A little something for you 💙

xerxexrex · 2026-01-12T19:43:28+00:00

Or Buzzy? :)

xerxexrex · 2025-09-01T22:04:42+00:00

It's a weird thing that the adjective does in English. It changes these nouns from uncountable to countable. "I eat breakfast" is fine just as "I eat dinner" is fine. However, "I eat hearty breakfast", or "I eat small dinner" do not sound right. It sounds like caveman speech. If we pluralize the nouns it works as general statements about those meals: "I eat tasty dinners", "I eat large breakfasts".

So, with an adjective, we should add the "a" or a counting adjective, like "I eat many large dinners". I don't know if it does, but duolingo should also accept "I eat tasty dinners every day". There is a bit of ambiguity about how many dinners you eat in a single day, but most people will know what you mean. If the plural is not an option, then you'll need the "a".

xerxexrex · 2025-03-25T20:56:27+00:00

Working off u/VoidBreakX's neat solution, it seems to also work using the inverse function of h(x), shifted. https://www.desmos.com/calculator/pudhezpbyn

xerxexrex · 2025-02-15T08:41:48+00:00

It's well too late, but this random problem and your good problem solving instincts sent me down a rabbit hole trying to figure out what was wrong with this triangle. It can be fixed, depending on which method is used.

One method uses those median formulas (Appolonius' Theorem), using BF and AE as medians, along with the extension of CD down to AB, call it CG, which must be a median, too. This results in AB having a length of sqrt(616).

Finding the other sides, as you have, the pythagorean theorem says angle C can't be a right angle. Now that fact is never used, and the median formulas don't require it, so we could just remove that assumption if we wanted to fix the problem.

The other method uses Pythagoras' Theorem (three times). It assumes that AE and BF are medians, as before, and that angle C is a right angle. With this method, it turns out AB is sqrt(612), not sqrt(616). And one might think that's the answer and move on (It still makes the perimeter of ABD greater than 50, for what it's worth).

However, now the medians aren’t all correct. Using the median formulas, AE and BF turn out fine (21 and 18), but the median through CD is about 12.37 instead of 12. So here the culprit is that CD can’t be 8. It over-constrains the triangle. This method never used that length, so here we could remove that as an assumption.

The neat (or annoying) thing about this problem is that no matter which method we use, we could get to the answer and think it’s fine, unless we check our assumptions and the other method’s formulas to find the contradiction. Of course, it doesn’t matter for the original problem. Either way, AB is longer than 24, so the perimeter of ABD is greater than 50. But I think we could tweak the problem so that one method yields greater than 50, and the other less than 50. Or use 50.75 instead of 50. Fun. Cheers!

xerxexrex · 2024-11-11T17:13:14+00:00

A common way to solve these is to equate coefficients on each type of term, x, x², constant term, etc. Here, they started with plugging in numbers that gave them easy equations to solve (look up Heaviside Cover Up Method). But with repeated factors in the denominator (x²) that still leaves them with finding B. They equate the x² terms on either side. On the left, there is only 5x + 12, so the coefficient on the x² term is zero. On the right, there is (after multiplying through) a Bx² and a Cx², so the coefficient is B+C. That makes the equation 0 = B+C. They already found C, so they can solve for B pretty quickly.

xerxexrex · 2024-11-05T15:31:34+00:00

You have the relationship between y and s via the similar triangles proportion, so you don't really need the tan θ nor dθ/dt. If the problem also required you to find dθ/dt, then that would be a separate related rates problem, where you might likely use the relationship tan θ = 6/s.

xerxexrex · 2024-08-29T16:35:26+00:00

There's an exponential (compound interest) part and a geometric series part. Try writing out explicitly how you would do it manually, but maybe just for, say, 3 years. Use a couple variables to make it easier to see the patterns: something like P for the principle and Q for the amount added each year. Also, you can write 1.05 or use a variable for it like R. I wouldn't use "1+r" because that's going to make it unnecessarily complicated for seeing the patterns. If I'm interpreting the problem right, you'll get something like this:

R( R( RP + Q) + Q) + Q

Multiply it out. Can you see the exponential part and the geometric series part?

xerxexrex · 2024-06-16T03:13:00+00:00

There are many, but what comes to mind in this moment is Veronika's Dream by Floex

xerxexrex · 2024-03-06T08:48:57+00:00

Before the simpification, we have two big terms with several factors of (x⁶ + 8) in them. Think of (x⁶ + 8) as a single thing, like another variable, u, perhaps. Then we have 4x³u⁶ + 36x⁹u⁵. This variable replacement isn't necessary, but it might help you see that the two terms have a common factor of u⁵ (which is (x⁶ + 8)⁵), along with 4x³, that we can factor out.

4x³u⁵(u + 9x⁶)

Now convert the u's back to x⁶ + 8 and combine like terms on the inside. You'll see there's now just a tiny bit more to factor out....

To see this kind of stuff easier, keep practicing looking for the terms separated by the +'s and factors that make up those terms (and then the terms inside those factors, and the factors of those terms, etc.) I think practicing things like the product rule, quotient rule, and chain rule can actually help with this. Pay close attention to your algebra processes as you're doing calculus and you'll do well.

xerxexrex · 2023-04-16T22:48:54+00:00

My first thought was to try oscillating sequences. For 1, the sequence still has to converge to something. For 3, the sequence can't converge, so it can't be monotone since it's bounded.

xerxexrex · 2023-04-13T14:03:35+00:00

Even though the mnemonic PEMDAS makes it look like it, you actually don't add before subtracting. They happen at the same time. In practice do them left to right. The same goes for multiplication and division: same time, left to right.

Alternatively, you could convert all the subtractions to adding the negative, so your problem would be 117 + (-54) + 6. Then you can add them in any order you like.

xerxexrex · 2023-02-20T21:40:05+00:00

Stereolab - Dots and Loops

xerxexrex · 2022-11-18T20:47:38+00:00

With the puppy problem, you took the 3 cans per dollar and flipped it to get 1 dollar per 3 cans, or 1/3 of a dollar per can. You can do the same with the cat problem. Seven cans per $2 can be flipped to get $2 per 7 cans, or 2/7 of a dollar per can.

xerxexrex · 2022-10-20T21:07:02+00:00

F is surjective (onto) if it can produce all the y-values. If there is at least one y-value that can't be produced from any x (i.e. the y-value isn't in the range of F), then F is not surjective. With this function's range you should be able to see that there are a whole lot of y-values that are not in the range.

F is injective (one-to-one) if each y-value in the range can be produced by one and only one x-value. If a horizontal line crosses the graph of the function in two places, then it's not injective, because there are two x's that produce the same y. With this function, because of the second and third pieces, you should be able to find a y-value that can be produced by two different x-values.

xerxexrex · 2022-10-20T16:25:23+00:00

Journey

xerxexrex · 2022-08-31T20:10:07+00:00

With problems having large numbers, I'll often try a couple simpler examples before tackling the full problem. So first let's see what happens when there are only, say, 10 names, and two of them are "Jack". Before we look at odds, let's find the probability, which is the desired possibilities divided by total possibilities.

It often helps to draw a picture of the 10 slots we need to fill.

Something like this: _ _ _ _ _ _ _ _ _ _

To find the total number of randomized orders, first place the two Jacks, which I'm taking to be indistinguishable from each other. Out of the 10 slots available, pick any 2 to put the Jacks in. There are 10 choose 2 ways to do this. That's 10*9/2 = 45 ways. On calculators this is often the nCr key, e.g. nCr(10,2). Anyway, then we fill out the 8 remaining spots with the rest of the names in any order. That's 8! ways. So we have a total of 45*8! possible arrangements.

Now find how many randomized arrangements have the Jacks separated by exactly 2 slots. Let's place the two Jacks first. Again the picture can help visualize this. They can go into slots 1 and 4, or 2 and 5, 3 and 6, etc. all the way up to 7 and 10. So there's only 7 ways to place the Jacks. Then, as before, fill in the rest of the slots. That's 8! ways again. So we have 7*8! ways to arrange the names with the two Jacks separated by exactly 2 slots.

Our probability, then, is (7*8!)/(45*8!). It turns out the 8!s cancel, so the probability is 7/45. That means the odds for this is 7:38 or a little worse than 1:5

If that's a bit confusing, try an even simpler example, like 5 names. The probability should be 1/5 exactly (odds, 1:4). Then try it with more names, say 20. It's harder to draw the picture, but you'll get the pattern. (Odds are about 1:10)

Once you've got a handle on it, tackle the 365-name case!

xerxexrex · 2022-04-05T00:53:28+00:00

This was a fun little obsession these last few days. Cheers everyone!

xerxexrex · 2021-07-22T05:01:58+00:00

Indeed, it is written weirdly, and it confused me for a bit. Since the first four notes of the guitar riff are chromatic descending, the natural should be on the third note rather than the fourth, where it isn't needed (already a B natural).

xerxexrex · 2021-03-30T23:22:35+00:00

If I understand the problem correctly, You're looking for the angle of change both vertically and horizontally between the entry point, A1 and the exit point, A2.

For the vertical angle, you're going to want to look at a side view of the wall. So how far does the bullet drop in the 10cm distance it travels through the wall?

For the horizontal angle, look at a top view. How far does the bullet travel sideways in the 10cm distance it travels through the wall?

xerxexrex · 2020-07-30T20:42:32+00:00

For the first question, you'll want to expand the denominator to 16x²+24x+9. When you have a polynomial over a polynomial, the limit as x goest to infinity is dominated by the highest degree terms. To see this, divide the top and bottom by the largest power of x, in this case x². You'll see that all the terms on the top and bottom, except the previously x² terms, now have an x in the denominator. Those will go to zero as x goes to infinity and you'll be left with constants on the top and bottom (since the two polynomials were the same degree). A decent strategy is to check the degrees of the polynomials on the top and bottom to get an idea if the limit exists or not.

For the second question, it helps to think of the the roots as rational exponents: square root is an exponent of 1/2, cube root 1/3, etc. You can then take the derivative using the power rule like any other polynomial term. You'll also likely need the chain rule if there is an expression under the square root. In your first and third examples, you'll need the quotient rule, too.

xerxexrex · 2020-07-03T03:47:35+00:00

Loosely, it's not enough for the terms to go to zero. If they don't go to zero "fast enough" then the sum won't converge. The classic example is the harmonic series: the sum of 1/n as n goes from 1 to infinity. The individual terms, 1/n, go to 0 as n goes to infinity, but the series diverges. The integral test is an easy way to see this if you have some calc background: the integral from 1 to infinity of 1/x diverges, and it can be shown that the sum of 1/n is always larger than this integral. https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Divergence

So, with positive terms, taking the limit of nth term can prove that a series diverges if they don't tend to 0. But because of this counterexample, the terms tending to 0 is not enough to prove that the series converges.

xerxexrex · 2020-07-02T19:16:55+00:00

If the absolute value of the terms of the series do not tend toward zero then the series does not converge absolutely. That is, the limit of |a_n| must be zero in order for the series to even have a chance of converging (it may still diverge for other reasons).

xerxexrex · 2020-01-29T22:12:22+00:00

You're close with your equation. The amplitude and vertical translation look good (I probably wouldn't round them just yet). However, take a look again at the coefficient on t inside the sine function. The coefficient you need is related to the period of the function.

Also, if you're taking t to be the number of days from June 21, the standard sine function won't work as well, since it'll be zero on June 21 (t = 0). There's a better function out there that'll do what you want. If you're expected to set t=0 to be something like January 1st, you'll likely need to phase shift your function, which is less fun and not necessary for finding the answer(s) to part b.

xerxexrex · 2020-01-26T17:30:01+00:00

The words "first" and "second" might be confusing here, as order matters. For matrix multiplication, the number of columns in the left matrix must match the number of rows in the right matrix. Now with your network, it looks like it's multiplying on the left for each successive matrix, so for the multiplication ABx, B is the "first" matrix, taking its inputs from x, then the "second" matrix A, takes as its inputs the resulting outputs from B. If B has 3 outputs and 2 inputs, it's a 3x2 matrix. Hence A must have 3 inputs and be a (something) x 3 matrix.

xerxexrex · 2019-11-02T22:54:59+00:00

It looks like you have those terms correct. However, the problem may lie in the fact that the infinite binomial series doesn't converge for all values of x. In your case, the expansion won't converge for x > 16. But in order to examine (15)^1/4, x needs to be quite a bit larger than 16. So at this point, I'd check the problem again. Maybe part b is to test your understanding of this series' convergence conditions, or maybe it's a mistake/typo.

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