i'm breaking all the rules just to say, if someone votes that they're in elementary school and drop-kicking them
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School work help
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Elementary, middle, or high school? (For the Australians and anyone else who uses this system)
Total votes: 35
we had a 9 yo on the forums once but never a 5 yo, lets try to break the record
Solution = 🚽😀Reyalp51 wrote:
we had a 9 yo on the forums once but never a 5 yo, lets try to break the record
But on a thread, middle school was best for me and probably will remain that way
There isn't my school grade here lol
Topic Starter
What grade u in?Corne2Plum3 wrote:
There isn't my school grade here lol
Topic Starter
To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:
\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]
Using the product rule, we have:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]
Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:
\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]
\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]
Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]
This would be the expression for \( \frac{\partial f}{\partial x} \).
\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]
Using the product rule, we have:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]
Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:
\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]
\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]
Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]
This would be the expression for \( \frac{\partial f}{\partial x} \).
did brother use chatgpt for this or did he actually work it outJebus1 wrote:
To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:
\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]
Using the product rule, we have:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]
Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:
\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]
\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]
Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]
This would be the expression for \( \frac{\partial f}{\partial x} \).
Topic Starter
I started working it out, got like halfway, and then just thought it'd be easier to get chatgpt to explain it 🤣sametdze wrote:
did brother use chatgpt for this or did he actually work it outJebus1 wrote:
To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:
\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]
Using the product rule, we have:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]
Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:
\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]
\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]
Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]
This would be the expression for \( \frac{\partial f}{\partial x} \).
And tbh it did a pretty good job of explaining 😅
Jangsoodlor
Jebus1 wrote:
I started working it out, got like halfway, and then just thought it'd be easier to get chatgpt to explain it 🤣sametdze wrote:
did brother use chatgpt for this or did he actually work it outJebus1 wrote:
To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:
\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]
Using the product rule, we have:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]
Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:
\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]
\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]
Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:
\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]
This would be the expression for \( \frac{\partial f}{\partial x} \).
And tbh it did a pretty good job of explaining 😅
ChatGPT usually failed to calculate maths correctly. Here is the example and Anoter example. Notice the inconsistencies of its answer.
the actual answer (if my calculation is correct) is zx^2(2xy) + (ycosz+yx^2)2xz
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
apparently if i base myself on the australian school system, i'm in "senior high school" (i'm in 11th grade)
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at itMistressRemilia wrote:
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
Definitely not physics, but maybe some biology yeahAnaxii wrote:
are you okay with physic and biology? because i suck at itMistressRemilia wrote:
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
i would like you to help me differentiate between the different leukocytes with your words 🙏MistressRemilia wrote:
Definitely not physics, but maybe some biology yeahAnaxii wrote:
are you okay with physic and biology? because i suck at itMistressRemilia wrote:
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
american moment[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
Topic Starter
💀💀, in Australia (ik I put middle school) we don't have middle school we just have primary and high, there is also kindy or kindergarten but they aren't really with the education system they're all independent plus I doubt any kindy kids are gonna be playing OSU!💀B0ii wrote:
american moment[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
different countries use different systems because education is not a solved science. It's not something like metric where you can get every country (except america for some reason) to agree to measure stuff on meters[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
So there's like lymphocytes which hang out mainly in the lymphatic system until they're needed. They include T-cells, B Cells, and the killer cells (I don't remember if they have a name). Part of what makes yersinia pestis (aka, the plague) so deadly is that it attacks the lymphatic system, hence why you get bubonic plague, where your lymph nodes swell up like nuts (there's other forms of plague, too).Anaxii wrote:
i would like you to help me differentiate between the different leukocytes with your words 🙏MistressRemilia wrote:
Definitely not physics, but maybe some biology yeahAnaxii wrote:
are you okay with physic and biology? because i suck at itMistressRemilia wrote:
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
B Cells are what makes your antibodies. They're part of your adaptive immune system and get activated after receiving signals. There's a bit of time between when your innate immune system notices a pathogen, when it activates the adaptive immune system through antigens, and when the B Cells have time to start making antibodies to fight an infection.
I forget exactly what T-Cells do... but they get hit by HIV. They're obviously important. Kurzgesagt has a very good video on the different bits of the immune system, so look that up and it'll probably tell you what T-Cells do.
So those above are the lymphocytes and are part of the adaptive immune system parts. T-Cells, B Cells, and natural killer cells.
There's also eosinophil, which attack parasites, and are also why you get allergies and such.
Basophils are responsible for inflammation, asthma, allergies, anaphylaxis... all the fun stuff like that. They're not well understood, and are the least common leukocytes in the blood stream. They release histamines.
Then there's monocytes, which are the largest white cells and go around eating shit up similar to neutrophils. They go around looking for invaders or trauma, and then go nom-nom-nom on infected cells. They'll call in neutrophils if they need help.
Neutrophils are also part of the innate immune system. iirc, they're the most common leukocyte in the blood, and for some reason, my body in particular tends to enjoy producing more neutrophils than normal. They go around killing things as well. When they need help, they release dendrytic cells. Those go find your adaptive immune system and kick it into action.
So you've got lymphocytes (B Cells that produce antibodies, T-Cells, and natural killer cells), eosinophils (kills parasites, does some stuff with inflammation and allergies), basophils (releases histamines to cause inflammation), neutrophils (KILL EVERYTHING EVEN WHEN IT MEANS KILLING MYSELF), and monocytes (nom nom nom dudes looking for trouble).
I think I have those correct... double-check with your textbooks and such in case I got something wrong.
Jangsoodlor
Topic Starter
so you've only got 9 years of education?Jangsoodlor wrote:
Here it's[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
Jangsoodlor
*Primary 6 Years I mis-typed kek.Jebus1 wrote:
so you've only got 9 years of education?Jangsoodlor wrote:
Here it's[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
Topic Starter
Ah gotchuJangsoodlor wrote:
*Primary 6 Years I mis-typed kek.Jebus1 wrote:
so you've only got 9 years of education?Jangsoodlor wrote:
Here it's[-Omni-] wrote:
for me, its
elementary: prek-5
middle: 6-8
high: 9-12
smh everywhere should use the same things it would be much easier😞
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
i think i understand a bit nowMistressRemilia wrote:
So there's like lymphocytes which hang out mainly in the lymphatic system until they're needed. They include T-cells, B Cells, and the killer cells (I don't remember if they have a name). Part of what makes yersinia pestis (aka, the plague) so deadly is that it attacks the lymphatic system, hence why you get bubonic plague, where your lymph nodes swell up like nuts (there's other forms of plague, too).Anaxii wrote:
i would like you to help me differentiate between the different leukocytes with your words 🙏MistressRemilia wrote:
Definitely not physics, but maybe some biology yeahAnaxii wrote:
are you okay with physic and biology? because i suck at itMistressRemilia wrote:
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
B Cells are what makes your antibodies. They're part of your adaptive immune system and get activated after receiving signals. There's a bit of time between when your innate immune system notices a pathogen, when it activates the adaptive immune system through antigens, and when the B Cells have time to start making antibodies to fight an infection.
I forget exactly what T-Cells do... but they get hit by HIV. They're obviously important. Kurzgesagt has a very good video on the different bits of the immune system, so look that up and it'll probably tell you what T-Cells do.
So those above are the lymphocytes and are part of the adaptive immune system parts. T-Cells, B Cells, and natural killer cells.
There's also eosinophil, which attack parasites, and are also why you get allergies and such.
Basophils are responsible for inflammation, asthma, allergies, anaphylaxis... all the fun stuff like that. They're not well understood, and are the least common leukocytes in the blood stream. They release histamines.
Then there's monocytes, which are the largest white cells and go around eating shit up similar to neutrophils. They go around looking for invaders or trauma, and then go nom-nom-nom on infected cells. They'll call in neutrophils if they need help.
Neutrophils are also part of the innate immune system. iirc, they're the most common leukocyte in the blood, and for some reason, my body in particular tends to enjoy producing more neutrophils than normal. They go around killing things as well. When they need help, they release dendrytic cells. Those go find your adaptive immune system and kick it into action.
So you've got lymphocytes (B Cells that produce antibodies, T-Cells, and natural killer cells), eosinophils (kills parasites, does some stuff with inflammation and allergies), basophils (releases histamines to cause inflammation), neutrophils (KILL EVERYTHING EVEN WHEN IT MEANS KILLING MYSELF), and monocytes (nom nom nom dudes looking for trouble).
I think I have those correct... double-check with your textbooks and such in case I got something wrong.
thank you for your help
Jangsoodlor wrote:
Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
Nekathe's solution
Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..
well i tried to solve this, but i got another start, different from yours Nekathe.
I am going to edit this according to what i've solved:
1. According to my old notes and the book: [1] N. Piskunov, "Capitulo XI: integral definida", Cálculo diferencial e integral, tercera edición, Editorial Mir, URSS, Editorial Mir Moscú, 1977, pp. 436-437. (IEEE style reference)
Here's the english version of my book: N. Piskunov differetial and integral calculus, i think it is the first edition but it has the information we use here. The page is 404-405 of the PDF, 403-404 of the book.
I'll first take out the 6 out of the 2 integrals to simplify the operation, thats what we used to do in math class.
2.I'll solve the exponential minus one in the power. -(2x+3y)=-2x-3y
now we have: e^(-2x-3y)
3. Well i comfirmed this with my notes and this vid:
Basically if you have for example: e^(a+b) you can separate into a product (e^a)(e^b)
Therefore: e^(-2x-3y)=(e^-2x)(e^-3y) and the first integral it is respect to y so we treat (e^-2x) as in the vid.
at this point i'll take (e^-2x) out multiplying the dy integral.
then acoding to [1] solve the definite integral from o to 1-x of (e^-3y) dy
this is wrong xd
leaving us with: this expresion: 6 ∫0_1 (e^-2x)∫0_1-x (e^-3y)dy dx
evaluate the integral from 0 to 1-x:
6 ∫0_1 (e^-2x) [(e^-3y)]0_1-xdx
evaluating:
6 ∫0_1 (e^-2x) [(e^-3⟨1-x⟩-e^-3⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-e^⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-1]dx
Multiplying (e^-2x) with the evaluated integral:
here we apply the power rule of the product a^m(a^n)=a^m+n.
6 ∫0_1 e^⟨-3+x⟩ - e^⟨-2x⟩dx
Here's almost done just solve this integral using the previous rules in [1]
6[e^⟨-3+x⟩ - e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨x⟩- e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨1⟩- e^⟨-2⟩-(e^⟨-3⟩.e^⟨0⟩- e^⟨-2.0⟩)]
6[e^⟨-2⟩-e^⟨-2⟩-(e^⟨-3⟩.1-1)]
6[e^⟨-2⟩-e^⟨-2⟩-e^⟨-3⟩+1)]
I am struggling here with the signs and the algebra maybe you can solve it to get the answer. I am going to sleep, rn heres midnight. hope u understand this. tomorrow hopefully i'll get the answer, it was really fun to solve this, thanks for this challenge
, i re-learned a few things. Salutations o/
evaluate the integral from 0 to 1-x:
6 ∫0_1 (e^-2x) [(e^-3y)]0_1-xdx
evaluating:
6 ∫0_1 (e^-2x) [(e^-3⟨1-x⟩-e^-3⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-e^⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-1]dx
Multiplying (e^-2x) with the evaluated integral:
here we apply the power rule of the product a^m(a^n)=a^m+n.
6 ∫0_1 e^⟨-3+x⟩ - e^⟨-2x⟩dx
Here's almost done just solve this integral using the previous rules in [1]
6[e^⟨-3+x⟩ - e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨x⟩- e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨1⟩- e^⟨-2⟩-(e^⟨-3⟩.e^⟨0⟩- e^⟨-2.0⟩)]
6[e^⟨-2⟩-e^⟨-2⟩-(e^⟨-3⟩.1-1)]
6[e^⟨-2⟩-e^⟨-2⟩-e^⟨-3⟩+1)]
I am struggling here with the signs and the algebra maybe you can solve it to get the answer. I am going to sleep, rn heres midnight. hope u understand this. tomorrow hopefully i'll get the answer, it was really fun to solve this, thanks for this challenge
, i re-learned a few things. Salutations o/26/02/24
i can't figure out the last part but i've remembered another few rules to make it easier
27/02/24
I figured this out i think, thanks to Joanna Angel at DC Mathematics server. With her help i figured out that i skipped a rule, a basic rule while integrating e powered to constants and variables greater than 1.
here is what i did:
Rules applied:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Jangsoodlor
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Well, here's the actual answer to that question
I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.Jangsoodlor wrote:
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Well, here's the actual answer to that question
I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Welp, i did embarrass myself. I solved it with “x-1” instead of “1-x”
I cant believe i made a mistake like that
yeaaah it be like that, but hey, don't be discouraged by that, you made all the process, just mechanize and solve it twice, like in a test, you not only need to be able to solve it, but to solve it fast and check it again. Salutations o/.Nekathe wrote:
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Welp, i did embarrass myself. I solved it with “x-1” instead of “1-x”
I cant believe i made a mistake like that
Jangsoodlor
Well, I just got introduced to multi-variable integral in Calculus II course last week so I don't know any techniques yet (except basic integration techniques from Calculus I like trigonometric substitution and stuffs). This problem is from another course I took, which technically requires only Calculus I, but the instructor mysteriously gave a problem that requires Calculus II knowledge.Hachiman Ryouta wrote:
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.Jangsoodlor wrote:
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Well, here's the actual answer to that question
I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
As for the deadline, it's already passed lol. Again, thanks for your help and maybe I'll post my attempt once I'm equipped with appropriate knownledge to solve this.
Here's the full problem btw
In order to find the Probability Distribution Function of f, given that event A happened, I need to solve for Probability of Event A happened first. Which is the area under that graph where x+y <=1. That's where Integration comes in. But I didn't expected it to be multi-variable lol
And here's the full solution
As for the book name, it's "Probability and Stochastic Process: A friendly introduction for Electrical and Computer Engineers" Second Edition by Roy D. Yates and David J. Goodman. It's a great book to read if you want to calculate the probability of you getting a certain character in Gacha games, then have literal mental breakdown and then questions your existence and sanity.
OMG probability and statistics wasn't the greatest subject back then when i took it in college the teacher was kind of inexperienced a i did not undertand it deep enough. anyways don't worry about the deadline, just solve the problem, because if you don't understand the first one, it is likely that you don't understand the next ones. Salutations o/Jangsoodlor wrote:
Well, I just got introduced to multi-variable integral in Calculus II course last week so I don't know any techniques yet (except basic integration techniques from Calculus I like trigonometric substitution and stuffs). This problem is from another course I took, which technically requires only Calculus I, but the instructor mysteriously gave a problem that requires Calculus II knowledge.Hachiman Ryouta wrote:
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.Jangsoodlor wrote:
Nekathe wrote:
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmkHachiman Ryouta wrote:
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..Nekathe wrote:
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.Jangsoodlor wrote:
Can anyone help me solve this double integral?
Time to embarrass myself.
first lets start with the inside integral.now the outside integral
- u sub, let u = 2x+3y
- since we are working with dy, we treat x as a constant when finding du. du/3 = dy
- change the bounds,
- top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
- bottom bounds: y=0, u then equals 2x
- the integral is now /int_2x^5x-3 (2e^-u)du
- solve the simple integral, -2e^(-5x+3)+2e^(-2x)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
- /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
- separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
- solve the first integral
- find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
- evaluate between the bounds, -(e^(-2)-e^-0)
- -e^(-2)+1
- now solve the second integral
- this one is similar to the previous one, indeterminant form yada yada
- indeterminant form, (-2/5)e^(-5x+3)
- evaluate between the bounds, (-2/5)(e^(-2)-e^3)
(consistent with mathway)
KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.
goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
well i tried to solve this, but i got another start, different from yours Nekathe.
Well, here's the actual answer to that question
I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
As for the deadline, it's already passed lol. Again, thanks for your help and maybe I'll post my attempt once I'm equipped with appropriate knownledge to solve this.
Here's the full problem btw
In order to find the Probability Distribution Function of f, given that event A happened, I need to solve for Probability of Event A happened first. Which is the area under that graph where x+y <=1. That's where Integration comes in. But I didn't expected it to be multi-variable lol
And here's the full solution
As for the book name, it's "Probability and Stochastic Process: A friendly introduction for Electrical and Computer Engineers" Second Edition by Roy D. Yates and David J. Goodman. It's a great book to read if you want to calculate the probability of you getting a certain character in Gacha games, then have literal mental breakdown and then questions your existence and sanity.
PD: aah i can't figure out the last part of the problem, i got it with an amplitude of 3 times the size of the solution and sings changed :S
Do you have any apps or techniques to improve in English?
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.NitroZz wrote:
Do you have any apps or techniques to improve in English?
honestly i learned english thanks to youtube videos and translators lolNitroZz wrote:
Do you have any apps or techniques to improve in English?
That's how I learned, playing games in english (sometimes because there was no translation), watching videos, movies etc.Hachiman Ryouta wrote:
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.NitroZz wrote:
Do you have any apps or techniques to improve in English?
Also being active on forums and chats (vc/text) helps a lot since you not only listen/read but also speak/write.
Thank you for the adviceHachiman Ryouta wrote:
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.NitroZz wrote:
Do you have any apps or techniques to improve in English?
fortunately, no one is in elementary school which is GOOD
Gsun
Maybe I'm thereAnaxii wrote:
fortunately, no one is in elementary school which is GOOD
i can help you kiddoNitroZz wrote:
Maybe I'm thereAnaxii wrote:
fortunately, no one is in elementary school which is GOOD
Good luck!! 🤞[-Omni-] wrote:
test in 30 mins wish me luck 😔
Best of luck[-Omni-] wrote:
test in 30 mins wish me luck 😔
Topic Starter
How'd the test go? 🫣[-Omni-] wrote:
test in 30 mins wish me luck 😔
I didn't know what emoji to use😭😭😭😭
they started these like new test rules and theyre like crazy strict but i think i got an A on itJebus1 wrote:
How'd the test go? 🫣[-Omni-] wrote:
test in 30 mins wish me luck 😔
I didn't know what emoji to use😭😭😭😭
(90 or above)
Topic Starter
Lessgo😎 gj[-Omni-] wrote:
they started these like new test rules and theyre like crazy strict but i think i got an A on itJebus1 wrote:
How'd the test go? 🫣[-Omni-] wrote:
test in 30 mins wish me luck 😔
I didn't know what emoji to use😭😭😭😭