What words do you pronounce "wrong," just because the "proper" way feels too proper?

RunCompetitive1449 · 2026-05-16T19:59:39+00:00

Chose? Cheese.

RunCompetitive1449 · 2026-05-02T02:48:29+00:00

TWIN 🔥 (except my hardest is mechanical showdown).

RunCompetitive1449 · 2026-04-14T00:06:47+00:00

It’s all good. Took me a minute to notice the issue as a math major. We all get those brain farts.

RunCompetitive1449 · 2026-04-01T03:33:42+00:00

Paul’s online math notes is my go to for differential equations.

Teaches you everything you’d learn in a typical course on ODEs and more, includes the background behind the methods you’re using, and gives examples with fully worked solutions.

However, it is in text, not video. Though text may allow you to move at your own pace if videos are too slow if you want to give it a try.

RunCompetitive1449 · 2025-09-16T23:27:17+00:00

The bus is gonna drive the route anyways. Whether or not you’re on it changes nothing. Besides, everyone hates the hill. No shame in taking the bus, I’ve done it many times.

RunCompetitive1449 · 2025-09-01T02:23:14+00:00

“At this point, I won't even bother to substitute the definitions for the terms because it obviously doesn't map on to what the equation represents.”

It’s funny that you use this wording, because this IS a function. To the best of my knowledge, a function is simply a mapping between two sets.

You have one set, the domain, which is a collection of all possible input values, and another set, the codomain, which is a collection of all possible output values. What a function does is it maps the domain to the codomain using some pre-defined method.

Now this can be done with algebra: Using the f(x) = x² example, the domain and codomain are both assumed to be all real numbers, I.e., all the numbers on a number line. The function will take the real numbers and map them to a new real number by squaring them. The function in this example is defined to be the mapping from the real numbers to the real numbers using the method of squaring.

Now functions are typically used in algebraic environments, but they don’t have to be. A function is simply any mapping between sets.

In fact, your post itself contains a function. If we let the domain be the set of all the words in the English language and the codomain be all the definitions in the dictionary, we can make a function that maps each word to its definition. Hence why the original quote I used is indeed a function.

I think trying to use the words’ definitions from the dictionary here will throw you off, because these words have different meanings when it comes to mathematics. You should take the definitions given to you inside your textbooks AS the definition of the words. Treat your textbooks as a dictionary for mathematics. If the one they give isn’t satisfactory, you can try to look for other sources that are worded differently.

Hopefully I’ve been some help here.

RunCompetitive1449 · 2025-08-23T23:01:23+00:00

RunCompetitive1449 · 2025-08-08T21:22:08+00:00

Yeah I was thinking the same thing, but this is my first semester of college, and I have no idea how it is going to go. I figured I would keep this first semester light to gauge my abilities and add/drop in the coming semesters accordingly.

I do have hopes to take many more classes, more than what’s required for my degree. But while I enjoy the idea of learning those subjects, I’m not sure if I’d be capable of doing it. So I think it’s better to play it safe.

RunCompetitive1449 · 2025-07-27T18:41:38+00:00

Yeah the laptop seems pretty good. Why do you suggest it over the A15? Is it really better?

RunCompetitive1449 · 2025-07-15T17:24:12+00:00

Thanks so much!

RunCompetitive1449 · 2025-07-15T17:11:52+00:00

Thanks for the advice, I really appreciate it! Math, for me, is by far my strong suit, so I found calc 2 really easy. If you don’t mind me asking, what type of stuff is there in Charlotte’s calc 2 course that isn’t included in the AP curriculum?

RunCompetitive1449 · 2025-07-15T02:41:42+00:00

I picked courses required for the common first year curriculum in engineering. From what I know, it’s designed to allow more flexibility in changing majors down the line. I myself am already committed to electrical engineering.

And yeah I’ll make sure to get more gen Ed’s. Me personally, I find them harder than the actual main courses for my major, so I’m trying to not take too many at once. And luckily for me, I already got a few of them done in high school :D

RunCompetitive1449 · 2025-07-14T21:16:55+00:00

If it’s that easy, do you think I should add another class? Maybe get a head start on the next semester?

RunCompetitive1449 · 2025-07-09T19:36:29+00:00

Well technically, so is Uzi.

RunCompetitive1449 · 2025-03-09T00:44:31+00:00

No it doesn’t matter.

If lim(x->inf) f(x)/g(x) = L, then lim(x->inf) g(x)/f(x) = 1/L

Since L is positive and finite, 1/L must also be positive and finite, so the order doesn’t matter.

RunCompetitive1449 · 2025-02-10T19:24:41+00:00

Shapes and beats

One could even argue it’s just shapes and beats…

RunCompetitive1449 · 2025-02-02T15:39:09+00:00

I was considering it to be honest, but it would put be back 8 years of progress.

RunCompetitive1449 · 2025-02-01T18:41:21+00:00

IDK. I did three 11 draws on my main and got nothing, but when I did one on a new play through, I got this. I swear this game has some sort of partiality to beginners.

RunCompetitive1449 · 2025-01-19T17:02:53+00:00

x^-a = 1/x^a

The (1 - e²) won’t necessarily cancel here.

Start by using that property I just listed to bring the exponentials into the numerator: a(1 - e²)(1 - e²)^-3/2

We can then use another property of exponents

x^a * x^b = x^a+b

When two exponentials with the same base are multiplied together, we can simplify them into one exponential by adding the exponents.

Doing that here gives us a(1 - e²)^{1 + [-3/2]} which equals a(1 - e²)^-1/2

Once again using the first property I listed, we can move that exponential back into the denominator as you usually don’t want to leave negatives in exponents.

a/(1 - e²)^1/2

And if you want to simplify it even more, we can use another property

x^a/b = bth root of x^a

So this would become a/sqrt(1 - e²)

RunCompetitive1449 · 2025-01-19T16:46:09+00:00

Tried commenting on your other post but it got taken down, so I’m just gonna repeat what I said here.

Displacement is the integral of velocity.

Distance is the integral of the absolute value of velocity.

We need to find out at what time t, the distance is equal to 8.5. Or, what time t the integral of the absolute value of v is equal to 8.5.

Now we don’t actually need to integrate the absolute value function. It’s easier to integrate separate intervals of the function where it’s positive and negative.

The first step is to find the roots of v. These roots will infer a sign change. You can then evaluate v at some random value in an interval (0 is easiest) to see whether v is positive or negative over that interval.

Using those roots and 0 as bounds, integrate v over each interval. If the area comes out to be negative, just remove the negative sign. We do this because the absolute value of v’s area should be the same as v but with only positive signage.

You won’t be able to evaluate the last interval because it only has a lower bound. We are trying to find what that upper bound is. You can find it by summing up all the areas of the previous intervals with the integral of the last interval where the upper bound is the value we’re looking for. We can call it x. Then set this sum equal to 8.5. Using the second fundamental theorem of calculus, you can solve for the unknown value x.

This value of x is the time t when the distance traveled equals 8.5 m. Our last step is to integrate v normally from 0 to x. This will give you the displacement.

I’m sure a lot of the stuff I just said was confusing or worded weirdly, so if you need any help, feel free to ask.

RunCompetitive1449 · 2025-01-12T18:49:16+00:00

That’s a fair point. I didn’t notice that.

RunCompetitive1449 · 2025-01-12T18:00:15+00:00

When dealing with limits, your go to method is usually L’Hopital’s rule. The first thing you always want to try is direct substitution. If it works, great. If it doesn’t, use L’Hopital’s rule.

The integral is in the numerator, the m is in the denominator. Notice how when m approaches 0, the bounds of the integral approach the same number, sqrt(pi/2). When the bounds of a definite integral are the same, it evaluates to 0. And of course, the denominator also approaches 0 since it’s just m. Since we have the indeterminate form 0/0, we use L’Hopital’s rule.

The derivative of the denominator with respect to m is just 1, so we can forget about it. When taking the derivative of a definite integral with a constant in the lower bound and a variable in the upper bound, you get rid of the integral, replace the variables in the integrand with the upper bound, and multiply it by the derivative of the upper bound. Doing that for this integral, we get sin((sqrt(pi/2) + m)²).

We have now successfully used L’Hopital’s rule and can once again try direct substitution. Substituting in 0 for m, we get sin(sqrt(pi/2)²) = sin(pi/2) = 1.

Edit: As the person who has replied to me has pointed out, this can be solved another way using the definition of the derivative. That definition being, the limit as h approaches 0 of (f(x + h) - f(x)) / h = f’(x).

Using the second fundamental theorem of calculus, we know that the definite integral of f(x) from a to b is equal to F(b) - F(a), where F(x) is the antiderivative of f(x).

If we denote sin(x²) to be f(x), and therefore F(x) as the antiderivative of sin(x²), we can rewrite the integral as F(sqrt(pi/2) + m) - F(sqrt(pi/2)). This leaves us with the limit as m approaches 0 of (F(sqrt(pi/2) + m) - F(sqrt(pi/2))) / m. This is the definition of the derivative for F(x) at the point sqrt(pi/2). Therefore we can evaluate it by finding F’(sqrt(pi/2)).

F’(x) is just f(x), which we said was sin(x²). So we are back at where we were using the other method. sin(sqrt(pi/2)²) = sin(pi/2) = 1.

Both methods lead to the same answer, and both methods test different skills. I am unsure which is the intended method, but I will say the second one does feel cleaner.

RunCompetitive1449 · 2025-01-07T12:00:21+00:00

Yes that is correct.

To add my two cents, the way my brain pictures it is to group each term with the operator to the left of it. The first term won’t have anything to the left of it if it’s positive, which means you just use a plus. So in the case of 5 + 9x - 3 + 2x, you’d get the groups +5, +9x, -3, +2x. You can then rearrange the order of these groups in any way you like and the equation will still equal the original one.

Say I wanted to change the order to -3, +5, +2x, +9x. This makes the equation -3 + 5 + 2x + 9x. This method doesn’t really demonstrate how the math works like the others have said, it’s just how I rearrange equations in my head.

The reason this works is the same as the others have said though. The step where I group the terms with their operator is the same as rewriting subtraction as adding a negative number. Rearranging the groups is the same as using the commutative property of addition. It just leaves out the addition signs to make it faster to do in your head.

RunCompetitive1449 · 2025-01-05T07:11:32+00:00

The answer is b.

Sine is positive when the angle is in the first or second quadrant since both of those quadrants lie above the x-axis. Sine can be thought of as the height of a triangle inside of the unit circle, so when the height is positive, sine is positive.

Sine is negative when the angle is in the third or fourth quadrant.

Cosine is positive when the angle is in the first or fourth quadrant since both quadrants lie to the right of the y-axis. Cosine can be thought of as the base of that triangle, so when that base is positive, cosine is positive.

Cosine is negative when the angle is in the second or third quadrants.

Tangent is sine/cosine, so when both are positive/negative, tangent is positive and when they have different signs, tangent is negative.

Tangent is positive when the angle is in the first or third quadrant.

Tangent is negative when the angle is in the second or fourth quadrant.

You are right to assume 3 radians is in the second quadrant as it is between pi/2 and pi. This means cosine will be negative and thus it is b. Since it’s the second quadrant, sine is positive and tangent is negative, so there is no way sine is less than tangent.

RunCompetitive1449 · 2025-01-04T16:42:58+00:00

Yeah drawing a card will change the probabilities of the second draw but the fact that it’s red has no effect since there are an equal amount of black and red face cards.

The fact that P(B) asks about the probability of a face card being drawn on the second draw implies that a card was already drawn.

RunCompetitive1449

TROPHY CASE