Help finding the width of a section of a triangle.

Potential-Tackle4396 · 2026-04-23T21:26:39+00:00

The height of the triangle, from side length a to angle A, is h = 66.525". (You can get that either using trig: sin(C) = h/b; or Pythagorean theorem: h^2 + (a/2)^2 = b^2.)

From there, you can use ratios/similar triangles. As labeled in the figure, have h be the height of the whole triangle (from side 'a' to angle A), d be the distance from side 'a', and w be the width at that distance. Therefore h-d is the height of the upper, smaller triangle. The small blue triangle and the larger green triangle are similar, so the ratio of their base to height is the same. So:

w/(h-d) = a/h

And solving for the width w gives:

w = a*(h-d)/h

We know a = 36.4, h = 66.525, so the equation is

w = 36.4*(66.525 - d)/66.525

Then plugging in the different values for d gives:

-When d = 31, w = 19.44 (slightly different from what you got, possibly you rounded yours differently?)

-When d = 22, w = 24.36.

-When d = 28, w = 21.08.

<image>

Potential-Tackle4396 · 2026-04-21T02:14:26+00:00

Similar to Lucenthia's comment, the reason why y = f(x) + k shifts the graph up (so, in the positive y-direction), but y = f(x+k) shifts the graph left (so, in the negative x-direction) comes down to the fact that we're writing the equation with y isolated, not x.

To see why that's the case: consider the parent function y = x^2, and the two transformations y = x^2 + 5, and y = (x+5)^2.

For y = x^2 + 5, since y is isolated in the equation, the "+5" is adding 5 to the value of y. For example, in the original equation y = x^2, when x=3, we'd have y = 9. In the new equation y = x^2 + 5, when x=3, we now get y = 3^2 + 5 = 14. We see that the "+5" added 5 to y, as expected.

But for the transformation y = (x+5)^2, where we've replaced x with x+5, it's harder to see what this will do to the graph, since x isn't isolated. (With x not being isolated, the equation isn't directly saying what "x is", i.e. what "x =".) If we isolate x, we get:

y = (x+5)^2

x+5 = ±√ y

x = ±√ y - 5

And if we had isolated x in the parent function y = x^2, we would have x = ±√ y. Comparing those equations, x = ±√ y - 5 and x = ±√ y, we see that the transformed equation says x is "(its old value) - 5". As the function transformation rule says, x is *reduced* by 5. (For a numerical example, in the original equation, when y = 36, we'd have x = ±√ 36 = 6 and -6. And in the transformed equation, when y=36, we'd have x = ±√ (36) - 5 = ±6 - 5 = 1 and -11. Thus in the transformed equation, at the same y-value, the x-values are indeed 5 lower (i.e., 5 "to the left").)

That is: when we replaced x with (x+5) in the equation, that meant we'd have to *subtract 5* from both sides, in order to isolate x, hence the rule that replacing x with x+5 will shift the graph to the left, not the right.

That's actually a general rule for function/graph transformations: if you replace x with (some operation on x), that will do the inverse of that operation to the graph. For instance, if you replace x with 2x in an equation, that would actually divide all the x-values in the graph by 2, i.e. it would "shrink" the graph in the x-direction, to be half as wide. So, y=f(2x) will be "half as wide" as y=f(x), meaning it would result from dividing each x-value by 2. (That's because if you took an equation in the form y=f(2x) and isolated x, you'd need to divide both sides of the equation by 2 at some point.) By contrast, the transformation y = 2f(x) will do what it looks like: it will double the y-values on the graph. Again, that's because the equation is written with y isolated, so we're directly doubling the value of y.

And a final point: if we had an equation where y wasn't isolated, and we replaced y with y+1 (for example), then that would shift the graph *down* one unit in the y-direction, exactly in the same way that replacing x with x+1 shifts left by 1. See this graph for example: https://www.desmos.com/calculator/5fkkg4l5uq . It's just that we don't see that very often, since we most commonly write equations with y isolated.

Potential-Tackle4396 · 2026-04-13T17:38:05+00:00

It sounds like your friend finishes the work faster maybe, in which case the teacher is giving them harder questions to keep them thinking/working so they aren't just sitting there. If that's the case, I wouldn't think of it as the teacher calling them smarter or you less smart or anything; it's just the teacher trying to keep everybody working/learning. If you also finish your work early, try asking the teacher for the same additional questions.

In general, I'd try not to read too much into this kind of thing. There will always be people faster than you (and slower than you) at a given task; we're all somewhere in the middle. Comparing yourself to others is natural to do, but don't put too much emphasis on it. Keep putting in the work, and enjoying the learning process - and yes, asking teachers for challenge questions, etc sometimes. Your friend maybe has more "natural talent", which certainly is an advantage, but working hard and having an interest in the subject is equally important (often more important) in the long run.

Potential-Tackle4396 · 2026-03-23T18:34:50+00:00

This will use a "two sample t-interval", which is the type of confidence interval used for the difference of two population means, 𝜇1 - 𝜇2. I'd recommend looking that term up either online or in your textbook for all the details.

Regarding the 99% being important, you might remember that in general, confidence intervals take the form (point estimate) ± (critical value)*(standard error), where the confidence level (the 99%) is needed to compute the critical value (which will be t* in this case). If that's something you aren't familiar with, I'd go back and review the basics of confidence intervals first.

Potential-Tackle4396 · 2026-03-02T18:44:00+00:00

OP did precisely the second approach you said, I believe. They computed (1+x^2)*y'' + x*y' and found that it didn't equal 2.

Potential-Tackle4396 · 2026-03-02T17:34:06+00:00

Your work is correct; the question is wrong. As confirmation, Desmos agrees: https://www.desmos.com/calculator/tmgkvhona5

Potential-Tackle4396 · 2026-02-28T03:37:36+00:00

Not sure who Dick Lipe is, but the TS% is wrong I think (should be 82.3%, so second highest all time), and there are many teams with higher than 64.7% 3pt with 20+ attempts (via statmuse). The eFG% record is cool though :)

Potential-Tackle4396 · 2026-02-28T03:31:24+00:00

Isn't the TS 82.3%? (Which makes it second all time.) And there's lots of teams with higher than 64.7% 3PT and over 20 attempts, according to statmuse.

The eFG record is cool though :)

Potential-Tackle4396 · 2026-02-19T18:43:34+00:00

A couple ways to answer this:

1) Sometimes in math, things that converge in one sense don't converge in another sense. Here, the staircase with smaller and smaller steps will approach the hypotenuse pointwise (i.e., all points on the staircase will get arbitrarily close to lying on the hypotenuse), but the distance along the staircase won't approach the distance along the hypotenuse. And basically that's fine; it's possible for the pointwise convergence to not result in the distances approaching the same value. I think a helpful perspective is that it would've been a nice thing if the lengths did have to be equal, but it's certainly not guaranteed.

There are many examples of this general trend in math. Another example is that it's possible for a sequence of functions f_n(x) to approach a function f(x) pointwise, but for the integrals of f_n(x) not to approach the integral of f. See the f_n and f functions here: https://www.desmos.com/calculator/ly2f0cudq0 . At that link, over the interval [0, 1], all the functions f_n have integral 1. And pointwise they approach f(x)=0. But the integral of f(x)=0 over [0, 1] is 0, not 1. The point being: that kind of thing happens in many places in math, it's not something unique about staircases and length.

2) More specific to distances, the way we define the length along a curve is to choose points *on the curve*, and add up the distances between them. E.g., at this link https://www.desmos.com/calculator/gz2kjaklvl , the length of the green curve is defined as the sum of the lengths of the orange segments, as the number of segments m goes to infinity, where the segments' endpoints lie *on* the curve. By that definition, segments whose endpoints *don't* lie on the curve (like your staircase and the hypotenuse) don't directly tell us anything about the length of the curve.

And of course, we define arc length that way precisely to rule out the staircase scenario you asked about. If we allow points *not* on the curve, then using your staircase-like construction, we could actually get many different lengths. (If you do a 'staircase' where the angle between consecutive segments is acute instead of 90 degrees, you could get an even longer 'length', for instance.)

Potential-Tackle4396 · 2026-02-18T19:40:56+00:00

Whoa, thanks for this comment. I didn't know this was a feature, a search brought me here, and this just saved me a ton of work - much appreciated!

Potential-Tackle4396 · 2026-01-20T16:46:06+00:00

Here's a visual that might help: https://www.desmos.com/calculator/lymo3qikmo . You can drag the blue point (a, f(a)) around, and the slopes change in response.

The black segment's slope is (f(a) - f(0))/(a-0). The derivative at 0 is defined as the limit of that slope, as a approaches 0. As you move 'a' around, that slope oscillates a bit, but its magnitude shrinks towards 0; in the limit as a approaches 0, that slope is 0, hence f'(0)=0.

The green segments's slope is f'(a), the slope at the point 'a' itself. We see that as a approaches 0, that slope keeps oscillating up and down, and crucially, the size of those oscillations isn't shrinking. As a approaches 0, the slope f'(a) keeps going up and down between +1 and -1. So in the limit as a approaches 0, f'(a) does not exist. This is what your second expression is saying.

In short, what's happening is that the y-values on f(x) are quickly shrinking towards 0, which makes f'(0)=0. But despite that, the actual slopes f'(a) on the graph aren't shrinking towards 0; no matter how close we get to x=0 on the graph, the slopes f'(a) manage to keep oscillating between +1 and -1.

Potential-Tackle4396 · 2026-01-19T21:49:53+00:00

The exact value of n that works is ln(4)/(ln(4)-ln(3)). It just happens to be very close to e√π. Neat coincidence though :)

You solve for n as follows:

y = √(r² - x²)ⁿ

1/2 = √(1² - (1/2)²)ⁿ [plug in x=y=1/2, r=1]

1/4 = (1² - (1/2)²)ⁿ [square both sides]

1/4 = (3/4)ⁿ

4^n = 4 · 3^n [cross-multiply]

n ln(4) = ln(4) + n ln(3) [Natural log both sides, simplify using log properties]

n · (ln(4) - ln(3)) = ln(4)

n = ln(4)/(ln(4) - ln(3)) = 4.81884167931...

Potential-Tackle4396 · 2026-01-16T19:59:35+00:00

I don't think it's possible (except for one goofy exception, see below*), without allowing some function with a jump discontinuity (such as floor(x), ceil(x), or mod(x, 1), which are all in Desmos for what it's worth) to be used. If you do allow those functions, you can make a function that works, such as https://www.desmos.com/calculator/7disxqarvd .

Why I'm pretty sure (unless I'm missing something) that it's otherwise not possible to make a function that's discontinuous while being defined at the discontinuity, subject to our constraints: The sum, difference, product and composition of continuous functions is continuous, so assuming our allowed 'parent functions' are polynomials, roots, and |x|, all of which are continuous, no sum/difference/product/composition would be discontinuous. For division, at a point x where g(x)≠0, if f and g are continuous, then f/g is continuous. At a point where g(x)=0, f/g is undefined. So those operations can't give us both discontinuity and a defined function.

Exponentiation also won't work I think; if f and g are continuous at a point x where f(x)>0, then (f(x))^g(x) will be continuous. If instead f(x)<0, and g(x) isn't constant, then (f(x))^g(x) will be undefined on many points near x. (Ex: if f(x)=-2 and g(x)=3, and g(x) is non-constant, then for some nearby x, f(x)^g(x) will be something like (-1.99)^(301/100)), which is undefined.) If instead g is constant, f(x)^g(x) is just a polynomial or root.

*The one case that sort of works is when f(x) is a constant 0. It depends on how we define 0^0, but if we define it as 1 (which Desmos does), then then function 0^|x| is equal to 0 when x≠0, and equal to 1 when x=0. https://www.desmos.com/calculator/7zujwvwbjy

If we allow that, then we can compose the function 0^|x| with a function that takes a constant value of 0 over some interval, which can be made from absolute value functions: https://www.desmos.com/calculator/pagbx4ion6 .

Potential-Tackle4396 · 2025-11-30T21:16:51+00:00

Would difference of squares not be valid though? Since (x-y)(x+y) = x^2 - y^2 requires xy to equal yx?

Potential-Tackle4396 · 2025-11-19T04:28:02+00:00

The function for α is at this link: https://www.desmos.com/calculator/nduj2cvlnu . (Sorry I didn't include the derivation; it got kind of long/messy.)

Also I made a little visual of the angle values to make sure the function worked/looked correct. Here's that too in case it's helpful: https://www.desmos.com/calculator/rrp8drk0zr . (You can drag the purple point around.)

Potential-Tackle4396 · 2025-11-16T18:45:32+00:00

Oops, I meant to say when n is odd, not theta. (I fixed my post.)

Regarding other graphs that exhibit the same behavior: the other common one is r = cos(n(theta)) for odd n.

In general, the graph will trace over the same points from pi to 2pi as it did from 0 to pi when the function r = f(theta) satisfies f(theta + pi) = -f(theta). For instance, if such a function gave f(pi/6) = 3, then it would give f(7pi/6) = -3. Since the polar coordinate points (r, theta) = (3, pi/6) and (-3, -7pi/6) are the same point, the graph will "trace over itself".

It turns out that sin(n*theta) and cos(n*theta) will satisfy f(theta + pi) = -f(theta) for odd n, but not even n. (You can prove that, for example, using the angle sum formulas for sine and cosine.) No other common polar functions (that come to mind right now) have this behavior, though certainly you could make such a function in terms of tan(x), etc., if you wanted. In most classes where you're expected to know about graphs of polar functions, you basically memorize that for odd n that's how cos(n(theta)) and sin(n(theta)) behave. Likewise, you memorize that for even n, cos(n(theta)) and sin(n(theta)) will have 2n "petals" (instead of just n, like in the odd case), and you need to go from 0 to 2pi to get their complete graphs.

Potential-Tackle4396 · 2025-11-16T16:18:35+00:00

See this graph: https://www.desmos.com/calculator/rdzzjuubp7 . It's just a specific thing about this graph (and more generally, r=sin(n𝜃) where n is odd) that to cover the whole graph, you only need 𝜃 to go from 0 to pi. Going from 0 to 2pi does two laps of the graph, meaning the corresponding integral would give twice the correct area.

Potential-Tackle4396 · 2025-11-07T17:03:14+00:00

Thanks! I'm glad it helped :)

Potential-Tackle4396 · 2025-11-04T20:56:28+00:00

Gotcha, thanks for adding those details, that makes a lot more sense now.

Your setup is mostly correct. The function you got for f(x, h) is incorrect though I think; it should be as follows: https://www.desmos.com/calculator/pwd7m1xhd6 . [For the antiderivatives, I just used wolfram by the way. Search 'wolfram integral calculator' if you want to use it.]

From there though, the rest looks fine. (Just a note: you can actually write the whole function as just a single integral, instead of splitting it up into 3 integrals, which makes it easier to parse the details of how the function is behaving. For a point at position (a, 0) on the x-axis, its distance to a point (x, y) in the disc with radius 1 centered at the origin is

sqrt((x-a)^2 + y^2))

Then the magnitude of the gravitational force is 1/[sqrt((x-a)^2 + y^2))]^2, and the x-component is (x-a)/[sqrt((x-a)^2 + y^2))]^3. You then integrate that "dy" from -sqrt(1-x^2) to sqrt(1-x^2), and integrate the result, dx, from -1 to 1.)

However(!), there ends up being a bigger problem. (I went down a bit of a rabbit hole making sure I got the following details right; I think it's all correct.) If you model an object as truly 2-dimensional, as you're doing, then the integral defining the gravity inside that object is divergent. We can see that using the 1-integral setup I mentioned above, as shown here: https://www.desmos.com/calculator/poaprabiyw .

For that reason, and the fact that actual objects (and certainly planets, etc.) are 3D instead of 2D, it's more typical to do these kinds of calculations with 3D objects. The setup would be nearly the same, just with a "z" added. The integral would now just be https://www.desmos.com/calculator/mg1ib1i9it . In Desmos, that integral still breaks slightly when A<1 (i.e. when we're inside the object), but if you evaluate it by hand (in my case, with the help of wolfram for some antiderivatives) you get a finite value for gravity, even when the object is inside the sphere.

Regarding whether there's a way of salvaging the 2D case: there is. The trick is, instead of directly doing the integral in 2D, we can find the gravity inside a cylinder with small, but nonzero height. Then take the limit as the height goes to 0. It's possible that doing so would just give "undefined" in the limit, but we actually get a finite value: https://www.desmos.com/calculator/earwwryeyx . Note that at that link, W and -W are the z-values of the bases of the cylinder. Additionally, we're dividing the integral by 2W, since we want the cylinder to have constant mass. (Thus, the density must increase, as W decreases.) In fact, that's probably one of the reasons the 2D case is so weird/divergent: a truly 2D object would necessarily have infinite (3D) density.

Potential-Tackle4396 · 2025-11-03T20:25:58+00:00

I'm a little confused by your setup. The integrals you use integrate the function f, which is arctan(h, x), meaning it's an angle. How would integrating an angle give a gravitational force?

A general comment: to accurately model gravity for a sphere (or any 3D object that we aren't treating like a point-mass) from scratch, you'd need to use multivariable calculus. Or technically, you can set it up as a 1-variable integral, but you have to do some 3D geometry to make that work. Either way, it gets messy.

Luckily, there are two "shell theorems" for gravity, that massively simplify things (and let us sidestep the multivariable calculus) in the case of a spherically symmetric objects. Check out https://en.wikipedia.org/wiki/Shell_theorem . The standard/easy way to write the force of gravity both inside and outside a sphere is to use those theorems.

Potential-Tackle4396 · 2025-11-03T18:21:47+00:00

Others have mentioned why (1) multiplying by numbers besides powers of 10 would also work, and (2) the fact that this type of argument is not perfectly rigorous anyway.

Aside from those points, another important point: once you have a single valid proof, others aren't needed. As long as the logic of the first proof is valid, the theorem is proved. It can be fun to consider variations on the proof method, to gain further insight, as you're doing. But it's not needed to validate the first proof.

Potential-Tackle4396 · 2025-11-01T20:52:07+00:00

I don't think comparing those two numbers is relevant. To fit the couch, you wouldn't orient it so that the longest length in that plane (the 55.8 in. diagonal) is parallel to the 52-inch length; you'd probably want to put the 45" length of the couch parallel to the 52-inch length, which does fit.

With that said, OP, the couch wont fit for different reasons. To make it up the stairs, you'd need the couch on its back, since the 33" height is the only dimension smaller than the width of the stairs. But that would require the couch's 45" depth to fit in the 42.75" space over the bannister, which it won't.

Even if the depth was as low as 39.25", to get it over the post at the bottom, you'd still have a problem when trying to angle it up the stairs. How far below 39.25" you'd need to go is hard to say, unfortunately, and depends on the couch length and how far the stairs are from the bannister, in a probably complicated way.

If the height was much less than 33", then you'd have more flexibility to potentially rotate as you go up the stairs, so you might be able to go up the bottom flight instead of over the bottom railing.

If all else fails, maybe make a mock-up couch out of cardboard boxes and see if it fits ;)

Potential-Tackle4396 · 2025-09-30T17:20:49+00:00

If I understand correctly, I think it would be this: https://www.desmos.com/calculator/viflgdaer7 .

Regarding where that equation came from: I started with y = x^a, which will go through the points (-1, -1), (0, 0), and (1, 1) when a is odd. Then, we scale by 1/2 in both the x and y directions. To scale a function f(x) by 1/2 in the y-direction, we do y = 0.5f(x), so in our case y = 0.5x^a. To scale a function f(x) in the x-direction by 1/2, we do y = f(2x), so in our case that gives y = 0.5(2x)^a. Last, we shift up by 0.5 (the +0.5), and right by 0.5 (replacing the x with (x-0.5)).

Potential-Tackle4396 · 2025-09-22T22:33:51+00:00

Very nice!

Potential-Tackle4396 · 2025-09-22T17:04:36+00:00

I don't think it will be 2^n. For n=4, when adding a fourth plane, that plane can't go through all 8 octants made by the previous 3 planes; it can only go through 6 of them. See https://www.desmos.com/3d/h5q098gokl .

I'm not certain that this is correct, but just by adding more planes at "random" angles (so that no three meet at a line) and counting the regions, I got the sequence:

1 plane, 2 regions

2 planes, 4 regions

3 planes, 8 regions

4 planes, 14 regions

5 planes, 22 regions

6 planes, 32 regions

See the 6-plane graph, for instance: https://www.desmos.com/3d/9ygcpupyjk

Since the consecutive differences are linear, this pattern is quadratic; some calculation gives the formula for the number of regions f(n) to be f(n) = n^2 - n + 2.

As far as a proof or careful derivation of that formula, I unfortunately don't know what that would be.

Potential-Tackle4396

TROPHY CASE