1. I feel like he’s talked about it plenty actually. He doesn’t seem interested in playing it but I remember he’s said he would be a very good player if he picked up (Not surprising he’d say that but he’s probably right)

  2. Yes, sin(0) is 1, but that's not a function, just a single value. In the same way x2 is a function but 32 is just a fancy way to say "9".

  3. So I'm still a little confused on that one because we definitely didn't learn that in class so I'm going to talk to my teacher tomorrow morning.

  4. Add 27 to both sides then take the cube root of both sides

  5. I looked through some of these comments and wow I thought my diagnosis was young! I was diagnosed at 19 and 22 now, and I actually just recently realized what most likely causes it! I got mono like a month before my symptoms started and I found out the Epstein-Barr virus that causes mono makes you more susceptible to like 7 different autoimmune diseases once you get it! And it was a double eff-you cause I don't even know how I got mono in the first place! Thankfully I've been on Orencia infusions for a year and a half now and I'm happy to say I'm in remission!

  6. That’s so great to hear! I’m happy for you

  7. First of all, I don’t think you got the bounds correct. The problem says 0 and 6 but you wrote 0 and 4. Which is fine but if you’re gonna do that you have to add a second integral to it that goes from 4 to 6. But the mistake in the actual evaluation was when you tried to take the antiderivative of (2x)1/3 . Two issues. It’s supposed to be divided by 2 and there shouldn’t be an extra x. I’ll show you with a u substitution. Let u = 2x. du = 2dx. du/2 = dx. So we now get

  8. If you have something like √(2x), or sin(2x), then yeah, you can. You don’t need to if you have enough intuition and a good enough understanding of the chain rule, but if you’re learning integrals for the first time then yes, it’s a good idea.

  9. Oh, you multiplied the numerators by 60? Hum... so you gotta multiply the numerators by the same value? Is that it?

  10. For this case, yes. It works out nicely because 20 is a multiple of 60. So when you multiply it to the numerator, they’ll end up canceling out pretty well with the denominator

  11. For integrals, it’s pretty important to have a decent understanding of derivatives. I’ll try to explain them assuming you have that knowledge. If not, then I highly recommend reviewing derivatives and then going back to integrals after that

  12. No, it isn't. The bounds are right, but why did you change 9?

  13. Op is in AP calculus AB which is basically calc 1

  14. Taking the square root won’t work here. What you probably meant is square both sides

  15. Squaring both sides is a good start. You just forgot how to square a binomial.

  16. I’d argue that it’s better to add sqrt(k) to both sides and then square both sides. That way you don’t have to square a binomial with two square roots

  17. Have you ever heard of Kiriko? She has this ability called cleanse that basically saves roadhog from nade or any other cc or condition

  18. one shot from range! If you’re standing center in hook range of the largest player in the game, yes.

  19. It’s two shot now but before yesterday it was not. It was really difficult to not get one shot as a squishy before the path even from range.

  20. You are given f(x). So, can you find f(6) and f(6+h)? If so, find them and just substitute into the limit given. Then, eliminate h from the denominator and finally, substitute the limit into whatever is left of the expression!

  21. I don’t really think that’s what they’re asking. OP’s title says that they’re confused on the concept of limits and limit of a difference quotient itself, not how to do it

  22. aha.. I not sure how to explain it. Basically I'll got a task same as this and I need to explain it in class. However i not sure how to start explaining to others.

  23. Can you show the full directions from the top of the picture? It would help us understand what we need to do

  24. it mentions this " for the following piecewise function, determine whether f(x) is continuous at x = 2"

  25. I see. So for f(x) to be continuous at x = 2, the limit as x approaches 2 from the left side must equal the limit as x approaches 2 from the right side. So what you should is evaluate the limit as x approaches 2 of f(x) from the both the left and right. When you evaluate it from the left, use the top part of the function because that’s when x is less than it equal to 2. And the when you do it from the right, use the bottom one because that’s when x is greater than 2. If they’re the same, that means f(x) is continuous at x = 2. If they’re not the same, then f(x) is not continuous at x = 2.

  26. When you say "component form" do you mean like the ones with the triangular brackets?

  27. Yes that’s what I mean. And I know but for some reason I didn’t think to just rewrite i, j, and k, in the component form. But I’ve run into a different issue now. k is a 3D vector, so how am I supposed to add it to the 2d vectors? Am I missing something? So far what I’ve done is rewrote i, j, and k into the component form then added them together, but then realized I was trying to add components of a 2d vector with a 3D vector and that doesn’t make sense. So how do I proceed from here?

  28. Both i and j are vectors in 3D space.

  29. But don’t they only have 2 components? And how do I do the cross product when they’re written in that form? Again I didn’t do the previous problems because they were not listed in my homework

  30. I didn’t include a pic but in Desmos when relating the two (like a system of equations) and setting one as the negative you get 2 intercects of the graphs at x = -0.648, 0.648.

  31. I don’t know what you mean when you say that you try to exprimées the term as a polynomial, but if you’re getting different zeros from doing that, it’s problem because it’s not the same function anymore. But the initial two zeros are correct. Another way to see it is graph the full function and look at where it crosses the x axis. You’ll get those same zeros you got when you compared them as two different functions. Those should be what you’re looking for. If you wanna solve this algebraically then I don’t know how you would do that

  32. Your friend on bedrock needs to open the port up if you are on the same network. If he is renting a server, he would need to open up the port and give you the public IP address to connect to the server. Geyser and floodgate plugins would work. The rest I wouldn't know much about since I'm mostly a Java-based player.

  33. I don’t think we are on the same network

  34. I’m on a pc they’re on mobile but we’re both using bedrock version

  35. This method is usually taught in calc 2 so I would guess that

  36. The method is, but the definite integral is sometimes given as a problem in calc 1, where the students are supposed to recognize that this is just the area of a quarter of the unit circle.

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