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October 10

class of 32 kids, 1 adult to supervise each group of 8. write algebraic expression to find out how many adults were needed.

well notice that you have 3 numbers, and note that there are 8 students, per adult thus 8x=32, x being the number of adults, needed to supervise. then divide both sides by 8 so that you get x=32/8 and that simplifies (reduces) to x=4

If 'a group' means 'a subset' then there are ${\displaystyle {\tbinom {32}{8}}=10518300}$ 'groups', so you need over ten million adults. Each kid belongs to ${\displaystyle {\tbinom {32-1}{8-1}}=2629575}$ groups, so every kid will be supervised by over two-and-half million adults. CiaPan 07:14, 10 October 2007 (UTC)[reply]

Possibly the question is about the following generalization (and abstraction): A set of n elements is partitioned into parts, each of which has a size not exceeding k. What is the least number of parts for which this is possible?

Calling the number of parts p, we have n ≤ pk, and the question can be rephrased as: give a

closed expression

P(n, k) such that the following two statements are equivalent:

n ≤ pk,

P(n, k) ≤ p.

This is solved, of course, by defining P(n, k) = n/k, but presumably we want to remain in the domain of

integers. Then the answer is found in the ceiling function

:

P(n, k) = ⌈n/k⌉.

This is a Galois connection. For the original question, we find P(32, 8) = ⌈32/8⌉ = ⌈4⌉ = 4. If we increase the class size by one but one adult can still supervise at most 8 kids, we have P(33, 8) = ⌈33/8⌉ = ⌈4.125⌉ = 5.  --Lambiam 13:36, 10 October 2007 (UTC)[reply]

Oh God, while this is hilarious, you realize that you've just made some poor high school freshman wet himself with fear of mathematics. OP: the first response is the one that you care about. Donald Hosek 18:11, 10 October 2007 (UTC)[reply]

LOL, it actually looks harder than it really is, hope that calms the OP down LOL A math-wiki 05:36, 11 October 2007 (UTC)[reply]

If I score 90/100 for my first exam and 70/100 for my second exam, my mean score should be 80/100. However, because my second exam is given a weightage of 75% and my first exam a weightage of 25%, my mean mark is 75/100.

When each value has a weightage in %, to get the mean of all the values, you have to multiply each value by its weightage before taking the mean of all the values.

That would be a

weighted mean. Confusing Manifestation 05:10, 10 October 2007 (UTC)[reply

]

Yes, a weighted mean ... often referred to as a "weighted" average. (Joseph A. Spadaro 22:10, 10 October 2007 (UTC))[reply]

I recently read up on Set Theory, Ordinals, and Cardinal Numbers, and I found the arguments involving different infinite sets to be illogical.

I think there is a much simpler answer, but first I need to point some facts about The Concept of Infinity. First of all, Infinity cannot be treated like a number or a variable, you can't calculate ${\displaystyle \infty +1}$. You might try and argue that since 1 gets arbitrary as ${\displaystyle x\to \infty }$, that ${\displaystyle \infty +1=\infty }$, but now you've inadvertantly violated the most fundamental principle in mathematics, equality, ${\displaystyle x\neq x+1}$. Instead you should write ${\displaystyle \infty \equiv \infty +1}$, because the concepts you implied by ${\displaystyle \infty }$ and ${\displaystyle \infty +1}$ are the same. The point infinity doesn't have numeric or algebraic value but it does have meaning. The basic idea is that it is equivalent conceptually, but not actually equal numerically.

What does this mean for infinite sets?

It means that the idea of compairing the 'size' of infinite sets is nonsensical. A much better way of handling the problem Cantor tried to address is as follows. Consider the sets ${\displaystyle \mathbb {N} }$ and ${\displaystyle \mathbb {Q} }$, the natural numbers, and the rational numbers. The problem as cantor saw it was that ${\displaystyle \mathbb {N} \subset \mathbb {Q} }$, but yet even ${\displaystyle \mathbb {N} }$ has an infinite number of elements. There really isn't a problem here because infinity has only conceptual value. A subset is a concept as well, it doesn't have any numeric value, not even an arbitrary (or variable) value, so there is no logical falacy. It is logical to look at the behavior of different infinite sets, and to use conceptual arguments and notation to categorize them, but any attempt to appliy numeric values to (such as aleph null, aleph one, etc.) is illogical. A good case to illustrate where my thinking diverges from Cantor's comes from the work I did above, consider another set ${\displaystyle \mathbb {K} }$ that is the set also a subset of ${\displaystyle \mathbb {Q} }$ and is defined by ${\displaystyle {\mathbb {Q} :X\notin \mathbb {N} }}$. Then we can logically conclude, ${\displaystyle \mathbb {N} \subset \mathbb {Q} }$, ${\displaystyle \mathbb {K} \subset \mathbb {Q} }$, ${\displaystyle \mathbb {N} \not \cup \mathbb {K} }$, and ${\displaystyle \mathbb {N} +\mathbb {K} \equiv \mathbb {Q} }$. These are also concepts (hence the equivalence). I would like to point out the infinite limits and infinite integrals specifically those that have finite outputs should also use and equivance sign, rather than an equals sign.

That is more or less like saying "Complex numbers are nonsensical, since they violate the most fundamental principle in mathematics, inequality, ${\displaystyle x^{2}\geq 0}$". First, While I agree that equality is a fundamental concept in mathematics, the existence of an x such that ${\displaystyle x=x+1}$ says more about addition (and the structure over which it is defined) than about equality. Second, just because for every real number we have ${\displaystyle x\neq x+1}$, doesn't mean that any mathematical structure that humanity will ever consider must satisfy this as well. The field of real numbers is just a drop in the mathematical ocean. We have plentiful structures where this does not hold, such as the real projective line, extended real number line, and of course, cardinal numbers. In the ordinal numbers, ${\displaystyle x\neq x+1}$ does hold but ${\displaystyle x\neq 1+x}$ does not.

In short, dismissing an entire and fundamental theory, only because it exhibits behavior somewhat different from our everyday finite numbers, is just silly. In fact, it would be quite surprising if infinite quantities behaved in exactly the same way as finite ones. -- Meni Rosenfeld (talk) 11:34, 10 October 2007 (UTC)[reply]

It would be more productive not to get hung up on concepts of numerical equivalence of infintities here I think.. eg If a set contains infinite subsets each with infinite members - then rather than attempting to 'count them' I would recommend using a method such as venn diagrams to get a simple understanding of how one infinite set can be 'bigger' than others..

Maybe someone else could recommend other methods of visualisation that simplify an overview of sets and set theory.87.102.79.56 15:42, 10 October 2007 (UTC)[reply]

Venn diagrams have little value (detrimental value, even) for understanding infinite cardinalities. -- Meni Rosenfeld (talk) 19:42, 10 October 2007 (UTC)[reply]

Nice counterarguments, I should mention that what is really confusing me is how cardinal number can compair the size of say ${\displaystyle \mathbb {N} }$ and ${\displaystyle \mathbb {K} }$ in my example scenario. Seeing as they have no elements in common how can cardinal numbers justify that ${\displaystyle \mathbb {K} >\mathbb {N} }$ —Preceding unsigned comment added by 69.54.140.201 (talk) 19:24, 10 October 2007 (UTC)[reply]

Actually, your set ${\displaystyle \mathbb {K} }$ above actually has the same size as ${\displaystyle \mathbb {N} }$, since ${\displaystyle \mathbb {Q} }$ is

countable

. The situation would be different if you took ${\displaystyle \mathbb {R} }$ instead. Regardless, I don't see any problem - does the fact that the sets ${\displaystyle \{1,2,3\}}$ and ${\displaystyle \{4,5\}}$ have no elements in common creates any difficulty in showing that the former is larger? You know how cardinals work (using bijections), right?

It is good that you have read up on set theory, but I can only suggest that you read up some more... -- Meni Rosenfeld (talk) 19:42, 10 October 2007 (UTC)[reply]

Uh, no and I'm guessing this is why cardinals seem illogical to me. It's hard to learn math from an encyclopedia LOL. —Preceding unsigned comment added by 69.54.140.201 (talk) 20:05, 10 October 2007 (UTC)[reply]

So I'll just say that ${\displaystyle \mathbb {R} }$ is larger than ${\displaystyle \mathbb {N} }$ because there is an injection from ${\displaystyle \mathbb {N} }$ to ${\displaystyle \mathbb {R} }$ but no injection from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {N} }$. Finding a good book will be helpful, though - I don't have one to recommend, but hopefully someone else here will. -- Meni Rosenfeld (talk) 20:36, 10 October 2007 (UTC)[reply]

Cantor's diagonal argument may or may not help you to understand this, just in case you haven't found it already. 130.88.47.27 13:00, 11 October 2007 (UTC)[reply]

Already found that (I have created an account fyi) A math-wiki 07:10, 12 October 2007 (UTC)[reply]

If you have n points evenly distributed on a circle of radius r, what is the average distance between one point and all the other points, i.e. the sum of the distance to from one point to each point in turn, divided by the number of other points. Thanks!

So far I got;

${\displaystyle D_{avg}={\frac {{\sqrt {2}}\sum _{k=1}^{n}(r{\sqrt {1-\cos({\frac {2k\pi }{n}})}})}{n}}}$

I think that works, but I really need something in the form of a function without summation series in it. Thank you. ΦΙΛ Κ 10:50, 10 October 2007 (UTC)[reply]

If you have points P and Q on a circle of radius r, centre O, and the angle POQ at the centre of the circle is 2kπ/n then I think the length of PQ is 2rsin(kπ/n) - so I don't see where the square root comes from in your sum. Also, if there are n points altogether, you should be summing n-1 terms and dividing by n-1 to find the average - otherwise when n=2 you get the answer r when you should expect to get 2r. Gandalf61 11:22, 10 October 2007 (UTC)[reply]

Actually, the expression for the distance is correct, it can be written as either ${\displaystyle r{\sqrt {2(1-\cos \theta )}}}$ (direct use of the cosine theorem) or ${\displaystyle 2\sin(\theta /2)r}$ (after simplifcation by trigonometric identities). Now, you may or may not include all n terms, depending on whether you want the distance from a point to itself to count. Whatever the case, I doubt the average can be given in closed form (though I'll try to look into it). Will an approximation help? -- Meni Rosenfeld (talk) 11:42, 10 October 2007 (UTC)[reply]

A good approximation (including the zero distance) is ${\displaystyle r\left({\frac {4}{\pi }}-{\frac {\pi }{3n^{2}}}\right)}$, and I suppose this can be improved with higher negative powers of n. -- Meni Rosenfeld (talk) 12:30, 10 October 2007 (UTC)[reply]

(OT: This was found empirically, with help from Plouffe's inverter; I was disappointed by its failure to find 3.819718634 being equal to ${\displaystyle 12/\pi }$.) -- Meni Rosenfeld (talk) 12:44, 10 October 2007 (UTC)[reply]

Your expression can be simplified. As remarked by Meni,

${\displaystyle 1-\cos {\tfrac {2k\pi }{n}}=2\sin ^{2}{\tfrac {k\pi }{n}},}$.

so

${\displaystyle {\sqrt {1-\cos {\tfrac {2k\pi }{n}}}}={\sqrt {2}}\left|\sin {\tfrac {k\pi }{n}}\right|.}$

If 0 ≤ k ≤ n, the value of the sine is nonnegative, so then we can omit taking the absolute value, and the summation becomes:

${\displaystyle \sum _{k=1}^{n}r{\sqrt {1-\cos {\tfrac {2k\pi }{n}}}}=\sum _{k=0}^{n-1}r{\sqrt {2}}\sin {\tfrac {k\pi }{n}}=r{\sqrt {2}}\sum _{k=0}^{n-1}\sin {\tfrac {k\pi }{n}}.}$

(Using the fact that the summand vanishes for both k = 0 and k = n, I've shifted the range of summation.) Since sin x = Im(e^ix), the last summation equals the imaginary part of

${\displaystyle \sum _{k=0}^{n-1}e^{{\frac {k}{n}}i\pi }={\frac {2}{1-e^{\frac {i\pi }{n}}}}=1+i{\frac {\sin {\tfrac {\pi }{n}}}{1-\cos {\tfrac {\pi }{n}}}}=1+i\cot {\tfrac {\pi }{2n}}.}$

Combining all this gives us that the sum of the distances between one point and the other points is equal to:

${\displaystyle 2r\cot {\tfrac {\pi }{2n}}.}$

I leave it to you whether you want to divide this by n or by n−1.  --Lambiam 14:42, 10 October 2007 (UTC)[reply]

I can't believe I haven't thought of that. I am so embarassed. I will give up on mathematics and start a floor-sweeping career. -- Meni Rosenfeld (talk) 19:46, 10 October 2007 (UTC)[reply]

I was more interested in the case where n → ∞. In that case we shift to the integral ${\displaystyle {\frac {r}{\pi {\sqrt {2}}}}\int _{0}^{2\pi }{\sqrt {1+\cos \theta }}d\theta }$ (the idea being that the average value of f(x) over [a,b] is ${\displaystyle {\frac {\int _{a}^{b}f(x)dx}{b-a}}}$. From there, it turns into a Calc II problem.

When using the limit from Lambiam's formula, remember that you're talking about ${\displaystyle {\frac {2r\cot {\tfrac {\pi }{2n}}}{n}}}$ where L'Hôpital's rule applies. You should get the same result either way, if not one of us made a mistake. Donald Hosek 20:32, 10 October 2007 (UTC)[reply]

So my friend had a multivariable calculus quiz recently, and one of the question was as follows:

“

Let ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ be differentiable functions of one variable. Find ${\displaystyle \partial z \over \partial x}$ if ${\displaystyle f(yz)=g(xz)}$.

”

Apparently her TA has no office hours and never goes over anything, so she's not expecting any help from there. I'm not entirely sure how to do this problem myself either. I can get as far as ${\displaystyle f'(yz)\cdot y{\partial z \over \partial x}}$ on the left, but that's about all (and to be honest, I'm not 100% sure I'm right, either). Perhaps someone could explain how to finish this problem? --M1ss1ontomars2k4 18:50, 10 October 2007 (UTC)[reply]

The question makes no sense as presented, since y appears in exactly one place and is not quantified. I suspect it should have been ${\displaystyle f(yz)=g(xy)}$ (for every y and x, where z is a function of x). In this case, you can take ${\displaystyle y={\frac {1}{x}}}$ and get ${\displaystyle f\left({\frac {z}{x}}\right)=g(1)}$, a constant. So, under some smoothness assumptions, as well as the assumption that f is not constant, we have ${\displaystyle z=\alpha x}$ for some ${\displaystyle \alpha }$. So we have ${\displaystyle g(x)=f(\alpha x)}$, so ${\displaystyle {\frac {dz}{dx}}=\alpha ={\frac {g'(0)}{f'(0)}}}$. -- Meni Rosenfeld (talk) 19:34, 10 October 2007 (UTC)[reply]

I'm not entirely sure what was meant by that y either. But my friend was very sure the quiz said ${\displaystyle g(zx)}$; I asked her multiple times to make sure. I guess you'd have to assume z is a function of x and y since the question asks for the partial derivative of z with respect to x. Either that or her TA is just nuts. Actually, from what she's told me, it's probably both. --M1ss1ontomars2k4 19:41, 10 October 2007 (UTC)[reply]

We may or may not need to assume that z is a function of both x and y, but that certainly doesn't fix the problem. The y issue has nothing to do with calculus, multivariate or otherwise. It has to do with much deeper aspects of the role variables play in mathematical statements. A variable must have some sort of quantification, be it either as a dummy ("for all y", "there exists a y") or as an item about which the statement contains information. But this question is not about y and y does not appear anywhere else in it, and any quantification would lead the given ${\displaystyle f(yz)=g(xz)}$ to be trivial in content (admittedly, this last observation does have something to do with calculus). -- Meni Rosenfeld (talk) 19:53, 10 October 2007 (UTC)[reply]

I had to think about this one for a awhile, I never actually studied this, but I did do stuff like this for fun while I was in AP Calculus and I'm pretty sure I know what you need to do. I believe your answer so far is correct. I don't know how to do ${\displaystyle {\partial z \over \partial x}}$ to ${\displaystyle g}$ though, forgot. I would guess that once you differentiate both, you can collect all the terms with ${\displaystyle {\partial z \over \partial x}}$ and then factor that out and divide what's left from that side to get ${\displaystyle {\partial z \over \partial x}=}$ —Preceding unsigned comment added by 69.54.140.201 (talk) 19:57, 10 October 2007 (UTC)[reply]

Ok I remember how to do g now, and I noticed your missing the other half of the product rule on f'

you have ${\displaystyle {\partial z \over \partial x}=f'(yz)y{\partial z \over \partial x}}$ but it should read

${\displaystyle {\partial z \over \partial x}f(yz)=f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}}$

${\displaystyle {\partial z \over \partial x}g(xz)=g'(xz){\frac {d}{dx}}[xz]}$

From the chain rule. Then use the product rule on

${\displaystyle {\frac {d}{dx}}[xz]}$

Which gives

${\displaystyle x{\partial z \over \partial x}+z}$

So you have

${\displaystyle {\partial z \over \partial x}g(xz)=g'(xz)x{\partial z \over \partial x}+g'(xz)z}$

Now remember that ${\displaystyle f(yz)=g(xz)}$, well this implies that ${\displaystyle f'(yz)=g'(xz)}$ thus

${\displaystyle f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}=g'(xz)x{\partial z \over \partial x}+g'(xz)z}$

and now we isolate ${\displaystyle {\partial z \over \partial x}}$

${\displaystyle {\partial z \over \partial x}={\frac {g'(xz)z+f'(yz)z{\frac {dy}{dx}}}{f'(yz)y-g'(xz)x}}}$

You have some things mixed up here. But that's irrelevant, you can't give a good answer to a bad question. -- Meni Rosenfeld (talk) 20:32, 10 October 2007 (UTC)[reply]

I think that fixes the mistakes, but more importantly we should consider that the TA in question probably didn't bother to think about the formal implications his lack of variable definitions gives us. He was probably testing whether his students, knew how to do it, not do they understand that this is technically unsound. So as long as dy/dx exists then my answer works, assuming I fixed all the mistakes too. —Preceding unsigned comment added by 69.54.140.201 (talk) 20:42, 10 October 2007 (UTC)[reply]

No, you did not fix all the mistakes, and the question cannot be solved, any more than the question "you are given that ${\displaystyle a=3}$. Find ${\displaystyle b^{2}}$" can be solved. -- Meni Rosenfeld (talk) 20:51, 10 October 2007 (UTC)[reply]

On second thought, if we do assume that y is meant to be a function of x, and that the solution should be given in terms of y, then there might be a correct solution - but I find this an extreme stretch of the question. -- Meni Rosenfeld (talk) 20:58, 10 October 2007 (UTC)[reply]

It would make more sense that z is a function of (x,y), and I'm curious what mistakes I've missed, need a good multivar Calculus review, been a year and a half since I did much Calculus. —Preceding unsigned comment added by 69.54.140.201 (talk) 21:11, 10 October 2007 (UTC)[reply]

We don't know much about f and g, but let us try some simple example and see where that leads us. Take the functions defined by

f(x) = 0 and g(x) = 0 (for all x).

Using these example function, the question reduces to:

Find ∂z/∂x if 0 = 0.

Something tells me that if we can't solve the problem for this simple example, we may have a problem solving it in general.  --Lambiam 21:15, 10 October 2007 (UTC)[reply]

[edit conflict, responding to anon] You can start with the treatment of ${\displaystyle {\partial z \over \partial x}g(xz)}$ as the derivative of ${\displaystyle g(xz)}$ with respect to to x, when it is actually the derivative of z with respect to x, multiplied by ${\displaystyle g(xz)}$.

I do take some of my earlier comments back, though. The question just might, after all, make some sense if we treat z as a function of x and y. I'll think about this some more. Note that in your solution, you do treat y as a function of x. -- Meni Rosenfeld (talk) 21:20, 10 October 2007 (UTC)[reply]

Some progress might be possible if we assume that f and g are nonconstant. -- Meni Rosenfeld (talk) 21:23, 10 October 2007 (UTC)[reply]

Looking at what I did more closely, I actually derivated both functions with respect to x. So my notation is wrong, not my thinking. Em I right??? Furthermore, if I recall correctly ${\displaystyle {d \over dx}xy}$ is worked out by the product rule. So i should remove the z on those two, which I have now done. That better? —Preceding unsigned comment added by 69.54.140.201 (talk) 21:41, 10 October 2007 (UTC)[reply]

Yes, see below. The improper notation makes it difficult for me to follow your derivation, but your result is almost correct (almost, because there is no ${\displaystyle {\frac {dy}{dx}}}$) so the thinking is probably right. -- Meni Rosenfeld (talk) 21:48, 10 October 2007 (UTC)[reply]

Okay, here goes. We are assuming that z is a function of x and y. Denote ${\displaystyle p(x,y)=f(yz(x,y))}$, ${\displaystyle q(x,y)=g(xz(x,y))}$. We are given ${\displaystyle f(yz)=g(xz)}$ so ${\displaystyle p(x,y)=q(x,y)}$ for every x and y, so their derivatives with respect to x are also equal:

${\displaystyle f'(yz)y{\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}p(x,y)={\frac {\partial }{\partial x}}q(x,y)=g'(xz)\left(z+x{\frac {\partial z}{\partial x}}\right)}$

${\displaystyle {\frac {\partial z}{\partial x}}(f'(yz)y-xg'(xz))=g'(xz)z}$

So wherever ${\displaystyle f'(yz)y\neq xg'(xz)}$, we have:

${\displaystyle {\frac {\partial z}{\partial x}}={\frac {g'(xz)z}{f'(yz)y-xg'(xz)}}}$

So other than the notation, anon got most of it right. Was that the intended solution? Only the TA knows. I do give this question the "most poorly worded question of the year" award. -- Meni Rosenfeld (talk) 21:38, 10 October 2007 (UTC)[reply]

If anyone is wondering why this question turn out to be solvable despite my "deeper aspects of the role variables play in mathematical statements"... That is because I did not account for notational shortcuts. Keeping in mind that z is shorthand for ${\displaystyle z(x,y)}$, y actually appears in the question 4 times (under the quantification "for all"). -- Meni Rosenfeld (talk) 21:44, 10 October 2007 (UTC)[reply]

Quick question, how is ${\displaystyle {\partial \over \partial x}f(yz)\not =f'(yz)y{\partial z \over \partial x}+f'(yz)z{\frac {dy}{dx}}}$ —Preceding unsigned comment added by 69.54.140.201 (talk) 21:48, 10 October 2007 (UTC)[reply]

Because when taking a partial derivative with respect to x, you treat y as a constant. So instead of using the product rule, you should use the ${\displaystyle (cf)'=cf'}$ rule. The expression you give is also correct if we agree that ${\displaystyle {\frac {dy}{dx}}=0}$. -- Meni Rosenfeld (talk) 21:55, 10 October 2007 (UTC)[reply]

See, ... I knew I was forgetting something LOL, ya I agree and thus the term goes to 0 and we don't bother writing it. Thanks

It should technically be ${\displaystyle {\partial y \over \partial x}}$ not ${\displaystyle {dy \over dx}}$ correct??? —Preceding unsigned comment added by 69.54.140.201 (talk) 23:29, 10 October 2007 (UTC)[reply]

It shouldn't be anything, really. If by y we mean then one-variable function which for every value of the argument returns the constant y, then it should be ${\displaystyle {\frac {dy}{dx}}}$. If by y we mean the function of two variables which returns the second argument, then its partial derivative with respect to the first variable will be ${\displaystyle {\frac {\partial y}{\partial x}}}$. But neither of these are applicable in our context. -- Meni Rosenfeld (talk) 23:36, 10 October 2007 (UTC)[reply]

I would have to disagree, I would say it would be ${\displaystyle {\partial y \over \partial x}}$ because we began by doing partial differentiation, and because y may equal a function of two variables but could be considered, F(X,Y)=y, where y is a constant. Thus the partial derivative is 0. The regular derivative would also yield zero in the one variable case, but since we started by doing partial differentiation, that's what the operation should stay until it is dropped because it equals zero. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:44, 10 October 2007 (UTC)[reply]

And I would have to maintain my position that, as y is a variable rather than a function, we have no business discussing its derivative in the first place. -- Meni Rosenfeld (talk) 23:49, 10 October 2007 (UTC)[reply]

Actually, you could argue that y is a constant, since by the definition of a partial derivative, all other variables (not functions like z) are held constant. Thus both of our arguments can coexist, it is fundamentally nonsensical to talk about y's derivative but in practice the notation comes up and it should be written with partials not ordinary derivatives since we started with PARTIAL differentiation not ordinary differentiation. —Preceding unsigned comment added by A math-wiki (talk • contribs) 00:31, 11 October 2007 (UTC)[reply]

I'm working with a probelm that states a jet ski company suggests a two hour minimum rental fee of $ 125.00 increased by another $ 55 bucks for every aditional hour surpassing the two hours. It says at the last few sentences of the probelm, Macy and Mya's rental was priced at the balance of $ 290.00. Number 20 (the probelm) asks me to write an equation to determine the n number of hours Macy and Mya utilized the jet skis.

As a rule preponderant in the wikipedia guidelines, I did my own homework.

First I wrote the probelm in terms of step 1, knowing the question.

1. The company charges a 120 minute renting of $ 125.00 plus a taxing 55 for use of ski exceeding the two hour minimizer.

2. Well, I thought, 125.00 plus a additional 55 dollars totals 180, one hundred ten less than what the two girls excerised their worth of usage to be.

But where I am stuck is where to ensue for there. We know what half of 150 and five more than half of that 200 dollars is, but where would I go subsequently?

--75.44.155.52 22:37, 10 October 2007 (UTC)[reply]

Am I missing something? The charge is 55$ for every hour beyond the first two. So a cost of 290$ means 3 hours more than 2 hours, that is, anywhere between 4 and 5 hours.

The question states you should write an equation; That would be ${\displaystyle {\frac {290-125}{55}}+2=5}$, that is, up to 5 hours. -- Meni Rosenfeld (talk) 22:56, 10 October 2007 (UTC)[reply]

The algebraic expression your looking for is ${\displaystyle 125+55(n-2)}$ for total hours ${\displaystyle n}$ so set that equal to 290 and solve for n. That's what I think your looking for. —Preceding unsigned comment added by 69.54.140.201 (talk • contribs)

By the way, SineBot is doing such a good job that I haven't bothered telling you this before, but... Please sign your posts by typing ~~~~ (4 tildes). You should also consider creating an account. -- Meni Rosenfeld (talk) 23:44, 10 October 2007 (UTC)[reply]

Agreed, I will mention the name here in a few minutes for you. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:46, 10 October 2007 (UTC)[reply]

Ok I got an account up now :) A math-wiki 23:55, 10 October 2007 (UTC)[reply]

Mazal Tov! -- Meni Rosenfeld (talk) 00:02, 11 October 2007 (UTC)[reply

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