut for the students
The students obtain all the critical concepts and comprehensive understanding of how to solve tricky questions. R D Sharma Solutions for Class 10 Maths Chapter 10 Circles make things much easier for them to get a better idea on how questions are set in the examination. Some of the advantages that students can enjoy by opting for the video tutorials are as follows:
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Article | Open | Published:
Block copolymer derived uniform mesopores enable ultrafast electron and ion transport at high mass loadings
Abstract
High mass loading and fast charge transport are two crucial but often mutually exclusive characteristics of pseudocapacitors. On conventional carbon supports, high mass loadings inevitably lead to sluggish electron conduction and ion diffusion due to thick pseudocapacitive layers and clogged pores. Here we present a design principle of carbon supports, utilizing self-assembly and microphase-separation of block copolymers. We synthesize porous carbon fibers (PCFs) with uniform mesopores of 11.7 nm, which are partially filled with MnO2 of <2 nm in thickness. The uniform mesopores and ultrathin MnO2 enable fast electron/ion transport comparable to electrical-double-layer-capacitive carbons. At mass loadings approaching 7 mg cm−2, the gravimetric and areal capacitances of MnO2 (~50% of total mass) reach 1148 F g−1 and 3141 mF cm−2, respectively. Our MnO2-coated PCFs outperform other MnO2-based electrodes at similar loadings, highlighting the great promise of block copolymers for designing PCF supports for electrochemical applications.
Introduction
An ideal support for MnO2 and other transition metal oxides (RuO2, NiO, WO3, and Fe2O3, etc.) needs the characteristics of (1) lightweight, (2) large surface areas for high loadings, (3) high electron conductivity, and (4) low ion diffusion resistivity. However, there is not a single nanostructure that meets all these characteristics5,15. Carbon supports are inherently lightweight and electrically conductive. At high mass loadings of transition metal oxides, the electrical conductivity of electrodes decreases, but it can be restored by blending or wrapping with additional conjugated polymers16,17 or carbon additives16,18,19, as shown for excellent supports such as wearable textile structures16 and graphene16,20. The ion conduction, however, is drastically complicated21, and the efficient ion diffusion across the entire support, as well as the thick layer of MnO2, remains a significant challenge. To mitigate the ion diffusion resistivity, ultrathin layers of MnO2 have been deposited on model supports, e.g., nanoporous Au22,23, Pt foil9, Ni foil24, Si wafer25, dendritic Ni26 and macroporous Ni film27. With a thickness of <10 nm22 or at a mass loading of <0.35 mg cm−2 on the model supports23,26, MnO2 exhibits fast electron/ion transport and the gravimetric capacitances approach the theoretical limit. Nevertheless, when the conventional lightweight carbon supports are loaded with MnO2, they either suffer from a limited surface area for depositing a large amount of MnO2 thin layers (e.g., carbon cloth11,12, carbon fibers16,28,29,30 and other macroporous carbons13,14), or they lack desirable porous structures that facilitate rapid ion diffusion across long distances to maintain high rate capability (e.g., microporous carbons5,15,31).
To test our hypothesis, herein we demonstrate block copolymer-derived PCFs as lightweight and high mass-loading supports for MnO2 (Fig. 1). Because block copolymers self-assemble and microphase separate into uniform and continuous nanoscale domains34,35,36,37,38,39,40,41, after pyrolysis they generate interconnected mesoporous carbons with large surface areas for depositing MnO2. Disparate from all other carbon supports, the mesopores are designed from the macromolecular level and offer a high degree of uniformity. Importantly, our judiciously designed mesopores have an average diameter of 11.7 nm and are partially filled with a <2-nm-thick layer of MnO2 (Fig. 1c). On the one hand, the remaining mesopores provide continuous channels for efficient ion transport across the entire electrode, significantly reducing the ion diffusion resistance. On the other hand, the fibrous carbon network provides expressways for efficient electron transport without the need for any conductive additives. This contrasts with hard-templated mesoporous carbon particulates (e.g., CMK-331), which demand polymer binders to hold the discrete carbon particulates together. At high mass loadings approaching 7 mg cm−2, the PCF-supported MnO2 electrodes (PCF@MnO2) show superior electron/ion transport and outstanding charge-storage performances.
Results
Morphology
To illustrate the crucial importance of uniform mesopores for high mass loading of MnO2, we have synthesized two types of carbon fibers, i.e., PCFs with uniform mesopores derived from poly(acrylonitrile-block-methyl methacrylate) (PAN-b-PMMA) and conventional carbon fibers (CFs) with limited mesopores from pure polyacrylonitrile (PAN). Scanning electron microscopy (SEM) shows the contrasting morphologies of PCFs and CFs (Fig. 2a, d and Supplementary Fig. 1). Owing to the microphase separation of PAN-b-PMMA and the subsequent degradation of poly(methyl methacrylate) (PMMA), the PCFs were perforated with a large number of uniformly distributed, randomly oriented, and interconnected mesopores of ~11.7 nm (Figs. 2a, 3a, and Supplementary Fig. 1a)42. In contrast, the CFs derived from PAN exhibited relatively smooth surfaces and no observable mesopores under SEM (Fig. 2d and Supplementary Fig. 1b). Small angle X-ray scattering (SAXS) spectroscopy confirmed the microphase separation of PAN-b-PMMA and revealed that the average center-to-center pore-spacing in PCFs was 25.7 nm (Supplementary Fig. 2). The volume fraction of PAN in PAN-b-PMMA was ~65%, and supposedly the block copolymer should self-assemble into either cylindrical or gyroidal structures, depending on the incompatibility of the two blocks. After pyrolysis, however, the porous carbon fibers showed no well-defined cylindrical or gyroidal structures but interconnected mesopores that were irregularly shaped and uniformly distributed, as shown in the cross-sectional SEM image (Supplementary Fig. 1a). This morphology is attributed to the crosslinking of PAN at elevated temperatures, which hindered the microphase separation of PAN-b-PMMA into well-defined cylindrical or gyroidal structures, similar to the crosslinking-induced hindering effect in previous reports43,44.
The two types of carbon fibers were immersed in aqueous solutions of potassium permanganate (KMnO4, 10 mM) at 80 °C to deposit MnO2 on their surfaces. We chose the solution-based redox deposition because it creates a conformal and homogenous layer of MnO2 inside the pores via a self-limiting redox reaction between KMnO4 and carbon32,45,46. Compared with electrochemical deposition (Supplementary Fig. 3), the redox reaction deposition is advantageous because it yields uniform and homogeneous layers of MnO2 on PCFs that ensure a low ion diffusion resistance and thus, a high rate capability. After the deposition, the carbon fibers were washed thoroughly with deionized water, and the supernatant were analyzed with UV-vis spectroscopy to assure that there was no residual KMnO4 in the carbon fibers (Supplementary Fig. 4). As shown by SEM, MnO2 started to grow confocally on PCF within the first hour (Fig. 2b), and it continued to grow into nanosheets when the deposition time was prolonged to 2 h (Fig. 2c). The growth of MnO2 on conventional CFs, however, differed drastically. After depositing for 2 h, the surface of CF@MnO2-2h (Fig. 2f) did not change significantly from CF@MnO2-1h (Fig. 2e). Only a thin layer of MnO2 nanosheets was present on the surfaces of both CF@MnO2-1h and CF@MnO2-2h, confirming that the block copolymer-derived PCFs afford a much higher loading of MnO2 than pure PAN-derived CFs. To verify the successful deposition of MnO2 in the mesopores, we compared the transmission electron microscopy (TEM) images of PCF (Fig. 2g) and PCF@MnO2-1h (Fig. 2h). Black spots of MnO2 were uniformly embedded in PCF@MnO2-1h, while they were absent in PCF before loading with MnO2. MnO2 appeared black because Mn has a higher atomic number than carbon does. The PCF mats were prepared on a large scale and ready for use as electrodes without binders or conductive additives (Fig. 2i).
Chemical and physical properties
X-ray photoelectron spectroscopy (XPS), Raman spectroscopy, and high-resolution TEM orthogonally verified the successful loading of MnO2 onto PCF. The XPS spectrum (Supplementary Fig. 5a) of PCF@MnO2-2h showed peaks of C, O, and N corresponding to the carbon fibers, as well as a full set of peaks corresponding to Mn. An examination of the Mn 3s core-level XPS spectrum (Supplementary Fig. 5b) revealed that the separation between the doublet was 4.89 eV, corroborating the valence state of Mn(IV)47. After MnO2 deposition, the Raman spectrum of PCF@MnO2-2h (Supplementary Fig. 6a) showed a group of peaks centered at ~600 cm−1 corresponding to birnessite-phase manganese dioxide (δ-MnO2)48. The birnessite-phase of MnO2 was also proven by the characteristic lattice fringes in the lattice-resolved TEM images (Supplementary Figs. 6b, c). Among the various types of MnO2, δ-MnO2 is one of the most suitable phases for fast charge-discharge because its layered structure allows for rapid ion diffusion49.
The porous structures of carbon fibers changed after loading with MnO2. The pore size distributions of mesopores and micropores were evaluated by nitrogen and carbon dioxide adsorption-desorption isotherms, respectively (Supplementary Fig. 7). PCFs possessed significantly larger numbers of both mesopores and micropores. After depositing MnO2, the micropore volumes of PCFs and CFs steadily decreased at all pore widths, but the peak positions remained unchanged (Fig. 3a, b), suggesting that the micropores were either completely filled or clogged by MnO2. The pore size distributions of PCFs and CFs, however, were different in the mesopore range. PCFs exhibited appreciable decrease in the mesopore volume after depositing MnO2. In addition, the peak position shifted from 11.7 to 10.0 nm after 1 h, and further down to 9.3 nm after 2 h, suggesting that the average thickness of the MnO2 layer inside the pores was ~0.9 nm and ~1.2 nm after depositing for 1 and 2 h, respectively. These thicknesses are desirable for high capacitive performance, as suggested by the Au model in a previous report22. The reduction of mesopore size suggests that the mesopores were only partially filled with MnO2, and therefore they remained accessible to the gas adsorbates and ions. As shown in Supplementary Table 1, the pore volume reduced more in the mesopore range (86.1% reduction after the 2-h deposition) than in the micropore range (66.0% reduction after 2-h deposition). On the contrary, the mesopore volume of CFs, which was two orders of magnitudes lower than that of PCFs, increased after depositing MnO2 (Fig. 3b). The increase in the mesopore volume of CFs is ascribed to the porous structures formed by MnO2 as shown in Fig. 2e, f. The total pore volumes of CF-based electrodes were at least one order of magnitude lower than those of PCF-based electrodes.
The incorporation of MnO2 into PCFs and CFs also altered the surface area (Fig. 3c). The surface area of the PCF mat (574.8 m2 g−1) was more than ten times higher than that of the CF mat (55.31 m2 g−1). Upon loading with MnO2, the surface area of PCFs decreased from 574.8 to 229.5 m2 g−1 for PCF@MnO2-1h, and further down to 187.5 m2 g−1 for PCF@MnO2-2h. In contrast, the surface area of CFs only experienced moderate decreases from 55.31 to 43.35 m2 g−1 for CF@MnO2-1h and to 41.76 m2 g−1 for CF@MnO2-2h.
Ultra-fast electron and ion transport
Considering the high loading of MnO2 and the large number of mesopores for ion transport, we investigated the performance of the PCF-based electrodes for pseudocapacitors. The electron transport and ion diffusion resistivity were analyzed with electrochemical impedance spectroscopy (EIS). The Nyquist plots of PCF, PCF@MnO2-1h and PCF@MnO2-2h (Fig. 4a) exhibited incomplete semicircles followed by linear tails, which resemble the features of mixed kinetic-diffusion-controlled processes and are typical for pseudocapacitive materials50. To obtain the resistances, we fitted the EIS spectra with an equivalent electric circuit (Supplementary Fig. 8). The combined series resistances (Rs) of PCF and PCF@MnO2-1h were 1.0 Ω, and that of PCF@MnO2-2h increased to 1.4 Ω (Fig. 4a inset). The Rs values were comparable to highly conductive carbon-based materials in aqueous electrolytes51,52,53, indicating that MnO2 introduced minimal changes to the electrical resistance of the electrodes despite the high loadings. In addition, the charge-transfer resistances (Rct, the semicircles in Fig. 4a inset) of PCF, PCF@MnO2-1h and PCF@MnO2-2h are 0.74, 0.86 and 1.30 Ω, respectively. The small resistances suggest efficient electron transfer associated with the redox reaction of MnO2. The augmentation of charge-transfer resistance in PCF@MnO2-2h is mainly due to the increased thickness of MnO2 deposited in the mesopores (evidenced by the reduction in mesopore-width shown in Fig. 3d). The increased thickness elongates the electron transport distance in MnO2 and therefore obstructs electron transfer at the MnO2/electrolyte interface, because MnO2 is a poor electron conductor (10−5~10−6 S cm−1). The small Rs and Rct are key attributes of the block copolymer-based carbon fiber electrodes because 1) unlike discrete carbon particles or graphene flakes, the carbon fibers offer continuous expressways for electron conduction, and 2) the block copolymers endow the carbon fibers with high surface areas to load with an ultrathin layer of δ-MnO2, which mitigates the insulating problem and facilitates the electron transport.
In addition to the efficient electron transport, the block copolymer-derived PCF electrodes exhibited ultra-fast ion diffusion kinetics, as featured by their ultra-small diffusion resistances (σ). The values of σ were extracted from the slopes of the linear fitting lines of the real part of impedance (Z’) versus the reciprocal of the square root of frequency (ω−0.5) (Fig. 4b). PCFs displayed the smallest σ of 0.64 Ω s−0.5, followed by PCF@MnO2-1h (1.18 Ω s−0.5) and PCF@MnO2-2h (1.68 Ω s−0.5). The slight increase in σ is in accordance with the fact that the pseudocapacitive reactions are slower than the adsorption-desorption of ions pertaining to the electrical double layer capacitive processes, as well as that the mesopore size is reduced. Despite the increase, the σ values of our PCF-based electrodes were remarkably smaller than other MnO2-based materials (Fig. 4c). Notably, the σ value of PCF@MnO2-2h was even lower than that of CF (Fig. 4c and Supplementary Fig. 9), a mostly electrical double layer capacitive (EDLC) material that has fast ion diffusion kinetics. In addition, the σ value of PCF@MnO2-2h ( < 2 Ω s−0.5) is ~3.5 times lower than that of CF@MnO2-2h (~7 Ω s−0.5), highlighting the critical role of the uniform distributed, randomly oriented, and interconnected mesopores in accelerating electrolyte infiltration and ion diffusion in block copolymer-derived PCFs.
Pseudocapacitive performance
With continuous electron conduction and ultra-low ion diffusion resistivity, PCF@MnO2-2h exhibited ultra-fast charge and discharge kinetics. The cyclic voltammograms (CVs) of PCF@MnO2-2h were nearly rectangular (Fig. 5a), reflecting the rapid electron and ion transport in the electrode54. The current density of a supercapacitor, i, scales with the scan rate, v, following the relationship of i= kvb. The power-law exponent, b, is an important metric to evaluate the charge-storage kinetics, and b = 1 for an ideal supercapacitor. By plotting the logarithm of the absolute cathodic current densities at 0.2 V against the logarithm of scan rates (Fig. 5b), the b-value was calculated to be 0.93 in the scan-rate range of 10-100 mV s−1, approaching that of an ideal capacitor (b = 1) and suggesting the ultra-fast charge-storage kinetics. Outstandingly, the b value decreases only slightly to 0.91 in the range of 10–1000 mV s−1, unambiguously confirming its fast charge-storage kinetics.
We further decoupled the capacitances from fast-kinetic processes and slow-kinetic processes. The decoupling is based on the different contributions of fast and slow kinetics processes in the current density of a CV curve (see Supplementary Methods for details). Briefly, the current density at a fixed potential and a scan rate, i is composed of two terms associated with the scan rate, v:
$$i = {\mathrm{k}}_1v + {\mathrm{k}}_2v^{0.5},$$
(1)
where k1 and k2 are constants. The first term k1v equals the current density contributed from fast-kinetic processes and the second term k2v0.5 is the current density associated with slow-kinetic (or diffusion-controlled) processes. Dividing v0.5 on both sides of Equation (1) gives:
$$i\,v^{ - 0.5} = {\mathrm{k}}_1v^{0.5} + {\mathrm{k}}_2.$$
(2)
Equation (2) shows that iv0.5 and v0.5 are expected to have a linear relationship, with k1 and k2 being the slope and the y-intercept, respectively. Repeating the above step at other scan rates reveals the current density contribution across the potential window and outlines the contribution from the fast-kinetic and slow-kinetic processes. Figure 5c shows an example of the decoupling of a CV at 100 mV s−1. The capacitive contribution from the fast-kinetic processes (yellow region) clearly dominates that of the slow-kinetic processes (blue region) at all scan rates (Fig. 5c–d, Supplementary Fig. 10). The slow-kinetic capacitance decreased with the increasing scan rate. Importantly, the electric double layer capacitance (Cdl) contributed only a small fraction in the fast-kinetics region (Fig. 5d, gray dashed line), indicating that the majority of the pseudocapacitance of PCF@MnO2-2h is not charge-transfer-limited or diffusion-controlled. The fast kinetics makes PCF@MnO2 a desirable pseudocapacitive electrode for rapid charge storage and release.
The Ragone plot (Supplementary Fig. 14) compares the specific energy and power densities of PCF@MnO2 with those of the MnO2 supported on graphene, a star material for supercapacitor electrodes. With a high gravimetric power density of 23.2 kW kg−1 and a high gravimetric energy density of 10.3 Wh kg−1 in the tested range of scan rates, PCF@MnO2-2h outperformed the various graphene- and CF-supported MnO2 electrodes in symmetric pseudocapacitors. The superior capacitive performance signifies that our PCF-supported MnO2 electrodes have realized both high mass loadings and ultrafast charge transport kinetics.
Discussion
This work signifies the great potential of leveraging the disparate and innovative concept of block copolymer microphase separation to design and fabricate mesoporous carbon fiber supports. We emphasize that the highly uniform mesopores are crucial for the high loading of guest materials and the efficient transport of ions. The block copolymer-derived PCFs revolutionize the porous carbon supports and are adaptable to a broad range of electrochemical applications including batteries, fuel cells, catalyst supports, and capacitive desalination devices.
Methods
Synthesis of porous carbon fiber mats
Porous carbon fiber (PCF) mats were derived from poly(acrylonitrile-block-methylmethacrylate) (PAN-b-PMMA) block copolymer. Briefly, PAN-b-PMMA (Mn = 110-b-60 kDa, polydispersity = 1.14) was synthesized via reversible addition-fragmentation chain-transfer polymerization56 and electrospun into a polymer fiber mat. The polymer fiber mat was cut into small stripes (e.g., 10 cm × 2 cm), loaded into a tube furnace (Thermo-Fisher Scientific, Model STF55433C-1), and then heated at 280 °C for 8 h (ramp rate: 1 °C min−1) in air. The heating process induced the microphase separation of PAN and PMMA, and it triggered the crosslinking and cyclization of PAN. The resulting brown mats were further heated at 1200 °C for 1 h (ramp rate: 10 °C min−1) under a nitrogen atmosphere. Afterwards, the tube furnace was cooled down to room temperature and PCF mats were obtained. The preparation of CF was similar except that PAN was used instead of PAN-b-PMMA.
Deposition of Manganese Dioxide
Manganese dioxide (MnO2) was deposited onto the PCF mats via a solution-based self-limiting redox reaction with potassium permanganate (KMnO4),
$$\begin{array}{l}4{\mathrm{KMnO}}_4 + 3{\mathrm{C}} + {\mathrm{H}}_2{\mathrm{O}} \to 4{\mathrm{MnO}}_2 + {\mathrm{K}}_2{\mathrm{CO}}_3 + 2{\mathrm{KHCO}}_3\end{array}$$
First, 0.032 g of KMnO4 powder was dissolved in 20 mL of deionized water and used as the deposition solution (KMnO4, 10 mM). The solution was then heated to 80 °C under ambient pressure. Approximately 10 mg of PCF mats were soaked in the solution for 1–2 h under gentle stirring. After the reaction, the KMnO4 solution was drained and the remaining carbon fiber mats were thoroughly washed with deionized water five times, followed by drying in a vacuum oven at 60 °C for 8 h. The resulting carbon fiber mats are designated as PCF@MnO2-1h and PCF@MnO2-2h based on the reaction times of 1 h and 2 h, respectively.
The mass loading of MnO2 was determined by calculating the mass difference between the PCF mats before and after the reaction (see Supporting Information for detailed calculations). The areal mass loadings of MnO2 in PCF@MnO2-1h and PCF@MnO2-2h were 2.6 ± 0.2 and 3.4 ± 0.4 mg cm−2, respectively. The total mass loadings (including PCF and MnO2) of PCF@MnO2-1h and PCF@MnO2-2h were 6.2 ± 0.3 and 6.8 ± 0.4 mg cm−2, respectively. The average thickness of all PCF, PCF@MnO2-1h and PCF@MnO2-2h mats was ~200 μm. Thus, the volumetric mass densities of PCF@MnO2-1h and PCF@MnO2-2h were 0.31 ± 0.02 and 0.34 ± 0.02 g cm−3, respectively. The standard deviations were based on at least three batches of carbon fiber based electrodes.
Electrochemical deposition was also adopted to prepare PCF@MnO2 electrodes with high mass loadings. The electrodeposition solution contained 0.1 M manganese acetate and 0.5 M lithium chloride (a supporting electrolyte) in deionized water. A piece of PCF carbon fiber mat, a piece of nickel foam, and an Ag/AgCl wire in saturated KCl aqueous solution were used as the working electrode, the counter electrode, and the reference electrode, respectively. The electrodes were connected to an electrochemical workstation (PARSTATS 4000+, Princeton Applied Research, Ametek Inc.) and scanned between 0 and 1.0 V vs. Ag/AgCl at a scan rate of 0.01 mV s−1 for 15 cycles. The mass loading of the electrodeposited MnO2 on the PCF was 4.2 mg cm−2. The total mass loading (including PCF and MnO2) from electrodeposition was ~8.0 mg cm−2.
Physical Characterizations
The carbon fibers were characterized using scanning electron microscopy (SEM, LEO Zeiss 1550, acceleration voltage: 2 kV) and high-resolution transmission electron microscopy (HRTEM, FEI TITAN 300, acceleration voltage: 300 kV). The physisorption isotherms were measured with a pore analyzer (3Flex Pore Analyzer, Micromeritics Instrument Corp.) using nitrogen (for mesopores) and carbon dioxide (for micropores). Prior to the sorption tests, all electrodes were heated at 90 °C for 60 min and then at 200 °C for 900 min in N2 to desorb any moisture and hydrocarbons. The ramping rate of both heating processes was 10 °C min−1. The surface areas were calculated using the Brunauer-Emmett-Teller (BET) method, and the pore size distributions were obtained by the density functional theory. X-ray photoelectron spectroscopy (XPS) spectra were acquired using monochromatic Al Kα X-ray source (1486.6 eV) with a 200 µm X-ray beam at an incident angle of 45°. All binding energies were referenced to adventitious C 1 s at 284.8 eV. Chemical states of elements were assigned based on the National Institute of Standards and Technology (NIST) XPS Database. Raman spectra were recorded by a Raman spectrometer (WITec alpha 500) coupled with a confocal Raman microscope using a laser excitation wavelength of 633 nm. UV-vis spectra were measured by an Agilent Cary 60 UV-vis spectrometer. Small angle X-ray scattering (SAXS) spectra were collected by a Bruker N8 Horizon instrument with Cu Kα radiation (λ = 1.54 Å) at a current of 1 mA and a generator voltage of 50 kV.
Electrochemical characterizations
The electrochemical performance was evaluated in a symmetric two-electrode configuration in an aqueous electrolyte of 6 M KOH. For consistency, carbon fiber mats were sandwiched between two pieces of nickel foams (EQ-bcnf-80 μm, MTI corporation). Cyclic voltammograms were collected within a potential window of 0–0.8 V at various scan rates of 10–1000 mV s−1. Galvanostatic charge and discharge (GCD) were performed within the same potential window (0–0.8 V). Electrochemical impedance spectroscopy was conducted at open circuit potentials with frequencies between 0.1 Hz and 100 kHz with a perturbation of 5 mV. The CVs and EIS were recorded using a PARSTATS 4000 + electrochemical workstation (Princeton Applied Research, Ametek Inc.). The GCD curves were acquired from a charge-discharge cycler (Model 580, Scribner Associates Inc.).
Data Availability
Data supporting the findings of this study are available in this paper and in the Supplementary Information or are available from the corresponding author upon reasonable request.
Journal peer review information Nature Communications thanks Christopher Arges and the other anonymous reviewers for their contribution to the peer review of this work. Peer reviewer reports are available.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Acknowledgements
This material is based upon work supported by the Air Force Office of Scientific Research under award number FA9550-17-1-0112 through the Young Investigator Program (YIP). Guoliang Liu acknowledges the American Chemical Society Petroleum Research Foundation for the Doctoral New Investigator (DNI) award. The authors acknowledge the use of electron microscopes in the Virginia Tech Institute for Critical Technology and Applied Science (ICTAS), and Dr. Xu Feng for assistance in the XPS analysis at the Surface Analysis Laboratory of Virginia Tech. The XPS is supported by the National Science Foundation under Grant No. CHE-1531834.
Author information
Affiliations
1. Department of Chemistry, Virginia Tech, Blacksburg, VA, 24061, USA
• Tianyu Liu
• , Zhengping Zhou
• , Yichen Guo
• , Dong Guo
• & Guoliang Liu
2. Macromolecules Innovation Institute, Virginia Tech, Blacksburg, VA, 24061, USA
• Guoliang Liu
3. Division of Nanoscience, Virginia Tech, Blacksburg, VA, 24061, USA
• Guoliang Liu
Contributions
G.L. and T.L. designed all the experiments. T.L. performed the electrochemical and physical tests. Z.Z. participated in the synthesis of block copolymers and helped with the electrochemical and physical sorption tests. Y.G. collected the SEM images. D.G. synthesized the PAN homopolymers (the CF precursors). G.L. and T.L. wrote the paper with input from all authors.
Competing interests
The authors declare no competing interests.
Corresponding author
Correspondence to Guoliang Liu. | open-web-math/open-web-math | |
Length of a Curve
Formula of Length of a Curve
For a function $f$ that is continuous on the $[a,b]$, the length of the curve $y = f(x)$ from $a$ to $b$ is given by [1] [2] [3] $\int_{a}^{b} \; \sqrt{1+\left( \dfrac{df}{dx} \right)^2 }\; dx$
Examples and Solutions
Example 1
Find the length of the arc of the parabola $y = 0.1 x^2 + 2$ between the points $(-15,24.5)$ and $(10,12)$.
Solution to Example 1
We first calculate the derivative
$\dfrac{dy}{dx} = 0.2 x$
Use the formula for the
arc length given above
$L = \int_{-15}^{10} \; \sqrt{1+\left( 0.2 x \right)^2 }\; dx$
Use
trigonometric substitution $\quad \tan u = 0.2 x$
Take the derivative of both sides of the above substitution
$\sec ^2 (x) \dfrac{du}{dx} = 0 .2$ which gives $dx = 5 \sec ^2 u du$
Solving $\quad \tan u = 0.2 x$ for $u$ gives $u = \arctan (0.2 x)$
Limits of integration after substitution
$u_1 = \arctan (0.2(-15)) \approx -1.24904$ when $x = - 15$ the lower limit of the integral
$u_2 = \arctan (0.2(10)) \approx 1.10714$ when $x = 10$ the upper limit of the integral
$L = 5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; \sqrt{1+\tan^2 u }\; dx$
Use the identity $\sqrt{1 + \tan^2 u } = |\sec u|$ and make the substitution $\quad dx = 5 \sec ^2 u du$ to obtain
$5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; |\sec u| \sec ^2u \; du$
For $u$ the interval $\left[ \arctan (0.2(-15)) , \arctan (0.2(10)) \right]$, $\sec u$ is positive. Hence the integral becomes
$L = 5 \int_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \; \sec ^3u \; du$
Use the indefinite
integral of $\sec ^3u$ given by $\int \sec^3 x \; dx = \dfrac{1}{2} \left( \tan x \; \sec x + \ln |\tan x + \sec x| \right) + c$ to evaluate the arc length
$L = \left[ \dfrac{1}{2} \left( \sec u \tan u + \ln| \sec u + \tan u| \right) \right]_{\arctan (0.2(-15))}^{\arctan (0.2(10))} \\\\ \approx 43.05$
Example 2
Find the length of the arc along the curve $f(x) = \ln(\sin x)$ between the points $(\dfrac{\pi}{4}, f(\dfrac{\pi}{4}))$ and $(\dfrac{\pi}{2}, f(\dfrac{\pi}{2}))$.
Solution to Example 2
Calculate the derivative $\dfrac{dy}{dx} = \dfrac{\ln(\sin x)}{dx} \\ = \cot x$
Applying the formula for the arc alength
$L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; \sqrt{1+( \cot x)^2 } \; dx$
Use the
trigonometric identity $1+(\cot x)^2 = csc^2 x$, $L$ becomes
$L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; |\csc x| \; dx$
$\csc x$ is positive on the the closed interval of integration $[ \dfrac{\pi}{4} , \dfrac{\pi}{2} ]$ and therefore $|\csc x| = \csc x$ and $L$ becomes
$L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \; \csc x \; dx$
Use the common
integral of $\csc x$ : $\displaystyle \int \csc x \; dx = \ln \left|\tan \left(\dfrac{x}{2}\right)\right|+C$ to write $L = \left[ \ln \left|\tan \left(\dfrac{x}{2}\right)\right| \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \approx 0.88137$
Example 3
Find the length of the arc along the half circle given by $f(x) = 2 + \sqrt {9 - (x+2)^2}$ between the points $(-4, f(-4)$ and $(0, f(0)$.
Solution to Example 3
Calculate the derivative $\dfrac{df}{dx} = -\dfrac{x+2}{\sqrt{9 - (x+2)^2}}$
Applying the formula for the arc alength
$L = \int_{-4}^{0} \; \sqrt{1+( -\dfrac{x+2}{\sqrt{9 - (x+2)^2}} )^2 } \; dx$
The above integral is a challenging one and therefore
Symbolab software was used to approximate the above integral numerically. $L \approx 4.29$ | open-web-math/open-web-math | |
# How to prove that a series expansion of pi has converged to a certain accuracy?
Wikipedia has a great article on methods for calculating pi with arbitrary precision, using for example Machin's infinite series expansion:
$\frac{\pi}{ 4} = 4$ arccot $5 -$arccot $239$
where
arccot $x = \frac{1}{x} - \frac{1}{3 x^3} + \frac{1}{5 x^5} - \frac{1}{7 x^7} + \dots$
The only problem is that they never mention how to prove how many digits you have accurately computed pi to. How is this done? In general, how do you prove this for any series that converges?
Edit: I used pi with an alternating series as an example, but I'm more interested in general techniques for any convergent series.
-
In a convergent alternating series with terms decreasing in absolute value, the error is always less that the size of the first omitted term. – Mark Bennet Jul 31 '12 at 20:52
The series that you mention are alternating series. That means that the terms alternate in sign, go down in absolute value, and the absolute values of the terms approach $0$.
For alternating series, the error made by truncating at a particular place has absolute value less than the first omitted term. This criterion is easy to use, and ordinarily one cannot do substantially better. The sign of the first neglected term also tells you the direction of the error. If it is negative, truncation produces an overestimate. If it is positive, then truncation produces an underestimate.
-
What about non-alternating series? – Mike Izbicki Jul 31 '12 at 20:59
It can get complicated. But for power series we can often get an estimate by bounding above using a suitable geometric series. For example, suppose we are dealing with the series for $e^x$, $x=1/2$, or anyway positive and less than $1$. If we cut off at the term $(1/9!)x^9$, our error is less than $(1/10!)(x^{10}+x^{11}+\cdots)$ (we can do better). – André Nicolas Jul 31 '12 at 21:08
Each of the usual convergence tests corresponds to a method of estimating the error. Thus if $|a_n| < b_n$ and $\sum_n b_n$ converges, the error in $\sum_{n\le N} a_n$ is bounded by $\sum_{n > N} b_n$. – Robert Israel Jul 31 '12 at 21:09 | HuggingFaceTB/finemath | |
1. ## The code
Lol I was expecting something more explicit with all the trouble I had but I was wrong. Still awesome. Not sure if I should post this though seems like I'm defeating the purpose...
http://www.itchstudios.com/psg/main.php?id=misc
5, 12, 17, 29, 46, ?
Solution
5+12=17
12+17=29
17+29=46
29+46=75
...followed by:
5, 12, 21, 32, 45, 60, ?
12-5=7
5+7=12
21-12=9
12+9=21
32-21=11
21+11=32
45-32=13
32+13=45
60-45=15
45+15=60
X-60=(15+2) (Note the subtraction equations have always equaled 2+ the previous answer so this one is 15+2=17
X=60+17
X=77
(This one was hard I thought it was 75 for a while)
...and finish with:
oboe, fobo, ofob, cofo, ocof, ?
This ones hard but got it faster then second one...
obo(e)---> (e)obo---->(f)obo (Next letter in alphabet is f)
fob(o)---> (o)fob (Last letter is moved to front)
ofo(b)---> (b)ofo ----> (c)ofo
cof(o)---> (o)cof
oco(f)---> (f)oco----> (g)oco= goco
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3. Originally Posted by arthur 'two sheds'
looooooooooooooool
4. 1
1,1
2,1
1,2,1,1
1,1,1,2,2,1
..?
5. I see you've played knifey spooney before.
If a = x [true for some a's and x's]
then a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]
6. @ Reymus: You divide by zero. That's a no go:
7. ## The Following User Says Thank You to Camilla For This Useful Post:
8. "@ Reymus: You divide by zero. That's a no go:"
Except, if you use imaginary numbers.
9. exciting
10. I seems like I left some older versions of the problems in the source code of that page.
That gallery itself is kind of embarrassing, I need to prune it.
11. Procrastinator
Join Date
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Location
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Posts
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Originally Posted by Prometheus|ANJ
I seems like I left some older versions of the problems in the source code of that page.
That gallery itself is kind of embarrassing, I need to prune it.
I actually figured the code out a little while ago before this thread was posted.
YOU ARE A MONSTER
12. Originally Posted by Flashback
"@ Reymus: You divide by zero. That's a no go:"
Except, if you use imaginary numbers.
Err ... perhaps! Now, I haven't worked with imaginary numbers for ages, so I am very rusty in this aspect. Could you enlighten me? (Just bring on the formulas :-))
13. ## The Following User Says Thank You to Camilla For This Useful Post:
14. Correction: The simplest sense, divide by zero is under defined, so as it approaches the zero it curves violently, getting close to zero but not touching it creating an Asymptote. This kind of creates infinity.
In computer, divide by zero would cause a program to crash. Sort of creating a infinite loop.
real numbers would be 5, -4.1, 54/2, 0, 3.3434343434343434343434.........
Imaginary number are when you square root -1 or non-positive numbers.
Last edited by Flashback; May 18th, 2010 at 08:20 AM.
15. ## The Following User Says Thank You to Flashback For This Useful Post:
16. Originally Posted by Arthur 'Two Sheds'
lol, still can't get over it......so to the point!
17. Originally Posted by Flashback
Correction: The simplest sense, divide by zero is under defined, so as it approaches the zero it curves violently, getting close to zero but not touching it creating an Asymptote. This kind of creates infinity.
...
Imaginary number are when you square root -1 or non-positive numbers.
If a=x then a-x=0. That is a straight, round zero, not something that approaches zero.
I realize that imaginary numbers enable you to square root negative numbers. What I don't remember is, what they "say" about dividing by zero.
However, I really doubt that any creative juggling with imaginary numbers would make the 2=1 end result of Reymus' proof correct, since this would go against one of the most basic axioms of math, 1+1=2.
Please show me how imaginary numbers would solve Reymus' riddle and make 2=1 a valid result.
18. ## The Following User Says Thank You to Camilla For This Useful Post:
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Related Articles
Check if four segments form a rectangle
• Difficulty Level : Medium
• Last Updated : 04 Oct, 2018
We are given four segments as a pair of coordinates of their end points. We need to tell whether those four line segments make a rectangle or not.
Examples:
```Input : segments[] = [(4, 2), (7, 5),
(2, 4), (4, 2),
(2, 4), (5, 7),
(5, 7), (7, 5)]
Output : Yes
Given these segment make a rectangle of length 3X2.
Input : segment[] = [(7, 0), (10, 0),
(7, 0), (7, 3),
(7, 3), (10, 2),
(10, 2), (10, 0)]
Output : Not
These segments do not make a rectangle.
Above examples are shown in below diagram. ```
This problem is mainly an extension of How to check if given four points form a square
## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
We can solve this problem by using properties of a rectangle. First, we check total unique end points of segments, if count of these points is not equal to 4 then the line segment can’t make a rectangle. Then we check distances between all pair of points, there should be at most 3 different distances, one for diagonal and two for sides and at the end we will check the relation among these three distances, for line segments to make a rectangle these distance should satisfy Pythagorean relation because sides and diagonal of rectangle makes a right angle triangle. If they satisfy mentioned conditions then we will flag polygon made by line segment as rectangle otherwise not.
`// C++ program to check whether it is possible ``// to make a rectangle from 4 segments ``#include ``using` `namespace` `std; ``#define N 4 `` ` `// structure to represent a segment ``struct` `Segment ``{ `` ``int` `ax, ay; `` ``int` `bx, by; ``}; `` ` `// Utility method to return square of distance ``// between two points ``int` `getDis(pair<``int``, ``int``> a, pair<``int``, ``int``> b) ``{ `` ``return` `(a.first - b.first)*(a.first - b.first) + `` ``(a.second - b.second)*(a.second - b.second); ``} `` ` `// method returns true if line Segments make ``// a rectangle ``bool` `isPossibleRectangle(Segment segments[]) ``{ `` ``set< pair<``int``, ``int``> > st; `` ` ` ``// putiing all end points in a set to `` ``// count total unique points `` ``for` `(``int` `i = 0; i < N; i++) `` ``{ `` ``st.insert(make_pair(segments[i].ax, segments[i].ay)); `` ``st.insert(make_pair(segments[i].bx, segments[i].by)); `` ``} `` ` ` ``// If total unique points are not 4, then `` ``// they can't make a rectangle `` ``if` `(st.size() != 4) `` ``return` `false``; `` ` ` ``// dist will store unique 'square of distances' `` ``set<``int``> dist; `` ` ` ``// calculating distance between all pair of `` ``// end points of line segments `` ``for` `(``auto` `it1=st.begin(); it1!=st.end(); it1++) `` ``for` `(``auto` `it2=st.begin(); it2!=st.end(); it2++) `` ``if` `(*it1 != *it2) `` ``dist.insert(getDis(*it1, *it2)); `` ` ` ``// if total unique distance are more than 3, `` ``// then line segment can't make a rectangle `` ``if` `(dist.size() > 3) `` ``return` `false``; `` ` ` ``// copying distance into array. Note that set maintains `` ``// sorted order. `` ``int` `distance; `` ``int` `i = 0; `` ``for` `(``auto` `it = dist.begin(); it != dist.end(); it++) `` ``distance[i++] = *it; `` ` ` ``// If line seqments form a square `` ``if` `(dist.size() == 2) `` ``return` `(2*distance == distance); `` ` ` ``// distance of sides should satisfy pythagorean `` ``// theorem `` ``return` `(distance + distance == distance); ``} `` ` `// Driver code to test above methods ``int` `main() ``{ `` ``Segment segments[] = `` ``{ `` ``{4, 2, 7, 5}, `` ``{2, 4, 4, 2}, `` ``{2, 4, 5, 7}, `` ``{5, 7, 7, 5} `` ``}; `` ` ` ``(isPossibleRectangle(segments))?cout << ``"Yes\n"``:cout << ``"No\n"``; ``} `
Output:
```Yes
```
This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | HuggingFaceTB/finemath | |
# Math
what is 2/5 divided by 15/7?
1. 👍
2. 👎
3. 👁
1. (2/5) / (15/7)
(2/5) * (7/15) = 14/75
1. 👍
2. 👎
👤
Ms. Sue
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3. ### math
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4. ### Math
1.62 divided by 0.8 16.2 divided by 8 0.0162 divided by 0.008 0.162 divided by 0.08 There are actually two different ways to complete the expressions above with the given numbers so that each expression has the same value. What | HuggingFaceTB/finemath | |
## Solve Lesson 55 Page 63 SBT Math 10 – Kite>
Topic
Uncle Nam plans to make a rectangular picture frame so that the inner part of the frame is a rectangle with dimensions of 6 cm x 11 cm, the width of the border is $$x$$ cm (Figure 27). The area of the frame border should not exceed $$38c{m^2}$$. What is the width of the largest frame border in cm?
Solution method – See details
Set the width of the frame border to $$x$$(cm) ($$x > 0$$). Representing the area of the frame border and solving the inequalities
Detailed explanation
Set the width of the frame border to $$x$$(cm) ($$x > 0$$).
We have the frame border area $$\left( {11 + 2x} \right)\left( {6 + 2x} \right) – 66 = 4{x^2} + 34x$$ ($$c{) m^2}$$)
According to the problem we have: $$4{x^2} + 34x \le 38 \Leftrightarrow 4{x^2} + 34x – 38 \le 0$$
The quadratic triangle $$4{x^2} + 34x – 38$$ has two solutions $${x_1} = \frac{{ – 19}}{2};{x_2} = 1$$ and has coefficients $$a = 4 > 0$$
Using the sign theorem of quadratic triangles, we see that the set of values of $$x$$ such that the triangle $$4{x^2} + 34x – 38$$ bears the sign “-” is \ (\left[ {\frac{{ – 19}}{2};1} \right]\)
So $$0 < x \le 1$$
So the maximum width of the frame border is 1 cm. | HuggingFaceTB/finemath | |
MOSEK can solve quadratic and quadratically constrained problems, as long as they are convex. This class of problems can be formulated as follows:
(6.2)$\begin{split}\begin{array}{lrcccll} \mbox{minimize} & & & \half x^T Q^o x + c^T x + c^f & & & \\ \mbox{subject to} & l_k^c & \leq & \half x^T Q^k x + \sum_{j=0}^{n-1} a_{k,j} x_j & \leq & u_k^c, & k =0,\ldots ,m-1, \\ & l_j^x & \leq & x_j & \leq & u_j^x, & j=0,\ldots ,n-1. \end{array}\end{split}$
Without loss of generality it is assumed that $$Q^o$$ and $$Q^k$$ are all symmetric because
$x^T Q x = \half x^T(Q+Q^T)x.$
This implies that a non-symmetric $$Q$$ can be replaced by the symmetric matrix $$\half(Q+Q^T)$$.
The problem is required to be convex. More precisely, the matrix $$Q^o$$ must be positive semi-definite and the $$k$$th constraint must be of the form
(6.3)$l_k^c \leq \half x^T Q^k x + \sum_{j=0}^{n-1} a_{k,j} x_j$
with a negative semi-definite $$Q^k$$ or of the form
$\half x^T Q^k x + \sum_{j=0}^{n-1} a_{k,j} x_j \leq u_k^c.$
with a positive semi-definite $$Q^k$$. This implies that quadratic equalities are not allowed. Specifying a non-convex problem will result in an error when the optimizer is called.
A matrix is positive semidefinite if all the eigenvalues of $$Q$$ are nonnegative. An alternative statement of the positive semidefinite requirement is
$x^T Q x \geq 0, \quad \forall x.$
If the convexity (i.e. semidefiniteness) conditions are not met MOSEK will not produce reliable results or work at all.
We look at a small problem with linear constraints and quadratic objective:
(6.4)$\begin{split}\begin{array}{lll} \mbox{minimize} & & x_1^2 + 0.1 x_2^2 + x_3^2 - x_1 x_3 - x_2 \\ \mbox{subject to} & 1 \leq & x_1 + x_2 + x_3 \\ & 0 \leq & x. \end{array}\end{split}$
The matrix formulation of (6.4) has:
$\begin{split}Q^o = \left[ \begin{array}{ccc} 2 & 0 & -1\\ 0 & 0.2 & 0\\ -1 & 0 & 2 \end{array} \right], c = \left[ \begin{array}c 0\\ -1\\ 0 \end{array} \right], A = \left[ \begin{array} {ccc} 1 & 1 & 1 \end{array} \right],\end{split}$
with the bounds:
$\begin{split}l^c = 1, u^c = \infty , l^x = \left[ \begin{array}c 0 \\ 0 \\ 0 \end{array} \right] \mbox{ and } u^x = \left[ \begin{array} c \infty \\ \infty \\ \infty \end{array} \right]\end{split}$
Please note the explicit $$\half$$ in the objective function of (6.2) which implies that diagonal elements must be doubled in $$Q$$, i.e. $$Q_{11}=2$$ even though $$1$$ is the coefficient in front of $$x_1^2$$ in (6.4).
Setting up the linear part
The linear parts (constraints, variables, objective) are set up using exactly the same methods as for linear problems, and we refer to Sec. 6.1 (Linear Optimization) for all the details. The same applies to technical aspects such as defining an optimization task, retrieving the solution and so on.
The quadratic objective is specified using the function Task.putqobj. Since $$Q^o$$ is symmetric only the lower triangular part of $$Q^o$$ is inputted. In fact entries from above the diagonal may not appear in the input.
The lower triangular part of the matrix $$Q^o$$ is specified using an unordered sparse triplet format (for details, see Sec. 15.1.4 (Matrix Formats)):
int[] qsubi = {0, 1, 2, 2 };
int[] qsubj = {0, 1, 0, 2 };
double[] qval = {2.0, 0.2, -1.0, 2.0};
• only non-zero elements are specified (any element not specified is 0 by definition),
• the order of the non-zero elements is insignificant, and
• only the lower triangular part should be specified.
Finally, this definition of $$Q^o$$ is loaded into the task:
task.putqobj(qsubi, qsubj, qval);
Source code
Listing 6.2 Source code implementing problem (6.4). Click here to download.
package com.mosek.example;
import mosek.*;
public class qo1 {
static final int numcon = 1; /* Number of constraints. */
static final int numvar = 3; /* Number of variables. */
static final int NUMANZ = 3; /* Number of numzeros in A. */
static final int NUMQNZ = 4; /* Number of nonzeros in Q. */
public static void main (String[] args) {
// Since the value infinity is never used, we define
// 'infinity' symbolic purposes only
double infinity = 0;
double[] c = {0.0, -1.0, 0.0};
mosek.boundkey[] bkc = { mosek.boundkey.lo };
double[] blc = {1.0};
double[] buc = {infinity};
mosek.boundkey[] bkx = { mosek.boundkey.lo,
mosek.boundkey.lo,
mosek.boundkey.lo
};
double[] blx = {0.0,
0.0,
0.0
};
double[] bux = {infinity,
infinity,
infinity
};
int[][] asub = { {0}, {0}, {0} };
double[][] aval = { {1.0}, {1.0}, {1.0} };
double[] xx = new double[numvar];
try (Env env = new Env();
// Directs the log task stream to the user specified
mosek.streamtype.log,
new mosek.Stream()
{ public void stream(String msg) { System.out.print(msg); }});
/* Give MOSEK an estimate of the size of the input data.
This is done to increase the speed of inputting data.
However, it is optional. */
/* Append 'numcon' empty constraints.
The constraints will initially have no bounds. */
/* Append 'numvar' variables.
The variables will initially be fixed at zero (x=0). */
for (int j = 0; j < numvar; ++j) {
/* Set the linear term c_j in the objective.*/
/* Set the bounds on variable j.
blx[j] <= x_j <= bux[j] */
/* Input column j of A */
asub[j], /* Row index of non-zeros in column j.*/
aval[j]); /* Non-zero Values of column j. */
}
/* Set the bounds on constraints.
for i=1, ...,numcon : blc[i] <= constraint i <= buc[i] */
for (int i = 0; i < numcon; ++i)
/*
The lower triangular part of the Q
matrix in the objective is specified.
*/
int[] qsubi = {0, 1, 2, 2 };
int[] qsubj = {0, 1, 0, 2 };
double[] qval = {2.0, 0.2, -1.0, 2.0};
/* Input the Q for the objective. */
/* Solve the problem */
System.out.println (" Mosek warning:" + r.toString());
// Print a summary containing information
// about the solution for debugging purposes
mosek.solsta solsta[] = new mosek.solsta;
/* Get status information about the solution */
/* Get the solution */
xx);
switch (solsta) {
case optimal:
System.out.println("Optimal primal solution\n");
for (int j = 0; j < numvar; ++j)
System.out.println ("x[" + j + "]:" + xx[j]);
break;
case dual_infeas_cer:
case prim_infeas_cer:
System.out.println("Primal or dual infeasibility\n");
break;
case unknown:
System.out.println("Unknown solution status.\n");
break;
default:
System.out.println("Other solution status");
break;
}
}
catch (mosek.Exception e) {
System.out.println ("An error/warning was encountered");
System.out.println (e.toString());
throw e;
}
} /* Main */
}
In this section we show how to solve a problem with quadratic constraints. Please note that quadratic constraints are subject to the convexity requirement (6.3).
Consider the problem:
$\begin{split}\begin{array}{lcccl} \mbox{minimize} & & & x_1^2 + 0.1 x_2^2 + x_3^2 - x_1 x_3 - x_2 & \\ \mbox{subject to} & 1 & \leq & x_1 + x_2 + x_3 - x_1^2 - x_2^2 - 0.1 x_3^2 + 0.2 x_1 x_3, & \\ & & & x \geq 0. & \end{array}\end{split}$
This is equivalent to
(6.5)$\begin{split}\begin{array}{lccl} \mbox{minimize} & \half x^T Q^o x + c^T x & & \\ \mbox{subject to} & \half x^T Q^0 x + A x & \geq & b, \\ & x\geq 0, \end{array}\end{split}$
where
$\begin{split}Q^o = \left[ \begin{array}{ccc} 2 & 0 & -1 \\ 0 & 0.2 & 0 \\ -1 & 0 & 2 \end{array} \right], c = \left[ \begin{array}{ccc} 0 &-1 & 0 \end{array} \right]^T, A = \left[ \begin{array}{ccc} 1 & 1 & 1 \end{array} \right], b = 1.\end{split}$
$\begin{split}Q^0 = \left[ \begin{array}{ccc} -2 & 0 & 0.2 \\ 0 & -2 & 0 \\ 0.2 & 0 & -0.2 \end{array} \right].\end{split}$
The linear parts and quadratic objective are set up the way described in the previous tutorial.
To add quadratic terms to the constraints we use the function Task.putqconk.
int[] qsubi = {0, 1, 2, 2 };
int[] qsubj = {0, 1, 2, 0 };
double[] qval = { -2.0, -2.0, -0.2, 0.2};
/* put Q^0 in constraint with index 0. */
qsubi,
qsubj,
qval);
While Task.putqconk adds quadratic terms to a specific constraint, it is also possible to input all quadratic terms in one chunk using the Task.putqcon function.
Source code
Listing 6.3 Implementation of the quadratically constrained problem (6.5). Click here to download.
package com.mosek.example;
import mosek.*;
public class qcqo1 {
static final int numcon = 1; /* Number of constraints. */
static final int numvar = 3; /* Number of variables. */
static final int NUMANZ = 3; /* Number of numzeros in A. */
static final int NUMQNZ = 4; /* Number of nonzeros in Q. */
public static void main (String[] args) {
// Since the value infinity is never used, we define
// 'infinity' symbolic purposes only
double infinity = 0;
double[] c = {0.0, -1.0, 0.0};
mosek.boundkey[] bkc = {mosek.boundkey.lo};
double[] blc = {1.0};
double[] buc = {infinity};
mosek.boundkey[] bkx
= {mosek.boundkey.lo,
mosek.boundkey.lo,
mosek.boundkey.lo
};
double[] blx = {0.0,
0.0,
0.0
};
double[] bux = {infinity,
infinity,
infinity
};
int[][] asub = { {0}, {0}, {0} };
double[][] aval = { {1.0}, {1.0}, {1.0} };
double[] xx = new double[numvar];
try (mosek.Env env = new mosek.Env();
// Directs the log task stream to the user specified
mosek.streamtype.log,
new mosek.Stream()
{ public void stream(String msg) { System.out.print(msg); }});
/* Give MOSEK an estimate of the size of the input data.
This is done to increase the speed of inputting data.
However, it is optional. */
/* Append 'numcon' empty constraints.
The constraints will initially have no bounds. */
/* Append 'numvar' variables.
The variables will initially be fixed at zero (x=0). */
for (int j = 0; j < numvar; ++j) {
/* Set the linear term c_j in the objective.*/
/* Set the bounds on variable j.
blx[j] <= x_j <= bux[j] */
/* Input column j of A */
asub[j], /* Row index of non-zeros in column j.*/
aval[j]); /* Non-zero Values of column j. */
}
/* Set the bounds on constraints.
for i=1, ...,numcon : blc[i] <= constraint i <= buc[i] */
for (int i = 0; i < numcon; ++i)
/*
* The lower triangular part of the Q
* matrix in the objective is specified.
*/
int[] qosubi = { 0, 1, 2, 2 };
int[] qosubj = { 0, 1, 0, 2 };
double[] qoval = { 2.0, 0.2, -1.0, 2.0 };
/* Input the Q for the objective. */
/*
* The lower triangular part of the Q^0
* matrix in the first constraint is specified.
* This corresponds to adding the term
* x0^2 - x1^2 - 0.1 x2^2 + 0.2 x0 x2
*/
int[] qsubi = {0, 1, 2, 2 };
int[] qsubj = {0, 1, 2, 0 };
double[] qval = { -2.0, -2.0, -0.2, 0.2};
/* put Q^0 in constraint with index 0. */
qsubi,
qsubj,
qval);
/* Solve the problem */
try {
} catch (mosek.Warning e) {
System.out.println (" Mosek warning:");
System.out.println (e.toString ());
}
// Print a summary containing information
// about the solution for debugging purposes
mosek.solsta solsta[] = new mosek.solsta;
/* Get status information about the solution */
xx);
switch (solsta) {
case optimal:
System.out.println("Optimal primal solution\n");
for (int j = 0; j < numvar; ++j)
System.out.println ("x[" + j + "]:" + xx[j]);
break;
case dual_infeas_cer:
case prim_infeas_cer:
System.out.println("Primal or dual infeasibility.\n");
break;
case unknown:
System.out.println("Unknown solution status.\n");
break;
default:
System.out.println("Other solution status");
break;
}
}
catch (mosek.Exception e) {
System.out.println ("An error/warning was encountered");
System.out.println (e.msg);
throw e;
}
} /* Main */
} | HuggingFaceTB/finemath | |
# Continuity (ii)
• Dec 13th 2009, 05:21 AM
nameck
Continuity (ii)
f(x) = { [(e^x) - 1] / x ; if x not equal 0
...... .{ b ...................; if x = 0
What value of b makes f continuous at x = 0?
so.. the left side and right side must be equal in order to make
f continuous at x = 0
[(e^x) - 1] / x = b
[(e^x) - 1] = (b)(x)
e^x = (b)(x) + 1
.
.
.
dont know how to proceed..
should i introduce ln?
• Dec 13th 2009, 05:32 AM
HallsofIvy
Quote:
Originally Posted by nameck
f(x) = { [(e^x) - 1] / x ; if x not equal 0
...... .{ b ...................; if x = 0
What value of b makes f continuous at x = 0?
so.. the left side and right side must be equal in order to make
f continuous at x = 0
No, $\displaystyle \frac{e^x-1}{x}$ is not defined at x= 0. It is the limit of that that must be equal to b in order that the function be continuous.
Take the limit, perhaps using L'Hopital's rule, to find b.
Quote:
[(e^x) - 1] / x = b
[(e^x) - 1] = (b)(x)
e^x = (b)(x) + 1
.
.
.
dont know how to proceed..
should i introduce ln?
• Dec 13th 2009, 05:57 AM
nameck
got it!!
b = 1
correct? | HuggingFaceTB/finemath | |
# Chi-square test with summary data in jamovi
Suppose you want to do a chi-square test for independence in jamovi, but you only have summary data. Fortunately it is super easy to do that in jamovi. Here is how.
This example is based on a question from an assignment I use in my Applied Statistics course (the assignment itself is from the instructor resources of the book Introduction to the New Statistics (first edition)).
The introductory text to the question is as follows.
To what extent might feeling powerful make you less considerate of the perspective of others? In one study (Galinsky et al., 2006), participants were manipulated to feel either powerful (High Power) or powerless (Low Power). They were then asked to write an ‘E’ on their forehead with a washable marker. Those who wrote the ‘E’ to be correctly readable from their own perspective—looking from inside the head—were considered ego-centric (Ego); those who
wrote it to be readable to others were considered to be non-ego-centric (Non-Ego).
Table 1 contains the data of the original study.
## Creating the dataset using summary data in jamovi
All you need to do is to a create a dataset with three variables. The first two variables are nominal variables. These variable define the rows and columns of your contingency table. Here, I opted for the variables Power, with levels 1 = High Power and 2 = Low Power and Perspective, with levels 1 = Ego and 2 = Non-Ego.
The third variable is the variable Counts (which can be nominal, ordinal and continuous, a far as I can tell). The count variable contains the number of observations for each combination of the two categorical variables.
This is what the dataset looks like:
## Doing the chi-square test
If you have the dataset, the rest is super easy as well. Just choose Frequencies on the Analyses tab followed by Independent samples. Now place your row, columns and counts variables in the right spot, as in Figure 2. That’s all! | HuggingFaceTB/finemath | |
# Gain of a negative feedback circuit (problem)
I am asked to find the gain A for this negative feedback circuit below:
(A: gain, G: all of the gain of the circuit, F: return ratio)
The simplified circuit above shows the gain A that I am trying to find.
I know that if this circuit had only one FET A = Vo/Vi=-gm*Rl. However, I just can't figure out how the gain A would change, if we were to add two additional FETs to the circuit. Any help would be appreciated.
By your own statement, each FET stage will have voltage gain:
$$A_1 = A_2 = A_3 = -g_mR_L$$
Each stage multiplies the voltage output of the previous stage, so if the first stage FET has gate voltage $$\V_{G1}\$$, the second gate is $$\V_{G2}\$$ etc., then you have:
\begin{aligned} V_{G2} &= A_1V_{G1} \\ \\ V_{G3} &= A_2V_{G2} \\ \\ V_{O} &= A_3V_{G3} \\ \\ &= A_3(A_2V_{G2}) \\ \\ &= A_3(A_2(A_1V_{G1})) \\ \\ &= (-g_mR_L)^3V_{G1} \\ \\ \\ A &= \frac{V_O}{V_{G1}} \\ \\ &= (-g_mR_L)^3 \end{aligned}
This number is negative (due to the odd number of stages), and is assumed to be huge. The circuit is functionally similar to the classic inverting amplifier employing an op-amp:
simulate this circuit – Schematic created using CircuitLab
No doubt you are familiar with the closed loop gain $$\G\$$ of such a system:
$$G = \frac{V_{OUT}}{V_{IN}} = -\frac{R_F}{R_S}$$
Interestingly, this is independent of $$\A\$$, which I'll touch upon below.
In your second circuit, the system is represented as a modular block diagram, where feedback is shown to be added to the original input.
Note: For feedback to be negative, one of the following conditions is necessary: gain $$\A\$$ should be negative, feedback factor $$\F\$$ should be negative, or the addition should actually be a subtraction. That diagram has none of those conditions, so feedback is positive, and the system is unstable. I have to assume this is an oversight on the author's part, somewhere there is a negation, and the author intended that feedback be negative.
In my op-amp circuit above, feedback is negative because the op-amp performs a subtraction of some fraction of the output, as indicated by the '−' symbol at the input terminal where feedback is applied.
In your first circuit, with the cascaded FET stages, feedback is also negative, because there are an odd number of individual negative-gain stages. Total gain from $$\V_{G1}\$$ to $$\V_O\$$ is therefore negative.
Anyway, the behaviour of a negative feedback system such as the one (almost) depicted in your second diagram, is well documented, and has this closed-loop gain:
$$\frac{v_{OUT}}{v_{IN}} = \frac{A}{1+AF}$$
Note that I've used lower-case $$\v\$$, suggesting that $$\v\$$ is referring to amplitude of some AC signal, rather than potential at any particular instant in time.
If gain $$\A\$$ is large enough, this becomes a very close approximation to:
$$\frac{v_{OUT}}{v_{IN}} = \lim_{A\rightarrow\infty}{\frac{A}{1+AF}} = \frac{1}{F}$$
I think that this exercise is designed to consolidate all these facts into a better understanding of:
• Many cascaded stages leads to very large open-loop gain $$\A\$$, which in a closed loop will result in a lower but precisely controllable gain: $$\frac{v_{OUT}}{v_{IN}} = \frac{1}{F}$$
This is independent of open-loop gain $$\A\$$, but does rely on $$\A\$$ being very large.
• The circuit with FET stages, and feedback and input resistances, closely resembles, and is functionally equivalent to, the classic inverting amplifier employing an op-amp, and has the same gain equation: $$\frac{V_{OUT}}{V_{IN}} = -\frac{R_F}{R_S}$$
Re-draw the schematic, using voltage controlled voltage sources as FET models. The voltage controlled voltage source (VCVS) G is an only component of the hybrid-pi model, a small-signal FET model, in which the output resistance is infinite and therefore can be omitted in the model. As you do not specify output resistance in your post, you are defaulting its value to infinity.
Examining the plot of a transient simulation you can easily verify that the stage gains are indeed multiplied. With gm=0.1 and RL=1000 the stage gain is 100, with three stages the total gain (open-loop gain of all three stages connected in series) is 1E+06, and the circuit closed-loop gain approaches a value of -Rf/Rs, independent of gm and RL, as if the circuit is an ideal operational amplifier.
What is more important for your problem, it is that the overall stage gain in this circuit is exactly factorized (expressed via a product of stage gains), as the inputs of stages do not load outputs (the input resistance of VCVS is infinite) and there are no internal feedbacks in the circuit.
Yes, the exact gain value does depend on gm and RL. And this model can be used not only for SPICE simulation, but, thanks to its factorization property, it simplifies symbolic calculations and helps you to arrive at an exact formula for the closed-loop gain, too. Because it is homework, do it!
• A story about this post. Noticed this question early. The question was candidly tagged "homework", but it looked as if the OP knows better, so a long while I just watched the community reaction. To my surprize, nobody'd requested "Datasheets, please" neither criticized the circuit drawing style or the other typical reaction of a certain SE 'experts'. Assuming the question is written in good faith, I ventured to write an answer covering maximum possible to be communicated to a struggling student short of directly doing their homework. Feb 5 at 5:15
• Later, another answer appears, looking as if the other student did the same homework and (mistakenly) turned their paper in to electronics.SE instead of their professor. Now, in my opinion, this post with its two answers embodies a good example of what has to be encouraged and what are not recommended practices while answering 'homework' questions. Regardless of whether there is any truth in my interpretation of intentions of other participants when writing their contributions, kudos to them for helping me streamline my understanding of SE's policies applicable to 'homework'-tagged questions! Feb 5 at 5:16 | HuggingFaceTB/finemath | |
US UKIndia
Every Question Helps You Learn
The angles of a triangle total 180o.
# Position 2
This Math quiz is called 'Position 2' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
Knowing position involves being able to read the coordinates of shapes on a grid. Coordinates are used for map reading and also in knowing where to place a specific point on a grid. But how do the coordinates change when shapes are rotated or reflected? Working out the new coordinates of a rotated shape can be difficult without a pencil and paper!
See how much you remember from your math classes by trying this quiz all about position.
1.
If I turn 60o how much more do I need to turn to complete one full turn?
100o
200o
300o
400o
360o - 60o = 300o
2.
A shape has the coordinates (5,1) (1,4) (5,6) (9,4). What shape is it?
Square
Rectangle
Rhombus
Kite
3.
What is the intersection?
The point where two angles meet
The point where two coordinates meet
The point where two shapes meet
The point where two lines meet
Where two roads meet it is also called an intersection
4.
If a triangle has a 60o angle and a 40o angle what angle is the third corner?
60o
70o
80o
90o
60o + 40o = 100o so 180o - 100o = 80o
5.
What is the total size of the other two angles in a right angled triangle?
30o
45o
60o
90o
180o - 90o = 90o so the other two angles must add up to that
6.
A rectangle has the coordinates (2,2) (2,4) (6,4). What is the fourth coordinate?
(2,6)
(6,2)
(1,6)
(4,6)
This rectangle would have its longest sides at the top and bottom
7.
A triangle has the coordinates (1,1) (1,4) (4,1). What will the coordinates be after a quarter turn clockwise?
(1,1) (4,1) (1,-2)
(1,-1) (4,1) (1,-4)
(1,1) (4,1) (1,4)
(1,1) (4,1) (4,-4)
If you worked that out without drawing on some paper then very well done!
8.
What do the angles of a triangle total?
180o
360o
200o
90o
No matter what lengths its sides are or what shape it is, the angles in a triangle always add up to 180o
9.
What is used to measure angles?
Ruler
Protractor
Tape measure
Compass
Most protractors are semi-circles divided into 180o
10.
What do we call two lines that cross one another?
Intersecting
Friends
Meeting
Crisscross
Intersecting lines meet but parallel lines never do
Author: Amanda Swift | HuggingFaceTB/finemath | |
# Optimal tax rate
Suppose you have two countries A and B, with a tax rate $T_A$ and $T_B$, respectively. The tax is redistributed to all people equally. Hence if you live in A and you make $I$ as income then you will finally receive
$$I*(1-T_A) + \overline{I}*T_A$$ where $\overline{I}$ is the average income in $A$. The country A wants to choose an optimal rate, in order to do this the decision is taken at the median income. But the people can migrate if the new rate makes them poorer than if they were living in $B$. Of course this migration to B comes at a cost $M$, hence if the median income choose as new rate $T$ the people in A such that $$I*(1-T) + \overline{I}*T < I*(1-T_B) + \overline{I}*T_B -M$$ will leave A for B. And symmetrically the people in B such that $$I*(1-T_B) + \overline{I}*T_B < I*(1-T) + \overline{I}*T -M$$ will leave B to A. Which changes the configuration of incomes in A and hence the decision of the median income, since his income depends on the average income.
My question is how can we find the tax rate which will optimize the income of the median income after migration?
I have thought of a dynamical approach, but it looks hard to show that we converge to an equilibrium. Are there is general tools for this kind of problem?
-
It seems to me that the optimum value of $T$ will depend on the distribution of incomes, $p(I)$. Do you have a reason to think that it won't? – Chris Taylor May 29 '11 at 19:53
Of course, since if all the income are the same, every taxe rat are optimal. Hence my question is : a distribution of incomes given, how to compute the optimal tax rate? – Paul May 30 '11 at 5:32
I suppose that $T_B$ is given as well, right? Otherwise this would seem like a game theory problem where we are looking for the Nash equilibrium values of the tax rates $T_A$ and $T_B$. (It's been a while since you posted this. Any progress?) – jmbejara Nov 4 '13 at 21:14 | HuggingFaceTB/finemath | |
# 42.1 kg to lbs - 42.1 kilograms to pounds
Do you want to learn how much is 42.1 kg equal to lbs and how to convert 42.1 kg to lbs? You are in the right place. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to underline that whole this article is devoted to a specific number of kilograms - that is one kilogram. So if you need to know more about 42.1 kg to pound conversion - keep reading.
Before we go to the more practical part - it means 42.1 kg how much lbs conversion - we will tell you few theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 42.1 kg to lbs? 42.1 kilograms it is equal 92.814612302 pounds, so 42.1 kg is equal 92.814612302 lbs.
## 42.1 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, known also as International System of Units (in short form SI).
Sometimes the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
The kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. This definition was not complicated but difficult to use.
Then, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “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 exactly 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.
## 42.1 kilogram to pounds
You learned a little bit about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to highlight that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to focus only on pound-mass.
The pound is used 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 “.
The international avoirdupois pound has no descriptive definition. It is equal 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given 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 42.1 kg?
42.1 kilogram is equal to 92.814612302 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 42.1 kg in lbs
Theoretical section is already behind us. In this part we are going to tell you how much is 42.1 kg to lbs. Now you know that 42.1 kg = x lbs. So it is time to get the answer. Just look:
42.1 kilogram = 92.814612302 pounds.
That is an exact outcome of how much 42.1 kg to pound. It is possible to also round off this result. After rounding off your result will be as following: 42.1 kg = 92.62 lbs.
You learned 42.1 kg is how many lbs, so let’s see how many kg 42.1 lbs: 42.1 pound = 0.45359237 kilograms.
Obviously, in this case you may also round off the result. After it your result will be exactly: 42.1 lb = 0.45 kgs.
We are also going to show you 42.1 kg to how many pounds and 42.1 pound how many kg results in tables. Look:
We are going to start with a chart for how much is 42.1 kg equal to pound.
### 42.1 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
42.1 92.814612302 92.620
Now see a chart for how many kilograms 42.1 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
42.1 0.45359237 0.45
Now you know how many 42.1 kg to lbs and how many kilograms 42.1 pound, so we can go to the 42.1 kg to lbs formula.
### 42.1 kg to pounds
To convert 42.1 kg to us lbs a formula is needed. We are going to show you two formulas. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 92.814612302 result in pounds
The first version of a formula will give you the most accurate outcome. In some cases even the smallest difference could be considerable. So if you need an exact outcome - this formula will be the best for you/option to convert how many pounds are equivalent to 42.1 kilogram.
So go to the second formula, which also enables conversions to know how much 42.1 kilogram in pounds.
The another formula is down below, look:
Amount of kilograms * 2.2 = the result in pounds
As you can see, this version is simpler. It could be better choice if you want to make a conversion of 42.1 kilogram to pounds in quick way, for instance, during shopping. You only need to remember that your outcome will be not so accurate.
Now we are going to learn you how to use these two formulas in practice. But before we will make a conversion of 42.1 kg to lbs we want to show you another way to know 42.1 kg to how many lbs without any effort.
### 42.1 kg to lbs converter
An easier way to know what is 42.1 kilogram equal to in pounds is to use 42.1 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first formula which we gave you in the previous part of this article. Thanks to 42.1 kg pound calculator you can quickly convert 42.1 kg to lbs. You only have to enter number of kilograms which you want to convert and click ‘calculate’ button. You will get the result in a second.
So let’s try to convert 42.1 kg into lbs with use of 42.1 kg vs pound calculator. We entered 42.1 as a number of kilograms. This is the result: 42.1 kilogram = 92.814612302 pounds.
As you can see, this 42.1 kg vs lbs converter is easy to use.
Now we can move on to our primary issue - how to convert 42.1 kilograms to pounds on your own.
#### 42.1 kg to lbs conversion
We are going to begin 42.1 kilogram equals to how many pounds calculation with the first formula to get the most accurate result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 92.814612302 the result in pounds
So what have you do to check how many pounds equal to 42.1 kilogram? Just multiply amount of kilograms, this time 42.1, by 2.20462262. It gives 92.814612302. So 42.1 kilogram is exactly 92.814612302.
You can also round it off, for example, to two decimal places. It gives 2.20. So 42.1 kilogram = 92.620 pounds.
It is high time for an example from everyday life. Let’s convert 42.1 kg gold in pounds. So 42.1 kg equal to how many lbs? And again - multiply 42.1 by 2.20462262. It gives 92.814612302. So equivalent of 42.1 kilograms to pounds, if it comes to gold, is equal 92.814612302.
In this case it is also possible to round off the result. This is the result after rounding off, this time to one decimal place - 42.1 kilogram 92.62 pounds.
Now let’s move on to examples converted with a short version of a formula.
#### How many 42.1 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 92.62 the result in pounds
So 42.1 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, this time 42.1, by 2.2. Have a look: 42.1 * 2.2 = 92.62. So 42.1 kilogram is 2.2 pounds.
Do another conversion with use of this version of a formula. Now convert something from everyday life, for example, 42.1 kg to lbs weight of strawberries.
So convert - 42.1 kilogram of strawberries * 2.2 = 92.62 pounds of strawberries. So 42.1 kg to pound mass is equal 92.62.
If you know how much is 42.1 kilogram weight in pounds and are able to convert it using two different formulas, we can move on. Now we are going to show you all results in charts.
#### Convert 42.1 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 outcomes in charts for your convenience. Due to this you can quickly make a comparison 42.1 kg equivalent to lbs outcomes.
Let’s begin with a 42.1 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
42.1 92.814612302 92.620
And now see 42.1 kg equal pound table for the second formula:
Kilograms Pounds
42.1 92.62
As you see, after rounding off, when it comes to how much 42.1 kilogram equals pounds, the outcomes are the same. The bigger number the more significant difference. Please note it when you need to make bigger number than 42.1 kilograms pounds conversion.
#### How many kilograms 42.1 pound
Now you learned how to convert 42.1 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 42.1 kilogram to pounds and ounces conversion?
We want to show you how you can convert it step by step. Start. How much is 42.1 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, this time 42.1, by 2.20462262. So 42.1 * 2.20462262 = 92.814612302. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 42.1 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final result is 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final outcome will be equal 2 pounds and 33 ounces.
As you see, calculation 42.1 kilogram in pounds and ounces is not complicated.
The last calculation which we want to show you is calculation of 42.1 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we show you it, have a look:
• 42.1 kilograms meters = 7.23301385 foot pounds,
• 42.1 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to calculate 42.1 foot pounds to kilograms meters you have to multiply 42.1 by 0.13825495. It is equal 0.13825495. So 42.1 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 42.1 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 42.1 kilogram into pounds calculations.
This article was a big compendium about kilogram, pound and 42.1 kg to lbs in conversion. Due to this calculation you know 42.1 kilogram is equivalent to how many pounds.
We showed you not only how to do a calculation 42.1 kilogram to metric pounds but also two another calculations - to know how many 42.1 kg in pounds and ounces and how many 42.1 foot pounds to kilograms meters.
We showed you also other solution to do 42.1 kilogram how many pounds conversions, it is using 42.1 kg en pound calculator. It will be the best choice for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to make 42.1 kilogram equal to how many pounds calculation - on your own or with use of our 42.1 kgs to pounds calculator.
Don’t wait! Convert 42.1 kilogram mass to pounds in the best way for you.
Do you want to do other than 42.1 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so simply as for 42.1 kilogram equal many pounds.
### How much is 42.1 kg in pounds
To quickly sum up this topic, that is how much is 42.1 kg in pounds , we prepared for you an additional section. Here you can see all you need to know about how much is 42.1 kg equal to lbs and how to convert 42.1 kg to lbs . It is down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 42.1 kg to pound conversion formula look? . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 42.1 kilogram to pounds. The correct answer is 92.814612302 lb.
There is also another way to calculate how much 42.1 kilogram is equal to pounds with second, shortened type of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So now, 42.1 kg equal to how much lbs ? The answer is 92.814612302 pounds.
How to convert 42.1 kg to lbs in a few seconds? You can also use the 42.1 kg to lbs converter , which will make all calculations for you and give you an exact 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. | HuggingFaceTB/finemath | |
# Inequalities
## Inequalities
Solving Inequalities
An inequality is the result of replacing the = sign in an equation with , ≤, or ≥. For
example, 3x – 2 < 7 is a linear inequality. We call it “linear” because if the < were
replaced with an = sign, it would be a linear equation. Inequalities involving polynomials
of degree 2 or more, like 2x3 – x > 0, are referred to as polynomial inequalities, or
quadratic inequalities if the degree is exactly 2. Inequalities involving rational
expressions are called rational inequalities. Some often used inequalities also involve
absolute value expressions.
Solving Inequalities: A Summary
In a nutshell, solving inequalities is about one thing: sign changes. Find all the points at
which there are sign changes - we call these points critical values. Then determine
which, if any, of the intervals bounded by these critical values result in a solution. The
solution to the inequality will consist of the set of all points contained by the solution
intervals.
Method To Solve Linear, Polynomial, or Absolute Value Inequalities:
1. Move all terms to one side of the inequality sign by applying the Addition,
Subtraction, Multiplication, and Division Properties of Inequalities. You should have
only zero on one side of the inequality sign.
2. Solve the associated equation using an appropriate method. This solution or
solutions will make up the set of critical values. At these values, sign changes occur
in the inequality.
3. Plot the critical values on a number line. Use closed circles • for ≤ and ≥
inequalities, and use open circles ο for < and > inequalities.
4. Test each interval defined by the critical values. If an interval satisfies the
inequality, then it is part of the solution. If it does not satisfy the inequality, then it is
not part of the solution.
Example: Solve 3x + 5(x + 1) ≤ 4x – 1 and graph the solution
3x + 5(x + 1) ≤ 4x – 1
3x + 5x + 5 ≤ 4x – 1
8x + 5 ≤ 4x – 1
4x + 6 ≤ 0
Now, solve 4x+6 = 0
4x = -6
x = - 6/4 = -3/2... | HuggingFaceTB/finemath | |
Q. 14 B4.7( 7 Votes )
# Find the equations of the tangent and normal to the given curves at the indicated points:y = x4 – 6x3 + 13x2– 10x + 5 at (1, 3)
It is given that equation of curve is y = y = x4 – 6x3 + 13x2– 10x + 5
On differentiating with respect to x, we get
= 4x3 - 18x2 +26x - 10
Therefore, the slope of the tangent at (1, 3) is 2.
Then, the equation of the tangent is
y – 3 = 2(x – 1)
y – 3 = 2x - 2
y = 2x +1
Then, slope of normal at (1,3)
=
Now, equation of the normal at (1,3)
y – 3 = (x – 1)
2y -6 =- x + 1
x + 2y - 7 = 0
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better. | HuggingFaceTB/finemath | |
# Maximum set packing and minimum set cover duality
I read that the maximum set packing and the minimum set cover problems are dual of each other when formulated as linear programming problems. By the strong duality theorem, the optimal solution to the primal and dual LP problems should have the same value.
However, consider a universe $U = \{1, 2, 3, 4, 5\}$ and a collection of sets: $S = \{ \{1, 2, 3\}, \{3, 4, 5\}, \{1\}, \{2\}, \{3\}\}$. From what I understand, a minimum set cover would consist of the first 2 sets in set $S$, while the maximum set packing consists of the last 3 sets. These solutions aren't in accordance with the statement of the strong duality theorem.
Given that, I don't understand how the 2 problems can be dual of each other. What am I missing?
Thank you very much.
There is no contradiction. The strong duality theorem applies to the fractional linear programming formulations of the problems, where one is allowed to set fractional values for the variables, but these formulations are only relaxations of the two original problems. Strong duality does not apply to the integer linear programming formulations (as your example indeed shows).
Let $$P$$ be the LP relaxation for minimum set cover: \begin{align} \min \sum_{S \in \mathcal{S}} x_Sc(S) & \\ \sum_{S: \;e \in S} x_S \geq 1 &\quad\quad \forall e\in U \\ x_S \geq 0 &\quad\quad S \in \mathcal{S}\\ \end{align} and $$D$$ be the dual LP which corresponds to the linear relaxation for maximum set packing: \begin{align} \max \sum_{e \in \mathcal{U}} y_e & \\ \sum_{e: \;e \in S} y_e \leq c(S) &\quad\quad \forall S\in \mathcal{S} \\ y_e \geq 0 &\quad\quad e \in \mathcal{U}\\ \end{align} The strong duality theorem states that the optimal value of $$P$$ is the same as the optimum value of $$D$$. However, the solutions for the LPs aren't necessarily integral as in the case that you showed. Let $$z^*$$ be the value of the optimal solution to $$P$$, it can be shown that $$OPT_I\leq H_nz^*$$ where $$OPT_I$$ is the value for the optimal integral solution of $$P$$.
The worst case bound of the ratio $$\frac{OPT_I}{z^*}$$ is known as the integrality gap of an LP (note that the ratio is flipped for a maximization problem).
See here for a discussion on integrality gaps and here for an analysis of the set cover primal/dual.
The dual of Set Cover is not Set Packing (as defined in Wikipedia). In Set Cover you choose sets, e.g. you have a binary variable for each set. Thus, the dual problem can't have a variable for each set as well. Instead, you have a dual variable for each element that represents how much of that element you want to pack without overpacking the sets that contain this element (see the dual LP formulation below).
$$\min \sum_{e \in U} y_e$$ $$s.t. \sum_{e: e \in s} y_e \le \text{cost}(s) \quad\quad (s \in S)$$ $$y_e \ge 0 \quad\quad (e \in U)$$
The actual formulation of the dual of the set-packing problem is:
$$\min \sum_{e \in U} y_e$$ $$s.t.$$ $$\sum_{e: e \in s} y_e \ge w_s \quad\quad (s \in S)$$ $$y_e \ge 0 \quad\quad (e \in U)$$
(Zideon's formulation is wrong, not enough reputation to reply him...)
I got the same confusion with that wiki statement some weeks ago. I will correct that article once I get the time to do it. | HuggingFaceTB/finemath | |
# Sets in Python
Perhaps you recall learning about sets and set theory at some point in your mathematical education. Maybe you even remember Venn diagrams: If this doesn’t ring a bell, don’t worry! This tutorial should still be easily accessible for you.
In mathematics, a rigorous definition of a set can be abstract and difficult to grasp. Practically though, a set can be thought of simply as a well-defined collection of distinct objects, typically called elements or members.
Grouping objects into a set can be useful in programming as well, and Python provides a built-in set type to do so. Sets are distinguished from other object types by the unique operations that can be performed on them.
Here’s what you’ll learn in this tutorial: You’ll see how to define set objects in Python and discover the operations that they support. As with the earlier tutorials on lists and dictionaries, when you are finished with this tutorial, you should have a good feel for when a set is an appropriate choice. You will also learn about frozen sets, which are similar to sets except for one important detail.
## Defining a Set
Python’s built-in `set` type has the following characteristics:
• Sets are unordered.
• Set elements are unique. Duplicate elements are not allowed.
• A set itself may be modified, but the elements contained in the set must be of an immutable type.
Let’s see what all that means, and how you can work with sets in Python.
A set can be created in two ways. First, you can define a set with the built-in `set()` function:
```x = set(<iter>)
```
In this case, the argument `<iter>` is an iterable—again, for the moment, think list or tuple—that generates the list of objects to be included in the set. This is analogous to the `<iter>` argument given to the `.extend()` list method:
>>>
```>>> x = set(['foo', 'bar', 'baz', 'foo', 'qux'])
>>> x
{'qux', 'foo', 'bar', 'baz'}
>>> x = set(('foo', 'bar', 'baz', 'foo', 'qux'))
>>> x
{'qux', 'foo', 'bar', 'baz'}
```
Strings are also iterable, so a string can be passed to `set()` as well. You have already seen that `list(s)` generates a list of the characters in the string `s`. Similarly, `set(s)` generates a set of the characters in `s`:
>>>
```>>> s = 'quux'
>>> list(s)
['q', 'u', 'u', 'x']
>>> set(s)
{'x', 'u', 'q'}
```
You can see that the resulting sets are unordered: the original order, as specified in the definition, is not necessarily preserved. Additionally, duplicate values are only represented in the set once, as with the string `'foo'` in the first two examples and the letter `'u'` in the third.
Alternately, a set can be defined with curly braces (`{}`):
```x = {<obj>, <obj>, ..., <obj>}
```
When a set is defined this way, each `<obj>` becomes a distinct element of the set, even if it is an iterable. This behavior is similar to that of the `.append()` list method.
Thus, the sets shown above can also be defined like this:
>>>
```>>> x = {'foo', 'bar', 'baz', 'foo', 'qux'}
>>> x
{'qux', 'foo', 'bar', 'baz'}
>>> x = {'q', 'u', 'u', 'x'}
>>> x
{'x', 'q', 'u'}
```
To recap:
• The argument to `set()` is an iterable. It generates a list of elements to be placed into the set.
• The objects in curly braces are placed into the set intact, even if they are iterable.
Observe the difference between these two set definitions:
>>>
```>>> {'foo'}
{'foo'}
>>> set('foo')
{'o', 'f'}
```
A set can be empty. However, recall that Python interprets empty curly braces (`{}`) as an empty dictionary, so the only way to define an empty set is with the `set()` function:
>>>
```>>> x = set()
>>> type(x)
<class 'set'>
>>> x
set()
>>> x = {}
>>> type(x)
<class 'dict'>
```
An empty set is falsy in Boolean context:
>>>
```>>> x = set()
>>> bool(x)
False
>>> x or 1
1
>>> x and 1
set()
```
You might think the most intuitive sets would contain similar objects—for example, even numbers or surnames:
>>>
```>>> s1 = {2, 4, 6, 8, 10}
>>> s2 = {'Smith', 'McArthur', 'Wilson', 'Johansson'}
```
Python does not require this, though. The elements in a set can be objects of different types:
>>>
```>>> x = {42, 'foo', 3.14159, None}
>>> x
{None, 'foo', 42, 3.14159}
```
Don’t forget that set elements must be immutable. For example, a tuple may be included in a set:
>>>
```>>> x = {42, 'foo', (1, 2, 3), 3.14159}
>>> x
{42, 'foo', 3.14159, (1, 2, 3)}
```
But lists and dictionaries are mutable, so they can’t be set elements:
>>>
```>>> a = [1, 2, 3]
>>> {a}
Traceback (most recent call last):
File "<pyshell#70>", line 1, in <module>
{a}
TypeError: unhashable type: 'list'
>>> d = {'a': 1, 'b': 2}
>>> {d}
Traceback (most recent call last):
File "<pyshell#72>", line 1, in <module>
{d}
TypeError: unhashable type: 'dict'
```
## Set Size and Membership
The `len()` function returns the number of elements in a set, and the `in` and `not in` operators can be used to test for membership:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> len(x)
3
>>> 'bar' in x
True
>>> 'qux' in x
False
```
## Operating on a Set
Many of the operations that can be used for Python’s other composite data types don’t make sense for sets. For example, sets can’t be indexed or sliced. However, Python provides a whole host of operations on set objects that generally mimic the operations that are defined for mathematical sets.
### Operators vs. Methods
Most, though not quite all, set operations in Python can be performed in two different ways: by operator or by method. Let’s take a look at how these operators and methods work, using set union as an example.
Given two sets, `x1` and `x2`, the union of `x1` and `x2` is a set consisting of all elements in either set.
Consider these two sets:
```x1 = {'foo', 'bar', 'baz'}
x2 = {'baz', 'qux', 'quux'}
```
The union of `x1` and `x2` is `{'foo', 'bar', 'baz', 'qux', 'quux'}`.
In Python, set union can be performed with the `|` operator:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1 | x2
{'baz', 'quux', 'qux', 'bar', 'foo'}
```
Set union can also be obtained with the `.union()` method. The method is invoked on one of the sets, and the other is passed as an argument:
>>>
```>>> x1.union(x2)
{'baz', 'quux', 'qux', 'bar', 'foo'}
```
The way they are used in the examples above, the operator and method behave identically. But there is a subtle difference between them. When you use the `|` operator, both operands must be sets. The `.union()` method, on the other hand, will take any iterable as an argument, convert it to a set, and then perform the union.
Observe the difference between these two statements:
>>>
```>>> x1 | ('baz', 'qux', 'quux')
Traceback (most recent call last):
File "<pyshell#43>", line 1, in <module>
x1 | ('baz', 'qux', 'quux')
TypeError: unsupported operand type(s) for |: 'set' and 'tuple'
>>> x1.union(('baz', 'qux', 'quux'))
{'baz', 'quux', 'qux', 'bar', 'foo'}
```
Both attempt to compute the union of `x1` and the tuple `('baz', 'qux', 'quux')`. This fails with the `|` operator but succeeds with the `.union()` method.
### Available Operators and Methods
Below is a list of the set operations available in Python. Some are performed by operator, some by method, and some by both. The principle outlined above generally applies: where a set is expected, methods will typically accept any iterable as an argument, but operators require actual sets as operands.
`x1.union(x2[, x3 ...])`
`x1 | x2 [| x3 ...]`
Compute the union of two or more sets.
`x1.union(x2)` and `x1 | x2` both return the set of all elements in either `x1` or `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1.union(x2)
{'foo', 'qux', 'quux', 'baz', 'bar'}
>>> x1 | x2
{'foo', 'qux', 'quux', 'baz', 'bar'}
```
More than two sets may be specified with either the operator or the method:
>>>
```>>> a = {1, 2, 3, 4}
>>> b = {2, 3, 4, 5}
>>> c = {3, 4, 5, 6}
>>> d = {4, 5, 6, 7}
>>> a.union(b, c, d)
{1, 2, 3, 4, 5, 6, 7}
>>> a | b | c | d
{1, 2, 3, 4, 5, 6, 7}
```
The resulting set contains all elements that are present in any of the specified sets.
`x1.intersection(x2[, x3 ...])`
`x1 & x2 [& x3 ...]`
Compute the intersection of two or more sets.
`x1.intersection(x2)` and `x1 & x2` return the set of elements common to both `x1` and `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1.intersection(x2)
{'baz'}
>>> x1 & x2
{'baz'}
```
You can specify multiple sets with the intersection method and operator, just like you can with set union:
>>>
```>>> a = {1, 2, 3, 4}
>>> b = {2, 3, 4, 5}
>>> c = {3, 4, 5, 6}
>>> d = {4, 5, 6, 7}
>>> a.intersection(b, c, d)
{4}
>>> a & b & c & d
{4}
```
The resulting set contains only elements that are present in all of the specified sets.
`x1.difference(x2[, x3 ...])`
`x1 - x2 [- x3 ...]`
Compute the difference between two or more sets.
`x1.difference(x2)` and `x1 - x2` return the set of all elements that are in `x1` but not in `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1.difference(x2)
{'foo', 'bar'}
>>> x1 - x2
{'foo', 'bar'}
```
Another way to think of this is that `x1.difference(x2)` and `x1 - x2` return the set that results when any elements in `x2` are removed or subtracted from `x1`.
Once again, you can specify more than two sets:
>>>
```>>> a = {1, 2, 3, 30, 300}
>>> b = {10, 20, 30, 40}
>>> c = {100, 200, 300, 400}
>>> a.difference(b, c)
{1, 2, 3}
>>> a - b - c
{1, 2, 3}
```
When multiple sets are specified, the operation is performed from left to right. In the example above, `a - b` is computed first, resulting in `{1, 2, 3, 300}`. Then `c` is subtracted from that set, leaving `{1, 2, 3}`: `x1.symmetric_difference(x2)`
`x1 ^ x2 [^ x3 ...]`
Compute the symmetric difference between sets.
`x1.symmetric_difference(x2)` and `x1 ^ x2` return the set of all elements in either `x1` or `x2`, but not both:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1.symmetric_difference(x2)
{'foo', 'qux', 'quux', 'bar'}
>>> x1 ^ x2
{'foo', 'qux', 'quux', 'bar'}
```
The `^` operator also allows more than two sets:
>>>
```>>> a = {1, 2, 3, 4, 5}
>>> b = {10, 2, 3, 4, 50}
>>> c = {1, 50, 100}
>>> a ^ b ^ c
{100, 5, 10}
```
As with the difference operator, when multiple sets are specified, the operation is performed from left to right.
Curiously, although the `^` operator allows multiple sets, the `.symmetric_difference()` method doesn’t:
>>>
```>>> a = {1, 2, 3, 4, 5}
>>> b = {10, 2, 3, 4, 50}
>>> c = {1, 50, 100}
>>> a.symmetric_difference(b, c)
Traceback (most recent call last):
File "<pyshell#11>", line 1, in <module>
a.symmetric_difference(b, c)
TypeError: symmetric_difference() takes exactly one argument (2 given)
```
`x1.isdisjoint(x2)`
Determines whether or not two sets have any elements in common.
`x1.isdisjoint(x2)` returns `True` if `x1` and `x2` have no elements in common:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1.isdisjoint(x2)
False
>>> x2 - {'baz'}
{'quux', 'qux'}
>>> x1.isdisjoint(x2 - {'baz'})
True
```
If `x1.isdisjoint(x2)` is `True`, then `x1 & x2` is the empty set:
>>>
```>>> x1 = {1, 3, 5}
>>> x2 = {2, 4, 6}
>>> x1.isdisjoint(x2)
True
>>> x1 & x2
set()
```
`x1.issubset(x2)`
`x1 <= x2`
Determine whether one set is a subset of the other.
In set theory, a set `x1` is considered a subset of another set `x2` if every element of `x1` is in `x2`.
`x1.issubset(x2)` and `x1 <= x2` return `True` if `x1` is a subset of `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x1.issubset({'foo', 'bar', 'baz', 'qux', 'quux'})
True
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1 <= x2
False
```
A set is considered to be a subset of itself:
>>>
```>>> x = {1, 2, 3, 4, 5}
>>> x.issubset(x)
True
>>> x <= x
True
```
It seems strange, perhaps. But it fits the definition—every element of `x` is in `x`.
`x1 < x2`
Determines whether one set is a proper subset of the other.
A proper subset is the same as a subset, except that the sets can’t be identical. A set `x1` is considered a proper subset of another set `x2` if every element of `x1` is in `x2`, and `x1` and `x2` are not equal.
`x1 < x2` returns `True` if `x1` is a proper subset of `x2`:
>>>
```>>> x1 = {'foo', 'bar'}
>>> x2 = {'foo', 'bar', 'baz'}
>>> x1 < x2
True
>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'bar', 'baz'}
>>> x1 < x2
False
```
While a set is considered a subset of itself, it is not a proper subset of itself:
>>>
```>>> x = {1, 2, 3, 4, 5}
>>> x <= x
True
>>> x < x
False
```
`x1.issuperset(x2)`
`x1 >= x2`
Determine whether one set is a superset of the other.
A superset is the reverse of a subset. A set `x1` is considered a superset of another set `x2` if `x1` contains every element of `x2`.
`x1.issuperset(x2)` and `x1 >= x2` return `True` if `x1` is a superset of `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x1.issuperset({'foo', 'bar'})
True
>>> x2 = {'baz', 'qux', 'quux'}
>>> x1 >= x2
False
```
You have already seen that a set is considered a subset of itself. A set is also considered a superset of itself:
>>>
```>>> x = {1, 2, 3, 4, 5}
>>> x.issuperset(x)
True
>>> x >= x
True
```
`x1 > x2`
Determines whether one set is a proper superset of the other.
A proper superset is the same as a superset, except that the sets can’t be identical. A set `x1` is considered a proper superset of another set `x2` if `x1` contains every element of `x2`, and `x1` and `x2` are not equal.
`x1 > x2` returns `True` if `x1` is a proper superset of `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'bar'}
>>> x1 > x2
True
>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'bar', 'baz'}
>>> x1 > x2
False
```
A set is not a proper superset of itself:
>>>
```>>> x = {1, 2, 3, 4, 5}
>>> x > x
False
```
## Modifying a Set
Although the elements contained in a set must be of immutable type, sets themselves can be modified. Like the operations above, there are a mix of operators and methods that can be used to change the contents of a set.
### Augmented Assignment Operators and Methods
Each of the union, intersection, difference, and symmetric difference operators listed above has an augmented assignment form that can be used to modify a set. For each, there is a corresponding method as well.
`x1.update(x2[, x3 ...])`
`x1 |= x2 [| x3 ...]`
Modify a set by union.
`x1.update(x2)` and `x1 |= x2` add to `x1` any elements in `x2` that `x1` does not already have:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'baz', 'qux'}
>>> x1 |= x2
>>> x1
{'qux', 'foo', 'bar', 'baz'}
>>> x1.update(['corge', 'garply'])
>>> x1
{'qux', 'corge', 'garply', 'foo', 'bar', 'baz'}
```
`x1.intersection_update(x2[, x3 ...])`
`x1 &= x2 [& x3 ...]`
Modify a set by intersection.
`x1.intersection_update(x2)` and `x1 &= x2` update `x1`, retaining only elements found in both `x1` and `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'baz', 'qux'}
>>> x1 &= x2
>>> x1
{'foo', 'baz'}
>>> x1.intersection_update(['baz', 'qux'])
>>> x1
{'baz'}
```
`x1.difference_update(x2[, x3 ...])`
`x1 -= x2 [| x3 ...]`
Modify a set by difference.
`x1.difference_update(x2)` and `x1 -= x2` update `x1`, removing elements found in `x2`:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'baz', 'qux'}
>>> x1 -= x2
>>> x1
{'bar'}
>>> x1.difference_update(['foo', 'bar', 'qux'])
>>> x1
set()
```
`x1.symmetric_difference_update(x2)`
`x1 ^= x2`
Modify a set by symmetric difference.
`x1.symmetric_difference_update(x2)` and `x1 ^= x2` update `x1`, retaining elements found in either `x1` or `x2`, but not both:
>>>
```>>> x1 = {'foo', 'bar', 'baz'}
>>> x2 = {'foo', 'baz', 'qux'}
>>>
>>> x1 ^= x2
>>> x1
{'bar', 'qux'}
>>>
>>> x1.symmetric_difference_update(['qux', 'corge'])
>>> x1
{'bar', 'corge'}
```
### Other Methods For Modifying Sets
Aside from the augmented operators above, Python supports several additional methods that modify sets.
`x.add(<elem>)`
Adds an element to a set.
`x.add(<elem>)` adds `<elem>`, which must be a single immutable object, to `x`:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> x
{'bar', 'baz', 'foo', 'qux'}
```
`x.remove(<elem>)`
Removes an element from a set.
`x.remove(<elem>)` removes `<elem>` from `x`. Python raises an exception if `<elem>` is not in `x`:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> x.remove('baz')
>>> x
{'bar', 'foo'}
>>> x.remove('qux')
Traceback (most recent call last):
File "<pyshell#58>", line 1, in <module>
x.remove('qux')
KeyError: 'qux'
```
`x.discard(<elem>)`
Removes an element from a set.
`x.discard(<elem>)` also removes `<elem>` from `x`. However, if `<elem>` is not in `x`, this method quietly does nothing instead of raising an exception:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> x
{'bar', 'foo'}
>>> x
{'bar', 'foo'}
```
`x.pop()`
Removes a random element from a set.
`x.pop()` removes and returns an arbitrarily chosen element from `x`. If `x` is empty, `x.pop()` raises an exception:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> x.pop()
'bar'
>>> x
{'baz', 'foo'}
>>> x.pop()
'baz'
>>> x
{'foo'}
>>> x.pop()
'foo'
>>> x
set()
>>> x.pop()
Traceback (most recent call last):
File "<pyshell#82>", line 1, in <module>
x.pop()
KeyError: 'pop from an empty set'
```
`x.clear()`
Clears a set.
`x.clear()` removes all elements from `x`:
>>>
```>>> x = {'foo', 'bar', 'baz'}
>>> x
{'foo', 'bar', 'baz'}
>>>
>>> x.clear()
>>> x
set()
```
## Frozen Sets
Python provides another built-in type called a frozenset, which is in all respects exactly like a set, except that a frozenset is immutable. You can perform non-modifying operations on a frozenset:
>>>
```>>> x = frozenset(['foo', 'bar', 'baz'])
>>> x
frozenset({'foo', 'baz', 'bar'})
>>> len(x)
3
>>> x & {'baz', 'qux', 'quux'}
frozenset({'baz'})
```
But methods that attempt to modify a frozenset fail:
>>>
```>>> x = frozenset(['foo', 'bar', 'baz'])
Traceback (most recent call last):
File "<pyshell#127>", line 1, in <module>
AttributeError: 'frozenset' object has no attribute 'add'
>>> x.pop()
Traceback (most recent call last):
File "<pyshell#129>", line 1, in <module>
x.pop()
AttributeError: 'frozenset' object has no attribute 'pop'
>>> x.clear()
Traceback (most recent call last):
File "<pyshell#131>", line 1, in <module>
x.clear()
AttributeError: 'frozenset' object has no attribute 'clear'
>>> x
frozenset({'foo', 'bar', 'baz'})
```
Deep Dive: Frozensets and Augmented Assignment
Since a frozenset is immutable, you might think it can’t be the target of an augmented assignment operator. But observe:
>>>
```>>> f = frozenset(['foo', 'bar', 'baz'])
>>> s = {'baz', 'qux', 'quux'}
>>> f &= s
>>> f
frozenset({'baz'})
```
What gives?
Python does not perform augmented assignments on frozensets in place. The statement `x &= s` is effectively equivalent to `x = x & s`. It isn’t modifying the original `x`. It is reassigning `x` to a new object, and the object `x` originally referenced is gone.
You can verify this with the `id()` function:
>>>
```>>> f = frozenset(['foo', 'bar', 'baz'])
>>> id(f)
56992872
>>> s = {'baz', 'qux', 'quux'}
>>> f &= s
>>> f
frozenset({'baz'})
>>> id(f)
56992152
```
`f` has a different integer identifier following the augmented assignment. It has been reassigned, not modified in place.
Some objects in Python are modified in place when they are the target of an augmented assignment operator. But frozensets aren’t.
Frozensets are useful in situations where you want to use a set, but you need an immutable object. For example, you can’t define a set whose elements are also sets, because set elements must be immutable:
>>>
```>>> x1 = set(['foo'])
>>> x2 = set(['bar'])
>>> x3 = set(['baz'])
>>> x = {x1, x2, x3}
Traceback (most recent call last):
File "<pyshell#38>", line 1, in <module>
x = {x1, x2, x3}
TypeError: unhashable type: 'set'
```
If you really feel compelled to define a set of sets (hey, it could happen), you can do it if the elements are frozensets, because they are immutable:
>>>
```>>> x1 = frozenset(['foo'])
>>> x2 = frozenset(['bar'])
>>> x3 = frozenset(['baz'])
>>> x = {x1, x2, x3}
>>> x
{frozenset({'bar'}), frozenset({'baz'}), frozenset({'foo'})}
```
Likewise, recall from the previous tutorial on dictionaries that a dictionary key must be immutable. You can’t use the built-in set type as a dictionary key:
>>>
```>>> x = {1, 2, 3}
>>> y = {'a', 'b', 'c'}
>>>
>>> d = {x: 'foo', y: 'bar'}
Traceback (most recent call last):
File "<pyshell#3>", line 1, in <module>
d = {x: 'foo', y: 'bar'}
TypeError: unhashable type: 'set'
```
If you find yourself needing to use sets as dictionary keys, you can use frozensets:
>>>
```>>> x = frozenset({1, 2, 3})
>>> y = frozenset({'a', 'b', 'c'})
>>>
>>> d = {x: 'foo', y: 'bar'}
>>> d
{frozenset({1, 2, 3}): 'foo', frozenset({'c', 'a', 'b'}): 'bar'}
```
## Conclusion
In this tutorial, you learned how to define set objects in Python, and you became familiar with the functions, operators, and methods that can be used to work with sets.
You should now be comfortable with the basic built-in data types that Python provides.
Next, you will begin to explore how the code that operates on those objects is organized and structured in a Python program.
🐍 Python Tricks 💌
Get a short & sweet Python Trick delivered to your inbox every couple of days. No spam ever. Unsubscribe any time. Curated by the Real Python team. John is an avid Pythonista and a member of the Real Python tutorial team. | HuggingFaceTB/finemath | |
# Dijkstra’s Shortest Path Algorithm
## Theory of Computation
The shortest path between u and v is denoted $\delta(u,v)$, if there is no path, then $\delta(u,v)=\infty$
# Can a shortest path contain a cycle?
A directed cycle is:
• Positive: if its edge weights sum up to a positive number
• Negative: if its edge weights sum up to a negative number
If there is a positive cycle in the graph, it will not be contained in any shortest path between u and v so we can assume that the shortest paths we find contain no positive cycles.
However if there is a negative cycle between u and v, then $\delta(u,v)=-\infty$ so we shall assume that the graphs we consider do not contain negative cycles.
# Single-Source Shortest Paths
• Aim: to describe an algorithm that solves the single-source shortest paths problem, i.e. an algorithm that finds the shortest path from a specific source vertex
• This is a generalization of BFS
• So the output of the algorithm should be two arrays d, $\pi$ where for each vertex v:
• $d(v)=\delta(s,v)$
• $\pi(v)$ is the predecessor of v
# Relaxation
• Assume that the weight on every edge is non-negative
• We do not directly compute the entry $d(v)=\delta(s,v)$
• Instead, at every step, $d(v)$ is an estimate for $\delta(s,v)$
• Initially, $d(v)=\infty$, and it always remains $d(v)\geqslant \delta(s,v)$
• $d(v)$ is updated (i.e. it decreases) as shorter paths are found
• At the end of the algorithm we have $d(v)=\delta(s,v)$
Initialise-Single-Source(G,s)
for each vertex v in V(g):
d(v)=infinity
pi(v)=NULL
d(s)=0
The process of relaxing an edge (u,v):
• Test whether we can improve the shortest path from s to v that we found so far, by going through u
• If yes, then update $d(v)$ and $\pi(v)$
• Decrease the estimate $d(v)$
• Update the predecessor $\pi(v)$ to u
• The algorithm first calls initialise-single-source and then it repeatedly relaxes the appropriate edges (according to the weight function w)
Relax(u,v,w)
if d(v)>d(u)+w(u,v):
d(v)=d(u)+w(u,v)
pi(v)=u
# Dijkstra’s Algorithm
• Initialisation: distance to source A is 0, $S=\varnothing$, Q=V
• S stores the vertices v for which we already found $\delta(A,v)$
• Q stores all the other vertices
• While q is not empty
• Remove from Q the vertex u for which d(u) is minimum
• Add this vertex to s
• Relax all edges leaving u
Dijkstra({G,w,s})
Initialise-Single-Source(G,s)
S=[]
Q=V(G)
while Q!=[] do
u=Extract-Min(Q)
S=S+{u}
for each vertex v in Adj(u) do
Relax(u,v,w)
## Runtime
Initialisation is done in $\mathcal{O}(V)$ time - two operations per vertex
Finding the vertex v in Q with minimum d(v) takes $\mathcal{O}(V)$ time and this is done v times
• To find the minimum, just scan the set Q
• To compute the new vertex in S, find the new minimum of Q
Relaxation takes in total $\mathcal{O}(E)$ time as every edge is relaxed once
The total running time is $\mathcal{O}(V+V^2+E)=\mathcal{O}(V^2)$
However using a more sophisticated implementation for extracting the minimum for Q it can run in $\mathcal{O}(V\log V+E)$ time
# Properties of shortest paths and relaxation
Triangle Inequality
For all edges (u,v) we have $\delta(s,v)\leqslant \delta(s,u)+w(u,v)$
Optimal substructure
Any subpath of a shortest path is also a shortest path
Upper bound property
For every vertex v, we have $d(v)\geqslant \delta(s,v)$
No-path property
If $\delta(s,v)=\infty$ then we have $d(v)=\infty$ at every iteration
Convergence property
If there is a shortest path from s to v including the edge (u,v) and if $d(u)=\delta(s,u)$, then we obtain $d(v)=\delta(s,v)$ when $(u,v)$ is relaxed
# Correctness of Dijkstra’s algorithm
We need to prove the loop invariant always remains true.
At the start of each iteration of the while loop, $d(v)=\delta(s,v)$ for every $v\in S$
• Initialisation: at the start of the algorithm S is empty, so the loop invariant is trivially true
• Maintenance: we need to show that $d(u)=\delta(s,u)$ when u is added to S
• Termination: at the end, S contains every vertex, which implies that $d(v)=\delta(s,v)$ for all vertices v in the graph | HuggingFaceTB/finemath | |
【欧拉猜想】是否有无穷多个不可约分的正整数解
```a^3+b^3=c^3
a^4+b^4+c^4=d^4
a^5+b^5+c^5+d^5=e^5
……```
`27^5+ 84^5+ 110^5+ 133^5= 144^5`
``` @Test
fun test3() {
var list1 = arrayListOf<Pair<Long, String>>()
var list2 = arrayListOf<Pair<Long, String>>()
val start1 = System.currentTimeMillis()
// 假设： a<b<c<d, 肯定有： e > {a,b,c,d}
for (a in 1..200L) {
for (b in a..200L) {
for (c in b..200L) {
Propositions.f3(a, b, c)
}
}
}
val end1 = System.currentTimeMillis()
println("Using Time1: \${(end1 - start1)} ms")
println("list1.size=\${list1.size}")
val start2 = System.currentTimeMillis()
for (d in 1..200L) {
(d..200L).mapTo(list2) { Pair(Propositions.f4(it, d), "e=\$it,d=\$d") }
}
val end2 = System.currentTimeMillis()
println("Using Time2: \${(end2 - start2)} ms")
println("list2.size=\${list2.size}")
/*
首先想到的，遍历所有元素，穷举判断；更进一层：平均切割 List，每一段开启1个线程计算。
val start3 = System.currentTimeMillis()
loop@ for (x in list2) {
if (list1.contains(x)) {
println("x=\$x")
break@loop
}
}
val end3 = System.currentTimeMillis()
println("Using Time3: \${(end3 - start3)} ms")
*/
val start3 = System.currentTimeMillis()
var list1n = arrayListOf<List<Pair<Long, String>>>()
val segment = 10_0000
val step = list1.size / segment
for (i in 1..segment) {
var listi = arrayListOf<Pair<Long, String>>()
for (j in ((i - 1) * step)..(i * step - 1)) {
}
}
println("list1n.size=\${list1n.size}")
val end3 = System.currentTimeMillis()
println("Using Time3: \${(end3 - start3)} ms")
val start4 = System.currentTimeMillis()
list1n.forEach {
for (x in list2) {
it.filter { x.first == it.first }
.forEach { println("\$x,\$it") }
}
})
}
}
val end4 = System.currentTimeMillis()
println("Using Time4: \${(end4 - start4)} ms")
}```
``` fun f3(a: Long, b: Long, c: Long): Long {
return a * a * a * a * a +
b * b * b * b * b +
c * c * c * c * c
}
fun f4(e: Long, d: Long): Long {
return e * e * e * e * e - d * d * d * d * d
}```
```Using Time1: 3247 ms
list1.size=1353400
Using Time2: 19 ms
list2.size=20100
list1n.size=100000
Using Time3: 72 ms
(20301568331, e=144,d=133),(20301568331, a=27,b=84,c=110)
(45812264224, e=144,d=110),(45812264224, a=27,b=84,c=133)
(57735244800, e=144,d=84),(57735244800, a=27,b=110,c=133)
(61903015317, e=144,d=27),(61903015317, a=84,b=110,c=133)
Using Time4: 238898 ms```
1988年，Noam Elkies找出一个对n= 4制造反例的方法。他给出的反例中最小的如下：
`2682440^4+ 15365639^4+ 18796760^4= 20615673^4`
Roger Frye以Elkies的技巧用电脑直接搜索，找出n= 4时最小的反例：
`95800^4+ 217519^4+ 414560^4= 422481^4`
image.png
1988年，哈佛大学的Noam Elkies在一次计算中，发现了这个等式：
image.png
【参考阅读】
1637年左右，“业余数学家之王”费马先生在阅读丢番图《算术》拉丁文译本时，曾在第11卷第8命题旁写道：“将一个立方数分成两个立方数之和，或一个四次幂分成两个四次幂之和，或者一般地将一个高于二次的幂分成两个同次幂之和，这是不可能的。关于此，我确信已发现了一种美妙的证法 ，可惜这里空白的地方太小，写不下。”
1872年，英国当时最著名的数学家凯利正式向伦敦数学学会提出了这个问题，于是四色猜想成了世界数学界关注的问题，于是又一个猜想引得无数一流数学家抛头颅洒热血。
1742年6月7日，哥德巴赫写信给欧拉，提出了一个猜想：任何一个奇数，比如77，可以把它写成三个素数之和，即77=53+17+7；又如461可以写成257+199+5，仍然是三个素数之和。即发现“任何大于5的奇数都是三个素数之和。”
1742年6月30日欧拉先生给哥德巴赫回信了：这个命题看来是正确的，但是暂给不出严格的证明。同时欧拉对上述命题做了修改：任何一个大于2的偶数都是两个素数之和。这个欧拉版本是现在常见的猜想陈述，当然，他到死也没能给予证明。
200多年过去了，至今没有完全解决。不过由此猜想带来的数学新方法则层出不穷，从另一方面促进数学自身的发展。
“马”失前蹄
17世纪法国著名的僧侣数学家马林•梅森（Mersenne）在欧几里得、费马等人有关研究的基础上对2p－1（数学界把这种数称为 “梅森数”，并以Mp记之。）作了大量的计算、验证，并于1644年在他的《物理数学随感》一书中断言：
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D is the set of all the multiples of 3 between 20 and 100. E is the se
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D is the set of all the multiples of 3 between 20 and 100. E is the se [#permalink]
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D is the set of all the multiples of 3 between 20 and 100. E is the set of all the factors of 400. Set D and Set E have how many numbers in common?
A. 0
B. 1
C. 3
D. 5
E. 12
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Originally posted by honchos on 09 Jul 2015, 23:58.
Last edited by Bunuel on 10 Jul 2015, 01:59, edited 1 time in total.
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D is the set of all the multiples of 3 between 20 and 100. E is the se [#permalink]
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10 Jul 2015, 02:03
4
honchos wrote:
D is the set of all the multiples of 3 between 20 and 100. E is the set of all the factors of 400. Set D and Set E have how many numbers in common?
A. 0
B. 1
C. 3
D. 5
E. 12
D is a set of multiples of 3.
400 is NOT a multiple of 3, hence none of its factors is a multiple of 3.
Therefore, those two sets won't have any overlap.
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Re: D is the set of all the multiples of 3 between 20 and 100. E is the se [#permalink]
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04 Dec 2017, 11:55
honchos wrote:
D is the set of all the multiples of 3 between 20 and 100. E is the set of all the factors of 400. Set D and Set E have how many numbers in common?
A. 0
B. 1
C. 3
D. 5
E. 12
We see that D = {21, 24, 27, …, 99} and E = {1, 400, 2, 200, 4, 100, 5, 80, 8, 50, 10, 40, 16, 25}.
We have listed all the elements in E; however, since none of these elements is a multiple of 3, set D and set E have no numbers in common.
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Re: D is the set of all the multiples of 3 between 20 and 100. E is the se [#permalink] 03 Feb 2019, 19:31
Display posts from previous: Sort by | HuggingFaceTB/finemath | |
Legend of the Lo Shu (= story of turtle Lo Wi)
Hello I am Lowi, a five thousand year old turtle. I live in the river Lo (the yellow river) in China. Here’s my story!
When I was young (in the year 2800 before Christ) I was a servant of the river god. When the river god was angry the river would overflow. The people of a village near by the river would place a gift by the bank of the river. They hoped the river god would accept the gift, and the river would not overflow again. Each time as the villagers placed a gift by the bank of the river I would come out of the river and walk around the gift.
One day there was a little boy near by the river. He looked at my shell and saw that my shell-pattern consisted of nine cells. The nine cells contain 1, 2, 3, 4, 5, 6, 7, 8 or 9 dots. He repeatedly counted the dots of 3 cells (horizontal, vertical or diagonal) in a row. Each time the little boy counted 15 dots.
The little boy went to the headman of the village and told him about the spots on my shell. The headman organised a meeting. The villagers deceided to place 15 gifts by the bank of the river.
I came out of the river and I walked 15 times around the presents. That’s a long time for a turtle! Then the river god appeared. He accepted the gifts and indicated to the villagers that the river would not overflow again.
So, that was a pretty exciting story, wasn’t it! Would you like to find out how to make the magic square from my story. There are eight different ways to make the 3x3 magic square. You will need the following 3 instructions for this:
[instruction 1] Always put the 5 in the middle of the square.
[instruction 2] Always put the 2, 4, 6 and 8 in one of the corners of the square.
[instruction 3] Always put the 2 and 8 and the 4 and 6 in the same diagonal (so
not in the same row or the same column).
5 5 5 5 5 5 5 5
3x3 magic square.xls | HuggingFaceTB/finemath | |
1. ## Peicewise Functions Continuity
I am doing some review, starting back at the fundamentals and I do not know why but piecewise confuse the hell out of me at times. I have the following piecewise function and have to tell if it is continuous or not on the interval [-1,1]
$f(x)=\frac{x}{|x|} x not equal to 0$
$f(x)= x = 0 when x = 0$
Sorry to be so abrupt, I know how to tell continuity with normal functions such as if it a rational fraction and long as the value that deems the function undefined is not in that interval it is continuous? also if the sign changes from left to right in a interval?
Any tips to help me with continuity
2. Originally Posted by The Power
Any tips to help me with continuity
Graph it out and make sure it doesn't have any gaps
3. We are told to evaluate it algebraically since we are not allowed to use calculators on a test. Thanks for the help though, it know its not terrible hard, its just that piecewise functions to through me off at times.
Edit: I think I finally recall my information and remove this mental block. A piecewise function is a whole function just has different rules for set intervals. So in sense if the interval for
$f(x)=\frac{x}{|x|}$ when x is not equal to 0,
Sorry I forgot latex tags. So if the rule was a tad different and allowed 0, this would cause the function to be undefined causing a "gap" or hole in the graph thus rendering it no longer continuous, however since we have the rule when x equals 0 the output is 0 this allow the graph to continue on the interval [-1,1] Please someone correct me if I am wrong.
4. $\frac{x}{{\left| x \right|}} = \left\{ {\begin{array}{rl}
{1,} & {x > 0} \\
{ - 1,} & {x < 0} \\
\end{array} } \right.$
5. As Plato said, if x< 0, |x|= -x so $\frac{|x|}{x}= -1$ for x< 1. If x> 0, |x|= x so $\frac{|x|}{x}= 1$ for x> 1.
In order for the function f(x) to be continuous at x= a, three things must be true:
1) f(a) is defined
2) $\lim_{x\rightarrow a} f(x)$ exists
3) $\lim_{x\rightarrow a} f(x)= f(a)$
(Since $\lim_{x\rightarrow a} f(x)= f(a)$ pretty much implies the two sides exist, usuallly we just state (3).)
Now, if $\lim_{x\rightarrow a} f(x)$ exists then the two "onesided limits", $\lim_{x\rightarrow a^-} f(x)$ and $\lim_{x\rightarrow a^+} f(x)$ must exist and be equal. Think about what that means here. | HuggingFaceTB/finemath | |
# A cable of span l and central dip d is subjected to uniform load w per unit horizontal length. The horizontal component of tension in the cable is
Free Practice With Testbook Mock Tests
## Options:
1. $$\frac{{{\rm{w}}{{\rm{l}}^2}}}{{4{\rm{d}}}}$$
2. $$\frac{{{\rm{w}}{{\rm{l}}^2}}}{{8{\rm{d}}}}$$
3. $$\frac{{{\rm{w}}{{\rm{l}}^2}}}{{12{\rm{d}}}}$$
4. $$\frac{{{\rm{w}}{{\rm{l}}^2}}}{{16{\rm{d}}}}$$
### Correct Answer: Option 2 (Solution Below)
This question was previously asked in
TN TRB Civil 2012 Paper
## Solution:
∑MA = 0,
$$\left( {{\rm{w}} \times {\rm{l}}} \right) \times \left( {{\rm{w}} \times {\rm{l}}} \right) \times \frac{{\rm{l}}}{2} - {{\rm{V}}_{\rm{B}}} \times {\rm{l}} = 0$$
∴ $${{\rm{V}}_{\rm{A}}} = \frac{{{\rm{Wl}}}}{2}\:and\:{{\rm{V}}_{\rm{B}}} = \frac{{{\rm{Wl}}}}{2}$$
∑Mc = 0
$$- \left( {{\rm{W}} \times \frac{{\rm{l}}}{2} \times \frac{{\rm{l}}}{4}} \right) + \left( {\frac{{{\rm{Wl}}}}{2} \times \frac{{\rm{l}}}{2}} \right) - {\rm{H}} \times {\rm{d}} = 0$$
∴ $${\rm{H}} = \frac{{{\rm{W}}{{\rm{l}}^2}}}{{8{\rm{d}}}}$$ | HuggingFaceTB/finemath | |
# Surface area of triangular prism?
Surface Area Of A Triangular Prism Step by step lesson to find the triangular prism surface area Below are the steps explaining how to find the surface area of a ... - Read more
Surface Area of a Triangular Prism by Scott Giomini ... Solving for Surface Area of Retangular prisms, Right Triangular prisms and Cylinders by ... - Read more
## Surface area of triangular prism? resources
### Surface Area of Triangular Prisms ( Read ) | Geometry | CK ...
What is the surface area of this triangular prism? ... A triangular prism has a triangular end with a base of 12 inches and a height of 9 inches.
### Triangular Prism Tutorial - Learn how to calculate volume ...
Triangular Prism Image/Diagram Triangular Prism Example : Case 1: Find the surface area and volume of a triangular prism with the given altitude 2, base 3, height 4 ...
### Volume and Surface Area of Triangular Prisms (A ...
The Volume and Surface Area of Triangular Prisms (A) math worksheet from the Measurement Worksheet page at Math-Drills.com.
### The surface area of a triangular prism - Math Central
Subject: surface area of a triangular prism Name: Amanda Who are you: Student. What is the formula for the surface area of a triangular prism?
### Surface area of a prism - Math Open Reference
Animated demonstration of the prism surface area ... Surface area of a right prism. ... Try this Change the height and dimensions of the triangular prism by dragging ...
### Surface Area of a Triangular Prism (tent)
Surface Area of a Triangular Prism (tent) • Calculate the separate areas of all surfaces and add up. • Since it is still a type of area, the answer will always be ...
### surface area of the triangular prism - Solving Math Problems
triangular prism - surface area Complete each step using the formula surface area equals
### The Surface Area of a Prism explained - Geometric Shapes ...
Understanding the surface area of a prism is essential ... the triangular faces. The surface area of ... the prism is a Regular Polygon. The surface area ...
### Surface Area Of A Triangular Prism | Surface Area Quizzes
This online quiz will test your knowledge of triangular prism surface area.
### Surface Area of Prisms - Teachers' Choice
The surface area of any prism equals the sum of the areas of its faces, ... Triangular based prism. Base shape: Triangle: base 'b', height 'h', ...
### Triangular Prism Surface Area
Lessons to find triangular prism surface area. Below are the key lessons to find the surface area of a triangular prism. We encourage the learner to go through the ...
### Surface Area of a Triangular Prism - YouTube
This is a video to help with finding the surface area of a triangular prism.
### How to Find Surface Area of a Triangular Prism: 14 Steps
Understand why this area equation does not match that of a standard triangle. When aiming to find the surface area of the equation, you should stop here.
### Online Conversion - Surface Area of a Triangular Prism
Calculate the surface area of a triangular prism. ... Did you find us useful? Please consider supporting the site with a small donation.
### How To Find The Surface Area Of A Triangular Prism (Math)
How To Find The Surface Area Of A Triangular Prism. John, a maths tutor and secondary maths teacher, shows viewers how to find the surface area of a triangular prism.
### How to Find the Surface Area of a Triangular Prism | The ...
The Classroom » Higher Education Prep » How to Find the Surface Area of a Triangular Prism; How to Find the Surface Area of a Triangular Prism by Chance E. Gartneer ...
### How to Get the Surface Area of Prisms | eHow
Sum the three doubled products to obtain the rectangular prism's surface area. ... of different surface areas and ... Surface Area of a Triangular Prism ...
### How to Find the Volume and Surface Area of a Triangular ...
Transcript: How to Find the Volume and Surface Area of a Triangular Prism. Hi, I am Eric Stone from South Burlington High School in South Burlington, Vermont, here ...
### How to Find the Surface Area of a Triangular Prism Easily ...
The surface area of any prism measures its complete exterior. The prism, a three-dimensional solid, has two identical bases, which are parallel to one another and ...
### Find surface area of triangular prisms - for teachers ...
In this lesson you will learn how to find the surface area of a triangular prism by analyzing the prism's faces, and computing their areas.
### Surface Area of Triangular Prism or Solid (with videos ...
Triangular Prisms: Given Dimensions, Find Surface Area and Volume 2 Problem: Use the dimensions to determine the surface area and volume of the box.
### Surface Area - Edmonton Public Schools
Surface Area. B = area of ... area, surface area, volume, right prism, triangular prism, rectangular prism, cube, cylinder, face, congruent, equilateral, net, prism.
### Surface Area of Triangular Prisms - Saylor
www.ck12.orgConcept 1. Surface Area of Triangular Prisms Guided Practice Here is one for you to try on your own. Find the surface area of this triangular prism.
### Surface Area of a Triangular Prism | Math@TutorNext.com
How to find the Surface Area of a Triangular Prism The surface area of a triangular prism includes the lateral surface area and the base areas.
### Surface Area and Volume of a Isosceles Triangular Prism
See next for the surface area and volume formulas for all common geometric solids.
### Surface Area of a Triangular Prism - Lesson
To find the surface area of a right triangular prism we must use this formula: SA = wh + lw + lh + ls. The variables in this formula stand for:
### Surface Area of a Triangular Prism Formula | Formulas ...
Some solved problems on the surface area of a triangular prism are given below:
### Interactivate: Finding the Surface Area of a Triangular Prism
Finding the Surface Area of a Triangular Prism Discussion: Mentor: Can anyone tell me what surface area means? Student 1: Surface area is the number of square ...
### Surface area of Triangular based prism - Mathematics
This triangular based prism shown above has a triangle base. The area of this triangular base s ½ bxh and the perimeter of this triangular base is s 1, s 2, s 3.
### Surface Area of a Triangular Prism Calculator | Calculator ...
Find the surface area of triangular prism whose sides are equilateral with 4cm, the prism height is 6 cm and base height is 3.4 cm. Step 1 : Given: The surface area ...
### Triangular Prism | Properties of a Triangular Prism | Math ...
Triangular prism consists of 2 triangular shapes and three ... Surface area of triangular prism is the sum of the lateral surface area and twice the base area ...
### Surface area of Prisms (with worked solutions & videos)
Example: Calculate the surface area of the following prism. Solution: There are 2 triangles with the base = 4 cm and height = 3 cm. Area of the 2 bases
### How Do You Find the Lateral and Surface Areas of a ...
Note: Want to know how the find the lateral and surface areas of a triangular prism? Then check out this tutorial! You'll see how to apply each formula to the given ...
Related Questions
Recent Questions | HuggingFaceTB/finemath | |
# [SOLVED] 33b-counting problem-players and line up
• Sep 5th 2009, 08:24 PM
yvonnehr
[SOLVED] 33b-counting problem-players and line up
I don't understand how to do this or which formulas to use.
A little league team has 15 players.
How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?
• Sep 5th 2009, 09:23 PM
yvonnehr
Quote:
Originally Posted by yvonnehr
I don't understand how to do this or which formulas to use.
A little league team has 15 players.
How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?
Well, I think the book's answer may be wrong on this one. Here is my solution. Is this correct?
(15 choose 9) x 9! = 181681894400
"15 choose 9" represents the number of ways you can pick 9 players from the 15 on the roster, and 9! represents the possible ordered lineups. I multiplied them together because because we are matching one set of outcomes to the other set of outcomes.
• Sep 5th 2009, 09:26 PM
Rapha
Hi yvonnehr!
Quote:
Originally Posted by yvonnehr
I don't understand how to do this or which formulas to use.
A little league team has 15 players.
How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?
To select 9 players out of 15 there are $\displaystyle \begin{pmatrix} 15\\ 9 \end{pmatrix}$ different ways.
After that there are 9 players and nine different positions (1,2,3,4...,8,9).
There are 9! ways tochoose the batting order.
So the solution of this problem is
$\displaystyle \begin{pmatrix} 15 \\ 9 \end{pmatrix}*9!$
regards
Rapha
Edit:
Quote:
Originally Posted by yvonnehr
Well, I think the book's answer may be wrong on this one. Here is my solution. Is this correct?
(15 choose 9) x 9! = 181681894400
"15 choose 9" represents the number of ways you can pick 9 players from the 15 on the roster, and 9! represents the possible ordered lineups. I multiplied them together because because we are matching one set of outcomes to the other set of outcomes.
Yes, that sounds good.
By the way what is the book's answer?
• Sep 5th 2009, 10:14 PM
garymarkhov
Quote:
Originally Posted by yvonnehr
A little league team has 15 players. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?
Are there two questions here? If you start by selecting 9 players for the starting lineup and don't care about their batting order, there are 15!/(6!*9!) = 5005 combinations. Then, once you start caring about batting order, there are 15!/6! = 1816214400 permutations.
By the way, I used to get confused by combinations and permutations all the time. No more after reading Easy Permutations and Combinations | BetterExplained | HuggingFaceTB/finemath | |
# How Many Days In 10 Years
How Many Days in 10 Years?
We’re all familiar with the fact that a year has 365 days, but did you know that 10 years comprises 3,650 days? That’s right. 10 years is equivalent to 3,650 days, providing you don’t take into account the leap years which occur every four years.
## What is a Leap Year?
Contents
A leap year is a calendar year that contains an additional day. This additional day is known as an intercalary day, or a leap day, and it is added to the calendar to keep calendar dates aligned with the astronomical or seasonal year. This is necessary because the Earth’s orbit around the sun takes approximately 365.242189 days to complete, whereas a calendar year is 365 days. This difference between the calendar year and the astronomical year is why we have leap years.
## How Many Leap Years Are There in 10 Years?
A leap year occurs every four years, and there are two leap years in every decade. Therefore, in a 10-year period, there are two leap years.
## What is the Difference if a 10-Year Period Includes Leap Years?
If a 10-year period includes two leap years, then the total number of days in the 10-year period will be 3,658 days instead of the standard 3,650 days.
## How Do We Calculate the Number of Days in 10 Years?
The easiest way to calculate the number of days in 10 years is to simply multiply the number of days in a year (365) by the number of years (10). This will give you a total of 3,650 days in 10 years.
## What is the Difference Between the Number of Days in 10 Years and the Number of Days in 10 Years During a Leap Year?
The difference between the number of days in a 10-year period during a leap year and a 10-year period without a leap year is 8 days. This is because the two leap years which take place during a 10-year period add an extra 8 days to the total number of days.
## Summary
In summary, 10 years is equivalent to 3,650 days, unless the 10-year period includes two leap years, in which case it is equivalent to 3,658 days. It is important to note that a leap year occurs every four years and that two leap years occur in every decade. | HuggingFaceTB/finemath | |
Friday, August 6, 2021
CBSE 12th Maths Question Paper and Solutions of Board Exam 2020
Download previous year question papers and answer keys of CBSE Plus Two Maths examination held on March 17, 2020. All set question papers together with their answer keys / marking scheme/ solutions are avaialble for download from the links given below
CBSE 12th Board Maths Exam 2020
The exam was said to be of moderate difficulty. We are giving question papers and solutions of all the series question papers together with the marking scheme, published by CBSE for the valuation of the answer keys.
Selected few questions
1. The total revenue received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5 in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.
2. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the window to admit maximum light through the whole opening. How having large windows help us in saving electricity and conserving environment ?
3. Three rotten apples are mixed with seven fresh apples. Find the probability distribution of the number of rotten apples, if three apples are drawn one by one with replacement. Find the mean of the number of rotten apples.
4. In a shop X, 30 tins of ghee of type A and 40 tins of ghee of type B which look alike, are kept for sale. While in shop Y, similar 50 tins of ghee of type A and 60 tins of ghee of type B are there. One tin of ghee is purchased from one of the randomly selected shop and is found to be of type B. Find the probability that it is purchased from shop Y.
5. Find the points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with both the axes. Also find the equation of the normals.
6. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. Given that total time for cutting is 3 hours 20 minutes and for assembling 4 hours. The profit for type A souvenir is 100 each and for type B souvenir, profit is Rs. 120 each. How many souvenirs of each type should the company manufacture in order to maximize the profit ? Formulate the problem as an LPP and solve it graphically.
7. A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of gold while that of B requires 1 g of silver and 2 g of gold. The company can use atmost 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs. 40 and that of type B Rs. 50, find the number of units of each type that the company should produce to maximize the profit. Formulate and solve graphically the LPP and find the maximum profit.
8. Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group. | HuggingFaceTB/finemath | |
CSC 161 Grinnell College Spring, 2010 Imperative Problem Solving and Data Structures
# Introduction to Loop Invariants
## Goals
This laboratory exercise introduces the concept of loop invariants and provides some practice in using loop invariants in developing programs.
## Approaches to Problem Solving
Problem: Read a (nonzero) number r from the keyboard and compute (and print) r0, r1, r2, r3, ..., r10.
Solutions: Program loop-invariants-1.c shows three correct solutions to this problem, illustrating that even a simple problem may be solved in a rather large number of ways. Since each code segment represents a somewhat different way of thinking about problem solving and loops, the following discussion considers each solution in some detail.
### Code Segment 1
printf ("First Solution\n");
prod = 1;
i = 0;
while (i <= 10) {
printf ("\t%6.2lf", prod);
prod *= r;
i++;
}
printf("\n");
In this first approach, at the top of the loop, i represents an exponent and prod represents an ri value that is ready to be printed. Initially, i=0 and prod=1, which sets up the values for the first pass through the loop.
Within the loop, the previously computed values are printed, and then both i and prod are updated for the next time.
To review, the code executes correctly because the following ideas work together properly.
1. At the start of each time through the loop, i represents a current exponent of r and prod represents ri — a value that has not yet been printed.
2. i and prod are correctly initialized for r0, so statement A is true when the loop first begins.
3. Both i and prod are updated correctly during each time through the loop, so statement A is true whenever the loop starts again.
4. The while-loop condition (i<=10) is consistent with the values i represents, so the loop terminates at the correct time.
### Jargon
In the statement while (i <= 10), the test (i <= 10) is sometimes called a loop-continue condition; the loop continues as long as this condition is true. In contrast, the negative of this expression i > 10 is sometimes call a exit condition.
In contrast, statement A above is called a loop invariant. A loop invariant is a statement about variables and relationships among them, where the statement is to be true every time the machine gets to the very top or bottom of a loop (whether the loop continues or not). For a while loop, therefore, a loop invariant is a statement that is true every time the Boolean expression is evaluated.
### Code Segment 2
printf ("Second Solution\n");
printf ("\t%6.2lf", 1.0);
prod = 1;
i = 0;
do {
i++;
prod *= r;
printf ("\t%6.2lf", prod);
}
while (i < 10);
printf("\n");
This approach illustrates a different meaning for the variables i and prod. Here, at the top of the loop, i represents an exponent and prod represents the ri value that has most recently been printed. Thus, in this approach, i and prod are initialized to 0 and 1=r0, and these values are printed before the loop. Then, within the loop, the i and prod are updated before the new values are printed. With this perspective of i, processing continues until i=10, since this is the last desired output value.
As with the previous code segments, we can summarize why the code works correctly with four statements.
1. At the start of each time through the loop, i represents a current exponent of r and prod represents ri — the value that has been printed most recently.
2. i and prod are correctly initialized for r0, and this value of r0 is printed, so statement A is true when the loop first begins.
3. Both i and prod are updated correctly during each time through the loop, so statement A is true whenever the loop starts again.
4. The loop-continue condition (i<10) is consistent with the values i represents, so the loop terminates at the correct time.
### Jargon, Revisited
Here, once again, Statement A is called a loop invariant, and that statement is true at the start, each time through the loop (at the top of the loop), and at the end. This notion of a loop invariant, therefore, applies to any type of loop construct, not just for while loops. In the current context, for a do-while loop, a loop invariant is a statement that is true every time the computer gets to the do and the statement is also true when execution of the loop is over.
### Code Segment 3
printf ("Third Solution\n");
printf ("\t%6.2lf", 1.0);
prod = r;
printf ("\t%6.2lf", prod);
i = 0;
while (i < 9) {
i++;
prod *= r;
printf ("\t%6.2lf", prod);
}
printf("\n");
A loop invariant for this Code Segment might be described as follows:
1. At the start of each time through the loop, i represents the number of multiplications of r with itself to obtain the product prod — thus prod = ri+1. Further, prod is the value that has been printed most recently.
From this perspective, the first two cases, r0 and r1, do not require any multiplications of r by itself and so are handled as part of initialization. Also, the final desired value, r10, requires only 9 multiplications, so processing should halt when this number of multiplications has been performed.
### Highlights
In developing correct code, our discussion has emphasized four points.
1. We write a statement of what each variable represents at the top of the loop. Such statements are called loop invariants. For example, "i represents a current exponent of r" could be a loop invariant.
2. We initialize variables before a loop, so that the loop invariants are true at the start of the loop.
3. We check the loop body updates variables appropriately, so the loop invariant remains true each time through the loop. I.e., if the variables have values prescribed by loop invariant statements at the start of the loop, then the loop body should update variables so the loop variant statements remain true when execution returns the next time to the beginning of the loop.
4. We check that the loop-continue condition stops the loop at the appropriate time -- using the loop invariants to understand clearly what the variables represent.
This laboratory exercise is based on part of an on-going project of introducing the concepts of assertions and loop invariants informally in CS1 and CS2 courses. Early funding for this work came, in part, from NSF Grant CDA 9214874, "Integrating Object-Oriented Programming and Formal Methods into the Computer Science Curriculum". Henry M. Walker worked as Senior Investigator on this portion of that effort.
This document is available on the World Wide Web as
http://www.cs.grinnell.edu/~walker/courses/161.sp10/readings/reading-loop-invariants.shtml
created 25 October 2007 last revised 27 February 2010 For more information, please contact Henry M. Walker at walker@cs.grinnell.edu. | HuggingFaceTB/finemath | |
# Mistakes for Learning Core Competancies
In my mistakes for learning assignment, I got the opportunity to re-try a question in physics that I had previously gotten wrong. While I re did it I used my personal awareness to reflect on where mistakes were made and how I will grow from them. I got to use this learning opportunity to analyze … Continue reading Mistakes for Learning Core Competancies
# Core Competencies
I used my creative thinking when we did the challenges project in Spanish. We got to choose any Spanish-speaking country and talk about an issue that they’re currently thinking about. We got to think creatively and outside of the box to think about how they’re affected especially in times like these. We also got to … Continue reading Core Competencies
# The Book Thief Museum
The Book Thief Museum: The Book Thief Museum Presentation: PHYSICAL DESIGN- This Museum shows many amazing artifacts from the novel “The Book Thief”. This Museum is designed to show the artifacts as a memorial of the events that occurred in the book, as well as artifacts that relate to the book, such as the movie “The Book Thief”. The main hall displays a sample … Continue reading The Book Thief Museum
# Unit Five Summary
In this unit we learned how to add, subtract, multiply, and divide rational expressions. We also learned how to solve rational equations and find restrictions. In this video I show what I learned about multiply and dividing rational expressions.
# Unit Four Summary
In this unit we learned different forms of writing out quadratic equations (factored form, general form, vertex/ standard form). We also learned how to graph those equations as parabolas. We now know new vocabulary words such as vertex, maximum/ minimum, congruent, parent function, and many more. Heres a video on how to Find the vertex … Continue reading Unit Four Summary
# Unit two summary
This unit in math we learned how to calculate radical equations. In this video I focused on how to divide with radicals. Video:
# Unit one Summary
The concept I chose to focus on from unit one is turning exponents into radicals. In the video I show how to turn even negative and decimal exponents into radicals. I also showed a couple important tricks to rememer when converting the exponents to radicals such as knowing “flower power” and that “exponents are lazy” … Continue reading Unit one Summary
# Dart Dash
Game design video: Core competancies: | HuggingFaceTB/finemath | |
# Celsius to Fahrenheit conversion.
Edit any of the fields below and get answer:
Celsius Fahrenheit =
How many degrees Fahrenheit in 36.6 degrees Celsius? How to convert 36.6 Celsius to Fahrenheit?
Use our Celsius to Fahrenheit converter to understand:
## What is the formula to convert 36.6 from °C to °F?
Celsius to Fahrenheit formula: [°F] = [°C] × 9⁄5 + 32
The final formula to convert 36.6 Celsius to Fahrenheit is: [°F] = 36.6 × 9⁄5 + 32 = 97.88
The 36.6 Celsius fever is unlikely to be harmful
### The Celsius scale and the Fahrenheit scale
The discovery of temperature scales is one of the most important pieces of science history in our human lives. It’s hard to imagine how life would be if we could not measure temperature, using temperature scale is so entwined with our day to day lives, deciding what to wear or when to adjust the room AC, in the manufacturing industries we monitor temperature changes using the scales and most of all the health industry relies on measuring body temperature changes for certain medical diagnostics and deductions. The interesting part in all this is that there are two major scales used to measure temperature, the Celsius scale, and the Fahrenheit scale. Many people do not know the difference between the two, countries that use the Celsius scale have troubles relating or converting to the Fahrenheit scale and the same is true for the countries that use the Fahrenheit scale. So let us look into the two temperature scales and see how they differ from each other, to do this a little background information on the two scales is necessary.
### The Celsius scale
The Celsius scale was developed by a very reputable astronomer from Sweden called Anders Celsius. Logically the scale was named after this great astronomer to honor his excellent work. It is the most used method of measuring temperature in the world and as such, it is the SI derived unit of choice scientifically. Most academic institutions teach this scale instead of the Fahrenheit scale.
Interesting fact: before the scale was renamed to honor the Swedish genius, it was call centigrade derived from Latin words centum and gradus which mean 100 and steps respectively.
When referring to a specific temperature on the scale, the term degree Celsius is adopted which is given the symbol notation °C.
The Celsius temperature scales measure temperature and is used as a value to appreciate how hot or cold something is. Normally the scale ranges from 0° for the freezing point of water and 100° for the boiling point of water. Negative values are used for temperatures below freezing point.
There is a relation between the Celsius scale and another one called the Kelvin scale, so often you will notice some conversions from either one of the scales to the other. This relationship is used a lot in thermodynamics chemistry. To convert between the Celsius scale and the Kelvin you use the relation 0 K= −273.15 °C.
### The Fahrenheit scale
From Sweden and astronomy, we go to Germany and physics, the Fahrenheit scale was developed by a German physicist called Daniel Fahrenheit and so the scale was named in honor. Similar to the Celsius scale, the Fahrenheit scale is expressed in degree Fahrenheit, symbolized as °F.
There are many theories on how Daniel Fahrenheit developed his scale. It is said that the physician used the temperature of a solution of brine and defined it as the lower point in the scale, 0 °F. He chose a solution of brine because brine is made up of the quantity of ice and salt. As he worked to develop the scale, he defined the melting point of ice to be 32 °F and he estimated the average human body temperature to be 96 °F, which was later modified by modern scientist to become 98°F when the scale was adjusted. Similar to the Celsius scale, the Fahrenheit scale is defined by two points, the freezing point of water set at 32 °F, and the boiling point of water set at 212 °F. The Fahrenheit scale was mostly used by the United States and its associated states. The rest of the world moved to the Celsius scale.
### The relation between the Fahrenheit scale and the Celsius scale
While both scales measure temperature, scientists concluded that the Celsius scale was more precise that the Fahrenheit scale which brought about the worldwide switch from degree Fahrenheit to degree Celsius. They developed a relation to converting between two, it is given by:
C = (5/9)*(F - 32)
Meaning to convert from one scale to another you would:
Celsius to Fahrenheit: Multiply by 9, divide by 5, and then add 32
Fahrenheit to Celsius: Subtract 32, then multiply by 5, then divide by 9
Interesting fact: The Fahrenheit scale and the Celsius scale are the same at -40 degrees.
This relation can also be used when converting from one scale to the other.
Convert 36.6 °C (Celsius) to °F (Fahrenheit)
Convert 36.7 °C (Celsius) to °F (Fahrenheit)
Convert 36.8 °C (Celsius) to °F (Fahrenheit)
Convert 36.9 °C (Celsius) to °F (Fahrenheit)
Convert 37 °C (Celsius) to °F (Fahrenheit)
Convert 37.1 °C (Celsius) to °F (Fahrenheit)
Convert 37.2 °C (Celsius) to °F (Fahrenheit)
Convert 37.3 °C (Celsius) to °F (Fahrenheit)
Convert 37.4 °C (Celsius) to °F (Fahrenheit)
Convert 37.5 °C (Celsius) to °F (Fahrenheit)
Convert 37.6 °C (Celsius) to °F (Fahrenheit) | HuggingFaceTB/finemath | |
# Radioactive substance has 10 ^8 nuclei, it`s half life is 30s the number of nuclei left after 15s is
5 years ago
Dear Student,
0.693/t1/2 = k
t1/2= half life
k=disintegration or rot consistent
=>0.693/30 = k
=>k=0.0231
k=2.303log(a/a-x)/t
a=initial sum
a-x= sum left after deterioration
t=time of rot
=>0.0231=2.303log(108/108-x)/15
=>108/108-x=1.41
=>Amount left=108-x=7,09,21,986 nuclei.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
2 years ago
A radioactive for the substance has 10^8 nuclei
k=0.693/t1/2
t1/2= half life (Its half life is 30s)
k=disintegration or rot consistent 0.693/30 = k
k=0.0231
k=2.303log(a/a-x)/t
a=initial sum
a-x= sum left after deterioration
t=time of rot
0.0231=2.303log(108/108-x)/15
108/108-x=1.41
Amount left=108-x=7,09,21,986 nuclei.
Therefore The no, of nuclei left after 15s is nearly = 7,09,21,986 nuclei | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Combinations similar to Ways of Summing Odd Numbers:
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Broad Topics > Decision Mathematics and Combinatorics > Combinations
### Flagging
##### Stage: 3 Challenge Level:
How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours?
### Brailler
##### Stage: 3 Challenge Level:
The machine I use to produce Braille messages is faulty and one of the pins that makes a raised dot is not working. I typed a short message in Braille. Can you work out what it really says?
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
##### Stage: 3 Challenge Level:
Is it possible to use all 28 dominoes arranging them in squares of four? What patterns can you see in the solution(s)?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Even Up
##### Stage: 3 Challenge Level:
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
### Pattern of Islands
##### Stage: 3 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Man Food
##### Stage: 2 and 3 Challenge Level:
Sam displays cans in 3 triangular stacks. With the same number he could make one large triangular stack or stack them all in a square based pyramid. How many cans are there how were they arranged?
### Sam Again
##### Stage: 3 Challenge Level:
Here is a collection of puzzles about Sam's shop sent in by club members. Perhaps you can make up more puzzles, find formulas or find general methods.
### Last Biscuit
##### Stage: 3 and 4 Challenge Level:
A game that demands a logical approach using systematic working to deduce a winning strategy
### Factoring a Million
##### Stage: 4 Challenge Level:
In how many ways can the number 1 000 000 be expressed as the product of three positive integers?
### Colour Building
##### Stage: 3 Challenge Level:
Using only the red and white rods, how many different ways are there to make up the other colours of rod?
### The Olympic Torch Tour
##### Stage: 4 Challenge Level:
Imagine you had to plan the tour for the Olympic Torch. Is there an efficient way of choosing the shortest possible route?
### Take Three from Five
##### Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### The Secret World of Codes and Code Breaking
##### Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Flight of the Flibbins
##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
##### Stage: 2, 3 and 4 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
##### Stage: 3 Challenge Level:
In a league of 5 football teams which play in a round robin tournament show that it is possible for all five teams to be league leaders.
### Dicing with Numbers
##### Stage: 3 Challenge Level:
In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Chances Are
##### Stage: 4 Challenge Level:
Which of these games would you play to give yourself the best possible chance of winning a prize?
### Coins
##### Stage: 3 Challenge Level:
A man has 5 coins in his pocket. Given the clues, can you work out what the coins are?
### Crossing the Bridge
##### Stage: 3 Challenge Level:
Four friends must cross a bridge. How can they all cross it in just 17 minutes?
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Coin Tossing Games
##### Stage: 4 Challenge Level:
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by. . . .
### Trominoes
##### Stage: 3 and 4 Challenge Level:
Can all but one square of an 8 by 8 Chessboard be covered by Trominoes?
### Nines and Tens
##### Stage: 3 Challenge Level:
Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice? | HuggingFaceTB/finemath | |
$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$
1 : Consider the function $f(x) = (\sage {p1f1}) + (\sage {p1f2}) + (\sage {p1f3}) + (\sage {p1f4})$. What are the maximum number of relative extrema that $f(x)$ could have? $\answer [id=c]{\sage {p1ans1}}$.
1.1 : What is the minimum number relative extrema that $f(x)$ could possibly have? $\answer [id=b]{\sage {p1ans2}}$
1.1.1 : Enter any number that could be a valid number of possible local extrema for $f(x)$.
$\answer [id=a]{\sage {p1ans1}}$
2 : Consider the function $f(x) = (\sage {p2f1}) + (\sage {p2f2}) + (\sage {p2f3}) + (\sage {p2f4})$. What are the maximum number of relative extrema that $f(x)$ could have? $\answer [id=cc]{\sage {p2ans1}}$.
2.1 : What is the minimum number relative extrema that $f(x)$ could possibly have? $\answer [id=bb]{\sage {p2ans2}}$
2.1.1 : Enter any number that could be a valid number of possible local extrema for $f(x)$.
$\answer [id=aa]{\sage {p2ans1}}$
3 : Consider the function $f(x) = (\sage {p3f1}) + (\sage {p3f2}) + (\sage {p3f3}) + (\sage {p3f4})$. What are the maximum number of relative extrema that $f(x)$ could have? $\answer [id=ccc]{\sage {p3ans1}}$.
3.1 : What is the minimum number relative extrema that $f(x)$ could possibly have? $\answer [id=bbb]{\sage {p3ans2}}$
3.1.1 : Enter any number that could be a valid number of possible local extrema for $f(x)$.
$\answer [id=aaa]{\sage {p3ans1}}$ | HuggingFaceTB/finemath | |
# Logistic
Chapter 3.7: Logistic
The Logistic reliability growth model has an S-shaped curve and is given by Kececioglu :
$R = \frac{1}{1+be^{-kt}}, b \gt 0, k \gt 0, T \simeq 0$
where $b\,\!$ and $k\,\!$ are parameters. Similar to the analysis given for the Gompertz curve, the following may be concluded:
1. The point of inflection is given by:
${{T}_{i}}=\frac{\ln (b)}{k}\,\!$
2. When $b\gt 1\,\!$, then ${{T}_{i}}\gt 0\,\!$ and an S-shaped curve will be generated. However, when $0\lt b\le 1\,\!$ , then ${{T}_{i}}\le 0\,\!$ and the Logistic reliability growth model will not be described by an S-shaped curve.
3. The value of $R\,\!$ is equal to 0.5 at the inflection point.
## Parameter Estimation
In this section, we will demonstrate the parameter estimation method for the Logistic model using three examples for different types of data.
## Example: Logistic for Reliability Data
Using the reliability growth data given in the table below, do the following:
1. Find a Gompertz curve that represents the data and plot it with the raw data.
2. Find a Logistic reliability growth curve that represents the data and plot it with the raw data.
Development Time vs. Observed Reliability data and Predicted Reliabilities
Time, months Raw Data Reliability (%) Gompertz Reliability (%) Logistic Reliablity (%)
0 31.00 24.85 22.73
1 35.50 38.48 38.14
2 49.30 51.95 56.37
3 70.10 63.82 73.02
4 83.00 73.49 85.01
5 92.20 80.95 92.24
6 96.40 86.51 96.14
7 98.60 90.54 98.12
8 99.00 93.41 99.09
Solution
1. The figure below shows the entered data and the estimated parameters using the standard Gompertz model.
Therefore:
\begin{align} & \widehat{a}= & 0.9999 \\ & \widehat{b}= & 0.2485 \\ & \widehat{c}= & 0.6858 \end{align}\,\!
$R=(0.9999){{(0.2485)}^{{{0.6858}^{T}}}}\,\!$
The values of the predicted reliabilities are plotted in the figure below.
Notice how the standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.
2. The least squares estimators of the Logistic growth curve parameters are given by Crow : where:
\begin{align} {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\ {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ \\ {{Y}_{i}}= & \ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ \\ \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \end{align}\,\!
In this example $N=9\,\!$, which gives:
\begin{align} \overline{Y}&=\frac{1}{9}\sum_{i=0}^{8}ln\left (\frac{1}{R_{i}}-1 \right ) \\ &= -1.7355 \\ \\ \overline{T}&=\frac{1}{9}\sum_{i=0}^{8}T_{i} = 4 \\ \sum_{i=0}^{8}T_{i}^{2} &= 204 \\ \sum_{i=0}^{8}T_{i}Y_{i} &= -106.8630 \end{align}\,\!
From the equations for $b_{i}\,\!$ and $\hat{b_{0}}\,\!$:
\begin{align} \hat{b_{1}} &= \frac{-106.8630 - 9(4)(-1.7355)}{204-9(4)_{2}} \\ & = 0.7398 \\ \hat{b_{0}} &= -1.7355 - (-0.7398)(4)\\ &= 1.2235 \end{align}\,\!
And from the least squares estimators for $\hat{b}\,\!$ and $\hat{k}\,\!$:
\begin{align} \widehat{b}= & {{e}^{1.2235}} \\ = & 3.3991 \\ \widehat{k}= & -(-0.7398) \\ = & 0.7398 \end{align}\,\!
Therefore, the Logistic reliability growth curve that represents this data set is given by:
$R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}\,\!$
The following figure shows the Reliability vs. Time plot. The plot shows that the observed data set is estimated well by the Logistic reliability growth curve, except in the region closely surrounding the inflection point of the observed reliability. This problem can be overcome by using the modified Gompertz model.
## Example: Logistic for Sequential Success/Failure Data
A prototype was tested under a success/failure pattern. The test consisted of 15 runs. The following table presents the data from the test. Find the Logistic model that best fits the data set, and plot it along with the reliability observed from the raw data.
Sequential Success/Failure Data with Observed Reliability Values
Time Result Observed Reliability
0 F 0.5000
1 F 0.3333
2 S 0.5000
3 S 0.6000
4 F 0.5000
5 S 0.5714
6 S 0.6250
7 S 0.6667
8 S 0.7000
9 F 0.6364
10 S 0.6667
11 S 0.6923
12 S 0.7143
13 S 0.7333
Solution
The first run is ignored because it was a success, and the reliability at that point was 100%. This failure will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The test essentially begins at time 1, and is now considered as time 0 with $N=14\,\!$. The observed reliability is shown in the last column of the table. Keep in mind that the observed reliability values still account for the initial suspension.
Therefore:
\begin{align} \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\ = & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ = & -0.43163 \end{align}\,\!
and:
\begin{align} \bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\ = & 6.5 \\ \underset{i=0}{\overset{13}{\mathop \sum }}\,T_{i}^{2}= & 819.0 \\ \underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69 \end{align}\,\!
Now, from the least squares estimators, the values are:
\begin{align} {{{\hat{b}}}_{1}}&= \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\ &= \frac{-61.69-14 \cdot 6.5 \cdot (-.43163)}{819.0-14 \cdot 6.5^{2}} \\ &= -0.0985 \\ \\ \hat{b_{0}}&= \overline{Y} - \hat{b_{1}}\overline{T} \\ &= (-.043163)-(-0.0985) \cdot 6.5 \\ &= 0.2087 \end{align}\,\!
Therefore:
\begin{align} \widehat{b}= & {{e}^{0.2087}} \\ = & 1.2321 \\ \widehat{k}= & -(-0.0985) \\ = & 0.0985 \end{align}\,\!
The Logistic reliability model that best fits the data is given by:
$R=\frac{1}{1+1.2321\cdot \ \,{{e}^{-0.0985T}}}\,\!$
The following figure shows the Reliability vs. Time plot.
### Example: Logistic for Grouped per Configuration Data
Some equipment underwent testing in different stages. The testing may have been performed in subsequent days, weeks or months with an unequal number of units tested every day. Each group was tested and several failures occurred. The data set is given in columns 1 and 2 of the following table. Find the Logistic model that best fits the data, and plot it along with the reliability observed from the raw data.
Grouped per Configuration Data
Number of Units Number of Failures $T_i\,\!$ Observed Reliability
10 5 0 0.5000
8 3 1 0.6250
9 3 2 0.6667
9 2 3 0.7778
10 2 4 0.8000
10 1 5 0.9000
10 1 6 0.9000
10 1 7 0.9000
10 1 8 0.9000
Solution
The observed reliability is $1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!$ and the last column of the table above shows the values for each group. With $N=9\,\!$, the least square estimator $\overline{Y}$ becomes:
\begin{align} \bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ = & -1.4036 \end{align}\,\!
and:
\begin{align} \bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\ = & 4 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,T_{i}^{2}= & 204 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 \end{align}\,\!
Now from the least squares estimators, $\hat{b_{i}}\,\!$ and $\hat{b_{0}}\,\!$, we have:
\begin{align} {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ = & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\ = & -0.2967 \\ & \\ {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ = & \left( -1.4036 \right)-\left( -0.2967 \right)\cdot 4.0 \\ = & -0.2168 \end{align}\,\!
Therefore:
\begin{align} \widehat{b}= & {{e}^{-0.2168}} \\ = & 0.8051 \\ \widehat{k}= & -(-0.2967) \\ = & 0.2967 \end{align}\,\!
The Logistic reliability model that best fits the data is given by:
$R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\!$
The figure below shows the Reliability vs. Time plot.
## Confidence Bounds
Least squares is used to estimate the parameters of the following Logistic model.
$\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}\,\!$
Thus, the confidence bounds on the parameter $b\,\!$ are given by:
$b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}\,\!$
where:
\begin{align} SE(\ln \hat{b})&=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}} \\ \\ \sigma &=\sqrt{SSE/(n-2)} \end{align}\,\!
and the confidence bounds on the parameter $k\,\!$ are:
$k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})\,\!$
where:
$SE(\hat{k})=\frac{\sigma }{\sqrt{{{S}_{xx}}}},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}\,\!$
Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability, which is:
$CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}\,\!$
### Example: Logistic Confidence Bounds
For the data given above for the reliability data example, calculate the 2-sided 90% confidence bounds under the Logistic model for the following:
1. The parameters $b\,\!$ and $k\,\!$.
2. Reliability at month 5.
Solution
1. The values of $\hat{b}\,\!$ and $\hat{k}\,\!$ that were estimated from the least squares analysis in the reliability data example are:
\begin{align} \widehat{b}= & 3.3991 \\ \widehat{\alpha }= & 0.7398 \end{align}\,\!
Thus, the 2-sided 90% confidence bounds on parameter $b\,\!$ are:
\begin{align} {{b}_{lower}}= & 2.5547 \\ {{b}_{upper}}= & 4.5225 \end{align}\,\!
The 2-sided 90% confidence bounds on parameter $k\,\!$ are:
\begin{align} {{k}_{lower}}= & 0.6798 \\ {{k}_{upper}}= & 0.7997 \end{align}\,\!
2. First, calculate the reliability estimation at month 5:
\begin{align} {{R}_{5}}= & \frac{1}{1+b{{e}^{-5k}}} \\ = & 0.9224 \end{align}\,\!
Thus, the 2-sided 90% confidence bounds on reliability at month 5 are:
\begin{align} {{[{{R}_{5}}]}_{lower}}= & 0.8493 \\ {{[{{R}_{5}}]}_{upper}}= & 0.9955 \end{align}\,\!
The next figure shows a graph of the reliability plotted with 2-sided 90% confidence bounds.
## More Examples
### Auto Transmission Reliability Data
The following table presents the reliabilities observed monthly for an automobile transmission that was tested for one year.
1. Find a Logistic reliability growth curve that best represents the data.
2. Plot it comparatively with the raw data.
3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99% be achieved?
4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of January the following year?
Reliability Data
Month Observed Reliability(%)
June 22
July 26
August 30
September 34
October 45
November 58
December 68
January 79
February 85
March 89
April 92
May 95
Solution
1. The next figure shows the entered data and the estimated parameters.
2. The next figure displays the Reliability vs. Time plot.
3. Using the QCP, the next figure displays, in months, when the reliability goal of 99% will be achieved.
4. The last figure shows the reliability at the end of January the following year (i.e., after 20 months of testing and development).
### Sequential Data from Missile Launch Test
The following table presents the results for a missile launch test. The test consisted of 20 attempts. If the missile launched, it was recorded as a success. If not, it was recorded as a failure. Note that, at this development stage, the test did not consider whether or not the target was destroyed.
1. Find a Logistic reliability growth curve that best represents the data.
2. Plot it comparatively with the raw data.
3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.5% with a 90% confidence level be achieved?
4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?
Sequential Success/Failure Data
Launch Number Result
1 F
2 F
3 S
4 F
5 F
6 S
7 S
8 S
9 F
10 S
11 F
12 S
13 S
14 S
15 S
16 S
17 S
18 S
19 S
20 S
Solution
1. The next figure shows the entered data and the estimated parameters.
2. The next figure displays the Reliability vs. Time plot.
3. The next figure displays the number of launches before the reliability goal of 99.5% will be achieved with a 90% confidence level.
4. The next figure displays the reliability achieved after the 35th launch.
### Sequential Data with Failure Modes
Consider the data given in the previous example. Now suppose that the engineers assigned failure modes to each failure and that the appropriate corrective actions were taken.
The table below presents the data.
1. Find the Logistic reliability growth curve that best represents the data.
2. Plot it comparatively with the raw data.
3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.50% be achieved?
4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?
Sequential Success/Failure Data with Modes
Launch Number Result Mode
1 F 2
2 F 1
3 S
4 F 3
5 F 3
6 S
7 S
8 S
9 F 2
10 S
11 F 1
12 S
13 S
14 S
15 S
16 S
17 S
18 S
19 S
20 S
Solution
1. The next figure shows the entered data and the estimated parameters.
2. The next figure displays the Reliability vs. Time plot.
3. The next figure displays the number of launches before the reliability goal of 99.5% will be achieved.
4. The last figure displays the reliability after the 35th launch. | HuggingFaceTB/finemath | |
# How do you solve ln (2x-5) = 3.78?
Jan 18, 2016
I found: $x = \frac{{e}^{3.78} + 5}{2} = 24.41$
#### Explanation:
I would use the definition of log to change it into an exponential as:
$2 x - 5 = {e}^{3.78}$
Rearranging:
$2 x = {e}^{3.78} + 5$
And:
$x = \frac{{e}^{3.78} + 5}{2} = 24.41$ | HuggingFaceTB/finemath | |
# Factors of 278
Factors of 278 are 1, 2, 139, and 278
#### How to find factors of a number
1. Find factors of 278 using Division Method 2. Find factors of 278 using Prime Factorization 3. Find factors of 278 in Pairs 4. How can factors be defined? 5. Frequently asked questions 6. Examples of factors
### Example: Find factors of 278
• Divide 278 by 1: 278 ÷ 1 : Remainder = 0
• Divide 278 by 2: 278 ÷ 2 : Remainder = 0
• Divide 278 by 139: 278 ÷ 139 : Remainder = 0
• Divide 278 by 278: 278 ÷ 278 : Remainder = 0
Hence, Factors of 278 are 1, 2, 139, and 278
#### 2. Steps to find factors of 278 using Prime Factorization
A prime number is a number that has exactly two factors, 1 and the number itself. Prime factorization of a number means breaking down of the number into the form of products of its prime factors.
There are two different methods that can be used for the prime factorization.
#### Method 1: Division Method
To find the primefactors of 278 using the division method, follow these steps:
• Step 1. Start dividing 278 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 278, which is 2. Divide 278 by 2 to obtain the quotient (139).
278 ÷ 2 = 139
• Step 3. Repeat step 1 with the obtained quotient (139).
139 ÷ 139 = 1
So, the prime factorization of 278 is, 278 = 2 x 139.
#### Method 2: Factor Tree Method
We can follow the same procedure using the factor tree of 278 as shown below:
So, the prime factorization of 278 is, 278 = 2 x 139.
#### 3. Find factors of 278 in Pairs
Pair factors of a number are any two numbers which, which on multiplying together, give that number as a result. The pair factors of 278 would be the two numbers which, when multiplied, give 278 as the result.
The following table represents the calculation of factors of 278 in pairs:
Factor Pair Pair Factorization
1 and 278 1 x 278 = 278
2 and 139 2 x 139 = 278
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 278. They are called negative pair factors.
Hence, the negative pairs of 278 would be ( -1 , -278 ) , ( -2 , -139 ) .
#### What is the definition of factors?
In mathematics, factors are number, algebraic expressions which when multiplied together produce desired product. A factor of a number can be positive or negative.
#### Properties of factors
• Each number is a factor of itself. Eg. 278 is a factor of itself.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 139, 278 are exact divisors of 278.
• 1 is a factor of every number. Eg. 1 is a factor of 278.
• Every number is a factor of zero (0), since 278 x 0 = 0.
• What is prime factorization of 278?
Prime factorization of 278 is 2 x 139.
• How do you find factors of a negative number? ( eg. -278 )?
Factors of -278 are -1, -2, -139, -278.
• What are the prime factors of 278?
The factors of 278 are 1, 2, 139, 278.
Prime factors of 278 are 2, 139.
• What are pair factors of 278?
The pair factors of 278 are (1,278), (2,139).
• What is the greatest prime factors of 278?
The greatest prime factor of 278 is 139.
• What are six multiples of 278?
First five multiples of 278 are 556, 834, 1112, 1390, 1668, 1946.
• What are factors of 278?
Factors of 278 are 1, 2, 139, 278.
• Which is the smallest prime factor of 278?
Smallest prime factor of 278 is 2.
• Is 278 a whole number?
Yes 278 is a whole number.
#### Examples of Factors
Kevin has been asked to write 3 factor(s) of 278. Can you predict the answer?
3 factor(s) of 278 are 1, 2, 139.
Sammy is puzzled while calculating the prime factors of 278. Can you help him find them?
Factors of 278 are 1, 2, 139, 278.
Prime factors of 278 are 2, 139
What is prime factorization of 278?
Prime factorization of 278 is 2 x 139 = 2 x 139.
Ariel has been assigned the task to find the product of all the prime factors of 278. Can you help her?
Prime factors of 278 are 2, 139.
Hence, the product of prime factors of 278.
Can you help Rubel to find out the product of the even factors of 278?
Factors of 278 are 1, 2, 139, 278.
Even factors of 278 are 2, 278.
Hence, product of even factors of 278 is; 2 x 278 = 556.
Joy wants to calculate mean of all the factors of 278. Help him in finding the mean of 278.
Factors of 278 are 1, 2, 139, 278.
To calculate the mean we need to calculate the sum of factors first. Sum of factors of 278 is 1 + 2 + 139 + 278 = 420.
Hence, the mean of factors of 278 is 420 ÷ 4 = 105.00.
Annie's mathematics teacher has asked her to find out all the positive and negative factors of 278? Help her in writing all the factors.
Positive factors are 1, 2, 139, 278.
Negative factors are -1, -2, -139, -278. | HuggingFaceTB/finemath | |
ENCYCLOPEDIA 4U .com
Web Encyclopedia4u.com
# Local ring
In abstract algebra, a local ring is a ring which has a unique maximal left ideal.
Some authors require that a local ring be (left and right) Noetherian, and the non-Noetherian rings are then called `quasi-local'. Wikipedia does not impose this requirement.
## Properties
Every local ring also has a unique maximal right ideal, and this right ideal is equal to the unique maximal left ideal (and equal to the ring's Jacobson radical) and hence a two-sided maximal ideal. (In the non-commutative case, this two-sided maximal ideal need not be the only one, however.)
The situation for commutative rings is simpler: a commutative ring is local if and only if it has a unique maximal ideal. This maximal ideal contains then precisely the non-units of the ring. In fact, a commutative ring is local if and only if the sum of two non-units is always again a non-unit.
## Examples
All fields and skew fields are local rings, since {0} is the only maximal ideal in these rings.
The kind of example that motivates the definition is the commutative ring R of real-valued continuous functions defined on some interval around 0 of the real line. The idea is that R will have a maximal ideal m consisting of all functions f in R with f(0) = 0. That m really is a maximal ideal follows easily from identifying the factor ring R/m with the field of real numbers.
To understand why R should just have this one maximal ideal, we translate that into the statement that any f in R outside m should be invertible, i.e. have a multiplicative inverse in R. This we can prove, by paying close attention to the characterisation of functions in R.
So assume f(0) is not 0 and define g by g(x) = 1/f(x) on some small interval around 0: this is a proper definition since f is continuous. We want to say that fg = 1. In fact it is 1 wherever it is defined. We have to understand that 1, the multiplicative identity in R. means a function taking the constant value 1 on some unspecified interval around 0. In order for that to work we must have 1.f = f, and that entails only considering the values of f near 0. Therefore we should identify two functions if they coincide on any interval containing 0. Then we do have a natural example of a local ring, which consists of functions (strictly, germs of functions) considered only in terms of their local behaviour at one point.
If we restricted to polynomials in R the definition would be easier, since two polynomials coinciding on a whole interval are identical. But to have the multiplicative inverses, we should make that rational functions. In that way we get the kind of example used in algebraic geometry.
Other examples of commutative local rings include the ring of rational numbers with odd denominator, and more generally the localization of any commutative ring at a prime ideal.
Non-commutative local rings arise naturally as endomorphism rings in the study of direct sum decompositions of modules over some other rings.
Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities. | HuggingFaceTB/finemath | |
# RS Aggarwal Solutions for Class 7 Maths Chapter 11 Profit and Loss
RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss are available here, so that students can check for the solutions whenever they are facing difficulty while solving the questions from RS Aggarwal Solutions for Class 7. These solutions for Chapter 11 are available in PDF format so that students can download it and learn offline as well. This book is one of the top materials when it comes to providing a question bank to practice from.
We at BYJU’S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 11 wherein, problems are formulated by our expert tutors to assist you with your exam preparation to attain good marks in Maths. Download pdf of Class 7 Chapter 11 in their respective links.
## Download PDF of RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss ### Also, access RS Aggarwal Solutions for Class 7 Chapter 11 Exercises
Exercise 11A
Exercise 11B
Exercise 11A
1. Find the SP when:
(i). CP = ₹ 950, gain = 6%
Solution:-
We have,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 6) /100) × 950)}
= {(106 /100) × 950}
= 100700/100
= ₹1007
(ii). CP = ₹ 9600, gain = 16(2/3) %
Solution:-
We have,
SP = {((100 + gain %) /100) × CP)}
= {((100 + (50/3)) /100) × 9600)}
= {(((300 + 50)/3)) /100) × 9600)}
= {((350/3) /100) × 9600}
= {((350/3) × (1/100)) × 9600}
= {(350/300) × 9600}
= {(350/3) × 96}
= {350 × 32}
= ₹11200
(iii). CP = ₹ 1540, loss = 4%
Solution:-
We have,
SP = {((100 – loss %) /100) × CP)}
= {((100 – 4) /100) × 1540)}
= {(96 /100) × 1540}
= 147840/100
= ₹1478.40
(iv). CP = ₹ 8640, loss = 12(1/2) %
Solution:-
We have,
SP = {((100 – loss %) /100) × CP)}
= {((100 – (25/2)) /100) × 8640)}
= {(((200 – 25)/2)) /100) × 8640)}
= {((175/2) /100) × 8640}
= {((175/2) × (1/100)) × 8640}
= {(175/200) × 8460}
= {1512000/200}
= ₹7560
2. Find the gain or loss percent when:
(i). CP = ₹ 2400 and SP = ₹ 2592
Solution:-
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (2592-2400)
= ₹ 192
Gain % = {(gain/CP) × 100}
= {(192/2400) × 100}
= {192/24}
= 8%
(ii). CP = ₹ 1650 and SP = ₹ 1452
Solution:-
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (1650 – 1452)
= ₹ 198
Loss % = {(Loss/CP) × 100}
= {(198/1650) × 100}
= {19800/1650}
= 12%
(iii). CP = ₹ 12000 and SP = ₹ 12800
Solution:-
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (12800 – 12000)
= ₹ 800
Gain % = {(gain/CP) × 100}
= {(800/12000) × 100}
= {800/120}
= 6(2/3) %
(iv). CP = ₹ 1800 and SP = ₹ 1611
Solution:-
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (1800 – 1611)
= ₹ 189
Loss % = {(Loss/CP) × 100}
= {(189/1800) × 100}
= {189/18}
= 10(1/2)%
3. Find the CP when:
(i). SP = ₹ 924, gain = 10%
Solution:-
By using the formula, we have:
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 10)) × 924}
= {(100/ 110) × 924}
= {92400/110}
= ₹ 840
(ii). SP = ₹ 1755, gain = 12(1/2) %
Solution:-
Gain = 12(1/2) = 25/2
By using the formula, we have:
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + (25/2))) × 1755}
= {(100/ ((200 + 25)/2)) × 1755}
= {(200/ 225) × 1755}
= {351000/225}
= ₹ 1560
(iii). SP = ₹ 8510, loss = 8%
Solution:-
By using the formula, we have:
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 8)) × 8510}
= {(100/ 92) × 8510}
= {851000/92}
= ₹ 9250
(iv). SP = ₹ 5600, loss = 6(2/3) %
Solution:-
Loss = 6(2/3) = 20/3
By using the formula, we have:
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – (20/3))) × 5600}
= {(100/ ((300 – 20)/3)) × 5600}
= {(300/ 280) × 5600}
= {168000/280}
= ₹ 6000
4. Sudhir bought an almirah for ₹ 13600 and spent ₹ 400 on its transportation. He sold it for ₹ 16800. Find his gain percent.
Solution:-
From the question,
Sudhir bought an almirah for = ₹ 13600 = cost price
Transportation cost = ₹ 400
The total cost price of almirah = ₹ (13600 + 400)
= ₹ 14000
He sold it for = ₹ 16800 = Selling price
By comparing SP and CP = SP > CP, so there is a gain
Gain = SP – CP
= 16800 – 14000
= ₹ 2800
Gain % = {(gain/CP) × 100}
= {(2800/14000) × 100}
= {2800/140}
= 20%
5. Ravi purchased an old house for ₹765000 and spent ₹115000 on its repairs. Then, he sold it a gain of 5%. How much did he get?
Solution:-
From the question,
Ravi purchased an old house for = ₹ 765000 = Cost price
He spent on its repairs = ₹ 115000
Total cost price of old house = (765000 + 115000)
= ₹ 880000
Then, he sold it at a gain of 5%
SP = {((100 + gain %) /100) × CP)}
= {((100 + 5) /100) × 880000)}
= {(105 /100) × 880000}
= 105 × 8800
= ₹ 924000
∴the selling price of the house is ₹ 924000
6. A vendor buys lemons at ₹25 per dozen and sells them at the rate of 5 for ₹ 12. Find his gain or loss percent.
Solution:-
Cost price of 12 lemons = ₹25
Then, cost price of 1 lemon = ₹ (25/12)
Cost price of 5 lemons = (25/12) × 5
= 125/12
= ₹ 10.42
He sold 5 lemons for = ₹12 =Selling price
By comparing SP and CP = SP > CP, so there is a gain
Gain = SP – CP
= 12 – 10.42
= ₹ 1.58
Gain % = {(gain/CP) × 100}
= {(1.58/10.42) × 100}
= {15800/1042}
= 15.2%
7. The selling price of 12 pens is equal to the cost price of 15 pens. Find the gain percent.
Solution:-
Let the cost price of 1 pen = ₹ 1
So, cost price of 12 pens = ₹ 12
SP of 15 pens = ₹ 15
From the question,
Selling price of 12 pens = cost price of 15 pens
Gain = SP – CP
= 15 -12
= ₹ 3
Gain % = {(gain/CP) × 100}
= {(3/12) × 100}
= {300/12}
= 25%
8. The selling price of 16 spoons is equal to the cost price of 15 spoons. Find the loss percent.
Solution:-
Let the cost price of 1 spoon = ₹ 1
So, cost price of 16 pens = ₹ 16
SP of 15 spoons = ₹ 15
From the question,
Selling price of 16 spoons = cost price of 15 spoons
Loss = (CP) – (SP)
= ₹ (16 – 15)
= ₹ 1
Loss % = {(Loss/CP) × 100}
= {(1/16) × 100}
= {100/16}
= 6.25%
= 6(1/4) %
9. Manoj purchased a video for ₹12000. He sold it to Rahul at a gain of 10%. If Rahul sells it to Rakesh at a loss of 5%, what did Rakesh pay for it?
Solution:-
From the question,
Manoj purchased a video for = ₹ 12000 = Cost price
He sold it to Rahul at a gain of = 10 %
Selling price of video from Manoj to Rahul,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 10) /100) × 12000)}
= {(110 /100) × 12000}
= 110 × 120
= ₹ 13200
∴Selling price of video from Manoj to Rahul is ₹ 13200
Then, Rahul purchase a video from Manoj at cost price of = ₹ 13200
Rahul sells it to Rakesh at Percentage of loss = 5%
Selling price of video when Rahul sells it to Rakesh,
SP = {((100 – loss %) /100) × CP)}
= {((100 – 5) /100) × 13200)}
= {(95 /100) × 13200}
= {95 × 132}
= ₹12540
∴Rakesh pay for a video is ₹12540
10. On selling a sofa-set for ₹ 21600, a dealer gains 8%. For how much did he purchase it?
Solution:-
From the question,
Dealer selling a sofa-set for = ₹21600 = Selling price
He gains on selling = 8%
Cost price of sofa-set,
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 8)) × 21600}
= {(100/ 108) × 21600}
= {2160000/108}
= ₹ 20000
11. On selling a watch for ₹ 11400, a shopkeeper loses 5%. For how much did he purchase it?
Solution:-
From the question,
Shopkeeper selling a watch for = ₹11400 = Selling price
He loses on selling = 5%
Cost price of watch,
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 5)) × 11400}
= {(100/ 95) × 11400}
= {11400} × 95
= ₹ 12000
Exercise 11B
Mark against the correct answer in each of the following:
1. A man buys a book for ₹ 80 and sells it for ₹ 100. His gain % is,
(a) 20% (b) 25% (c) 120 (d) 125%
Solution:-
(b) 25%
Because,
Cost price of book = ₹ 80
Selling price of book = ₹ 100
Since (SP) > (CP), so there is a gain
Gain = (SP) – (CP)
= ₹ (100 – 80)
= ₹ 20
Gain % = {(gain/CP) × 100}
= {(20/80) × 100}
= {(20/20) × 25}
= 25%
2. A football is bought for ₹ 120 and sold for ₹ 105. The loss % is
(a) 12(1/2)% (b) 14(2/7)% (c) 16(2/3)% (d) 13(1/3)%
Solution:-
(a) 12(1/2)%
Because,
Cost price of football = ₹ 120
Selling price of football = ₹ 105
Since (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (120 – 105)
= ₹ 15
Loss % = {(Loss/CP) × 100}
= {(15/120) × 100}
= {(15/12) ×10}
= {150/12}
=12.5%
= 12(1/2)%
3. On selling a bat for ₹ 100, a man gains ₹20. His gain % is
(a) 20% (b) 25% (c) 18% (d) 22%
Solution:-
(b) 25%
Because,
Selling price of bat = ₹ 100
Amount gain by selling bat = ₹20
Cost price of the bat = (100 – 20)
= ₹ 80
Gain % = {(gain/CP) × 100}
= {(20/80) × 100}
= {(20/20) × 25}
= 25%
4. On selling a racket for ₹198, a shopkeeper gains 10%. The cost price of the racket is
(a) ₹180 (b) ₹178.20 (c) ₹217.80 (d) ₹212.50
Solution:-
(a) ₹180
Because,
Selling price of racket = ₹ 198
Percentage gain by selling racket = 10%
Cost price of the racket,
CP = ₹ {(100/ (100 + gain %)) × SP}
= {(100/ (100 + 10)) × 198}
= {(100/ 110) × 198}
= {19800/110}
= ₹ 180
5. On selling a jug for ₹ 144, a man loses (1/7) of his outlay. If it is sold for ₹ 189, what is the gain %?
(a) 12.5% (b)25% (c) 30% (d)50%
Solution:-
(a) 12.5%
Because,
Let the CP be, ₹ x
Then, x – (1/7)x = 144
= (7x –x) = (144 × 7)
= x = (144 ×7)/6
= x = 168
∴ CP = ₹ 168,
New SP = ₹189
Gain = SP –CP
= 189 – 168
= 21
Gain % = {(gain/CP) × 100}
= {(21/168) × 100}
= {2100/168)
= 12.5%
6. On selling a pen for ₹ 48, a shopkeeper loses 20%. In order to gain 20% what would be the selling price?
(a) ₹ 52 (b) ₹ 56 (c) ₹ 68 (d) ₹ 72
Solution:-
(d) ₹ 72
Because,
Selling price of pen = ₹ 48
Shopkeeper loses = 20%
Cost price of pen =
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 20)) × 48}
= {(100/ 80) × 48}
= {4800/80}
= ₹ 60
In order to gain 20%,
SP = {((100 + gain %) /100) × CP)}
= {((100 + 20) /100) × 60)}
= {(120 /100) × 60}
= ₹ 72
## RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss
Chapter 11 – Profit and Loss contains 2 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.
• Some Terms Related to Profit and Loss
• To Find SO When CP and Gain% or Loss% are Given
• To Find CP when SP and Gain% or Loss% are Given
### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss
RS Aggarwal Solutions for Class 7 Maths Chapter 11 – Profit and Loss. The money paid by the shopkeeper to buy the products from wholesalers is called Cost Price. The rate at which the shopkeeper sells the products is called Selling Price. Profit and Loss are entirely based on cost price and selling price. In this chapter, students are going to learn the percentage gain and also percentage loss. | HuggingFaceTB/finemath | |
# After deriving a new coordinate via sequential linear transforms, how can I map translations back to the original coordinates?
I have a Bezier curve, defined with coordinates $P_1, P_2, P_3, P_4$ and apply a sequence of linear tranformations that turn these into $(0,0), (0,s), (s,s)$ and a fourth "free" coordinate (all coordinates are extended with a $z=1$ as a shortcut to do full 2D transforms)
To effect this, I'm applying the following operations:
1. translate by ${P_1}_x, {P_1}_y$ to set the curve origin to (0,0) using matrix:
$$\left [ \begin{matrix} 1 & 0 & -{P_1}_x \\ 0 & 1 & -{P_1}_y \\ 0 & 0 & 1 \end{matrix} \right ]$$
1. This generates a new set of coordinates $U_1...U_4$, which I then X-shear to align the new point 2 with x=0:
$$\left [ \begin{matrix} 1 & -\frac{{U_2}_x}{{U_2}_y} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right ]$$
1. This generates a new set of coordinates $V_1...V_4$, which get scaled so that the final points 1 through 3 will lie on (0,0), (0,s) and (s,s), where $s$ will usually be 1, but might not be so is kept symbolic:
$$\left [ \begin{matrix} \frac{s}{{V_3}_x} & 0 & 0 \\ 0 & \frac{s}{{V_2}_y} & 0 \\ 0 & 0 & 1 \end{matrix} \right ]$$
1. This generates a new set of coordinates $W_1...W_4$, which get Y-sheared to effect the final alignment of point 3 onto (s,s):
$$\left [ \begin{matrix} 1 & 0 & 0 \\ \frac{s - {W_3}_y}{{W_3}_x} & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right ]$$
Which generates a the final set of four coordinates, of which the first three are the fixed coordinates (0,0), (0,s) and (s,s), and the fourth is the "free" coordinate $F$:
$$F = \left ( \begin{matrix} \frac { s \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \\ \frac{s(-y_1+y_4)}{-y_1+y_2} + \frac { \left ( s - \frac{s(-y_1+y_3)}{-y_1+y_2} \right ) \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \end{matrix} \right )$$
Which works brilliantly to directly compute the mapped fourth coordinate. However, I'm also interested in manipulating this "resultant coordinate" and update the original point 4 accordingly, and have no idea how to invert the mapping.
Given a translated mapped coordinate (by some distance $(d'_x,d'_y)$):
$$F' = \left ( \begin{matrix} d'_x + \frac { s \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \\ d'_y + \frac{s(-y_1+y_4)}{-y_1+y_2} + \frac { \left ( s - \frac{s(-y_1+y_3)}{-y_1+y_2} \right ) \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \end{matrix} \right )$$
I'd love to know how to map that back to a $(d_x,d_y)$ value for the original point 4.
$$F' = \left ( \begin{matrix} d'_x + \frac { s \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \\ d'_y + \frac{s(-y_1+y_4)}{-y_1+y_2} + \frac { \left ( s - \frac{s(-y_1+y_3)}{-y_1+y_2} \right ) \left ( -x_1 + x_4 - \frac{(-x_1+x_2)(-y_1+y_4)}{-y_1+y_2} \right ) } { -x_1+x_3-\frac{(-x_1+x_2)(-y_1+y_3)}{-y_1+y_2} } \end{matrix} \right )$$
$$T_1^{-1} \cdot T_2^{-1} \cdot T_3^{-1} \cdot T_4^{-1} \cdot F'$$
$$\left ( \begin{matrix} \frac{-dy \cdot x_1 + (-dx+dy)x_2 + dx \cdot x_3 + s \cdot x_4)}{s} \\ \frac{-dy \cdot y_1 + (-dx+dy)y_2 + dx \cdot y_3 + s \cdot y_4)}{s} \end{matrix} \right )$$ | HuggingFaceTB/finemath | |
## Algebra 1
$p^{8}-18p^{4}q^{2}+81q^{4}$
$(p^{4}-9q^{2})^{2}$ Using the rule of $(a-b)^{2}$ = $a^{2}$-2ab+$b^{2}$ In this case, the a= $p^{4}$ and b= $-9q^{2}$ as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $(p^{4})^{2}-2(p^{4})(-9q^{2})+(-9q^{2})^{2}$ $p^{8}-18p^{4}q^{2}+81q^{4}$ | HuggingFaceTB/finemath | |
# proof of Rodrigues’ rotation formula
Let $[\mathbf{x},\mathbf{y},\mathbf{z}]$ be a frame of right-handed orthonormal vectors in $\mathbb{R}^{3}$, and let $\mathbf{v}=a\mathbf{x}+b\mathbf{y}+c\mathbf{z}$ (with $a,b,c\in\mathbb{R}$) be any vector to be rotated on the $\mathbf{z}$ axis, by an angle $\theta$ counter-clockwise.
The image vector $\mathbf{v}^{\prime}$ is the vector $\mathbf{v}$ with its component in the $\mathbf{x},\mathbf{y}$ plane rotated, so we can write
$\mathbf{v}^{\prime}=a\mathbf{x}^{\prime}+b\mathbf{y}^{\prime}+c\mathbf{z}\,,\\$
where $\mathbf{x}^{\prime}$ and $\mathbf{y}^{\prime}$ are the rotations by angle $\theta$ of the $\mathbf{x}$ and $\mathbf{y}$ vectors in the $\mathbf{x},\mathbf{y}$ plane. By the rotation formula in two dimensions, we have
$\displaystyle\mathbf{x}^{\prime}$ $\displaystyle=\cos\theta\,\mathbf{x}+\sin\theta\,\mathbf{y}\,,$ $\displaystyle\mathbf{y}^{\prime}$ $\displaystyle=-\sin\theta\,\mathbf{x}+\cos\theta\,\mathbf{y}\,.$
So
$\mathbf{v}^{\prime}=\cos\theta(a\mathbf{x}+b\mathbf{y})+\sin\theta(a\mathbf{y}% -b\mathbf{x})+c\mathbf{z}\,.$
The vector $a\mathbf{x}+b\mathbf{y}$ is the projection of $\mathbf{v}$ onto the $\mathbf{x},\mathbf{y}$ plane, and $a\mathbf{y}-b\mathbf{x}$ is its rotation by $90^{\circ}$. So these two vectors form an orthogonal frame in the $\mathbf{x},\mathbf{y}$ plane, although they are not necessarily unit vectors. Alternate expressions for these vectors are easily derived — especially with the help of the picture:
$\displaystyle\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z}$ $\displaystyle=\mathbf{v}-c\mathbf{z}=a\mathbf{x}+b\mathbf{y}\,,$ $\displaystyle\mathbf{z}\times\mathbf{v}$ $\displaystyle=a(\mathbf{z}\times\mathbf{x})+b(\mathbf{z}\times\mathbf{y})+c(% \mathbf{z}\times\mathbf{z})=a\mathbf{y}-b\mathbf{x}\,.$
Substituting these into the expression for $\mathbf{v}^{\prime}$:
$\mathbf{v}^{\prime}=\cos\theta(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z% })+\sin\theta(\mathbf{z}\times\mathbf{v})+c\mathbf{z}\,,$
which could also have been derived directly if we had first considered the frame $[\mathbf{v}-(\mathbf{v}\cdot\mathbf{z}),\mathbf{z}\times\mathbf{v}]$ instead of $[\mathbf{x},\mathbf{y}]$.
We attempt to simplify further:
$\mathbf{v}^{\prime}=\mathbf{v}+\sin\theta(\mathbf{z}\times\mathbf{v})+(\cos% \theta-1)(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z})\,.$
Since $\mathbf{z}\times\mathbf{v}$ is linear in $\mathbf{v}$, this transformation is represented by a linear operator $A$. Under a right-handed orthonormal basis, the matrix representation of $A$ is directly computed to be
$A\mathbf{v}=\mathbf{z}\times\mathbf{v}=\begin{bmatrix}0&-z_{3}&z_{2}\\ z_{3}&0&-z_{1}\\ -z_{2}&z_{1}&0\end{bmatrix}\,\begin{bmatrix}v_{1}\\ v_{2}\\ v_{3}\end{bmatrix}\,.$
We also have
$\displaystyle-(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z})$ $\displaystyle=-a\mathbf{x}-b\mathbf{y}$ (rotate $a\mathbf{x}+b\mathbf{y}$ by $180^{\circ}$) $\displaystyle=\mathbf{z}\times(a\mathbf{y}-b\mathbf{x})$ (rotate $a\mathbf{y}-b\mathbf{x}$ by $90^{\circ}$) $\displaystyle=\mathbf{z}\times(\mathbf{z}\times(a\mathbf{x}+b\mathbf{y}+c% \mathbf{z}))$ $\displaystyle=A^{2}\,\mathbf{v}\,.$
So
$\mathbf{v}^{\prime}=I\mathbf{v}+\sin\theta\,A\mathbf{v}+(1-\cos\theta)A^{2}\,% \mathbf{v}\,,$
proving Rodrigues’ rotation formula.
## Relation with the matrix exponential
Here is a curious fact. Notice that the matrix11If we want to use coordinate-free , then in this section, “matrix” should be replaced by “linear operator” and transposes should be replaced by the adjoint operation. $A$ is skew-symmetric. This is not a coincidence — for any skew-symmetric matrix $B$, we have ${(e^{B})}^{\textrm{t}}=e^{{B}^{\textrm{t}}}=e^{-B}=(e^{B})^{-1}$, and $\det e^{B}=e^{\operatorname{tr}B}=e^{0}=1$, so $e^{B}$ is always a rotation. It is in fact the case that:
$\displaystyle I+\sin\theta\,A+(1-\cos\theta)A^{2}=e^{\theta A}$
for the matrix $A$ we had above! To prove this, observe that powers of $A$ cycle like so:
$\displaystyle I,A,A^{2},-A,-A^{2},A,A^{2},-A,-A^{2},\ldots$
Then
$\displaystyle\sin\theta\,A$ $\displaystyle=\sum_{k=0}^{\infty}\frac{(-1)^{k}\theta^{2k+1}A}{(2k+1)!}=\sum_{% k=0}^{\infty}\frac{\theta^{2k+1}A^{2k+1}}{(2k+1)!}=\sum_{k\textrm{ odd}}\frac{% (\theta A)^{k}}{k!}$ $\displaystyle(1-\cos\theta)\,A^{2}$ $\displaystyle=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\theta^{2k}A^{2}}{(2k)!}=\sum% _{k=1}^{\infty}\frac{\theta^{2k}A^{2k}}{(2k)!}=\sum_{k\geq 2\textrm{ even}}% \frac{(\theta A)^{k}}{k!}\,.$
Adding $\sin\theta\,A$, $(1-\cos\theta)A^{2}$ and $I$ together, we obtain the power series for $e^{\theta A}$.
Second proof: If we regard $\theta$ as time, and differentiate the equation $\mathbf{v}^{\prime}=a\mathbf{x}^{\prime}+b\mathbf{y}^{\prime}+c\mathbf{z}$ with respect to $\theta$, we obtain $d\mathbf{v}^{\prime}/d\theta=a\mathbf{y}^{\prime}-b\mathbf{x}^{\prime}=\mathbf% {z}\times\mathbf{v}^{\prime}=A\mathbf{v}^{\prime}$, whence the solution (to this linear ODE) is $\mathbf{v}^{\prime}=e^{\theta A}\mathbf{v}$.
Remark: The operator $e^{\theta A}$, as $\theta$ ranges over $\mathbb{R}$, is a one-parameter subgroup of $\mathrm{SO}(3)$. In higher dimensions $n$, every rotation in $\mathrm{SO}(n)$ is of the form $e^{A}$ for a skew-symmetric $A$, and the second proof above can be modified to prove this more general fact.
Title proof of Rodrigues’ rotation formula ProofOfRodriguesRotationFormula 2013-03-22 15:23:25 2013-03-22 15:23:25 stevecheng (10074) stevecheng (10074) 14 stevecheng (10074) Proof msc 51-00 msc 15-00 DimensionOfTheSpecialOrthogonalGroup | open-web-math/open-web-math | |
Mixed Fraction A
Mixed Fraction B
A + B = 453/35
GENERATE WORK
## Mixed Numbers Addition - work with steps
Input Data :
Mixed Fraction A = 5 8/7
Mixed Fraction B = 6 4/5
Objective :
Find the sum of two given mixed numbers?
Solution :
5 8/7 + 5 8/7 = ?
Convert Mixed fraction into fration
5 8/7 = ((5\times7)+8)/7 = (35 + 8)/7 = 43/7
6 4/5 = ((6\times5)+4)/5 = (30 + 4)/5 = 34/5
denominator of the two fraction is different. Therefore, find lcm for two denominators (7, 5) = 35
Multiply lcm with both numerator & denominator
43/7 + 34/5 = (43\times35)/(7\times35) + (34\times35)/(5\times35)
= (43\times5)/(35) + (34\times7)/(35)
= (215)/(35) + (238)/(35)
Add two numerator of the fraction
(215)/(35) + (238)/(35) = (215 + 238)/35 = 453/35
5 8/7 + 5 8/7 = 453/35
Mixed numbers addition calculator uses two mixed numbers, i.e. two numbers in terms of a whole numbers and proper fractions, $A\frac{a}{b}$ and $B\frac{c}{d}$ for positive integers $a,b,c$ and $d$, and calculates their sum. It is an online tool for finding the sum in the simplest form of two mixed numbers and give the step by step procedure for adding two mixed numbers.
It is necessary to follow the next steps:
1. Enter two mixed numbers $A\frac{a}{b}$ and $B\frac{c}{d}$ in the box. These numbers must be in terms of whole numbers and proper fractions. The numerators and denominators in the proper fractions must be positive integers.
2. Press the "GENERATE WORK" button to make the computation;
3. Mixed numbers addition calculator will give the sum of two numbers represented as mixed numbers.
Input : Two mixed numbers
Output : A fraction in the simplest form or decimal number
Conversion of Mixed Number to Improper Fraction Rule: The mixed number $A\frac{a}{b}$ for $a,b>0$ can be rewritten as improper fraction by the following formula $$A\frac{a}{b}=\frac{A\times b+a}{b},\quad \mbox{for}\;a,b>0$$ Mixed Numbers Adding Rule: The sum of two mixed numbers $A\frac{a}{b}$ and $B\frac{c}{d}$ is determined by the following formula
• If denominators of proper fractions of mixed numbers are equal, $b=d$: $$A\frac{a}{b}+B\frac{c}{b}=\frac{A\times b+a}{b}+\frac{B\times b+c}{b}=A+B+\frac{a+c}{b},\quad \mbox{for}\;b\ne0$$
• If denominators of proper fractions of mixed numbers are different, $b\ne d$: $$A\frac{a}{b}+B\frac{c}{d}=\frac{A\times b+a}{b}+\frac{B\times d+c}{d}=\frac{(A\times b+a)\times d+(B\times d+c)\times b}{b\times d},\quad \mbox{for}\;b,d\ne0$$ or equivalently, $$A\frac{a}{b}+B\frac{c}{d}=\frac{(A\times b+a)\times \frac{LCM(b,d)}{b}+(B\times d+c)\times \frac{LCM(b,d)}{d}}{LCM(b,d)},\quad \mbox{for}\;b,d\ne0$$ where $LCM(b,d)$ is the least common multiple of $b$ and $d$
• Conversion of Improper Fraction to Mixed Number Rule:The improper fraction $\frac{a}{b}, a>b$ can be rewritten as mixed number by the following formula $$\frac{a}{b}=\Big[\frac a b\Big]\frac{a-\Big[\frac a b\Big]\times b}{b},\quad \mbox{for}\;b\ne0$$ where square brackets $[\; ]$ mean round down to the nearest integer
## How to Add Two Mixed Numbers?
A mixed number $A\frac ab$ or sometimes called a \underline{mixed fraction} represents the sum of a nonzero integer number $A$ and a proper fraction $\frac ab$. The numerator $a$ and denominator $b$ of the proper fraction must be positive integers. In the notation of mixed numbers, the sum does not explicitly use operator plus. For example, two pizza and one-third of another pizza is denoted by $2\frac 13$ instead of $2+\frac 13$. Negative mixed number, for example $-2\frac 13$ represents the sum $-(2+\frac 13)$. Mixed numbers can also be written as decimals, for example, $2\frac 12=2.5$.
Improper fractions are rational numbers where the numerator is greater than the denominator. Improper fractions can be rewritten as a mixed number in the following way:
• Divide the numerator by the denominator;
• The whole part of the quotient is the whole number of the mixed number;
• The reminder is the new numerator of the proper fraction;
• The denominator of the proper fraction is equal to the denominator of the improper fraction.
More precisely, the improper fraction $\frac{a}{b}, a>b,$ can be rewritten as a mixed number in the following way $$\frac{a}{b}=\Big[\frac a b\Big]\frac{a-\Big[\frac a b\Big]\times b}{b},\quad \mbox{for}\;b\ne0,$$ where square brackets $[\; ]$ mean round down to the nearest integer. For example, $\frac 8 5$ is equal to $1\frac 35$.
To rewrite a mixed number to an improper fraction follow the next steps:
• Multiply the denominator of the proper fraction by the whole number in the mixed number and add it to its numerator;
• The denominator of the improper fraction is equal to the denominator of the proper fraction of the mixed number.
This means, a mixed number $A\frac{a}{b}$ for $a,b>0$ can be rewritten as improper fraction in the following way $$A\frac{a}{b}=\frac{A\times b+a}{b},\quad \mbox{for}\;a,b>0$$ For example, $$10\frac 35=\frac{10\times5+3}{5}=\frac{53}5$$ When we deal with mixed numbers, there are two types of addition:
• When denominators of proper fractions of mixed numbers are equal
When denominators of proper fractions of mixed numbers are equal, then the sum of two mixed numbers can be expressed in the following way: $$A\frac{a}{b}+B\frac{c}{b}=\frac{A\times b+a}{b}+\frac{B\times b+c}{b}=A+B+\frac{a+c}{b},\quad \mbox{for}\;b\ne0$$
• When denominators of proper fractions of mixed numbers are different
When denominators of proper fractions of mixed numbers are different, to add two or more mixed numbers, it is necessary to follow the next steps:
1. Convert mixed numbers to corresponding improper fractions;
2. Find the LCM of denominators of derived improper fractions;
3. Rewrite these fractions over the LCM;
5. The result is the sum of numerators over the LCM;
6. Simplify the result if needed.
This method can be expressed algebraically: $$A\frac{a}{b}+B\frac{c}{d}=\frac{(A\times b+a)\times \frac{LCM(b,d)}{b}+(B\times d+c)\times \frac{LCM(b,d)}{d}}{LCM(b,d)},\quad \mbox{for}\;b,d\ne0$$ If $LCM(b,d)=b\times d$, then the previous formula becomes $$A\frac{a}{b}+B\frac{c}{d}=\frac{A\times b+a}{b}+\frac{B\times d+c}{d}=\frac{(A\times b+a)\times d+(B\times d+c)\times b}{b\times d},\quad \mbox{for}\;b,d\ne0$$ To add two or more mixed numbers, convert them to improper fractions then add the fractions. For example, let us find the sum for $5\frac 37$ and $6\frac 45$. After converting these numbers to improper fractions, we obtain $$5\frac 37+6\frac 45=\frac {5\times 7+3}{7}+\frac {6\times 5+4}{5} =\frac {38}7+\frac {34}5$$ Since $LCM(7,5)=7\times 5=35$, then $$5\frac 37+6\frac 45=\frac {38\times 5+34\times 7}{5\times 7}=\frac {428}{35}$$ To write the sum in simplest form, find the GCF of the numerator and denominator of the sum. Because $428$ and $35$ are relatively prime numbers, the final result is $\frac{428}{35}$. To write the sum as a mixed number, we use the above mentioned conversion from an improper fraction to a mixed number: $$\frac{428}{35}=12\frac{8}{35}$$ The similar consideration can be applied in addition of algebraic expressions.
The mixed numbers addition work with steps shows the complete step-by-step calculation for finding the sum of two mixed numbers $5\frac{3}{7}$ and $6\frac{4}{5}$ using the mixed numbers addition rule. For any other mixed numbers, just supply two mixed numbers in terms of a whole number and proper fraction and click on the "GENERATE WORK" button. The grade school students may use this adding mixed numbers calculator to generate the work, verify the results of adding numbers derived by hand, or do their homework problems efficiently.
### Real World Problems Using Mixed Numbers Addition
Mixed numbers are useful in counting whole things and parts of these things together. It is used primarily in measurement. Especially of interest are mixed numbers whose denominator of the fractional part is a power of two. They are commonly used with U.S. customary units such as inches, pounds, etc. For instance, $1\; {\rm inch}=2\frac{54}{100}\; \rm{cm}$.
### Practice Problems for Mixed Numbers Addition
Practice Problem 1:
The biggest watermelon from Joe's farm is $13\frac 13$ kilograms, which is $1\frac 23$ kilograms more than the average weight of watermelon from Ann's. What is the average weight of watermelon from Ann's farm?
Practice Problem 2:
A market opens for $3\frac 13$ hours in the morning and $4\frac 17$ hours in the afternoon. How long has the market opened for the day?
The mixed numbers addition calculator, formula, step by step calculation, real world problems and practice problems would be very useful for grade school students (K-12 education) to understand the addition of two or more numbers represented as mixed numbers. Using this concept they can be able to solve complex algebraic problems and equations. | HuggingFaceTB/finemath | |
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## Class 10 (Foundation)
### Course: Class 10 (Foundation)>Unit 3
Lesson 2: Factorising polynomials
# Strategy in factoring quadratics (part 2 of 2)
There are a lot of methods to factor quadratics, which apply on different occasions and conditions. After learning all of them in separate, let's think strategically about which method is useful for a given quadratic expression we want to factor.
## Want to join the conversation?
• At , how did he get the +1 to add to the equation when factoring out (x+3)?
• If you have a expression inside of a parentheses without any number in front, it is an invisible 1, so (x+3) is the same as 1(x+3) - distributing a 1 does not change anything. So when Sal factors, he needs the invisible 1 to make sense of the expression.
• Lets just pull a zero out of no where... ugh.
• The identity property of addition says that we can add 0 to any number without changing its value. At this point in learning to factor, students are used to seeing quadratics as trinomials (3 terms). So, putting in the missing middle term with a 0x helps students understand how the difference of two squares can be factored if you don't remember the pattern for the special product.
• I understand how to do this, but I am struggling with this one equation and cannot find anything to help me with it. y=x^2+4x-12
• While factoring quadratics, you have 2 options; grouping and the criss cross method. I personally prefer the criss-cross method for its simplicity. In this question, you'll want to ask yourself: What multiples to -12 but adds up to 4? In this case, you would know that one number would have to be positive while the other is negative, and the positive number will be larger than the absolute negative number. The answer in this question will be -2 and 6 because they fit the requirements.
This is how it would look like in the criss-cross method:
x -2 (Criss cross means the top left multiplying the bottom
x 6 right and adding it up with the top right multiplying with the bottom left)
Therefore, your solution would be (x-2)(x+6). There are methods on Khan Academy and I suggest you learn all of them, especially the criss cross method. I hope this helped!
• question guys;
ex: 1+12x+36x^2
so i did the method of grouping... here are my steps..
36^2+12x+1
(36x^2+6x)(6x+1)
6x(6x+1)(6x+1)
what did i do wrong there..
i know there are other ways of doing this problem but the way i did it, by grouping, where did i go wrong..
i know the answer is (6x+1)^2
• The error is in the second step. There should be a plus sign between the parentheses
(36x^2+6x)+(6x+1)
You should have 4 terms (items being added / subtract when you group the pairs.
Then, you find the common factor in each pair
The common factor for: (36x^2+6x) is 6x. Factor it out and you get: 6x(6x+1)+(6x+1)
Then, we need to find the common factor in the 2nd pair. It is 1. Factor it out and you get:
6x(6x+1)+1(6x+1)
We now have 2 terms. The common factor is the binomial (6x+1). Factor it out and you get: (6x+1)(6x+1) or (6x+1)^2
Hope this helps.
• In the form ax^2 + bx + c, where "a" isn't equal to one in the last example, don't you still have to divide by "a" after you're done factoring? Thanks.
(1 vote)
• Technically, no. For example you could factor that equation into: (ax+b)(x+c/b)
(1 vote)
• What about equations like V^4+v^4x^4?
• Do you mean that you would have an exponent exponented? I don't quite follow you.
• Sort of, sometimes the quadratic formula tells us there are no real solutions, so that would be your answer. we generally do not solve in the imaginary domain,
• Can 7(x^2-9) be 7(x-3)^2?
(1 vote)
• No quite because it is the difference of perfect squares, it would be 7(x-3)(x+3). | HuggingFaceTB/finemath | |
# 11260 (number)
11,260 (eleven thousand two hundred sixty) is an even five-digits composite number following 11259 and preceding 11261. In scientific notation, it is written as 1.126 × 104. The sum of its digits is 10. It has a total of 4 prime factors and 12 positive divisors. There are 4,496 positive integers (up to 11260) that are relatively prime to 11260.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 10
• Digital Root 1
## Name
Short name 11 thousand 260 eleven thousand two hundred sixty
## Notation
Scientific notation 1.126 × 104 11.26 × 103
## Prime Factorization of 11260
Prime Factorization 22 × 5 × 563
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 5630 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 11,260 is 22 × 5 × 563. Since it has a total of 4 prime factors, 11,260 is a composite number.
## Divisors of 11260
1, 2, 4, 5, 10, 20, 563, 1126, 2252, 2815, 5630, 11260
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 23688 Sum of all the positive divisors of n s(n) 12428 Sum of the proper positive divisors of n A(n) 1974 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 106.113 Returns the nth root of the product of n divisors H(n) 5.70415 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 11,260 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 11,260) is 23,688, the average is 1,974.
## Other Arithmetic Functions (n = 11260)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 4496 Total number of positive integers not greater than n that are coprime to n λ(n) 1124 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1365 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 4,496 positive integers (less than 11,260) that are coprime with 11,260. And there are approximately 1,365 prime numbers less than or equal to 11,260.
## Divisibility of 11260
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 4 4 1
The number 11,260 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
## Base conversion (11260)
Base System Value
2 Binary 10101111111100
3 Ternary 120110001
4 Quaternary 2233330
5 Quinary 330020
6 Senary 124044
8 Octal 25774
10 Decimal 11260
12 Duodecimal 6624
20 Vigesimal 1830
36 Base36 8os
## Basic calculations (n = 11260)
### Multiplication
n×y
n×2 22520 33780 45040 56300
### Division
n÷y
n÷2 5630 3753.33 2815 2252
### Exponentiation
ny
n2 126787600 1427628376000 16075095513760000 181005575484937600000
### Nth Root
y√n
2√n 106.113 22.4137 10.3011 6.46112
## 11260 as geometric shapes
### Circle
Diameter 22520 70748.7 3.98315e+08
### Sphere
Volume 5.98004e+12 1.59326e+09 70748.7
### Square
Length = n
Perimeter 45040 1.26788e+08 15924
### Cube
Length = n
Surface area 7.60726e+08 1.42763e+12 19502.9
### Equilateral Triangle
Length = n
Perimeter 33780 5.49006e+07 9751.45
### Triangular Pyramid
Length = n
Surface area 2.19603e+08 1.68248e+11 9193.75 | HuggingFaceTB/finemath | |
# Search Printable Operations and Algebraic Thinking Subtraction Worksheets
152 filtered results
152 filtered results
Subtraction
Operations and Algebraic Thinking
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Let's Practice Subtraction! 1–5
Worksheet
Let's Practice Subtraction! 1–5
Preschoolers will count to subtract birds, bugs, eggs, and more in this pictorial springtime math worksheet.
Preschool
Math
Worksheet
Color by Number: Garden Plants
Worksheet
Color by Number: Garden Plants
Children use their math skills to complete a garden tableau in this sweet color-by-number puzzle.
Kindergarten
Math
Worksheet
Let's Practice Subtraction! 1–10
Worksheet
Let's Practice Subtraction! 1–10
Children will subtract bees, hens, daffodils, and more in this playful spring-themed math worksheet.
Kindergarten
Math
Worksheet
Number Line Subtraction
Worksheet
Number Line Subtraction
Put those counting skills to the test! Use the number line to help your child understand the basics of subtraction. She'll practice subtracting one.
Kindergarten
Math
Worksheet
Practice with Ten Frames
Worksheet
Practice with Ten Frames
Each ten frame has ten boxes. Some boxes have red chips and others don't. If the top half of the ten frame is filled with red chips, how many are left? 10-5=5.
Math
Worksheet
Subtraction Color By Number: Color the Fish!
Worksheet
Subtraction Color By Number: Color the Fish!
Solve the equations and then color by number to reveal a colorful underwater friend! You'll get some great subtraction and coloring practice while you're at it.
Kindergarten
Math
Worksheet
Halloween Math: Simple Subtraction 1
Worksheet
Halloween Math: Simple Subtraction 1
Here's a way for your child to learn subtraction and practice her numbers. The Halloween theme keeps counting and math fun!
Kindergarten
Math
Worksheet
Thanksgiving Math: Simple Subtraction #2
Worksheet
Thanksgiving Math: Simple Subtraction #2
Children will practice one-digit subtraction in this Thanksgiving-themed worksheet.
Kindergarten
Math
Worksheet
Practice Adding and Subtracting with Ten Frames
Worksheet
Practice Adding and Subtracting with Ten Frames
Help your little one practice her adding subtracting skills with this game of ten frames! Each ten frame has a set of ten boxes.
Math
Worksheet
Spring Math Assessment Subtraction within 20
Worksheet
Spring Math Assessment Subtraction within 20
Use these fun toy themed math problems to assess your students’ ability to solve addition and subtraction problems within 20.
Kindergarten
Math
Worksheet
Subtraction for Kids
Worksheet
Subtraction for Kids
Ready, set... go! Start off subtraction with some picture equations, perfect for helping your child understand a new math concept.
Kindergarten
Math
Worksheet
Outer Space Math
Worksheet
Outer Space Math
Get little ones excited about math with some stellar subtraction practice!
Kindergarten
Math
Worksheet
Beginning Word Problems
Worksheet
Beginning Word Problems
Bug out on word problems! These simple story problems are perfect for helping beginners master subtraction.
Kindergarten
Math
Worksheet
Subtracting in Space
Worksheet
Subtracting in Space
Get ready for a galactic expedition! How many stars do you see? Is that a rocketship? Help your kindergartener practice their subtraction skills as they discover what is happening in space.
Kindergarten
Math
Worksheet
Subtraction on a Number Line
Worksheet
Subtraction on a Number Line
Hop, hop! Children learn to subtract with the help of a number line in this math worksheet.
Kindergarten
Math
Worksheet
Preschool Math: Take Away the Bees
Worksheet
Preschool Math: Take Away the Bees
Preschool Math: Take Away the Bees introduces subtraction to preschoolers using a fun and simple approach!
Preschool
Math
Worksheet
Spring Kindergarten Math Assessment Subtraction within 20
Worksheet
Spring Kindergarten Math Assessment Subtraction within 20
Use these fun fruit and vegetable themed math problems to assess your students’ ability to solve subtraction problems within 20.
Kindergarten
Math
Worksheet
Subtract One Assessment
Worksheet
Subtract One Assessment
Help kids grasp the concept of subtraction by taking one away in this worksheet.
Preschool
Math
Worksheet
Preschool Math: Stellar Subtraction
Worksheet
Preschool Math: Stellar Subtraction
Get your little space explorer ready for launch into kindergarten! Give him a head start in math with this stellar subtraction worksheet.
Preschool
Math
Worksheet
Subtraction on the Farm
Worksheet
Subtraction on the Farm
Students will practice their subtraction skills as they figure out how many animals are left in each problem.
Kindergarten
Math
Worksheet
Counting Up Subtraction
Worksheet
Counting Up Subtraction
For each problem on this first grade math worksheet, kids subtract single-digit numbers to see how many cheese pieces are left after the mouse eats its snack.
Kindergarten
Math
Worksheet
End of Year Math Assessment Addition and Subtraction Word Problems within 10
Worksheet
End of Year Math Assessment Addition and Subtraction Word Problems within 10
It’s almost the end of kindergarten! Use this helpful math worksheet to assess whether your students are able to distinguish between addition and subtraction equations in word problems.
Kindergarten
Math
Worksheet
Mid-Year Kindergarten Math Assessment: Subtraction within 10
Worksheet
Mid-Year Kindergarten Math Assessment: Subtraction within 10
Use these animal-themed math problems to assess your students’ ability to solve subtraction problems within 10.
Kindergarten
Math
Worksheet
Thanksgiving Math: Simple Subtraction #3
Worksheet
Thanksgiving Math: Simple Subtraction #3
Children practice simple subtraction in this holiday-themed worksheet.
Kindergarten
Math
Worksheet
Halloween Math: Simple Subtraction 2
Worksheet
Halloween Math: Simple Subtraction 2
It's not exactly the Itsy Bitsy Spider on this worksheet but the Halloween theme keeps learning subtraction fun so you can mix math practice with the holiday!
Kindergarten
Math
Worksheet | HuggingFaceTB/finemath | |
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# Top college graduates are having more difficulty
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20 Jul 2008, 23:12
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Top college graduates are having more difficulty demonstrating their superiority to prospective employers than did the top students of twenty years ago when an honors degree was distinction enough. Today’s employers are less impressed with the honors degree. Twenty years ago no more than 10 percent of a given class graduated with honors. Today, however, because of grade inflation, the honors degree goes to more than 50 percent of a graduating class. Therefore, to restore confidence in the degrees they award, colleges must take steps to control grade inflation.
Which one of the following is an assumption that, if true, would support the conclusion in the passage?
(A) Today’s students are not higher achievers than the students of twenty years ago.
(B) Awarding too many honors degrees causes colleges to inflate grades.
(C) Today’s employers rely on honors ranking in making their hiring decisions.
(D) It is not easy for students with low grades to obtain jobs.
(E) Colleges must make employers aware of the criteria used to determine who receives an honors degree
[Reveal] Spoiler: OA
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Re: CR: Top College Students [#permalink]
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20 Jul 2008, 23:15
b)
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 00:19
x97agarwal wrote:
Top college graduates are having more difficulty demonstrating their superiority to prospective employers than did the top students of twenty years ago when an honors degree was distinction enough. Today’s employers are less impressed with the honors degree. Twenty years ago no more than 10 percent of a given class graduated with honors. Today, however, because of grade inflation, the honors degree goes to more than 50 percent of a graduating class. Therefore, to restore confidence in the degrees they award, colleges must take steps to control grade inflation.
Which one of the following is an assumption that, if true, would support the conclusion in the passage?
(B) Awarding too many honors degrees causes colleges to inflate grades. The casuality is reverse , that is simply degrees cause honors so that is invalid
(C) Today’s employers rely on honors ranking in making their hiring decisions. The passage mentions about there are so many honoured students today so this is criterion is out of fashion. Today they rely on something else
(D) It is not easy for students with low grades to obtain jobs. Out of the scope of conclusion
(E) Colleges must make employers aware of the criteria used to determine who receives an honors degreeOut of scope, nothing has mentioned in the passage about criteria.
Conclusion is ''colleges must take steps to control grade inflation so in order to contro grade inflation, employers have to gain confidence for honoured students. If the criteria is known than employers will gain confidence against colleges. The answer is E What is the orginal answer?
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 01:19
2
KUDOS
It is asking for an assumption would support the conclusion in the passage?
The conclusion is to reduce grade inflation : grade inflation is only incorrect if students are at the same standard as before. if they were better, given them honors is fine. therefore, my vote is for
(A) Today’s students are not higher achievers than the students of twenty years ago.
OA plz
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 01:50
I agree with A.
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 02:57
IMO A
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 03:46
A for me as well.
A basically says that the grades are being generously given out rather than actually earned by students ... basically, students are no smarter now than they were twenty years ago
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 05:19
A
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 07:41
Sorry there was no OA available.
IMO A as well.
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 08:19
A as well.
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Re: CR: Top College Students [#permalink]
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21 Jul 2008, 08:26
C IMO
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Re: CR: Top College Students [#permalink]
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16 Aug 2010, 06:17
1
KUDOS
IMO A..
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Re: CR: Top College Students [#permalink]
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16 Aug 2010, 20:44
10000% A. A prevents the alternate cause.
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Re: CR: Top College Students [#permalink]
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17 Aug 2010, 00:56
Its A.
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Re: CR: Top College Students [#permalink]
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18 Aug 2010, 01:23
A
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Re: CR: Top College Students [#permalink]
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16 Oct 2010, 11:17
WHATS WRONG WITH B
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Re: CR: Top College Students [#permalink]
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29 Nov 2010, 06:33
B is not an assumption.
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Re: CR: Top College Students [#permalink]
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29 Nov 2010, 06:53
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Expert's post
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This post was
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mrinal2100 wrote:
WHATS WRONG WITH B
B - Awarding too many honors degrees causes colleges to inflate grades.
Actually it is the reverse. Inflated grades lead to too many honors degrees. And this is already mentioned in the stimulus.
An assumption is a necessary premise that is missing from the stimulus. It strengthens the conclusion. If the assumption is negated, the conclusion breaks apart.
Conclusion here is: to restore confidence in the degrees they award, colleges must take steps to control grade inflation.
The author is assuming that grades are inflated. That today’s students are not higher achievers than the students of twenty years ago. If it is true, then his conclusion strengthens. Colleges must take steps to control grade inflation is they want to restore confidence in their degrees.
Let's negate the assumption (A)
If today's students are actually higher achievers and that is the reason why 50% of them get honors degrees, then author's conclusion - to restore confidence in the degrees they award, colleges must take steps to control grade inflation - has no merit. Then the grades are not inflated.
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Re: CR: Top College Students [#permalink]
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29 Nov 2010, 07:56
3
KUDOS
x97agarwal wrote:
Top college graduates are having more difficulty demonstrating their superiority to prospective employers than did the top students of twenty years ago when an honors degree was distinction enough. Today’s employers are less impressed with the honors degree. Twenty years ago no more than 10 percent of a given class graduated with honors. Today, however, because of grade inflation, the honors degree goes to more than 50 percent of a graduating class. Therefore, to restore confidence in the degrees they award, colleges must take steps to control grade inflation.
Which one of the following is an assumption that, if true, would support the conclusion in the passage?
(A) Today’s students are not higher achievers than the students of twenty years ago. - correct
(B) Awarding too many honors degrees causes colleges to inflate grades. - against the paragraph, since grade inflation has caused the increase in honors degree awardees, not the vice-versa
(C) Today’s employers rely on honors ranking in making their hiring decisions. - goes against the paragraph
(D) It is not easy for students with low grades to obtain jobs. - Irrelevant..
(E) Colleges must make employers aware of the criteria used to determine who receives an honors degree - goes against the conclusion, coz if college were to make the employers aware, the college would not need to control grade inflation
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Re: CR: Top College Students [#permalink]
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19 Dec 2010, 16:58
+1 A =)
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# 3 Switches 3 Fans puzzle – Genius Logic puzzle for Interview
### Can you solve this genius logic puzzle ever? The Three switches new puzzle – Only for Genius with answer
The best logic and brain teasers puzzle asked in interviews. We always looking for genius level puzzles or interesting & logics puzzles, which gives us little bit challenge.
To solve this puzzle, you will have to think like Sherlock Holmes. Check out the terms for this logic puzzle:
## The Three Switches and Three Fans Puzzle Question:
There are three on/off switches in front of a seminar hall and the door is closed.
They all three switches are connected to three ceiling fans inside the hall.
You have to find out which switch operates which fan before you open the door.
And there is no way to check inside what’s going on, before open the door.
You can flip the switches on and off as many time as you want. But once you open the door, you will not allow to touch the switches again.
You must figure out which switch control’s which fan.
So, how do you find it, which switch connected to which fan?
Found the solution? Comment blow, let see what logic you found.
Need help? check hint blow.
Hint:
This puzzle is all about logic, if you want to solve this puzzle then think logically. The point is, what happens when you turn on the switches. Get the hint from this image blow and let see if you can solve the puzzle this time. Comment your options or solution in the comment box. If you want to confirm, you can check the answer at the end of the post.
Connect with us on facebook for more interesting & best interviews puzzles updates.
Turns ON the first two switches for 30 seconds. Now switch OFF the second one, open the door and check. You will found, one fan is ON which is connected to the first switch and then look for the fan which one is rotating slowly, this one is connected to the second switch and the last one is for last switch. According to image; 1st switch to C, 2nd to A, 3rd to B. Simple 😉
Search items:
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# Factors and primes
In this lesson, we will be investigating factors, factor pairs and prime numbers.
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Factors and Primes
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What is the definition of a factor?
1/5
Q2.What is a prime number?
Select one (1) boxes
2/5
Q3.Which of these is not a prime number?
3/5
Q4.Which of these is not a factor of 16?
4/5
Q5.Why do square numbers have an odd number of factors?
5/5
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Factors and Primes
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What is the definition of a factor?
1/5
Q2.What is a prime number?
Select one (1) boxes
2/5
Q3.Which of these is not a prime number?
3/5
Q4.Which of these is not a factor of 16?
4/5
Q5.Why do square numbers have an odd number of factors?
5/5
# Lesson summary: Factors and primes
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga | HuggingFaceTB/finemath | |
Third edition of Artificial Intelligence: foundations of computational agents, Cambridge University Press, 2023 is now available (including the full text).
## 4.2 Possible Worlds, Variables, and Constraints
To keep the formalism simple and general, we develop the notion of features without considering time explicitly. Constraint satisfaction problems will be described in terms of possible worlds.
When we are not modeling change, there is a direct one-to-one correspondence between features and variables, and between states and possible worlds. A possible world is a possible way the world (the real world or some imaginary world) could be. For example, when representing a crossword puzzle, the possible worlds correspond to the ways the crossword could be filled out. In the electrical environment, a possible world specifies the position of every switch and the status of every component.
Possible worlds are described by algebraic variables. An algebraic variable is a symbol used to denote features of possible worlds. Algebraic variables will be written starting with an upper-case letter. Each algebraic variable V has an associated domain, dom(V), which is the set of values the variable can take on.
For this chapter, we refer to an algebraic variable simply as a variable. These algebraic variables are different from the variables used in logic, which are discussed in Chapter 12. Algebraic variables are the same as the random variables used in probability theory, which are discussed in Chapter 6.
Symbols and Semantics
Algebraic variables are symbols.
Internal to a computer, a symbol is just a sequence of bits that can be distinguished from other symbols. Some symbols have a fixed interpretation, for example, symbols that represent numbers and symbols that represent characters. Symbols that do not have fixed meaning appear in many programming languages. In Java, starting from Java 1.5, they are called enumeration types. Lisp refers to them as atoms. Usually, they are implemented as indexes into a symbol table that gives the name to print out. The only operation performed on these symbols is equality to determine if two symbols are the same or not.
To a user of a computer, symbols have meanings. A person who inputs constraints or interprets the output associates meanings with the symbols that make up the constraints or the outputs. He or she associates a symbol with some concept or object in the world. For example, the variable HarrysHeight, to the computer, is just a sequence of bits. It has no relationship to HarrysWeight or SuesHeight. To a person, this variable may mean the height, in particular units, of a particular person at a particular time.
The meaning associated with a variable-value pair must satisfy the clarity principle: an omniscient agent - a fictitious agent who knows the truth and the meanings associated with all of the symbols - should be able to determine the value of each variable. For example, the height of Harry only satisfies the clarity principle if the particular person being referred to and the particular time are specified as well as the units. For example, we may want to reason about the height, in centimeters, of Harry Potter at the start of the second movie of J. K. Rowling's book. This is different from the height, in inches, of Harry Potter at the end of the same movie (although they are, of course, related). If you want to refer to Harry's height at two different times, you must have two different variables.
You should have a consistent meaning. When stating constraints, you must have the same meaning for the same variable and the same values, and you can use this meaning to interpret the output.
The bottom line is that symbols can have meanings because we give them meanings. For this chapter, assume that the computer does not know what the symbols mean. A computer can only know what a symbol means if it can perceive and manipulate the environment.
A discrete variable is one whose domain is finite or countably infinite. One particular case of a discrete variable is a Boolean variable, which is a variable with domain {true, false}. If X is a Boolean variable, we write X=true as its lower-case equivalent, x, and write X=false as ¬x. We can also have variables that are not discrete; for example, a variable whose domain corresponds to a subset of the real line is a continuous variable.
Example 4.2: The variable Class_time may denote the starting time for a particular class. The domain of Class_time may be the following set of possible times:
dom(Class_time)={8, 9, 10, 11, 12, 1, 2, 3, 4, 5}.
The variable Height_joe may refer to the height of a particular person at a particular time and have as its domain the set of real numbers, in some range, that represent the height in centimeters. Raining may be a Boolean random variable with value true if it is raining at a particular time.
Example 4.3: Consider the electrical domain depicted in Figure 1.8.
• S1_pos may be a discrete binary variable denoting the position of switch s1 with domain {up, down}, where S1_pos=up means switch s1 is up, and S1_pos=down means switch s1 is down.
• S1_st may be a variable denoting the status of switch s1 with domain {ok, upside_down, short, intermittent, broken}, where S1_st=ok means switch s1 is working normally, S1_st=upside_down means switch s1 is installed upside down, S1_st=short means switch s1 is shorted and acting as a wire, S1_st=intermittent means switch S1 is working intermittently, and S1_st=broken means switch s1 is broken and does not allow electricity to flow.
• Number_of_broken_switches may be an integer-valued variable denoting the number of switches that are broken.
• Current_w1 may be a real-valued variable denoting the current, in amps, flowing through wire w1. Current_w1=1.3 means there are 1.3 amps flowing through wire w1. We also allow inequalities between variables and constants as Boolean features; for example, Current_w1 ≥ 1.3 is true when there are at least 1.3 amps flowing through wire w1.
Example 4.4: A classic example of a constraint satisfaction problem is a crossword puzzle. There are two different representations of crossword puzzles in terms of variables:
1. In one representation, the variables are the numbered squares with the direction of the word (down or across), and the domains are the set of possible words that can be put in. A possible world corresponds to an assignment of a word for each of the variables.
2. In another representation of a crossword, the variables are the individual squares and the domain of each variable is the set of letters in the alphabet. A possible world corresponds to an assignment of a letter to each square.
Possible worlds can be defined in terms of variables or variables can be defined in terms of possible worlds:
• Variables can be primitive and a possible world corresponds to a total assignment of a value to each variable.
• Worlds can be primitive and a variable is a function from possible worlds into the domain of the variable; given a possible world, the function returns the value of that variable in that possible world.
Example 4.5: If there are two variables, A with domain {0,1,2} and B with domain {true,false}, there are six possible worlds, which you can name w0,..., w5. One possible arrangement of variables and possible worlds is
• w0: A=0 and B=true
• w1: A=0 and B=false
• w2: A=1 and B=true
• w3: A=1 and B=false
• w4: A=2 and B=true
• w5: A=2 and B=false
Example 4.6: The trading agent, in planning a trip for a group of tourists, may be required to schedule a given set of activities. There can be two variables for each activity: one for the date, for which the domain is the set of possible days for the activity, and one for the location, for which the domain is the set of possible towns where it may occur. A possible world corresponds to an assignment of a date and a town for each activity. | HuggingFaceTB/finemath | |
For a given sequence \{a_n\} , a_p+a_q=a_{p+q} where p,q∈N^{*}. Prove a_n=2n when a_1=2.
#### Question
For a given sequence \{a_n\} , a_p+a_q=a_{p+q} where p,q∈N^{*}. Prove a_n=2n when a_1=2.
Collected in the board: Arithmetic sequence
Steven Zheng posted 6 months ago
Let p=1 and q=n
a_p+a_q=a_{p+q}
becomes
a_1+a_n=a_{n+1}
which is the recursive formulas for an arithmetic sequence.
When a_1=2, using the formula for the general term of an arithmetic sequence
a_n = a_1 + (n-1) d
in which d is the common difference of two consecutive terms of the sequence.
Therefore,
a_n=2+2n-2=2n
Steven Zheng posted 6 months ago
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# Solution for Estimate the Speed with Which Electrons Emitted from a Heated Emitter of an Evacuated Tube Impinge on the Collector Maintained at a Potential Difference of 500 V with Respect to the Emitter. - CBSE (Science) Class 12 - Physics
ConceptExperimental Study of Photoelectric Effect
#### Question
Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg−1.
#### Solution
Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
KE = 1/2 mv^2 = eV
:. v = ((2eV)/m)^(1/2) = (2V xx e/m)^(1/2)
= (2 xx 500 xx 1.76 xx 10^11)^(1/2) = 1.327 xx 10^ "m/s"
Therefore, the speed of each emitted electron is 1.327 xx 10^7 "m/s"
Is there an error in this question or solution?
#### APPEARS IN
NCERT Physics Textbook for Class 12 Part 2 (with solutions)
Chapter 11: Dual Nature of Radiation and Matter
Q: 20.1 | Page no. 409
Solution for question: Estimate the Speed with Which Electrons Emitted from a Heated Emitter of an Evacuated Tube Impinge on the Collector Maintained at a Potential Difference of 500 V with Respect to the Emitter. concept: Experimental Study of Photoelectric Effect. For the course CBSE (Science)
S | HuggingFaceTB/finemath | |
# TI-82: Statistics Mode
## 1-Variable Statistics
There is one routine on the calculator, called 1-Variable Statistics which will find
• the sample mean
• the sum of the data values
• the sum of the squares of the data values
• the sample standard deviation
• the population standard deviation
• the sample size
• the minimum value
• the first quartile
• the median
• the third quartile
• the maximum value
It also saves all of these values to variables so they can be used in other calculations without having to re-enter the information.
### Obtaining 1-Variable Statistics
1. Enter the data. If the raw data is used, it goes into one list only. If a frequency distribution is to be entered, then put the data values into one list and the corresponding frequency into another list.
2. Calculate the 1-Variable Statistics (STATS CALC 1)
3. Be sure to specify which list your data is in when you do the 1-Variable Statistics.
#### Examples
1-VAR STATS L1
This will find the 1-Variable statistics for the data contained in list 1. Each element in list 1 is assumed to have a frequency of 1.
1-VAR STATS L1,L2
This will find the 1-Variable statistics for the data contained in list 1. Each element in list 1 is assumed to have a frequency of the corresponding element in list 2.
### Using the 1-Variable Statistics
The results of the 1-Variable Statistics (and any other statistical calculations done later) are under the VARS key, menu choice number 5: STATISTICS. Go ahead and press that now "VARS 5"
#### Variance
Since the calculator gives the standard deviations instead of the variances, the first thing often found is the variance. The sample standard deviation is number 3 and the population standard deviation is number 4.
``` VARS 5 3 x2 - sample variance
VARS 5 4 x2 - population variance```
#### Variation
The variation is used as a step in finding the variance and ultimately the standard deviation when calculating those values by hand. Since we are going to concentrate on understanding the statistics and letting the calculator concentrate on finding the statistics, we aren't going to use the variation very often.
One place that the variation will come into use is when we're working with Analysis of Variance (ANOVA). The total variation can be found by multiplying the degrees of freedom by the sample variance or the sample size by the population variance.
``` ( VARS 5 1 - 1 ) * VARS 5 3 x2
VARS 5 1 * VARS 5 4 x2```
#### Standard Scores
The way to compute the z-score is to subtract the mean from each value and then divide by the standard deviation. Since the mean and standard deviation are stored in variables, you can use them in the formula directly. This expression will convert the entire list at once. If you wish to find the z-score for a single value, then replace the L1 with that value.
` ( L1 - Xbar ) / Sx`
To generate the Xbar, press: VARS 5 2
To generate the Sx, press: VARS 5 3
#### Interquartile Range
The quartiles are stored under VARS STATISTICS(5) BOX.
` Q3 - Q1`
or
` VARS 5 BOX 3 - VARS 5 BOX 1`
#### Range
` MAXX - MINX`
or
` VARS 5 9 - VARS 5 8`
#### Midrange
` ( MAXX + MINX ) / 2`
or
` ( VARS 5 9 - VARS 5 8 ) / 2`
#### Cutoff for mild outliers
` Q1 - 1.5 * (Q3 - Q1)`
and
` Q3 + 1.5 * (Q3 - Q1)`
#### Cutoff for extreme outliers
` Q1 - 3 * (Q3 - Q1)`
and
` Q3 + 3 * (Q3 - Q1)` | HuggingFaceTB/finemath | |
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multiplier effect formula
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## multiplier effect formula
This formula has established a straight connection with investments in economy and job creation. Theoretically, this leads to a money (supply) reserve multiplier formula of: MSRM=1RRRwhere:MSRM=Money supply reserve multiplier\begin{aligned}&\text{MSRM}=\frac{1}{\text{RRR}}\\&\textbf{where:}\\&\text{MSRM}=\text{Money supply reserve multiplier}\\&\text{RRR}=\text{Reserve requirement ratio}\end{aligned}MSRM=RRR1where:MSRM=Money supply reserve multiplier. If consumers save 20% of new income and spend 80% of new income then their marginal propensity to consume (MPC) is 0.8. This formula works on a premise that the expenditure of one person is the income for another person apart from that portion which is being saved by the earning person. For example, when looking at banks with the highest required reserve requirement ratio, which was 10% prior to Covid-19, their money supply reserve multiplier would be 10 (1/.10). This is a requirement determined by the country's central bank, which in the United States is the Federal Reserve. The multiplier ratio is a ratio between change in real income and the initial injection. As banks hold more in reserves, less individuals and business receive loans restricting the amount of cash available in the economy. If the marginal propensity to consume is 0.8 or 80% then calculate the multiplier in this case. If the marginal propensity to consume is 0.8 or 80% then calculate the multiplier in this case. Total reserves = required reserves + excess reserves. The money supply multiplier effect can be seen in a country's banking system. Monetary policy refers to the actions undertaken by a nation's central bank to control money supply and achieve sustainable economic growth. This means that the multiplier effect was 2 ($200,000/$100,000). We can calculate a multiplier and a total effect on GDP. Looking at the example below provides some additional insight. Holding a thought on the occurrence of multiplier process and effect, let’s see this scenario to make it more understandable. For the calculation of the multiplier formula in economics, the formula used is. Thus, depending on the type of investment, it may have widespread effects on the economy at large. In general, the multiplier used in gauging the multiplier effect is calculated as follows: Multiplier = Change in Income ÷ Change in Spending The multiplier effect can be observed in several different types of situations and used by a variety of different analysts when examining and estimating expectations for new capital investments. So that is injection / 0.2. The Federal Reserve watches this effect closely to see what banks are holding in reserves. There is an assumption that the expenditure of one person is another person’s income in the form of profit, wages, salaries, etc and that person, in turn, spend it again majorly on the consumption front. Important Terms: 1. Figure 1 Diagram to show Multiplier effect. A decrease in government expenditures decreases GDP by a multiple in the same fashion. In a closed economy we ignore exports and imports. The government wants to build hundreds of hospital in a near town, he gives a call to all the builders and injects about $500 million in that project. Simply put, every$1 of investment produced an extra 2 of income. Syllabus: Use the multiplier to calculate the effect on GDP of a change in an injection in investment, government spending or exports (I,G,X). In general, the multiplier used in gauging the multiplier effect is calculated as follows: Multiplier=Change in IncomeChange in Spending\begin{aligned}\text{Multiplier}=\frac{\text{Change in Income}}{\text{Change in Spending}}\end{aligned}Multiplier=Change in SpendingChange in Income. If banks are lending less, then their multiplier will be lower and the money supply will also be lower. The multiplier effect refers to the proportional amount of increase, or decrease, in final income that results from an injection, or withdrawal, of spending. In the above-mentioned example, the government has invested 100,000 in the economy for infrastructure development and after the application of Multiplier effect ( k ) which resulted in multiples of 5, the real GDP would increase to 5,00,000. Reserve Ratio = 5.5% Therefore, the calculation of money multiplier will be as follows, Money Multiplier will be – =1 / 0.055 = 18.18 Hence, the money multiplier would be 18.18 Accessed July 27, 2020. Login details for this Free course will be emailed to you, This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. "Reserve Requirements." Public spending triggers a domino or ripple effect, with additional spending occurring throughout the whole economy. If we assume that the injection of government expenditures for example is 100 billion yen. The first level, dubbed M1, refers to all of the physical currency in circulation within an economy. Using an MPC multiplier, the equation would be: MPC Multiplier=11−MPC=11−0.8=5where:\begin{aligned}&\text{MPC Multiplier}=\frac{1}{1-\text{MPC}}=\frac{1}{1-0.8}=5\\&\textbf{where:}\\&\text{MPC}=\text{Marginal propensity to consume}\end{aligned}MPC Multiplier=1−MPC1=1−0.81=5where:. Sound knowledge of this concept helps to judge and understand various cycles of business as it entails an understanding which is fairly accurate. From the investment of government let’s see how multiplier effect occurs. Economists and bankers often look at a multiplier effect from the perspective of banking and money supply. Answer. MPS – Marginal Propensity to Save 3. Therefore, the multiplier is 5 – which means the initial1 million investment would provide a $5 million stimulus to the wider economy. The marginal propensity to consume is 0.9 which will remain constant over the period of time. Image by Sabrina Jiang © Investopedia 2020, What Is the Money Supply? Let’s assume that the govt. = 1/( 1 – 0.8) 3. For example, suppose that investment demand increases by one. The multiplier’s value can be calculated by using the following formula: Muliplier = 1 ÷ (1 – marginal propensity to consume) If the marginal propensity to consume are 0.25 or 0.50, the multipliers are 1.33 and 2 respectively. It is important to remember that when income is spent, this spending becomes someone else’s income, and so on. Some economists also like to factor in estimates for savings and consumption. This also closely affect… Investment Banking Training (117 Courses, 25+ Projects), 117 Courses | 25+ Projects | 600+ Hours | Full Lifetime Access | Certificate of Completion, Change in Real GDP = Investment * Multiplier. Is it Important?" MRT – Marginal Rate of Taxation 4. This additional income would follow the pattern of marginal propensity to save and consume. That is the injection to the economy divided by 1- 0.8. On the contrary, in periods of overfull employment, a decline in investment will have a serious effect on the levels of income and employment where the MPS is high (or MPC is low). We got the following data for the calculation of the multiplier effect. The government will spend$25 bn and there will be $10 bn increase in taxes collected. This multiplier is called the money supply multiplier or just the money multiplier. You can learn more from the following articles –, Copyright © 2021. So effect on the budget:$10 – $25 =$-15 bn; Also, I remember while preparing for the IB Economics exam there was one question in one of the maths papers. The size of the multiplier depends on the percentage of deposits that banks are required to hold as reserves. Even though the multiplier formula in economics has various limitations, it has a far-reaching impact on economic decisions and policymaking of the nation. Multiplier formula denotes an effect which initiates because of increase in the investments (from the government or corporate levels) causing the proportional increase in the overall income of the economy, and it is also observed that this phenomenon works in the opposite direction too (the decrease in income effects a decrease in the overall spending). I.e. What is the definition of multiplier effect?More broadly, this concept is simply the expansion of economic activity due to the increase of one single activity. If a second borrower subsequently deposits funds received from the lending institution, this raises the value of money supply even though no additional physical currency actually exists to support the new amount. Calculation of multiplier effect formula is as follows –. Many economists believe that new investments can go far beyond just the effects of a company’s income. Generally, economists are usually the most interested in how capital infusions positively affect income. It is also described as the number of times the change in income exceeds the injection which caused the autonomous change in the first place. Most economists view the money multiplier in terms of reserve dollars and that is what the money multiplier formula is based on. Reserve requirements refer to the amount of cash that banks must hold in reserve against deposits made by their customers. Focus on lending and the money multiplier in this video I explain how to calculate the spending multiplier a... Total effect on GDP be expressed as ∆C/∆Y, which is a change real... Saving versus spending helps one to understand the amount of expected money supply multiplier or ( )... With the examples and downloadable excel sheet the company ’ s income increases by one along with the and... 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Means that every$ 1 of new income will generate $2 of extra income the Covid-19 pandemic eliminating! The larger the MPC of a standard multiplier, using a money multiplier analyze... The MPC the larger the multiplier effect importance of the multiplier ratio is a requirement determined by the 's..., unbiased content in our this changed as the Fed responded to the amount cash. To factor in estimates for savings and consumption, economists are usually the most interested in how capital positively... This case white papers, government data, original reporting, and public private... Accounts for a diverse candidate the period of recession or economic uncertainty, when unemployment of labor is and. Exploited by governments attempting to use fiscal stimulus policies to increase the general level of economic activity this has a. And public or private investors to remain invested and increasing investments in the private sector of physical... 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Of business as it entails an understanding which is fairly accurate the spending multiplier it entails an understanding is... Copyright © 2021 – 1 has a far-reaching impact on economic decisions and policymaking of the nation 1 investment... Lend out or invest economic uncertainty, when unemployment of labor is high and other resources are.. Please keep in mind that these clips are not designed to teach the... By 0.2, which is a change in the same fashion, when unemployment of labor is high and resources! Syllabus: Draw a Keynesian AD/AS diagram to show the impact of the physical currency in circulation an! On an on till the saving gets equal to the economy at.... Spending multiplier gauge the economic condition of the country 's central bank control... Income, and interviews with industry experts video I explain how to calculate the multiplier ; the multiplier.! Within an economy you can learn more about the standards we follow in producing accurate, unbiased content in.... What banks are lending more than their reserve requirement is 10 %, also. Yen divided by 1- 0.8 increased means that every $1 of investment an... Some economists also like to factor in estimates for savings and consumption cycles of business as it entails understanding. Maximum capacity, the company ’ s income increases by one see what banks are holding in reserves in! By the simple spending multiplier monetary policy refers to all of the nation and heavily... When unemployment of labor is high and other resources are underutilized example 100. We also reference original research from other reputable publishers where appropriate the builders demands for$ 500 million of to! ( 2 marks ) those funds view the money multiplier formula in has... Increasing investments in economy and job creation period of recession or economic,., creating income for other firms and individuals supply reserve multiplier increases and versa. Be $10 bn increase in bank lending should translate to an input... 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Is also another variation of a standard multiplier, using a money multiplier in this case measure how of... We follow in producing accurate, unbiased content in our production dollar creates extra of! Creating more money supply requirement decreases the money supply below provides some additional insight the occurrence multiplier! Endorse, Promote, or Warrant the Accuracy or Quality of WallStreetMojo 1 / ( –. The formula that economists use to describe this consumption, economists are the... Or Warrant the Accuracy or Quality of WallStreetMojo an exogenous increase in demand of withdrawal income are! And a total effect on national income and the money multiplier the same fashion in income loans the. And money supply and achieve sustainable economic growth investments in economy and job.. Extra income of the injection of new income will generate $2 of.... Diagram to show the impact of the nation demand increases by$ 200,000 holding a thought the... The private sector of the multiplier effect from the following data for the multiplier in this example every! Look at a multiplier refers to the town of Ceelo, where Margie has just inherited 1,000,000. Key concepts, simply multiply the initial injection … the multiplier effect formula is as follows –.. Suppose that investment demand increases by $200,000 was 2 ($ 200,000/ $100,000 ) ;. Investment demand increases by$ 200,000 by governments attempting to use primary sources to support their work the of! Input that amplifies the effect of some other variable government spending private investors remain! Assume the MPC the larger the MPC the larger the multiplier process also operates in.. Formula has established a straight connection with investments in economy and job creation effects on the percentage of deposits banks! Economy at large consume is 0.8 or 80 % then calculate the formula. The infrastructure project in the private sector of the multiplier ratio is 5.5 prevailing! Uses and importance of the multiplier will be \$ 10 in money supply by a multiple the! To gauge the economic condition of the economy the saving gets equal to the of. At maximum capacity, the company ’ s the formula for the calculation of multiplier a between! A nation this changed as the Fed responded to the Covid-19 pandemic by eliminating these requirements to up... Helps to gauge the economic condition of the multiplier effect occurs / ( 1 – MPC ) 2 lending generated! National policymaking economic decisions and policymaking of the nation ripple effect, with additional spending occurring throughout the whole.... See what banks are lending less, then calculate the multiplier formula is based on taxation and money! Reference original research from other reputable publishers where appropriate understand various cycles of as. | open-web-math/open-web-math | |
# Can we make the set of all non-negative integers a field? [duplicate]
Can we define any kind of addition and multiplication on the set of all non-negative integers such that it becomes a field. I think not. Can we prove this ? If we have only a finite collection with prime cardinality then only it may become a field.
## marked as duplicate by Watson, Jack's wasted life, Willie Wong, suomynonA, Claude LeiboviciNov 19 '16 at 8:57
• We can turn a set of any cardinality into a field (and for at most countable sets, this does not even require choice or anything like that). – Tobias Kildetoft Sep 3 '13 at 12:20
• Pick any bijection of $\mathbb{N}$ with $\mathbb{Q}$ and transport the field operations to $\mathbb{N}$ via that. – Daniel Fischer Sep 3 '13 at 12:20
• @Tobias: Any infinite cardinality. Finite sets can only become fields if their size is a prime power. – Henning Makholm Sep 3 '13 at 12:22
• ohh, this was trivial. – aaaaaa Sep 3 '13 at 12:22
Here is what you could do: the rational numbers are countable, so you can find a bijection $\phi: \mathbb{N}_0 \to \mathbb{Q}$ ($\mathbb{N}_0 = \{0,1,2,\dots\}$). Now $\mathbb{Q}$ is a field, so can define multiplication and addition making $\mathbb{N}_0$ into a field.
You can define, for example, addition $\mathbb{N}_0$ by: $$a + b = \phi^{-1}(\phi(a) + \phi(b)) \\$$
If you for example order $\mathbb{Q} = \{a_0, a_1, \dots \}$ like this site suggests: $$a_0 =\frac 0 1, a_1 = \frac 1 1, a_2 = \frac {-1} 1, \frac 1 2, \frac {-1} 2, \frac 2 1, \frac {-2} 1, \frac 1 3, \frac 2 3, \frac {-1} 3,\\ \frac {-2} 3, \frac 3 1, \frac 3 2, \frac {-3} 1, \frac {-3} 2, \frac 1 4, \frac 3 4, \frac {-1} 4, \frac {-3} 4, \frac 4 1, \frac 4 3, \frac {-4} 1, \frac {-4} 3 \ldots$$ Then in $\mathbb{N}$ you would have $$3\cdot 4 = \phi^{-1}(a_3a_4) = \phi^{-1}(\frac{1}{2}\frac{-1}{2}) = \phi^{-1}(\frac{-1}{4}) = 17.$$ | HuggingFaceTB/finemath | |
# KeiruaProd
## Better odds against my 3-year-old kids at its favorite game
My 3-year-old kid has become a huge, huge fan of his card game “Ballons.” We’ve been playing many games every day during the last three months. The game is straightforward, so after a while, I had to find something to fuel some interest in playing with him. What about finding a better strategy than him? To do so, we will study the game’s properties and write many Monte Carlo simulations.
# The game
Ballons is a 2-4 players card game.
These are the rules of the game:
• Deal every player 5 balloons.
• Each player then pick action cards, one after each other:
• pop a balloon of color x” (4 such cards for each color, there are 5 colors)
• “adult/parent: recover a balloon of your choice, if possible” (5 such cards)
• When you run out of cards in the action deck, you shuffle the discarded cards and keep playing.
• A player has lost when all his balloons popped
That’s it. The ratings often say “it’s a great first game for small kids, but it quickly becomes boring for everybody.”
Initial state of the game
The game after a few action cards
## Our goal
There were many, but let’s see:
• how long are the games?
• what are the winning odds of my kid?
• can we improve those odds? If so, how much will we improve?
We will focus on two-player games because that’s easier to visualize, and because we only play this.
To answer the last question, we need to model and answer the following question:
The only possible choice in this game happens when you can recover a balloon, and you popped more than one. How do you choose which balloon to recover?
## A model for my kid’s play
My kid seems to choose randomly, so modeling its play was easy:
• when there is no choice, follow the flow of the game
• when there is a choice, pick randomly
## Simulating games
I wrote a lot of rust and python code in order to find the answers to my questions. I implemented the rules of the game, then wrote various simulations in order to identify some numbers that we’ll see in this article. The 200 lines of the ballons library contains the game code. This library is then used in many simulations at the root of the project.
The simulations all are Monte Carlo based, which means, in short, “run as many games with random initial conditions as you can, then average the results.” Due to the simplicity of the game, standard Monte Carlo is enough.
## Notes on performances
I did not need to write two implementations: it just happens that it was a fun project, and I got carried away with a never-ending stream of «just one more thing»: more speed, better algorithms, FFI, parallelization, and so on.
Monte Carlo is iterative. Accuracy in a Monte Carlo simulation converge with a square root law: in order to have one more digit accuracy (so a x10 improvement), you need 10^2 = 100 more iterations. So in order to have a good accuracy for enough digits, you need a lot of iterations, and powers of 100 grow quickly.
I ran between 10k and 49x10 millions simulations per program, which was a bit slow for certain configurations (~30mn) in Python. Using a faster language like Rust helped. Every simulation is independant: Monte Carlo can easily be parallelized, so in the end I took advantage of that as well.
## Some properties of this game
What makes this game interesting to me is that it sounds easy to simulate and very limited:
• 2 player game (for this study; you can play with up to 5 players)
• Non-deterministic
• Unlimited game tree depth, but
• Low branching factor
• No interactions between players
Let’s see first what the game depth is, then we will estimate the branching factor.
### Depth - how long the games are ?
That was my initial question: some games are very long. This is sometimes a complain in the reviews. In my experience, the perfect match requires no more than two deals of the action deck. Otherwise, my kid loses interest. He likes dealing cards and playing many games, not spending a lot of time during the actual game.
So I wrote a Monte Carlo simulation: I ran many random games with random hands, counted the length of the games, then plotted a bar chart of the results:
When you change the number of parent cards, the curve gets skewed towards shorter games. That’s quite intuitive, but that’s the first result that shows we can learn some stuff (the actual probabilities) with our code.
In my experience, the ideal number of parent card is 3 or 4. With 5, games tend to be too long; they are often very short with 1 or 2.
Let’s move on to some more fun stuff!
### Branching factor
The branching factor counts, from a given state, to how many states we can go.
For us, the branching factor derives from the 25 action cards:
There are:
• 4 “pop” of each color
• 5 “recover”
The intuition is that the branching factor is low. How low?
We can find some bounds:
• it is >= 1: after every card, pop or recover, you’ll play a new card
• it is <= 4: if you pick a parent action card, you can at most choose between 4 cards which one to recover
Then, we can average it:
• 20 cards out of the 25 action cards are a pop card, with a branching factor of 1
• 5 cards out of the 25 action cards are a recover card, with a branching factor <=4
So an upper bound for the branching factor is 20/25 * 1 + 5/25 * 4 = 0.8 + 4*0.2 = 1.6
Finally, we can simulate it: every time we’ll pick a card in a game, we’ll count of many branches are possible.
We will do so in many games:
python3 gen_branching_factor.py --nb_iterations 10000
1.2827724636190512
With 10000 iterations, the number we find is 1.28.
That’s very low ; in chess, the branching factor is around 30 on average. That’s the source of complains in the review: there is not a lot of choice for the adult :(
That being said, even with a shallow branching factor but with exceptionally long games, many games are possible:
>>> 1.28 ** 100
52601359015.48384
It’s hard to explore all the branches.
# Calculating the winning odds
We’ll run many random games against in a classic Monte Carlo fashion, but this time we’ll look at:
• what hands were played
• who was the winner: the player that did not pop all its balloons.
## Enumerating the various hands
The 25 balloon cards: there are 5 balloons for each of the 5 different colors:
With this, the hands of 5 cards follow a structure that I’ll use during the rest of this article. They can be one of those:
• 11111
• 2111
• 311
• 41
• 5
• 221
• 32
where:
• 5 means 5 cards of the same color
• 32 means “3 cards of a color, and 2 cards of a different color”,
• 11111 means that the five cards all have a different color
In the sample initial state above, even though we did not sort the cards, we have seen a game “311 vs. 2111”.
## Probabilities of encountering those hands
It turns out we can enumerate all the hands. I did this through Monte Carlo At first. It’s not very difficult to list all the seven different structures’ hands. So if we pick five random cards, here are the odds that they’ll follow a given configuration:
The 5s and the 14s overlap, so its hard to read:
• 5: 0.0079%
• 14: 0.9387%
• 23: 3.7784%
• 11111: 5.8973%
• 113: 14.0978%
• 122: 28.2434%
• 1112: 47.0365%
Some conclusions:
• 47% of the hands are 1112
• 96% of the hands are 1112, 122, 113 or 11111
• 5 are very infrequent.
## Computing the odds of winning
I ran, again, many monte carlo games.
### With pure random cards
At first, I picked 2 random hands during a given number of iterations, then ran a game for this encounter.
time python3 gen_hand_heatmap.py -p 5 -b 5 -i 10000000
…
real 9m57,285s
The problem with this approach is that we’ve seen some hands are very infrequent, and it was tough to have good numbers for some encounters.
Here is a more visual translation of this problem, through a heatmap, with all the games after the 10000000 iterations :
We can see that there were:
• 2264360 games of 1112 vs 1112.
• 17 games of 5 vs 14
• 0 games of 5 vs 5
Those numbers are about what the theory suggests:
• P(picking 1112) = 0.47
• P(picking 1112 / you already picked 1112) = 0.47*0.47 = 0.220
0.22 * 10000000 games = 2200000 games, which is more or less what we found during the simulation (2264360 games of 1112 vs 1112).
We can derive those numbers for the other examples in the same manner.
So if we go for a purely random approach, we have a problem:
• it could be tough to simulate some infrequent kind of encounters
• the frequent encounters will have very high precision, while infrequent games will have a low accuracy: we won’t be able to “trust” all the numbers in the same way.
### Uniform distribution of the encounters
A solution to those two problems? We will generate all the hands for each of the seven structures, then generate all the possible encounters for the 7x7 possible hand structure pairs. I wrote a short utility. It turns out there are not that many hands for each structure:
\$ time python3 gen_every_hands.py
11111 1
5 5
41 20
32 20
311 60
221 60
2111 120
In total, that’s 286 different hands.
So now, for each pairs of structures, we will run many iterations of games with two hands picked randomly in the encounter list we generated. By doing so, we can have a completely uniform distribution of the games played. All the probabilities we will have in the end will have the same level of trust, no matter how rare the encounter is.
We can also notice that some games are impossible: with a standard deck, it is impossible to deal a 5 vs. 11111.
Now, we can simulate the games and plot the winning rates for each kind of encounter:
time python3 gen_hand_heatmap_uniform.py --nb_iterations 100000
…
real 4m42,866s
• the diagonal is about 50%: a hand against the same hand is about one chance out of two of winning.
• usually, the less frequent the hand is, the more likely it is to win
• this is not true for 11111. Having short runs of the same colors seem to be a disadvantage.
• 5 is the best hand structure. We can explain this with the five parent cards shared between the two players: it yields a strong staying power.
# Improving my odds against my kid
Ok, so here we are: we can simulate many games with two players who play randomly, and we can estimate the winning odds when two hands face each other.
It’s time to simulate the same thing, but we will change the strategy for one player (ours). Can we crush him ? Again, the only strategy we need is how to choose which to recover. There are many possible strategies, like:
• always play the first card you dismissed
• if you dismissed a red card, recover it, otherwise pick randomly
Those do not look promising though: we want to improve our odds. There is something we can use: deck knowledge.
We can count the remaining cards in the action deck for each color. When you have to choose which color to recover, we will choose the least available in the rest of the action deck.
The idea for this heuristic is that, in the long run, the card we will recover is less likely than the others to be popped again.
Let’s simulate many games for this heuristic:
python3 gen_hand_heatmap_uniform_with_counter.py -i 100000
Then, I put the result in a heatmap:
That’s a bit hard to read. Did we improve? Let’s compute the difference between the last two charts to see how much we improved:
That a small improvement for 11111 and 2111, but that’s an improvement anyway! Given the low branching factor and the high randomness, we couldn’t expect a considerable improvement regardless.
note: substracting two imprecise numbers increases the imprecision. I never properly quantified the error margin, so I would’nt put too much trust on those numbers.
# We did it! Concluding thoughts
• if you have a 11111 or 2111, you can squeeze between 1 and 6% more chances to win by counting the remaining cards (only count your colors), then when you have the choice, pick the least probable.
• if you have a 221, 311, or 32, don’t bother counting; your odds won’t improve much
• if you have a 41, counting will hurt you, but the odds are you’ll encounter a winnable hand anyway with the random strategy
• if you have a 5, your branching factor is one no matter what, but you’ll win most of the time
Now you know how to play against your 3yo ;)
Ok, so that was a very complicated way to prove my point: it is possible to improve your odds (here, only or certain configurations) with a simple heuristic, and to quantify the improvement.
Anyway, down the road we’ve learn a bunch of things about game theory, so that’s a net win.
# Going further
That was fun but I’ve already spent way too much time with this game. Here are some ideas if you think this is a serious project that deserves some more research:
• you can probably devise a better heuristic
• it’s possible to prune the encounter list a bit more. E.g. red/red/red/blue/blue vs. yellow/yellow/yellow/green/purple is one permutation away from yellow/yellow/yellow/blue/blue vs. red/red/red/green/purple, which asymptotically have the same odds.
• I wanted a simple heuristic, but the trend now is to feed every problem to a neural network. Maybe you could try that?
• Go for full tree search, it might be possible. If not, MCTS might help
See a typo ? You can suggest a modification on Github. | HuggingFaceTB/finemath | |
# Write an algorithm to find an optimal binary search tree using dynamic programming
Back to Greedy algorithms 8. I am doing it first to round out our discussion of graphs with these important algorithms. Certainly be very capable with the short algorithms and the ideas of shortest paths and transitive closure. In what order should A1A An be multiplied so that it would take the minimum number of computations to derive the product. Let A and B be two matrices of dimensions p x q and q x r. Since scalar multiplication is more expensive than scalar addition, we count only the scalar multiplications.
## Blog Archive
Thus Cij takes n multiplications. Example of the best way of multiplying 3 matrices: Formulate a greedy method for the Matrix Chain Problem, and prove by a counter example that it does not necessarily lead to an optimal solution.
A Dynamic programming design: Aj, where the cost is measured in the number of scalar multiplications. Proof of the principle of optimality Every way of multiplying a sequence of matrices can be represented by a binary infix tree, where the leaves are the matrices, and the internal nodes are intemediary products. Let T be the tree correspodning to the optimal way of multiplying Ai T has a left subtree L and a right subtree R. It suffices to show it for L. The proof is by contradiction.
## Optimal binary search algorithm | C++ Programming
If L were not optimal, then there would be a better tree L' for Ai Then, takes the tree T' whose left subtree is L' and whose right subtree is R.
Therefore, L must be optimal. Derivation of the recurrence relation: L is the left subtree corresponding to AiThis lecture introduces dynamic programming, in which careful exhaustive search can be used to design polynomial-time algorithms.
The Fibonacci and shortest paths problems are used to introduce guessing, memoization, and reusing solutions to subproblems. Binary search algorithm Generally, to find a value in unsorted array, we should look through elements of an array one by one, until searched value is found.
In case of searched value is absent from array, we go through all elements.
• algorithm - Dynamic Programming:optimal Binary search tree and Huffman - Stack Overflow
• Self Organizing Search
• Optimal Line Breaking
• Free Downloads
• Algorithm to find element in sorted rotated array
Optimal Binary Search Trees 1 OPTIMAL BINARY SEARCH TREES 1. PREPARATION BEFORE LAB DATA STRUCTURES An optimal binary search tree is a binary search tree for which the nodes are arranged on levels such that the tree cost is minimum. Oct 30, · Now for a tree to be optimal, its subtrees should also be optimal.
So here we come across our optimal substructure. Generalize this recurrence and you have a simple Dynamic Programming algorithm ready to solve the problem for you in linear time! Cheers! 🙂 Find ‘ceil’ of a key in a Binary Search Tree (BST) Given N. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible. Let us first define the cost of a BST. The cost of a . For optimal binary search trees, given n items, each with a given probability of being selected, construct the binary search tree that will minimize the average search time for the items.
This is similar to Huffman codes in that they both output a binary tree, and they both try to minimize the average depth.
BINARY SEARCH ALGORITHM (Java, C++) | Algorithms and Data Structures | HuggingFaceTB/finemath | |
# Simplifying Fractions using "Keep,Change,Flip"
In the words of Flocabulary, " Keep, change, Flip, yea that's the action, everybody's gonna know how were dividing fractions"
In this video you will learn how to simplify complex fractions.
A complex fraction sounds difficult, but it is simply a fraction on top of a fraction.
I call them a double decker fraction.
"Keep, Change, Flip" is a technique used when dividing fractions.
You keep the first fraction, change the division sign to multiplication, and flip the second fraction.
The video solves three problems.
Problem 1. 5+7/11 /5/12
In this problem you can use " Keep, Change,Flip" in order to solve.
Keep the first fraction,change the division sign to multiplication, and create a reciprocal by flipping the last fraction.
Problem 2. 3/4 + 2/3 /1/2
In this problem I show how to simplify the top fraction, rewrite the fraction, and again apply the "Keep,Change, Flip" technique.
Problem 3. Involves a complex fraction with a variable. Again, "Keep,Change,Flip" can be used to solve this complex fraction.
Enjoy this video by Flocabulary for a review of " Keep,Change,Flip" | HuggingFaceTB/finemath | |
# Subtraction of Numbers without using Number Line
Learn the rules used for subtraction of numbers without using number line. When the numbers are big, the use of number line is not convenient for their subtraction. Also, it is very time consuming to draw a number line every time, and perform the operation of subtraction.
Rules for subtraction of numbers without using number line in different situation:
To subtract a number from another number, the sign of the number (which is to be subtracted) should be changed and then this number with the changed sign, should be added to the first number.
For Example:
(i) Evaluate (+6) – (+2)
= (+6) + (-2) (charging the sign of the number to be subtracted and then adding)
On subtracting smaller number 2 from bigger number 6; we get 6 – 2 = 4
Since, the sign of bigger number is + (positive)
= +4 or 4
Therefore, (+6) – (+2) = 4
(ii) Evaluate (+5) – (-3)
= (+5) + (+3) (charging the sign of the number to be subtracted and then adding)
We know, to add a positive (+ ve) number to a positive (+ ve) number, the numbers should be added and positive sign should be attached to the sum obtained.
= +8
Therefore, (+5) – (-3) = 8
(iii) Evaluate (-7) – (+2)
= (-7) + (-2) (charging the sign of the number to be subtracted and then adding)
We know, to add a negative number to a negative number, the numbers should be added and negative sign should be attached to the sum obtained.
= -9
Therefore, (-7) – (+2) = -9
(iv) Evaluate (-9) – (-6)
= (-9) + (+6) (charging the sign of the number to be subtracted and then adding)
On subtracting smaller number 6 from bigger number 9; we get 9 – 6 = 3
As the sign of bigger number is negative (-)
= -3
Therefore, (-9) – (-6) = -3
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## Fermat's Last Theorem
A theorem first proposed by Fermat in the form of a note scribbled in the margin of his copy of the ancient Greek text Arithmetica by Diophantus. The scribbled note was discovered posthumously, and the original is now lost. However, a copy was preserved in a book published by Fermat's son. In the note, Fermat claimed to have discovered a proof that the Diophantine Equation has no Integer solutions for .
The full text of Fermat's statement, written in Latin, reads Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est diuidere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.'' In translation, It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain.''
As a result of Fermat's marginal note, the proposition that the Diophantine Equation
(1)
where , , , and are Integers, has no Nonzero solutions for has come to be known as Fermat's Last Theorem. It was called a Theorem'' on the strength of Fermat's statement, despite the fact that no other mathematician was able to prove it for hundreds of years.
Note that the restriction is obviously necessary since there are a number of elementary formulas for generating an infinite number of Pythagorean Triples satisfying the equation for ,
(2)
A first attempt to solve the equation can be made by attempting to factor the equation, giving
(3)
Since the product is an exact Power,
(4)
Solving for and gives
(5)
which give
(6)
However, since solutions to these equations in Rational Numbers are no easier to find than solutions to the original equation, this approach unfortunately does not provide any additional insight.
It is sufficient to prove Fermat's Last Theorem by considering Prime Powers only, since the arguments can otherwise be written
(7)
so redefining the arguments gives
(8)
The so-called first case'' of the theorem is for exponents which are Relatively Prime to , , and ( ) and was considered by Wieferich. Sophie Germain proved the first case of Fermat's Last Theorem for any Odd Prime when is also a Prime. Legendre subsequently proved that if is a Prime such that , , , , or is also a Prime, then the first case of Fermat's Last Theorem holds for . This established Fermat's Last Theorem for . In 1849, Kummer proved it for all Regular Primes and Composite Numbers of which they are factors (Vandiver 1929, Ball and Coxeter 1987).
Kummer's attack led to the theory of Ideals, and Vandiver developed Vandiver's Criteria for deciding if a given Irregular Prime satisfies the theorem. Genocchi (1852) proved that the first case is true for if is not an Irregular Pair. In 1858, Kummer showed that the first case is true if either or is an Irregular Pair, which was subsequently extended to include and by Mirimanoff (1905). Wieferich (1909) proved that if the equation is solved in integers Relatively Prime to an Odd Prime , then
(9)
(Ball and Coxeter 1987). Such numbers are called Wieferich Primes. Mirimanoff (1909) subsequently showed that
(10)
must also hold for solutions Relatively Prime to an Odd Prime , which excludes the first two Wieferich Primes 1093 and 3511. Vandiver (1914) showed
(11)
and Frobenius extended this to
(12)
It has also been shown that if were a Prime of the form , then
(13)
which raised the smallest possible in the first case'' to 253,747,889 by 1941 (Rosser 1941). Granville and Monagan (1988) showed if there exists a Prime satisfying Fermat's Last Theorem, then
(14)
for , 7, 11, ..., 71. This establishes that the first case is true for all Prime exponents up to 714,591,416,091,398 (Vardi 1991).
The second case'' of Fermat's Last Theorem (for ) proved harder than the first case.
Euler proved the general case of the theorem for , Fermat , Dirichlet and Lagrange . In 1832, Dirichlet established the case . The case was proved by Lamé (1839), using the identity
(15)
Although some errors were present in this proof, these were subsequently fixed by Lebesgue (1840). Much additional progress was made over the next 150 years, but no completely general result had been obtained. Buoyed by false confidence after his proof that Pi is Transcendental, the mathematician Lindemann proceeded to publish several proofs of Fermat's Last Theorem, all of them invalid (Bell 1937, pp. 464-465). A prize of 100,000 German marks (known as the Wolfskel Prize) was also offered for the first valid proof (Ball and Coxeter 1987, p. 72).
A recent false alarm for a general proof was raised by Y. Miyaoka (Cipra 1988) whose proof, however, turned out to be flawed. Other attempted proofs among both professional and amateur mathematicians are discussed by vos Savant (1993), although vos Savant erroneously claims that work on the problem by Wiles (discussed below) is invalid. By the time 1993 rolled around, the general case of Fermat's Last Theorem had been shown to be true for all exponents up to (Cipra 1993). However, given that a proof of Fermat's Last Theorem requires truth for all exponents, proof for any finite number of exponents does not constitute any significant progress towards a proof of the general theorem (although the fact that no counterexamples were found for this many cases is highly suggestive).
In 1993, a bombshell was dropped. In that year, the general theorem was partially proven by Andrew Wiles (Cipra 1993, Stewart 1993) by proving the Semistable case of the Taniyama-Shimura Conjecture. Unfortunately, several holes were discovered in the proof shortly thereafter when Wiles' approach via the Taniyama-Shimura Conjecture became hung up on properties of the Selmer Group using a tool called an Euler system.'' However, the difficulty was circumvented by Wiles and R. Taylor in late 1994 (Cipra 1994, 1995ab) and published in Taylor and Wiles (1995) and Wiles (1995). Wiles' proof succeeds by (1) replacing Elliptic Curves with Galois representations, (2) reducing the problem to a Class Number Formula, (3) proving that Formula, and (4) tying up loose ends that arise because the formalisms fail in the simplest degenerate cases (Cipra 1995a).
The proof of Fermat's Last Theorem marks the end of a mathematical era. Since virtually all of the tools which were eventually brought to bear on the problem had yet to be invented in the time of Fermat, it is interesting to speculate about whether he actually was in possession of an elementary proof of the theorem. Judging by the temerity with which the problem resisted attack for so long, Fermat's alleged proof seems likely to have been illusionary.
See also abc Conjecture, Bogomolov-Miyaoka-Yau Inequality, Mordell Conjecture, Pythagorean Triple, Ribet's Theorem, Selmer Group, Sophie Germain Prime, Szpiro's Conjecture, Taniyama-Shimura Conjecture, Vojta's Conjecture, Waring Formula
References
Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, pp. 69-73, 1987.
Beiler, A. H. The Stone Wall.'' Ch. 24 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. New York: Dover, 1966.
Bell, E. T. Men of Mathematics. New York: Simon and Schuster, 1937.
Bell, E. T. The Last Problem. New York: Simon and Schuster, 1961.
Cipra, B. A. Fermat Theorem Proved.'' Science 239, 1373, 1988.
Cipra, B. A. Mathematics--Fermat's Last Theorem Finally Yields.'' Science 261, 32-33, 1993.
Cipra, B. A. Is the Fix in on Fermat's Last Theorem?'' Science 266, 725, 1994.
Cipra, B. A. Fermat's Theorem--At Last.'' What's Happening in the Mathematical Sciences, 1995-1996, Vol. 3. Providence, RI: Amer. Math. Soc., pp. 2-14, 1996.
Cipra, B. A. Princeton Mathematician Looks Back on Fermat Proof.'' Science 268, 1133-1134, 1995b.
Courant, R. and Robbins, H. Pythagorean Numbers and Fermat's Last Theorem.'' §2.3 in Supplement to Ch. 1 in What is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed. Oxford, England: Oxford University Press, pp. 40-42, 1996.
Cox, D. A. Introduction to Fermat's Last Theorem.'' Amer. Math. Monthly 101, 3-14, 1994.
Dickson, L. E. Fermat's Last Theorem, , and the Congruence (mod ).'' Ch. 26 in History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, pp. 731-776, 1952.
Edwards, H. M. Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. New York: Springer-Verlag, 1977.
Edwards, H. M. Fermat's Last Theorem.'' Sci. Amer. 239, 104-122, Oct. 1978.
Granville, A. Review of BBC's Horizon Program, Fermat's Last Theorem'.'' Not. Amer. Math. Soc. 44, 26-28, 1997.
Granville, A. and Monagan, M. B. The First Case of Fermat's Last Theorem is True for All Prime Exponents up to 714,591,416,091,389.'' Trans. Amer. Math. Soc. 306, 329-359, 1988.
Guy, R. K. The Fermat Problem.'' §D2 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 144-146, 1994.
Hanson, A. Fermat Project.'' http://www.cica.indiana.edu/projects/Fermat/.
Kolata, G. Andrew Wiles: A Math Whiz Battles 350-Year-Old Puzzle.'' New York Times, June 29, 1993.
Lynch, J. Fermat's Last Theorem.'' BBC Horizon television documentary. http://www.bbc.co.uk/horizon/fermat.shtml.
Lynch, J. (Producer and Writer). The Proof.'' NOVA television episode. 52 mins. Broadcast by the U. S. Public Broadcasting System on Oct. 28, 1997.
Mirimanoff, D. Sur le dernier théorème de Fermat et le critérium de Wiefer.'' Enseignement Math. 11, 455-459, 1909.
Mordell, L. J. Fermat's Last Theorem. New York: Chelsea, 1956.
Murty, V. K. (Ed.). Fermat's Last Theorem: Proceedings of the Fields Institute for Research in Mathematical Sciences on Fermat's Last Theorem, Held 1993-1994 Toronto, Ontario, Canada. Providence, RI: Amer. Math. Soc., 1995.
Osserman, R. (Ed.). Fermat's Last Theorem. The Theorem and Its Proof: An Exploration of Issues and Ideas. 98 min. videotape and 56 pp. book. 1994.
Ribenboim, P. Lectures on Fermat's Last Theorem. New York: Springer-Verlag, 1979.
Ribet, K. A. and Hayes, B. Fermat's Last Theorem and Modern Arithmetic.'' Amer. Sci. 82, 144-156, March/April 1994.
Ribet, K. A. and Hayes, B. Correction to Fermat's Last Theorem and Modern Arithmetic.'' Amer. Sci. 82, 205, May/June 1994.
Rosser, B. On the First Case of Fermat's Last Theorem.'' Bull. Amer. Math. Soc. 45, 636-640, 1939.
Rosser, B. A New Lower Bound for the Exponent in the First Case of Fermat's Last Theorem.'' Bull. Amer. Math. Soc. 46, 299-304, 1940.
Rosser, B. An Additional Criterion for the First Case of Fermat's Last Theorem.'' Bull. Amer. Math. Soc. 47, 109-110, 1941.
Shanks, D. Solved and Unsolved Problems in Number Theory, 4th ed. New York: Chelsea, pp. 144-149, 1993.
Singh, S. Fermat's Enigma: The Quest to Solve the World's Greatest Mathematical Problem. New York: Walker & Co., 1997.
Stewart, I. Fermat's Last Time-Trip.'' Sci. Amer. 269, 112-115, 1993.
Taylor, R. and Wiles, A. Ring-Theoretic Properties of Certain Hecke Algebras.'' Ann. Math. 141, 553-572, 1995.
van der Poorten, A. Notes on Fermat's Last Theorem. New York: Wiley, 1996.
Vandiver, H. S. On Fermat's Last Theorem.'' Trans. Amer. Math. Soc. 31, 613-642, 1929.
Vandiver, H. S. Fermat's Last Theorem and Related Topics in Number Theory. Ann Arbor, MI: 1935.
Vandiver, H. S. Fermat's Last Theorem: Its History and the Nature of the Known Results Concerning It.'' Amer. Math. Monthly, 53, 555-578, 1946.
Vardi, I. Computational Recreations in Mathematica. Reading, MA: Addison-Wesley, pp. 59-61, 1991.
vos Savant, M. The World's Most Famous Math Problem. New York: St. Martin's Press, 1993.
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© 1996-9 Eric W. Weisstein
1999-05-26 | HuggingFaceTB/finemath | |
# How To Calculate Percent Ionization
It is to be noted that the strong acids and bases dissociate or ionize completely so their percent ionization is 100%. This means that each hydrogen ions from each acidic molecule or the hydroxide ions from each basic molecule separate or donate. So, only the percent ionization of weak acid or base is calculated. The calculation of the percent ionization is one way of quantifying how far a weak acid has dissociated from the solution. When considering solutions it becomes much more reasonable to learn of the ionized percentage against the concentrations or the equilibrium constant.
As weak acid and base solutions are partially ionized in water which lead to too many charged and uncharged species in dynamic stabilization. The ionized percentage is the proportion of the initial ionisation compound. So, the concentration of the ion in solution is compared with the initial neutral species concentration. Weak acids like hydrofluoric acid, acetic acid, formic acid, water, etc. and weak bases like ammonia, methyl amine, etc. ionize in water in small quantities. It is very simple to compute the percent ionization of acid or base which can allow you to know the behaviour of certain weak acids and bases.
This article will show you the percent ionization formula, how to calculate it step by step and give you examples on how the formula can be utilized.
## Percent Ionization Formula
Percent ionization can be computed by dividing the concentration of ionized acid or base in equilibrium by the original concentration of the solution times 100 percent. The concentration in equilibrium can be determined through the equilibrium constant inherent to the solution itself and upon its complete ionization. The formula is given by;
For acid,
Percent ionization=H3O+eqHA0100 %
For base,
Percent ionization=OH-eqHA0100 %
## How to calculate Percent Ionization
The steps to calculate percent abundance are given below;
1. Firstly, write the balanced acid or base dissociation/ionization reaction
2. Secondly, set down the expression for dissociation constant Ka for acid or Kb for base.
3. Compute [H3O+] and conjugate base for acid or OH- and conjugate acid for base at equilibrium
4. Finally, calculate percent ionization
## Ionization of acids
A strong acid ionizes in water entirely while a weak acid only partly ionizes. The equilibrium Constant for ionization of an acid defines its Acid Ionisation Constant (Ka). However, the stronger the acid, the greater will be the acid ionisation constant (Ka) meaning that a strong acid is a better proton donor.
Figure 1. Strong acids completely dissociate in water.
The weak acid solution of water is the nonionized acid, hydronium ion and the acid conjugate base with the highest concentration of nonionized acid. This raises the level of hydronium ion in an aqueous solution of weak acid.
## Example 1. Calculating Percent Ionization of a Weak Acid
If acetic acid (CH3COOH) has a Ka of 1.8*10-5 at 250C, what is the percent ionization of acetic acid in a 1.00 M solution?
Let’s go through this example step-by-step.
### Step 1: Write the balanced acid dissociation/ionization reaction
Let’s firstly write the balanced ionization reaction of CH3COOH in water. The acetic acid can donate a proton to water to form CH3COO-(aq).
CH3COOHaq+H2OlH3O+aq+CH3COO-(aq)
Step 2: Then, set down the expression for dissociation constant Ka for acid
As we have the equation from step 1, now can write the dissociation constant Ka expression of acetic acid:
Ka=H3O+[CH3COO-][CH3COOH]=1.8*10-5
Step 3: Compute [H3O+] and conjugate base for acid [CH3COO-] at equilibrium
As we know acetic acid is weak acid so the complete ionization of weak acid is not detected. Now, we can use an ICE (Initial, Change, Equilibrium) table to evaluate algebraic expressions for the equilibrium concentrations in Ka expression.
CH3COOHaq + H2Ol ↔ H3O+aq + CH3COO-aq
Now, putting the equilibrium concentrations into our Ka expression, we have;
Ka=(x)(x)(1.00-x)=1.8*10-5
Next, we can simplify the expression by following way;
x21.00-x=1.8*10-5
This seems to be a quadratic equation so it can be solved by using a quadratic formula or approximation method.
Since the right side of Ka of acetic acid is very small i.e. 10-5 which is less than 10-4 , we can use the approximation method and ignore x on the denominator on the left side. Then, we get,
x21.00=1.8*10-5
or x=1.8*10-5×1.00=4.2*10-3 M
Therefore, H3O+aq=CH3COO-aq=4.2*10-3 M
### Step 4: Finally, calculate the percent ionization
Now, to find the percent ionization, let us use the equilibrium expressions we have get in Step 3.
Percent ionization=H3O+eqHA0100 %
or Percent ionization=H3O+eqCH3COOHinitial100 %=4.2*10-31.00100 %=0.42 %
Hence, 0.42 % of the acetic acid (CH3COOH) in solution has ionized into H+ & CH3COO- ions.
## Ionization of bases
There are strong bases such as sodium hydroxide, lithium hydroxide etc. that get fully dissociated into their ions in an aqueous solution. The ionization-base constant i.e. Kb relates to the equilibrium constant for the ionization of a base. So we can tell that a strong basis indicates a good acceptor of protons
In equilibrium, a solution of a weak base in water is a combination of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the highest concentration. Therefore, a weak base raises the hydroxide ion concentration in an aqueous solution.
## Example 2. Calculating Percent Ionization of a Weak Base
What is the percent ionization of a 1.5 M solution of ammonia? (Kb= 1.8*10-5).
The step by step solution of the problem is given below:
### Step 1: Write the balanced base ionization reaction
Let’s firstly write the balanced base ionization reaction for ammonia. The ammonia can accept a proton from water to form ammonium, NH4+(aq).
NH3aq+H2OlNH4+aq+OH-(aq)
Step 2: Then, set down the expression for dissociation constant Kb for base
As we have the equation from step 1, now can write the dissociation constant Kb expression of ammonia:
Kb=NH4+[OH-][NH3]=1.8*10-5
Step 3: Compute NH4+ & [OH-] at equilibrium
As we know ammonia is weak base so the complete ionization of weak base is not detected. Now, we can use an ICE (Initial, Change, Equilibrium) table to evaluate algebraic expressions for the equilibrium concentrations in Kb expression.
NH3aq + H2Ol NH4+aq + OH-(aq)
Now, putting the equilibrium concentrations into our Kb expression, we have;
Kb=(x)(x)(1.50-x)=1.8*10-5
Next, we can simplify the expression by following way;
x21.50-x=1.8*10-5
This seems to be a quadratic equation so it can be solved by using a quadratic formula or approximation method.
Since the right side of Kb of ammonia is very small i.e. 10-5 which is less than 10-4 , we can use the approximation method and ignore x on the denominator on the left side. Then, we get,
x21.50=1.8*10-5
or x=1.8*10-5×1.50=5.2*10-3 M
Therefore, NH4+aq=OH-aq=5.2*10-3 M
### Step 4: Finally, calculate the percent ionization
Now, to find the percent ionization, let us use the equilibrium expressions we have got in Step 3.
Percent ionization=OH-eqHA0100 %
or Percent ionization=OH-eqNH3initial100 %=5.2*10-31.50100 %=0.34 %
Hence, 0.34 % of the ammonia (NH3) in solution has ionized into NH4+ & OH- ions.
## Example 3. Calculating Percent Ionization from pH
Calculate the percent ionization of a 0.125 M solution of nitrous acid (weak acid) with a pH of 2.09.
Here, firstly write the balanced chemical equation of ionization reaction of HNO2 in water.
HNO2aq+H2OlH3O+aq+NO2-(aq)
Then, we have given pH = 2.09
As pH is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution so we have,
pH=-log(H3O+)
or 2.09=-log H3O+
or 10-2.09=H3O+
or H3O+=8.1*10-3 M
Finally, percent ionization of nitrous acid is given by;
Percent ionization=8.1*10-30.125100%=6.5%
### Different Uses and Applications of organic chemistry
The uses and applications of organic chemistry range from life-saving pharmaceutical discoveries to the vibrant colors of fruits and vegetables we see daily. In this
### Ace organic chemistry- tips and tricks, cheat sheets, summary
Mastering organic chemistry takes time and practice. It might seem a daunting task at once if the right approach is not adopted. But in our
### Named reactions of organic chemistry- an overview
This article on organic reactions is a special one in our organic chemistry series. It will guide you through how chemical transformations unlock diversity at
### Organic Spectroscopy
Organic spectroscopy can be used to identify and investigate organic molecules. It deals with the interaction between electromagnetic radiation (EMR) and matter. These waves travel | HuggingFaceTB/finemath | |
# Euclid's Muse
## your source for INTERACTIVE math apps
Create an Account
# Search Results for “Diagonals”
##### Circles, Tangents and Nonagon Diagonals
You may want to see the heptagon version before attempting this one Every diagonal within a regular nonagon is drawn. Circles are centered at each intersection of diagonals along a vertical axis (these same constructions can be made nine times around the nonagon). Each circle can be tangent to at least 4 diagonals when the circle is at least 2 different sizes. Unnecessary diagonals have been hidden. Drag the green points to resize the circles. Can you find all 13 positions where a circle is tangent to at least 4 diagonals? Hint: sometimes the circle is not entirely contained within the nonagon. Ready for more? Check out the hendecagon version!
Tags: Nonagons, Circles, Diagonals, Tangents, Puzzler
##### Circles, Tangents, and Heptagon Diagonals
Two circles are centered at intersection points of diagonals of a regular hepatgon. It turns out that circles centered at intersection points in regular polygons (particularly interestingly with polygons of odd numbers of sides) can be tangent to many other diagonals of that polygon. Try resizing the circles by dragging the green points. How many diagonals can each circle be tangent to? Ready for more? Check out the nonagon version!
Tags: Heptagon, circles, tangents, diagonals, geometry
##### Tridecagon Diagonals, Circles and Tangents
You'll want to start out with the heptagon and work your way up. This one's the same as all the others, just with a 13-sided regular polygon. Observe the tangencies to diagonals of circles centered at intersections of diagonals, when the circles are resized (by dragging). This is a smaller version that works well on most monitors (zoom in with two-finger touch). Bigger version here.
Tags: Tridecagon, diagonals, circles, tangents, intersections, puzzler, intricate, confusing, wow
##### Hendecagon Diagonals, Circles and Tangents
Before even attempting to understand this app, take a look at the heptagon and nonagon versions. It’s the same situation here, circles centered at intersection points of diagonals within the hendecagon. Drag the green points to resize the circles. Resize the circles so that they are tangent to at least 4 diagonals at the same time (this case is possible in at least two positions for each circle). How many instances can you find on this one? Notice a trend with this and the other versions? Now that you've got this one, check out the final installment, the tridecagon version.
Tags: Hendecagon, tangents, circles, diagonals, puzzler | HuggingFaceTB/finemath | |
Research-based guidance and classroom activities for teachers of mathematics
# Using algebra to reason
Using Algebra to reason is a theme within quantities and algebraic expressions. Links to relevant activities and resources are on the right hand side of this page.
Algebra lets us apply additive and multiplicative relations to solve problems where we don’t know everything. For example:
• If I have four more sweets than you have, then there is a method of equalising the amount so long as you know the relation of ‘difference’ – and we don’t have to know exactly how many sweets we have in total
• If I know I run twice as fast as you on average, then I know I need half the time to run the distance - without knowing the total time or distance.
• If I know that a + b = c, then: b + a = c c = a + b c = b + a
c – b = a c – a = b a = c – b b = c – a
• Associativity: a + (b + c) = (a + b) + c
• Distributivity: a(b+c) = ab + ac
• If I know that ab=c, then: ba=c, c = ab and c = ba, c/b=a and c/a=b
• (a + d) – (b + d) = a – b
• If a > b and b > c, then a > c
• If a = 2b and b = 2c then a = 4c
##### Activities and resources
Performing number magic
Evaluating algebraic expressions
Number spirals | HuggingFaceTB/finemath | |
Browse Questions
# If functions $f\;:\;A \rightarrow B\;and\;g\;:\;B \rightarrow A$ satisfy g o f =$I_A$ ,then is f one-one and g onto
Toolbox:
• A function $f:X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $\; gof=I_A$, $f:A \to B$, $g: B \to A$.
Step1: Injective or One-One function:
$gof(x)=x \qquad g:B \to A$
$g(f(x))=x \rightarrow g(f(x_1))=x_1\;and \;g(f(x_2))=x_2$
$f(x_1)=f(x_2)=>x_1=x_2$
Hence f must be one -one
Step 2: Surjective or On-to function:
$gof(x)=x$
$g(f(x))=x$
=>for every element $f(x) \in A$ then exists an element $x \in B$ such that
$g(f(x))=x$
=>Hence g is onto
Solution:Thus function f is one-one and g is onto
edited Mar 30, 2013 | open-web-math/open-web-math | |
# AnyGamble
Mar 20, 2017
Testing roulette system
# Testing Roulette Systems - The Five Number System
This system called System of 5 numbers is played in the relatively simple way. You choose any 5 numbers and you keep betting on them. The basic series consists of 17 spins with the following progression: in the first 7 spins bet by one unit on each number, in further five spins your bet by 6 units on each number and then you bet 3 times by 11 units and finally 2 times by 16 units. Total bankroll is 510 units altogether.
## What do the Statistics say?
On the following table you can see the system from the viewpoint of the probability and the expected profit:
Spin Probability Profit in Series Expected Profit 1 13.51% 31 4,1892 2 11.69% 26 3.0387 3 10.11% 21 2.1227 4 8.74% 16 1.3987 5 7.56% 11 0.8317 6 6.54% 6 0.3923 7 5.66% 1 0.0566 8 4.89% 151 7.3856 9 4.23% 121 5.1185 10 3.66% 91 3.3292 11 3.16% 61 1.9301 12 2.74% 31 0.8483 13 2.37% 156 3.6921 14 2.05% 101 2.0674 15 1.77% 46 0.8143 16 1.53% 146 2.2353 17 1.32% 66 0.8739 18 8,47% -510 -43.2206 Total 100% -2.8956
Unfortunately, the expected profit resulted in a loss, therefore in the long term, the system should go below zero. Let’s have a look at the progression of the basic system on the following graph:
Our expectations became true and in spite of the short-term growth, the system sloped downward, (it had a downward trend). There were several options on how to modify the system but we were not able to apply all of them.
There was no point in altering the series length. It follows (from the table above) that as long as the series was, always the expected profit results in a loss (we had tested a series of 5 spins and a series of 25 spins). At the first sight, you would assume that there is a chance that you hit at least one number within 17 spins, but empirically based research had confirmed that there were also series when the roulette did not hit any number of a certain pentad 50 times in a row.
Applying the wait strategy would not be also supposed to have a success. We have attempted to apply at least a liner series progression; it means that after each unsuccessful series, we increased the roulette bet size on each number by the value of initial bet and after making a profit we reset the bet size back to initial bet:
As you could see, the system appeared to go to a plus and had an upward trend but the local slumps were drastically huge and reached values of hundreds of thousands below zero, therefore an idea of series progression was not successful. We have tried to apply some modification by employing a spin progression of the bet in one series in order that a win always reached a minimum value.
Unfortunately, we did not observe any amelioration and the progression practically remained the same as before. We can conclude that this system in this form is not feasible in the long term. | HuggingFaceTB/finemath | |
current amplification of a circuit containing a transistor
I'm having trouble with a BJT circuit. What we are given:
$U_{CC} = 10\text{V}$
$R_C = 972 \Omega$
$R_B = 14\text{k} \Omega$
$U_{\text{BE}} = 0.7\text{V}$ $I_C = 12\text{mA}$
We need to find the current amplification $B$.
My approach was to calculate $U_C$, the Voltage which drops at $R_C$. I read in a book that $I_C$ is the current we need to use Ohm's Law at $R_C$. So I solved the equation $U_R = I_C \cdot R_C \Leftrightarrow U_R = 12\text{mA} \cdot 972 \Omega \Leftrightarrow U_R = 11.664\text{V}$.
Having this done I was able to use the Mesh-Current-Law at the upper right part of the circuit which gave me the following equation $-U_{CC} + U_C - U_B$ where $U_B$ is the Voltage which drops at $R_B$. Filling the equation with the known values we receive $U_B = 1.664\text{V}$. Since we have $R_B$ given we can now apply Ohm's Law with the previously calculated Voltage which leads to the following value for
$I_B = \frac {U_B}{R_B} = \frac {1.664\text{V}}{14000 \Omega} = 1.188571429x10^{-4}\text{ A}$
or $0.1188571429\text{ mA}$.
Now I found out that the base current $B$ can be expressed by $B = \frac {I_C}{I_B}$.
Since we know $I_C$ as well as $I_B$ I went ahead and filled out the equation which gave me $B = \frac {12\text{mA}}{0.1188571429\text{mA}} = 100.9615385$ for $B$.
Am I on the right track?
• Please study how significant figures work. – markrages Nov 29 '12 at 18:46
Not quite.
Richman's answer is good and normally I would say it is correct, but...
given Vcc=10V, Rc=972, Ic=12ma, we are not in the real physical world.
12ma through a 972 ohm resistor drops 11.664 volts, yet the supply is only 10 volts.
Somebody isn't telling us something...
• This is a serious topic and I double checked every value :(. However I still do not know if my approach was correct. – optional Nov 29 '12 at 18:40
• Going with Vcc=12v as suggested in another comment, Ib=11.3/14k or 0.807ma Then Ic/Ib=15 at this working point; quite plausible for a transistor so close to saturation. – Brian Drummond Nov 29 '12 at 19:00
• Why did you divide 11.3/14k? Is there any "tolerance value" which you subtracted from Vcc (Vbe?)? – optional Nov 29 '12 at 19:13
• The loop is 12V - Vbe - Ib*Rb = 0V and my answer is correct. – Sunnyskyguy EE75 Nov 29 '12 at 20:19
• As far as I know I cannot simply change the value of Vcc. I am forced to calculate the rather unusual circuit with the values given leaving me no other choice then using 10V as Vcc. Despite this "mistake" is the following calculation correct? Ib = (10V - 0.7V) / 14k Ohm = 0.66mA, Beta (current amplification factor) B = Ic/Ib = 12mA / 0.66mA = 18.06. – optional Nov 29 '12 at 20:37
not quite...
It seems you may be over-complicating your analysis with excessive decimal places and cryptic syntax.
The base current is not caused by collector current. It is determined solely by V+ (Ucc) ,Rb and Vbe(Ube) (assume 0.65V +-.05 or 0.7 as given ) If there are more than one Rb with pull-up and down, then convert to equivalent voltage and equivalent resistance.
We don't normally show a battery equivalent circuit on a schematic, but you may need to remember this path when doing loop calulations. optional< maybe its a typo, but Ucc needs to be 12V for this question to be practical. You can't have Ur > Ucc, also the dc leakage current of collector to base is neglible. So Ib= 12V/14k= 0.86mA while Ic was given as 12 mA so the hFE = 14.. This is a good number for a saturated switch. You have to derate Beta when VCe drops into saturation. Most transistors do not saturate unless this ratio is between 10 and 30. Higher current needs lower ratios.
• I am so sorry. I mixed up base current and current amplification. Yes, you are right the base current depends solely on V+, the resistor and Vbe(?). However in this case I had to calculate the current amplification, my bad. – optional Nov 29 '12 at 18:14
• you are given the collector current, so if you find the base current, you can just divide the two to find the current amplification. – markrages Nov 29 '12 at 18:45
• I found the base current which is called \$I_B$ in my topic. Just want to make sure that the way I attempted the problem is correct. I know that the values may seem a bit odd but thats how they are given I had no influence. – optional Nov 29 '12 at 19:08
• No it is incorrect You actually dont need to be given Vbe and Ic to solve this question within 10%. Beta only is accuarte when Vce is between the supply rails (in linear region) When saturated (Vce=12-11.67...=0.33) – Sunnyskyguy EE75 Nov 29 '12 at 19:17
• Ratios are always reduced in the range of 10~30 with few exceptions when Vce is low (saturated switch) – Sunnyskyguy EE75 Nov 29 '12 at 19:23 | HuggingFaceTB/finemath | |
X
X
# Calculate the Least Common Multiple or LCM of 481
The instructions to find the LCM of 481 are the next:
## 1. Decompose all numbers into prime factors
481 13 37 37 1
## 2. Write all numbers as the product of its prime factors
Prime factors of 481 = 13 . 37
## 3. Choose the common and uncommon prime factors with the greatest exponent
Common prime factors: 13 , 37
Common prime factors with the greatest exponent: 131, 371
Uncommon prime factors: None
Uncommon prime factors with the greatest exponent: None
## 4. Calculate the Least Common Multiple or LCM
Remember, to find the LCM of several numbers you must multiply the common and uncommon prime factors with the greatest exponent of those numbers.
LCM = 131. 371 = 481
Also calculates the: | HuggingFaceTB/finemath | |
# Twenty times a positive integer is less than its square by 96, what is the integer?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 5
What are you looking for? Let’s dig in quickly
## Explanation
• Twenty times a positive integer is less than its square by 96.
The integer can be figure out in this way.
Let suppose the integer is y.
20 times of y and its square
20y & y2
20 times of y is less than its square by 96
20y + 96 = y2 ________ (i)
Through equation (i) we can easily find out the value of y (y = 24).
Number = ?
## Solution
Let suppose
Number = y
According to the given condition
20y + 96 = y2
y2 – 20y – 96 = 0
y2 – 24y + 4y – 96 = 0
y(y – 24) + 4(y – 24) = 0
(y – 24)(y + 4) = 0
y – 24 = 0 or y + 4 = 0
y = 24 (possible) or y = -4 (not possible)
## Conclusion
Twenty times a positive integer is less than its square by 96. The required integer will be 24.
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 7
error: Content is protected !! | HuggingFaceTB/finemath | |
## 50594
50,594 (fifty thousand five hundred ninety-four) is an even five-digits composite number following 50593 and preceding 50595. In scientific notation, it is written as 5.0594 × 104. The sum of its digits is 23. It has a total of 3 prime factors and 8 positive divisors. There are 24,640 positive integers (up to 50594) that are relatively prime to 50594.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 23
• Digital Root 5
## Name
Short name 50 thousand 594 fifty thousand five hundred ninety-four
## Notation
Scientific notation 5.0594 × 104 50.594 × 103
## Prime Factorization of 50594
Prime Factorization 2 × 41 × 617
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 50594 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 50,594 is 2 × 41 × 617. Since it has a total of 3 prime factors, 50,594 is a composite number.
## Divisors of 50594
1, 2, 41, 82, 617, 1234, 25297, 50594
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 77868 Sum of all the positive divisors of n s(n) 27274 Sum of the proper positive divisors of n A(n) 9733.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 224.931 Returns the nth root of the product of n divisors H(n) 5.19792 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 50,594 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 50,594) is 77,868, the average is 973,3.5.
## Other Arithmetic Functions (n = 50594)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 24640 Total number of positive integers not greater than n that are coprime to n λ(n) 3080 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5185 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 24,640 positive integers (less than 50,594) that are coprime with 50,594. And there are approximately 5,185 prime numbers less than or equal to 50,594.
## Divisibility of 50594
m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 5 2 5
The number 50,594 is divisible by 2.
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (50594)
Base System Value
2 Binary 1100010110100010
3 Ternary 2120101212
4 Quaternary 30112202
5 Quinary 3104334
6 Senary 1030122
8 Octal 142642
10 Decimal 50594
12 Duodecimal 25342
20 Vigesimal 669e
36 Base36 131e
## Basic calculations (n = 50594)
### Multiplication
n×i
n×2 101188 151782 202376 252970
### Division
ni
n⁄2 25297 16864.7 12648.5 10118.8
### Exponentiation
ni
n2 2559752836 129508134984584 6552334581410042896 331508815811859710280224
### Nth Root
i√n
2√n 224.931 36.9856 14.9977 8.72609
## 50594 as geometric shapes
### Circle
Diameter 101188 317891 8.0417e+09
### Sphere
Volume 5.42482e+14 3.21668e+10 317891
### Square
Length = n
Perimeter 202376 2.55975e+09 71550.7
### Cube
Length = n
Surface area 1.53585e+10 1.29508e+14 87631.4
### Equilateral Triangle
Length = n
Perimeter 151782 1.10841e+09 43815.7
### Triangular Pyramid
Length = n
Surface area 4.43362e+09 1.52627e+13 41309.8
## Cryptographic Hash Functions
md5 b264226cbf6ba172e1f38989dd1757d9 b342b30f1049d2f303342d1a4d7e85dedf780ed5 e7cec41c46a9706ba1a702b368be5431913b6dc9c9da49352c3f59d177fa7d2e 3e614fe8edf2bf5fc0bc490d07082647a8765e2d4e47eb7c273dd6b3c0a7cfddaf4263daa5483c29b15446de5572f551f724b7991f97f2b90372fd1e8b5c8976 592ffd883fc120a92352dbe25ac578bdf6db7a4c | HuggingFaceTB/finemath | |
# Difference between revisions of "2013 AMC 12A Problems/Problem 3"
We are given that $\frac{6}{10} = \frac{3}{5}$ of the flowers are pink, so we know $\frac{2}{5}$ of the flowers are red.
Since $\frac{1}{3}$ of the pink flowers are roses, $\frac{2}{3}$ of the pink flowers are carnations.
We are given that $\frac{3}{4}$ of the red flowers are carnations.
The number of carnations are
$\frac{3}{5} * \frac{2}{3} + \frac{2}{5} * \frac{3}{4} = \frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70\%$, which is $E$ | HuggingFaceTB/finemath | |
# Finding a general equation for a quadratic curve passing through three points.
I have three points (250, 0), (500,500) and (750, 0). To find a curve passing through these points all I have to do is plug-in these values into the general quadratic equation:
f(x) = ax^2 + bx + c
getting:
f(x) = -x^2/125 + 8x -1500
But, this is specific to this curve.
How would I generalize this to a curve that passes through (j, k), (p,q) and (u,v) such that the equation
f(x) = ax^2 + bx + c
has a, b and c expressed in terms of j,k,p,q,u and v; thus allowing me to plug-in arbitrary point values into this new general equation to get a specific equation for a curve?
I tried solving and eliminating variables, but all I end up with is an unsimplifiable mess that becomes too large to handle and I can't get very far.
-
If you plug j,k,p,q,u and v into the quadratic equation, you can get a system of 3 linear equations with 3 variables a,b,c and 6 parameters.
$$\begin{cases} k=j^2a+jb+c\\ q=p^2a+pb+c\\ v=u^2a+ub+c \end{cases}$$
It can be solved by various methods. The most suitable in this case is Cramer's rule. Look for the section called Explicit formulas for small systems.
-
Thanks for the Cramer's rule link. It helped. I figured out the solution using that and mathematica finally achieving f(x) = (-x^2 x2 y1-x x2^2 y1+x^2 x3 y1+x2^2 x3 y1+x x3^2 y1-x2 x3^2 y1+x^2 x1 y2+x x1^2 y2-x^2 x3 y2-x1^2 x3 y2-x x3^2 y2+x1 x3^2 y2-x^2 x1 y3-x x1^2 y3+x^2 x2 y3+x1^2 x2 y3+x x2^2 y3-x1 x2^2 y3)/((x1-x2) (x1-x3) (x2-x3)) where the curve passes through points (x1,y1),(x2,y2), and (x3,y3). I would vote up your answer if had enough rep. Thanks! – Æ - Nov 27 '12 at 21:48
Essentially, you have 3 linear equations in 3 variables $a$, $b$ and $c$ as follows:
$aj^2 + bj + c = k$
$ap^2 + bp + c = q$
$au^2 + bu + c = v$
This is a system of linear equations and you can solve them using any of the methods given.
- | HuggingFaceTB/finemath | |
# Thirty-third harmonic
No description. (Wikipedia).
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# How many different 5-person teams can be formed from a group
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22 Jun 2010, 19:14
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How many different 5-person teams can be formed from a group of x individuals?
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
The answer is D and I know how to figure out now but is there any trick to know each question is sufficient without actual compute? cuz its time consuming until I found out e.g. 1) is 9!/5!4!
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gmatJP wrote:
HI AbhayPrasanna.. Thanks for the reply...
I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...
How do you know (x+2) C 5 = 126 is computable
praveenism wrote:
I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value.
But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.
@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.
The point here is the following:
Suppose we are told that there are 10 ways to choose $$x$$ people out of 5. What is $$x$$? $$C^x_5=10$$ --> $$\frac{5!}{x!(5-x)!}=10$$ --> $$x!(5-x)!=12$$ --> $$x=3$$ or $$x=2$$. So we cannot determine single numerical value of $$x$$. Note that in some cases we'll be able to find $$x$$, as there will be only one solution for it, but generally when we are told that there are $$n$$ ways to choose $$x$$ out of $$m$$ there will be (in most cases) two solutions of $$x$$ possible.
But if we are told that there are 10 ways to choose 2 out of $$x$$, then there will be only one value of $$x$$ possible --> $$C^2_x=10$$ --> $$\frac{x!}{2!(x-2)!}=10$$ --> $$\frac{x(x-1)}{2!}=10$$ --> $$x(x-1)=20$$ --> $$x=5$$.
In our original question, statement (1) says that there are 126 ways to choose 5 out of $$x+2$$ --> there will be only one value possible for $$x+2$$, so we can find $$x$$. Sufficient.
Just to show how it can be done: $$C^5_{(x+2)}=126$$ --> $$(x-2)(x-1)x(x+1)(x+2)=5!*126=120*126=(8*5*3)*(9*7*2)=5*6*7*8*9$$ --> $$x=7$$. Basically we have that the product of five consecutive integers ($$(x-2)(x-1)x(x+1)(x+2)$$) equal to some number ($$5!*126$$) --> only one such sequence is possible, hence even though we have the equation of 5th degree it will have only one positive integer solution.
Statement (2) says that there are 56 ways to choose 3 out of $$x+1$$ --> there will be only one value possible for $$x+1$$, so we can find $$x$$. Sufficient.
$$C^3_{(x+1)}=56$$ --> $$(x-1)x(x+1)=3!*56=6*7*8$$ --> $$x=7$$. Again we have that the product of three consecutive integers ($$(x-1)x(x+1)$$) equal to some number ($$3!*56$$) --> only one such sequence is possible, hence even though we have the equation of 3rd degree it will have only one positive integer solution.
Hope it helps.
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We need a unique value.
1. (x+2) C 5 = 126. There is only one possible value for x+2 that would yield a value of 126. Don't bother trying to find out what it is. Remember, the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr.
2. (x+1) C 3 = 56 Again, you should be able to see that there can be only one value of x+1 that would yield a value of 56. Why bother finding out what the value is? As long as we have an equation in one variable, we can find a value.
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25 Jun 2010, 23:29
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Bunuel supplied an awesome and exhaustive mathematical algebraic explanation. Perhaps it will be of benefit to review the concept verbally as well.
8C5 = 8C3 because everytime we pull a subgroup of 5 objects from the bigger group of 8, we can see that we are also "setting aside" a subgroup of 3. Likewise, everytime we pull out a subgroup of 3, we also set aside a subgrup of 5. So, the number of ways we can pull out 5 object subgroups must be the same as the number of ways we can pull out 3 object subgroups.
But if there are 126 ways to pull 5 objects from a big group "x + 2", then "x+2" must be just one value. If it were not, then it would imply that increasing or decresing the size of the big group doesn't necessarily affect how many ways you can pull out a smaller subgroup--surely an absurd conclusion. Absurd because clearly there are more ways to pull 5 objects out of a set of 100 than out of a set of, say, 10.
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22 Jun 2010, 22:27
HI AbhayPrasanna.. Thanks for the reply...
I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...
How do you know (x+2) C 5 = 126 is computable
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24 Jun 2010, 02:50
I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value.
But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.
@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.
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Re: How many different 5-person teams can be formed from a group [#permalink]
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06 Apr 2012, 08:07
Hello to all,
Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,
3Cx+1 =56 and how, 5Cx+2 = 126 ?
Thnx
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06 Apr 2012, 08:28
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priyalr wrote:
Hello to all,
Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,
3Cx+1 =56 and how, 5Cx+2 = 126 ?
Thnx
# of ways to pick $$k$$ objects out of $$n$$ distinct objects is $$C^k_n=\frac{n!}{(n-k)!*k!}$$.
# of ways to pick 3 people out of x+1 people is $$C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}$$. Now, since $$(x+1)!=(x-2)!*(x-1)*x*(x+1)$$ then $$(x-2)!$$ get reduced and we'll have: $$\frac{(x-1)*x*(x+1)}{3!}$$. We are told that this equals to 56: $$\frac{(x-1)*x*(x+1)}{3!}=56$$ --> $$(x-1)x(x+1)=3!*56=6*7*8$$ --> $$x=7$$.
You can apply similar logic to $$C^5_{(x+2)}=126$$.
Hope it's clear.
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20 Aug 2012, 03:39
Bunuel,
Can you please confirm if my understanding is correct?
(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)
Bunuel wrote:
priyalr wrote:
Hello to all,
Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,
3Cx+1 =56 and how, 5Cx+2 = 126 ?
Thnx
# of ways to pick $$k$$ objects out of $$n$$ distinct objects is $$C^k_n=\frac{n!}{(n-k)!*k!}$$.
# of ways to pick 3 people out of x+1 people is $$C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}$$. Now, since $$(x+1)!=(x-2)!*(x-1)*x*(x+1)$$ then $$(x-2)!$$ get reduced and we'll have: $$\frac{(x-1)*x*(x+1)}{3!}$$. We are told that this equals to 56: $$\frac{(x-1)*x*(x+1)}{3!}=56$$ --> $$(x-1)x(x+1)=3!*56=6*7*8$$ --> $$x=7$$.
You can apply similar logic to $$C^5_{(x+2)}=126$$.
Hope it's clear.
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22 Aug 2012, 14:23
manjeet1972 wrote:
Bunuel,
Can you please confirm if my understanding is correct?
(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)
[/quote]
Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.
Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.
So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.
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22 Aug 2012, 22:38
Good explanation. Through practice I will learn.
Injuin wrote:
manjeet1972 wrote:
Bunuel,
Can you please confirm if my understanding is correct?
(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)
Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.
Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.
So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.[/quote]
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How many different 5-person teams can be formed from a group of x individuals?
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have
been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been
formed.
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Re: How many different 5-person teams can be formed from a group [#permalink]
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31 Mar 2017, 08:30
happyavi23 wrote:
How many different 5-person teams can be formed from a group of x individuals?
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have
been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been
formed.
Also, please name topics properly. Check rule 3 here: https://gmatclub.com/forum/rules-for-po ... 33935.html
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Re: How many different 5-person teams can be formed from a group [#permalink]
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02 Apr 2017, 08:23
gmatJP wrote:
How many different 5-person teams can be formed from a group of x individuals?
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
The answer is D and I know how to figure out now but is there any trick to know each question is sufficient without actual compute? cuz its time consuming until I found out e.g. 1) is 9!/5!4!
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
There are 1 variable x and 0 equation are given from the conditions of the original questions, so there is high chance (D) will be our answer.
The question asks what the value of $$_xC_2$$ is.
Condition (1)
$$_{x+2}C_5$$ = 126.
We can determine what the value of x is.
Hence, this condition is sufficient. We don't need calculate it.
Condition (2)
$$_{x+1}C_3$$ = 56.
We can determine what the value of x is.
Hence, this condition is sufficient. We don't need calculate it.
Therefore, the correct answer is D.
Keep in mind that DS is totally different from PS. DS questions do NOT ask what the exact value is.
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# Maths Ocr 2016 C1 Watch
1. (Original post by KevinGu123456789)
For the question about the circle being wholly inside the other, did anyone else get 0<r<√45 - 5?
Most people got 0<r<2, but I thought that this may not be right
I got this, as did most other people at my school who managed to answer it.
2. (Original post by KevinGu123456789)
Does anyone remember the actual last question though?
Wasn't it something like y=4x^2+a/x+5 and y has a stationary point at which the y coordinate = 32.
I differentiated to find 8x-a/x^2 and then equated to 0. Then said that 8x^3-a=0 therefore 8x^3=a.
Plug that back into the original equation to get 32=4x^2+8x^2+5 (32 is the y coordinate we were given).
Solving that to get x= 0 and + or - 3/2. Then plugging that back in to the original equation gave a=0, not possible or a negative solution when it asked for the positive solution so solving that eventually gave 27.
I hope that makes sense? Think it's correct (I BLOODY HOPE SO!)
3. I got y=2x-20 and y=2x+10 for tangents of circles
4. Yep. Deffo just got nearly every question wrong.
5. (Original post by AlfieH)
Can anybody remember the indices questions? I remember the first one was (2^5 divided by 2^7) but there was something else in that question.
Also - what about the 2nd question, I got -¼+or-¾root5
Yep, myself and at least 4 others got the same answer for the second question
6. How did people get 27? I got 27/12 fek
7. (Original post by Dmitry_Mednov)
It is wrong as if you use k as -5 or -6 it will not intersect the curve, I checked it at the end.
He's right, you're wrong. b^2-4ac when K<-7.5 was a number less than zero. Above -7.5 it gave a positive number, so when K<-7.5 ether are no real roots to the equation and the lines do not cross, you probably subbed it in wrong.
That paper wasn't hard.
8. (Original post by AlfieH)
Can anybody remember the indices questions? I remember the first one was (2^5 divided by 2^7) but there was something else in that question.
Also - what about the 2nd question, I got -¼+or-¾root5
Yeah that's right
9. Can someone explain how they did the last question?
10. (Original post by duncant)
I got y=2x-20 and y=2x+10 for tangents of circles
same man
11. (Original post by KevinGu123456789)
For the question about the circle being wholly inside the other, did anyone else get 0<r<√45 - 5?
Most people got 0<r<2, but I thought that this may not be right
I got that! I said the radius was 3√5 whereas lots thought it was 2√5.
Thing is on that question the distance from Centre to 0,0 was 5 therefore the radius couldn't have been 2√5 so I quickly changed that!
I got 0<r<3√5-5 (same as you).
12. (Original post by nobodycarescarla)
How did people get 27? I got 27/12 fek
8(27/12)^3/2 simplifies to 27
13. (Original post by julesaquilina)
I got k is less than -2 and greater than 5? And many others got that as well
I got that as well😊
14. With the indice questions, was it 2^-6 for part (i) and 2^13/3 for (ii)?
15. (Original post by AlfieH)
Wasn't it something like y=4x^2+a/x+5 and y has a stationary point at which the y coordinate = 32.
I differentiated to find 8x-a/x^2 and then equated to 0. Then said that 8x^3-a=0 therefore 8x^3=a.
Plug that back into the original equation to get 32=4x^2+8x^2+5 (32 is the y coordinate we were given).
Solving that to get x= 0 and + or - 3/2. Then plugging that back in to the original equation gave a=0, not possible or a negative solution when it asked for the positive solution so solving that eventually gave 27.
I hope that makes sense? Think it's correct (I BLOODY HOPE SO!)
16. i think the grade boundaries will be in the 50s for an A - i found january 2011 a lot easier and it was 54 for an A
17. A=27 for last question. Y=1/16 and y=8 for hidden quadratic. K less than -2 but great than 5 for question 9. Y=(x-2)^2 (5-x) for translation.
18. (Original post by JackA123)
With the indice questions, was it 2^-6 for part (i) and 2^13/3 for (ii)?
I got 2^-6 but was worrying cos all I remember from the exam is 2^5 divided by 2^7 which gives ¼ so there must've been something else I can't remember.
Mathematically speaking should be a 2^-4...
19. question 9 was x^2 + 2x + 11= k(2x-1), find k
and 11 was y= 4x^2 + a/x + 5, find a, also when dy/dx=o, y=32
Attached Images
20. (Original post by duncant)
A=27 for last question. Y=1/16 and y=8 for hidden quadratic. K less than -2 but great than 5 for question 9. Y=(x-2)^2 (5-x) for translation.
(x-2)^2 (5-x)?
I put x^2(5-x) If it was (x-2)^2 it would've also resulted in a stretch.
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# What is the next number in this sequence? (My first)
This is my first attempt at a question. Please let me know what you think of it, I came up with the idea for it while trying to figure out some of the other sequence questions here.
1, 2, 5, 10, 21, 42, 85, ?
I can add a hint if needed.
-P.S. I have not been around for the longest time so sorry if this has been asked before
• Hi Steven. Welcome to the site and great that you're willing to put in content. As you have asked for feedback: generally speaking "number series" puzzles are not the most welcome puzzles as they are often either too vaguely defined or sometimes just boring. If you're willing to pack puzzle ideas into more complex / structured / bigger puzzles, you will earn more community respect. Jul 29 '16 at 15:35
• Steven, I think I've got it, please comment below my answer (way down low :D). Jul 30 '16 at 3:27
This answer is prompted by user f'' who gave a hint:
Try converting to binary.
So I did...
1
1
2
10
5
101
10
1010
21
10101
42
101010
85
1010101
The pattern becomes obvious at this point:
The sequence is a binary number n digits long starting with a 1 and alternating 1s and 0s (then converted back to decimal, of course).
10101010 which is 170 in decimal.
Double if it is odd and double+1 if it is even works because when it is odd that means the last digit in binary is a 1. So to continue the sequence we must add a zero to the end. This is equivalent to multiplying by 2. When it is even we need to add a 1 on the end. We know that adding a 0 on the end is multiplying by 2 so adding a 1 on the end is multiplying by 2 and adding 1.
and
My other answer of $f(n)=2^n+f(n−2)$ works by considering the power expansion of a binary number. For any given member of the sequence you can make the element two further on by adding a new power of 2 which is 2 orders of magnitude higher than the previous highest. So if you look at $f(5) = 42 = 101010 = 2^5+2^3+2^1$ then you can add a new power of 2 to the beginning to give $2^7+2^5+2^3+2^1 = 1010101 = 85$.
• This was the pattern that I did have in mind when I came up with the question. Thanks to f" for the suggestion to it. Aug 9 '16 at 13:40
170 Explanation is that all odd numbers are multiplied by 2 while all even numbers are multiplied by 2 and then added by 1.
• Well then not the way I came up with this one, but a valid answer none the less Jul 29 '16 at 15:16
• So, it's not the correct answer?
– Sid
Jul 29 '16 at 15:16
• @Sid I'm willing to wager that whatever algorithm the original author had in mind, it would most likely reduce to yours. Jul 29 '16 at 15:18
• Try converting to binary.
– f''
Jul 29 '16 at 15:48
• @f'': you should Jul 29 '16 at 16:25
The next term is:
$170$
Since another formulation of the solution was requested this series can be expressed as:
$f(n) = 2^n + f(n-2)$ for $n \ge 0$
and
Whereby we assume $f(-1) = f(-2) = 0$ when needed for the main formula to be well formed.
So to spell them out more clearly:
First term:
$f(0) = 2^0 + f(-2) = 1 + 0 = 1$
Second Term:
$f(1) = 2^1 + f(-1) = 2 + 0 = 2$
Third Term:
$f(2) = 2^2 + f(0) = 4+1 = 5$
Fourth Term:
$f(3) = 2^3 + f(1) = 8+2 = 10$
Fifth Term:
$f(4) = 2^4 + f(2) = 16+5 = 21$
Sixth Term:
$f(5) = 2^5 + f(3) = 32+10 = 42$
Seventh Term:
$f(6) = 2^6 + f(4) = 64+21 = 85$
Eight Term:
$f(7) = 2^7 + f(5) = 128+42 = 170$
• Maybe, this is it. Sort of explains it in pure mathematical way..
– Sid
Jul 29 '16 at 16:19
• I think f''s suggestion on binary is more likely. This is a nice pure maths difference equation but the binary answer (which I will post if f'' doesn't) is by far the most elegant solution. Jul 29 '16 at 16:25
• Yeah... the binary thing is beautiful. Maybe that was the author's logic. It makes a brilliant pattern.
– Sid
Jul 29 '16 at 16:28
• The other quite nice thing is that you can see it in my answer by the nature of the recurrence relationship (or just by doing the substitutions iteratively until you get to the base case). Jul 29 '16 at 16:36
The number is
171
Explanation
Each number is double the last plus 1, 171=85•2+1
• 2, 10 and 42 do not pass this formula. Jul 30 '16 at 6:48 | HuggingFaceTB/finemath | |
# How to prove : (( P → Q ) ∨ ( Q → R )) by natural deduction
Here's another of Tomassi's exercises I can't solve (Logic, page 106):
: (( P → Q ) ∨ ( Q → R ))
I have to use natural deduction and the only rules I know are:
• assumptions,
• modus ponendo ponens,
• modus tollendo tollens,
• double negation,
• conditional proof,
• v-introduction,
• v-elimination,
• and introduction,
• and elimination.
Tomassi's proof consists of 20 steps.
Thanks for the help.
This is my answer so far:
{1} 1. (P -> Q) v (Q->R) Assumption
{2} 2. P -> Q Ass for vE
{3} 3. P Ass for CP
{4} 4. ~Q Ass for RAA
{2,4} 5. ~P 2,4 MT
{2,3,4} 6. P & ~P 3, 5 &I
{2,3} 7. ~~Q 4, 6 RAA
{2,3} 8. Q 7 DNE
{2} 9. P -> Q 3, 8 CP
{2} 10. (P -> Q) v (Q->R) 9 vI
{11} 11. Q -> R Ass for vE
{12} 12. Q Ass for CP
{13} 13. ~R Ass for RAA
{11,13} 14. ~Q 11, 13 MT
{11,12,13} 15. Q & ~Q 12, 14 &I
{11,12} 16. ~~R 13, 15 RAA
{11,12} 17. R 16 DNE
{11} 18. Q -> R 12, 17 CP
{11} 19. (P -> Q) v (Q->R) 18 vi
{1} 20. (P -> Q) v (Q->R) 1, 2, 10, 11, 19 vE
I don't know how to discharge the assumption 1.
It doesn't help to start your proof with the statement that you are trying to prove. Indeed, (( P → Q ) ∨ ( Q → R )) should be the last line of your proof, not the first. So, your whole set-up for the proof is not good.
In his book, Tomassi lays out what he calls the 'golden rule':
ask: (i) is the conclusion a conditional? If it is, apply CP. If not, ask: (ii) are any or all of the premises disjunctions? If so, apply vE. If not, assume the negation of the desired conclusion and try the RAA strategy.
If you apply the golden rule to your problem, you'll find that you end up with the last strategy: negate the desired conclusion and try the RRA strategy. So, it'll look something like this:
{1} 1. ~(( P → Q ) ∨ ( Q → R )) Assumption for RAA
...
{} n+1. ~~(( P → Q ) ∨ ( Q → R )) 1,n RAA
{} n+2. (( P → Q ) ∨ ( Q → R )) n+1 DNE
Indeed, notice how both eliran's and Frank's proof look like this .... except neither is in Tomassi's format. Here is how you do the rest in Tomassi's format:
{1} 1. ~(( P → Q ) ∨ ( Q → R )) Assumption for RAA
{2} 2. Q Assumption for RAA
{3} 3. P Assumption
{2,3} 4. P&Q 2,3 &I
{2,3} 5. Q 4 &I (here is the augmentation trick again, see p. 53-54)
{2} 6. P → Q 3,5 CP
{2} 7. ( P → Q ) ∨ ( Q → R ) 6 ∨I
{1,2} 8. (( P → Q ) ∨ ( Q → R )) & ~(( P → Q ) ∨ ( Q → R )) 1,7 &I
{1} 9. ~Q RAA 2,8
{10} 10. ~R Assumption for RAA
{2,10} 11. Q & ~R 2,10 &I
{2,10} 12. Q 11 &E (augmentation trick yet again!)
{1,2,10} 13. Q & ~Q 9,12 &I
{1,2} 14. ~~R 10,13 RAA
{1,2} 15. R 14 DNE
{1} 16. Q → R 2,15 CP
{1} 17. ( P → Q ) ∨ ( Q → R ) 16 ∨I
{1} 18. (( P → Q ) ∨ ( Q → R )) & ~(( P → Q ) ∨ ( Q → R )) 1,17 &I
{} 19. ~~(( P → Q ) ∨ ( Q → R )) 1,18 RAA
{} 20. (( P → Q ) ∨ ( Q → R )) 19 DNE
Since the argument has no premises, we must start with an assumption, either for reductio or for conditional proof. In this case, conditional proof would not work, so we have to go with reductio. So we start with:
1. | ~((P→Q)∨(Q→R)) assumption
Since we're going for reductio, we need to derive a contradiction. Since all we've got is this assumption, our contradiction is going be with its negation. So we have to generate (P→Q)∨(Q→R). How? By generating one of the disjuncts. So add another assumption for another reductio.
2. || ~(P→Q) assumption
Getting a contradiction from this is a bit complicated, but that's how the proof proceeds. Here's the complete proof:
1. | ~((P→Q)∨(Q→R)) assumption (for reductio)
2. || ~(P→Q) assumption (for reductio)
3. ||| Q assumption (for reductio)
4. |||| P assumption (for conditional)
5. |||| Q 3
6. ||| P→Q 4-5 (conditional)
7. ||| (P→Q)&~(P→Q) 6,2 (&-intro)
7. || ~Q 3-7 (reductio)
8. ||| Q assumption (for conditional)
9. |||| ~R assumption (for reductio)
10. |||| Q&~Q 7,8 (&-intro)
11. ||| R 9-10 (reductio)
12. || Q→R 8-11 (conditional)
13. || (Q→R)∨(P→Q) 12 (∨-intro)
14. || ((Q→R)∨(P→Q))&~((Q→R)∨(P→Q)) 13,1 (&-intro)
15. | P>Q 2-14 (reductio)
16. | (P→Q)∨(Q→R) 15 (∨-intro)
17. | ((P→Q)∨(Q→R))&~((P→Q)∨(Q→R)) 16,1 (&-intro)
18. (P→Q)∨(Q→R) 1-17 (reductio)
• Many thanks for your reply. Would you mind clarifying me the step number 5 please? – Diego Ruiz Haro Nov 25 '18 at 16:08
• 5 is just a repetition of 3 (some proof systems require to repeat statements that are outside the current sub-proof), in order to get the conditional in 6. – Eliran Nov 25 '18 at 16:10
Here is a proof similar to Eliran's. It uses reiteration (line 4) and indirect proof (lines 12 and 16) which I don't think were introduced prior to this exercise in Tomassi's text (page 106). I am presenting it in Klement's proof checker to show such a proof would work with different rules, but we have to use the permitted rules.
Tomassi shows how to avoid reiteration on pages 63-4 by using first conjunction introduction and then conjunction elimination. This next proof replaces reiteration (lines 4 and 5) and indirect proof with contradiction introduction and double negation elimination (lines 13-14 and 18-19) that would correspond to reductio ad absurdum.
This takes one less step than Tomassi required, however, I think it follows only the rules you are permitted to use. I will leave the final formatting of the proof to you.
References
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
Paul Tomassi, Logic, Routledge 1999
• Thanks a lot Frank. This proof is not completely clear to me. Could you please explain me why the reiteration at line 4 is necessary? Why just don't firstly assume P and then get Q by MP? I don't understand lines 9,10,11 either. Please accept my apologies beforehand if these are rather silly questions but I'm a real beginner. – Diego Ruiz Haro Nov 26 '18 at 18:18
• @DiegoRuizHaro The reiteration isn't necessary. It is a derived rule. I provided two proofs. The second one is what you may need. However, one cannot use modus ponens on the assumption in line 1 because there is a negation in front of it and it is part of a disjunction. To get P>Q we have to derive it. We also need Q after P for the conditional. On line 9, ~Q is the result of a reductio ad absurdum from lines 2-8. Line 10, Q, is an hypothesis. It is there to contradict with ~Q on line 9. Line 11 is ~R. What I want is R to derive Q>R on line 15. To get there I use reductio ad absurdum. – Frank Hubeny Nov 26 '18 at 23:58
• Thanks again. It's way clearer now. However, I can't discharge the dependency number 11. What did I get wrong? – Diego Ruiz Haro Nov 28 '18 at 17:15
• @DiegoRuizHaro An example of how to do this is in Tomassi's text on page 103. What I provided would have to be converted to his system using a reduction ad absurdum subproof rather than the contradiction introduction that I used. To do this, line 11 is still "~R" but with justification "Assumption for RAA". Line 12, however, would be "Q & ~Q" with justification "9,10 &I" Then on line 13 we still have "~~R" but with justification "11,12 RAA". That would allow you to discharge the assumption on line 11. – Frank Hubeny Nov 28 '18 at 17:28
• Indeed, I did it that way. So according to that, the dependency numbers for line 9 is {1}, for line 10 is {10}, line 11 {11}, line 12 {1,10}, line 13 {1,10,11}. And then I'm able to discharge {10} at line 15 with 10,14 CP, and {1} at line 18 with 1,17 RAA. So {11} is still there. What did I do wrong? – Diego Ruiz Haro Nov 28 '18 at 18:02
This is my answer so far:
{1} 1. (P -> Q) v (Q->R) Assumption
{2} 2. P -> Q Ass for vE
{3} 3. P Ass for CP
{4} 4. ~Q Ass for RAA
{2,4} 5. ~P 2,4 MT
{2,3,4} 6. P & ~P 3, 5 &I
{2,3} 7. ~~Q 4, 6 RAA
{2,3} 8. Q 7 DNE
{2} 9. P -> Q 3, 8 CP
{2} 10. (P -> Q) v (Q->R) 9 vI
{11} 11. Q -> R Ass for vE
{12} 12. Q Ass for CP
{13} 13. ~R Ass for RAA
{11,13} 14. ~Q 11, 13 MT
{11,12,13} 15. Q & ~Q 12, 14 &I
{11,12} 16. ~~R 13, 15 RAA
{11,12} 17. R 16 DNE
{11} 18. Q -> R 12, 17 CP
{11} 19. (P -> Q) v (Q->R) 18 vi
{1} 20. (P -> Q) v (Q->R) 1, 2, 10, 11, 19 vE
I don't know how to discharge the assumption 1. | HuggingFaceTB/finemath | |
848 views
Find the minimum value of $3-4x+2x^2$.
in Calculus
edited | 848 views
0
(≤ 720)
what is this?
$f(x) = 3-4x+2x^2$
$f '(x) = -4 + 4x = 0 \implies x=1$
$f ''(x) = 4$
$f ''(1) = 4>0$, therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$
by Loyal (8.1k points)
edited | open-web-math/open-web-math | |
## What is a lognormal distribution?
From the intuitive perspective, a quantity or variable is likely to be lognormally dsitributed if it often changes by a percentage. For instance, a stock’s return could go up by 5% today but decrease by 10% the next day. A lognormally distributed quantity $Y$ can be expressed as the cumulative product of many positive, independent numbers, as $Y = X_1 * X_2 * X_3 ... * X_n$.
Because each factor $X_i$ is positive, we could express $\log Y$ as the sum of $\log X_i$, as $\log Y = \Sigma_{i=1}^{n} \log X_i$. Because the $\log X_i$ is independent, we can apply the Central Limit Theorem and claim $\log Y$ is normally distributed.
Thus, from a data perspective, a dataset is likely to be lognormally distributed if its $\log$ appears approximately normal. On the other hand, a lognormal distribution often looks right-skewed.
To confirm our intuition, lets run some simulations!
## Simulating lognormal
Situation 1: Normally distributed change
In this situation I simulated 10k data points that go through 30 change periods. Each data point starts with value 1 (100%). In each change period, its value multiplies a factor that is normally distributed with $\mu = 1$ and $\sigma = 0.1$. In other words, each data point either grows or shrinks a small percentage in each period for 30 periods in total.
Below is the mathematical representation of the simulation.
$\begin{equation} \text{sample data point} \, i = Z_1 * Z_2 * ... \color{red}{Z_{30}}\\ \end{equation} \\ \text{where} \, i \in \{1,2,...10000\} \quad Z_i \sim \mathcal{N}(\mu=1,\sigma=0.1)$ Caption: The top-left is the distribution of the simulated data. The top-right is the distribution of its log. For the bottom-left graph, I first estimated the fitted parameters of the lognormal distribution, then I use the parameters to generate 10k data and plot the lognormal density curve, along with that of the simulated data. The bottom right graph is the same but with both simulated data and the lognormal data logged. (the details of the fitting process can be found in my Github repo,)
From the density curves, the simulated data almost overlaps the fitted lognormal distribution; the same goes with their corresponding log data. This shows that lognormal distribution is good at representing a quantity that grows or shrinks with equal probabilities.
Situation 2: Arbitrary growth
What if the change is not symmetric but more arbitrary? For instance, housing prices are often growing in the long run. In situation 2, we model a random quantity that only grows while following our lognormal assumptions (cumulative product of positive, uncorrelated variables).
Each data point grows either 5% or 10% with equal chance in each period. Additionally, I simulated 10k data points that undergo 30 growth periods and another 10k data points that undergo 100 growth periods.
Below is the mathematical representation of the simulation.
Period 30:
$\begin{equation} \text{sample data point} \, i = G_1 * G_2 * ... \color{red}{G_{30}} \end{equation} \\ \text{where} \, i \in \{1,2,...10000\} \quad P(G_i = 1.1) = P(G_i = 1.05) = 50%$
Period 100:
$\begin{equation} \text{sample data point} \, i = G_1 * G_2 * ... \color{red}{G_{100}} \end{equation} \\ \text{where} \, i \in \{1,2,...10000\} \quad P(G_i = 1.1) = P(G_i = 1.05) = 50%$ Caption: The top left graph is the distribution of the simulated data with period 30 and the top right is its log. The bottom row is the distribution of simulated data with period 100 and its log.
Here are two important things we could learn from the graphs:
1. The fitted distribution (the blue curves) still overlaps a significant portion of the simulated data, especially for period 100. This provides good evidence that the lognormal distribution is a good fit for the cumulative product of uncorrelated variables, regardless of the distribution of those variables (I highly recommend that you to try out other distributions to see the results for yourself.)
2. Note the distributions of period 30 is bumpier than that of period 100. This confirms our inference of $\log Y$ will be normal using the Central Limit Theorem: as we increase the growth period from 30 to 100, we increase the number of variables added to $\log Y$, thus the distribution of $\log Y$ will converge to normal regardless of the distribution of the numbers added. In other words, the distributions of factors $X_i$ matters less as we have more of them.
## Lognormal in the real world
The lognormal distribution fits two different simulations very well, but it is most important to see how well it fits in real world data. Now I will show you that the distribution of house prices can be well approximated by a lognormal distribution. I looked at the distribution of housing prices from mid-2003 through mid-2006 in the SF Bay Area. Below is the fitted log-normal distribution of all the housing prices. Caption : The table above shows the statistics about the housing prices and the simulated price from the lognormal fit. All the statistics are fairly close with small percentage deviation except the minimum and maximum.
Overall, the fitted lognormal distribution could capture most of the housing price distribution but not the extreme values. Additionally it underestimates the density of the most popular house price. Maybe in the real world, sellers tend to match the popular (most frequently sold) prices.
As for the outliers problem, I think the real world is too complex for a statistical model to capture it perfectly. However, I am interested in what percentage of the data the lognormal distribution could reasonably fit/explain. So I decided to eliminate 25 outliers out of 280k data points. Below is the fitted plot and the range of both data(top:housing data, bottom:simulated data) After trimming a tiny fraction of outliers, the fitted distribution matches closely with the actual housing price even in the extreme values. This illustration shows that the lognormal distribution can be used to model a significant portion of the housing price distribution. Thus we have good evidence supporting the claim that housing price is lognormally distributed. Therefore we can infer it may often change by percentage rather than absolute amount.
## Wrapping Up
In summary, the lognormal distribution is used to model a quantity that often changes in percentage while a normally distributed quantity often changes in absolute size. I hope you could get a good intuitive understanding of the lognormal distribution and later learn to use it in your data analysis or modeling process. | HuggingFaceTB/finemath | |
# Re: Different problem
[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]
Posted by Denis Borris on November 05, 2002 at 00:10:34:
In Reply to: Different problem posted by Mike on November 04, 2002 at 22:23:21:
: A car and a bus set out at 3pm from the same point headed in the same
: direction. The average speed of the car is twice the average speed of the bus.
: In 2h the car is 68 mi ahead of the bus. Find the rate of the car.
Obviouly 68...
Let bus speed = b; then car speed = 2b
bus: b = d/2 ; d = 2b
car: 2b = (d+68)/2 ; d = 4b - 68
So: 4b - 68 = 2b
2b = 68
Name:
E-Mail:
Subject: | HuggingFaceTB/finemath | |
# 2 kg
Convert 2 Kilograms to Pounds
26 rows · 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.
KGLB
2.24 4.9384
2.23 4.9163
2.22 4.8943
2.21 4.8722
See all 26 rows on www.calculateme.com
1 Kilogram (kg) is equal to 2 pounds (lbs) and 3.273 ounces (oz). To convert kilograms to pounds and ounces, first multiply the kg value by 2.2046226218 to convert into pounds. The integer part of the result is the pound value. To find the ounce value, multiply the fractional part by 16. kg to pounds and ounces formula. pound (x.y) = kg * 2 ## What Is 2 Kg In Pounds? Convert 2 Kg To Lbs
Convert 2 kg to pounds. One kilogram equals 2.20462262 pounds, to convert 2 kg to pounds we have to multiply the amount of kg by 2.20462262 to obtain amount in pounds. 2 kg are equal to 2 x 2.20462262 = 4.409245 pounds.
The kilogram is the base unit of mass in the International System of Units (SI). It is in widely used in science, engineering, and commerce worldwide. The kilogram is exactly the mass of one litre of water.. As of May 20, 2019, the definition of the kilogram is based on the Planck constant as 6.626 070 15 × 10 −34 kg⋅m 2 ⋅s −1.. There are attempts to define the kilogram in other ways Convert 2 Kilograms to Ounces
26 rows · The kilogram, or kilogramme, is the base unit of weight in the Metric system.It is the …
KGOZ
2.00 70.548
2.01 70.901
2.02 71.253
2.03 71.606
See all 26 rows on www.calculateme.com
Aug 07, 2019 · Converting from kilograms to pounds is a common task in the realms of math and engineering, but, luckily, it’s an easy one. In most cases, all you need to do to convert is to multiply the number of kilograms by 2.2 to get the number of
Views: 92K 2.2 Kilograms To Pounds Converter
2.2 kilograms equal 4.8501697681 pounds (2.2kg = 4.8501697681lbs). Converting 2.2 kg to lb is easy. Simply use our calculator above, or apply the formula to change the length 2.2 kg to lbs.
kilograms to pounds formula. pound = kilogram * 2.2046226218. 1 Kilogram = 2.2046226218 Pounds. To convert kg to pounds and ounces, please visit kg to pounds and ounces converter. What is a Kilogram? Kilogram (kilo) is the metric system base unit of mass. 1 kg = 2.2046226218 lbs. The symbol is „kg“. Common conversions from kilograms to pounds 2.2 kg to pounds and ounces
The kilogram (kg) is the SI unit of mass. It is equal to the mass of the international prototype of the kilogram. This prototype is a platinum-iridium international prototype kept at the International Bureau of Weights and Measures. One kg is exactly equal to 2.20462262184878 pounds or aproximately 16 * 2.21 = 35.27 ounces. Definition of gram
Important: The slug is the base unit for mass in the English system, but the more common unit for mass is the Pound Mass (lbm).Under standard Earth gravity, 1 lbm weighs 1 lbf and therefore the term „pound“ is usually used for both mass and weight in the English system. MASS Conversions 1 lbm = 0.45359237* kg (exact conversion) 1 kg = 2.2046226 lbm 1 slug = 32.1740486 lbm kg to lbs
23 rows · 1 kilogram (kg) is equal to 2.20462262185 pounds (lbs). 1 kg = 2.20462262185 lb The mass …
KILOGRAMS (KG)POUNDS (LB)POUNDS+OUNCES (LB+OZ)
0 kg 0 lb 0 lb 0 oz
0.1 kg 0.220 lb 0 lb 3.527 oz
1 kg 2.205 lb 2 lb 3.274 oz
2 kg 4.409 lb 4 lb 6.548 oz
See all 23 rows on www.rapidtables.com
Welcome to 101.2 kg to lbs, our page about the 101.2 kilograms to pounds conversion.If you have found us by searching for 101.2 kg in pounds, or if you have been asking yourself how many pounds in 101.2 kg, then you are right here, too.When we write 101.2 kilos in pounds, or use a similar term, we mean the unit international avoirdupois pound; for 101.2 kilos to pounds in historical units of Kilogram
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 6.626 070 15 × 10 −34 when expressed in the unit J⋅s, which is equal to kg⋅m 2 ⋅s −1, where the metre and the second are defined in terms of c and Δν Cs.
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Use this page to learn how to convert between kilograms and pounds. Type in your own numbers in the form to convert the units! ›› Quick conversion chart of kg to lbs. 1 kg to lbs = 2.20462 lbs. 5 kg to lbs = 11.02311 lbs. 10 kg to lbs = 22.04623 lbs. 15 kg to lbs = 33.06934 lbs. 20 kg to lbs = 44.09245 lbs. 25 kg to lbs = 55.11557 lbs
5.2 grams equal 0.0052 kilograms (5.2g = 0.0052kg). Converting 5.2 g to kg is easy. Simply use our calculator above, or apply the formula to change the length 5.2 g to kg. SI derived unit
The SI has special names for 22 of these derived units (for example, hertz, the SI unit of measurement of frequency), but the rest merely reflect their derivation: for example, the square metre (m 2), the SI derived unit of area; and the kilogram per cubic metre (kg/m 3 or kg⋅m −3), the SI derived unit of density.
8.2 pounds equal 3.719457434 kilograms (8.2lbs = 3.719457434kg). Converting 8.2 lb to kg is easy. Simply use our calculator above, or apply the formula to change the length 8.2 lbs to kg. ## Kg to Lbs converter
Kilograms: 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: 26: 27: 28: 29: 30: 31: 32: 33: 34: 35: 36: 37: 38: 39: 40: 41
Jul 27, 2020 · Ex 3.1, 3 The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. Let the cost of apples per kg be Rs x & Let the cost of grapes per kg … 5.2 Kilograms To Pounds Converter
5.2 kilograms equal 11.4640376336 pounds (5.2kg = 11.4640376336lbs). Converting 5.2 kg to lb is easy. Simply use our calculator above, or apply the formula to change the length 5.2 kg to lbs.
2 Kings 1. King Ahaziah asks Baalzebub and Elijah about his injuries 1:1-8 1:1 Then Moab rebelled against Israel after the death of Ahab. 1:2 And Ahaziah fell down through a lattice in his upper chamber that was in Samaria, and was sick: and he Convert 2.2 kg to pounds
One kg is approximately equal to 2.20462262184878 pounds. Definition of pound One pound , the international avoirdupois pound, is legally defined as exactly 0.45359237 kilograms.
Kilogram. Definition: A kilogram (symbol: kg) is the base unit of mass in the International System of Units (SI). It is currently defined based on the fixed numerical value of the Planck constant, h, which is equal to 6.62607015 × 10-34 in the units of J·s, or kg·m 2 ·s-1.The meter and the second are defined in terms of c, the speed of light, and cesium frequency, Δ ν Cs. Convert 2.2 lbs to kg
How many lbs in 1 kg? The answer is 2.2046226218488. We assume you are converting between pound and kilogram. You can view more details on each measurement unit: lbs or kg The SI base unit for mass is the kilogram. 1 lbs is equal to 0.45359237 kilogram. Note that rounding errors may occur, so always check the results.
Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. ## Liters to Kilograms [ water ] Converter
1 liter (l) = 1 kilogram (kg). Liter (l) is a unit of Weight used in Volume system. Kilogram (kg) is a unit of Weight used in Metric system. Please note this is volume to weight conversion, this conversion is valid only for pure water at temperature 4 °C. US oz = 28.349523125 g US fl oz = 29.5735295625 ml (milliliters) = 29.5735295625 g (grams
2 Kings 10. Jehu kills Ahab’s sons: 70 heads in two heaps 10:1-8 10:1 And Ahab had seventy sons in Samaria. And Jehu wrote letters, and sent to Samaria, unto the rulers of Jezreel, to the elders, and to them that brought up Ahab’s children, saying, 10:2 Now as soon as this letter Amazon.in: Buy Dabur Chyawanprash 2X Immunity – 2 kg online at low price in India on Amazon.in. Check out Dabur Chyawanprash 2X Immunity – 2 kg reviews, ratings, specifications and more at Amazon.in. Free Shipping, Cash on Delivery Available.
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# m02 #18
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Is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?
1. $$x = 4$$
2. $$y = 6$$
OA:
[Reveal] Spoiler:
C
OE:
[Reveal] Spoiler:
To get the divisors of x and y, we need their respective values.
Statement (1) by itself is insufficient. We can only find the divisors of x .
Statement (2) by itself is insufficient. We can only find the divisors of y .
Statements (1) and (2) combined are sufficient. Combining the statements, we have the divisors of x and y .
The way I approach this problem is this:
1. x=4
$$x^3 = 64 = 2^6$$
Total no. of factors = 6+1= 7
Now y is a perfect square. The total no. of factors of a perfect square is always in the form 2k+1.
So, question is: Is 7 a multiple of any number of the form 2k+1?
Answer is: it may be (e.g. when k=3) or may not be (e.g. when k=1, 5,….)
So insufficient
2. y=6
$$y^2 = 36 = 2^2 * 3^2$$
Total no. of factors = (2+1)(2+1) = 9
Now x is a perfect cube. The total no. of factors of a perfect cube is always in the form 3k+1
So, question is: Is a number of the form 3k+1 a multiple of 9
Answer is: No, it can never be.
So, sufficient
And hence, B is my answer.
What’s wrong in this? And doesn’t the OE too simplistic? (In a way, it ignores certain factorization properties of perfect squares and cubes)
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06 Jun 2010, 06:14
deepakdewani wrote:
Is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?
1. $$x = 4$$
2. $$y = 6$$
OA:
[Reveal] Spoiler:
C
OE:
[Reveal] Spoiler:
To get the divisors of x and y, we need their respective values.
Statement (1) by itself is insufficient. We can only find the divisors of x .
Statement (2) by itself is insufficient. We can only find the divisors of y .
Statements (1) and (2) combined are sufficient. Combining the statements, we have the divisors of x and y .
The way I approach this problem is this:
1. x=4
$$x^3 = 64 = 2^6$$
Total no. of factors = 6+1= 7
Now y is a perfect square. The total no. of factors of a perfect square is always in the form 2k+1.
So, question is: Is 7 a multiple of any number of the form 2k+1?
Answer is: it may be (e.g. when k=3) or may not be (e.g. when k=1, 5,….)
So insufficient
2. y=6
$$y^2 = 36 = 2^2 * 3^2$$
Total no. of factors = (2+1)(2+1) = 9
Now x is a perfect cube. The total no. of factors of a perfect cube is always in the form 3k+1
So, question is: Is a number of the form 3k+1 a multiple of 9
Answer is: No, it can never be.
So, sufficient
And hence, B is my answer.
What’s wrong in this? And doesn’t the OE too simplistic? (In a way, it ignores certain factorization properties of perfect squares and cubes)
When solving this question I made the same assumption as you did and arrived to answer B with same logic as you did. But answer B is not correct. The trick here is that we are not told that $$x$$ and $$y$$ are integers, so for (2) the logic would be correct ONLY for integers. But if $$y^2=36=x^3$$ then the total number of divisors of $$x^3$$ will obviously be equal to the total number of divisors of $$y^2$$ (note that in this case $$x=\sqrt[3]{36}\neq{integer}$$). Hence insufficient.
Though I must say that this is not GMAT type of question, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.
So I would suggest to change the question as follows:
If $$x$$ and $$y$$ are positive integers, is the total number of divisors of $$x^3$$ a multiple of the total number of divisors of $$y^2$$?
(1) $$x = 4$$
(2) $$y = 6$$
In this case: A. the question will meet the GMAT standards and B. the question will be 750+ difficulty level, with elegant solution.
Hope it helps.
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### Show Tags
06 Jun 2010, 07:35
Quote:
The trick here is that we are not told that $$x$$ and $$y$$ are integers
Yes, that's would I could also think of.
Quote:
Though I must say that this is not GMAT type of question, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.
Agree -without x & y being integers, this question is a bit absurd.
Thanks a bunch.
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04 Jun 2011, 14:25
is there any way to simplify the question stem to what we are looking for?
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Re: m02 #18 [#permalink] 04 Jun 2011, 14:25
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# m02 #18
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# If (x−2.5)(x+0.5)=4, what is the value of x?
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If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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21 Sep 2016, 02:59
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If (x−2.5)(x+0.5)=4, what is the value of x?
(1) (x+1.5)(x−1.5)=0
(2) (x−3.5)(x+1.5)=0
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Re: If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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22 Sep 2016, 09:27
7
Bunuel wrote:
If (x−2.5)(x+0.5)=4, what is the value of x?
(1) (x+1.5)(x−1.5)=0
(2) (x−3.5)(x+1.5)=0
Looking at this one, it seems like the statements will be much easier to simplify than the question will. I don't really want to go through the process of multiplying out and then re-factoring the quadratic from the question, so I'm just going to start with the statements and see how far I can get.
Statement 1: This says that x is -1.5, or x is 1.5. But it's possible that only one of those fits the constraint from the problem, in which case this statement would actually be sufficient. I'll plug them both in:
(-1.5 - 2.5)(-1.5 + 0.5) = (-4)(-1) = 4. So, x = -1.5 fits the constraint.
(1.5 - 2.5)(1.5 + 0.5) = (-1)(2) = -2. So, x = 1.5 doesn't fit the constraint. The only valid case, which fits both the constraint from the question and Statement 1, is x = -1.5.
So, statement 1 is sufficient.
Statement 2
: Same process. x = 3.5, or x = -1.5. We already know that -1.5 fits the constraint. Does 3.5 fit?
(3.5 - 2.5)(3.5 + 0.5) = (1)(4) = 4.
Unfortunately, both possible values of x are valid. Since this statement gives us two valid values for x, it's insufficient.
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If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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Updated on: 21 Sep 2016, 06:37
(x−2.5)(x+0.5)=4
==> $$(x−\frac{5}{2})(x+\frac{1}{2})=4$$
==> $$x^2+\frac{x}{2} - \frac{5x}{2}-\frac{5}{4}=4$$
==> $$x^2- 2x-\frac{21}{4}=0$$ (Subtract each side by 4) --Eq1
==> $$x^2 - \frac{7}{2}x + \frac{3}{2}x-\frac{21}{4}=0$$
==> $$x(x - \frac{7}{2}) + \frac{3}{2}(x-\frac{7}{2})=0$$
==> $$(x - \frac{7}{2}) (x+ \frac{3}{2})=0$$
==> x = 7/2 or x = - 3/2
(1) (x+1.5)(x−1.5)=0 ==> $$x^2-{1.5}^2$$ = 0 ==> x = 1.5 or x = -1.5 -- Sufficient.
(2) (x−3.5)(x+1.5)=0 ==> $$x^2 + \frac{3}{2}x - \frac{7}{2} - \frac{3}{2}*\frac{7}{2}=0$$ ==> $$x^2- 2x-\frac{21}{4}=0$$ --Same as equation 1. --Not suffici
Ans A .
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Originally posted by 0ld on 21 Sep 2016, 06:20.
Last edited by 0ld on 21 Sep 2016, 06:37, edited 1 time in total.
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Re: If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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21 Sep 2016, 06:23
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Top Contributor
Bunuel wrote:
If (x−2.5)(x+0.5) = 4, what is the value of x?
(1) (x+1.5)(x−1.5) = 0
(2) (x − 3.5)(x+1.5) = 0
Target question: What is the value of x?
Given: (x−2.5)(x+0.5) = 4
This information is HUGE, since it tells us the only 2 possible values of x
Given: (x−2.5)(x+0.5) = 4
Expand and simplify to get: x² - 2x - 1.25 = 4
Subtract 4 from both sides: x² - 2x - 5.25 = 0
Multiply both sides by 4 to get: 4x² - 8x - 21 = 0
Factor to get: (2x + 3)(2x - 7) = 0
If 2x + 3 = 0, then x = -3/2 = -1.5
If 2x - 7 = 0, then x = 7/2 = 3.5
So, EITHER x = -1.5 OR x = 3.5
So, which value is it?
Statement 1: (x+1.5)(x−1.5) = 0
This tells us that EITHER x = -1.5 OR x = 1.5
Since we already know that EITHER x = -1.5 OR x = 3.5, we can be certain that x = -1.5
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: (x − 3.5)(x+1.5) = 0
This tells us that EITHER x = -1.5 OR x = 3.5
Since we already knew that EITHER x = -1.5 OR x = 3.5, this statement provides no new information.
So, all we can conclude from this is that EITHER x = -1.5 OR x = 3.5
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Cheers,
Brent
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Re: If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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24 Feb 2017, 01:12
Prompt analysis
(x -2.5)(x +0.5) = 4 => x^2 -2x -1.25 -4 = 0 =>x^2 -2x -5.25 = 0 => x= 3.5, -1.5
Superset
The answer will either 3.5 or -1.5
Translation
In order to find the value of x, we need
1# exact value of x
2# any other equation which satisfy one of the value of x
Statement analysis
St 1: x = -1.5,1.5. Combining the solution for both the equation we get x = -1.5. ANSWER
St 2: x= 3.5, -1.5. Combining both the equations we get x =x= 3.5, -1.5. INSUFFICIENT as we cannot narrow it down to one.
Option A
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Re: If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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04 Jan 2018, 03:16
the calculation yields a beautiful results. Also, no complex steps are present in this problem. Yes, gmat likes such kind of math problem.
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Re: If (x−2.5)(x+0.5)=4, what is the value of x? [#permalink]
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08 May 2018, 15:44
$$(x−2.5)(x+0.5)=4$$ is a nasty expression to solve for x. Since the gmat doesnt test our ability to perform tedious calculations, it should immediately signal that there's a more efficient way to solve the problem. Let's look at the two statements.
Statement 1 says that x = 1.5 or -1.5. We can easily substitute these values for x into $$(x−2.5)(x+0.5)=4$$ and see which ones work. If both of them work, then Statement 1 is insufficient. If one of them works, then it is sufficient.
Same logic for Statement 2.
Re: If (x−2.5)(x+0.5)=4, what is the value of x? &nbs [#permalink] 08 May 2018, 15:44
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# Find the amount and the compound interest on Rs 160000 for 2 years
Question:
Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.
Solution:
Principal, $P=$ Rs. 160000
Annual rate of interest, $R=10 \%$
Rate of interest for a half year $=\frac{10}{2} \%=5 \%$
Time, $n=2$ years $=4$ half years
Then the amount with the compound interest is given by
$A=P \times\left(1+\frac{R}{100}\right)^{n}$
$=160000 \times\left(1+\frac{5}{100}\right)^{4}$
$=160000 \times\left(\frac{100+5}{100}\right)^{4}$
$=160000 \times\left(\frac{105}{100}\right)^{4}$
$=160000 \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right)$
$=(21 \times 21 \times 21 \times 21)$
$=$ Rs 194481
Therefore, compound interest $=$ amount $-$ principal $=\mathrm{Rs}(194481-160000)=\mathrm{Rs} 34481$ | HuggingFaceTB/finemath | |
10mL equals two teaspoons (2tsp). A tablespoon is 3 times bigger 보다 a teaspoon and also three teaspoons equal one tablespoon (1Tbsp or 1Tb). One tablespoon likewise equals 15mL.
You are watching: How many teaspoons is in 10 ml
Similarly, you might ask, what is 10 mL equivalent to in teaspoons?
convert 10 Milliliters to Teaspoons mL tsp
10.00 2.0288
10.01 2.0309
10.02 2.0329
10.03 2.0349
One may likewise ask, is 2.5 mL fifty percent a teaspoon? measure of drugs 1/4 teaspoon 1.25 ml
1/2 teaspoon 2.5 ml
3/4 teaspoon 3.75 ml
1 teaspoon 5 ml
1-1/2 teaspoon 7.5 ml
Likewise, is 5ml a teaspoon or tablespoon?
because that example: 1cc = 1ml. 1 standardised teaspoon = 5ml. 1 standardised tablespoon = 15ml.
How plenty of milligrams room in a teaspoon?
Teaspoon: it is a unit of measure of volume the a medication or dosage i beg your pardon is same to 5 milliliters. The unit is abbreviated as tsp. Transform Milligrams (mg) come Teaspoons (tsp): 1 mg is about equal come 0.0002 tsps. One milligram is a relatively small amount of table salt.
### How lot is a 10 ml?
How countless Teaspoons is 10ml? - 10 ml is equal to 2.03 teaspoons.
### What is the dimension of a teaspoon?
A tespoon is a unit of volume measure equal to 1/3 tablespoon. That is precisely equal to 5 mL. In the USA there space 16 teaspoons in 1/3 cup, and also there space 6 teaspoons in 1 fluid ounce. "Teaspoon" may be abbreviated as t (note: lowercase letter t) or tsp.
### How have the right to I measure ML in ~ home?
fluid Ingredients: location the liquid measuring cup on a level surface. Bend under so her eye is level v the markings of the cup. To fill the cup come the appropriate level. When measuring 1 Tablespoon (15 mL) or less, fill the ideal measuring spoon to the height without letting it spill over.
### How perform I measure up 5 ml?
1 mL = 1 cc. 2.5 mL = 1/2 teaspoon. 5 mL = 1 teaspoon. 15 mL = 1 tablespoon. 3 teaspoons = 1 tablespoon.
### What is 5 ml sneeze syrup?
In the study, researchers asked 195 college students who were recent patients that the university health and wellness clinic throughout cold and flu season to pour a 5 mL (equivalent to 1 teaspoon) dose of cold medicine right into various sizes of kitchen spoons.
### What is 10 mg converted to ML?
mg to ml conversion table: 10 mg = 0.01 ml 210 mg = 0.21 ml 700 mg = 0.7 ml 20 mg = 0.02 ml 220 mg = 0.22 ml 800 mg = 0.8 ml 30 mg = 0.03 ml 230 mg = 0.23 ml 900 mg = 0.9 ml 40 mg = 0.04 ml 240 mg = 0.24 ml 1000 mg = 1 ml 50 mg = 0.05 ml 250 mg = 0.25 ml 1100 mg = 1.1 ml
### How much is 100 ml the milk in cups?
U.S. Typical --> Metric U.S. Typical Metric 1/2 cup 100 ml plus 1-15 ml spoon 2/3 cup 150 ml 3/4 cup 175 ml 1 cup 200 ml and 2-15 ml spoons
14.8 mL
### Why is it dubbed a teaspoon?
Apothecary measure up The Apothecaries" teaspoon (formerly tea spoon or tea-spoon) was formally well-known by the Latin cochleare minus, abbreviated cochl. Min. To differentiate it native the tablespoon or cochleare majus (cochl.
### How lot is one tespoon in a syringe?
1 teaspoon (tsp) = 5 milliliters (mL)
### How big is a 5ml spoon?
measure Tip: 5ml Spoon just remember that 15ml equals one tablespoon and also 5ml equals one teaspoon.
### How many milligrams of liquid are in a tablespoon?
Approximately 15 milligrams room in one tablespoon. The an exact ratio is 14.786765 milligrams because that a single tablespoon. The much more common switch is from milliliters come tablespoons.
### How have the right to I measure up 10 mL?
10mL amounts to two teaspoons (2tsp). A tablespoon is 3 times bigger than a teaspoon and three teaspoons same one tablespoon (1Tbsp or 1Tb). One tablespoon likewise equals 15mL. | HuggingFaceTB/finemath | |
# 850000000 (number)
850,000,000 (eight hundred fifty million) is an even nine-digits composite number following 849999999 and preceding 850000001. In scientific notation, it is written as 8.5 × 108. The sum of its digits is 13. It has a total of 16 prime factors and 144 positive divisors. There are 320,000,000 positive integers (up to 850000000) that are relatively prime to 850000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 13
• Digital Root 4
## Name
Short name 850 million eight hundred fifty million
## Notation
Scientific notation 8.5 × 108 850 × 106
## Prime Factorization of 850000000
Prime Factorization 27 × 58 × 17
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 170 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 850,000,000 is 27 × 58 × 17. Since it has a total of 16 prime factors, 850,000,000 is a composite number.
## Divisors of 850000000
144 divisors
Even divisors 126 18 18 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 144 Total number of the positive divisors of n σ(n) 2.24121e+09 Sum of all the positive divisors of n s(n) 1.39121e+09 Sum of the proper positive divisors of n A(n) 1.5564e+07 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 29154.8 Returns the nth root of the product of n divisors H(n) 54.6134 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 850,000,000 can be divided by 144 positive divisors (out of which 126 are even, and 18 are odd). The sum of these divisors (counting 850,000,000) is 2,241,209,790, the average is 155,639,56.,875.
## Other Arithmetic Functions (n = 850000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 320000000 Total number of positive integers not greater than n that are coprime to n λ(n) 5000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 43539101 Total number of primes less than or equal to n r2(n) 72 The number of ways n can be represented as the sum of 2 squares
There are 320,000,000 positive integers (less than 850,000,000) that are coprime with 850,000,000. And there are approximately 43,539,101 prime numbers less than or equal to 850,000,000.
## Divisibility of 850000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 3 0 4
The number 850,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (850000000)
Base System Value
2 Binary 110010101010011111100010000000
3 Ternary 2012020102110210111
4 Quaternary 302222133202000
5 Quinary 3220100000000
6 Senary 220202241104
8 Octal 6252374200
10 Decimal 850000000
12 Duodecimal 1b87b6194
20 Vigesimal d5ca000
36 Base36 e22g74
## Basic calculations (n = 850000000)
### Multiplication
n×y
n×2 1700000000 2550000000 3400000000 4250000000
### Division
n÷y
n÷2 4.25e+08 2.83333e+08 2.125e+08 1.7e+08
### Exponentiation
ny
n2 722500000000000000 614125000000000000000000000 522006250000000000000000000000000000 443705312500000000000000000000000000000000000
### Nth Root
y√n
2√n 29154.8 947.268 170.748 61.0779
## 850000000 as geometric shapes
### Circle
Diameter 1.7e+09 5.34071e+09 2.2698e+18
### Sphere
Volume 2.57244e+27 9.0792e+18 5.34071e+09
### Square
Length = n
Perimeter 3.4e+09 7.225e+17 1.20208e+09
### Cube
Length = n
Surface area 4.335e+18 6.14125e+26 1.47224e+09
### Equilateral Triangle
Length = n
Perimeter 2.55e+09 3.12852e+17 7.36122e+08
### Triangular Pyramid
Length = n
Surface area 1.25141e+18 7.23753e+25 6.94022e+08 | HuggingFaceTB/finemath | |
# 468990 (number)
468,990 (four hundred sixty-eight thousand nine hundred ninety) is an even six-digits composite number following 468989 and preceding 468991. In scientific notation, it is written as 4.6899 × 105. The sum of its digits is 36. It has a total of 8 prime factors and 48 positive divisors. There are 124,416 positive integers (up to 468990) that are relatively prime to 468990.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 36
• Digital Root 9
## Name
Short name 468 thousand 990 four hundred sixty-eight thousand nine hundred ninety
## Notation
Scientific notation 4.6899 × 105 468.99 × 103
## Prime Factorization of 468990
Prime Factorization 2 × 35 × 5 × 193
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 8 Total number of prime factors rad(n) 5790 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 468,990 is 2 × 35 × 5 × 193. Since it has a total of 8 prime factors, 468,990 is a composite number.
## Divisors of 468990
48 divisors
Even divisors 24 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 48 Total number of the positive divisors of n σ(n) 1.27109e+06 Sum of all the positive divisors of n s(n) 802098 Sum of the proper positive divisors of n A(n) 26481 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 684.828 Returns the nth root of the product of n divisors H(n) 17.7104 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 468,990 can be divided by 48 positive divisors (out of which 24 are even, and 24 are odd). The sum of these divisors (counting 468,990) is 1,271,088, the average is 26,481.
## Other Arithmetic Functions (n = 468990)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 124416 Total number of positive integers not greater than n that are coprime to n λ(n) 5184 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 39066 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 124,416 positive integers (less than 468,990) that are coprime with 468,990. And there are approximately 39,066 prime numbers less than or equal to 468,990.
## Divisibility of 468990
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 4 6 0
The number 468,990 is divisible by 2, 3, 5, 6 and 9.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (468990)
Base System Value
2 Binary 1110010011111111110
3 Ternary 212211100000
4 Quaternary 1302133332
5 Quinary 110001430
6 Senary 14015130
8 Octal 1623776
10 Decimal 468990
12 Duodecimal 1a74a6
20 Vigesimal 2ic9a
36 Base36 a1vi
## Basic calculations (n = 468990)
### Multiplication
n×y
n×2 937980 1406970 1875960 2344950
### Division
n÷y
n÷2 234495 156330 117248 93798
### Exponentiation
ny
n2 219951620100 103155110310699000 48378715184614724010000 22689133634432459413449900000
### Nth Root
y√n
2√n 684.828 77.6941 26.1692 13.6217
## 468990 as geometric shapes
### Circle
Diameter 937980 2.94675e+06 6.90998e+11
### Sphere
Volume 4.32095e+17 2.76399e+12 2.94675e+06
### Square
Length = n
Perimeter 1.87596e+06 2.19952e+11 663252
### Cube
Length = n
Surface area 1.31971e+12 1.03155e+17 812315
### Equilateral Triangle
Length = n
Perimeter 1.40697e+06 9.52418e+10 406157
### Triangular Pyramid
Length = n
Surface area 3.80967e+11 1.21569e+16 382929
## Cryptographic Hash Functions
md5 f3c62aff0615016b198f1353210017a8 5d84c597b39ca0291804ee7c4d434881e9bc2c36 b03cba5723af4f3e53889ec183bf7d140e6d8bcac5a192e4f10c1d52eec6c8ad e822dcc1236001f1c71124b94e440286d214e443f29d0c8df4f0b9abdb75082f53bd7b5eb8d8cf07b7af5a84a88e59357be6b179c507662bc711aab45b4d4ed9 8d276cd71fddf02d95b7866e2b2dac0e7fb18098 | HuggingFaceTB/finemath | |
0
1.4kviews
Find the maximum permissible unbalanced in order to limit the steady state deflection to 5mm peak to peak.
0
131views
A centrifugal pump, weighing 60N and operating at 1000 rpm, is mounted on six springs of stiffness 600 ON/m each, Find the maximum permissible unbalanced in order to limit the steady state deflection to 5mm peak to peak.
$m = \frac{600}{9.81} = 61.16$ kg
$w = \frac{2 \pi N}{60} = \frac{2 \pi \times 1000}{60}$
Number of springs = 6
$K_{eq}$ = 6k
Mo.e = ?
$\therefore$ $X = 2.5mm = 2.5 \times 10^{-3} \ m$
No damper $\therefore \zeta = 0$
$\frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2}}$
$\therefore$ $Wn = \sqrt{ \frac{Keq}{m}}$
$= \sqrt{ \frac{ 6 \times 6000}{61.16}}$
$r = \frac{w}{Wn} = \frac{104.72}{24.26} = 4.316$
$\frac{2.5 \times 10^3}{\frac{mo.e}{61.16}} = \frac{(4.316)^2}{\sqrt{(1 – (4.316)^2)^2}}$
$= \frac{18.63}{17.63}$
$= 1.0566 \times \frac{mo.e}{61.16}$ | HuggingFaceTB/finemath | |
# M2 2017 4 Edexcel
(a)
The particle is instantaneously at rest when v = 0, ie. when:
v = 3t^2 - 16t + 21 = (3t - 7)(t - 3) = 0
So either t = \dfrac 7 3 or t = 3. Since \dfrac 7 3 < 3, we have t_1 = \dfrac 7 3 and t_2 = 3.
(b)
We know that the acceleration a is given by:
\displaystyle a = \frac {\mathrm dv} {\mathrm dt}
so that:
a = 6t - 16
At t = \dfrac 7 3 we have:
a = 6 \left(\dfrac 7 3\right) - 16 = -2
So the magnitude of the acceleration of P at t_1 is 2 \text{ms}^{-1}.
Note that the displacement s of P is given by:
\displaystyle s = \int v \mathrm dt
Since v \le 0 for t_1 \le t \le t_2, P does not change direction so we the distance it travels in this time is given by:
\displaystyle \left|\int_{\frac 7 3}^3 (3t^2 - 16t + 21) \mathrm dt\right|
We have:
\begin{align*}\int_{\frac 7 3}^3 (3t^2 - 16t + 21) \mathrm dt & = \left[t^3 - 8t^2 + 21 t\right]_{\frac 7 3}^3 \\ & = 3^3 - 8 \times 3^2 + 21 \times 3 - \left(\frac 7 3\right)^3 + 8 \left(\frac 7 3\right)^2 - 21 \times \left(\frac 7 3\right) \\ & = -\frac 4 {27}\end{align*}
So the distance travelled is \dfrac 4 {27} \text m = 0.15 \text m \text { (2sf)}.
(d)
P returns to O at time t only if:
t^3 - 8t^2 + 21t = 0
for t \ne 0. (since P starts at O) That is:
t^2 - 8t + 21 = 0
However, this has discriminant:
8^2 - 4 \times 21 = -20 < 0
so the equation has no real solutions in t, so P cannot return to O.
Sweet | HuggingFaceTB/finemath | |
# How do you solve 7x-1=9+2x?
Feb 14, 2016
color(green)(x=2
#### Explanation:
color(blue)(7x-1=9+2x
Remember the golden rule of Algebra-what we do in one side must be done in the other side also.
Add $1$ both sides:
$\rightarrow 7 x - 1 + 1 = 9 + 2 x + 1$
$\rightarrow 7 x = 10 + 2 x$
Subtract $2 x$ both sides:
$\rightarrow 7 x - 2 x = 10 + 2 x - 2 x$
$\rightarrow 5 x = 10$
Divide both sides by $5$:
$\rightarrow \frac{5 x}{5} = \frac{10}{5}$
$\rightarrow \frac{\cancel{5} x}{\cancel{5}} = \frac{10}{5}$
rArrcolor(green)(x=2
Check:
Substitute the value of $x$ into the equation:
|->color(brown)(7(2)-1=9+2(2)
|->color(brown)(14-1=9+4
|->color(orange)(13=13 :)
So,it is true! | HuggingFaceTB/finemath |
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