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0

The correct answer is $22/5$: $$ \frac{\frac{2}{3}+\frac{4}{5}}{\frac{5}{6}-\frac{1}{2}}= \frac{\frac{10+12}{15}}{\frac{5-3}{6}}= \frac{\frac{22}{15}}{\frac{2}{6}}= \frac{\frac{22}{15}}{\frac{1}{3}}= \frac{22}{15}\cdot \frac{3}{1}=\frac{22}{5}. $$


1

Well I'm not a programmer, but I'm pretty sure that you need math to program like the air you breathe. Especially if you want to become very, very good at it; I mean if you want to understand algorithm running time, do analysis on the complexity of a specified algorithm or trying to find a new one who performs better, you really need to be confident in ...


0

I quite enjoyed How Not to be Wrong, which seems like it might be what you're looking for. From the Amazon summary: The math we learn in school can seem like a dull set of rules, laid down by the ancients and not to be questioned. In How Not to Be Wrong, Jordan Ellenberg shows us how terribly limiting this view is: Math isn't confined to abstract ...


0

"What is Mathematics" by Courant and Robbins. Note that mathematics could be considered as a part of reality. In some sense one not need looking for applications to real life because math is a part of that reality.


1

To continue with your argument: you have the property that $r_{i,u}+r_{i,p−u−1}−p$ and $r_{i,u′}+r_{i,p−u′−1}−p$ have the same sign for all $i$. This is equivalent to saying that $r_{i,u} - r_{i,u+1}$ and $r_{i,u'} - r_{i,u'+1}$ have the same sign for all $i$. This difference is always $p-i$ or $-i$, so this means that $$ r_{i,u} - r_{i,u+1} = r_{i,u'} - ...


0

This means solving the differential equation $$ \ddot{x}(t) = a = 5 \mbox{m}/\mbox{s}^2 $$ with the inital conditions $x(0) = x_0 = 0\, \mbox{m}$, $\dot{x}(0) = v_0 = 25 \mbox{m}/\mbox{s}$ and inverting $x(t) = 1500 \mbox{m}$ for $t$. Integration gives $$ \dot{x}(t) = a t + v_0 $$ and integrating again gives $$ x(t) = \frac{a}{2} t^2 + v_0 t + x_0 $$ ...


0

Use the kinematics equation $\Delta x = v_i t+\frac{1}{2}at^2$. Substitute everything you know and solve for $t$.


2

I could be wrong, but I believe it can be read here simply as "the $j$ such that $z_{jt}$ is at its maximum value".


3

It is true there is a lot of high level math in physics, some of which is done by physicists and some by mathematicians. For example the professor I am working with works with (and writes papers with) physicists very often, but the work he does is really math because it comes from a field called algebraic geometry which depends highly on pure math fields ...


1

$\newcommand{\Reals}{\mathbf{R}}$Even in one dimension, a "sphere" has equation $x^{2} = r^{2}$, i.e., $x = \pm r$. If I understand, your friend's question is this: Let $n$ be a positive integer, $x_{1}$, ..., $x_{n}$ Cartesian coordinates, and define the sphere of suidar $r$ to be the set of points in $\Reals^{n}$ satisfying $$ x_{1}^{2} + \dots + ...


1

We know that geometric mean is $\sqrt{ab}$ And its given that the area will be equal iff geometric mean of the sides of the rectangle equal the side of the square. So equating what we know till now we get :$$\sqrt{ab}=a$$$$ab=a^2$$ Which implies $a=b$. I think you can carry on after this


0

Hint: The geometric mean of two numbers a and b is $\sqrt{ab}$. What happens when you equate areas of square of side $a$ and rectangle of length $l$ and breadth $b$?


1

Your problem is the different units. Remember that "acre" is a measure of area and we should convert to another measure of area such as square feet. One acre is $43560\, \mbox{ft}^2$ Just multiply this by the given volume to get the volume in $\mbox{ft}^3$ and then divide by the net flow rate that you calculated. This will give the time to empty in ...


1

Consider this You know, $2 \lt 3$ right? But you also know $-2>-3$ by following the number line, having 0 at the centre, as -2 is closer to 0 than -3. Let's divide the number line into positive space $\forall x\gt0$ and negative space $\forall x \lt 0$. Just remember, when you move from one space to another the equality sign will alter, become ...


0

Experiment with different values around $\pm1/3$ and conclude: $$\begin{align} &x=-2/3&\to-3x&=2&\color{green}\ge-1\\ &x=-1/3&\to-3x&=1&\color{green}\ge-1\\ &x=-1/6&\to-3x&=1/2&\color{green}\ge-1\\ &x=1/6&\to-3x&=-1/2&\color{green}\ge-1\\ &x=1/3&\to-3x&=-1&\color{green}\ge-1\\ ...


0

If you multiply or divide both sides of an inequality by a negative number then you have to flip the sign around, e.g. $$-3x\ge-1\implies x\le-1/(-3) \implies x\le 1/3$$. If you are adding or subtracting however, you do not need to flip the inequality sign, and this can often make you less uncertain that you are correct by only adding/subtracting rather ...


1

For the second equation, the right side can be rewritten as $$\log_2x+4=\log_2x+\log_216=\log_216x$$ at which point you can eliminate the logarithms. As for the first equation, I'd recommend expressing both sides as a power of $2$.


1

$$\log_2 y=\log_2x+4$$ $$y=2^4x$$ $$y=16x$$ Plug into first equation $$8^{16x}=4^{2x+3}$$ $$4^{24x}=4^{2x+3}$$ $$\vdots$$ Can you take it from there?


3

Reading, in pretty much any form, will almost always benefit you. Chances are you won't have a solid grasp on the material if you don't complete the problems, but this sort of reading will allow you to become informed about the larger ideas at play in various areas of mathematics and will help inform your future learning. It sounds like you'll end up taking ...


0

You need to use a shooting method. Let's abstract it a bit. Let's define $F(x_{ic})$ to be a function that takes your initial conditions as its argument and outputs the value of your solution at $x = 151900$. In other words, $F$ is a solution trajectory of your ODE, and we might think of $F$ as a single run of ode45, executed with initial conditions ...


0

As one who was taught in the old school of euclidean plane geometry, the whole apparatus of proposition and proof, gave a basis to the understanding of shape which more modern students do not have: the idea of a proof is something foreign: they are equal because they lkook it, does not answer, and the basis of calculus ( differentiation from first principals ...


0

I think the only way would be to find a different proof that holds against the main concern of the typical objections: Does the constructed number actually exist? I'll try to illustrate a proof that can address this concern. Let's consider lists of real numbers written in base 10 and build a number from the list using this rule: If the n^th digit of ...


2

I will give some suggestions and you tell me what you think about them. Then i can suggest others. First of all, some revision: Topology------ Introduction to metric and topological spaces by Sutherland. Very good book with an excellent motivation for compactness (with an example that appears in analysis) Real Analysis and Advanced Calculus---David ...


0

ISO 80000-2:2009 entitled "Quantities and units — Part 2: Mathematical signs and symbols to be used in the natural sciences and technology" should cover them.


4

Maybe instead of handling your example, because the context is not always relevant, let's look at possible groupings of the symbols. Equality $=$ is usually used for equality. $\equiv$ is occasionally used for "identically equal to," which is in a sense stronger than equality, by denoting that the thing on the left and the thing on the right are equal in ...


6

If your example is about speed, then this a question for physicists and not for logicians. The most important thing is that these symbols are precisely that - symbols. They can be defined to mean whatever you want them to. Of course, we try to keep some common basis of definitions across a particular topic. But even the standard equal sign can be used for ...


6

Many of the symbols you list, unless they have meanings I'm unaware of, are meaningless in this context. In general: $=$ means, as I assume you know, is 'equal to' (there is a certain subtlety here, but I'm hoping you don't want to go into it). $\approx$ means 'approximately equal to.' $\sim$ means, in the contexts I'm aware of, 'is asymptotic to,' ...


1

Dummit and Foote's $\textit{Abstract Algebra}$ text is definitely not lacking in terms of mathematical content. Although the comments above are correct when they say there is almost no category theory, the number of topics covered in the text makes up for it. I have the text in front of me and it totals 7 pages of category theory. On the other hand, I have ...


0

If you're interested in commutative algebra, then I would recommend Atiyah & MacDonald's Introduction to Commutative Algebra which should be accessible with some basic knowledge of rings. For algebraic geometry, I'd strongly advise learning some commutative algebra first and then a couple of options are Hassett's Introduction to Algebraic or Ideals, ...


1

Why not just forget about the different variations of the formula (e.g. point slope, 2 points, slope intercept, etc. ) and just memorize this one. $$\color{Tomato}{m=\frac{\Delta y}{\Delta x}}$$ If the slope ($m$) and a single point ($x_0, y_0$) are known, then this forumla becomes $$\begin{align} m&=\frac{\Delta y}{\Delta x}\\ ...


1

Perhaps you can use colours and simple objects like eggs, which students in grade 6 can easily acquaint with. Suppose a basket A has 5 eggs: a red egg, a blue egg, a green egg, a yellow egg, and a white egg. Also suppose a basket B has no eggs. Asking which eggs belong in both baskets shall illustrate how the two sets intersects (if at all, since we are ...


3

Assuming he already knows what an empty set is, it should not be too hard. Usually, a good way to represent sets is bags with things in them, and an empty set is just a bag with nothing in it. Now, you can represent an intersection of sets with picking common things. So if one bag has eggs, flour and butter and the other has butter, milk and bread, the ...


2

You almost have it with $y-4=(3x/4)+(6/4)$. Just get the 4 to the right hand side $y=\frac34x+\frac32+4$. So $$y=\frac34x+\frac{11}2$$


2

(5/6)-(1/2) <- for this one you could just multiply the numerator and denominator of the second fraction by 3 instead of 6 and it'll be (5/6)-(3/6) Then you'll get (22/15)/(2/6) , Then, to divide two fractions you take the second one and flip it then multiply. So, this (22/15)/(2/6) becomes this (22/15)*(6/2) To make your life easier see if you can ...


-1

You will get $\frac{12*22}{15*4} = \frac{22}{5}$.


0

Simply notice that $x=1$ is a root. Hence by Ruffini your polynomial $x^3-x^2-x+1$ is divisible by $x-1$. Thus there exists a second degree polynomial $p(x)$ (you can find by a simple polynomial division) such that $x^3-x^2-x+1=(x-1)p(x)$. Being $p$ of degree $2$ you can easily find its roots with the well known $\Delta$ formula, in order to factorize ...


0

\begin{align}x^3-x^2-x+1&=x^3-x^2-x+1+(3x^2-3x^2)+(3x-3x)\\ &=x^3+3x^2+3x+1-x^2-x-3x^2-3x\\ &=(x+1)^3-4x^2-4x\\ &=(x+1)^3-4x(x+1)\\ &=(x+1)\left[(x+1)^2-4x\right]\\ &=(x+1)(x-1)^2\end{align}


3

Let $f(x)=x^3-x^2-x+1$. By Rational Root Theorem, any rational root $\frac{p}{q}$ of $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ has $p\mid a_0$ and $q\mid a_n$ (i.e. $q$ divides $a_n$). Proof is simple: substitute $x=\frac{p}{q}$, multiply both sides by $q^n$, observe divisibility. Since $f(1)=0$, we know $x^3-x^2-x+1=(x-1)P(x)$ for some $P(x)$, ...


3

$$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$ $$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$


0

we want to factor the polynomial $\mathbb{P}=x^{3}-x^{2}-x+1$ factor out $x^2$ to start as follows; $\mathbb{P}=x^2(x-1)-(x-1)$ This makes sense because if you expand above you would get, $x^3-x^2-(x-1)=x^3-x^2-x+1$ which is of the form you had at the start. Now factor out $(x-1)$ $$=(x-1)(x^2-1^2)$$ It may be helpful to see it as, $(x-1)(x^2-1)$ say ...


1

Depending on the size, you can ask questions. Questions are always welcome in smaller talks. If it's bigger, just listen, make notes of things you might want to discuss afterward either with other people there or the speaker. Why talks are so great is that you're actually THERE with the person who's supposedly an expert.


0

I won't say it is (or isn't) the best choice, but I will say it's possible, though maybe a bit challenging, with Sage: This image is from this link. I can confirm that Sage plays nicely with OS X (better than with Windows, I believe), although I've never tried to animate anything; it may have quite a learning curve. You can download and run everything on ...


1

Just do the usual math: $$ n+\frac{1-n}{n^2+1}=\frac{n(n^2+1)+1-n}{n^2+1}= \frac{n^3+n+1-n}{n^2+1}=\frac{n^3+1}{n^2+1} $$ If you want to go the other way around: $$ \frac{n^3+1}{n^2+1}=\frac{n^3+n-n+1}{n^2+1}= \frac{n^3+n}{n^2+1}+\frac{1-n}{n^2+1}= n+\frac{1-n}{n^2+1} $$


1

by using the long division ..................


-1

Equation 1 is At = x(h-y)/2 Equation 2 is AT = (y/2)(x + b) Equation 3 is AT2 = bh/4 Equation 4 is At = x(h-y)/2 AT2 = bh/4 At = AT2 x(h-y)/2 = bh/4 2x(h-y) = bh 2xh – 2xy = bh x(2h – 2y) = bh x = bh/(2h – 2y)


1

One answer is that the "algebraic" statement allows for a generalization to arbitrary rings (rather than just $\mathbb Z$), see further down in the Wikipedia article. This is useful is various contexts. The two most direct such generalizations are on the one hand principal ideal domains, where the CRT helps classify finitely generated modules, and on the ...


3

All mathematics is "proof based" because all mathematics are supported by a logical structure. We can focus on the structures to varying degrees. When I think of "proof-based mathematics" the first thing that comes to mind is a "classical" high school geometry class. Except for top private and public schools, in the US most students no longer take such a ...


2

I think the answer to this question relies heavily on the CW structure imposed on $S^1$. I can realise $S^1$ with an arbitrary number of $1$-cells.


3

Why do we need to rationalize fractions? To put it simply, you don't. Not in practical life, anyway. Rationalizing or not, the value of the number is the same. However, there is an advantage of rationalizing. Consider, for example, the number $1/\sqrt{2}$. It is known that $\sqrt{2} \approx 1,4142...$. I'll sort of answer your question with another one: ...


2

There is no mathematical reason, it is one of "penmanship." My hypothesis is that it makes grading easier for them, since requiring this will standardize the answers they get into one form, allowing them to mark the answers that match that form "correct" and those that don't "incorrect" without actually numerically evaluating the answers that may be ...



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