help-49

[winged wedge]

wouldnt a still be discontinous

cuz ur gonna have a jump discontinuity

[cedar vortex]

no

what's the limit as x -> -1 of (x^2-1)/(x+1)

[winged wedge]

-2?

[cedar vortex]

ye

and we say f(-1) = -2

for a

[worthy wing]

What conditions must hold for a function to be continuous? @winged wedge

At a point

[winged wedge]

it existing?

[worthy wing]

That is one

So if x=d

First condition is f(d) exist

Second condition is the limit of f(x) as this approaches to d exists

And the third condition is that f(d) = lim x->d of f(x)

Do these conditions hold for yoir exercise at x=-1

[winged wedge]

ok

yr

ye

[worthy wing]

You can also see visually

,w plot (x^2-1)/(x+1)

[grand pond]

Results provided by WolframAlpha

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

Requested by _arkimond_

[winged wedge]

so

is there no hole?

[worthy wing]

As you can see, as x approaches to -1, f(x) approaches to -2

No because at f(-1)

The function is defined to be -2

[winged wedge]

ohh

i get it now

i thoight there was a jump discontinuity

causing it to be discontinuous

[worthy wing]

If that were true

Then the third condition would not have hold

Cause the limit could not be equal to f(d)

[midnight plank]

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[midnight plank]

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[ancient dragon]

Looking for help with a math question, not even really sure where to start. My teacher talked about how to find domain, but none of here examples were anywhere near as complicated as this one

[worthy wing]

What difficulties do you see?

What happens when t=2?

What happens when t=-1/2?

@ancient dragon

[neat canyon]

hi Samuel

[ancient dragon]

I figured it out

sorry for no reply

was deep in thought

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[worthy kestrel]

a sylow p group of say 2^2

[worthy kestrel]

does it mean we have a group that is Z4 or does it just mean we have a group of order 2^2

but we are not sure of the structure

i believe its the later. just after i wrote out this example..

[hidden hedge]

do anyone know why no one is coming to help in my channel its been like 20mins

[uncut cloud]

A (p)-group is just a group of order (p^k) for some (k). A Sylow (p)-\textbf{sub}group is a maximal (p)-subgroup

[grand pond]

Invariance

[uncut cloud]

there's no such thing as a "Sylow 2-group" on its own

it's Sylow because it's a subgroup

and a Sylow 2-group of order 4 can be any group of order 4

[worthy kestrel]

. close

ty

.close

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[naive flint]

hi

is my work right so far

i boxed where i left off

is cm^3 to m^3 valid thing to do

[sharp coral]

you can convert between cm^3 and m^3

[thin crow]

hi, i am brazilian (so sorry if I write something wrong because I'm not fluent in English) and i don`t no if this server help with problemes like this "A car manufacturer announced that it offers its customers more than 1,000 different car configurations, varying the model, engine, options and color of the vehicle. Currently, it offers 7 car models with 2 types of engines: 1.0 and 1.6. Regarding options, there are 3 possible choices: multimedia center, alloy wheels and leather seats, with the customer being able to choose to include one, two, three or none of the available options.

To be faithful to the disclosure made, the minimum number of colors that the automaker must make available to its customers is
" i don`t understood anything please help me.

[sharp coral]

!occupied

[midnight plank]

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).

[thin crow]

okk

[naive flint]

oh i am so sry

Oh i thought i occupied

is it good so far?

[midnight plank]

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[mortal blaze]

pls help I'm not entirely sure what to do

[opaque fjord]

uh

im jsut gonna give you the conversions

1N = 10^5 dynes

and 1m = 100cm

so just plug that in

[mortal blaze]

ohhh I see thanks

[midnight plank]

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[dusty stream]

I think I just proved the opposite of what the question asks 😭

[sharp coral]

if x < 0 then 0 < e^x < 1

[dusty stream]

but then if y=-1/1-e^x, and if 0<e^x<1, then y would be a negative number

[sharp coral]

the last line is wrong, y = -1/(e^x-1) does not simplify to -1/(1-e^x)

[dusty stream]

oh wait why 😭

[sharp coral]

you multiplied the denominator but not the numerator by -1

[dusty stream]

OH

wait omg thank you

🙏

I think I got it then

thank you so much I didn't catch that 😭

.close

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[last slate]

no 😩 just plug in this y now

and verify that it works

[dusty stream]

girl I'm so sorry I was confused but I figured it out now 😭

[last slate]

calling me girl and all

i'm done

[midnight plank]

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[vast bone]

wait i need help on this again bc i skipped doing it earlier since it would take too long

[vast bone]

okay so i know the derivative when x is positive is 1

for the numerator

and then the bottom its just 1/2(2-x^2)^-0.5(-2x)

Then use quotient rule

am i doing anything wrong

i did it like twice and i got diff answers than the answer key 😭

126

[dawn dagger]

oh that one again

[vast bone]

LMFAO

you're always active

ye but i need help 🤲

[dawn dagger]

So you wanna apply quotient

We only consider the positive derivative

[vast bone]

wait i got the same thing except multiply -2x

because isnt the derivative -x^2

[grand pond]

bacc

[dawn dagger]

yep

good catch

so simplifying it a bit

[grand pond]

bacc

[vast bone]

wait how idd it become x^2

[dawn dagger]

x*x

[vast bone]

ohhhh

ic

[dawn dagger]

but you are evaluating f'(1) right

[vast bone]

yeah

[dawn dagger]

hmm

[vast bone]

can i just plug it in at this point

or do i have to simplify further

[grand pond]

bacc

[dawn dagger]

no we can do this

1+1

divided by 1

so 2

funny thing is

even without the x^2 but only x you should have gotten the same answer

[vast bone]

answer key says the function is y=2x-1

[dawn dagger]

function?

[vast bone]

yeah

[dawn dagger]

what is the actual task

I thought to find f'(1)

[vast bone]

find equation of the tangent line

[dawn dagger]

oh

then you are half way there

we just got the slope

2

y = 2x + c

[vast bone]

OHHH

[dawn dagger]

Now it goes through the point (1,1)

[vast bone]

bro im actually slow

[dawn dagger]

nah

i like your pfp

[vast bone]

i thought it was the equation 😭

[dawn dagger]

it's always so happy

[vast bone]

LMAO

im opposite

esp when doing math 💀

[dawn dagger]

😂

[vast bone]

wait so you can just use point slope right

to find equation

[dawn dagger]

ye

essentially the same

i sometimes avoid cause people might feel overwhlemed

[grand pond]

bacc

[dawn dagger]

tangent line at a

It's actually straightfoward but some people get scared

[vast bone]

okay i see

yay thank you

[novel lion]

nice

[midnight plank]

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[lofty cedar]

Did i do 1a and 1b correctly?

[midnight plank]

@lofty cedar Has your question been resolved?

[midnight plank]

@lofty cedar Has your question been resolved?

[lofty cedar]

<@&286206848099549185>

[urban cairn]

yeah?

[lofty cedar]

could you check over my work

<@&286206848099549185>

[urban cairn]

bruh shut up

[flat veldt]

<@&268886789983436800> troll

[lofty cedar]

bro got the wrong role

[urban cairn]

mb

im doing bio rn

[misty gorge]

please don’t post off-topic messages in someone else’s help channel.

[urban cairn]

🤓☝️

alr

i won’t

see

bio

[fallow scarab]

<@&268886789983436800> spammer

[midnight plank]

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[last slate]

wow

using the ln or the root as dv it still is a huge expression

is there any method easier to do it?

[feral sedge]

after the u-substitution ax + b, this becomes an integration by parts problem

[last slate]

well, i thought i could use 1/(ax+b)^1/2

[midnight plank]

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[fleet flume]

Whats the probability of neither a or b when: P(a) = 0.01, P(b) = 0.5?

[late roost]

are they independent?

[fleet flume]

Should I use: 1 - p(a and b)

[fleet flume]

yes

inconditional probability

[late roost]

not really, that would be the probability of not a or not b
you seem to want not a AND not b
could just be miscommunication though

[fleet flume]

can I do something like: 1 - P(a or b)

Where the or is NOT exclusive

[late roost]

yep

also, you really don't need to specify that the or isn't exclusive. That's pretty much the default assumption when it comes to or

[fleet flume]

thanks

.close

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[dense bay]

quik notation q: if I have f(t) = M * (dv/dt) - ie newtons second law - and I wanted to represent integrating both sides on interval (-inf, t] for some t>0, should I swap the variables inside to something else (e.g. tau)?

[dense bay]

followup, after I integrate say $ \int^{t}_{-\infty} \frac{dv}{d\tau} d\tau$, should I write the result as $v(\tau)$?

[grand pond]

jack

[sharp coral]

no, the tau gets integrated out and the result will be v(t) because t is in the bounds

[dense bay]

ah i see, forgor that tau gets replaced by the real values for a sec ty!

.close

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[amber island]

can I just write 4x=-5

do I have to write the fraction?

[feral sedge]

no, presumably the problem asks you to solve for x and “4x = -5” does not answer that

[amber island]

Makes sense
Thanks!

.close

[midnight plank]

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[amber island]

.reopen

[midnight plank]

✅

[amber island]

wait

@feral sedge

question

how do I turn 4x=-5 into a fraction

[late roost]

divide both sides by 4

@amber island

[amber island]

4 cancels out

so -5?

what

[late roost]

what

[visual tiger]

4 cancels out on the left

and on the right, if you divide by 4 as well?

[amber island]

-5/4

-1.25

[visual tiger]

-1.25

[amber island]

WAIT

1.25 as a fraction

opps

wait

[amber island]

-1.25 as a fraction is -1 1/4

[visual tiger]

"-1 1/4" is a bit misleading

it deserves at least some parentheses

"-(1 1/4)" would cut it

[amber island]

😀

it still does not solve

[visual tiger]

wdym?

[amber island]

x = -5/4

[visual tiger]

we got x = -5/4

so we found x

[amber island]

im getting

this

[amber island]

not this

[visual tiger]

1 1/4 and 5/4 are the same thing

1 1/4 = 4/4 + 1/4 = 5/4

-(1 1/4) = -5/4

[amber island]

if i put 1 1/4 on a test that would be correct?

[visual tiger]

with parentheses and a minus sign it COULD work

[visual tiger]

but definitely pick this way of writing fractions

[visual tiger]

over this one

[amber island]

im still so confused

how do I get this way

[visual tiger]

how did you get 1 1/4 in the first place?

[amber island]

from the decimal

[visual tiger]

bruh

you're not supposed to compute through decimal

4x = -5

[amber island]

well

[visual tiger]

divide by 4

[amber island]

this

yeah

[cedar canyon]

hey, whats hey

[visual tiger]

$\frac{4x}{4} = \frac{-5}{4}$

[grand pond]

rafilou2003

[visual tiger]

as you said

the 4s on the left cancel

[amber island]

WAIT

SO

[visual tiger]

so $x = \frac{-5}{4}$

[grand pond]

rafilou2003

[amber island]

BECAUSE RHE 4 CANCELS

[cedar canyon]

answer is -5/4

[amber island]

we have 5 left

so

-5/4

wooo

praying i get into honors geometry

gonna be easy

thanks guys

[midnight plank]

@amber island Has your question been resolved?

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[fading shadow]

Sorry for my bad handwriting

Any help at all would be greatly appreciated

[last slate]

by the segment addition postulate, GE + EO = GO, and EO + OM = EM, by substitution, GO = EM

[fading shadow]

Thanks

@last slate could you possibly help guide me on another one

[last slate]

ok

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[last slate]

im sorry

[fading shadow]

Does this seem right

[last slate]

what does the 2nd line say

[fading shadow]

angle 2 and angle 3 form a vertical angle

[last slate]

do you mean angle 3 and angle 4

[fading shadow]

Yeah sorry 😅

[last slate]

idk

what i would do would be to use the correspondign angles converse to prove that the two lines are parallel

then by the correspodning angles theorem angle 2 is congreunt to angle 4

[fading shadow]

Ok

[fading shadow]

One last one sorry

[last slate]

i cant see it

[midnight plank]

@fading shadow Has your question been resolved?

[midnight plank]

@fading shadow Has your question been resolved?

[midnight plank]

@fading shadow Has your question been resolved?

[midnight plank]

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[cold plank]

How do I start number 20

[gaunt nimbus]

conjugate

[cold plank]

Oh okay

I forgot sorry

Can someone check if I did this right

oh wait nvm

solved

.close

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[pearl hull]

Wat have you tried?

[rugged bay]

Ok so what do u know about an open dot on a number line like that

[remote owl]

it dosent include that number

[rugged bay]

Good

So which of the signs will u use

What can u cancel out

[remote owl]

equal to ones

[rugged bay]

Good

So it’ll be just the normal looking signs

[remote owl]

Yes

I dont understand how to make it the inequality

like what number were

[rugged bay]

Ok so if u were to turn this into an equality what do you think it’ll look like give it an attempt

[remote owl]

5 > -5

?

x > -5?

[rugged bay]

No so u would have to include an absolute value sign

On the x

It would be

5 < |x|

[remote owl]

ok thats right

thanks

can you help me with another its about graphing @rugged bay

[rugged bay]

Sure

[rugged bay]

Ok so

What do u know about the signs which have a dash under them firstly

[remote owl]

include the number

solid dot

[rugged bay]

Ok good

Now

When ur solving an absolute value

There’s going to be 2 answers for x

[remote owl]

oh yes case 1 case 2

[rugged bay]

So first answer u can solve for just by solving normally

So pretend the absolute value wasn’t there and solve through

[remote owl]

-5 and 5?

[rugged bay]

No no

[rugged bay]

Do this

Just solve normally one time

U should get just one answer

And say whole thing like t> 29 or whatever

What did u get

[remote owl]

-5?

-9?

[rugged bay]

Ok good good but

What are the signs going to be

[rugged bay]

Write it fully out

Like this

[remote owl]

-5 > -9?

sorry im kinda stupid

[rugged bay]

No ur good but u need to put those two separately and u forgot the x

So lemme show u the work

[remote owl]

ok

thanks

[rugged bay]

,r

,rcw

[rugged bay]

Ok u see what I did

In the first way I kept everything as positive

And solved through

[remote owl]

I understood the way 1 at the beginning but way 2 stumps me

I get what you did though

[rugged bay]

Yeah so that’s the case because absolute values have 2 answers that will work

Yeah and u have to make it negative

And do u see what u did with the sign

The arrow point

I flipped it

Do you know why

[remote owl]

I remember the teacher talking about why but I forgot

[rugged bay]

It’s because when u multiply it divide by a negative across an inequality

The sign switches

Doesn’t matter if it’s an equal to one or not

[remote owl]

oh ua

[rugged bay]

But it’s only switching if u MULTIPLY or DIVIDE

U can add and subtract however u want

[rugged bay]

Just switches if u do these by a NEGATIVE specifically

Hope that makes sense

[remote owl]

got it

so how would I graph it

I ploted -9 -5 closed circle

[rugged bay]

Ok so u have these two answers

[rugged bay]

Yes good

So if u were reading these out loud

[remote owl]

away from eachother?

[rugged bay]

How would u say them verbally

[rugged bay]

Yes but u have to know why

[remote owl]

the signs are different

[rugged bay]

The first one is said “x is GREATER than or equal to -5”
And second one is said “x is LESS than or equal to -9”

So it’ll be numbers -5 AND all numbers bigger than it

[remote owl]

ok

[rugged bay]

While second one is -9 and all numbers LESS than it

[remote owl]

ohhh

[rugged bay]

Another hack is to just go the way the tail is pointing of the inequality sign

If it’s pointing left go left

[remote owl]

ohhhh

that helpss

[rugged bay]

Yes

It only works when x is on the left tho

[remote owl]

@rugged bay thank you

.close

[midnight plank]

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[meager ore]

If (\sum a_n) is absolutely convergent and (\lim_{n \to \infty} b_n = 0), show that

[
c_n = \sum_{k=1}^{n} a_k b_{n-k} \to 0
]

[grand pond]

Halex

[meager ore]

as a_n is absolutely convergent, lim |a_n| = 0

[fallow scarab]

|b_n| >= 1 only finitely many n

[meager ore]

why is that?

[meager ore]

but would't that mean that |b_n| < 1?

[junior flower]

eventually yea

[meager ore]

🤔 how can that help me prove c_n converges to 0?

[junior flower]

sorry that wasn't right

(deleted)

say that for all i >= N, |b_i| < 1. say you choose a large n. the first N terms can be made arbitrarily small (by picking a large enough n) because lim a_n = 0. then the rest of the terms are bounded in abs value by sum |a_n|

now this isn't very helpful

but

if you pick a small epsilon and then say that for all i >= N, |b_i| < epsilon

then...?

[midnight plank]

@meager ore Has your question been resolved?

[junior flower]

also do you have a simpler proof ? that wasn't as helpful as i initially thought because we need to prove that c_n converges to 0, not just converges

[meager ore]

this problem looked easier from the very start haha

[fallow scarab]

c_n can be bounded by a constant + sum_n a_n

[fallow scarab]

This is part of that constant

[midnight plank]

@meager ore Has your question been resolved?

[junior flower]

hm this is incorrect but i'm not going to bother editing again

i think it was right at some point in the past

fix $\varepsilon_1>0$. since $\lim_{n\to\infty} b_n = 0$, there exists $N$ such that for all $i \geq N$, $|b_i| < \varepsilon_1$.
to write this in long form to make it easier, for every $n$ we have $$c_n = a_1 b_{n-1} + a_2 b_{n-2} + a_3 b_{n-3} + .. + a_{n-1} b_{1} + a_n b_{0}.$$
we can pick $n$ so large that
$$|a_1 b_{n-1}| + |a_2 b_{n-2}| + .. + |a_{n-N} b_{N}| < \varepsilon_1 \sum |a_n|.$$
now fix $\varepsilon_2 > 0$. since $\lim_{n\to\infty} a_n = 0$, we can also pick $n$ so large that
$$|a_{n-N+1} b_{N-1}| + ... + |a_n b_{0}| < \varepsilon_2 \sum_{n=0}^{N-1} |b_n|$$

[grand pond]

layla is not harper

[junior flower]

that is my edit

"also pick n" might sound a little sus because the N from earlier depends on n

but it's ok. the same N will work for a larger n

you wouldn't write the proof like this. but it's an idea

well it pretty much is the proof, other than some details not being proven

[midnight plank]

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[green grail]

How is it wrong

Is it capital i

Instead

[worthy kestrel]

chemistry :(

[fallen aurora]

you're given some directions under the FEEDBACK, check that thing

maybe you'll find the problem

[midnight plank]

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[brisk crag]

I need help so bad I got to -root 7, 4- root 7, -4 root 7, and 8 as the remainder but idk how to write the answer

[slim steeple]

What method did you use? Long division?
Also if you can show your work that would be better

[brisk crag]

ah sorry sending it now

but my handwriting might be a little messy

[slim steeple]

I will try to understand

[slim steeple]

What's your remainder?

Also what method is this ?

[brisk crag]

synthetic

8/x+root 7

[slim steeple]

That's your r(x)

[brisk crag]

oh waut

just the remainder is 8

[slim steeple]

(x²+(4-√7)....) is q(x)

[slim steeple]

Ohhh

[slim steeple]

What is this then

[brisk crag]

I thought you were referring to the right of the bubke

bubble

[slim steeple]

Well if your remainder is 8 then r(x) will be 8

[slim steeple]

This is q(x)

[brisk crag]

I never heard of r(x)

What’s that stand for

Remainder of x?

[slim steeple]

Remainder in term of x, but you don't have x term in remainder

So it will simply be 8

[slim steeple]

Suppose you had 2x+3/x
So it would have been x * 2 +3
3 would have been r(x) in this case

2 as q(x)

[brisk crag]

I see

Hold up lemme try again

Yeah that should be correct I think

[slim steeple]

[midnight plank]

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[graceful ferry]

,rotate

[grand pond]

Couldn't find an attached image in the last 10 messages.

Sep 24, 2024, 7:46 AM

[graceful ferry]

,rotate

[graceful ferry]

Help please

[midnight plank]

@graceful ferry Has your question been resolved?

[graceful ferry]

<@&286206848099549185>

[visual tiger]

you want to find another linear combination of those vectors that gives 0, other than the null 0+0+...+0 combination

so suppose you did have one

so $\lambda_1,...,\lambda_6$ being those coefficients, not all of them being 0

[grand pond]

rafilou2003

[visual tiger]

start by showing that it's impossible for $\lambda_1 = 0$

[grand pond]

rafilou2003

[visual tiger]

and then without loss of generality, you can assume $\lambda_1 = 1$

[grand pond]

rafilou2003

[visual tiger]

(just divide by lambda_1 if it's not the case)

then find the values of the other coefficients...

[graceful ferry]

Why ?

I also don't understand why we are multiplying the vectors by lambdas when i need to find the values for alpha

[visual tiger]

Do you know what it means for a group of vectors to be linearly dependent?

[graceful ferry]

Can you please tell me i'm not sure actually

Theres many different definitions just sayimg the same thing

[visual tiger]

so a group of vectors is linearly dependent if

there exists a linear combination of those vectors that gives the 0 vector

and not every coefficient in that linear combination is 0

for example, the group of vectors ${\vec 0,u,v}$ is linearly dependent because $1\cdot \vec 0 + 0\cdot u + 0\cdot v = \vec 0$

[grand pond]

rafilou2003

[graceful ferry]

Okay

[visual tiger]

so

[visual tiger]

IF this group were to be linearly dependent

[visual tiger]

there would exist such coefficients

such that lambda1 * (1+alphax) + lambda2 * (x+alphax^2) + ... = 0

now our objective is

supposing that they do exist

what would they be, and what value (or values) of alpha allows that?

[visual tiger]

so this is what I recommend first

[visual tiger]

and then assuming lambda1 = 1 will simplify your computations

[midnight plank]

@graceful ferry Has your question been resolved?

[midnight plank]

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[visual swan]

can somebody help me with the first question?

it is systems of equations algebra 2

[midnight plank]

@visual swan Has your question been resolved?

[visual swan]

<@&286206848099549185>

i need help with second one

[mighty plaza]

its ez

[visual swan]

I dont know how to do it

i got my two equations for the second one

8v+9b=228

4v+5b=124

[deft mortar]

so you did the first one but can't do the second?

what is the problem according to you?

[quasi dragon]

It's simultaneous equations right?

[visual swan]

I think so

[deft mortar]

yeah, he just needs to link stuff up and take out values, but let him figure it out himself

[visual swan]

so I am multiplying the bottom equation by negative two

and I now got -8v-10b=-248

8v+9b=228

[deft mortar]

hm

[visual swan]

and now I know I got b=20

[deft mortar]

if you are confident, then you can add those 2 to seprate v out

[mighty plaza]

maybe subtract both equation like 8v+9b-(4v+5b)=228-124

[visual swan]

is that right?

[deft mortar]

so where are you stuck then?

[visual swan]

I am not sure how to find v my teacher said there is a fast way but I forgot it.

[mighty plaza]

that will work also

[mighty plaza]

use elimination mehod

[last slate]

Substitution is my go to method for system of equations

Elimination might worj

Work*

[mighty plaza]

but it can get legthy

[last slate]

Ehhhhh

[mighty plaza]

try elimination first

[visual swan]

ok

I think I got it guys

thx

[mighty plaza]

np

thanks boh

cause he got the idea to sue the elimintaion method

use

[deft mortar]

well, if your problem was to find the right method, apply it and see if you can finish up

[visual swan]

yeah im putting it back into the equation right now

[quasi dragon]

8v + 9b = 228
4v + 5b = 124

4v = 124 - 5b (choose the simpler one first)
v = 31 - 5b/4 (remember this equation for later)

put v in the first equation

8(31 - 5b/4) + 9b = 228
243 - 10b + 9b = 228
243 - 228 = b
b = 15

now use the equation i mentioned before. you do not have to start from beginning for v.

v = 31 - 5b/4
v = 31 - 5*15/4

ok, i think i did a mistake. i am getting a decimal: 12.25

[midnight plank]

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[pine vine]

hello, i am trying to come up with a way to calculate pi and so far I've succeeded but i have a few questions

İ tried to Turn f(n-h) 's into just N's doing some kind of substitution but i am not sure if i can even do that according to limits

and another question is

whenever i try to put in sqrt(1-x²) in place

the answer is always wrong

At least according to the way i made c# calculate it

and another thing is

i tried to use this formula as a means of getting the length of a function over a range but Wolfram alpha doesn't evaluate this

[fallow scarab]

What even is this supposed to be

What's f in terms of h

[pine vine]

integrals of f(x) minus f(x)(1-h)

[fallow scarab]

Integral with respect to what variable

[pine vine]

f is meant to be any function, I'm trying to find the length of functions over a period

[pine vine]

oh

dx

[fallow scarab]

Then this is just zero for all continuous f(x) on [a, b]

[pine vine]

but it's supposed to be the area of f(x) minus almost itself but not

which is supposed to give a line

why doesn't it work?

[midnight plank]

@pine vine Has your question been resolved?

[midnight plank]

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[tight jasper]

how do i do this

[tribal temple]

,calc 3 - 13

[grand pond]

Result:

-10

[tribal temple]

[novel lion]

https://cdn.discordapp.com/emojis/888818793239945216.webp?size=80&quality=lossless

[tight jasper]

shut up

@everyone

[novel lion]

[tribal temple]

It happens to everyone sometimes

[celest steeple]

True

[tight jasper]

.close

[midnight plank]

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[feral sedge]

🫵

[midnight plank]

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[buoyant prism]

how do you find the conjugate of this denominator?

do i just inverse every sign?

[pliant spindle]

ig so

[last slate]

for me

[pliant spindle]

maybe i'm wrong

[last slate]

id factor first

[buoyant prism]

oh

okay ill try that first

i factored it and got (rt2 - 1)(rt3 + 1)

so the conjugate is

(1 - rt2) and (-1 - rt3) ?

what about the rt6

will that stay positive

or negative

[pliant spindle]

honestly i'm lost id better shut up 😭

[last slate]

$(\sqrt{6} + \sqrt{2} + 1) - \sqrt{3}$ Therefore conjugate is $(\sqrt{6} + \sqrt{2} + 1) + \sqrt{3}$

sorry u cant really factor

[grand pond]

icannotdoanymorecauchy

[buoyant prism]

wait how did you get that

[lethal path]

you sure it's $+1$ instead of $-1$?

cause $\sqrt{3} \sqrt{2} - \sqrt{3} + \sqrt{2} - 1 = (\sqrt{3} + 1)(\sqrt{2} - 1)$

[grand pond]

higher's secret brother

[buoyant prism]

yeah i made a mistake there

[last slate]

oh u're right

mb

[lethal path]

wow

[buoyant prism]

so how did you get this?

[last slate]

ignore that

[last slate]

this is easier to do

the conjugates are $(\sqrt{3} - 1)(\sqrt{2}+1)$

[grand pond]

icannotdoanymorecauchy

[last slate]

${(a-b)(a+b) = a^2 - b^2}$

[grand pond]

icannotdoanymorecauchy

[last slate]

im sure u know which brackets to match and what to o next

plz get a new channel

[fallow scarab]

also delete your messages

[midnight plank]

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[lethal hound]

Does anyone know what to study for a test to go to algebra 1? The test is mostly on pre-algebra. But I was doing math 7 last year. So I don't know any Big part in pre-algebra that I have to study well? The test ( kind of similar to final exam) is tomorrow. So I am studying right now. Any kind of help is appreciated.

[fallow scarab]

Nobody here would know your specific region and school district. Just go watch Khan academy algebra 1 lessons

[sinful trout]

the organic chemistry tutor is a great youtuber as well

[lethal hound]

????

I need help on like what to study.

Because I was in math 7 last year

[sinful trout]

do you not have a syllabus?

[lethal hound]

and now I am on math 8

[sinful trout]

my guy we won't know the specifics of your curriculum

[lethal hound]

And I want to go to algebra 1 and I will do a test tmr

[sinful trout]

you must have been provided with a syllabus

[gloomy mulch]

ask yo teacher bruh

[lethal hound]

Yo how the hell is my teacher suppose to know I asked her twice

[sinful trout]

she's supposed to know because she's your teacher?

[lethal hound]

She doesnt make the questions or do the test

she is just teaching math 8

[fallow scarab]

surprise, nobody here does either

go study yourself and come back when you work on a concrete math problem and you get stuck

[lethal hound]

k k

.close

[midnight plank]

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[novel gyro]

How do you do this question? It wants me to find T and V but I can’t get the answer I think I did something wrong while plugging in the values but I don’t know what

[midnight plank]

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[novel gyro]

.close

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[keen solstice]

How can I use the degree rules to evaluate the limit as this function approaches infinity?

I'm specifically referring to this rule:

[twin forge]

The numerator isn't visble

[grim vector]

Degree of e^x ?

[keen solstice]

the numerator is 2e^x

Specifically I've never been exposed to degrees that are other than the form ax^n + bx^n

[shell wigeon]

That rule doesn't match your question

You have e^x, not x^n

[grim vector]

You can factor and use propreties calculate the limit

[keen solstice]

So that rule only works with polynomials which have variables?

[grim vector]

Yes, with polynomials you can do this

But it doesn't say much about the limit value

Just that its converging or not

[keen solstice]

ahh i see

I'm assuming in this case I would have to now just factor out e^x from the top and bottom?

and then directly evaluate the limit as x approaches infinity?

[grim vector]

Indeed

[keen solstice]

gotcha!

thanks, that answers my question

[grim vector]

You're welcome

[keen solstice]

.close

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[keen solstice]

How do I use the degree rules for evaluating functions as x approaches infinity for this: t^(1/3)+12t-2t^2

[keen solstice]

I first rewrote the function in standard form: -2t^2+12t+t^(1/3) then I applied the rules and got infinity

[twin forge]

Theres no denominator?

[keen solstice]

But the answer is -infinity

No denominator

I just said the denominator was 1x^0

[twin forge]

Then its -ve infinity

[keen solstice]

how though?

I thought the degree of the numerator was greater than the degree of the denominator, thus causing it to be infinity?

[twin forge]

Yes and the degree of the denomnitor is 0

[keen solstice]

what's causing it to be -ve infinity?

[twin forge]

The largest power is on -2t^2

The largest power is negative

[keen solstice]

so in addition to that rule, if the coefficient of the largest degree is negative, the answer is negative?

[twin forge]

Yes

Thinking of it like this

(Infinity)^2 is greater than Infinity

So we if subtract (Infinity)^2 we'll get a negative answer

[keen solstice]

Ahh I see

That makes sense

Are there any other rules I should keep in mind?

[grim vector]

With polynomials, the limit value will be the limit of the greatest monomial (the terms with the biggest power)

Since its -2t^2

t^2 goes to +infty

Multiply by -2 and you get a negative infinity

[grim vector]

Think about derivatives definition if you deals with 0/0 form

With sqrt you can sometimes use conjugates

[keen solstice]

I haven't learned derivs yet lol

[grim vector]

Ah then you good to go with factorisation and basic rules for now

[keen solstice]

yep

by the way, the degree rules apply for both pos and neg infinity, correct?

[grim vector]

The degree rule say nothing about the sign of the infinity

So yeah

Just a method to know whether diverging or converging function

[keen solstice]

so if this was approaching -infinity instead of infinity, it would still be -infinity, correct?

[keen solstice]

^^^ this is just in response to what u said here

[grim vector]

Yes

(You could have do it by factor out the t^2 from the whole expression and use proprety)

It would have lead to same result

[grim vector]

Its a parabola in fact

So yeah it goes down both sides

[keen solstice]

got it, thanks a lot

i understand now

[midnight plank]

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[midnight plank]

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[lavish venture]

conjugate

[lapis fern]

yeah so 6-9i

i did it but I get the wrong answe

[lavish venture]

show your work

[lavish venture]

well i’m looking at your final answer and seeing something wrong immediately

[fathom tendon]

how the 9 in the top row turn into 6

of the denom

6(6-9i) + 6[?]i(6-9i)

[lapis fern]

what what

oh

[lavish venture]

did you just get green or did you have the not very people role?

[fathom tendon]

I think just now

[lavish venture]

welcome

🤝🏻

[lapis fern]

what do you mean by this

[lavish venture]

how is the denominator 90?

the denominator should be a difference of two squares: 36-(-81)

from the conjugate

[lapis fern]

how did you get 81

[lavish venture]

(9i)^2

for some reason you wrote 54

[lapis fern]

oh yeah I wrote 6i instead of 9i

-9i

oops

[lavish venture]

yep

[lapis fern]

my answer is still incorrect?

[lavish venture]

how’d you get 72i?

[lapis fern]

OH

should the final answer me like that or separate

[lavish venture]

do they want it in a+bi form

both would be correct

[lapis fern]

standard form so yeah

[lavish venture]

it’s standard to write it in a+bi though

[lapis fern]

so the one above ?

or

[midnight plank]

@lapis fern Has your question been resolved?

[midnight plank]

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[hushed glen]

i need desperate physics help

[hushed glen]

This spool is attached to hung from the wall

The only information thats given is the mass of the spool, and g. Using that, I am to calculate the normal force, and the tension in the cable. But really all I need help with is the tension.

[hollow dome]

Normal force exists because of spools horizontal part doesn't it?

[hushed glen]

yes

it'd be opposite of the x-component for tension

[hollow dome]

Yes

[hushed glen]

but the issue is calculating tension

[hollow dome]

You know the angle, can't you just create a 30-60-90 triangle

[hushed glen]

I thought about doing that, but I wouldn't be able to find any of the sides since the part of the cable connected to the spool is not directly above the middle of the spool.

[hollow dome]

That makes me uncomfortable too yeah

[hushed glen]

i just dont even know where to really begin

[sharp coral]

it doesn't really matter where the cable is attached for the purpose of calculating the force exerted by the cable.. all we need to know is that the tension in the cable is in the same direction as the cable itself

[hushed glen]

but how does that help determine the magnitude

[sharp coral]

if the magnitude is T, then one of the components should be T*sin(theta) and the other should be T*cos(theta) for the appropriate directions and an appropriately chosen theta

[hushed glen]

Understood, but figuring out the value of T is the problem at hand

[sharp coral]

well if you balance forces, T should be the only unknown in one of the directions

[hushed glen]

In the x direction theres the normal force and the x-component of T

in the y direction there is gravity, friction, and the y component of T

the only one I've been able to calculate is gravity by just doing mass(g)

[hollow dome]

I'd say friction is 0 if there is no limit how much spool can lift

[hushed glen]

That doesn't make sense either since one of the things the question is asking for is also the coefficient of friction

I simplified that by saying im just attempting to compute the normal force, because after that its easy

[sharp coral]

well the torques from the spool and friction would also have to cancel

[hushed glen]

But to go about calculating the torque, id also need to figure out where the spool is connected to the cable relative to A since its not directly above

[sharp coral]

it would have to be tangent to the circle / perpendicular to the position vector

[hushed glen]

position vector?

[hushed glen]

ahh

i think i see

so your making a relative coordinate system rotated 60 degrees from where it normally is

I'm lost on finding the torque due to gravity with that model tough

[sharp coral]

gravity has to come from the center of mass

any vector pointing directly to or from the center must exert 0 torque

[hushed glen]

then what torque is balancing out the torque from tension

[sharp coral]

well normal force is a similar story

so the only other force exerting torque is friction

[hushed glen]

but I'm not given friction

[sharp coral]

well it's also perpendicular

[hushed glen]

i dont follow

friction would be vertical, no?

[sharp coral]

friction points straight up, the line pointing from the center to the point of contact is horizontal

[hushed glen]

I dont see how I can find that torque without the magnitude of friction

[sharp coral]

you can find the torque exerted by both forces in terms of their magnitudes

[hushed glen]

I see that, but those very magnitudes is what im trying to find

[sharp coral]

3 equations & 3 unknowns at this point

[hushed glen]

I am so lost.

<@&286206848099549185> ?

[strange moss]

what’s the question asking

torque of gravity?

[hushed glen]

its asking for coefficient of friction, and tension

but really they go hand in hand

so just tension

[strange moss]

the torques have to be equal right

[hushed glen]

yes

[strange moss]

torque of gravity is 0

[hushed glen]

but how do i find the torque from the other forces?

[strange moss]

i know how to do it but i don’t know the math

if that makes sense

[hushed glen]

i know what you mean

thats where im at

[strange moss]

okay so the force of tension upward has to be equal to the force of gravity

[hushed glen]

but i dont know where to start for even finding tension

i'd have to do something like

cos(60)T to find the vertical component

but I dont have T

[strange moss]

it would equal my

mg

[hushed glen]

but what about friction?

cause it would equal mg - friction

[strange moss]

friction only creating torque right

[hushed glen]

i don't know

the question wants the coefficient of friction

[strange moss]

yea

tension creates normal force

against the wall

[hushed glen]

yes

but the only thing im given is the weight and that angle

and the measurements of the spool

[strange moss]

yeah so just assume the friction is only creating a torque

then you can find tension

[hushed glen]

i think i see

wait

but id need the magnitude of either of the 2 torques

to figure out tension

[strange moss]

oh i’m finding it without using torque

just force in the y direction

but honestly idk if that would work

[hushed glen]

ive tried that

it doesnt

aaa

its okay

i think

ive spent enough time on this problem for tghe day

maybe i should talk to my prof lol

[strange moss]

it’s a good problem

[hushed glen]

it is

but im done for now

.close

[midnight plank]

Channel closed

Closed by @hushed glen

Use .reopen if this was a mistake.

[strange moss]

i’ll lyk if i can figure it out

i’m not at a desk or anything en

rn

[midnight plank]

Available help channel!

Send your question here to claim the channel.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[grave whale]

guys help

[grave moth]

4x^2/y^8

[peak rune]

that light

asf

[grave whale]

i dont want the answer i wanna know how to do it

i have a test and i dont wanna fail

[lavish venture]

you need to review your algebra rules

[grave whale]

i know how to do i just dont know where to start

[peak rune]

cmon bruh

[grave whale]

i need help guys.

<@&286206848099549185>

[grave moth]

Cross out 3x^3from both numerator and denominator. Move y^-1and y^-5to the denominator, flip the sign in their exponent, giving:
4x^2/y^5y^2y^1 = 4x^2/y^(5+2+1) = 4x^2/y^8

[midnight plank]

@grave whale Has your question been resolved?