Klizard Posted December 20, 2015 Share Posted December 20, 2015 Hello to everyone on the forums! Kenn here! (finally) While I am by no means new to Warframe I am very much new to posting on the forums due to being a combination of nervous and lazy... but no more! I've been playing Warframe for a little over two years and I don't plan to slow down about as much as DE themselves do. I'm a huge fan of the company and the dynamic itself and being an aspiring game programmer and designer I watch them pretty closely and with bated breath. While I have been working on concepts for quite some time I just recently started putting them up in the actual forums, do check them out if you have time and see what you think. Learning the coding on the forums has been interesting to attempt and by no means do I dare use any of it here. I'm a prominent Frost player and am usually around and willing to lend a helping hand (or in some cases straight up carry) to anyone who asks nicely! As my topic states, Greetings and Solutions to all! Link to comment Share on other sites More sharing options...
Omega-Shadowblade Posted December 20, 2015 Share Posted December 20, 2015 Solutions? you mean salutations? Link to comment Share on other sites More sharing options...
Klizard Posted December 21, 2015 Author Share Posted December 21, 2015 Solutions? you mean salutations? Nope. I mean solutions! As in solving problems, queries, or equations! Link to comment Share on other sites More sharing options...
ShadowQueen_X Posted December 21, 2015 Share Posted December 21, 2015 Hi ! Welcome to the forums, glad you overcame your nervousness / laziness :) Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 21, 2015 Share Posted December 21, 2015 Nope. I mean solutions! As in solving problems, queries, or equations! awesome! maybe you can help with this??? Let and let g(x) be an antiderivative of f(x). Then if g(5) = 7 find g(1) Link to comment Share on other sites More sharing options...
Iceheart125 Posted December 21, 2015 Share Posted December 21, 2015 This guy- this guy's legit. i was in his clan for an age, it was quiet, but he's a ton of fun to run with. he was the first one to help me out when i was new, and whenever he's on and his internet doesnt suck, he carries me, regardless of my setup, needing carried or not. if he'd join an alliiiance, so there were more peeeopleeee, id be back to the bang in a flash! Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 21, 2015 Share Posted December 21, 2015 id be back to the bang in a flash! that's what she said Link to comment Share on other sites More sharing options...
Klizard Posted December 21, 2015 Author Share Posted December 21, 2015 awesome! maybe you can help with this??? Let and let g(x) be an antiderivative of f(x). Then if g(5) = 7 find g(1) Sadly... I cannot equate. I wish solutions upon you with that one. Just none will come from me! This guy- this guy's legit. i was in his clan for an age, it was quiet, but he's a ton of fun to run with. he was the first one to help me out when i was new, and whenever he's on and his internet doesnt suck, he carries me, regardless of my setup, needing carried or not. if he'd join an alliiiance, so there were more peeeopleeee, id be back to the bang in a flash! Excuses! But thanks for the kind words! Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 21, 2015 Share Posted December 21, 2015 Sadly... I cannot equate. I wish solutions upon you with that one. Just none will come from me! nuuuuuuu I thought you could do equations ;-; #heartbroken Link to comment Share on other sites More sharing options...
Oakencrown Posted December 21, 2015 Share Posted December 21, 2015 awesome! maybe you can help with this??? Let and let g(x) be an antiderivative of f(x). Then if g(5) = 7 find g(1) oh no, OH NO! ITS MATH GUYZ RUN!! Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 21, 2015 Share Posted December 21, 2015 oh no, OH NO! ITS MATH GUYZ RUN!! #halp Link to comment Share on other sites More sharing options...
(PSN)AngelShur Posted December 21, 2015 Share Posted December 21, 2015 Welcome Tenno, do you know Wolfram Alpha? Link to comment Share on other sites More sharing options...
-PH-Samael Posted December 21, 2015 Share Posted December 21, 2015 Hi and welcome to the Forums! :-) Link to comment Share on other sites More sharing options...
(PSN)CowboyJeff72 Posted December 21, 2015 Share Posted December 21, 2015 Welcome aboard Tenno! Link to comment Share on other sites More sharing options...
--MZez-- Posted December 21, 2015 Share Posted December 21, 2015 Welcome to the forums Link to comment Share on other sites More sharing options...
Klizard Posted December 21, 2015 Author Share Posted December 21, 2015 nuuuuuuu I thought you could do equations ;-; #heartbroken Tell ya what, I'll do some research. I just haven't been taught math of a high level... so I need to check into it. Welcome Tenno, do you know Wolfram Alpha? This one I do not. Inform me! Thank the rest for the general welcoming invitations! Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 21, 2015 Share Posted December 21, 2015 Tell ya what, I'll do some research. I just haven't been taught math of a high level... so I need to check into it. <3 Link to comment Share on other sites More sharing options...
trunks013 Posted December 21, 2015 Share Posted December 21, 2015 solultion like this ? Welcome to the forum BTW ^.^ Link to comment Share on other sites More sharing options...
Klizard Posted December 22, 2015 Author Share Posted December 22, 2015 solultion like this ? Welcome to the forum BTW ^.^ Hahaha, you just may understand. And thanks for the welcome! Link to comment Share on other sites More sharing options...
Klizard Posted December 22, 2015 Author Share Posted December 22, 2015 awesome! maybe you can help with this??? Let and let g(x) be an antiderivative of f(x). Then if g(5) = 7 find g(1) Wait... if g(5) = 7 doesn't g(1) = 7/5? making g = 1.4 I know nothing of calculus, but that's just algebra right there isn't it? Did you try to trick me or did I miss the point? Link to comment Share on other sites More sharing options...
Klizard Posted December 22, 2015 Author Share Posted December 22, 2015 Welcome Tenno, do you know Wolfram Alpha? I do now, I actually just stumbled across it. Pretty intense. Link to comment Share on other sites More sharing options...
(PSN)AngelShur Posted December 22, 2015 Share Posted December 22, 2015 Please google it. They also have a handy app... Check it out, I won't spoil it! *whispering* please don't tell others, like your teachers *end of whisper* :) I do now, I actually just stumbled across it. Pretty intense. We use nothing else on University here (Netherlands). Make it your friend! :) Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 22, 2015 Share Posted December 22, 2015 Wait... if g(5) = 7 doesn't g(1) = 7/5? making g = 1.4 I know nothing of calculus, but that's just algebra right there isn't it? Did you try to trick me or did I miss the point? Bazinga http://ltcconline.net/greenl/courses/105/antiderivatives/DifficultProblemSolution.htm Link to comment Share on other sites More sharing options...
Yzjdriel Posted December 22, 2015 Share Posted December 22, 2015 (edited) awesome! maybe you can help with this??? Let and let g(x) be an antiderivative of f(x). Then if g(5) = 7 find g(1) Ahh, calculus. It's wonderful. This one's a bizatch. See below. Wait... if g(5) = 7 doesn't g(1) = 7/5? making g = 1.4 I know nothing of calculus, but that's just algebra right there isn't it? Did you try to trick me or did I miss the point? That's not quite how calculus works. Here ya go, Firefly: You can't actually integrate this by hand, and I'm sure you've tried wolfram alpha or you wouldn't be here of all places... So first off we know that g'(x) = f(x) and g(5) = 7 We can now create a third function , h(x) = g(x) - 7 Thus h'(x) = f(x) and h(5) = 0 We now construct the integral from 5 to x of the cube root of (u2+4u) (du) If we evaluate this for h(5), we get 0 as previously stated. Now we simply evaluate it for 1. The integral from five to 1 of the cube root of (u2+4u) (du) is about -10.882. The problem is asking for g(1), and we stated above that h(x) = g(x) - 7, so just add the 7 back. You end up with about -3.882. EDIT: Ha you troll XP. Well there went ten minutes of my life working this out and typing it up... Oh well. Edited December 22, 2015 by Yzjdriel Link to comment Share on other sites More sharing options...
-Kryptyk- Posted December 22, 2015 Share Posted December 22, 2015 (edited) Ahh, calculus. It's wonderful. This one's a bizatch. See below. That's not quite how calculus works. Here ya go, Firefly: You can't actually integrate this by hand, and I'm sure you've tried wolfram alpha or you wouldn't be here of all places... So first off we know that g'(x) = f(x) and g(5) = 7 We can now create a third function , h(x) = g(x) - 7 Thus h'(x) = f(x) and h(5) = 0 We now construct the integral from 5 to x of the cube root of (u2+4u) (du) If we evaluate this for h(5), we get 0 as previously stated. Now we simply evaluate it for 1. The integral from five to 1 of the cube root of (u2+4u) (du) is about -10.882. The problem is asking for g(1), and we stated above that h(x) = g(x) - 7, so just add the 7 back. You end up with about -3.882. EDIT: Ha you troll XP. Well there went ten minutes of my life working this out and typing it up... Oh well. HAHAHA LOL just doing what I do best xD but nice job working that one out and taking ten minutes out of your life just for me #truelove :') Edited December 22, 2015 by Firefly0037 1 Link to comment Share on other sites More sharing options...
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