help-27

[lost laurel]

the left side

[surreal shell]

again i multiplied by congigate already?

[lost laurel]

oh

didn't notice

sorry

[surreal shell]

should I attempt something like that for the other side

[lost laurel]

wait

just make the LHS have a common denominator instead

[wind marten]

Hint: (a+b)^2 on top
Hint 2: Factor bottom with difference of squares

[surreal shell]

oh yea

[lost laurel]

that would give you $\frac{cos(x)-cos(x)sin(x)+cos(x)+cos(x)sin(x)}{cos^2(x)}$

[surreal shell]

the (a+b)(a-b)

[woven radish]

A dense set(Ping when reply)

[surreal shell]

ok thanks

.close

[devout snow]

Channel closed

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[wind marten]

Wait pause DONT factor the bottom just leave it as cos^2 lmao

[surreal shell]

yea

i wasn't going to touch that side

.reopen

[devout snow]

✅

[surreal shell]

I got lost somewhere along the way

[lost laurel]

simplify this

[surreal shell]

doesnt that give0/cos^2(x)

[lost laurel]

no?

$\frac{2cos(x)}{cos^2{x}}$

[woven radish]

A dense set(Ping when reply)

[surreal shell]

oh and then you just ok

yea thanks

.close

[devout snow]

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[devout snow]

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[cedar pike]

Help pls

[cedar pike]

Pls pls pls pls

[royal radish]

i am less likely inclined to help you now that you said pls 4 more times

[solar goblet]

,rccw

[cedar pike]

Help

[devout snow]

@cedar pike Has your question been resolved?

[devout snow]

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[wispy geyser]

am i right with C?

[faint zinc]

,w expand (x+3)(x-3)(x-4)

[woven radish]

Results provided by WolframAlpha

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

Requested by omnipotententity

[faint zinc]

Yes

[wispy geyser]

your a legend man

thank you sm

.close

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[twin temple]

Can someone check if my work looks correct?

These are the original questions for reference

[devout snow]

@twin temple Has your question been resolved?

[twin temple]

<@&286206848099549185>

[devout snow]

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[twin temple]

.reopen

[devout snow]

✅

@twin temple Has your question been resolved?

[twin temple]

.close

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[devout snow]

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[devout snow]

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[urban silo]

Hi

Can anyone help me with this

[tall knoll]

!msgdel

[devout snow]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[restive river]

tl;dr: open a new channel, can't use this one no more

[devout snow]

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[odd harness]

Can’t get this right no matter what I try

[wicked turtle]

what's the vertical distance between the highest point and the lowest?

[odd harness]

9

[wicked turtle]

2 - (-6)

is not 9

[urban harbor]

3*

oh nm 2

[sinful rover]

what is going on

[urban harbor]

i misread that 2 as a 3 too for some reason :p

[odd harness]

8

[wicked turtle]

i just counted grid lines haha

[wicked turtle]

yea

[sinful rover]

yesss

[wicked turtle]

so what's the amplitude?

[odd harness]

I’m cooking

[wicked turtle]

(hint: it's not 5)

[odd harness]

8

[wicked turtle]

no..

[sinful rover]

nooooo

8/2?

[odd harness]

4

[sinful rover]

yessssssss

[wicked turtle]

ok so that's one issue

[sinful rover]

its half of the vertical length

the next....

[wicked turtle]

now what's the period of this function?

[odd harness]

2pi/pi/5

[sinful rover]

huh

[wicked turtle]

no there should not be any pi involved

what's the horizontal distance between two peaks, for example

[odd harness]

8?

[wicked turtle]

how did you find that?

can you list two x values where a peak occurs?

[odd harness]

Peaks at -6 on X coordinate

And 2

[wicked turtle]

2?

x=2?

[odd harness]

0

Oops

[wicked turtle]

yea, that's better

[odd harness]

So 6

[wicked turtle]

so period = 6 yes?

yep

[odd harness]

Yeee

[wicked turtle]

now how do you use the period

when writing the formula for the function

[odd harness]

2pi/K

[wicked turtle]

yep

so it should be cos(what?)

[odd harness]

2pi/6

[wicked turtle]

yep

ok, so that just leaves the C

[odd harness]

4 Cos (2pi/6)

Is what we got so far

[wicked turtle]

yep

now what should C be?

how do you find it?

[odd harness]

+2?

Y intercepts

[wicked turtle]

+2 will take cosine and shift it up by 2

this one looks like it is shifted down

[odd harness]

So -2

[wicked turtle]

yea

so last question, how do you know if it's cos or sin

[odd harness]

You tell me

[wicked turtle]

well if we just have cos(x) and sin(x)

what's the difference between them, how can you tell from the graph which it is?

[odd harness]

Sine starts at 0

[wicked turtle]

right

and cosine has a peak at 0

[odd harness]

Yea

So it’s cos

[wicked turtle]

and your function has a peak at 0 so yep

cosine it is

so i think you have all the ingredients right now

[odd harness]

Looks like this on Desmond

[wicked turtle]

,rotate

[wicked turtle]

wait, don't you want cos?

and why is it just plotting a straight line

oh you forgot the x

inside the cos

[odd harness]

So this?

[wicked turtle]

how did you get 5 in the denominator

didn't we find it was 6?

[odd harness]

Bingo got it right

[wicked turtle]

Bungo

haha

[odd harness]

I got 3 more

[wicked turtle]

oh man

[odd harness]

Or should I just Chat GPT

[wicked turtle]

i'll let someone else help with that, it's like 3:30am here

no don't trust chatgpt for math, it makes a lot of mistakes

[odd harness]

Let’s do one more

Bungo

[wicked turtle]

just close this channel and open a new one

so it will go to the top of the list

[odd harness]

How about we knock another one out

[wicked turtle]

nah man i'm tired haha

[odd harness]

It’s word problem

It’s easier

[wicked turtle]

there are other helpers online, someone will help i'm sure

[odd harness]

Nah

#L mans

#close

.close

[devout snow]

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[celest shore]

Hello, I know its probably really primitive question. Currently, Iam learning how to integrate functions. Whats wrong with my solution of the bottom (3.) problem?

[solar goblet]

hehe "primitive"

[lost laurel]

is that $\int \frac{dx}{2+3x^2}$?

[woven radish]

A dense set(Ping when reply)

[celest shore]

Yeah

[lost laurel]

Well, what do you think

Does this remind of a standard derivative

[celest shore]

I wanted to adjust it to formula 1/1+x^2

[midnight echo]

Before applying the arctan integral at the end of line (3), you substitute in u = sqrt(6)x. This is correct, but you also need to adjust the dx (as you always do in a u sub)

[zenith narwhal]

You can use atan formula

[celest shore]

So All I need is to sub √6 * x = u?

[midnight echo]

That's already what you did. However you forgot the du/dx = sqrt(6)

In order to convert the differential (dx) into du (recall regular u substitution), you have to do dx = 1/sqrt(6) du

ie. divide by sqrt(6)

[celest shore]

Thank you very much

[devout snow]

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[solid osprey]

ABC is a right angle, BD=CD=DE=1 and DE is the angle bisector of ADC, find AD

[solid osprey]

rough sketch by me, dont expect it to be accurate in the slightest bit

[urban harbor]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[urban harbor]

oh nm, i thought that was an equilateral triangle in the bottom right, which wouldve made it impossible, guess it isnt

[sudden willow]

what is this app

[solid osprey]

notes

[fossil locust]

I misread titik as titit

[solid osprey]

wierdly the topic is power of a point??

[fossil locust]

hmmm

[feral agate]

interesting

[hollow ice]

AB and AE are tangents to circumcircle BEC

[feral agate]

i think you can let AD=x and get some equations in x using pythagoras' theorem and angle bisector theorem

seems a bit tedious

[feral agate]

[hollow ice]

thats the only circle I see to use power of point

[solid osprey]

sceptic on that AE

[hollow ice]

BEC is 90

[solid osprey]

doesent that imply AED=90 and this question immediately crumbles

[hollow ice]

wait, why?

[solid osprey]

wait am i stupid

AE isnt tangent it just passes the circle

[hollow ice]

Ohhh right

I am stupid

So, AE.AC = AB^2

Ughhh, its geometry... Why do I even try 😭

[solid osprey]

IKR FUCK GEOMETRY

i had extras while exam and the only one i didnt go to was geometry cause FUCK that

[hollow ice]

Well, AD^2 = 1 + AB^2 = 1 + AE*AC

[fossil locust]

also triangle BEC must be similar to triangle ABC

(right triangles and both share angle C)

[solid osprey]

no way i actually have short term memory loss

[fossil locust]

if angles DBE = BED = x

yeah what am I thinking angle BAC = x

[hollow ice]

all that seems consistent

[fossil locust]

DC/DA = CE/AE means 1/DA = CE/AE

or 1/(DA + 1) = CE/CA = CE/(BC * BC/CE) = (CE/BC)^2 = CE^2 / 4

if we can somehow find CE

[hollow ice]

idk why you nah

[fossil locust]

oh wait lol

[solid osprey]

yeah but you said ABC≡BEC so that means BAC=EBC

[fossil locust]

also if I'm not tripping, BE perpendicularly bisecting AD means that AEDB is a kite

or that AB = AE

[hollow ice]

You are tripping the same way I did

[fossil locust]

really?

[solid osprey]

BEA is a right angle not DEA

[fossil locust]

ah

[solid osprey]

dw i also thought that and found a contradiction then scratched my head for like 5 mins

[fossil locust]

ah yeah

[finite briar]

Sooo ur not done right??

[solid osprey]

no

[finite briar]

Hmm i remember reading a formula for finding ADB

Gimme a minute

[hollow ice]

EDC=180-2C, so ADC is twice that, and ADB = 180-ADC

[finite briar]

Bruh no in terms of cosine and stuff

[hollow ice]

Oh lol

[finite briar]

Nvm it was for angle bisector

[hollow ice]

BD * sin(ADB) = 1/2 * BE * sin(ADC) coz areas are equal

this is the only one I can think of

[finite briar]

Hmm?

What are u equating again?

[hollow ice]

area = 0.5 a b sinA

[finite briar]

This figure seems cursed, i almost thought AD perpendicular to BE

[hollow ice]

We want length AD

[finite briar]

Area of which two triangles that's what I am asking

[hollow ice]

the ones split by median, ADB and ADC

[finite briar]

U missed a half on the rhs btw

[finite briar]

And this is always true because sin(180-x)=sin(x)

[hollow ice]

Yea, thats pointless, I realised that later

Now Im just gonna lurk to see what yall come up with. I just enjoy good geometric solutions. I cant do them myself for most part

[solid osprey]

i have no idea how to make an accurate diagram :/

[finite briar]

I rederived Pythagoras 💀

[solid osprey]

peak

[finite briar]

Welp I got to the lengths of AE,EC,BE in terms of AB

[finite briar]

Wait since this is power of a point..what if the construction is to extend AD to F such that A,B,C,F are concyclic?

[solid osprey]

lemme try to find a way to cook up an accurate diagram hold on

[finite briar]

Okay

[solid osprey]

ok this is really close (like 5 sigfigs close)

km does this do anything

[finite briar]

I see

Welp

I think I got it

Answers sqrt(5)?

[solid osprey]

no clue

[finite briar]

Hmm but uhh my solution isn't geometric

[solid osprey]

although apparently its in the form a+sqrtb and were finding a+b

[finite briar]

Its coordinate geometry+trigonometry

[finite briar]

I hope it's 5

[solid osprey]

although it could be a red herring

[finite briar]

Where did u even find this

[solid osprey]

uhhhh my teacher?

although i doubt its sqrt5

[finite briar]

What where did I mess up.?

[solid osprey]

i have no idea

[finite briar]

Uhh should I tell you what I did?

[solid osprey]

i mean sure ig

[finite briar]

So I let A be (0,x)

From there I found the slope of AC

[solid osprey]

its not rly graded my teacher just gives us like 15 questions but we only got the time to go trough like 7 so the rest are practice

[finite briar]

That's tan(theta)

[finite briar]

Hmm

Okay nvm found my mistake

[solid osprey]

theta is BCA right

[finite briar]

Yesh

I ended up doing ADC=180-2theta when in reality EDC = 180-2theta

[devout snow]

@solid osprey Has your question been resolved?

[solid osprey]

.close im gonna snooze gn

[devout snow]

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[finite briar]

Gn @solid osprey

[devout snow]

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[spark hound]

Can some1 explain this PLEASE

[winter patrol]

which part do you need expaining

[spark hound]

Why is sin(-t) = -y, but cos(-t) = x?

Like I understand -t means going clockwise

[winter patrol]

sin gives the y coordinate,
cos gives the x-coordinate

[spark hound]

If we go to Q3, then cos will be negative tho no?

[winter patrol]

no

cos gives the x-coord

the point is (x,-y)

you're still on the right side of the plane, where the x-values are positive

[spark hound]

So is it just saying -t being in Q1 or Q4?

[winter patrol]

no

[spark hound]

So cos(any negative t) = x

?

[winter patrol]

no

its saying if cos(t) = x, if t is in Q1,
then cos(-t) = x

[spark hound]

Ahh

Okayh

Okayyy

Ty

.solved

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[north roost]

?

[devout snow]

@steep shard Has your question been resolved?

[steep shard]

Nope

[north roost]

cant u find the solution to this

@steep shard

u have the initial values

u can find the solution first, then apply differentiate it, no?

[steep shard]

I can do but after finding the complimentary function I get stuck to find c1 and c2 for the solution

[north roost]

how so?

\begin{align}
c_1 + c_2 &= 2\
\frac{c_1 + ec_2}{e^3} &= -\frac{1-3e}{e^3}
\end{align}

[woven radish]

k

[steep shard]

Sorry I mean complimentary function

I get y(0) = 3

[north roost]

can i see ur work

[steep shard]

So if I differentiate y'(0) which is equal to zero...

[steep shard]

Yess pls

[north roost]

u solved for

c1 and c2 wrong

its simpler than that

@steep shard

[steep shard]

Hey I got the answer thnx mn it took 1 hrs

But I am satisfied thnx to help anyways sir

[north roost]

👍

[steep shard]

Thnx for helping me sir

[steep shard]

Haa I am wrong ig

[north roost]

?

[steep shard]

How is this one done? The step can you explain pls

The answer is -3 at last and your right

[devout snow]

@steep shard Has your question been resolved?

[devout snow]

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[north glen]

what is phi here?

[north glen]

how can I calculate it?

[verbal beacon]

I'm not sure if this is correct, but phi should go from 0 to pi/2 and theta should go from pi to 3pi/2

[north glen]

both are right, but idk how to get phi

I have the result

but idk how to get phi

[verbal beacon]

r u talking abt the bounds of phi?

[north glen]

yes

[verbal beacon]

well i think phi has to make a vector above the xy plane since z>=0

[runic folio]

should be 0 to pi/2 i believe

[verbal beacon]

yess ok thx

can u explain im not rlly good i think

[runic folio]

x^2+y^2+z^2 is a sphere centered at origin

with the restrictions on x,y and z we know that we only have to consider 1/8th part of sphere

you can visualise a vector which follows a spherical curve from the origin to one of the axes

[narrow bloom]

yeah this, and the question becomes how do we parametrize the correct 1/8th of the sphere

following this convention https://en.wikipedia.org/wiki/Spherical_coordinate_system#Integration_and_differentiation_in_spherical_coordinates you get that you should find values of φ such that (cos φ, sin φ) lies in the region where x≤0 and y≤0; see the picture there

[devout snow]

@north glen Has your question been resolved?

[north glen]

lemme read

😄

so for example this would be 1/4 pi?

using that "vector"

I'm absolutely cooked with sphere coords

[devout snow]

@north glen Has your question been resolved?

[devout snow]

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[spring jungle]

Need someone to check my work

for Row 3 i have: 0, 3, 6, 9, 12, 15, 2, 5, 8, 11, 14, 1, 4, 7, 10, 13

Row 4: 0, 4, 8, 12, 0, 4, 8, 12, 0, 4, 8, 12, 0, 4, 8, 12

b) I have x = 3, y = 11
because 3 x 11 = 33 \equiv 1 (mod 16)

c) 15x \equiv 7 mod 16
15 \equiv -1 mod 16
-x \equiv -1 mod 16
-1 (-x) \equiv -1 (7) mod 16

x \equiv -7 \equiv 9 mod 16

x = 9

d) Only numbers relatively prime to 16 have inverses mod 16 (and thus a 1 in their row). All even numbers share a factor of 2 with 16, so they are not invertible.
The rows without a 1 are the even numbers: {2, 4, 6, 8, 10, 12, 14}

[gritty badger]

bro

just copy ur pic on paint

it would be easier to read

for us

[spring jungle]

oh

[devout snow]

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[spring jungle]

omg it closed again

if you click the picture its still harder to read?

[devout snow]

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[tawdry burrow]

Hey, just curious on how to solve this question and need some pointers on what to do. I'm unsure of what it means by a "linear graph"

[acoustic leaf]

you have the graph of y = x^2 - 6x + 5 there. how might you rearrange the given equation to involve that?

[tawdry burrow]

By adding x and subtracting 2?

[acoustic leaf]

yes

[tawdry burrow]

Subtract 4*

So x^2-6x+5=x-4

[acoustic leaf]

yes

[tawdry burrow]

Ok, so what do I do after that ?

[acoustic leaf]

if you have an equation like that, the solutions will be the intersection of the two curves

[tawdry burrow]

So the intersection of x-4 and x^2-6x+5?

[acoustic leaf]

yes

[tawdry burrow]

(4,-3)

But how does that help give solutions?

Ohhh

So the 2 quadratic equations also intersect at 4,-3

[acoustic leaf]

can you show how you got those intersections?

[tawdry burrow]

On the graph?

Or just like, generally

[acoustic leaf]

yes

on the graph

[acoustic leaf]

you should draw the line y = x - 4

[tawdry burrow]

I think i see it now

I calculate the points of intersection and that should be the answer?

[acoustic leaf]

yes, the x-coordinates of the points of intersection are your answers

[tawdry burrow]

Ok

I'll quickly solve that then

I've forgotten how to do it

I presume you solve them simultaneously

[acoustic leaf]

the point is not to actually solve it algebraically, just come up with an approximate x-coordinate from the graph

[tawdry burrow]

Oh

I was going to say doing it simultaneously would just loop back

I got 1.7 and 5.3

[acoustic leaf]

that works

[tawdry burrow]

Ok, thank you for the help!

.close

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[stiff quartz]

how do I start?

<@&286206848099549185>

[devout snow]

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[hot viper]

<@&286206848099549185>

[sinful rover]

find the integral of f(x)

[hot viper]

how do i set up the integral?

[sinful rover]

and evaluate from -4 to 4

[hot viper]

ok

[sinful rover]

integral from -4 to 4 of (20x) +x^2 - x^3 dx

@hot viper you got it?

[hot viper]

i split the integral

[sinful rover]

why?

youre finding the area not the total

there is negative area in that problem

[hot viper]

∫[g(x)-f(x)]dx+∫(f(x)-g(x)]dx?

but in between -4,4

[sinful rover]

why would you do g(x) first?

oh i see

yea

[hot viper]

I believe this is the formula for the sum of two integrals

[sinful rover]

sorry i thought you were trying to find the absolute area

yes youre right

[hot viper]

okiiii

[lusty sapphire]

!15m

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[sinful rover]

when g(x) is over f(x) thats where the first integral is to when it ends and then f(x) is over g(x) is when the second integral starts and ends

[sinful rover]

so what are your bounds for this

for each individually

[hot viper]

sorry i am having trouble understanding your question.

do you mean the shaded area?

because you add both and get the A

[sinful rover]

yes

what are the boudns for each of the integrals

you have the integrals right but you need bounds

[hot viper]

it would be

x(x-5)(x+4)=0

[sinful rover]

um what

[hot viper]

5,0,-4?

[sinful rover]

no

for what values of x is the shaded area begining and ending

[hot viper]

oh

∫0,-4, ∫4,0

[sinful rover]

im having a hard time understand your notation, is the first number going on the top or bottom?

[hot viper]

firs number on top yes

[sinful rover]

ok yes those bounds are right

good job

[hot viper]

now i just set up the integrals, solve them, then add?

[sinful rover]

yes

you had the equations right before so the integral of g(x) - f(x) and then the second was f(x) - g(x)

plug in your numbers and solve 🙂

[hot viper]

∫0,-4(0-(20x+x^2-x^3))dx+∫4,0((20x+x^2-x^3)-0)dx

[sinful rover]

yesssss

[hot viper]

[-10x^2 - 1/3 x^3 + -1/4 x^4]0,-4 + [10x^2 + 1/3 x^3 - 1/4 x^4]4,0

[sinful rover]

it should be 1/3 x^4, the negatives cancle

in the first one

but other than that yes and then evaluate

[hot viper]

736/3+-224/3

512/3?

[sinful rover]

not sure plug into a calc

but the equation is right

[hot viper]

ill just check it on pearson.

i got it wrong 😦

ok my calc says 192

[supple knot]

,w int -4 to 4 abs(20x+x^2-x^3)

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[hot viper]

okay! got that thanks for the help

[sinful rover]

youre welcome!

❤️

@hot viper close channel with .close

[hot viper]

.close

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[azure portal]

If anyone are familiar with Khan Academy, would you mind telling me if there is harder stuff on the website than the app? I heard it on the internet but I’m unsure.

[dense jay]

you could just look at the website

[devout snow]

@azure portal Has your question been resolved?

[azure portal]

Yea

[devout snow]

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[granite anchor]

Idek how to start these

[versed jetty]

for b my first instinct is to calculae the mass sof water involved

[granite anchor]

How do I do that

This is calc 2 btw

Unit is called “work”

[versed jetty]

well u have the volume

so multiply that by the density

[granite anchor]

okay so 30000 is the mass

idk what next

like is there some sort of forula

im supposed to use?

to solve work problems

💔

okay so i found out

Work = Force * distance

and Force = mass * acceleration(gravity)

[granite anchor]

but what ist he distance for b)?

is it the length of the tank?

im so confused i thought i had to integrate somewhere for this too

a calc 2 question without integration sounds sus

<@&286206848099549185>

[sinful rover]

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[granite anchor]

@cosmic quarry ban her

[sinful rover]

what did i do?

lmfao

[granite anchor]

not help me 💔

[sinful rover]

omgg momyyyyy

[granite anchor]

helper pls help

im dead lost

[devout snow]

@granite anchor Has your question been resolved?

[granite anchor]

BRO

<@&286206848099549185>

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[strong igloo]

is this a typo? There's no test case for f'(t)

[winter patrol]

did you try solving with the given info

[strong igloo]

yeah, 4e^t -3sin(t) + Ct + D

oh wait

am i meant to solve a system of equations

[winter patrol]

yes

[strong igloo]

wild

thanks

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[restive cairn]

why does quotient rule give me m = -1/3?

[restive cairn]

the answer key states that its -1/9 and idk how they got that

but like they didnt use quotient rule

[acoustic leaf]

could you show your work applying the quotient rule?

[restive cairn]

oh okay

It’s a bit messy but I’m doing it on paper

But it’s at the bottom

[acoustic leaf]

,rccw

[acoustic leaf]

1. note that the derivative of 3 - x is -1, you did not use this result
2. - (6x - 2x²) ≠ -6x - 2x², you did not distribute the - properly

[restive cairn]

ohh ok i see

i usually neglect the derivative if its just x

😔

thank you

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[young nymph]

I need help with integrating the volume between two curves about the y - axis. I have functions y = 2 - x and y = x^2 and I need to find their volume from [-2, 1]. I tried using the shell method but I get a negative number.

[devout snow]

@young nymph Has your question been resolved?

Show your work, and if possible, explain where you are stuck.

[young nymph]

Sorry for poor handwriting

and i get -9pi/2

<@&286206848099549185>

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[median depot]

i dont remember when to look at the restricted domain vs the restricted range for these types of problems

[median depot]

ik how to solve its just i know we restrict something but i dont remember where

nvm

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[limber ermine]

how can i write 3^2(n-1) as this , like what is the formula

[past mango]

if that says $\frac{9^n - 1}{8}$ then you can't. $3^{2(n-1)} \neq \frac{9^n - 1}{8}$

[woven radish]

Everbound

[devout snow]

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[restive river]

can someone help with C programming?

[restive river]

i'm trying to storage numbers in an array with a while loop and want it to stop when I input 0

my first attempt did not work:

i tried changing my code but without success, tried asking people... eventually had to resort to Gemini

and this worked:

I know what the do-while loop does, what I'm asking is someone to tell me what is the difference here

in the first code, the loop ignored when i input zero

[static ember]

because you first stored at index i, then incremented i and then checked the array at the new index i

effectively storing at i, while checking i+1 instead

[restive river]

what do you mean? isn't i=0 in the first iteration?

[static ember]

i = 0, you put something into seq[0], then i = 1 and you check if seq[1] is not 0

[restive river]

i see, by that logic wouldn't the loop stop anyways since the elements in the array are already zero?

when i declare seq[5] and seq[0]=1 it's the array [1 0 0 0 0] correct?

[static ember]

yes

[restive river]

in the first iteration i store a number let's say X, then in the next iteration the while should verify for [X 0 0 0 0 ] ?

it should stop because the other numbers already 0

[static ember]

I would use an if check after the scan instead, with a break if it's 0

[restive river]

lemme try that

i also thought of that

[static ember]

well the array isnt guaranteed to be 0 initialized

it contains garbage values

[restive river]

oh

[static ember]

memset it before

[restive river]

I think this is the problem

[static ember]

it is a problem but not the problem

[restive river]

it still goes through

[static ember]

you need the if check after the scan

[restive river]

WORKED!

I feel stupid now lol

[devout snow]

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[limber ermine]

can this be used if x=A wich A is a matrix?

[fossil locust]

absolutely!

you just have to be a bit careful cause the order of matrix multiplication matters

I can find you a proper derivation

[limber ermine]

ye

it will be smth like S-SA= A-A^n+1

how can i get S in front

or S(1-A)=S-SA

[limber ermine]

?

[fossil locust]

do you not know this

$AB \ne BA$ for matrices $A, B$

[woven radish]

south

[limber ermine]

yes

[fossil locust]

so when you are doing say $S(I - B)$ or something, where $I$ is the identity matrix

[woven radish]

south

[limber ermine]

i asked this

[fossil locust]

it matters

[limber ermine]

its 1 not I

[limber ermine]

ye but where S is put in this case

thats what i wanna know

in front or behind B

[fossil locust]

S(I - B) = S - SB

[limber ermine]

thx

[fossil locust]

so you have to repeat the proof again with (I - B)S = S - BS

to make it work

[fossil locust]

false, the identity element for a matrix is the identity matrix

[limber ermine]

so for this sum

[fossil locust]

not the scalar 1

[limber ermine]

i will have to do it twice right?

[fossil locust]

yep!

[restive river]

just for aesthetics in the terminal, is it possible to input two numbers in the same row with scanf?

for example scanf(%d %d, &number1,&number2);

[static ember]

well, did you try that?

[restive river]

yes it automatically changes the row

does it depend on the IDE?

[static ember]

press enter after inputting the two numbers

and space inbetween

[restive river]

ah ok makes sense

yeah

sorry for bothering with dumb questions

[devout snow]

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[rancid abyss]

can somebody help me with this

[rancid abyss]

it's a no-calculator que

[stable cloak]

Integrate with respect to time from 0 to 2

[rancid abyss]

i tried

im just not able to do it

this is the markscheme

[stable cloak]

The answer seems probable enough

[fossil locust]

REVISION VILLAGE

hello Ib student

[fossil locust]

distance is not displacement

[mild sorrel]

hello ┬─┬ノ( º _ ºノ)

[fossil locust]

just saying that, but actually here v > 0 cause e^(anything) is never negative

[rancid abyss]

so?

i dont think the value can ever be negative

[fossil locust]

so it doesn't matter

[rancid abyss]

yup

[fossil locust]

just wanted to make a point for other questions

[mild sorrel]

:<

[fossil locust]

https://www.youtube.com/watch?v=2I-_SV8cwsw

try watching this

[mild sorrel]

Never learn from a youtuber

[fossil locust]

ah but it's the trickier case since you need to stop at the integral of t e^(-3t^2) dt

[rancid abyss]

"VERY EASY"

ok beta

[fossil locust]

clickbait

I know

[rancid abyss]

watch me fail

[fossil locust]

you Indian bruv

[rancid abyss]

yeh lmao

its a joke

[fossil locust]

ah i see

[mild sorrel]

beta = son

[fossil locust]

to use u-sub

[fossil locust]

that's how I know

[fossil locust]

but honestly this working is perfectly clear

it's just spotting you need u = t^2

and not u = t^3

[rancid abyss]

yeh alright

ill try to solve it properly in a bit

[fossil locust]

maybe try another question that has this technique, where you need to differentiate e^(-3t^2) or the equivalent

[rancid abyss]

ok ty

[fossil locust]

doing that first makes you realise that if the reverse chain rule has t e^(-3t^2), you need to multiply by t^2

yeah I mean try and best understand what I said

but the real test is spotting the reverse chain rule for IBP

for another question

it's always much more beneficial if you try yourself

for this question yeah you're kind of spent after you see the full sol

np!

[rancid abyss]

i will

i feel like im just exhausted rn

[fossil locust]

IB is draining agreeed

[rancid abyss]

ive all 3 aahl papers tom

im so fucking cooked

[fossil locust]

ah you're dp2? wow

brutal mock schedule

just try to survive, yeah it's hard I know

[rancid abyss]

sem 2 exams

mocks are usually around march

[fossil locust]

right yeah that makes more sense

so yeah you still have time before the final mock

basically try to cheer yourself up I guess, like in every paper there are some easy questions

so I would recommend doing section A first

no point trying to grind out section B if you aren't warmed up properly

and you should start each exam with a fresh mind, don't work beforehand

also try to sleep the night before

[rancid abyss]

im not too worried about the exam tbh

i already hv a 7 predicted

i just want a higher score than my friend

[devout snow]

@rancid abyss Has your question been resolved?

[severe prairie]

THE GOATED YOUTUBER

HE SAVED MY LFIE

[fossil locust]

bruh.

thought you were grinding for 7

why waste your energy on competing with your friend then

[devout snow]

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[pseudo rampart]

i have no idea how to solve this

[mild sorrel]

wdym you have 'no idea'

The diameter of the bigger circle is: sum of diameters of the smaller circles

[sand dove]

you can find the radius of the big circle by adding the radii of the circles inside (as it should be on the graph)

then find a way to express the shaded area

in terms of the areas of all the circles

(formula for area of circle/disk of radius r: pi * r^2)

[pseudo rampart]

ohhh ok i got it so i find the area for both of the circles

the small circle is 1pi

and the bigger one is 2^2 which is 4 pi

[sand dove]

yeah

and the shaded area?

[pseudo rampart]

uhh no idea

id assume it would be 3

since we add the big and small circle to find its radius

[sand dove]

uhhhh

[pseudo rampart]

3^2

9?

[sand dove]

first of all

area of big circle?

maybe is easier

what's the radius of the big circle

[pseudo rampart]

i think 3

[sand dove]

yes

[sand dove]

so what's the area of the big circle

[pseudo rampart]

9 pi

[sand dove]

yes

[sand dove]

but we're not exactly after the area of the big circle

we're after the shaded area

what's different between the shaded area and the area of the big circle?

[pseudo rampart]

the shaded area isnt the whole circle

so do we minus it by the area of the small and big circle

4pi+1pi =5pi

9pi-5pi= 4pi?

[sand dove]

yep that's the shaded area

so now

[sand dove]

can we answer the question

[pseudo rampart]

since it says small circle yeah it would be 1/4

thank u smm it makes perfect sense now

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[autumn trout]

solved this by u-sub while the correct answer wanted me to use long division

[autumn trout]

I can see that the differnce is only a constant so is my answer also correct?

[winter patrol]

the -3 + C, combine into another constant

[autumn trout]

yeah true

[winter patrol]

work is fine

note that in the step of splitting the fraction, you effectively did the same thing as the long division

[autumn trout]

oh thats cool i didnt think of it like that

thanks

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[limber ermine]

any ideas how to find X?

[devout snow]

@limber ermine Has your question been resolved?

[limber ermine]

<@&286206848099549185>

[valid silo]

this is true for which n?

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[devout snow]

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[limber ermine]

n is real

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[spring oasis]

\textbf{11.} Let $f$ be a function that satisfies:
[
\frac{f'(x)}{f^2(x)}= \cos\left(\frac{1}{3}x\right), \quad \text{with } f(0) = \frac{1}{5}.
]
Then, $f(x)$ is:
\begin{enumerate}[label=\alph*)]
\item $\frac{1}{-3\sin\left(\frac{1}{3}x\right) + 5}$
\item $\frac{1}{\sin\left(\frac{1}{3}x\right) + 5}$
\item $\frac{1}{3\sin\left(\frac{1}{3}x\right) + 5}$
\item $\frac{1}{-\sin\left(\frac{1}{3}x\right) + 5}$
\end{enumerate}

[woven radish]

xbz

[noble mirage]

Where are you stuck at?

[spring oasis]

starting out

[devout snow]

@spring oasis Has your question been resolved?

[noble mirage]

it's ez
Use separation of variables

[spring oasis]

.solved

[devout snow]

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[cursive saffron]

I just don’t get the first step that’s it

I don’t what is it

[supple knot]

6/2 = 3

[cursive saffron]

Know*

[cursive saffron]

Huh

But why

[supple knot]

(1/2)*6 is the same as 6/2

,calc (1/2) * 6

[woven radish]

Result:

3

[supple knot]

,calc 6/2

[woven radish]

Result:

3

[cursive saffron]

Oh

Thanks

[sinful rover]

can you show a whole picture of the trapezoid?

[cursive saffron]

I get it all though except the first step of the equation

But yes

Wait

[sinful rover]

so you understand the equation just not the multiplying?

the equation for a trapezoid is 1/2 h (b+b)

thats the left side

and the otherside is the rectangle

[cursive saffron]

Yeah just the first step I didn’t get what they did

[sinful rover]

they just distributed the 1/2 and 6

or like did 6 x 1/2 so you get 3

[cursive saffron]

How do you distribute 2 number at the same time?

[sinful rover]

they didnt, i corrected myself^

[cursive saffron]

Oh

[sinful rover]

yea

[cursive saffron]

Thanks

[sinful rover]

youre welcome 🙂

you can close the chat with .close if your question has been answered have a good day ❤️

[cursive saffron]

Close.

.close

[devout snow]

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[smoky nimbus]

Multiplication is commutative so (1/2)(x + 12)(6) is the same as (1/2) * 6 * (x + 12)

Then you can do (1/2) * 6 = 3

[devout snow]

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[small urchin]

what does it mean to integrate on a closed surface?

[night moat]

No open boundaries

[pastel pasture]

do you mean like a surface integral?

[small urchin]

Yes

[pastel pasture]

yeah basically you can get from one point of the surface to any other point of the surface without leaving it

[small urchin]

I mean what the final value represents

[pastel pasture]

or are you asking like a physical interpretration

ah

[small urchin]

Yes

[pastel pasture]

I mean

physically it doesn't mean anything

its sort of like a 4-dimensional height

Maybe you could think of having some sort of temperature function T, and you want to integrate this over the entire earth

[small urchin]

Mmm

[pastel pasture]

its like a line integral in 3d now

[small urchin]

For example, what does it mean to integrate a vector point by point on a 2D surface

I don't know how to explain

[pastel pasture]

I mean

These all have various applications in physics

finding work, flux, etc

[small urchin]

I know but I don't know if I can go into too much detail

[pastel pasture]

[small urchin]

For example

The integral of a function represents the area subtended by it

But does this also apply if a vector is integrated?

[pastel pasture]

I mean

Kind of

If you have a vector line integral

you can sort of think of it as the thin sheet of area that this vector projects onto the ground

as it travels

[small urchin]

?

I'm not understanding

I mean would be the area?

[pastel pasture]

idk my drawing is kind of horrible

the blue is a vector function

you can think of it as a roller coaster and integrating gives that area under the roller coaster

[small urchin]

Yes

and if the vector is in 2d?

It would be the area

[pastel pasture]

yeah I guess

[pastel pasture]

btw this is technically 2d

eh

its weird

[small urchin]

Do you know where I can find the meanings of the integrals in these applications?

[pastel pasture]

any standard textbook should have it

there's also that trevor guy on youtube

[small urchin]

Okay

Thanks very much

.close

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[fast river]

i was just wondering where the ln C came from

i get where C come from \

but how come they wrote it as ln c

shouldnt it have been just ln |x|+C

[urban spindle]

I think here the idea is that for every C, there's a C', so that ln(C')=C

it's not exactly the same C you'd put if you wrote +C.

[fast river]

so in defreential equations

[north roost]

I think the point is that it’s easier to simplify

[urban spindle]

It's also conveniently picked this way, so you can further solve it.

[fast river]

ya

[north roost]

So the right side only include one term

[fast river]

so in all defferntial equations will it be multiplied by C ?

or is it just this case with ln

[urban spindle]

So the integral in general is solved by adding a constant C

you could consider it +C, then try this: Take C' so that ln(C')=C. then C'=e^C, right?

[north roost]

Solutions with e^x is another example I think it takes form of Ce^f(x)

[fast river]

oh yaaa

i have seen that

i was stuck on it yesterday

[urban spindle]

So you could add instead of C, ln(e^C), for same result

now, this e^C is also a constant

So, we also note it...as C 😄

[fast river]

oh god thats anoyying

[urban spindle]

also worth noting you can't transform the constant in any way. for example, you cannot write it as +1/C

[fast river]

like thsi question