help-13

[crimson sedge]

theta becomes 90

cos 90 = 0

prov ( u - prov)

let me show my work...

prov = (u.v/|v|^2)v)

(u.v/|v|^2)v).(u-(u.v/|v|^2)v))

(u.v/|v|^2)v).u - |(u.v/|v|^2)v)^2|```

im stuck at this last step

[dire geode]

What's prov

[crimson sedge]

thats projection of u vector

on v

[dire geode]

v.u=u.v

[crimson sedge]

yes

[dire geode]

You can simplify that number on the first term

[crimson sedge]

number?

[dire geode]

Yes the dot product returns a scalar

[crimson sedge]

(u.v/|v|^2)v).u

are u referring to this

first term

[dire geode]

Yes that's the first term

[crimson sedge]

(u.v/|v|^2) u.v

oh

|u.v|^2 / |v|^2

(|u|cosx)^2

[dire geode]

You also have a u.v squared term in the right

[crimson sedge]

|(u.v/|v|^2)v)^2|

[dire geode]

But you have a mistake somewhere on the second term

Because you're squaring a vector

Probably should be v.v or something similar

[crimson sedge]

i did like prov is a vector so prov.prov = |prov|^2

i should keep it in dot product

ig

(u.v/|v|^2)v).(u.v/|v|^2)v)

u can take scalars out

oh wait

lmao

u.v . u.v

= |u.v|^2?

what am i even doing

😅

[dire geode]

.

[crimson sedge]

Mhm

[dire geode]

Write u.v = c since it's confusing you

[crimson sedge]

(u.v/|v|^2)v).(u.v/|v|^2)v)

= (u.v/|v|^2)(v.v)

right

(c/|v|^2)v).(c/|v|^2)v)

v.v = |v|^2

[cedar kiln]

@crimson sedge Has your question been resolved?

[dire geode]

Are you still stuck?

What's your expression after you did all the simplifying

[crimson sedge]

(|u|cosx)^2 this is the first term

idk what to even do about second

lmao

(c/|v|^2)v).(c/|v|^2)v)

should i take the scalar value (c/|v|^2))

out and do v.v

or what

(c/|v|^2) v.v => c/|v|^2 *|v|^2
=> c

[cedar kiln]

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[keen carbon]

im not sure if this is even right I js tried it? but how do I get slope of j

[keen carbon]

it's perpendicular so it's supposed to meet and then never again bur how

i think I fixed the slip but Idk how to get slope j

slope*

I'm not sure if this is right at all?

.close

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[undone epoch]

I'm wondering how this happend?

[undone epoch]

Since it would be $$x^2=e^{2i\pi/3}$$ and $$x^2=e^{4i\pi/3}$$ after quadratic formula.

[wraith dagger]

Good

[dire geode]

how what happened

[undone epoch]

Like how did $e^{i(\pi/3+k\pi)}$ became a solution

[wraith dagger]

Good

[undone epoch]

Since the solutions are these ones, which only gives $\theta=4i\pi/3, 2i\pi/3, i\pi/3$.

[wraith dagger]

Good

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[crimson sedge]

How to resolve $e^{23/(3*2)^2+kq\pi}$

[wraith dagger]

Tharkys

[dire geode]

What does resolve mean

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[left gull]

Could you check my proof structure ? The identity and inverse I did in a differently formatted way. Should I have listed it out like I did for the inverse or maybe this listing isn't even necessary at all

[solid juniper]

it’s ok

[cedar kiln]

@left gull Has your question been resolved?

[left gull]

Anything I can do to make it better? or is it pretty much good?

[solid juniper]

you could probably be more concise but it’s fine

[left gull]

Okay thank you : )

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[wary sage]

I'm trying to prove Fermat's Little Theorem. I think I did it, but I'm not sure if the proof is valid.

[wary sage]

The reason I'm not sure, is that I don't know if simply declaring that the sum is an integer is valid.

[cedar kiln]

@wary sage Has your question been resolved?

[near snow]

Yes, it's correct

[wary sage]

So I don't have to prove that the sum is an integer? That makes my life a lot easier then

.close

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[fossil coral]

um

[fossil coral]

...

is this problem, kinda like this one?

[frank shale]

yea since its all to the same power if it was higher u do long divison i believe

[fossil coral]

So basically, Im missing a lot of coefficients then?

[frank shale]

this a weird one the powers all hlla high

[gritty viper]

You're looking for a remainder

The answer should be a single number, not a polynomial

[fossil coral]

k

[gritty viper]

But yes you can't just stick the coefficients into synthetic division like that

You would need a 0 for every power you skipped

[fossil coral]

frack

[gritty viper]

So it would actually be 2 0 0 0 ... 0 0 -39 0 0 0 ...

with hundreds of zeros between each non-zero number

[fossil coral]

would -5 be it?

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[fossil coral]

Is this how I solve this problem?

[fossil coral]

Used mathway to type it out quicker

[gritty viper]

All that will tell you is that -4 is in fact a zero

Because when you press enter you'll get 0

[fossil coral]

so dont do that,

[gritty viper]

right

[fossil coral]

what to do instead?

[gritty viper]

-4 is a zero

so what do you know about the factorization

[fossil coral]

x+4=0

GCFs?

1

[gritty viper]

x+4 should be a factor

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[agile citrus]

i need someone to confirm my work

[agile citrus]

this is the question

Does this satisfy part a ?

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

only need confirmation

nothing else

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[agile citrus]

<@&286206848099549185>

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[open vine]

Did i do this right? A 0.14% chance seems kinda low

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[crimson sedge]

Calculate the area of the shaded region

I think the height is π/2

And the base π?

[livid hound]

incorrect, can you show your work

[crimson sedge]

I am doing it in my mind

base π/2?

[crimson sedge]

range restrictions

[livid hound]

how is that leading to pi/2 for height

[crimson sedge]

the minimum value for arc(x)= 0 is when cos(0) = 1 = x, so sen^4x+ cos^4x = 1, so x can be 0 or π/2 or π...

[crimson sedge]

because the maximun is pi/2

for range

[livid hound]

how?

[crimson sedge]

to be biyective

and inyective

[lusty grotto]

sen is sin?

[livid hound]

don't just throw words around

[crimson sedge]

I know what u mean

I'll think more

[livid hound]

for what values of k will
arccos(k) be pi/2?

[crimson sedge]

0

[livid hound]

can you get that value from sin^4(x) + cos^4(x)?

[crimson sedge]

mm right, I can't

I can get maximum 1/2?

so π/3

is the height?

[livid hound]

yes

[crimson sedge]

or minimum?

[livid hound]

misread what you typed
yeh, min of sin^4(x) + cos^4(x) will be 1/2

[crimson sedge]

max is 1

?

[livid hound]

yes

[crimson sedge]

so base is π?

bc first 0 is π/2

[livid hound]

base of the triangle, no

[crimson sedge]

well, consider the triangle

whole

like

all the floor

for now

[livid hound]

the whole graph? so ignore the triangle?

[crimson sedge]

I will subtract large triangle

and small triangle

[livid hound]

what?

[crimson sedge]

the tackit for getting the are under the shaded region

is like

that

also subtracting the big base minus the small base

but

idk if u get me

[livid hound]

i don't

[crimson sedge]

like well π - π/4 = 3π/4

[livid hound]

oh i see what you mean

[crimson sedge]

is that the base of the triangle?

[livid hound]

but you can just apply the area formula once directly

yes

[crimson sedge]

ik, I just wanted to calculate large triangle area minus small one

(3π/4) (π/3) (1/2) = π^2 / 8

like this?

no

what am I doing wrong

[livid hound]

yes

[crimson sedge]

no...

do u know what's wrong?

[livid hound]

someone messed up in writing the answer options

[crimson sedge]

yep

thanks 👍

.close

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[errant vessel]

It’s a problem of correlation using Karl’s method…. Can you tell me what’s with the no of blind per lakh column…. I don’t understand why and how they are doing it..

[cedar kiln]

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[errant vessel]

<@&286206848099549185>

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[rose flax]

how can i approach this question best?

[crimson sedge]

u can separate the sum in 3 sum

[rose flax]

so sum of 1/2^r add sum of 1/3^r and so on ...

[crimson sedge]

yea

[rose flax]

okay ty, do i do the sum to infinity for each and then add them up?

[crimson sedge]

yes

[rose flax]

awesome ty ill try that now

[rose flax]

hm i didnt do it quite right

this is the mark scheme im not sure why they did B= A - ....

[clever venture]

you forgot thar r = 2

not 0

[rose flax]

this is the mark scheme im not sure what they did

[clever venture]

then they probably did it wrong

whoever it was

and didn't notice the 2 there

[rose flax]

.close

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[crimson sedge]

so

Im so confused with piecewise functions

Like insane

Also

I wanna ask if this is right

[undone epoch]

Looks good

[bleak viper]

i agree

looks Good indeed

[steep badge]

very good

[crimson sedge]

Ohh wait

I caught a mistake

Its graphing the postitive variations

but its negative

[cedar kiln]

@crimson sedge Has your question been resolved?

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@crimson sedge Has your question been resolved?

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[rare heron]

is this actually just -1x/(x-1)(x^2-x)

and done?

[oblique flare]

no

[rare heron]

how to do this problem

factoring rational expressions

[oblique flare]

first off when adding or subtracting fractions we need common denominators

You can find it for algebraic expressions using LCM

[rare heron]

isn't the LCM literally 1

[oblique flare]

no

[steep badge]

try to simplify (x^2 -x)

[rare heron]

x(x-1)

[steep badge]

yea

to make denominator equal for the left term

multiply and divide by x

[limber dawn]

btw this would be if you multiplied fractions

[steep badge]

only for left term

[rare heron]

@

[steep badge]

this is to make the denominator equal

[rare heron]

oh I see

[steep badge]

thats only for the left term

not the whole equation

[rare heron]

what do I do next

[steep badge]

since denominator is same

you can just subtract the terms in numerator

[rare heron]

I'm not sure what you mean by that

[steep badge]

x^2 +(-1) = x^2 -1

you have this

merge it

same denominator ^

[rare heron]

so its just x-1?

and ✅

done

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[dry fossil]

$\sum^{\infty}_{n=1} (n+1) y^n$

[wraith dagger]

alee

[dry fossil]

To sum this series, do I have to start from the geometric series and then derive it?

[versed siren]

this is an AGP, Arithmetico Geometrico Progression

the method to solve these is

to let this be S

[dry fossil]

Ok

[versed siren]

then divide by the ratio, here y

to get

S*y

then

S - S*y

leave the first term of S alone

start from second term of S

and the first term of S*y

wait

I think I made a mistake

it'd be

S * y

not S/y

[dry fossil]

Where

[versed siren]

mb

[versed siren]

every where

Sy = sum (n+1)*y^(n+1)

wait lemme write it on paper

[versed siren]

@dry fossil This

[dry fossil]

Thanks guys!

[versed siren]

I'm one person

my guy

[dry fossil]

🙂

[versed siren]

Rishabh is also me

[dry fossil]

Ah ok hahaha

Thank you very much then!

[versed siren]

you can evaluate the sum in last line?

[versed siren]

np

[dry fossil]

It would be Taylor's series

y^n = 1/1-x

For n = 0

[versed siren]

its a Geometric progression

[versed siren]

yea

[dry fossil]

For n = 1 1/(1-x) -1

[versed siren]

that works too

!done

[cedar kiln]

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[dry fossil]

.close

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[quasi pewter]

<@&286206848099549185>

[languid pendant]

Let me nderstand

wait

[past wave]

do you know about alternte interior angles

and all?

[quasi pewter]

yh

they are equal

[languid pendant]

yes

[past wave]

so use them to get an eqn between A and F

[quasi pewter]

but a and b are non parallel

[languid pendant]

aight

[past wave]

bro is it your doubt?

[languid pendant]

no I'm clear

[past wave]

kk

[past wave]

its given that CBE + EFC = 180

use that

[languid pendant]

If ∠EFC = 3x+50° then
y = 150°-3x

[quasi pewter]

y=130-3x

[past wave]

bro let him do it on his own

[past wave]

bro then?

[quasi pewter]

is it 94

[past wave]

y?

[quasi pewter]

yh

[past wave]

yes

94

[quasi pewter]

k ty

[past wave]

wlcm

[quasi pewter]

btw what are u studying

[past wave]

im in school

[quasi pewter]

which grade

and country

[past wave]

11th India

[languid pendant]

If ∠A = 2x+70° then
∠CBE = 2x+70°
So ATQ,
5x = 60°

[quasi pewter]

i am in india too

[past wave]

nicee

[quasi pewter]

9th grade

[past wave]

makes sense

state?

[quasi pewter]

prepping for tallentex

[languid pendant]

x=12°

[quasi pewter]

tamil nadu

[past wave]

ohh

Chandigarh

[quasi pewter]

hmmm

[languid pendant]

So, y = 180° - 86°

Done?

[quasi pewter]

yh i got it long back

ty

[languid pendant]

And I tried it all by myself (although I did a tiny mistake before :P)

[cedar kiln]

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[quasi pewter]

@gleaming path

<@&286206848099549185>

[cedar kiln]

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[winged hemlock]

I want to figure out the coefficients of the quotient for this division, and I’ve set it up as follows. I’d like to know if it's set up correctly this way

[winged hemlock]

If I ask ChatGPT to solve it using the long division method, it arrives at a different result than mine

here my resolution

Here is the result and the verification

This is the result that ChatGPT arrives at using the 'long division method'

what do you think 🤷‍♂️

Is the form of my C(x) and R(x) expressed correctly?

[past wave]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[winged hemlock]

ok then..

[past wave]

ik op posted it but lol

[winged hemlock]

lol?

???

[past wave]

gimme a min

[winged hemlock]

okey

[past wave]

thats usual for helpers who use chat gpt to answer other ppls answers

op means original poster

[winged hemlock]

I only use it to compare my results and their solutions

It's true that sometimes it gives errors

[past wave]

,w [2x^2 + (3/4)x][4x^2-7] + 14x^2 + (13/4)x + 1

[winged hemlock]

y esto?

[past wave]

,w eval [2x^2 + (3/4)x][4x^2-7] + 14x^2 + (13/4)x + 1

[winged hemlock]

😵‍💫

[past wave]

,w [2x^2 + (3/4)x][4x^2-7] + 14x^2 + (13/4)x + 1

hey i found the problem

@winged hemlock

[winged hemlock]

TELL MEE

🤔

[past wave]

see the remainder chat gpt got

it is quad

it should be linear

[winged hemlock]

this is wrong you say?

[past wave]

do you remeber that when we write

Divident = Divisor*Quotient + Remainder

remainder has to be less than divisor or quotient

i dont rember

similarly in polynomials, the degree must be less

[winged hemlock]

ooh yeees

[past wave]

gpt prob forgot that lol

[winged hemlock]

it should be less than x¨2

[past wave]

less than is not the correct term

we cant say

the degree should be less than 2 in this case

[winged hemlock]

If I wanted to put that, it would take me forever to respond to you, my friend. Sorry, my native language is Spanish, and I have to translate everything, haha

Were my propositions for C(x) and R(x) correct?

[past wave]

ohh my bad i didnt know

[past wave]

yes extremely

i wasted my time checking your

should have directly checked gpts

[winged hemlock]

🤣

[winged hemlock]

Well, now I will try to solve it using long division by myself. How do I close this? Thank you very much for taking the time

[past wave]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[past wave]

wlcm bro

[winged hemlock]

okey

.close

[cedar kiln]

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[frank sleet]

whats the integral of 1/((2x+1)(sqrt(4x^2+4x)))

[lean lotus]

like this?

[frank sleet]

yes

i was trying to complete the square

but the problem is no matter what i do it will not be as same as 2x+1

as this ai did

[crimson sedge]

$\int \frac{1}{(2x+1)(\sqrt{(4x^2+4x)}}$

[wraith dagger]

Sukiyaki

[lean lotus]

u make u = 2x+1

and du would be 2

[frank sleet]

okay?

and then?

[lean lotus]

1/2 * integral

and i think u need to u sub again

[cedar kiln]

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[frank sleet]

Thank tou so much

[cedar kiln]

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[bold pond]

what is tangential force?

[crimson sedge]

I am assuming

(assuming be the key word here)

that the force acts along the surface of the ramp

wait lemme draw an FBD for ya

1. the actual forces have been drawn in black arrows

2. the component forces have been drawn in dotted arrows

3. NOTE: the length of the arrows is not to scale or anything (cause obviously, mg, can't be smaller than mg cos theta since that is a component)

[past wave]

is the wedge fixed?

[crimson sedge]

fixed as in?

not moving?

[past wave]

the wedge can move on the ground too

yes

[crimson sedge]

yeah probably fixed

[past wave]

if that is the case then we should use center of mass

[crimson sedge]

otherwise it would be stated

[past wave]

its not mentioned

[crimson sedge]

probably is mate

[slender ginkgo]

if the thing can move on the ground

[past wave]

it should be mentioned if its fixed not the other way around imo

[slender ginkgo]

then we cant really hold the block in place can we

[crimson sedge]

fasho

[past wave]

oh

[crimson sedge]

that is a good point

I missed that

[past wave]

didnt red that

[bold pond]

well it says in the question the block is being held stationary

[crimson sedge]

just equate $$mg \sin \theta = F_{\text{tangential}}$$

[wraith dagger]

Edmund Cloudsley

[crimson sedge]

and then solve for f tangential

[bold pond]

so

[past wave]

why am i not able to react to that message?

[crimson sedge]

Illuminati 🔺 👁️

[past wave]

hmmmmm

[bold pond]

60.5sin(66.3)

= Ft

[crimson sedge]

yup yup

[bold pond]

ok let me see

oh true

cause that is preventing it from moving

lol

wow that is really simple

[crimson sedge]

A great tip I could share for these kinds of problems (cause somtimes they can become really flipping complex) is that you can draw these forces as arrows on a cartesian plane

[bold pond]

can't belive i overlooked it

[crimson sedge]

Just put the point in the middle and then arrow surrounding it representing the forces

and if forces in a certain direction are equal

equate teh componenents of the forces in those directions

Like something like this this

[bold pond]

i got around 55.40N

[crimson sedge]

Cool

is that the correct answer?

[bold pond]

now its asking this

there is a similar question my profesor did in a video but the mass was a wedge is this ok to draw like this or do we need OUR mass to be a cube or rectangle/

[bold pond]

i have to submit it when i complete everything i will find out

[crimson sedge]

I think (not completely sure) either way works

Cause the end of the day we are considering it as a point mas

[bold pond]

yea

[crimson sedge]

Ok so horizontally right?

[bold pond]

yea

[crimson sedge]

What does that mean

Let’s think of it mathematically

[bold pond]

i think we need componenets for it

[crimson sedge]

Horizontally means in line with the x axis

[crimson sedge]

Yup

[past wave]

yeah

[crimson sedge]

Decompose always to components

Put in a Cartesian plane

And solve

Tangentially means same gradient as the derivative at the point

Horizontally is gradient = 0

[bold pond]

so cos(0)?

[crimson sedge]

Not exactly

[bold pond]

which direction is the tangentially force from the first question being applied?

[crimson sedge]

⬆️ it’s labelled as F in this diagram

[bold pond]

isnt that friction?

ohh nm

ok

it is just basically in the same angle as the ramp

[crimson sedge]

There is no friction in this question. The ramp is frictionless

Even if there was, F_tangential would still point the same way

[crimson sedge]

*same slope as the ramp to be exact

[bold pond]

i need to take a 10-20 min break

[crimson sedge]

Alright

Mate just put these forces on a Cartesian plane

Really it will make it much easier

[cedar kiln]

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@bold pond Has your question been resolved?

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[hard valley]

oh hello

[hard valley]

why a+b is not a+b a a+2ab+b

[warped hornet]

uh?

can you send a pic of your ques as its unclear

[arctic marsh]

guys I need help with like whats the root square of x

[warped hornet]

!occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[arctic marsh]

Oh ok

CEEEE YAA

[warped hornet]

byee

i will remember and wait for you for eternity

[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.

[past wave]

btw feels like (a+b)^2

but my guy forgot square everywhere

[hard valley]

im not your guy

i have a boyfriend!!

[past wave]

i meant my guy as a friend lmao

[crimson sedge]

next level delusion

[cedar kiln]

@hard valley Has your question been resolved?

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@hard valley Has your question been resolved?

[hard valley]

Not yet

[cedar kiln]

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[astral mulch]

I got one but the graph doesn’t approach 1

it’s significantly lower than one

[fallen moat]

,rccw

[astral mulch]

.close

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[cedar kiln]

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[hoary plume]

Why is this not a bernoulli in y

[gritty viper]

How could it be Bernoulli in y without being Bernoulli in x

It's perfectly symmetrical between x and y

[cedar kiln]

@hoary plume Has your question been resolved?

[hoary plume]

oh wait either in that case

<@&286206848099549185>

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[civic elk]

Is it fair to assume what i circled in red?

[ancient shadow]

yes if they're forces

wait no

not the y1= = 40

but the x1 = x2 should be fine

[civic elk]

i was thinking y1 would prob be y1=40+y1

but wasnt sure

[ancient shadow]

yh by N2L

if u take it vertically

Y1 = Y2 + 40

oh and if the system is equilibrium this holds otherwise

it doesn't

[civic elk]

yeah otherwise its a bit harder

[ancient shadow]

"bit"

[civic elk]

yep just a bit

thanks

.close

[cedar kiln]

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[civic elk]

.reopen

[cedar kiln]

✅

[civic elk]

so this makes sense, but im now confused on where to even start the problem?

[cedar kiln]

@civic elk Has your question been resolved?

[cedar kiln]

@civic elk Has your question been resolved?

[ancient shadow]

Wait

What's the problem

Work out the forces?

Oh I see the problem

And where u might be confused

Can u think of a relationship between X1 and Y1 and similarly with X2 and Y2

Because if you can you might be able to eliminate two variables and then get a set of simultaneous equations

[ancient shadow]

Maybe redraw this with T and all

[civic elk]

Wouldn’t it be the Pythagoras theorem?

[ancient shadow]

I was gonna discount that but ig that does work

But the thing is u don't know the hypotenuse value

So you'd struggle there

[civic elk]

What were u thinking?

[ancient shadow]

And just add another variable

[ancient shadow]

U have angles and right angled triangles

[civic elk]

Yeah

[ancient shadow]

Trig

Resolve the forces w trig

[civic elk]

I tried it but came empty handed

I’m at work rn, but when I get home ima try trig again see if I can get it

[ancient shadow]

👍

[civic elk]

Ima close this for now tho

.close

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[plucky mirage]

Can someone double check my work?

[plucky mirage]

Im not too sure if tan is the right to use for the angle either

[open owl]

Off topic but your handwriting is very clean.

[plucky mirage]

Hahahaha thank you

I was actually wondering if someone would comment on it bc it happens sometimes at school too

[cedar kiln]

@plucky mirage Has your question been resolved?

[plucky mirage]

<@&286206848099549185>

[plush wind]

@plucky mirage Hello

[plucky mirage]

I just wanted someone to double check my work

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[plush coyote]

having trouble with this problem im not sure what to do

[slate lintel]

well i guess the last one might want an approximation but i can't tell what's wrong with the first

[plush coyote]

nah even that doesnt work

[cedar kiln]

@plush coyote Has your question been resolved?

[plush coyote]

<@&286206848099549185>

[plush wind]

the result is that you express x and y by r.

do you understand>

?

[plush coyote]

but in this case what is r

[plush wind]

r is random.

r > 0

r => 0-

Hi

[cedar kiln]

@plush coyote Has your question been resolved?

[plush coyote]

ok but then how am i supposed to find the sin and tan data

[plush wind]

Hmmm, Look the picture.

There is the values for you.

sin a = -0.5, cos a = 0.87 = 3^0.5 / 2, tan a = 1 / 3^0.5

So a = 210deg

do you understand

?

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[manic acorn]

Do i directly use these results?

Or do i have to calculate seperately for the Z(1/n+1) because the result is valid for n=0, but in the question n=0 wont be valid?

Something like this?

[vernal shell]

is that a 2 or a z

lol

I imagine 2 right

no it's a z

ok

[manic acorn]

Its a z

[cedar kiln]

@manic acorn Has your question been resolved?

[cedar kiln]

@manic acorn Has your question been resolved?

[cedar kiln]

@manic acorn Has your question been resolved?

[cedar kiln]

@manic acorn Has your question been resolved?

[cedar kiln]

@manic acorn Has your question been resolved?

[nimble mountain]

is this a call for help

whats the original question

[twilit summit]

lol

[nimble mountain]

not sure what is happening for the rest but yea u should handle n=0 separately

[twilit summit]

I don't even know what the context is. Is this an abstract algebra question

Looks like number theory. but I don't know

[upper laurel]

no
the same problem happened for Z{1/n}, so since you ignored the problem for Z{1/n}, you also ignore the problem here

[upper laurel]

this works

[cedar kiln]

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[shrewd niche]

i am genuinely so lost with part b and c. can someone help?

[dire geode]

Solve for y means y= f(x)

Find f(x)

[shrewd niche]

so i would just get y by itself?

[modest storm]

ya

[shrewd niche]

i did that and it said it was wrong

is it not 5 radical x(7-4x^7) ?

[modest storm]

reread the question

[dire geode]

Probably better if you don't factor

[shrewd niche]

am i getting further away from doing this the right way?

no wait i got it

i'm supposed to get y by itself and the differentiate what's left which is f(x), right?

but what the heck does part c mean?

[dire geode]

Plug this y into a

[shrewd niche]

what

[dire geode]

Should be the same as part b

[dire geode]

Did you do this

[shrewd niche]

yes

[dire geode]

This is f(x), right?

[shrewd niche]

yea, but i did what you said and un-factored it to be 5 radical 7x-4x^8

[dire geode]

Good

Do you see a y in your answer to a

[shrewd niche]

yes

put my f(x) into where the y is?

oh my god

is it supposed to be this long??

[shrewd niche]

thanks for ur help🫡

.close

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[desert badge]

Yo can l get some help?

[desert badge]

This problem here

[pastel vault]

hint: the tangent at the point where the quarter-circle and the semicircle meet

is perpendicular to both the radius of the quarter-circle, and the semicircle

same thing for the quarter-circle and the circle actually

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[pastel vault]

use partial fractions

[crimson sedge]

pog

yes i am stupid

thanks

lemme try

i got the answe 😄

no tby partial frac tho

[pastel vault]

oh nice no worries then

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[chrome flax]

Can anyone help with programming in R for probability and stats

[chrome flax]

Please

[cedar kiln]

@chrome flax Has your question been resolved?

[cedar kiln]

@chrome flax Has your question been resolved?

[upper ruin]

Ask in #computing-software

[cedar kiln]

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[lapis valve]

Does anyone know where I went wrong? Sorry for the atrocious handwriting

I know I can use the reduction formula but wanted to learn to do without in case I forget the formula

[mental trail]

I think use of parentheses is important

the 27 was distributed but not the minus

[lapis valve]

I am always forgetting to distribute the sign

Thank you Rafilou!!

.close

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[mental trail]

I get this

from WA

changing sec^2 into tan^2-1 I think we get what you got

[mental trail]

accounting for sign

[lapis valve]

Yes I just switched the - to plus and it looks good I think

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[split belfry]

i just got introduced to limits and i have some problems solving this
prove that lim((2n^2+3n)/(n^2+4n+1)=2 using epsilon
can anyone here help me out?

[mighty shuttle]

$\varepsilon-\delta$?

[wraith dagger]

Veni, vidi, perii is not f(wai)

[split belfry]

not sure

[undone epoch]

Limit to what?

[split belfry]

b)

i have to prove that the limit is 2

[undone epoch]

I don’t think you need epsilon-delta. You could just use the variable with the biggest exponent.

And divide it with all terms.

[split belfry]

i tried that

[undone epoch]

Steps?

[split belfry]

wait a second

[undone epoch]

Might be I’m who’s wrong, if it’s epsilon-delta proof you’re doing.

[split belfry]

basically i have to solve for n

can you help me with that

[undone epoch]

Yep, I’m who’s wrong. You should wait for other person.

[delicate roost]

that works for evaluating the linit iges

they wanna prove it

[split belfry]

can you help me?

[delicate roost]

idts, you can do @helpers in like 2 mins

[split belfry]

ok

<@&286206848099549185>

T_T

tommorow i will go to my maths teacher and he il be like: this is so easy come to the whiteboard and he il magically solve it

i was able to prove some easier ones

ok i give up, thanks anyway

i will remain on the server cause it s cool here

.close

[cedar kiln]

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[bronze pivot]

what does this mean

[hot rampart]

x is, by definition, a

[bronze pivot]

oh ok lol

.close

[cedar kiln]

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[cedar kiln]

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[gilded mesa]

How this limit applies that sequence is not cauchy

[dire geode]

what's the sequence

b_N?

[cedar kiln]

@gilded mesa Has your question been resolved?

[gilded mesa]

It is sequence of partial sums of series

[dire geode]

the sequence is either b_N, a_k, or r_k. which is it

[gilded mesa]

Here is rn

[dire geode]

the answer is b_N

[gilded mesa]

I am saying why b_N is not cauchy

Answer not b_N, we have to find that series is convergent or divergent

[dire geode]

b_N diverges because the tail of the sum is greater than 1

[dire geode]

answer to my own question

[gilded mesa]

What is tail mean

[dire geode]

$\sum_{k=1}^\y \frac{a_k}{r_k} = b_N+$ tail

[wraith dagger]

riemann

[gilded mesa]

Ok I get this now

[dire geode]

tail $=\sum_{k=N+1}^\y \frac{a_k}{r_k}$

[gilded mesa]

But how tail>1 do with cauchy

[wraith dagger]

riemann

[gilded mesa]

I understanding terminology but not getting idea why this limit has to do with cauchy is it going to infinity or what

[dire geode]

do this a couple ways. either you prove cauchy sequences are convergent sequences and tails of convergent sequences go to zero. or you can prove a sequence is cauchy implies its tail converges to 0

[gilded mesa]

So you saying for each N we get a tail and that sequence of tails converges to 0 than it is cauchy

[dire geode]

absolutely not

[dire geode]

the sequence is b_N

write the definition of cauchy for b_N and what it means

[gilded mesa]

What you mean here that tail of convergent sequence goes to zero
We talking about sequence right ,so what is tail ?

[gilded mesa]

Here if N=1
Tail convergence to zero so series on right converge to b1

And we are getting different tails for each N what is it mean please help

[dire geode]

this is the tail when N = m-1

[cedar kiln]

@gilded mesa Has your question been resolved?

[cedar kiln]

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[rocky osprey]

where is the angle from stage 2 (II) located? the one with 28,03 degrees

is that the other sharp angle?

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@rocky osprey Has your question been resolved?

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@rocky osprey Has your question been resolved?

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[odd sierra]

i did this yesterday but smhow forgot

Like

i got to 12.6 seconds

for 1.90 m/s and 150 distance

using first equation

but idk how to get the total here

[dusk goblet]

total what

[odd sierra]

total time for travel

like idk what to do

with the 2.95

[dusk goblet]

well if you’re using equation 1 then you’ll need the displacement yes

[odd sierra]

Ok

[dusk goblet]

can you tell me what the initial velocity would be

from the point where it starts to decelerate

[odd sierra]

for 2.95?

[dusk goblet]

to when it comes to rest

[dusk goblet]

that’s acceleration

[odd sierra]

oh

[dusk goblet]

and it would be -2.95

[odd sierra]

yes

[dusk goblet]

you have to use the first numbers to find vo

[odd sierra]

oh

so i found t

12.6

now i need to use them to find

Vo

[dusk goblet]

don’t tell me these numbers

[odd sierra]

then get total time

ok

[dusk goblet]

give me equations

not numbers

[odd sierra]

ok i work on it

so

the 2nd equation

from my pic

will help find the v0?

first one helped w/ the time

[dusk goblet]

let’s look at the first scenario

before it starts to slow down

in that scenario vo = 0

a = 1.9

[odd sierra]

yep

[dusk goblet]

x = 150

so we can use v^2 = vo^2 + 2ax

because what we want is the final velocity

[odd sierra]

oh

[dusk goblet]

which will in fact be the initial velocity for the second scenario where it’s decelerating

v^2 = 2(1.9)(150)

,calc sqrt(2(1.9)(150))

[wraith dagger]

Result:

23.874672772627

[dusk goblet]

that’s our new vo for the second scenario

now we can use x = vot + 1/2 at^2

or we still the displacement actually

we only have a and v and vo

so again let’s use v^2 = vo^2 + 2ax

[odd sierra]

ohok

[dusk goblet]

v = 0

because it comes to rest

[dusk goblet]

vo =^

a = -2.95

and we’re solving for x

thus x = (2)(1.9)(150)/((2)(2.95))