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5th Grade Math Worksheet with Answers - Practice questions in a form of quiz

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5th Grade Math Problems with Answers - Practice questions with options and get answer in one click

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Problem 752

Alice tossed a fair coin until a head occurred. Then Bob tossed the coin until a head occurred. Suppose that the total number of tosses for Alice and Bob was $7$.

Assuming that each toss is independent of each other, what is the probability that Alice tossed the coin exactly three times?

Hint.

Let $X$ be the number of Alice’s tosses until she got a head.

Then $X$ is a geometric random variable with parameter $p=1/2$.
Recall that a geometric random variable can be interpreted as the number of trials until getting the first success and each trial is successful with probability $p$ and fails with probability $q=1-p$.

Note that
\[P(X=k)= q^{k-1} p = (1-p)^{k-1} p.\]

Solution.

Let $X$ be the number of Alice’s tosses until she got a head. Also, let $Y$ be the number of Bob’s tosses until he got a head. As each toss is an independent event and the coin is fair, the random variables $X$ and $Y$ are geometric with parameter $p=1/2$ (the probability of success(head) is $p=1/2$ and the probability of failure(tail) is $1-p=1/2$).

This question asks to find the conditional probability
\[P(X=3 \mid X+Y=7) = \frac{P\left((X=3) \cap (X+Y=7) \right)}{P(X+Y=7)}.\]

Let us first compute the numerator as follows.
\begin{align*}
P\left((X=3) \cap (X+Y=7) \right) &= P\left((X=3) \cap (Y=7-X) \right) \\
&= P\left((X=3) \cap (Y=4) \right) \\
&= P(X=3) \cdot P(Y=4)
\end{align*}
since $X$ and $Y$ are independent.

Now, as $X$ and $Y$ follow the geometric distribution with parameter $p$, we have
\begin{align*}
P(X=2) &= \left(\frac{1}{2}\right)^2 \frac{1}{2} = \frac{1}{2^3}\\
P(Y=4) &= \left(\frac{1}{2}\right)^3 \frac{1}{2} = \frac{1}{2^4}.
\end{align*}
The product of these gives the numerator $\frac{1}{2^7}$.

---

Next, let us calculate the denominator $P(X+Y=7)$.
The law of total probability yields
\begin{align*}
P(X+Y=7) = \sum_{k=1}^6 P(X=k) P(X+Y = 7 \mid X=k).
\end{align*}

As before, since $X$ is a geometric random variable with parameter $p=1/2$, we have
\[P(X=k) = \left(\frac{1}{2}\right)^{k-1} \frac{1}{2} = \frac{1}{2^k}.\]

The second factor $P(X+Y \mid X= k)$ can be computed as follows.
\begin{align*}
&P(X+Y = 7 \mid X= k)\\
&= P(Y = 7 – X \mid X = k)\\
&= P(Y = 7 – k \mid X = k)\\
&= P(Y=7 – k)\\
&= \left(\frac{1}{2}\right)^{(7-k)-1} \frac{1}{2} = \frac{1}{2^{7-k}}
\end{align*}
Here, the third equality follows because $X$ and $Y$ are independent.

---

It follows that the denominator becomes
\begin{align*}
P(X+Y=7) &= \sum_{k=1}^6 \frac{1}{2^k} \cdot \frac{1}{2^{7-k}}\\
&= \sum_{k=1}^6 \frac{1}{2^7}\\
&= \frac{6}{2^7}.
\end{align*}

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Given that $$\int_{1/\phi}^{1/\phi^2}{ \dfrac{\ln(1-x)}{x}}dx=\dfrac{\pi^2}{30}$$ Find the value of $$\int_{1/\phi}^{1/\phi^2} \left(\dfrac{\ln(1-x)}{x}\right)^2 dx$$ in terms of $\phi$ and $\pi$. Where $\phi=\frac{1+\sqrt 5}{2}$ is the golden ratio.

I have tried to do it by taylor series, also tried integration by parts but it is getting ugly and too many terms are coming.

Source: Made by Prof. Raghava.

from Hot Weekly Questions - Mathematics Stack Exchange

It is obvious that $\sum_{p \in S_n} 1=n!$ because it is just counting how many permutations there are of $n$ symbols.

But I have also observed that $\sum_{p \in S_n} 2^{c(p)}=(n+1)!$, where $c(p)$ is the number of cycles of $p$.

What is the combinatorial interpretation of this identity?

An example. In $S_3$ we have one permutation with 3 cycles, three permutations with 2 cycles and two with 1 cycle. Then $1\times 2^3+3\times 2^2+2\times 2^1=24=4!$

from Hot Weekly Questions - Mathematics Stack Exchange

This workshop was on “Computational and algorithmic aspects” (of groups, representations and applications, presumably). In my opinion, this was the week when the programme really took off.

There were many good talks, so as ever I shall just select a few. First, two contrasting styles to actually use the computer to solve problems.

Richard Parker gave a talk on “10 years of meataxe development”, the Meataxe being Richard’s program for decomposing modules. He described it as “a rant”; he works very close to the metal in an attempt to squeeze more speed out of the hardware. He said (and this echoed comments at our Big Data symposium a few years ago) that data movement is more expensive than actually doing the work. (So, of the three Rs, arithmetic is cheap and quick, while reading and writing are much slower and more expensive.) He claimed that programmers mostly ignore modern hardware design. Typically there are three levels of cache with very differing speeds well above RAM access, and if you don’t use them you are taking a factor of hundreds longer than you need. But things are not made easier by the lack of user-friendly assemblers for today’s very complex chips. His wish list was a scheduler, a more friendly assembler, and a solution to the synchronization problem.

At the other end was Nicolas Thiéry. He has a new PhD student, who is going to work on representation theory of semigroups. The software he needs is scattered over many different systems (GAP, Magma, Singular, many others); with some effort he can call everything he needs from Sage, but it is considerably harder than it should be because of differing formats. This is a software engineering problem.

But there was much more to Nicolas’ talk. He was much more concerned to give us insight and motivation than to write down detailed definitions. What is the difference between groups and semigroups? He explained, rather briefly, the J-classes of a semigroup, and how these might be structured round a group. Why do you want to do representation theory of semigroups? Group theory has been applied to the analysis of Markov chains by Diaconis and others; but in this imperfect world, things mightn’t be reversible, and semigroups are like groups where the operations can’t always be reversed. (His picture of a group is a perfect circle.) His example was the Tsetlin library, rephrased in more homely terms: you pull the shirt you want out of the pile, and when it is washed it is put back on top.

I took away the impression that moving from groups to semigroups is not unlike moving from ordinary to modular representation theory of finite groups: things are no longer completely reducible, and you have to deal with the radical.

There was a surprising amount of algebraic geometry; some of it I found very interesting. I’ll mention two. Mohammed Barakat talked about “Chevalley’s Theorem on constructible images made constructive”. In a topological space, a set is locally closed if it is the intersection of an open set and a closed set, and is constructible if it is a finite union of locally closed sets. Chevalley’s theorem says that the image of a rational map of affine varieties is constructible (in the Zariski topology). If this sounds hard, his simple example made it clear. We take the map between two-dimensional affine spaces taking the point (x,y) to x,xy). The image of this map is the whole plane with the Y-axis removed and the origin put back, obviously a constructible set. He has a new proof of this which produces a kind of canonical decomposition into locally closed sets, and which can be implemented on a computer, not just over a field but over rings like the integers too. There were several nice applications.

Tobias Rossman took us on an absolute roller-coaster ride connecting group theory, combinatorics, graph theory, and algebraic geometry. To cut to the chase, given a finite graph Γ, he defines a group scheme G_Γ where you have a generator for each vertex, two generators commute if the vertices are non-adjacent (I am not certain I have this right), and commutators are central (so the group is nilpotent of class at most 2). The object of interest is the number of conjugacy classes of this group, over some given field or ring. This has a zeta-function with an Euler product, and the factors are bivariate rational functions of p and p^-s. For some special graphs (a subclass of the cographs or N-free graphs), these are simple combinations of translates of the Riemann zeta function. The proof covers a huge amount of ground (as proofs in this area tend to do, in my experience).

Joanna Fawcett and Melissa Lee gave us an update on the base size 2 project: which primitive groups have a base of size 2, or perhaps easier, which ones don’t? (The base size is the minimal number of points whose pointwise stabiliser is the identity.) There is a nice graph, the Saxl graph, associated with groups of base size 2: very simply, the edges are the bases. This is conjectured to have nice properties resembling those of the generating graph of a finite simple group.

Madeleine Whybrow talked about dihedral axial algebras. It would take a while to explain this, but let’s say that the Griess algebra for the Monster is the motiviating example of an axial algebra, though others arise in Jordan algebras. There are now quite simple axioms for axial algebras, though it has taken a while to reach this point, and (the point of the talk) good computational tools to explore them. “Dihedral” just means “two generators”. One can put in values for various numbers that occur in the multiplication table of the idempotents. What has emerged is that the values that occur in the Griess algebra (1, 1/4 and 1/32), which seemed rather artificial at first, really are the sweet spot.

Finally, a lovely talk by Eilidh McKemmie on invariable generation. A group is invariably generated by a set of elements if you can replace any of them by arbitrary conjugates and still have a generating set. It is a result due to Pemantle, Peres and Rivin in one direction and Eberhard, Ford and Green in the other, that four random elements invariably generate the symmetric group S_n with probability bounded away from zero, but three random elements do not. Eilidh has proved that exactly the same holds in classical groups. (There is a slight complication here: classical groups have two parameters, and we may take them to infinity in either order or together, but I will skip over this.)

All rather exciting and exhausting!

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Given some function involving two variables, $\mathit p(x,y)$, is the formula $$\mathit \exists!x\exists yp(x,y)$$ equivalent to $$\mathit\exists y\exists!xp(x,y)$$ I have tried writing out the formal definition for the unique existential quantifier, but I get stuck simplifying the statement when the unique quantifier is first. I have also tried giving $\mathit p(x,y)$ some meaning and talking through it, but I can't decide if they mean the same thing. Any help is appreciated!

from Hot Weekly Questions - Mathematics Stack Exchange

Assume we have the following stochastic process:

$$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$

where $(B)_{t\geq 0}$ is a Brownian Motion.

I have to show that $X_t$ is not a martingale.

I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process is a martingale, this makes me think that $X_t$ is actually a local martingale, but I don't see how to prove that it's not a proper martingale.

Thanks in advace.

EDIT:

After some calculation I have reduced the problem to show that

$$\mathbb E^{\mathcal F_u}\bigg(\int_{B(u)}^{B(t)}e^{s^2} ds-\int_u^t B(s)e^{B(s)^2}ds\bigg)\neq0$$

Still, I haven't been able to go on from here, maybe someone could find this illuminating.

from Hot Weekly Questions - Mathematics Stack Exchange

Most people can relate to (or feel) the frustration caused by being stuck in traffic, waiting in a queue to board a plane, or circling the parking lot to find a space. Routes that could take 30 minutes can turn … Continue reading →

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The paper “A geometric diffuse-interface method for droplet spreading“, co-authored by Darryl Holm (Imperial College), Lennon Ó Náraigh (University College Dublin) and Cesare Tronci, has been published in the Proceedings A of the Royal Society of London. The paper exploits the theory of geometric gradient flows to introduce an alternative regularization of the thin-film equation valid in the case of large-scale droplet spreading. A link to the published version is here, and the first lines of the introduction are shown below.

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When dinosaurs ruled the earth, one of my assignments in a Problem Seminar in undergraduate was to devise and prove a conjecture about antiprimes, and this was my attempt:

An antiprime (also called a highly composite number) is a positive integer that has more divisors than any number less than it. The first few antiprimes are $$1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360,...$$ Conjecture: For every antiprime $n>1$, there is a prime $p$ such that $p\mid n$ and $n/p$ is an antiprime.

Anyways, I never found a proof for that, and about ten years ago I asked the xkcd math forums if they could guide me. Instead, someone posted a counterexample that was quite enormous.

A question on MESE is asking about relatively elementary mathematical conjectures whose smallest counterexamples are large numbers. I'd like to suggest my problem, but the xkcd forums went down five months ago over a data breach and my thread wasn't cached by Google or the Wayback Machine.

Can someone find that counterexample? The person who posted didn't indicate if they came up with their number through mathematical reasoning or programming. I discovered this morning that OEIS has a list of the first ten thousand antiprimes, so in theory it may just come down to finding the prime decomposition of each of them. But, if possible, is there a mathematical argument that would lead one to the correct number?

from Hot Weekly Questions - Mathematics Stack Exchange

In my previous post, I wrote out a proof (that an even number is an odd number plus 1) that included the following counterintuitive steps: A common reaction that I get from students, who are taking their first steps in learning how to write mathematical proofs, is that they don’t think they could produce steps […]

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In the Aperiodical’s Big Internet Math-Off 2019, Becky Warren posted an entry about Geogebra’s ‘reflect object in circle’ tool (it’s the second article in the post). I enjoyed playing with the tool and, after making a few colourful designs, it occurred to me that one of them would make a great cake for the MathsJam bake-off. It would only work if the curves were accurate; sadly this would be beyond my drawing abilities, and definitely beyond my piping abilities. But with some help from 3D printing I thought I might be able to manage it.

Here are the steps I used to transfer the design to a cake.

Step 1 (Geogebra) – Save as an image

I started by simplifying the design. Squares are straightforward, so I only needed the part within the centre circle. I hid all the squares and removed all the colours, then thickened the lines to make them easier to see. Finally, I took a screenshot and saved it as a bitmap image.

Step 2 (Inkscape) – Convert image to path and generate scad

The next step was to use Inkscape to convert the image into an OpenSCAD file that can be used for 3D printing. If you google ‘Inkscape OpenSCAD’ you’ll find several extensions that will do this; I used Paths2OpenSCAD.

To generate the scad, start by importing the image into Inkscape. Then, with the image selected, go to Path > Trace Bitmap. This opens up a window with lots of options; accept the default and click OK, then close the window. Inkscape has now converted the image into a vector path; this means that it has coordinates for each point on the path.

Now go to Extensions > Generate from Path > Paths to OpenSCAD. This opens up another window. In the output file box, enter the location to save the scad file, then click OK. Now you can open the file in OpenSCAD.

Step 3 (OpenSCAD) – Scale design

The generated OpenSCAD file starts like this.

It has a method called ‘poly_path854’ (a new number is generated for each file). This method uses the coordinates of the points on the path from Inkscape to make a polygon in OpenSCAD.

Pressing F5 to preview shows a replica of the image, rendered as a 3D model.

At the very end of the file is a line that calls the ‘poly_path854’ method, passing in a number. The number is the height of the object when printed – you can change this to suit your needs.

The part that I found difficult was to work out the size the model would be when it was printed. The axes on the OpenSCAD preview window give some indication, but it’s not possible to read accurate values from it. I wanted my model sized appropriately to match a square cutter that I already owned. In the end, I printed out a 1 mm high model so that I could measure the original size without wasting too much plastic, then I scaled the entire model to match the square cutter. The final line of the file became

scale([6/(4.2*sqrt(2)),6/(4.2*sqrt(2)),1]){
     poly_path854(5);
 }

Step 4 – Print

I sent the model to the printer, and a few hours later I had my cutter!

Step 5 – Imprint on cake

Once I had baked the cake and covered it in fondant, I used the cutter to make an imprint of the design on the icing.

Now it only remained to cut the squares around the edges and give it some colour. I used a small amount of vodka to thin out paste food colouring so that I could paint on the fondant icing. (Honestly, it was a tiny amount of vodka – less than 5 ml over the entire cake.)

That’s it! Geogebra to cake in five steps. Here’s the finished cake, alongside the original design in Geogebra.

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5th Grade Math Worksheet with Answer - Practice questions in the form of test and get solution in one click

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Once we had a country and thought it so fair
If you look through the mirror you still find it there
But now our great country is broken and torn
And all of its promise and liberties worn

Procol Harum, “An Old English Dream”

In 2003 Procol Harum released their 11th album, The Well’s On Fire. That name, with hindsight, has a prophetic ring to it. I didn’t listen to this until several years later, at which time it was clear to me that two songs on the album, “The Blink of an Eye” and “Wall Street Blues”, could be read as foreseeing the global financial crisis of 2008.

Meanwhile I hadn’t noticed the prophetic nature of the verse from the opening track, quoted above.

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Problem 751

There are two boxes containing red and blue balls. Let us call the boxes Box A and Box B. Each box contains the same number of red and blue balls. More specifically, Box A has 5 red balls and 5 blue balls. Box B has 20 red balls and 20 blue balls. You choose one box. Then draw two balls randomly from the chosen box without replacement, that is, you will not return the first ball into the box before picking up the second ball.

If you draw two balls with the same color, then you win. Otherwise, you lose. To maximize the chance of winning, which box should you pick?

Solution.

Let us consider the probability that you draw two red balls from the box you choose.
Since both boxes contain the same number of red and blue balls, the probability that your first ball is red is $1/2$ no matter which box you choose. More rigorously, if you choose Box A, then the probability that the first ball is red is $5/10 = 1/2$. Also, if you choose Box B, then the probability is $20/40=1/2$.

Thus, the difference (if any) comes from the probability of the second ball. Suppose that you choose Box A. Assume that your first ball was red. Then Box A currently contains 4 red balls and 5 blue ball. Hence the probability that the second ball is red is $4/9$.
Similarly, if you choose Box B and the first ball is red, then the probability that the second ball is red is $19/39$ since there are 19 red balls and 20 blue balls in Box B.

So, which probability is bigger? As we have
\[4/9 = 0.444\dots \text{ and } 19/39 = 0.487\cdots\] The probability for Box B is bigger. Thus, the probability of getting two red ball is higher when you choose Box B. The same argument applies for two blue balls. Therefore, to maximize the chance of winning, you should select Box B.

Remark.

Let us give an intuitive solution. This approach is not mathematically rigorous but it might help you reach the conclusion quicker.
First, you may think the actual numbers $5$ and $20$ are not so importance. What is importance is which number is bigger. So, instead, we consider extremely $2$ and $2,000,000,000$ (any gigantic number is fine). So Box A contains only 2 red balls and 2 blue balls. On the other hand Box B contains $2,000,000,000$ balls for each color.

As before the probability of getting the first red ball is the same regardless of your choice of a box. For the second ball, however, there is only one red ball remained in Box A, which is a half of the number of blue balls in Box A. On the other hand, Box B still contains 1,999,999,999 red balls and 2,000,000,000 blue balls. The ratio of two colors in Box B is almost 1. Thus, you have a better chance of getting the second red ball if you choose Box B.

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A conversation about mathematics including fractals inspired by a Romanesco Broccoli. Presented by Katie Steckles and Peter Rowlett, idea suggested by John Read (thanks John!).

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Online Math Worksheet for Grade 5 - Practice questions in a form of quiz and get solution in one click

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Let $x,y,z>0$. Prove that: $$3\geq \frac{(x+ y)^{2}x^{2}}{(x^{2}+ y^{2})^{2}}+ \frac{(y+ z)^{2}y^{2}}{(y^{2}+ z^{2})^{2}}+ \frac{(z+ x)^{2}z^{2}}{(z^{2}+ x^{2})^{2}}$$

I need to the hints and hope to see the Buffalo Way help here! Thanks a lot!

My idea is as follows: Because this inequality is cyclic. So, it's enough to prove this inequality in two cases: $x\leq y\leq z$ and $x\geq y\geq z$. I can prove it with $x\leq y\leq z$ but with $x\geq y\geq z$, I can't.

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Online Free Math Worksheet for Grade 5 - Practice questions with solution

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$$\lim_{n\to\infty}\frac{2^n}{n^{\sqrt{n}}}$$

The way I see it is that denominator is super-exponential and the numerator is only exponential, so this must be $0$ but wolfram is saying it is infinite.

How do I evaluate this limit?

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In Borceux and Bourn's book "Mal'cev, Protomodular, Homological and Semi-Abelian Categories", the authors introduce two similar metatheorems that are used throughout the book to greatly simplify many proofs. Contrary to the well-known metatheorems resulting from embedding theorems such as Freyd-Mitchell's or Barr's, which pose strict conditions to the categories they can be applied to, but then allow to transfer large amounts of statements to easier categories, these metatheorems pose strict conditions to the structure of the statements, but can be applied to very general types of categories (arbitrary categories for the first metatheorem, and pointed categories for the second).

For those who don't have immediate access to the book, here are both theorems:

Let $\mathcal{P}$ be a statement of the form $\varphi \Rightarrow \psi$, where $\varphi$ and $\psi$ can be expressed as conjunctions of properties in the following list:

1. (just for pointed categories) some arrow is a zero arrow;
2. some finite diagram is commutative;
3. some morphism is a monomorphism;
4. some morphism is an isomorphism;
5. some finite diagram is a limit diagram
6. some arrow $f: A \to B$ factors (of course, uniquely) through some specified monomorphism $s: S \hookrightarrow B$.

If this statement $\mathcal{P}$ is valid in the category of (pointed) sets, it is valid in every (pointed) category.

The proof proceeds via the Yoneda embedding, and then asserts that each of the above statements is valid in a presheaf category iff it is valid pointwise (i.e. in $\mathbf{Set}$ or $\mathbf{Set}_*$ respectively).

My problem is though, that in the proof the demanded structure of the statement $\mathcal{P}$ (implication of conjunctions) is ignored. It is only mentioned that the Yoneda embedding "preserves and reflects all ingredients which appear in $\mathcal{P}$" (i.e. the items on the list), and that "... all the properties listed in our statement are valid in $[\mathcal{E}^\text{op}, \mathbf{Set}]$ precisely when they are valid pointwise in $\mathbf{Set}$."

Even more irritatingly, the authors note afterwards, that "the list of properties in our metatheorem is not exhaustive. A better way to express the metatheorem would have been to state it for Horn sentences and the "unique existential quantifier $\exists!$", but we do not want to enter those considerations here.".

More generally, I wonder why I can't find this metatheorem anywhere else, since it is so useful. Is there an alternative formulation, e.g. via the internal logic of categories, which I do not recognize as equivalent yet?

Thus my questions are the following:

• How does a complete proof of the metatheorem proceed?
• How does the "better way" noted above work? Is there another source that elaborates this?
• Is there a more systematic way to formulate the theorems?

from Hot Weekly Questions - Mathematics Stack Exchange

      There are geniuses (like Srinivasa Ramanujan) who learn mathematics quickly -- but I am not one of them.  In the following poem I reflect on how I learn . . . 

       Reflection     by JoAnne Growney

       I read and I did not understand.
       Less than a page.
       I read slowly and I did not understand.
       I read and took notes.
       My notes were three times as long as what I had read.

       I rewrote what I read in my own words.
       I reread it and inserted extra clarifying words.  
Read more »

from Intersections -- Poetry with Mathematics

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For the fourth year in a row, MIND Research Institute was given a special end-of-year gift of collaborating with our amazing partner, Rockwell Automation, at their 2019 Automation Fair®! Automation Fair is the premier industrial automation event offering unmeasurable knowledge and skill-building opportunities. It was held November 20 and 21 in Chicago and brought together a record-breaking attendance of over 19,000 makers, builders, and innovators from across the globe.

We were invited to exhibit at the Engineering our Future booth along with another leading STEM-based organization, FIRST® (For Inspiration and Recognition of Science and Technology) Robotics. Our exhibit showcased the programs Rockwell Automation invests in to deeply engage students in problem solving and the fun of STEM! We featured our flagship program, ST Math®, and our storytelling board games, MathMINDs Games: South of the Sahara, funded by Rockwell Automation.  

If you are at @AutomationFair make sure to stop by the #EngineeringOurFuture booth to learn more about how you can help change the world by inspiring young minds to keep doing cool things with #STEM!

Special thanks to @FIRSTweets & @MIND_Research #CuriousPeopleWonder #ROKLive pic.twitter.com/EHORghx2ur

— Jay Flores (@JayFlores2032) November 21, 2019

Showcasing Math Skills With Our Students

Prior to the event, I was charged with finding ST Math student ambassadors to showcase their math skills during the fair. With so many ST Math partner schools in the Chicago area, it was difficult selecting only a handful of students to join us. It was actually my toughest task—but the most rewarding. We chose ambassadors from Amelia Earheart School and Pershing Magnet School.

The students were very familiar with ST Math as most of them were top of their class in program completion. What they were not familiar with were our South of the Sahara games. However, just like our research shows, the storybooks that accompany each game proved intuitive. In no time, our ambassadors understood what needed to be done to play the games and challenge guests!

Our STMath Ambassadors are rocking it. #ROKLive #STMath #MathMINDs pic.twitter.com/E4K1br51Xl

— Brandon D Smith (@bds_math) November 20, 2019

The students were incredible ambassadors who took their roles very seriously. They really were the stars of the show and didn’t hesitate to approach all of the professional engineers in attendance.  These young STEM leaders and mathematicians would ask questions like:

• Do you like math? 
• How did you learn math? 
• Can I show you a way you wish you would have learned math?

The answer to that last one was always, yes. We had guests who would stay over ten minutes playing games with the students! That’s impressive in an expansive exhibit hall.

Highlight of @ROKLive! 🤩🥰@ROKAutomation CEO Blake Moret learning @STMath from our Chicago 4th grade students! @MIND_Research #STEMeducation #leadership @JayFlores2032 pic.twitter.com/YJXkqZfKJn

— Karin Wu (@KarinCWu) November 21, 2019

There were a significant amount of guests that recognized our penguin mascot, JiJi. Visitors would remark that they’d seen that penguin in their own homes before! Yes, many of the adults enjoying Automation Fair also had children who were learning math at school and through ST Math homework.

One of the most impactful moments came when our young female students noticed that there were very few women walking around the exhibit hall—and even fewer who were minorities. That was very eye-opening for all of us. One student said to me, “Now I really want to be an engineer so other little girls like me have someone to look up to.” I loved hearing those words because that is exactly what MIND and Rockwell Automation aim to do: empower our students.

Bold Conversations with Rockwell Automation

During the event, Rockwell Automation also hosted many different conversations about STEM, culture, and the future. I made sure to join the “Bold Conversations” panel where inclusion and diversity were discussed. These weren’t just surface conversations either—these discussions dove deep. The panelists, which included Patricia Contreras, Director of Global Community Relations and Contributions at Rockwell Automation, shared stories and encouraged authentic dialogue. Patricia spoke about the challenges of creating a culture that differentiates a company for all the right reasons.

The amazing @prcontrerasmke! Thank you for all you do for #WomenInSTEM! @STMath @MIND_Research #ROKLive @ROKAutomation pic.twitter.com/iIsm7LSIGC

— Karin Wu (@KarinCWu) November 21, 2019

With leaders like Patricia, I know students will not only aspire to fill big roles in STEM, but will thrive in those roles! Thank you, Patricia, for using your time on stage to include everyone—especially our young students—in these bold conversations.

One other very special opportunity that Rockwell Automation gave us during Automation Fair was the chance to interview in the #ROKLive studio. Karin Wu, our Executive Vice President and Director of Social Impact, had a great conversation about STEM with Becky House, Senior Vice President of General Counsel at Rockwell Automation and MIND Board Member, and Dr. Brenda DeMar-Williams, Principal of Amelia Earhart School. You can watch the whole interview below (starting at 2:52). 

Our partnership with Rockwell Automation extends far beyond the classroom. Together, we’re able to show students and community members the importance of early math skills. The goal of our partnership was reflected throughout Automation Fair as our students toured exhibits, asked questions, and saw the reality of the future of STEM. That future needs diverse people like our students to solve the world’s most challenging problems.

Thank you to Rockwell Automation for having us, and we hope to see you at the 2020 Automation Fair, November 18 and 19, here in our very own backyard, Anaheim, California!

Students from @earharteagles @ROKLive say thank you @ROKAutomation! @STMath is helping ALL students aim for careers in #STEM! @MIND_Research pic.twitter.com/J4rIYZsFyx

— Karin Wu (@KarinCWu) November 21, 2019

Check out our Facebook videos page to see more coverage and behind the scenes moments from the event.

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Congratulations to Jonah Varney for passing his PhD confirmation examination. It was held today, Thursday 30 January, and the examiners were Jon Bevan and Jonathan Deane. The title of Jonah’s project is “Marginal stability rates for the joint spectral radius“. The project supervisor are Ian Morris.

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Check out this Problem of the Week.
Be sure to let us know how you solved it in the comments below or on social media!

Solution below.



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What does research suggest about early development of spatial skills?

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Grade 5 Math Worksheet Free - Practice questions in a form of test and also get solution in one click

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from math https://ift.tt/2GBFX6U

Here we have a simple, but a very hard problem: How many different figures (or graphs) can be formed with a regular polygon of n vertices and a number d of diagonals of this polygon?

Consider T(n,d) the total number of distinct figures formed by the polygon of n vertices and d diagonals. The question is: how to determine T(n,d)?

The figure below shows some examples for the pentagon (n=5) and hexagon (n=6), to better understand the problem:

I thought about using Polya´s Enumeration Theory, but I didn't understand how I can do that. There seems to be a similarity with the method of counting graphs by Polya´s Counting Theory, or still similar to the circular coloring problem solved by Polya.

Does anyone have any ideas or methods to solve this problem?

from Hot Weekly Questions - Mathematics Stack Exchange

from Maths & Physics News

$$\int_{0}^{\infty} \frac{x^2+x+1}{x^6+x^4+1} dx $$

Wolfram says $1.80276\ldots $

Calculating this seems complicated to me because the residues are pretty hard to find and I have tried the infinite residue method but it didn't work out.

from Hot Weekly Questions - Mathematics Stack Exchange

I was reading the Wikipedia article for complex affine spaces, which says the following:

Affine geometry, broadly speaking, is the study of the geometrical properties of lines, planes, and their higher dimensional analogs, in which a notion of "parallel" is retained, but no metrical notions of distance or angle are. Affine spaces differ from linear spaces (that is, vector spaces) in that they do not have a distinguished choice of origin. So, in the words of Marcel Berger, "An affine space is nothing more than a vector space whose origin we try to forget about, by adding translations to the linear maps."[1] Accordingly, a complex affine space, that is an affine space over the complex numbers, is like a complex vector space, but without a distinguished point to serve as the origin.

What Marcel Berger said is what interests me:

... "An affine space is nothing more than a vector space whose origin we try to forget about, by adding translations to the linear maps."

What is meant here by "forgetting about the origin by adding translations to the linear maps"? Can someone please explain, using basic linear algebra and geometry, and with mathematics included, what this means?

from Hot Weekly Questions - Mathematics Stack Exchange

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     Today's posting features work by Punya Mishra, professor and administrator at Arizona State University -- a writer who offers explanations of mathematical concepts in his poems, explanations that can appeal to students!   Mishra's website features several mathy poems and in our email correspondence he said that "The Mathematical i" is his favorite.  Here are are several stanzas:

from  The Mathematical i     by Punya Mishra

       The negative numbers were full of dismay
       We have no roots, they were heard to  say
       What, they went on, would be the fruit
       of trying to find our square root?

       Matters seem to be getting out of hand
       Since the negatives have taken a stand,
       On the fact that positives have two roots, while they have none.
       They plead, would it have killed anybody to give us just one?
       The square roots of 4 are + and – 2! As for -4 ?  How unfair,
       He has none! None at all. Do the math gods even care?
Read more »

from Intersections -- Poetry with Mathematics

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from Maths & Physics News

Basically the question is proving that you can create all integers with binary but instead using $-2$ as the base to be able to create negative integers.

Exact question:

Prove that every integer (positive, negative, or zero) can be written as the sum of distinct powers of $−2$.

I somewhat get how you can induct upon increasing powers for $2^0+2^1+2^2$ etc and prove that it will always hold for the next number but I'm not sure how this will work with negative integers since If I induct upwards I can't go down and I can't start at $-\infty$.

from Hot Weekly Questions - Mathematics Stack Exchange

What do you wonder?  Why is this string of digits special?  What will be the next number like this one?

February2020.pdf

CCSS:  4.OA, 6.NS, 7.NS

For members we have an editable Word docx, a PowerPoint of this activity, and solutions.

February22020.docx    02022020.pptx    February22020-solution.pdf

from Yummy Math

Limit[Sum[2(2k)^(1/(2k))-k^(1/k),{k,n+1,2n}]-n, n -> ∞]

to solve by hand, $$\sqrt[y]y=e^{\frac{\ln y}{y}}\sim1+\frac{\ln y}{y}$$ $$2\sqrt[2k]{2k}-\sqrt[k]k\sim1+\frac{\ln2}{k}$$ $$\sum_{k=n+1}^{2n}\left(2\sqrt[2k]{2k}-\sqrt[k]k\right)\sim n+\ln2\sum_{k=n+1}^{2n}\frac1k\sim n+\ln 2 \, \int_{n+1}^{2n}\dfrac{1}{t}dt$$ $$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2\sqrt[2k]{2k}-\sqrt[k]k\right)-n\right)=\ln^22$$

from Hot Weekly Questions - Mathematics Stack Exchange

Free Online Math Worksheet for 5th Grade - Practice questions n a form of test

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Consider $p(x) = (x^5-1)(x^2+1)$. Then, its splitting field is $\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)$.

Thus, $f\in \text{Gal}(\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)/\mathbb{Q})$ maps $\omega = e^{\frac{2\pi i}{5}}$ to any of $\omega^k$ for $k=1,...,4$ and $i$ to $\pm i$.

In that way, I can conclude that $|\text{Gal}(\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)/\mathbb{Q})|= 4 \cdot 2 = 8$.

Now, how do I know which of the groups of order $8$ it is? It might be $\mathbb{Z}_2 \times \mathbb{Z}_4$ because of element orders but I am not sure.

from Hot Weekly Questions - Mathematics Stack Exchange

Parallel Perpendicular and Intersecting Lines - Concept - Examples with step by step explanation

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Slopes of Parallel and Perpendicular Lines - Concept - Examples with step by step explanation

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Parallel and Perpendicular Lines Worksheet - Concept - Problems with step by step explanation

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Parallel and Perpendicular Lines - Concept - Examples with step by step explanation

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Equation with Infinite and No Solution - Concept - Examples with step by step explanation

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Equations with No Solution and Identity Worksheet - Concept - Problems with step by step explanation

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Let $a=\sqrt{3+2\sqrt{3}}$. Then \begin{align*} &a=\sqrt{3+2\sqrt{3}}\\ &\implies a^2=3+2\sqrt{3}\\ &\implies a^2-3 = 2\sqrt{3}\\ &\implies (a^2-3)^2 = 4\cdot 3=12\\ &\implies (a^2-3)^2-12=0. \end{align*}

So, $a$ a root of $$(x^2-3)^2-12=x^4-6x^2-3.$$

By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.

---

However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.

What is wrong with my attempt?

from Hot Weekly Questions - Mathematics Stack Exchange

So in order to understand Stokes' theorem, I tried to prove it for a parallelogram $P$ with parameterized boundary $R(t)=(x(t),y(t))$ spanned by two vectors $\mathbf{u}$ and $\mathbf{v}$.

The parallelogram :

So that for some covector field "1-form" $\omega$ : $$\int_{\partial P}\omega = \int_{P}d\omega$$ with the edges having $t=0,1,2,3$ I think this reduces to : $$\omega_{R(0)}(\mathbf{u})+\omega_{R(1)}(\mathbf{v})-\omega_{R(2)}(\mathbf{u})-\omega_{R(3)}(\mathbf{v})=d\omega(\mathbf{u}\wedge\mathbf{v})$$ and this is my try : $$\mathbf{u}=R(1)-R(0)=R(2)-R(3)\\ \mathbf{v}=R(2)-R(1)=R(3)-R(0)$$ and let $ω_t=ω_{R(t)}$ so : $$\omega_{0}(\mathbf{u})+\omega_{1}(\mathbf{v})-\omega_{2}(\mathbf{u})-\omega_{3}(\mathbf{v})=(ω_0-ω_2)(\mathbf{u})+(ω_1-ω_3)(\mathbf{v})$$ let : $$ω_1-ω_0=ω_2-ω_3= \Delta_{\mathbf{u}}~ω~,~ω_2-ω_1=ω_3-ω_0= \Delta_{\mathbf{v}}~ω~,\\\Delta_{\mathbf{u}}~ω(\mathbf{v})=\Deltaω(\mathbf{u},\mathbf{v})$$ so : $$(ω_0-ω_2)(\mathbf{u})+(ω_1-ω_3)(\mathbf{v})=\\(-\Delta_{\mathbf{u}} ~ω-\Delta_{\mathbf{v}}~ω)(\mathbf{u})+(\Delta_{\mathbf{u}} ~ω-\Delta_{\mathbf{v}}~ω)(\mathbf{v})=\\ -\Delta_{\mathbf{u}} ~ω(\mathbf{u})-\Delta_{\mathbf{v}}~ω(\mathbf{u})+\Delta_{\mathbf{u}} ~ω(\mathbf{v})-\Delta_{\mathbf{v}}~ω(\mathbf{v})=\\- \Deltaω(\mathbf{u},{\mathbf{u}})-\Deltaω(\mathbf{v},{\mathbf{u}})+\Delta ω(\mathbf{u},{\mathbf{v}})-\Deltaω(\mathbf{v},{\mathbf{v}})$$ I'm not quite sure how to continue from here but if $-\Deltaω(\mathbf{v},{\mathbf{v}})-\Deltaω(\mathbf{u},{\mathbf{u}}) =0$ so let : $$dω(\mathbf{u},\mathbf{v})=\Delta ω(\mathbf{u},{\mathbf{v}})-\Deltaω(\mathbf{v},{\mathbf{u}})$$ it's easy to proof that $d\omega$ is multi-linear and anti-symmetric so : $$d\omega(\mathbf{u},\mathbf{v})=d\omega(\mathbf{u}\wedge \mathbf{v})$$

so I know I probably made a lot of mistakes but Is the core of this way to proof it right? Is there any papers who proof stokes theorem in general by proofing it for a parallelogram or parallelepiped spanned by vectors? because it makes much sense and it's a beautiful approach, and what are my exact mistakes?

from Hot Weekly Questions - Mathematics Stack Exchange

This year’s JMM have come and gone, but many related posts are still available. Here are just a few that I recommend. “To be or not to be there: Conferencing in the age of flygskam” In this piece for the … Continue reading →

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Check out this Advanced Knowledge Problem of the Week.
Be sure to let us know how you solved it in the comments below or on social media!

Solution below.



from The Center of Math Blog https://ift.tt/36zUHxA

In each episode of the Inside Our MIND podcast, we take a look at issues and challenges facing education that we are working to address through research, technology and strategic initiatives.

In our latest episode, Brian welcomes Senior Director of Product Management, Edie Perkins, to the show to talk about the recently announced new version of our flagship program, ST Math. Edie provides an overview of the enhancements and new features that will help students, teachers, and administrators get even more out of the program.  Over the next several months, we’ll be taking a deeper dive into those new features and enhancements on the podcast and our blog. You can learn more today on stmath.com, which now has a “What’s New” page and new ST Math games and resources for you to experience. 

You can listen to the episode in the player below:

Topics:
0:45 Intro
4:15 What's not changing
5:00 The goal of making more learning happen
6:00 The new student experience
10:45 The new teacher experience
15:45 The new administrator experience

Thanks for listening to the podcast! Please leave us a review on iTunes, Google Podcasts, Spotify, Spreaker or wherever you are listening to the show. Subscribe to get future episodes as soon as they are released!  
 

Additional Resources:

• Video: Introducing the New ST Math
• Press Release: MIND Research Institute Unveils All-New Version of ST Math

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