help-13

[fallen heath]

try to find questions which pique your basics

[gentle lintel]

where

[fallen heath]

I'm not the best person to ask that Q

I found sufficient matter for practice/discussion provided to me during my calc. course

maybe look out for books?

[jaunty mural]

standard level 👀

[gentle lintel]

can't find advanced ones

[jaunty mural]

The further math calculus isnt very hard, no, unlike what ansh is feeding you

[gentle lintel]

without getting redirected to shady or quetionable websites

[jaunty mural]

yh sure rip

i havent saved any resources myself so cant do much

Maybe you could look at past STEP questions?

They are of decent difficulty and are available free online from legit source

[jaunty mural]

the calculus Qs on step II and III should be of decent difficulty

slightly easier than 1st yr uni

but harder than high school

[gentle lintel]

this?

https://maths.org/step/sites/maths.org.step/files/s2s3/Calc_questions_2019_0.pdf

[gentle lintel]

maybe i can try MAT too

[jaunty mural]

in looking rn

yh either

[gentle lintel]

going to uk 3 weeks later uwu

[jaunty mural]

i cant remember where all the step papers are

possibly their official website

[fallen heath]

oiiii =_=

[jaunty mural]

ur feeding good Qs

nothing.

Its just u wont find this sht at hs lvl lol

@gentle lintel https://www.admissionstesting.org/for-test-takers/step/preparing-for-step/

their official website has all the pasts on it

[fallen heath]

this was my module for hs calc tho, meant for prepping for an entrance exam

and right. this isn't what you need either! All you need are questions that shape your insights in calc

[gentle lintel]

i'll just try MAT lol

fits my situation better

[jaunty mural]

Well maybe chrom needs like a proof writing kinda course

idk

for analysis at least

[fallen heath]

Yeahhh Chromium

If you're interested in starting Analysis

[gentle lintel]

oh so i should start analysis?

[fallen heath]

Just do it nowww

Lol

Sure you can

[gentle lintel]

(should i start la -> anal -> mutivar)

[jaunty mural]

yh and just make sure u can do highschool/early uni level calculus stuff

[fallen heath]

[jaunty mural]

lag and analysis can be taken together

[fallen heath]

save multivar for 6 months later

[gentle lintel]

yea i'll be patient

[fallen heath]

after you're done with analysis(single variable limit, continuity, differentiability, inf/sup stuff) and Linear Algebra

me happy cuz analysis beginning is just calc revisited but more formal 💀

[jaunty mural]

For now tho, try do the calculus Qs in step II and III

if you can do most of them, ur definitely ready to move on from this stuff

[fallen heath]

Yeah if I'm paid I'm willing to share some stuff 👀 (lol)

[jaunty mural]

step III has hyperbolic trig functions, either learn them (takes little time, u just need defns) or skip those

[gentle lintel]

oh so step goes from I to III?

III is hardest?

[jaunty mural]

I is easy

[fallen heath]

For example: the right sub to do here would've been $$x = \alpha \cos^2 \theta + \beta \sin^2 \theta$$

[jaunty mural]

II and III are hard

[gentle lintel]

!?

[fallen heath]

where alpha = 3, beta = 2

[jaunty mural]

man i wouldnt have known this ansh lmao

probably wouldve taken me a while to try that one

thats a nice Q

[fallen heath]

or more generally for situations like: $$\int \sqrt{\frac{x - \alpha}{\beta - x}} \dd{x}$$ or $$\int \sqrt{(x - \alpha)(\beta - x)} \dd{x}$$

[fallen heath]

you just sub this and get away with it 😶

👀

welp, I can share some of these everyday but it feels like too heavy a responsibility guiding someone step by step 🏃‍♂️

so at most, I can only suggest some pointers. As to how to enhance your abilities, it's on you

[jaunty mural]

ping me when u do

I never rlly got much better than standard stuff

[fallen heath]

well conclusion of the discussion:

-> feel free to start with Analysis n Linear Alg. (if you're not going for a hs exam anytime soon or idk how self-tutoring works)

-> meanwhile, deal with couple calc questions everyday. A book specifically meant for calculus practice would be nice

I have a "Textbook for calculus" by K.C. Sinha and it's a hugeee book but hugeee => manyyyy practice problems so it's fun

[gentle lintel]

can i have the pdf

[fallen heath]

This book is strictly for practice purposes though lol

[fallen heath]

btw also, it'd be nice not to skip parts which you feel are a bit physics intended cuz it might as well just be a couple mathematical function definition preset upon well defined variables

[gentle lintel]

hm so

what problems should i try

[fallen heath]

@jaunty mural >_< the game we played lol

[jaunty mural]

hi hi

If I'm not mistaken

well hmmm

idk, I feel like you can guess the correct delta using the 1st order taylor expansion...

Maybe not quite......... only when epsilon is small enough

hmmmmmmmm

<random thoughts>

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[narrow gull]

I cant see to figure out how to determine a_n and b_n from the given Fourier series. I found that a_0 = 89.6

[dense wing]

I mean n=4 to n=100 is trivial

[fresh geyser]

mash potato

s

[narrow gull]

huh?

[dense wing]

finding a_n and/or b_n is just a case of looking at the n values and comparing the coefficients

for example if you have 52.5cos(x), that's a term when n=1, so a_1=52.5

[narrow gull]

I see

but what if I wanted to find erm smtg else

like a_7 or a_50

do I start with a_n? or straight from a_1

[dense wing]

you can't really do a_n as an explicit rule

you have to compare when n=stuff

[narrow gull]

I dont get what you mean by n=stuff

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[main jay]

i thought it would be B because on the unit circle, 1 is the maximum value. but yeah tan acts differently compared to sin/cos in that retrospect... i was just wondering if there was a good way to figure this out?

[dense wing]

tan(theta) b/w -inf and -1

[proven comet]

Is inverted sorry

But can someone reply with the answer like it needs to be 20% ik but idk how to get 20%

[cedar kiln]

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[fallen heath]

consider the graph for tan x

[cedar kiln]

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[fallen heath]

oh :o gg

[main jay]

mosh's explanation makes sense 🙂

[fallen heath]

!!

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[feral haven]

I need help with how to go about solving a certain type of differential equation. I know how to solve a directly integrable, separable, and linear first order ODE but I've been presented with an equation of a similar form to dy/dx=2y^{1/2} as far as I can see, this isn't any of the forms I previously mentioned. If it is, please tell me which type so I know roughly how to proceed

[wraith dagger]

$\dv{y}{x} = 2y^{\frac{1}{2}}$

[fallen heath]

Does this not look like a separable ODE?

[feral haven]

I mean I guess so. It's thrown me because there's no function of x

(explicitly involving x)

[fallen heath]

don't worry. it's fine

you could just pull over the 2√y to the left

[feral haven]

so separate by dividing by root(y)

[fallen heath]

and notice [√y]' = 1/2√y

[feral haven]

okay got it, I'll figure it out from there, thank you 🙂

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[forest dust]

Consider functions f:{1,2,3,4,5}→{1,2,3,4,5,6,7,8,9,10}.
How many functions are there total?
100000
How many functions are injective?
30240
How many of the injective functions are increasing? To be increasing means that if a<b then f(a)<f(b), or in other words, the outputs get larger as the inputs get larger.

``` I was able to find the first to questions relatively easy but i dont know how to get specifically the number of increasing injective functions

[fallen heath]

,calc 0

[wraith dagger]

Result:

0

[fallen heath]

hmm

[fallen heath]

You can pick any pair of 5 digits and there's only one way in which they can be increasing right?

[forest dust]

it doesnt specify how many digits you must choose

[fallen heath]

You do though-

I mean, the domain consists of 5 values so

the range should be 5 for injective as well right?

[forest dust]

oh yeah

youre right yes my bad

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[crimson sedge]

Create a fractions question that includes the following elements:

• involves exactly four proper fractions in the question
• involves exactly two integers in the question
• involves at least one step that requires cancelling
• involves at least one step that requires a common denominator
• involves an answer that is a negative mixed fraction

[sharp widget]

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[crimson sedge]

I asked before and I didn't get an answer, so I'll ask again: I need to determine the "trigonometric functions of an angle." (NOTE: This is a right triangle.) One of those angles is 90 degrees. Can I just pull those from a table? Ie, the sin of a 90 degree angle is ALWAYS 1?

[bold vine]

sin(90°) = 1. Yes

[crimson sedge]

Okay, so cos and tan and the others are also always the same. Thank you! I'll just memorize them.

[gaunt hamlet]

The sin of any angle never changes. sin(30) is always the same. cos(163) is always the same.

[bold vine]

Correct

[crimson sedge]

But for an angle of 22.33 degrees, I do need to calculate it the "traditional" way. Makes sense.

[gaunt hamlet]

You have to calculate every angle. It's just that some angles have nice answers

[crimson sedge]

Right.

Well, that's all! Thank you for replying.

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[dim mist]

pls help me

and explain me by showing the work

pls

,ti

[wraith dagger]

The current time for Napoleon is 01:37 AM (+06) on Tue, 22/02/2022.

[dim mist]

I have an exam tmrw at 8 pm but I need to understand it and also I have school in the morning, so pls help me

as fast as one can

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[grim dome]

To find column X, do i just do A^-1 * b ?

[flint plinth]

You could but that's way more work than just doing gaussian elimination.

[grim dome]

oh i forgot i could do that

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[wicked fossil]

How can I get started for this? Thank you!

[lethal jackal]

this is a binomial distribution with p = 0.98 and n = 125

look up "Poisson approximation to binomial" and "normal approximation to binomial"

[wicked fossil]

Thank you!

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[visual socket]

I tried doing sin(50)/a = sin(28/27) and after I did the math for that I got 44 but it was wrong. What did I do wrong?

[trail jackal]

i think you didn't round to the nearest tenth...

[visual socket]

Oh right it's supposed to be 44.1

[trail jackal]

yes

[visual socket]

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[toxic garden]

Was just reviewin

and wanted to ask why

where did the numerator 5+2 come from

[hoary notch]

$$a^x * a^y = a^{x+y}$$

[wraith dagger]

Tronsi

[hoary notch]

So you're finding 1/2 + 1/5

[toxic garden]

ye

[hoary notch]

Do you know how to add fractions?

[toxic garden]

uhhh yea

[hoary notch]

So then what is 1/2 + 1/5?

[toxic garden]

7/10?

[hoary notch]

List out your steps

[toxic garden]

uhh

I totally did not use a calculator

[hoary notch]

Well do you know how to add fractions by hand? Set a common denominator and add the numerators?

$$\frac{1}{2}+\frac{1}{3}=\frac{13}{23}+\frac{12}{32}$$

[wraith dagger]

Tronsi

[cedar kiln]

@toxic garden Has your question been resolved?

[toxic garden]

I might have forgotten

[hoary notch]

https://www.youtube.com/watch?v=82pSWisvTpo

[toxic garden]

wait ill check it out later

caus im in class rn

ahh I see

5/10 + 2/10?

@hoary notch

[hoary notch]

Yes

[toxic garden]

whats next

[hoary notch]

Well do you have another question?

[toxic garden]

that becomes 7/10 right

oh wrait

nvm im dum

thanks again man

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[formal nest]

how is this true ?

[humble moss]

limitssss

[formal nest]

it becomes inf -inf

so :/

[humble moss]

its gonna become 1/2 - inf + inf - inf + inf

[formal nest]

yhea

[humble moss]

the -inf + inf -inf .... is gonna = 0

[formal nest]

how do you know

[humble moss]

• inf + inf = 0

[formal nest]

no

[humble moss]

keep doing that infinite times

[formal nest]

thats not true

[humble moss]

how?

[formal nest]

well its kinda true but thats unsure'

[humble moss]

that like saying -100000 +100000 = 0

[formal nest]

no

inf is not a value

[humble moss]

from what i learned n calc

whenever we had 1/inf we considered it as 0

[formal nest]

thats not my case

[humble moss]

or we learned it as 1 over some really big number

which is considered 0

[formal nest]

mine is inf/number

[humble moss]

same as inf / whatever number

some really big number over 10 = infinity

[formal nest]

my exam is thursday

im gonna fail

[humble moss]

essentially that equation is saying the limit as x -> inf is = 1/2

because its 1/2 -inf + inf ....

everything to the right of the 1/2 is gonna equal 0

to in reality its just 1/2 - 0

[formal nest]

usually when you have inf - inf you have to modify the equation to not get inf- inf

[humble moss]

its the same as -inf + inf

inf - inf = -inf + inf

you are just moving the -inf to be first instead of 2nd

[humble moss]

inf - inf is same as -inf + inf

[formal nest]

ok ?

[humble moss]

its like 1-1 = -1 + 1

which is why that whole right section is gonna become 0

doing it your way you can rearrange it so its lim x-> inf (1/2 - 1/4x^2 - 1/3x - 1/6x^4-1/5x^3...)

[formal nest]

what youre saying is that this should equal 0

but its not

and lim x-->inf of x/2 = inf not 0

<@&286206848099549185>

[humble moss]

that for the steps they show some weird thing i never learned

never learned to factor out the x

[formal nest]

in what grade are you in ?

[humble moss]

12

[formal nest]

alr

[humble moss]

taking ap calc ab

[formal nest]

so you just started learning limits ?

[humble moss]

learned it a few months ago

[fallen heath]

TF?

[formal nest]

ik LMAO

XD

[fallen heath]

( T_T)＼(^-^ )

[formal nest]

ahahhaha

[fallen heath]

$\lim_{x \to \infty} \qty(\frac{1}{2} - \frac{1}{3}x + \frac{1}{4}x^2 - \frac{1}{5}x^3 + \ldots)$

[humble moss]

interesting

[fallen heath]

Is that the limit?

[formal nest]

yhea

looks like

6 is a 5

[humble moss]

^

[fallen heath]

well, this works now ( T_T)＼(^-^ )

[humble moss]

what is that emoji suppose to mean

[fallen heath]

$= \lim_{x \to \infty} \frac{1}{x^2} \qty(\frac{1}{2}x^2 - \frac{1}{3}x^3 + \frac{1}{4}x^4 - \frac{1}{5}x^5 + \ldots)$

[formal nest]

o.O

[humble moss]

so its gonna = 0?

since 1/x^2 = 0

[fallen heath]

$= \lim_{x \to \infty} \frac{1}{x^2} \qty(e^{-x} - 1 + x)$

[formal nest]

what

[humble moss]

how you get e^-x?

[formal nest]

how did you find the series

[fallen heath]

btw 👀 I'm not sure your limit really works or evaluates to 1/2

[formal nest]

https://www.youtube.com/watch?v=oSmyE7sj3js

this is where i found it

[fallen heath]

timestamp?

[formal nest]

the end

its a 2 min video

[fallen heath]

Lmao

Nvm she made a mistake

[formal nest]

o:

[fallen heath]

please spare her (ヘ･_･)ヘ┳━┳

[formal nest]

are you sure ?

[fallen heath]

that was a lim (x approaches 0)

Yeah

[formal nest]

omg

omg

all of this

for a stupid mistake

in all the videos ive wachted of her SHE ALWAYS makes a mistake

[fallen heath]

,w lim (e^(-x) - 1 + x)/x^2 as x approaches infty

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Ansh#3502

[formal nest]

thanks Ansh youre the best

[fallen heath]

btw

[formal nest]

ye ?

[fallen heath]

Did you notice the small tab she opened in the top right corner?

[formal nest]

yhea

[fallen heath]

it had something written like "Taylor Series"

[formal nest]

yhea..

[fallen heath]

That's where the series was written

e^x = 1 + x + x^2/2 + x^3/3 + ...

[formal nest]

oh

i see

aight thanks

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[placid tree]

help

[dense wing]

answer is 7

Glad I could help

[placid tree]

bruh

i just need help with 6 and 7

[dense wing]

,rotate

[frigid canyon]

for 7th you'll need to use Leibniz integral rule

[frigid canyon]

for sixth, what is y'?

[placid tree]

wtf is lebniz

[placid tree]

y' means the derivatie

[frigid canyon]

yes

[frigid canyon]

well, instead you can do this

use integration by parts and proceed

it would be longer but it shld work

[placid tree]

lol what

[gaunt hamlet]

Just use the first part of FTC

For 7

[placid tree]

could you elaborate?

[gaunt hamlet]

It might actually be better to use the second part since the top bound isn't just x

We want the derivative of that integral

[placid tree]

yeah so

heres what i have rn

[placid tree]

ik

[gaunt hamlet]

That's correct

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[gentle lintel]

why is $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

[wraith dagger]

Chromium

[gentle lintel]

the $dt$'s don't cancel i suppose

[wraith dagger]

Chromium

[fallen heath]

cancel all the D's and the T's and you have y/x = y/x is why :o

[gentle lintel]

[fallen heath]

Consider $y = x^2 + 2$, then $$\dv{y}{x} = \frac{\dv{y}{t}}{\dv{x}{t}} = \frac{\dv{(x^2 + 2)}{t}}{\dv{x}{t}} = \frac{0}{0}$$

[fallen heath]

da heck

[gentle lintel]

how to parameterise this.

[fallen heath]

it's a parabola >_<

[gentle lintel]

x = t, y = t² + 2 ig

[fallen heath]

mhmm

[gentle lintel]

then dy/dx = 2t?

[fallen heath]

= 2x = dy/dx

[gentle lintel]

anyway why ;-;

holy

@jajdjsjjdjkdkdncndkd

[fallen heath]

lmfao @ helpers ghost ping by chromium

[gentle lintel]

@gentle lintel

[gentle lintel]

you gotta help me dude

@gentle lintel @gentle lintel @gentle lintel

[fallen heath]

Just use $y = f(x) \implies y(t) = f(x(t))$ and chain rule

[fallen heath]

screams

[gentle lintel]

but it’s not guaranteed that y is a function of x

or x is a function of y

[gentle lintel]

here we have ellipse

[fallen heath]

both are functions of "t"

[gentle lintel]

but y = f(x) implies y is a function of x

or do we only consider the neighbourhood around (x(t), y(t))?

implicit function theorem fuckery

[fallen heath]

nvm mb

consider $f(x,y) = f(x(t), y(t))$

[fallen heath]

evil laughter

[fallen heath]

in this case however, you'll be umm.. in a pinch

how do you even parametrize $x^3y + xy^3 = 1$

[cedar kiln]

@gentle lintel Has your question been resolved?

[gentle lintel]

<@&286206848099549185>

[gentle lintel]

this is my best guess

[jaunty mural]

Youre a bit late to ask about the chain rule now aren't you?

(dy/dt)(dt/dx)

You should be perfectly happy with dy/dt, so it just remains to ask whether dt/dx makes sense ig

[jaunty mural]

.

Well like you said it does of course

[gentle lintel]

similar to the implicit differentiation talk ig

[jaunty mural]

because why should you care about anything outside the neighborhood when differentiating

[gentle lintel]

okays

[jaunty mural]

Differentiating is linearisation of the neighborhood

👋 bai bai

[gentle lintel]

.close

[cedar kiln]

Channel closed

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cedar kiln]

Channel closed

Closed due to the original message being deleted

[fallen heath]

I understood absolutely nothing ._.

[inland tartan]

here's a nice, visual proof that the sum of odd numbers are squares

[crimson sedge]

Yup

[upper light]

Thanks

.close

.close

[fallen heath]

bot doesn't believe you understood that

[upper light]

I'll just delete the message

[fallen heath]

lol

[inland tartan]

bot went back online

short outage

try again

[upper light]

.close

[fallen heath]

Lol it's not that.. they deleted the original msg so bot already closed this one

[inland tartan]

LOL HOW

[crimson sedge]

Lol

[upper light]

I deleted original message

Guess I'll wait for timeout so I'll stop sending messages here from now on

[cedar kiln]

Available help channel!

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Remember:
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[final frost]

AX = {-1}
{1/2}
{2}

r = 2

[final frost]

Is there a formula to find r?

[soft cloak]

that's it

that's the answer

[final frost]

Yes but is there a formula for that

[soft cloak]

r is what's known as an eigenvalue

[final frost]

Or you just need to trial and error it

[soft cloak]

Whether eigenvalues exist and how to find them is rather involved, and it takes up a few chapters in classes usually

perhaps you'll see it in the future

But in the meantime, don't worry too much about it, do the calculation and find the constant multiple

[final frost]

Ah ok ok thank you

.close

[cedar kiln]

Channel closed

Closed by @final frost

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[covert island]

so far I know that

I need to find the derivative of both sides

so i have 1/2x^1/2 and 1/2y^1/2

for the lfs

and the rhs I have 0

[long fulcrum]

1/ 2sqrt(y) * dy/dx

[covert island]

$1/ 2sqrt(y) * dy/dx$

[wraith dagger]

c0nc3ptsz

[covert island]

wait isnt the derivative of y^1/2

[long fulcrum]

$\frac{1}{2 \sqrt(y)} \cdot \frac{\text{dy}}{\text{dx}}$

[covert island]

1/2(y)^1/2

oh wait

same thing

sorry

fuzzy brain lol

[wraith dagger]

Serena!

[covert island]

right or u correct?

[long fulcrum]

what do you mean?

[covert island]

y could be subbed out as u right?

cuz i always do chain rule as the outter variable is u

and the inner is whatever the variable is

so like (x+4)^2

the inner equation is x+4

the outter is u^2

right?

or did i lose my marbles?

[long fulcrum]

i mean the derivative of (x+4)^2 would follow the chain rule yeah so you'd get f' g(x) * g'(x) so 2(x+4) * d/dx ( x+4)

which is just 2(x+4)

for this question you can sub in u, but i mean it's just a detour(?)

derivative of y = 1/(2 sqrt(y)) like you mentioned but since we're derivative it with respect to x we apply chain rule

[covert island]

mhm

so once we do that

this is where things get fuzzy for me

[long fulcrum]

so derivative of y wrt x would be dy/dx so to synthesize the root(y) portion's derivative would be 1/(2 sqrt(y)) * dy/dx

[covert island]

wrt?

[long fulcrum]

with respect to

[covert island]

oh yeah

so

1/2root(y'x)

works to right

cuz i wrote that part down

[long fulcrum]

what is y'x

[covert island]

dy/dx = y'x

no?

cuz its the derivative of y with respects to x

y'x?

[zealous crystal]

dy/dx=y'

[long fulcrum]

no i mean like why is it inside the bracket?

[zealous crystal]

Not y'x

[covert island]

uhhh im not sure actually

ok wait

so then

we have

1/2sqrt(y')

is that correct so far?

[long fulcrum]

no

[covert island]

1/2sqrt(y)*(y')

?

[long fulcrum]

[f(g(x))] ' = f'g(x) * g'(x)

[long fulcrum]

yes

[covert island]

my bad

ok cool

ok so from here

I assume

we isolate it

so y' is alone

[long fulcrum]

yes.

[covert island]

correct?

im supoer close

i think its an algerbaic error

can u check if i have this part right

$1/(y') = -2rad(y)/2rad(x)$

[wraith dagger]

c0nc3ptsz

[long fulcrum]

i mean the answer suggests it isn't? lol how did it become 1/y'

$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot y' = 0\
y' = -\frac{1 \cdot 2\sqrt{y}}{2\sqrt{x}}$

you can cancel out all the ones and two's ofc i just put it there for clarity since you were struggling with the algebra

[wraith dagger]

Serena!

[covert island]

ok cool!

I got it

Thanks for your help!

I really appreciate it

!

[long fulcrum]

Sure, gl!

[covert island]

.close

[cedar kiln]

Channel closed

Closed by @covert island

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[thin blaze]

Hello. I have to solve monotony of 6^n(im on highschool). How do i do it?

[thin blaze]

i know the formulas: a_n < a_n+1 for increasing and a_n > a_n+1 for decreasing

[dense wing]

I mean, trivially $6^k<6\cdot 6^k=6^{k+1}$

[wraith dagger]

Mosh

[thin blaze]

6^n < 6^n+1
6^n - 6^n+1 < 0
6^n(1-6)<0
6^n(-5)<0

so, how do i continue here?

[dense wing]

that's always true

so you've shown 6^n<6^(n+1) for all n

[thin blaze]

so?

[south tundra]

So you're basically done

[dense wing]

You finished the proof...?

[thin blaze]

ah okay, thank you

much loveee

.close

[cedar kiln]

Channel closed

Closed by @thin blaze

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[cedar kiln]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson sedge]

<@&286206848099549185>

[elfin hemlock]

#❓how-to-get-help

(Or just read the box above)

[crimson sedge]

may i ask what did i do wrong..?

or

oh

its bc

they deleted my channel before

so

thats why

[elfin hemlock]

Ask a single question at a time

[crimson sedge]

oh.

[elfin hemlock]

Show/say what you have tried

[crimson sedge]

im sorrt

[elfin hemlock]

Don’t ping helpers

[crimson sedge]

A trapezoid is shown. The lower base measures 17.5 centimeters and the upper base measures 12.5 centimeters. A dashed line segment that measures 6 centimeters connects the 2 bases at or near their midpoints.
A. 90 cm2
B. 118.75 cm2
C. 130 cm2

[crimson sedge]

oh

ok

A trapezoid is shown. The lower base measures 17.5 centimeters and the upper base measures 12.5 centimeters. A dashed line segment that measures 6 centimeters connects the 2 bases at or near their midpoints.
A. 90 cm2
B. 118.75 cm2
C. 130 cm2

[elfin hemlock]

That isn’t even a question

[elfin hemlock]

And

[crimson sedge]

...?

[elfin hemlock]

I don’t see a question in what you posted, do you?

[elfin hemlock]

And

[crimson sedge]

they said that

all i need to do

it get a room

and post my questions

..

[frigid canyon]

nope

you have to tell what you've done

[crimson sedge]

oh

[elfin hemlock]

This is not get answers for your homework thing

You are meant to try yourself

[crimson sedge]

ik

[elfin hemlock]

And again

[crimson sedge]

i just need 2.

well, this is for my brother

not me....

[frigid canyon]

you doing his hw?

[crimson sedge]

no

he needs help

well...

kinda

bc like he helped me do sum

[elfin hemlock]

You haven’t even asked a question lol

[frigid canyon]

dude you can send him then

he can chat

[crimson sedge]

where...

[frigid canyon]

here ofc

[crimson sedge]

ok

ima get him

[frigid canyon]

k sure

[crimson sedge]

ok

hello..?

[frigid canyon]

yea

[crimson sedge]

any questions?

[frigid canyon]

umm, dude i think you are here to ask questions

[crimson sedge]

oh,

Find the area of the triangle.

A triangle is shown. The base of the triangle is labeled 28 yards. At approximately the center of this base, a vertical dashed line segment is drawn to the opposite vertex. A small square is located at the intersection of the dashed line with the side of the triangle. The dashed line segment is labeled 7.1 yards.
A. 198 yd2
B. 106 yd2
C. 99.4 yd2
D. 32.2 yd

so

[frigid canyon]

where's the diagram?

[crimson sedge]

uhm

[bold vine]

Start by drawing a diagram

[frigid canyon]

bruh

it says it's given

[bold vine]

Ah - my bad

[elfin hemlock]

“Test”

[crimson sedge]

its says part 1

so its a practice

[elfin hemlock]

What? A test can easily have multiple parts

[bold vine]

I'm not sure that's how that works

[crimson sedge]

back

[crimson sedge]

no, it got named wrong

even the teacher told him..

[crimson sedge]

ive also been stuck on this for 2 hours

:/

ofc

they ghost me

again.

[elfin hemlock]

Still not convinced that it isn’t a test

[crimson sedge]

:/

[elfin hemlock]

And

[obsidian coral]

Very suspicious that all signs points to a test and the only proof given was a message saying it's not

[crimson sedge]

ok,

[cedar kiln]

@tawdry quarry Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cedar kiln]

Please don't occupy multiple help channels.

[dire geode]

.close

[cedar kiln]

Channel closed

Closed by @dire geode

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Available help channel!

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Remember:
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[rancid arch]

X investigated access to television. Making a comparison of households by available television model, it was found that 81.6% had access to thin-screen television and 26% to tube television. how many households had access in 2019 only to thin-screen television?

[cedar kiln]

@rancid arch Has your question been resolved?

[cedar kiln]

@rancid arch Has your question been resolved?

[cedar kiln]

@rancid arch Has your question been resolved?

[zealous crystal]

is this question complete?

[rancid arch]

Yes, the original text of this topic is a foreign language, the translation may be wrong, the meaning of this topic is that there are some people use two kinds of TV, for example, the teacher asked who likes math, you raise your hand, the teacher then asked who likes geography, you also raise your hand, that is, you like 2 subjects, back to the original question, this topic you have to find out how many people only use thin screen TV, equivalent to how many people only like math, this question I I can not think of, the It's too hard.

I think it's better to close the channel if no one has helped yet

[zealous crystal]

well i got this stuff, i meant there is no data given like how many ppl were investigated or something like that?

[rancid arch]

NO

😶‍🌫️

[rancid arch]

.close

[cedar kiln]

Channel closed

Closed by @rancid arch

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cedar kiln]

@crimson crane Has your question been resolved?

[cedar kiln]

@crimson crane Has your question been resolved?

[cedar kiln]

@crimson crane Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

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[crimson sedge]

uh I am really stuck on this I worked on it yesterday for half an hour but couldn't figure it out

$$ a, a + r, a + 2r$$ so b = a + r and. c = a + 2r$$

[wraith dagger]

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

[crimson sedge]

but for the GP idk what to do

[hoary notch]

You posted this just 10 minutes ago, don't post it multiple times #❓how-to-get-help

[crimson sedge]

no I didnt

so you have $$a \quad b = a+r \quad c = a+2r$$ and $$a \quad c = xa \quad b = xc = x^2 a$$

[wraith dagger]

rustyy

[crimson sedge]

@crimson sedge ok

I understand that where x is the ratio in the Gp and r is the ratio in the Ap

[fallen heath]

@crimson sedge you

wow, that second person pronoun sounds so violent at times lmao

Anyways

[crimson sedge]

HI

HOW ARE U

[fallen heath]

When given three terms in A.P.

[crimson sedge]

yes

[fallen heath]

there's a common trick you could do to save yourself from a headache

[crimson sedge]

and that is?

[fallen heath]

Let the terms be $(x - a), x, (x + a)$

[fallen heath]

now, by their construction, they're already in A.P.

now, a, c, b being in geometric progression just means that: $$ab = c^2$$

[fallen heath]

plug these values in there

and get an expression for a, b, c in terms of one variable

[crimson sedge]

ab = c^2

dont really understand that

[jaunty mural]

a, ar, arr

a(arr) = (ar)^2

aarr = arar

[fallen heath]

rawr

[crimson sedge]

xd

[fallen heath]

Lol

[jaunty mural]

now?

[fallen heath]

for any three terms in G.P., their common ratio is same

[crimson sedge]

uh

[jaunty mural]

rr is just me writing r^2 fast

[crimson sedge]

so the first and last term of an arithmetric sequence is the middle squared???

[jaunty mural]

indeed

wait no

[crimson sedge]

is that a rule

[fallen heath]

so, umm $$\frac{b}{c} = \frac{c}{a} \implies ab = c^2$$

[jaunty mural]

wait i guess yes?

check it if you like

[fallen heath]

no lmao

[jaunty mural]

think about AP. The middle term is the mean

GP, the middle term is the geometric mean...

[fallen heath]

twice the middle term of an A.P. equals the sum of first and last terms

while

square the middle term of a G.P. equals the product of first and third terms

[crimson sedge]

hmmm

thats cool

[jaunty mural]

first and last also

[fallen heath]

not cool if you're just accepting it as a matter of fact

[jaunty mural]

but here we have 3 terms

[fallen heath]

can you reason why it's like that?

[jaunty mural]

yes prove it 👀

[crimson sedge]

ummm

[jaunty mural]

Call the middle one x

[fallen heath]

Let a, b, c be three consecutive terms of a G.P.

[crimson sedge]

wait yeah I get it

[fallen heath]

prove that b^2 = ac

[crimson sedge]

so A, A(R), A(R)(R)

so theres 2 A and 2 R in the product of first and last which is the middle squared

[fallen heath]

i-

well that works as well

rawr

[crimson sedge]

oke

[fallen heath]

so now,

[crimson sedge]

so back to problem

[fallen heath]

you have three terms in G.P.

a

[crimson sedge]

si

[fallen heath]

c

and

b

a, c, b are in G.P.

so what's the relation between the three?

[crimson sedge]

$$ ab = c^2 $$

[fallen heath]

[crimson sedge]

[fallen heath]

perfect

[wraith dagger]

痛苦

[crimson sedge]

ok now that works wew

[fallen heath]

what is a, b, c as we constructed when we said it was an A.P.?

surely you forgot(!?)

[crimson sedge]

this

[fallen heath]

bruh

you just

had

[crimson sedge]

so a, b, c or
a, a+r, a + 2r

[fallen heath]

to make your life

more hell

[crimson sedge]

bruh bruh bruh

[fallen heath]

than it already is

[crimson sedge]

[fallen heath]

okay!! hear me out

[crimson sedge]

oke

[fallen heath]

let's start with one of the conditions

Terms in A.P. or terms in G.P.

which?

[crimson sedge]

terms in

GP

[fallen heath]

GF

[crimson sedge]

[fallen heath]

ait

So the three terms in GP

[crimson sedge]

ab = c^2

[fallen heath]

let the terms be $a, ar, ar^2$

Do you see that these terms are in G.P.?

[crimson sedge]

yes

I do

I'm not that dumb

[fallen heath]

kek

now, this means

We've assumed: $$a = a, c = ar, b = ar^2$$

[crimson sedge]

?

[fallen heath]

[crimson sedge]

yes

ok

[fallen heath]

Do you understand that!?

[crimson sedge]

YES

I DO

[fallen heath]

now

which three terms are in A.P.?

[crimson sedge]

UM

a, a + r, a + 2r

[fallen heath]

tf

[crimson sedge]

?

a, b, c

[fallen heath]

Question says a, b, c are in A.P.

YES

[crimson sedge]

ye

[fallen heath]

and what is our a, b, c

in terms of a and r

like we just previously calculated

[crimson sedge]

hi

a, a + r, a + 2r

[fallen heath]

bruhhhh

[crimson sedge]

WHAT

[fallen heath]

$$a = a, c = ar, b = ar^2$$

[crimson sedge]

I thought u were talking about AP

not GP

[fallen heath]

let the terms be $a, ar, ar^2$

[crimson sedge]

how to prove rectangle is a square

[fallen heath]

We ARE TALKING about a, b, c being in AP

[crimson sedge]

this channel to occupied

[fallen heath]

but we already know what a, b, c are

in terms of a, r

[crimson sedge]

ok

[fallen heath]

SO WHY NOT use that frickin condition to avoid using another variable

[crimson sedge]

OK

I WILL DO THAT

[fallen heath]

FIRST, WE HAVE a, c, b in a GP

so we let $a, c, b \equiv a, ar, ar^2$

[fallen heath]

which automatically assigns $$a = a, b = ar^2, c = ar$$

[crimson sedge]

so in the AP, a + c = 2b

??

[fallen heath]

YES

$$a + ar = 2ar^2$

[crimson sedge]

oh was that what u were trying to get me to say

[wraith dagger]

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

[crimson sedge]

[fallen heath]

more like everytime you said "I do", I was being forced to imagine a wedding going on (ヘ･_･)ヘ┳━┳

[crimson sedge]

💒

[fallen heath]

Lmao

So do you understand

[crimson sedge]

ofc

[fallen heath]

$$a + ar = 2ar^2$$

[crimson sedge]

now what

[crimson sedge]

[fallen heath]

solve

for r

eliminate a

and you have some possible values of r

[crimson sedge]

$$1 + r = 2r^2$$

[wraith dagger]

痛苦

[fallen heath]

,w 1 + x = 2x^2

[wraith dagger]

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Requested by Ansh#3502

[fallen heath]

r = 1, -1/2

which one is appropriate

[crimson sedge]

its positive

um

uh

I think

hmmMMMM

[fallen heath]

[crimson sedge]

negative

[fallen heath]

whicH ONE

-1? -2? -0.1111?

[crimson sedge]

OK OK POSITIvE posisiITIVEIE POksiITVEIIIEIE

[fallen heath]

NO

[crimson sedge]

WHAT

[fallen heath]

It's the -1/2

[crimson sedge]

HOW

WHEN

WHAT

[fallen heath]

[crimson sedge]

WHY

WTF

WHHHATTT

[fallen heath]

well, you're given a < b < c so

[crimson sedge]

it cant be negative then the domain will be undefined

it has to be positive

[fallen heath]

wdym-

[crimson sedge]

cause look

[fallen heath]

what do I look at-

r = 1 just gives you a = b = c

but your question clearly suggests a < b < c

why'd I avoid that trivial condition

[crimson sedge]

ok yeah we talked about this 2 days ago

like since c is in the middle

[fallen heath]

[crimson sedge]

we can differentiate that its switching from neg to pos

[fallen heath]

NOW

TIME TO PLUG THOSE BACK

what is a

[crimson sedge]

ahem

a is a

?

[fallen heath]

perfect !

what is b

[crimson sedge]

-1/2a

[fallen heath]

[crimson sedge]

?

[fallen heath]

[fallen heath]

$b = a\mathbf{r^2}$

[crimson sedge]

its ar and r = -1/2

[fallen heath]

Omg

[crimson sedge]

OOOHHH the GP

nvm srry

[fallen heath]

:SCREAMS:

[crimson sedge]

its the 3rd term

not 2nd

[fallen heath]

YES

[crimson sedge]

[fallen heath]

imagining a not so dumb, dumb hooman

an airhead?

what's c?

[crimson sedge]

c is

-1/2a

👏

[fallen heath]

perfect

[crimson sedge]

yAY

so then

now we need to find smallest value for c

[fallen heath]

Now, your terms are $$a, \frac{a}{4}, -\frac{a}{2}$$

[fallen heath]

AND THEY NEED TO BE INTEGERS