  Jayden Mckay 2022-05-13 Answered

### Equivalence Relation on $\mathbb{N}\setminus \left\{1\right\}$Determine which one of the three properties are satisfied:1) $\left(2,2\right)\notin {A}_{1}$. So it is not reflexive.2) $\left(2,3\right)\to \left(3,2\right)\in {A}_{1}$. Is symmetric.3) (2,3) and $\left(3,4\right)$. Is not transitive.That means, ${A}_{1}$ is not an equivalence relation. Is it okay? Thanks in advance! Lexi Chandler 2022-05-13 Answered

### How to show $f:{\mathbb{Z}}^{+}\to {\mathbb{Z}}^{+},f\left(n\right)=n!$ is one-to-one?How to show $f:{\mathbb{Z}}^{+}\to {\mathbb{Z}}^{+},$, $f\left(n\right)=n!$ is one-to-one?I'm quite sure the function is one-to-one as 0 is not an element of the domain, so $f\left(0\right)=f\left(1\right)$ is not a concern. However doing something like setting $f\left(m\right)=f\left(v\right)$ where m,v are arbitrary elements of the domain doesn't really work, since n! doesn't have an inverse.I was able to get something working algebraically by also doing $f\left(m+1\right)=f\left(v+1\right)$, but I'm quite sure it is circular to do something like that.Any nudge in the right direction would be greatly appreciated. poklanima5lqp3 2022-05-13 Answered

### One cinema has 3 rooms (A,B,C). Room A needs 6 workers inside, B needs 5 workers and C needs 9 workers. How many ways can the owner of the cinema choose 3 teams of workers given that he has 50 workers available? Matthew Hubbard 2022-05-13 Answered

### Prove that successive differences between square roots of integers decreases without using limits or derivativesI have been trying to prove that $\sqrt{n+1}-\sqrt{n}>\sqrt{n+2}-\sqrt{n+1}$ for ALL $n=1,2,3,...$ without success. Jaime Coleman 2022-05-13 Answered

### Upperbound for difference between chromatic number $\chi \left(G\right)$ and list-chromatic number ${\chi }_{L}\left(G\right)$ Kazeljkaml5n9y 2022-05-13 Answered

### From homogeneous to non-homogeneous linear recurrence relationI'm trying to do the following exercise:Find a non-homogeneous recurrence relation for the sequence whose general term is${a}_{n}=\frac{1}{2}{3}^{n}-\frac{2}{5}{7}^{n}$From this expression we can obtain the roots of the characteristic polynomial P(x), which are 3 and 7, so $P\left(x\right)={x}^{2}-10x+21$ and ${a}_{n}=10{a}_{n-1}-21{a}_{n-2}\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\phantom{\rule{thickmathspace}{0ex}}n\ge 2,\phantom{\rule{thickmathspace}{0ex}}{a}_{0}=\frac{1}{10},\phantom{\rule{thickmathspace}{0ex}}{a}_{1}=-\frac{13}{10}$Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation. Edith Mayer 2022-05-13 Answered

### Recurrence relation and closed formulaI have the following problem:I must find the closed formula for${a}_{n}={b}_{n}+{\lambda }_{1}{a}_{n-1}+{\lambda }_{2}{a}_{n-2}$where ${b}_{n}=\frac{\mathrm{\Gamma }\left(n+d\right)}{\mathrm{\Gamma }\left(d\right)\mathrm{\Gamma }\left(n+1\right)},\phantom{\rule{1em}{0ex}}d\in \mathbb{R}.$.I don't have any experience in recurrence relations or discrete math, I'm trying to find some book recommendations to study this. I know that recurrences like${a}_{n}={\lambda }_{1}{a}_{n-1}+{\lambda }_{2}{a}_{n-2}$have closed formulas, but I'm stuck with the ${b}_{n}$ term.In a more general way, I'm searching a closed formula for a relation like${a}_{n}={b}_{n}+\sum _{i=1}^{p}{\lambda }_{i}{a}_{n-i}.$If any condition is not clear, please tell me so I can adjust it. Thanks! britesoulusjhq 2022-05-12 Answered

### How many strictly increasing functions $\left[4\right]\to \left[12\right]$ precede (2,3,4,5)?I'm having some trouble with the following question:Notation: Let [n] denote the set $\left\{1,...,n\right\}$ and we will represent a function $\left[k\right]\to \left[n\right]$ as a list: $\left(f\left(1\right),f\left(2\right),...,f\left(k\right)\right)$Consider all strictly increasing functions $\left[4\right]\to \left[14\right]$ and order them with the natural lexicographic order induced by the order in . How many functions precede the function (2,3,4,5)?- $\left(\genfrac{}{}{0}{}{13}{4}\right)$$\left(\genfrac{}{}{0}{}{13}{3}\right)$$\left(\genfrac{}{}{0}{}{15}{3}\right)$$\left(\genfrac{}{}{0}{}{14}{3}\right)$ spazzter08dyk2n 2022-05-11 Answered

### Discrete Mathematics Division Algorithm proofI'm not quite sure how to do this problem if anyone can do a step by step to help me understand it I would appreciate it a lot.Let a and b be positive integers with $b>a$, and suppose that the division algorithm yields $b=a\cdot q+r$, with $0\le r. (note: its a zero)Prove that $\mathrm{l}\mathrm{c}\mathrm{m}\left(a,b\right)-\mathrm{l}\mathrm{c}\mathrm{m}\left(a,r\right)=\frac{{a}^{2}\cdot q}{gcd\left(a,b\right)}.$ pevljivuaosyc 2022-05-11 Answered

### Prove that $\bigcap _{x\in \left[-1,0\right]}\left[x,1+x\right]=\left\{0\right\}$Answer:So my approach is a little bit different than the usual, that $\left\{0\right\}\subset \bigcap _{x\in \left[-1,0\right]}\left[x,1+x\right]$ and that $\bigcap _{x\in \left[-1,0\right]}\left[x,1+x\right]\subset 0$.So I tried to do it that we have $a\in \left[x,1+x\right],$, $x\in \left[-1,0\right]\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}a\in \left[-1,1\right]$, and is a real number.So now if we look at $x=-1+ϵ$, where $ϵ>0$ and is a small real number, we have some ${a}_{0}\in \left[-1+ϵ,ϵ\right]$ and if we look at $x=-ϵ$, we have ${a}_{1}\in \left[-ϵ,1-ϵ\right]$.Ok, so what I want to show is that if ${a}_{1},{a}_{0}\in \bigcap _{x\in \left[-1,0\right]}\left[x,1+x\right]$ iff ${a}_{1}={a}_{0}$.Now if we look at the inequalities (we say that $\left\{a\in \mathbb{R}|\mathrm{\forall }x\in \left[-1,0\right]:a\in \bigcap \left[x,1+x\right]\right\}$):$-1+ϵ\le a\le ϵ\phantom{\rule{1em}{0ex}}\wedge \phantom{\rule{1em}{0ex}}-ϵ\le a\le 1-ϵ$$-ϵ\le a\le ϵ$and as $ϵ\to 0$, we get: $0\le a\le 0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}a=0={a}_{1}={a}_{0}$Is my proof correct? How would you show it the usual way (with subset going both ways)? omar aljaradi 2022-05-09

### Identify each of the following statements as true or false in relation to confidence intervals (CIs). Note: 0.5 marks will be taken away for each incorrect answer. The minimum score is 0.    TrueFalseA 95% CI is a numerical interval within which we are 95% confident that the true mean μ$\mu$ lies.  A 95% CI is a numerical interval within which we are 95% confident that the sample mean x¯¯¯$\overline{x}$ lies.  The true mean μ$\mu$ is always inside the corresponding confidence interval.  For a sample size n=29$n=29$, the number of degrees of freedom is n=30$n=30$.  If we repeat an experiment 100 times (with 100 different samples) and construct a 95% CI each time, then approximately 5 of those 100 CIs would not$not$ contain the true mean 𝜇. Markeroladipo1 2022-04-16

### How many strings are there of lowercase letters of length four or less, not counting the empty string? Niraj Prajapati 2022-04-15

### Air flows into a compressor in a jet engine at 0.03 m3/s and 247 kPa.  During the process the temperature of the air increases from -35 °C to 444 °C.  Determine the power input required in kW to 2 decimal places.  Take the gas constant R=0.287 (kPam3)/(kgK) and the specific heat at constant pressure of the air to be 1.018 kJ/(kgK).of air to be 1.009 kJ/(kgK). Erika Bernard 2022-04-15 Answered

### discrete math functions pigeonhole princple How many multiples of 11 are in the set $\left\{1,2,3,..,901\right\}$ ? Dylan Yoder 2022-04-14 Answered

### discrete math functions pigeonhole principleHow many numbers in the set $\left\{1,2,3,\dots ,346\right\}$ are divisible by 5 or 7 ? Kali Thomas 2022-04-14 Answered

### discrete math functions pegeonhole principleHow many multiples of 6 are in the set $\left\{171,172,173,..,286\right\}$ ? robinmarian9nhn8 2022-04-12 Answered

### Predicate Logic and QuantifiersAssume that $\mathrm{\exists }x\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall }y\phantom{\rule{thinmathspace}{0ex}}P\left(x,y\right)$ is True and that domain of discourse is nonempty. If the statement is true, explain your answer; otherwise, give a counter example.From the question, we know that $\mathrm{\exists }x\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall }y\phantom{\rule{thinmathspace}{0ex}}P\left(x,y\right)$ means there is an x for which P(x,y) is true for every y.So the question wants me to prove $\mathrm{\forall }y\phantom{\rule{thinmathspace}{0ex}}\mathrm{\exists }x\phantom{\rule{thinmathspace}{0ex}}\mathrm{¬}P\left(x,y\right)$ that it is true, otherwise provide a counter example.How do i start to answer this question? Do i start by using rules of inferences to proof? deformere692qr 2022-04-12 Answered

### Chromatic index of complete graph ${K}_{n}$ after removing one edge measgachyx5q9 2022-04-12 Answered

### Let Q(x) be a quantifer for the universe ${\mathbb{Z}}^{+}$. I want to check whether $\left(\mathrm{\forall }y\left(Q\left(y\right)\vee \mathrm{\exists }xQ\left(x\right)\right)\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists }yQ\left(y\right)$. lurtzslikgtgjd 2022-04-10 Answered

### Riordan numbers recurrenceLet be ${C}_{n}$ the ${n}^{th}$ Catalan's number. Well, I have the following relation:$f\left(n\right)=\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right){C}_{k}\text{.}$I would like to know, if there is a way to obtain the recurrence:$f\left(n\right)=\frac{n-1}{n+1}\left(2f\left(n-1\right)+3f\left(n-2\right)\right)$ just by the first identity.

Students pursuing advanced Math are constantly dealing with advanced Math equations that are mostly used in space engineering, programming, and construction of AI-based solutions that we can see daily as we are turning to automation that helps us to find the answers to our challenges. If it sounds overly complex with subjects like exponential growth and decay, don’t let advanced math problems frighten you because these must be approached through the lens of advanced Math questions and answers. Regardless if you are dealing with simple equations or more complex ones, just break things down into several chunks as it will help you to find the answers.