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[wraith dagger]

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Requested by sherif_player

[blazing dune]

i just said that u = x here

i don't know about if u was equla something before

so it just like a general term

you can say that x here is the same as u

then subsitute the function of u

htere your taking the integration with respect to du

yes

and another mistake you did here

$$\frac{d}{du} (\frac{u}{2}) = \frac{1}{2}$$

[wraith dagger]

Sherif Player

[blazing dune]

not 2/2

the derivative of the function u/2 is 1/2

about the whole thing of the integration of that expression
it is considered a general rule that
$$\int\frac1{x^2+a^2}dx = \frac{1}{a} \arctan(\frac{x}{a}) + c$$

[wraith dagger]

Sherif Player

[blazing dune]

no problem, anything else?

[cedar kiln]

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[crimson sedge]

I really need help with an optimization problem. That is it. Where do I go for help with this problem?

[crimson sedge]

Im not a helper but do you mind if I ask what course this is?

like a university course?

yes

why?

[crimson sedge]

optimization 6463 or something

alright thanks.

I just wanted to know since ill start university in about 2 years

this is a grad level course. I do not know it drives me mad

grad? is that like post university?

yes

so you are about to end university then? or are you advanced?=

I graduated, and I am in grad school to get my masters. I guess advanced, I do not know.

oh wow

masters huh? isn't that really hard?

nvm. good luck on your masters

[cedar kiln]

@crimson sedge Has your question been resolved?

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@crimson sedge Has your question been resolved?

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@crimson sedge Has your question been resolved?

[crimson sedge]

jsut end this one I opened a different one

[wanton sail]

.close

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[neat jasper]

Is the 'density' of primes mod n uniform for each value (besides 0)?

[upper abyss]

For any value that can, be a prime, yes.

Pretending that 2 isn't a prime

[idle tusk]

https://en.wikipedia.org/wiki/Prime_number_theorem might be something you want to consult

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[ionic wolf]

The answer is 206 right

33k-9=90

So k equals 3

And so 53k-5 +mCAD=360

53k-5 equals 154

So measure of major arc CAD is 206

[past wind]

i don’t think so

we can tell that it is at least 270 by observation (3 right angles)

it should just be 360 - m<CPD

And since you know k it should be easy

[ionic wolf]

oh I left out arc DE

[past wind]

yep

[ionic wolf]

296 is the answer I think

[past wind]

sounds correct

[ionic wolf]

Ty

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[noble knot]

.open

It's actually from the last one I opened because now I realized there is one more thing I don't understand

[noble knot]

Basically how to prove 9(137²+119²)(137⁴+119⁴) can be divided by 2²

[nova glacier]

show that (137²+119²) is divided by 2 and so as (137⁴+119⁴)

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[noble knot]

Oh yeah that makes sense thanks

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[pliant orbit]

Why don’t they FOIL??

In the first line after =

1 + cos(2x)/2

All of that is squared

The top isn’t foiled, why?

[dire geode]

There are multiple ways to do the same problem

[livid hound]

more efficient simplification, cancellation of common factor
instead of unnecessary multiplication

also the their use of () is so cursed

[pliant orbit]

I don’t get why it isn’t foiled

Doesnt exponent come first

I get the cancel the left denominator with the right numerator

But where is the foiling

They can’t just ignore the ^2?

[livid hound]

order of operations doesn't necessarily say that you must do exponentiation first whenver its present

[dire geode]

There isn't any

[pliant orbit]

pEmdas

/ s

[livid hound]

but rather in situations like
pq^2 you don't multiply the p and q together before squaring

[pliant orbit]

What identity

[livid hound]

by definition of exponentiation
p^2 = p * p
and there is nothing saying that you must explicitly expand that

[pliant orbit]

What happened to (1 + cos2(2x))^2

[dire geode]

Got cancelled

[livid hound]

p^2/p = p

[pliant orbit]

I see what you guys mean

[livid hound]

(if p isn't 0)

[pliant orbit]

So p * p

One of the p’s cancel

🧠

[pliant orbit]

Damn thats an awakening

I didn’t think of not expanding

Always gotta FOIL

If I have cos(6x) instead of cos(2x)

How would I write cos(6x)?

Cos(2x) has three identities(?) like cos^ - sin^ but this is 6x

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[violet flume]

what have you tried?

[haughty flume]

i think i figured it out

sketching the parabola helped

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[teal turtle]

Uhm how do i get/find the cotermal angle of a decimal? And convert it inti radian form?

[cedar kiln]

@teal turtle Has your question been resolved?

[dire geode]

do you have an actual problem

[cedar kiln]

@teal turtle Has your question been resolved?

[midnight bane]

Coterminal Angle?

Coterminal Angles are angles who share the same initial side and terminal sides. Finding coterminal angles is as simple as adding or subtracting 360°
https://www.chino.k12.ca.us/site/handlers/filedownload.ashx?moduleinstanceid=5575&dataid=2608&FileName=NOTES COTERMINAL ANGLES.pdf

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[midnight bramble]

in the equation ax^3+bx^2-6x+c the roots are 1/2 ; (p^10+1)/(p+1) ; (p^10+1)/(p^10-p), what is the value of c if P belongs to real numbers except 0 -1 and 1

[digital cliff]

ax^3+bx^2+cx+d
roots x,y,z
x+y+z=-b/a
xy+xz+yz=c/a
xyz=-d/a

probably need a combo of the second and third one

[midnight bramble]

I mean, if you do that p^10 would make it really hard to operate, am I wrong?

[digital cliff]

why

[midnight bramble]

it is elevated ^10

[digital cliff]

so?

[midnight bramble]

to reduce that to a value that makes c a number like 1 2 or 3 (which are in the options) would make it hard to operate

[digital cliff]

you wont know till you try

[midnight bramble]

I'm saying it because I tried

at the end 1/2 multiplied by the other roots would make a and b depend on p^10, which on the equation would end up just getting nowhere on finding c

what I did was using 1/2 and making the equation equal to 0, but then finding a and b was the problem itself

this is basically that

[digital cliff]

that bottom one is -c/a, not c/a
however, you need to get what a is, and that would be by using -6/a=...

you dont need to find what b is at all

[midnight bramble]

so just find a in -6/a and replace it in -c/a?

[digital cliff]

pretty much, its the same a

-6/a=xy+xz+yz where x,y,z are your roots

[midnight bramble]

this is not guaranteed to be reduced though...

[digital cliff]

then youll have to work it, combine stuff

i never said it would be simple

[midnight bramble]

it is suppossed to be simple, it is a question that should be done in 3 min or less (from an exam)

[digital cliff]

blimey, now i wanna try

gimme a few minutes

[midnight bramble]

I'll try too

[digital cliff]

i got 2

lots of stuff cancels very nicely

if i did something dodgy

forgive me, its 5am

[midnight bramble]

yeah, I got it too, maybe I was a bit too lazy to do all the process in that short amount of time

anyways, thanks for that lol

[digital cliff]

np

fun exercise

[midnight bramble]

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[lilac nacelle]

hey there, how would i work out 2 ii? for part i i got thid

[lilac nacelle]

$\24\times\frac{100+a}{100}=b\times\frac{100-24}{100}

[muted timber]

$24\times\frac{100+a}{100}=b\times\frac{100-24}{100}$

[wraith dagger]

LordFelix

[lilac nacelle]

ty

[muted timber]

you're missing the last dollar sign, and had a backslash at the start that shouldnt be there

[lilac nacelle]

ok

[muted timber]

i would expect them to want you to simplify that equality

[lilac nacelle]

i looked in the answer book and they want a number

@muted timber

[muted timber]

well, yes

but the whole point is you calculating it. And you should start by simplifying the equality

[cedar kiln]

@lilac nacelle Has your question been resolved?

[deep wagon]

@lilac nacelle solve for either b or a, then plug it into (19b - 6a), and show that results in an integer (k)

[lilac nacelle]

!close

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[cobalt plover]

In the parallelogram ABCD, M is the middle point of BC, N is the middle point of CD. P is the intersection of AN and BD, Q is the intersection of AM and BD. Find the value of Area of CPQ / Area of ABCD.

Also theres a note that says i shouldnt use triangle similarity, so I guess midpoint theorem helps

<@&286206848099549185>

[crimson sedge]

hm

i cant solve this but i think i can try to think

[tardy sandal]

do this problem need exact numbers or just factor of sides?

[crimson sedge]

idk why but i get a feeling that, you can consider ABCD as 1 and then keep finding the triangles smaller than that so it reduces exactly by 1/2 everytime

[crimson sedge]

that is if they are symmetric in a way

[cobalt plover]

Like uh

I need to write it as a fractikn

Fraction

Eg. 1/8

[crimson sedge]

yep

[tardy sandal]

ratio expression right?

[crimson sedge]

use this

[crimson sedge]

yes yes

[tardy sandal]

using thales theoriem or sth we got mn/bd is 1/2

bad spelling

[crimson sedge]

is the answer 1/6? @cobalt plover

[cobalt plover]

I dont know...

I got stuck

[crimson sedge]

i never read this ☠️

[crimson sedge]

oh well, i hope someone solves it man

[cobalt plover]

Yeah

@vague rapids

Mind helping? I went on with other exerciees yesterday, sorry.

[cedar kiln]

@cobalt plover Has your question been resolved?

[tardy sandal]

ok im free for 10 mins

lemme see

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[clever yoke]

Where do I start with this one?

[verbal flume]

wait what is the original question?

[karmic pagoda]

Thats the whole question

[verbal flume]

Q is so poorly written omfg

[verbal flume]

💀

[karmic pagoda]

💀

https://tenor.com/view/ib-ibdp-stressed-school-high-school-gif-25672922

[cedar kiln]

@clever yoke Has your question been resolved?

[cobalt plover]

Bruh?

[gritty viper]

dx/dy = (dy/dx)^-1

[clever yoke]

And then do we find the equivalent of the second derivative?

[gritty viper]

ig so

maybe it's more doable converting to newton notation

[clever yoke]

What is that

Is that the one with the dot

I think the teacher wants us to do it with dy/dx cuz we havent learnt newton

[gritty viper]

f''(x) + f'(x)^3 * g''(y) = 0

where g is the inverse of f

Have you seen this notation

[clever yoke]

Oh ye

ok

Maybe that works

[gritty viper]

then f''(x) + f'(x)^3 * g''(f(x)) = 0

y=f(x)

[clever yoke]

Ok

This was on our latest exam and nobody could do it😭

[gritty viper]

India?

[karmic pagoda]

Australia

[clever yoke]

Year 11

[gritty viper]

That's insane

[clever yoke]

Our school has extremely hard topics and stuff but it tops the state every year

which is good

But this shit is hard af📉📉📉

[gritty viper]

Yeah

The core of the problem is figuring out how to rewrite $\f{d^2x}{dy^2}$

[wraith dagger]

thewizardofOU

[clever yoke]

Oh yea

I did that part on the test

But i forgot how

I did it

My brain

[gritty viper]

This much is at least doable

unfortunately for the second derivative I doubt it's so simple

[clever yoke]

hmm

Oh

We can just d/dx both sides right

Wait

Other way

d/dy both sides

And left side is the second derivative

And right side we use implicit, chain rule, and quotient rule maybe?

[gritty viper]

Both sides of this?

[clever yoke]

Yeah

[gritty viper]

I see

d/dy 1/(dy/dx)

$= \f{-\f{d}{dy}\f{dy}{dx}}{(\f{dy}{dx})^2}$

[clever yoke]

💀

[wraith dagger]

thewizardofOU

[gritty viper]

unfortunately the top does not simplify to -d/dx

[clever yoke]

I think i did something like that on the test

[gritty viper]

lol understandable

unfortunately type error

dy/dx is a function

d/dy applied to dy/dx is also a function

but d/dx is an operator

[clever yoke]

Ok

Big brain

I think on the test i cancelled out the “d”s on top and bottom

And the teacher wrote “you cant do that”😂😂😂

[gritty viper]

Ah rip

Yeah the ds never cancel

dx is always a single object, never d times x

not even d applied to x really

idk

somehow we need to show $\f{d}{dy}(\f{dy}{dx}) = \f{d^2y}{dx^2}*\f{dx}{dy}$

[wraith dagger]

thewizardofOU

[clever yoke]

ok

[gritty viper]

if we say z = dy/dx

It becomes $\f{dz}{dy} = \f{dz}{dx}*\f{dx}{dy}$

[wraith dagger]

thewizardofOU

[gritty viper]

ig thats QED

[clever yoke]

Wiat

Wait

Omg

Hold on

I think you cooked

[gritty viper]

eyyyy pog

[clever yoke]

https://tenor.com/view/desmondpacito-cooking-chef-gif-12378159

Thanks bro

Legend

[gritty viper]

np that was a fun challenge

[clever yoke]

👍

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[tranquil arch]

how do you show that $x^4 + y^2 = z^4$ has no positive integral solns

[wraith dagger]

CoolShot

[gritty viper]

How hard is Fermat's last theorem when n=4?

Because this is harder

Maybe use the characterization of Pythagorean triples to show x^2,y,z^2 can't be a Pythagorean triple?

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[sinful field]

this is probably a dumb question but here i go

[sinful field]

i know i need to use derivatives but what kind of formula should i use for this ?

btw the images has some flaws since i translated it, 1000 cm^3, 0.2 cm/s,

this doesnt help in solving it i assume though

[livid hound]

set up the volume equation
isolate h
differentiate both sides wrt t

[sinful field]

V = L * H * W ?

[livid hound]

yeh,
use the expressions you have in the problem

[sinful field]

alright thank you

.close

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[cedar kiln]

@serene vine Has your question been resolved?

[cedar kiln]

@serene vine Has your question been resolved?

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[abstract wolf]

Hey I’m really struggling with this.

I know a = 9.8 t=0.7 and I have to find u but idk how to get there

[abstract wolf]

Do I have to assume the final velocity is 0?

[umbral dew]

no

[slow jewel]

Initial is 0

[umbral dew]

this

[abstract wolf]

Oh my@god it is

Because they just dropped it

So how can it have any speed

[slow jewel]

Ye

[abstract wolf]

Mechanics is going to drive me insane

[slow jewel]

Final won't be 0 coz it's accelerating uniformly

[umbral dew]

kinematics is easy tho

[abstract wolf]

This is my first time ever doing it

[umbral dew]

nah i saw that

[slow jewel]

Ik

[abstract wolf]

Thank you so much, have a nice day and thanks again

.close

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[serene vine]

.

[sleek raft]

could any confirm this integral? Im gettign everything the same except the ln is inversed so I wanted to make sure it wasnt an error on my profs side

[fair mortar]

,, \cosh^2 x - \sinh^2 x = 1

[wraith dagger]

Bettim

[cedar kiln]

@sleek raft Has your question been resolved?

[cedar kiln]

@sleek raft Has your question been resolved?

[sleek raft]

just a confirmation 🥹 <@&286206848099549185>

[jagged wing]

!15m

[cedar kiln]

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.

[sleek raft]

45m?

[cedar kiln]

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[crimson sedge]

can someone help me understand why he did this

[crimson sedge]

i am assuming he put it into standard form of an ellipse

woah

whole lot of new words

what the hell is a foci

.close

[cedar kiln]

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[dusk shadow]

im struggling with a problem about liars and truth tellers:
"a" says everyone is telling the truth
"b" says a is a liar and c or d are liars
"c" says a or exclusively d are telling the truth
"d" says if b is telling the truth then d is lying

[dusk shadow]

i know im supposed to use a table but with 16 different outcomes theres no way ill have time for other problems

[cedar kiln]

@dusk shadow Has your question been resolved?

[cedar kiln]

@dusk shadow Has your question been resolved?

[cedar kiln]

@dusk shadow Has your question been resolved?

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[stark bronze]

[\int sin(x)^5 cos(x)^4\text{dx}]
[\int sin(x)^4cos(x)^4sin(x)\text{dx}]
[-\int(1-u^4)u^4\text{du}]
[-\int(u^4 - u^8)\text{du}]
[-(\frac{u^5}{5} - \frac{u^4}{8}+c)]
[-(\frac{cosx^5}{5} - \frac{cosx^9}{9}+c)]

[wraith dagger]

dopediscorduser

[stark bronze]

What am I doing wrong?

[rare vault]

why is it (u^4)/8 in the evaluation step?

oh typo

[stark bronze]

Yeah sorry should be u9/9

[rare vault]

the steps here look right then, can you explicitly write out the usub

,w integral of sin(x)^5 cos(x)^4 dx

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tuzis

[stark bronze]

Yeah see that has more terms than mine

All of the possible answer choices in the book for this have 3 terms as well

Thats why I thought there was a mistake

[plucky owl]

So you did u = cos?

[stark bronze]

Yes

-du = sin

[plucky owl]

How did you replace sin(x)^4 to be 1?

[rare vault]

well he converted it to 1-cos^2(x)

[plucky owl]

Or what was your process for sin(x)^4cos(x)^4 to be 1 - u^4

[rare vault]

mistake is surely in the usub but I don't have paper

[stark bronze]

My process was to use pythag to have sin = 1-cos2

[stark bronze]

I'm honestly just confused as to how I could mess up the sub in such a way that I would be missing additional terms at the end?

[plucky owl]

You mean sin^2 = 1 - cos^2?

[stark bronze]

Yes sorry

[plucky owl]

Because sin(x)^4 = (sin^2(x))^2

Therefore (1 - cos^2)^2

Then you expanded?

Because $(a + b)^2 \neq a^2 + b^2$

[wraith dagger]

CaptainNova22

[stark bronze]

Thats where i went wrong then?

Expanding the exponential?

[plucky owl]

Yeah

If you expanded (1 - u^2)^2 to be (1 - u^4)

(1 - u^2)^2 = (1 - u^2) (1 - u^2)

[stark bronze]

Yeah let me rework real quick

[-\frac{cos(x)^5}{5} + \frac{2cosx^7}{7}-\frac{cosx^9}{9}+c]

[wraith dagger]

dopediscorduser

[stark bronze]

This is what I got

[plucky owl]

,w integral of sin(x)^5 cos(x)^4 dx

[wraith dagger]

Results provided by WolframAlpha

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

Requested by captainnova22

[plucky owl]

Looks good

[cedar kiln]

@stark bronze Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Hi

[crimson sedge]

Two questions. First, why wolframalpha says (-2)^2/3 != cubic root of (-2)^2? Second, why there's a negative branch for y = x^2/3 in desmos? Thank you

sorry, i meant just another branch

[surreal cave]

because Desmos cannot do complex numbers consider: $(-2)^\frac{2}{3}=2^{\frac{2}{3}}e^{\frac{2i\pi}{3}}$

[wraith dagger]

PajamaMamaLlama

[crimson sedge]

Okay, thank you. Btw, if I'm asked for the graph of this function, should I assume -(-1+x)x > 0 for dom(f), just like Wolframalpha does, or -(-1+x)x != 0 ?

[pastel vault]

just $-(-1+x)x \ne 0$

[wraith dagger]

south

[pastel vault]

$u^{2/3}$ can be interpreted as $(u^{1/3})^2$

[wraith dagger]

south

[pastel vault]

so there's always going to be a real value of that

[crimson sedge]

Thank you

[pastel vault]

no worries

[crimson sedge]

.close

[cedar kiln]

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[pastel vault]

cheers

[cedar kiln]

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[acoustic ermine]

I can't solve this problem. It is supposed to use the L'Hopital's rule, but I can't get the derivatives right. The answer is supposed to be -2.

[cedar kiln]

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

[acoustic ermine]

I get it to here, and after my tries were all wrong and I have just crossed them out

[cedar kiln]

@acoustic ermine Has your question been resolved?

[neat dune]

Try calculating limits of ln(1 - x) /cot(pix) and tan(pix/2)/cot(pix) individually

Also write cot as cos/sin and tan as sin/cos

[cedar kiln]

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[fresh niche]

I need help for diffeq

[fresh niche]

This is my work

@uncut veldt

[cedar kiln]

@fresh niche Has your question been resolved?

[fresh niche]

No

@uncut veldt

@slate niche

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@fresh niche Has your question been resolved?

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@fresh niche Has your question been resolved?

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@fresh niche Has your question been resolved?

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[stray cairn]

(prefacing this by saying i am utterly terrible at math to the point even the most simple problems confuse me) I need help understanding how to even begin to start calculating a graph problem, I’m given the information that a trip would cost 3450$ for a 7 day package on the trip and 4500$ for 10 day package on the trip. How would i figure out the cost for the other days?

[stray cairn]

I’ve attempted trying to figure out the base cost by dividing 4500/10 or 3450/7 but it doesn’t lead me anywhere

[cerulean star]

!original

[cedar kiln]

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.

[cerulean star]

presumably (not necessarily realistically), this is an arithmetic or geometric progression

[stray cairn]

Are you talking about the scratched out stuff or more context to it i’m sorry

[cerulean star]

but that is an assumption

[cerulean star]

both

I would like to see the problem as it was originally stated

[stray cairn]

I genuinely don’t know how to give more context as that’s all i’m given for this and i just scratched out the silly titles lolol

[cerulean star]

there is more context than you are providing (probably)

[stray cairn]

i was just told to figure out the base price of the day packages and the rest of the days

[cerulean star]

did your teacher recently teach you about arithmetic or geometric sequences?

what level of Math is your class?

[stray cairn]

i really wish i could share what it would be

AFDA

algebra functions and data analysis

[cerulean star]

Chilli Willie's is an arithmetic sequence.

d = 450.

[stray cairn]

a step down from algebra 2

i was able to get the amount of 450 i just don’t know how i got it lolol

[cerulean star]

for Chilli Willie's?

[stray cairn]

Yes

[cerulean star]

it's already filled out and the "common difference" between each entry is 450.

now, Penguin Tours has three steps between the only known entries.

but if we assume it is also an arithmetic sequence, then
4500 - 3450 = 3d

[stray cairn]

what does the 3d stand for?

[cerulean star]

(three steps of the common difference between consecutive terms)

to go from 3450 to 4500 in an arithmetic sequence, we need to add the common difference d three times

(count the rows from 3450 to 4500..."one, two, three")

[stray cairn]

so 750*3?

[cerulean star]

no

solve for d

[stray cairn]

wait is it 3450 or 3750

for the penguin one

[cerulean star]

the only two values you listed for Penguin tours are 3450 and 4500

oh

fixed

sorry

@stray cairn

[stray cairn]

so the d would be 1050?

[cerulean star]

no, do the algebra

simplify what is given and solve for d

4500 - 3450 = 3d

1050 = 3d

[stray cairn]

350?

[cerulean star]

you tell me

count up by 350 from 3450

does it work?

[stray cairn]

yes, so in equation it’s 3(350) ?

[cerulean star]

yes

[stray cairn]

and to find the rest you just keep going down?

[cerulean star]

yeah

but this is Excel, right?

highlight the data you filled in

and drag the top of the selection up

to automatically fill in the remaining cells

oh it's google sheets

same thing

[stray cairn]

Oh I had no idea you could do that thank you very much for the help

[cerulean star]

you need either 2 or 3 consecutive cells filled in like an arithmetic progression to enable that, iirc

1,2,3

2,4,6

etc

5, 8, 11

[stray cairn]

Ah alr thank you^^

[cerulean star]

np

[stray cairn]

i’m not entirely sure how to end help request in here…

[cerulean star]

.close

[stray cairn]

.close

[cedar kiln]

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[inner elbow]

The expression M has to be simplified

[inner elbow]

I’m stuck until here

Any idea how that can be simplified further?

[cedar kiln]

@inner elbow Has your question been resolved?

[inner elbow]

<@&286206848099549185>

[cedar kiln]

@inner elbow Has your question been resolved?

[devout glade]

Hi. Let me have a look

So, in the red part, you factored the numertor on the right into a 2 perfect squares, but that's wrong. The numarator has a^2+ab+b^2 not 2ab, so it can't be factored like that

And then the blue was confusing me, but it may have followed from the previous line being wrong. Do you want to look that over and see if you can continue from there?

[inner elbow]

Ohhh I see it now thanks

I’ll write back when I retry it

[devout glade]

Sure thing!

[inner elbow]

So I got to here

[devout glade]

Almost there! You just made a typo

In the denominator on the left, it should be a^1/2 -b^1/2

Not a^2-b^1/2

That's right at the top, but you carried the mistake down all the way

[inner elbow]

But wouldn’t this be true?

[devout glade]

Close, but no. When you factor out the 1/2, you're subtracting 1/2 from each power, not dividing by it

If you distribute the 1/2 again, you'll end up with a^3/2 instead of a^1

[inner elbow]

So the question is about a^1/2 multiplied by what gives a

[devout glade]

Exactly!

[inner elbow]

Ohh

Wait I think I had multiplied the powers instead of adding them

I thought a^1/2.a^1/2 couldn’t make a because it would be a^1/4 but I realise it’s actually a^(1/2 + 1/2)

[devout glade]

Yeah, 😆.
It happens to everyone though, don't worry

See if that simplifies things a bit more?

[inner elbow]

Okay it worked!

Is that the solution?

[devout glade]

How did you simplify the denominator from the first to the second line?

And your two became an 'a' between those lines also, but I think that's just a typo

[inner elbow]

Wait, I’ve messed up that formula too

Sorry for the ridiculous amount of mistakes, I’m just very sleep deprived and decimal powers are the bane of my existence

[devout glade]

There might be one more mistake, but gimme a moment

[devout glade]

Yeah, lol. No stress though

[inner elbow]

At least the first line of this is true, right?

[devout glade]

It's no problem. I get you lol

[devout glade]

It looks okay yeah

[inner elbow]

Hmmm

Then I’m not sure what more can be done

Or wait I think I thought of something

[devout glade]

One more thing. You made one more typo

In the denominator on the right, it should be (ab)^1/4

[devout glade]

Go for it and let me know!

[devout glade]

So so close!

You used the difference of squares incorrectly in the last line

But otherwise it's perfect

It should just be [2(a^1/2-b^1/2)*b^1/4]/a^1/2[a^1/2-b^1/4]

And that just leaves (2b^1/4)/a^1/2

Sorry, typo

[inner elbow]

Oh I understood what you meant and got that exact answer

Thanks a lot for all the help!

[devout glade]

Perfect!

Anytime!

[inner elbow]

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

How do I show the steps of this solution of an recurrence relation in Maple? Like I'm wondering how they got T(0)

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[elder swallow]

Do you have T(1) ?

[crimson sedge]

no the recurrence relation is T(n) = T(n-1) + c

[half zenith]

Without knowing at least 1 term of the sequence, you can’t solve the recurrence relation

[crimson sedge]

I see, the way I'm doing it in hand not in Maple is through telescoping

[cedar kiln]

@crimson sedge Has your question been resolved?

[cerulean sail]

I guess noting that $T(n) = T(n-1) + 1$, they repeatedly apply that with $T(n-1) = T(n-2) + 1$ and keep going until they get to $T(0)$, which I guess is what they think is the initial value

[wraith dagger]

@cerulean sail

[cedar kiln]

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[steep bane]

I just dont know where to start, I don’t know what to do with the angle

[fierce sleet]

U know that tangent of circle make right angle with radius of circle ?

Tangents means line just touching on circle

[fickle trellis]

What does the question ask, archon?

[steep bane]

Yea but I dont really know how to use the angle to find the small and big arcs

[steep bane]

The angle i at the top of the triangle inside the circle

Its a bit pixelated

[fickle trellis]

I see.

[steep bane]

But its an isosceles triangle

[steep bane]

Would the first step be to double the angle for the smaller arc?

Im very lost sorry

[supple flume]

do u know theorems to do with inscribed angles on triangle

triangles

on circles

i mean

[sour mango]

Anything else given or JUST that ?

[steep bane]

Yes

[supple flume]

u can solve it with just that

[steep bane]

Its just that angle

[supple flume]

let me draw something for u

it might make it obvious

i cant draw circles on microsoft paint lol

just gonna get my book out

[steep bane]

Go ahead dude take ur time

[supple flume]

If you have trouble my number one tip is to draw it if it’s geometrical and make observations like tangent lines and other stuff

I didn’t draw it to scale but still works

[steep bane]

So does that mean the 72 and the radius would be supplementary?

[supple flume]

Not the radius

Just the angle

@steep bane

Thats probably what u mean but you use the wrong word

[steep bane]

Oh yeah thanks

What ive gotten so far is that the angle between the tangent and the diameter is 54 from (180-72)/2

And that the smaller arc between the two tangents is 108 from 54 x 2

[supple flume]

remember that the triangle is isoceles

(probably misspelled that)

[steep bane]

I think i got it now

I got the bottom right angle on the triangle as 54 from 108/2

And since there isnt a congruet sign on the opposite side its not one of the congruent angles

So (180-54)/2 should get the other 2 angls

I hope

[supple flume]

yes

[steep bane]

Thank you so much for your help

Oh my god

[supple flume]

just draw out a diagram when you get stuck

[steep bane]

Will di

Do*

[supple flume]

you solved that almost entirely on your own using the diagram

just remember tangent lines are one 90 degree angle lol

[steep bane]

I think it was just a visualization thing

Thank u again bro

[supple flume]

i almost always do the maths in my head but cant do geometry in my head

its just too hard to keep track of every angle and stuff

[cedar kiln]

@steep bane Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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

[crimson sedge]

1

[supple flume]

Well first you could expand the left side

[crimson sedge]

?

[supple flume]

Times each number in the brackets by 1/3

If you dont know how to do that i dont think your teacher should be giving you these questions

[crimson sedge]

I know how to do that

i'm doing it

okay done now what

[supple flume]

Now you want to move the x’s to one side and move the numbers to the other. Show me the equations you do so i can make sure you dont make a mistake

[crimson sedge]

okay

[crimson sedge]

Not sure how to show my work but i got it down i think

[supple flume]

Just type it

[crimson sedge]

Ok

-1/9 =-4/3

[supple flume]

Do you mean -x/9

[crimson sedge]

-1/9x

[supple flume]

Its not -1/9x

[crimson sedge]

uh

[supple flume]

Its -x/9

How did you get -1/9x

Just wanna see so you dont make the same mistake on a test you have or something

[crimson sedge]

Yes i will show you my process wait

[supple flume]

That is correct

Shouldve put the 1/9 in brackets

[crimson sedge]

Sorry

[supple flume]

Anyway from there you can change it to -x/9 since a(1/b) = a/b

Now that - symbol is annoying. Do you think we can times each side by -1?

[supple flume]

Dw

[crimson sedge]

I'm not sure if it would be right to do that

[supple flume]

Why wouldnt it?

If you do the same thing to both sides it stays the same right?

Because they are already =

[crimson sedge]

Oh right right

[supple flume]

So what do you think happens when we multiply both sides by -1

[crimson sedge]

the negative signs are gone

[supple flume]

Yes so this becomes x/9 = 4/3

Now how will we make the x/9 into just an x

[crimson sedge]

reciprocal?

Like

[supple flume]

What does the / mean

[supple flume]

Do you know the opposite of dividing?

[crimson sedge]

multiplying

[supple flume]

Yes

So if we divide x by 9 how do we get it back to an x

[crimson sedge]

multiplying by 9?

[supple flume]

Yes

Now we do the same thing to both sides

[crimson sedge]

ok

[supple flume]

So now we have x = 9 times 4/3

https://youtu.be/kWOTmyoaWJg?feature=shared

If you would like a video to watch with more problems with explanations of the answer you can watch this

[supple flume]

You just need to make sure you expand the equation by multiplying everything in the brackets by the number (or letter) outside of it then solve it like this video explains

[crimson sedge]

we got 12

[crimson sedge]

Okay i'll watch it

thank you!

[supple flume]

Correct

https://youtu.be/Z-ZkmpQBIFo?feature=shared

This video might be better as it is harder questions similar to the one you just did involving some brackets too

[crimson sedge]

Will watch it thank you

.close

[cedar kiln]

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[cedar kiln]

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[livid tundra]

question 29

when multiplying both the e^x and sin x

Yes

It's not hard

how did we end up with x+x^2+1/3 x^3 - 1/30...

[dire geode]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dire geode]

Did you multiply the two series together?

[livid tundra]

Do I have to ? 😦

[dire geode]

Use (a+b)(c+d) = ac + ad + bc + cd

Also called FOIL in some places

[livid tundra]

but e^x is (1+x+x^2/2! + x^3/3!)

(a+b+c+d)(e+f+g+h)

?

where e,f,g,h are the values of sinx

[dire geode]

Multiply however many terms you need to get this result

[livid tundra]

so FOIL works this way?

.

[livid tundra]

.

[dire geode]

Works what way

[livid tundra]

ae + bf + cg + dh

[dire geode]

No

[livid tundra]

I don't think i will be able to multiply all this without a single mistake

what can i do

<@&286206848099549185>

[scenic chasm]

Just write Everyone in line then calc it 🤷

Let me send it

[livid tundra]

Thank you

[dire geode]

Do it systematically
https://www.cuemath.com/foil-formula/

[livid tundra]

Yes but it's so long no ?

I have to do it 25 times

[dire geode]

So?

[dire geode]

No you don't

[livid tundra]

how

[dire geode]

Do just 3 terms

Instead of 4

[livid tundra]

ae + af + ag

then b

instead of "a"

[scenic chasm]

You see where x³? This vertical line?

[livid tundra]

x^3 / 2!?

[scenic chasm]

Go and calculate Thier numbers
So you will grt -1/3! = 1/6
And in same this vertical lineyou have also 1/ 2! = 1/2

Make addition, so you will get 1/3

That why i write x³/3

[scenic chasm]

Yeah This one and also the one above it(their is in same vertical line) sorry for my bad hand write

[scenic chasm]

Now do the same with the other verticals lines

For x⁴ it's+x⁴/3! - x⁴/3! So it's 0

@livid tundra

[livid tundra]

Got it

[scenic chasm]

For x⁵:
1/5! = 1/120
-1/2!3!= -1/12
1/4! = 1/24
Do addition= - 1/30

[cedar kiln]

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[lime dawn]

x ^ 3 = 3 ^ x

I know 1 answer, but i dont know how to get there and i dont know if thats the only.

I got a few steps, but i dont know if they were the correct steps.
If someone could write all the steps for me that would be great

[silver parcel]

notice what happens if you replace x by 3

[lime dawn]

This was all i could get, and im not even sure if this is correct

[lime dawn]

Yes i know that 3 is an answer, but what are the steps to get there, how do i show my work?

And i dont know if 3 is the only answer or not

[silver parcel]

there's no work really, this kind of equation is usually hard to solve when there are no solution you can guess, it's by seeing how 3^3 = 3^3 that you find a sol
but it's not the only one over the reals

there's another solution close to 2.4

but I think they asked you only over the integers so that 3 is the only sol asked bc you won't find the others in general

[crimson sedge]

Unless you know the raphsonnewton method

I don't think you can

Another way is to like draw a graph

Just sketch both of the curves

Points of intersection would are ur desired solutions

[lime dawn]

So i dont actually need to know all these different methods and just need to see it?

Thank you

.close

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[civic stirrup]

Hey there, I need help with this problem.

[civic stirrup]

I really have no idea how to do this, and would appreciate it if someone showed me how

I know we normally have to show our work, and I kind of know what complementary slackness is

But I have no idea how to solve this

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[bright sorrel]

need help finding fundamental system of solutions

[bright sorrel]

fundamental solution set*

,rotate

[cedar kiln]

@bright sorrel Has your question been resolved?

[bright sorrel]

<@&286206848099549185>

<@&268886789983436800>

[wanton sail]

what have you tried so far?

it's not really clear what you're asking, which is probably why you haven't gotten any responses yet

can you post the original question, preferably a screenshot?

[bright sorrel]

wdym? you have a system of equations and I have to find the fundamental solution set

[cedar kiln]

@bright sorrel Has your question been resolved?

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[quartz dust]

when integrating my function from -inf to 0 i got the result -25, but when visualizing it looks like it will never reach zero. Is te integral then -inf or can it still be -25

[quartz dust]

okay (:

.close

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[silver arrow]

Hello

[silver arrow]

I'm in need of help, where most people here couldn't figure out how to solve it properly.

Who can solve this task?

a, b and c are done. I only need serisweet only d

We need to do this:
Set up D and E in the equation of the circle and calculate them as two equations with two unknowns. You can calculate that in maple.

<@&286206848099549185>

[cedar kiln]

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[acoustic sigil]

Cam anyone help me with integral 10.26, this îs the method that I need to ușe but I dont understand it

[zealous compass]

What do you not understand about the method?

[cedar kiln]

@acoustic sigil Has your question been resolved?

[acoustic sigil]

I theoretically understand it

But can't figire out how to use it for the integral

[zealous compass]

Explain the method then

[acoustic sigil]

You need to ușe fractions to transform the top part of the integral into the derivative of the lower part, so that we can ușe substition and solve it

Right?

[zealous compass]

Yes

Specifically we want to use the 3x

Any other constants can be solved with arcsin

Or the other form if it’s like sqrt(1+x^2)

[acoustic sigil]

I've understood that already

But i cant figure it out in this particular case

I need to transform 3x-4 into 4x-6

[zealous compass]

You just need to transform 3x into 4x

so make it 4x and put 3/4 out the front

[acoustic sigil]

What about the -4 and 6 then?

Do i add and substract to make it the same as the derivative?

[zealous compass]

Yes

^

And you have a leftover constant

[acoustic sigil]

That constant îs really fucking annoying...

[cedar kiln]

@acoustic sigil Has your question been resolved?

[cedar kiln]

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[pearl lintel]

help

[scenic chasm]

Ask

[pearl lintel]

who are you

bros not even a tutor but yah can you help me

ngl I got 5$ if you do 10 questions for me 🙏 cash app

[broken mist]

You don't need to be a tutor or helper to provide assistance in this server

You also aren't allowed to solicit school work

[pearl lintel]

oh dang

my bad

Wel can u help me out bro

idk how to do that ngl

[broken mist]

What do you know about writing exponential expressions?

[pearl lintel]

0

nothing

[broken mist]

Do you know what an expression like 2^4 would mean?

$2^4$

[wraith dagger]

M. Frost

[pearl lintel]

yah

2x2x2x2

wait

[broken mist]

Exactly

[pearl lintel]

oh yah

[broken mist]

Now can you tell me what $2^3 \text{, ] 2^2$, and $2^1$ would be?

[wraith dagger]

M. Frost
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[broken mist]

oop broke the bot

$2^3, 2^2, 2^1$

[wraith dagger]

M. Frost

[pearl lintel]

3x3, 2x2, 1x1 ?

[broken mist]

Not quite, you have it flipped around

[pearl lintel]

oh

2x2x2

[broken mist]

2^4 is 2 multiplied 4 times, so the others would be

[pearl lintel]

2x2

and im guessing 2^1 is just 2? because 2x1 is just 2 idk

[broken mist]

Exactly

Now I don't really have an intuitive way of putting it, but radical expressions can be represented with exponentials too

An expression like $\sqrt{2}$ is the same as $2^{\frac12}$

[wraith dagger]

M. Frost

[pearl lintel]

0_0

hmmm

[broken mist]

Or more broadly, $\sqrt[a]{x^b} = x^{\frac{b}{a}}$

[wraith dagger]

M. Frost

[pearl lintel]

mhmmmm

[broken mist]

When you have a fraction in the exponent, it turns the denominator of the fraction into a root of that same number

So the cube root of 18, $\sqrt[3]{18}$, is the same as $18^{\frac13}$

[wraith dagger]

M. Frost

[broken mist]

Using that information, what do you suppose $\sqrt[6]{u}$ is?

[wraith dagger]

M. Frost

[pearl lintel]

mhm so how would I Solve for an exponent if its a fraction?

[pearl lintel]

uh U^1/6?

[broken mist]

Exactly

That's all that problem is asking for, for it to be changed from radical form to exponential form

[pearl lintel]

did i do it wrong?

[broken mist]

Try writing it as u^(1/6)

With the ^ and parentheses

Or u^ then hit that fraction button to the right

[pearl lintel]

it worked

🙏 Thanks that really helps bro

[broken mist]

When typing things like exponents in computers you need to be very specific with ^ and ()

[pearl lintel]

ahh ok noted

[broken mist]

Because computers can't tell if what you're typing is supposed to be right

In general, $\sqrt[a]{x} = x^{\frac{1}{a}}$

[wraith dagger]

M. Frost

[broken mist]

And if the x inside of the radical is raised to a power, you replace the 1 with that power

[pearl lintel]

imma just screenshot that real quick

[broken mist]

So $\sqrt[2]{x^n} = x^{\frac{n}{2}}$

[wraith dagger]

M. Frost

[pearl lintel]

thank you

.close

[cedar kiln]

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[deep otter]

Why is 1/2 • (2x + 1)^(-1/2) considered dy/dx? Don’t we differentiate both sides by t to find dy/dt?

[hollow trail]

we used the chain rule: y is a function of x, x is a function of t, so
dy/dt = dy/dx * dx/dt

dy/dx is found with normal differentiation methods here, we then multiply by dx/dt

[deep otter]

would the • 2 to the left of the • 15 be dx/dx?

if that makes sense

[hollow trail]

the 2 is d/dx (2x)

[deep otter]

what's the empty d though

[hollow trail]

that's an alternative way of writing
d(2x)/dx
or:
z = 2x
dz/dx = 2

either way it's because of the chain rule

[deep otter]

I guess I am having difficulty understanding it conceptually

so would it be like a very small change in 2x divided by a very small change in x