help-19

[slim kettle]

Yes

[terse coral]

and I just display it as 10/4 pi

[slim kettle]

Yeah, you can simplify 10/4 as 5/2 tho

[terse coral]

oh yeah

[slim kettle]

But it's still correct

[terse coral]

The circumfrence is the arch right?

or is it the sector

[slim kettle]

The arc that's it

[terse coral]

nice okay

[slim kettle]

Just that curve part of circle that's it

[terse coral]

ohh

okay, great

I have even more questions unless you wanna go now lol

Oh, and I promise you you're not doing my homework or anything, this is over the progression of about 40 pages

[slim kettle]

Oh lol

It's fine, you may clear your conceptual doubts

And for the above question, there is a formula, lemme send it

[terse coral]

oh there is?

okay

[slim kettle]

I mean 90 degrees was very obvious cuz, it is 1/4th of circle

But for tougher/random angles, you would need this

[terse coral]

hmm, okay, let me paste this down

the 0

its just 0/360:

??

[slim kettle]

Bro lmao

It's theta

[terse coral]

wth is that

[slim kettle]

The denotion of an angle

A variable

Like x

Like theta can be any angle

[terse coral]

crap, I've been walking around thinking its just some "O" placement variable

[slim kettle]

Lmfao

It's like for any angle, say theta......

Just a variable, nothing much

You may use any alphabet or symbol of your choice

But theta is used mostly in maths for angles

[terse coral]

oh, I see

Im actively saving screenshots rn, hold up

[slim kettle]

ez, I have no hurry at all

Take your time

[terse coral]

thats crazy for no hurry at all

[slim kettle]

Because my schooling is completed

Just waiting for college to start

I am in home doing timepass all day

So I got no hurry

[terse coral]

I see, Im sure you got into a good college with all this math brain lmao

[slim kettle]

Hah, an okaish one

[terse coral]

don't disclose me anything you don't wanna lmao, Ik its personal

[slim kettle]

Cuz this is 0 level maths in india

[terse coral]

Oh, I'm competing against a guy that went to India

well he's indian

[slim kettle]

Yeah

[terse coral]

Seems like Im screwed

[slim kettle]

You not competing mate

You just clarifying your things

[terse coral]

Im absolutley flamed, his ma has a whole phd in math and everything

Im saying this but, I admire him a lot too lmao

Heres the story problem

I don't even know what happened in the work here

[slim kettle]

This is the simplest one

Of all you asked till

[terse coral]

wait what

well, I mean, I am pretty much lost

[slim kettle]

It's just adding up 2pi(s)

Lemme explain you, a min

[terse coral]

okk

[slim kettle]

It's 11pi + 50 degrees

I mean (11 x 180)+50 degrees

And you can convert this into radians

mb I am giving answer without explanation

Uhmm, I am just trying to figure out a way to explain this to you

Okay,

So, we have to find the initial point

The point from where the movement began

Near the centre

At 0 degrees

From where the curve began

Got it?

[terse coral]

yes

[slim kettle]

Now, we just need to count how many times we perform a complete circle or revolve back to starting point

[terse coral]

oh, so I just needed to count

[slim kettle]

So, we go round and round and then return to the same level 5 times which is 5 x 2pi = 5 x 2 x 180 degrees

But after we done coming around for fifth time, we perform a half circle which is pi = 180 degrees

And then we travel for 50 more degrees and stop

And by adding all these angles, we get to the answer I said above

[terse coral]

oh I see

[slim kettle]

This

[terse coral]

so I just needed to count from the original 0

[slim kettle]

Yeah

[terse coral]

then convert it to pi for the circles made round

[slim kettle]

That how you intitate

Yea

[terse coral]

and then I can convert percentages and stuff okay

I'll ask one last example and I'll probably head to bed

[slim kettle]

Not percentages lmao, radians and degrees

Okay sure

[terse coral]

oh yeah

I've already done this and saw this but I feel like I don't actually remember/know the core of it

wait wait, hold up

the acute angle thingy

[slim kettle]

Hah this one ez too, but you gotta know a formula for this one

[terse coral]

THERES A FORMULA TOO??

[slim kettle]

Yessir

[terse coral]

how the hell is there a formula for this too T-T

[slim kettle]

You'll know

Now

[terse coral]

I lowkey don't know if the American system is messing with me by not including it or not

[slim kettle]

Hah, just a min, I piss and come

Don't go anywhere

Yeah, so...

[terse coral]

crazy way to word it but, I won't run off

[slim kettle]

lol my English sucks 😦

[terse coral]

nah, I understand you fine lol

[slim kettle]

So the formula here is...

[terse coral]

Okay I screwed sm up bc I went with the flow and got squareroot 6

for one side of this triangle

and that doesn't seem remotely right

[slim kettle]

For any acute angle in a right angled triangle,

[slim kettle]

Not even close

It's hard to type, I wrote and send you formula pic

[terse coral]

kk

[slim kettle]

Now since you know value of cos, we can find sin using this formula

[terse coral]

wait wait wait wait

so

SO THE WHOLE TIME WHEN THERE ARE ACUTES ITS A WHOLE FORMULA??

[slim kettle]

No, this is just a property for acute angles

Useful in some cases like here

[terse coral]

ohh okk

[slim kettle]

When we are given cos or sin and have to find the other

[terse coral]

Because when I refrence it to this its not like that so I was confused

[slim kettle]

Well, all these questions need this formula

Can't be solved otherwise

[terse coral]

oh

[slim kettle]

So, back yo out question

We were given cos

Now we will find sin with the formula

Now since we know sin and cos

We can find tan, as tan = sin/cos

And once we have sin, cos and tan, we can find cosec, sec and cot

Because....

Cosec = 1/sin

Sec = 1/cos

Cot = 1/tan

Clear?

[terse coral]

ye

that was shown to me

last time

[slim kettle]

Hmm

Any more doubts?

[terse coral]

nope, Im just gonna stop here today

I'll probably look back on this tmr

its like 4am rn

[slim kettle]

I forgot to ask you, what grade you are in?

[terse coral]

this is the honors algebra2 course for highschoolers in America

its a lot different than India I'd assume but

[slim kettle]

Oh, so is it school?

[terse coral]

yeah

[slim kettle]

Yeah it is

[terse coral]

America they usually in high school (9-12)

the normal course is

geometry - algebra1 - algebra2 - precalculus

wait no

yeah

[slim kettle]

So, you are in high school now?

[terse coral]

yup

[slim kettle]

9, 10, 11 or 12?

I assume 9

[terse coral]

yeah

Im trying to test out of it

[slim kettle]

Hmm

These are like very basics

You gotta get it all in your head for lifetime

[terse coral]

from what you've been saying I can very much see lol

[terse coral]

I will try but most American kids don't

Im not saying majority but

[slim kettle]

Oh nvm, in india, it is

Aisans built different XD

[terse coral]

lol yeah

though Indians do get a lot of slander

Its like racism but nobody really bats an eye

[slim kettle]

idk, just not into that bs

[terse coral]

I get it lol, a lot of ppl just be like that for no reason

[slim kettle]

Alright mate, you gotta sleep ig

[terse coral]

ofc yeah, cya

(I hope not honestly) (not in a bad way)

.close

[odd edge]

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[slim kettle]

Cya

Good luck

[terse coral]

thank you

[slim kettle]

Never mention

[odd edge]

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[carmine idol]

i am supposed to use the fact that (e^(x^2))’ = 2xe^(x^2) to solve integral(xe^(x^2))dx.
i figured the antiderivative of xe^(x^2) is (1/2)*e^(x^2) given this fact.
why is the solution putting 1/2 in front of the integral, instead of putting it as part of the integral? this gives me a different answer.

[slim kettle]

Because...

Since it multiplied the given equation by 2,

It gotta be divided by 2

I mean you put that 2 inside

It came out of nowhere

To balance it out, we multiplied 1/2 there

[odd edge]

@carmine idol Has your question been resolved?

[carmine idol]

right, that makes sense

[slim kettle]

Yeah, just manipulation

[carmine idol]

appreciate the help

[odd edge]

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[jolly halo]

hey there , i was trying this integral and couldnt figure out what was wrong in the first method i thought of
i tried it again a bit later and was able to solve it , but curious on what went wrong in the image i attached

[inland rover]

i would have approached a different method

[jolly halo]

yeah i used partial fraction in one , and in one i tried directly using hyperbolic , both worked out

i was new to integrals so wanted to make sure i dont repeat the mistake

cause this was the first method which popped up and i wanted to know for sure what i do wrong

[inland rover]

i think your method of integrating 1/x^2-4 is wrong

[wanton bison]

partial fraction decomposition is the way

[fervent hawk]

hello there~ @jolly halo

[charred summit]

uve substituted it wrong i guess

[fervent hawk]

there are 2 things

1. x²-4 = -4cos²(theta)
2. sec(theta) = 2/√(4-x²)

[charred summit]

I think $\int \dfrac{dx}{x^2-4} = \frac{1}{4}\ln\left|\dfrac{x-2}{x+2}\right|+C$

[clever fjord]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[wanton bison]

💀

[fervent hawk]

if you fixed those, you'll get the same final answer
3ln|x-2| + 2ln|x+2| +C
via some algebra

[jolly halo]

hm yeah cool

[jolly halo]

OH yeahh

damn thanks missed that

[fervent hawk]

for reference

[jolly halo]

yeeeeee omg thanks a lot!!!

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[wide drift]

What 'form' is velocity in if you get me

Is it a vector or a number

Like do I literally just do sqrt(x'(t)^2 + y'(t)^2)

Or is it x'(t) above y'(t) (in vector form, not fraction)

[amber junco]

u can intercovert the forms

like

which form do u need it in?

[wide drift]

Idk thats what I was wondering

From the question what form are they asking for

[amber junco]

u have the answer to the question?

[wide drift]

Yeah let me check

But wait also

What do we know about point A

[plush rose]

aha a Dutch source I see

[wide drift]

Like its at x = 0

[amber junco]

we know the time when it is at point A

[wide drift]

Jaaa 😄

[amber junco]

yup

[wide drift]

But we dont know y

[amber junco]

ok listen

I'll explain

so y = 0

then t comes out to be -1 from the second equation

but if x = 0 then t = +1, -1

from 1st equation

so point O is 0,0

so both x and y are zeero

so at time -1( common for both equations) = point O

and point A is at t = +1 seconds

[wide drift]

Wait let me read that a few times just to understand

[amber junco]

yup

[wide drift]

then t comes out to be -1 from the second equation
Does it?

[amber junco]

yup

[plush rose]

The velocity is a number ("snelheid").
If they want a vector they'd ask for a "snelheidsvector" (velocity vector)

[amber junco]

put y(t) = 0

and equate

[wide drift]

The derivatives of x(t) and y(y) are -2t and 2t + 2

[amber junco]

no no

the points A and O are positions

[wide drift]

Ahhh oke

[amber junco]

not velocities

so don't differentiate first

get t = -1

[wide drift]

Oh

[amber junco]

so after u get t = +1 for point A then u can differentiate

and get velocity in either vector form

or however u like

[wide drift]

How do you know that O is t = -1 and A is t = 1?

I get where -1 and 1 come from

But why do you know which is which

[amber junco]

because O = 0,0

so both x and y are zero

and both equations have common solution -1

so O = -1

and the +1 then has to be A

[wide drift]

solution to what sry?

[amber junco]

um so

for x(t) = 0

we get -1,+1

and y(t)= 0

we get only -1

this means both x and y positions go zero simultaneously at -1

[wide drift]

Okkkkkkk

I get it now

[amber junco]

since it is common in both y(t) = 0 and x(t) = 0

[wide drift]

Yeyeyeye I see

Thanks you explained it well

[amber junco]

yup

[amber junco]

np

[wide drift]

But ok

We want it in the form of a number

So I assume we use sqrt(x'(t)^2 + y'(t)^2)

[amber junco]

yea

u do

that

[wide drift]

And t = 1

at point A

[amber junco]

yup

[wide drift]

Awesome

[amber junco]

happy to help!

[wide drift]

Yeah I think I can figure out the rest haha

Thank you so much!!

To both of you haha

❤️

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[wide drift]

Im doing Boswell beta VWO B, but in english 😝

I want to go to university of Maastricht haha

But need to pass a maths exam first to make up for my deficit

[plush rose]

I'm doing vwo wiskunde B (the regular one, but a year ahead)

[wide drift]

Whats the regular one?

[plush rose]

Just what they teach you in class in the Netherlands

[amber junco]

are the courses hard guys?

[plush rose]

nah

[wide drift]

For me yes 😭

[plush rose]

my average score in 5 vwo was a 9.7

[amber junco]

lmao

[wide drift]

.reopen

[odd edge]

✅

[wide drift]

https://www.boswell-beta.nl/vwo/math-b/

This is the one I am doing

[amber junco]

nice

[wide drift]

Thats insane

Hahaha can I borrow your brain for this exam

[plush rose]

yh sure that'll be tan(90) euros

[wide drift]

Bro its worth

I need to get accepted this year, I failed last year 😭

But ok, back to study then haha

Thanks again guys

❤️

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[eager cypress]

Someone can guide me this question?

[swift lake]

u know median?

[odd edge]

@eager cypress Has your question been resolved?

[eager cypress]

yea it median at the age of 12

<@&286206848099549185>

<@&286206848099549185>

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[fluid falcon]

can someone help with this

[fluid falcon]

trigonometric Identities

[silk quail]

what exactly do you need help with?

[fluid falcon]

this part

[silk quail]

like do you not understand the process, or do you need help applying the identities

ok

so $\sin^2x+\cos^2x=1$, so $\cos^2x=1-\sin^2x$

[clever fjord]

fish

[silk quail]

thus you get $2\sin^2x-1$

[fluid falcon]

ohh

[clever fjord]

fish

[fluid falcon]

thank you fish

[silk quail]

you're welcome

[fluid falcon]

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[olive sun]

I did A and B with reasonable answers, but i'm struggling with C

You can't plug numbers in, multiplying by the reciprocal does nothing

What else is there?

I know there's something stupidly obvious i'm missing

somehow always is

[odd edge]

@olive sun Has your question been resolved?

[olive sun]

ah, forget it.

i'm sorry for wasting time.

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[noble pelican]

So, I'm solving this ODE:
dy/dx = (y-x)/(y+x)
It's homogeneous, so you can use the substitution y=ux or x=vy.
When I use the first substitution I get this:
ln|y^2+x^2| + 2arctan(y/x)=C
Which is the book's answer, but when I use the other substitution I get:
ln|y^2+x^2| - 2arctan(x/y)=C
Am I missing something or there are two possible answers to this ODE?

[odd edge]

@noble pelican Has your question been resolved?

[uneven osprey]

See, arctan(a) = arccot(1/a)

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[uneven osprey]

huh?

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[mystic saffron]

How do I solve this question because I thought you wouldn’t have to divide and itd be 0*0 unless the question is written incorrectly

[lusty helm]

tan pi/2 is infinite (undefined), so it's infinite*0 form

[mystic saffron]

What is the answer

[lusty helm]

lemme get pen paper wair

[mystic saffron]

Alr thanks you

[mystic saffron]

BRO

Howf U get that

How did

TYSM

Im struggling with this trig identity roo

Too

[lusty helm]

cot theta will cancel out if you write it cos theta/ sin theta

[mystic saffron]

Could u help me out with one more

[lusty helm]

did it directly

yea

send here

[mystic saffron]

Here is what I have so fsr

Can you write out every step please

[lusty helm]

sure

[mystic saffron]

I just have that but I dont know what to do next

[mystic saffron]

Huh

[lusty helm]

a²-b² = (a-b)(a+b)

[mystic saffron]

so wait

Ohh

[lusty helm]

yup

ignore confusion between x and theta XD

[mystic saffron]

So itd be

Cosx+sinx

[lusty helm]

yes

[mystic saffron]

omg

dude

i feel dumb

[lusty helm]

and then x = pi/4 or 45°

[mystic saffron]

im taking ap bc calc the upcoming school year

I dont even have the unit circle memorized

Yk what bc calc is right

[lusty helm]

yea

[mystic saffron]

I am nervous

[lusty helm]

dw you will be fine too

[mystic saffron]

Are you sure

Like i feel really lost

[lusty helm]

yu0

[mystic saffron]

Rn

She assigned us khan academt

For limits

[lusty helm]

see making mistakes doesn't matter

[mystic saffron]

And limits are lowkey annoying cuz its so much to remember

[lusty helm]

but repeating mistakes doee

just don't repeat it next time

[mystic saffron]

True

Its my first time doing this

So eyah

yea

[lusty helm]

exactly

a normal human like us might face difficulties but yeah you will turn out fine once you get enough practice

[mystic saffron]

Okay

My friend said that limits was the easiest topic

And Im struggling with it

I started it today

So…

Im quite worired abt the future

[lusty helm]

never believe what friend says

they are just.....

[mystic saffron]

Lol

Yeah alright

thank you

how do I close the channel

[lusty helm]

np

do .close

[mystic saffron]

.close

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[mystic saffron]

A person is walking on an inclined plane, which forms an angle e with respect to the horizontal, and observes a lighthouse at the highest part of the inclined plane with an angle of elevation of 1.50. If he advances 2 m towards the lighthouse, the new angle of elevation would be 20. Calculate the height of the lighthouse.

[mystic saffron]

what step should I do next?

[odd edge]

@mystic saffron Has your question been resolved?

[lusty helm]

2m?

you have taken 20m

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[high jay]

ive got 3 points A, B, C

and i need to find the coordinates of D. knowing that the vector OD = 1/2(vector AB + vector AC)

i got that the length of the segments AB and AC are both sqrt(13)

so then OD = sqrt(13)

meaning that the point can be D2, no?

but the marking scheme says it's D

[uneven osprey]

huh?

[high jay]

look on the graph

[uneven osprey]

That's not how vector addition works

[high jay]

what is wrong?

oh yehh

i just remembered that its done differently lol

[uneven osprey]

fr

[high jay]

so ab + ac = 6

1/2 of 6 is 3

makes sense

thanks

.close

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[odd edge]

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[dry frigate]

how could you tell the difference between a point discontinuity and a infinite discontinuity without looking at the graph and just examining the equation of the function?

[ember oak]

You could do the formal definition of a limit to test if it is an infinite discontinuity

[dry frigate]

could you explain it a little bit more

[ember oak]

do you know what the formal definition of a limit is?

[dry frigate]

not really

this is my AP calc AB summer work

[ember oak]

You should learn formal limits in AP calc I thought

[dry frigate]

yes but

I havent taken the class yet

im like

doing the summer work FOR the class

[ember oak]

oh

[dry frigate]

at my school they give work before you take the class

[lusty helm]

[ember oak]

then I guess you could just plug in points near where you think it's discontinuous, then see if it's going to the same value, going to different values, or going to infinity

[ember oak]

It's a standard practice for AP courses

[dry frigate]

yea ik

was just saying

lel

but ty

[ember oak]

I had to memorize 200 chemical names for a course

[dry frigate]

ah hell nah

this calc packet is 40 questions 😭

grind is on

[ember oak]

not too bad

5 per day

[dry frigate]

it was assigned last week

i was out of town

due thursday

LMFAO

[ember oak]

20 per day

[dry frigate]

.close

[odd edge]

Channel closed

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[dry frigate]

.reopen

[odd edge]

✅

[ember oak]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dry frigate]

I sort of know but like

not really where to begin

so 1 ig

[ember oak]

Hint: recall that sin²+cos²=1

[lusty helm]

use property
sin²x + cos²x = 1

[dry frigate]

ok ok

[lusty helm]

& sec²x-tan²x=1

[dry frigate]

1-sin²x = cos²x

1+tan²x=sec²x

[lusty helm]

Charlie, i suggest you watch: https://youtu.be/T7D1W1oD8wo?si=LSjFdX3YgFU7AB_j

[dry frigate]

so cos²x - sec²x = 1

?

[lusty helm]

no

it's in ✖️

[dry frigate]

okay okay ill give it a try

[lusty helm]

multiplication

so (cos²x)(sec²x)

[dry frigate]

ohh

[lusty helm]

and sec x = 1/cos x

[dry frigate]

1/x * x

okay that makes sense

i get it

cancel out the x and your left with 1

[lusty helm]

it's a short trick

used myself

[dry frigate]

okay okay

bet

[lusty helm]

you can close by doing .close

[dry frigate]

okay

thanks guys

.close

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[compact steppe]

hi, can someone provide a hint for why this limit evaluates to 1. x -> +- infty

[tacit wasp]

Make a substitution: $t = \frac{1}{x}$

[clever fjord]

Alberto Z.

[ember oak]

You'll either need l'hopital or limit definition of derivative of e (along with derivative of ln x and inverse derivative identity)

[tacit wasp]

In this way you should get a "well known" limit

[compact steppe]

thanks to both, I see, imma try it

[proper solstice]

is the answer 0?

oh

1

[lusty helm]

1

[uneven osprey]

fr that seems like 10

[compact steppe]

ah I considered l'hopital but I didn't see it resolves nicely

[lusty helm]

Atif, you can refer it
whenever the value of limit approaches to 0, you can replace like shown in the above notes

Make sure it approaches 0 and is in product (won't work with addition/sub)

[compact steppe]

thank you vansh

[proper solstice]

,w graph (e^1/x-1)/1/x

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by anshulyolo121

[compact steppe]

.close

[odd edge]

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[proper solstice]

tHanK you

[lusty helm]

ywc

[odd edge]

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[lucid sonnet]

Could anyone confirm whether this way of solving a first-order differential equation is correct?

[cold sage]

does this seem right to you?

[lucid sonnet]

$= -e^{\frac{1}{2}x}$

[clever fjord]

Tomi

[lucid sonnet]

@cold sage Im not good at using correct mathematics but that is what i wanted to use there

<@&286206848099549185>

[cold sage]

im guessing you were going for
$$\frac{d}{dx} \left[e^{-\frac{x}{2}}y\right]=\frac{e^{\frac{x}{2}}}{2}\cdot e^{-\frac{x}{2}}$$

oops

that?

oh wait no

one moment

[clever fjord]

AℤØ

[cold sage]

that?

[cold sage]

the - should be in the exponent

[lucid sonnet]

I was using
$y = \frac{1}{I(x)} \cdot \int I(x) Q(x) dx +C$

[clever fjord]

Tomi

[cold sage]

aha, yeah thats what would appear after this

[lucid sonnet]

that is right

[cold sage]

after integrating both sides

the idea of how youre doing it is right, you just need to be a bit more careful

[lucid sonnet]

the solution is supposed to be

now, this is not really what im getting

[cold sage]

$$y=e^{\frac{x}{2}} \int{\frac{1}{2}dx}$$
$$=e^{\frac{x}{2}} \cdot [\frac{x}{2}+C]$$

[clever fjord]

AℤØ

[cold sage]

if you fix this youll get the same thing

[lucid sonnet]

where is this coming from exactly?

[lucid sonnet]

okay let me see

[cold sage]

correct it in this line and then continue

[cold sage]

its coming straight after this

after integrating both sides

and dividing by I(x)

I(x)Q(x) should simplify to 1/2

[lucid sonnet]

$e^{-\frac{1}{2}x} \cdot e^{\frac{x}{2}}$

[cold sage]

-1/2(x)

not just -1/2

[lucid sonnet]

ah

my bad

[clever fjord]

Tomi

[lucid sonnet]

so they cancel out?
$e^{-\frac{x}{2} + \frac{x}{2}}$

[clever fjord]

Tomi

[cold sage]

yeah

youre left with 1/2

since it was 1/2 e^(x/2) * e^(-x/2)

[lucid sonnet]

$\frac{1}{e^{-\frac{1}{2}x}} == \frac{1}{2} e^{\frac{x}{2}}$

[clever fjord]

Tomi

[cold sage]

this isnt true

[lucid sonnet]

then my solution is still not correct

[cold sage]

where did that 1/2 come from

[lucid sonnet]

from the left side?

[cold sage]

?

a coefficient cant just appear from nowhere

youre applying the idea that 1/a=a^(-1)

[lucid sonnet]

that is 1/2x from the right handside sorry, that is from the solution

[cold sage]

i dont follow

[lucid sonnet]

I ended up with
$\frac{1}{e^{-\frac{1}{2}x}} + \frac{C}{e^{-\frac{1}{2}x}}$

[clever fjord]

Tomi

[cold sage]

that isnt right

how did you get there?

[lucid sonnet]

I(x) = 1/e^{-1/2x} right?

[cold sage]

I(x)=e^{-1/2 x}

[lucid sonnet]

okay but I need to divide by 1 over I(x)

[cold sage]

from what i can tell, you some how have the integral of 1/2 as 1

[lucid sonnet]

what is I(x) supposed to be? weve already passed thru that part

[cold sage]

this is I(x)

[cold sage]

this is 1/I(x)

[lucid sonnet]

1 over I(x) here

[lucid sonnet]

yes?

[cold sage]

we were previously at this top line

but you have your integral coming out to 1+C for some reason

[lucid sonnet]

$\frac{1}{e^{-\frac{1}{2}x}} [1 + C]$

[clever fjord]

Tomi

[cold sage]

its not 1+C

we were integrating 1/2

an integral cant come out as 1 even if you integrate 0 youd get an unknown constant, just the C

[lucid sonnet]

I corrected this already

$\frac{1}{e^{-\frac{1}{2}x}} [\int e^{-\frac{1}{2}x} \cdot e^{\frac{x}{2}} dx + c]$

[clever fjord]

Tomi

[cold sage]

first off, youre missing a half

[lucid sonnet]

:/

[cold sage]

Q(x) was 1/2 e^(x/2) and the half has disappeared

second off, the integral of 1, is not 1

[lucid sonnet]

ah yes that should be x + c

[cold sage]

yes, now resummon the half

from the void

[lucid sonnet]

isnt it minus 1/2 here?

[cold sage]

no

[cold sage]

^

[cold sage]

to this

was always positive

see

[lucid sonnet]

oh okay wait

[lucid sonnet]

so now I end up with
$\frac{x}{2e^{\frac{-1}{2}x}} + \frac{C}{2e^{-\frac{1}{2}x}}$

[clever fjord]

Tomi

[lucid sonnet]

this should not work out i think

[cold sage]

still wrong

there is no C/2

though i guess its still a constant

but how you got there is not right

[lucid sonnet]

ah caused i pulled out the one half

should not have done that

cause that only relates to one term

[cold sage]

you have
$\frac{1}{e^{-\frac{x}{2}}} \left(\frac{x}{2}+C\right)$

oops

[clever fjord]

AℤØ

[cold sage]

yeah, that

[lucid sonnet]

yes it makes sense, I will clean up the whole thing and make sure that i got it right

[lucid sonnet]

okay i got to this poitn now, but to get what the books says, you need to flip the sign to move them up in the numerator

[cold sage]

sure

1/a=a^(-1)

[lucid sonnet]

okay got it now, thank you @cold sage

.close

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[mystic saffron]

How should I begin?

[mystic saffron]

ABCD is a square

Calculate that value

I know their distances are equal

of the points

but Idk how I could put the expressions as equal in order to find the values

[mystic saffron]

I have already tried

and it's a mess

[steep mantle]

is B (12,5)?

[mystic saffron]

Yes

[verbal tinsel]

Yd = 5

[mystic saffron]

Mm they don't say it's in the same level

[verbal tinsel]

Equate BD = aroot(2) by Pythagoras theorem

[mystic saffron]

I would be assuming that

[verbal tinsel]

Oh

[mystic saffron]

and if I follow the same logic, then Xa = Xc = 6
and 12/15 is not the answer

[verbal tinsel]

how about equating length of sides with distance formula

We might get something

or equating the slopes

[mystic saffron]

Mm I get it

[mystic saffron]

not for this topic

In this exercise I am not supposed to use that

[mystic saffron]

well that's what I tried

before

[verbal tinsel]

How many equations you ended up with?

[mystic saffron]

I would need to organize it, but there would be 4

[verbal tinsel]

Then you're doing it right

[mystic saffron]

but values squared

[mystic saffron]

yeah, it's the most intuitive to do

[verbal tinsel]

There could be easier way by solving with slopes though

[mystic saffron]

m perhaps, but I can't solve it like that

it's the next topic

I am catching up

in some topics

that I know

but not so well

maybe

idk

like this problem

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

<@&286206848099549185>

[torpid edge]

Yc=Yd+5

let me try to solve it

idk either

[mystic saffron]

mm not sure

[torpid edge]

wait why not

[mystic saffron]

.

[mystic saffron]

why do you say this

?

[torpid edge]

the translation is the same

[mystic saffron]

I suppose that you assumed they are in the same line

[torpid edge]

or basically slope

[torpid edge]

what is in the same line

[mystic saffron]

horizontal

[torpid edge]

no

Yc-Yd = 5-0

do you see that

[mystic saffron]

oh I get it

yes

We can use that

I thought you assumed Yd was 5

[torpid edge]

oh

yea mb

didn't explain clearly

ok

so you can do the same thing:
Xc-0=12-Xa

so Xa+Xc=12

[torpid edge]

what are the answer options btw

Xc,Yc?

[mystic saffron]

it's b

[torpid edge]

what do they mean

[mystic saffron]

mm

Xa+Xc=12

[torpid edge]

what does 0.63 mean

[mystic saffron]

this

is the answer of that

[torpid edge]

but it's a fraction

[mystic saffron]

yes

[torpid edge]

is it in decimal

[mystic saffron]

yes

[torpid edge]

bruh what

ok

let me try to solve

[torpid edge]

12/19

drop the altitude from B

let origin be O

let the altitude from B to x axis be E

triangle BEA is congruent to triangle AOD

it's pretty easy from there

[mystic saffron]

Yd = 7 and Yc = 12
then Yd + Yc = 19

yes

Thanks a lot

(:

.close

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[rapid mortar]

can someone explain to me this math concept

[torpid edge]

do you see any common terms in both the numerator and denominator

[odd edge]

@rapid mortar Has your question been resolved?

[rapid mortar]

judging from the answers, I see that there's 8 and 4

[torpid edge]

ig you haven't done similar practice problems like this so I'll give a brief solution

so you see that you can factor out y in numerator and denominator:
y(56y-28)/y(35y+21)

make sure you understand this

since it's both in numerator and denominator

you can cancel the y

to get 56y-28/35y+21

you see that there's also a multiple of 7 you can factor out

which gives you 8y-4/5y+3

[odd edge]

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[fresh sphinx]

how do i start this question? i tried just finding a bunch of derivatives but they keep getting longer and longer so im assuming that youre supposed to do some rearranging of f(x) first

[fresh sphinx]

like is there a strategy or something i cn use for this question

[orchid torrent]

Complete the square and then use the binomial theorem for fractional values

[odd edge]

@fresh sphinx Has your question been resolved?

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[wanton mist]

I need help with inequalities

[torpid edge]

,rotate

[odd edge]

@wanton mist Has your question been resolved?

[elder vault]

Shit

Fk

It is tough

[mystic saffron]

make ab-b² the largest since its always positive

you should have a function with only a after that

[elder vault]

How

[mystic saffron]

-(b-1/2a)²+1/4a²

largest when b = 1/2a

[elder vault]

What is it?

[mystic saffron]

?

[elder vault]

I cannot see how you get the maximum

[mystic saffron]

thats unfortunate

[umbral minnow]

can you tell us the way u transformed the denominator to this?

[mystic saffron]

complete the square

[umbral minnow]

ah

[elder vault]

What about the sqrt(2)*a^3 term

How do you determine the minimum occurs when ab-b^2 has it’s maximum

[mystic saffron]

differentiation

ig

[elder vault]

Partial derivative?

[mystic saffron]

no

[mystic saffron]

1

[elder vault]

I’m confused, because there are two variables

[odd edge]

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[elder vault]

That did not explain my question

No

[elder vault]

No!

.reopen

[mystic saffron]

whafs ur question

[odd edge]

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[zealous venture]

Questions regarding vectors
so I have so values (x, y) and then using sqrt(x^2 + y^2) I get the magnitude
now if I have a center and do not know the (x,y) how do i draw the vector?

[odd edge]

@zealous venture Has your question been resolved?

[odd edge]

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[dusky locust]

can somebody explain this to me? Why is L the solution to c(a) and M the solution to b(a)?

[odd edge]

@dusky locust Has your question been resolved?

[odd edge]

@dusky locust Has your question been resolved?

[odd edge]

@dusky locust Has your question been resolved?

[odd edge]

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[plush nest]

.close

[odd edge]

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[wooden mica]

Please help, I'm getting the wrong answer according to the university but I'm using all the rules correctly in my view and cant get another answer but obviously I'm wrong. Can someone kind just please point out in which step I do something wrong because I cant see it. Here is the picture of my calculation and the question.

[uneven osprey]

in the second last step

it'll be 24 in numerator

[wooden mica]

Wait do you mean -(1/6) - 4 is not equal to -(37/6)?

[steady tide]

classic arithmetic error

4 = 24/6

[wooden mica]

Omg

Oh my god

Thanks so much

My head is like suddenly locked on 6*4 = 36

How could my head over and over justify this lol

[steady tide]

we all been through this

[wooden mica]

I litterally went over this exact number a couple of times and was like yea 6*4 = 36

.close

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[wanton merlin]

,tex $y''-2y'+y' = (4x-4)e^{-x}$

[clever fjord]

Sliden

[wanton merlin]

I can't solve for y_p

I have found the compliment answer

,tex $y_c = (Ax+B)e^{x}$

[clever fjord]

Sliden

[odd edge]

@wanton merlin Has your question been resolved?

[wanton merlin]

.close

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[odd edge]

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[thin palm]

very sorry for this dumb question but

[thin palm]

how do I integrate the area of a triangle?

[tacit wasp]

What do you mean?

Do you have an example or an exercise?

[thin palm]

yes

what I did is

I put this on a graph

[umbral obsidian]

you cant "integrate" the area of a triangle

integration helps you find the area of a triangle

[thin palm]

my bad sorry for the misunderstanding

[umbral obsidian]

you dont need integration to find the area of this triangle

[thin kelp]

yo that triangle

[umbral obsidian]

and plus it cant exist

[thin kelp]

doesnt exist

[umbral obsidian]

because 3 and 2

is smaller than 6

[thin palm]

oh I didn’t know I just put random side lengths

what triangle would work?

[umbral obsidian]

any 3 sides a, b, c
where a + b > c

wait

and a + c > b

and b + c > a

[thin palm]

would this work?

[thin palm]

I know but i’d want to use it

[tacit wasp]

Ok so you need a reference frame

And wrote coordinates of the three vertexes (or at least the equations of the lines that the three sides belong to)

[odd edge]

@thin palm Has your question been resolved?

[odd edge]

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[wide drift]

Part b

I think getting the derivative of f(x), subbing in x = 0 and y = 1 will give me the slope right? And that should be the negative reciprical of 4 right, which is the slope of g(x), right?

Is my method correct?

[late dust]

Not quite, the derivative of f is undefined at 0

[wide drift]

Hi Nel!!

[wide drift]

Oh

So what should I do

[late dust]

Well they gave you h for a reason

[wide drift]

Hmmmmm

[wide drift]

Do I literally just do the same thing I described above, but with h(x)?

And then say some shit like 'f(x) is equal to h(x) therefore the slopes are the same therefore they are both perpendicular to g(x) at (0, 1)'

[late dust]

Kinda

The point is that you can't compute the derivative of f at 0 or the derivative of g at 0

But since h = f for all x =/= 0, and k = g for all x =/= 0

You can still say that they are perpendicular at that point

[wide drift]

Hows that different from what I said? Not asking in a rude way haha I just genuinely dont see the difference 😅

(except that you wrote it nicer than me)

[late dust]

Simply because "f(x) is equal to h(x)" is false

(and that you also need to use k instead of g)

[wide drift]

Ah ok... so i would lose marks for that in an exam?

[wide drift]

whats k?

[late dust]

Depends on your teacher

[late dust]

It's in the question

[wide drift]

We dont have an equation for k tho right?

Oh it 'coincides'

How is 'coinciding' different to being the same as something?

[late dust]

It's not, but the question also states "for x =/= 0"

It's like the most important detail of the question

[wide drift]

So what does coinciding mean sry?

In maths context

[late dust]

It does mean the same

I was saying it's not different

[wide drift]

Oh

Ok well I think that answers all of my questions

Thank you!!!

❤️

.close