help-0

[coral cipher]

yeah

[subtle birch]

By definition, corresponding angles are the angles which are formed in matching corners or corresponding corners with the transversal when two parallel lines are intersected by a transversal

Clearly there are no parallel lines

and thus no corresponding angles

so, the question is wrong or the 'corresponding' has some different meaning here

[coral cipher]

Yep, the question is wrong here

ty

.close

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[vital brook]

can I get help with this please

[alpine sable]

You can see the teachers as 1 person and the students as 1 person

Now you just have 4 people of which 2 people can't be next to each other

So suppose you fix the place for the group of teachers.

Then how many places are there for the group of students?

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@vital brook Has your question been resolved?

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[cobalt solar]

Math problem help

[peak bough]

As much as it would be convenient, people here aren't mind readers. You need to post your question or people won't be able to help you

[abstract mantle]

This is a telescopic series apparently, I just dont know what to do

[lone heart]

@cobalt solar Has your question been resolved?

[gray isle]

consider k = k + 1 - 1

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[ivory olive]

Hello intelligent individuals!

[abstract mantle]

K = k ?

[rancid trail]

Bro the channel was closed and is being used by another person

[gray isle]

make a new channel if you still need assistance

[ivory olive]

So I believe I know how to start this one at least. But I am more concerned about the integral being huge

[gray isle]

try to know whether it'll actually be huge instead of worrying/speculating

[ivory olive]

Well, that's why I am working it out step by step. But I feel like after the 0 is plugged in from the integral it will not be so bad. Just the actual integral itself is scaring me a tad

[gray isle]

you didn't update the bounds

also don't be scared,

do you know what techniques would be useful to integrate something like this?

[ivory olive]

Besides substitution, I figured integration by parts. But I thought I would have to substitute first.

[gray isle]

sub wasn't necessary,
but yeh ibp would be useful here

[ivory olive]

Is substitution not necessary because $e^(-x/2)$ is one of the special functions?

[ocean seal]

madibe97

[ivory olive]

I also tried to use the bot & still trying to learn to use it to help with math talk... 😅

[gray isle]

sub isn't needed as the application of chain rule to integrate e^(-x/2) is simple enough without substitution

[ivory olive]

Reteaching myself the chain rule... that would be e being f(g(x)) and then -x/2 being my g(x)?

And then use the integration by parts to also be able to do the x^3 * e^ (-x/2)?

[lone heart]

@ivory olive Has your question been resolved?

[ivory olive]

Not yet, but I am working on it!

Still working... will post what I came up with so it can be checked by the helpers

[ivory olive]

<@&286206848099549185>
I am about to post what my game plan is and I just want to be sure I am on the right track with it

From here I need to take this derivative again after taking out the constants?

I am going to keep on with this mindset and come back if I need further assistance. Thanks!

.close

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[keen plinth]

you are looking for integer solutions

the expression you put for x should produce an integer for every value of k

the one you put does not

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[alpine sable]

if we can factorise (x^4 - y^4) to be (x^2 + y^2)(x^2 - y^2)

then x^4 and y^4 must be square numbers

so how do you prove x^4 is a square number by itself?

[wind cloak]

(x^2)^2

[alpine sable]

but square numbers are meant to be integer arent they?

[alpine sable]

or is that irrelevant

look at the closure property of multiplication of integers

so if x is an integer value, then x^2 will also be an integer

[sharp sail]

wdym by square numbers, perfect square of integers?

[alpine sable]

that is the definition of square numbers I think

[sharp sail]

if that's the case, then (x⁴-y⁴)=(x²+y²)(x²-y²) doesn't require x⁴ and y⁴ to be square numbers

Example (1.1⁴-7.9⁴) = (1.1²+7.9²)(1.1²-7.9²)
I don't think 1.1⁴ and 7.9⁴ are square numbers.

[alpine sable]

but square numbers are defined as the squares of integers

no?

[sharp sail]

yea but then why do x⁴ and y⁴ have to be square numbers? They don't

[alpine sable]

yeah, cos as Neon said

x^4 = (x^2)^2

let x^2 = y

so x^4 = y^2

[sharp sail]

x⁴ is not necessary a square number, say 1.1⁴

[alpine sable]

as i said, x can only be integer values

square numbers are defined as the square of integers

[sharp sail]

oh, then x⁴ is obviously a square number

[alpine sable]

yeah I get why you thought differently

should we close the channel?

[sharp sail]

I read the question, it says if (the identity in question) is true, then x⁴ and y⁴ has to be square numbers and he asked to prove why

[alpine sable]

he asked to prove why x^4 is a square number

[sharp sail]

he didn't mention x is integer, so I got wrong

[alpine sable]

for future reference, that's just implicit

[sharp sail]

Actually that has no relation to the identity mentioned in the question

[alpine sable]

yeah lmao

[sharp sail]

lol

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@alpine sable Has your question been resolved?

[lone heart]

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[alpine sable]

.reopen

[lone heart]

✅

[sharp sail]

what?

[tall talon]

What is your question? (Sadly I am not a mind reader)

[alpine sable]

ohhhh

[sharp sail]

Thankfully I am a mind reader

[tall talon]

Life is so unfair

[sharp sail]

if it's unfair to everyone it's fair to everyone

[alpine sable]

an expression in the form x^2 is not necessarily a square number

it's only a square number if x is an integer

[tall talon]

yes

[sharp sail]

yea great you understand it

[alpine sable]

thats why u can factorise (x^4 - y^4)

not because x^4 and y^4 are square numbers

but because its in the form (x^2 - y^2)

since x^4 = (x^2)^2

[sharp sail]

nice

you got the point

[alpine sable]

ye

@sharp sail@alpine sable@tall talonthank you for ur time :)

👋

.close

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[shut parcel]

My prof. gave us the answer key for an assignment. However, I’m not sure how he got those answers.

[shut parcel]

Is there a specific formula to fill the blanks?

[tardy stag]

it's a little confusing; the stuff in the right column is like "what percentage of accounts have this much or less in them" while the column on the left is "how many accounts have this much in them"

[shut parcel]

Ah, I see! How would I fill the blanks and conclude to those answers

[tardy stag]

you could use the intermediate step of computing local relative frequency i.e. "what percentage of accounts have this much in them"

[shut parcel]

Is this where I do 2^k greater than or equal to 200?

[tardy stag]

no, that's for like compound interest things

let's look at the first row

the first row is the simplest because "under $1000, or less" is just the same thing as "under $1000"

[shut parcel]

Yes

Under $1000

[tardy stag]

yeah so if we know that 10% of accounts have under $1000, how many accounts is that?

[shut parcel]

Ohhh

1000*0.10

Omg I was overthinking

[shut parcel]

1000*0.10 = 100

[tardy stag]

uhhh pause

10% of accounts are in this box, how many accounts is that?
how many accounts do we have total?

[shut parcel]

Wouldn’t you do 1000*0.10

Wait

100%-10% = 90%

[shut parcel]

Is that right😭

But then, 1000*0.90 = 900

[tardy stag]

if we multiply $1000 (the amount of money these accounts have) by 10%, we'll get $100, like 100 dollars
that's not really an answer that makes sense when the question is "how many accounts"

we're trying to answer the question of "how many accounts are in this category", and we know that 10% of them are
so how many accounts do we have in total? that's in the question text

[shut parcel]

I’m so confused

I’m sorry if I’m asking dumb questions😭

I get the 1000*0.10 = $100

So after we do 1000*0.10, how do we get 20

[shut parcel]

I don’t understand what you mean

[tardy stag]

this step doesn't make any sense

we have 200 accounts total, right?

so if 10% of them are in this box, how many accounts are in this box?

[shut parcel]

Yes, we have 200 accounts in total

OH

[tardy stag]

if i have 200 kittens and 10% of them are orange etc

[shut parcel]

200*0.10 = 20

[tardy stag]

yehhhh

[shut parcel]

OMG

That was dumb of me lolll

[tardy stag]

no hay problema, we all are silly sometimes

[shut parcel]

Thank you so muchhh

And for the second row, would it be the opposite?

[tardy stag]

for the second row it's the opposite yeah but you'll have to go through the intermediate step of "what percent of accounts are in this box"
because the right column is like "what percent of accounts are in this box or any previous box"

[shut parcel]

By opposite, do we divide

[tardy stag]

yeah

[shut parcel]

30/200, right

I got 0.15

[shut parcel]

I got 15% instead of 25%

[tardy stag]

yes, 15% is how many are in that box

[shut parcel]

15+10 = 25%

[tardy stag]

but the column on the right is asking how many are in that box or any previous box

yeah

[shut parcel]

So basically, we add from the previous

[tardy stag]

yeah

[shut parcel]

For the 5000 under 10000, I got a diff answer

I did 0.60*200

[tardy stag]

the 0.60 is the percentage in that box or the previous boxes so you'll have to subtract 25% first

[shut parcel]

Ooh!! 60% - 25% = 35%. So, 0.35*200

And for the over 10,000, what would I do if both are blank

[shut parcel]

Will the “over $” percentage be always 100%?

[tardy stag]

yeah

[tardy stag]

you'd have to fill it in from the rest of it - whatever's left over ig

[shut parcel]

So do I total it?

[tardy stag]

the stuff on the left should total to 200

and yes the "over $" will always be 100%

[shut parcel]

Thank you!!!

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[alpine sable]

I know that A. B =B.A

But confused about how this mean dv/dt . dr =dr/dt .dv

<@&286206848099549185>

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[remote tusk]

@cold spire

[remote tusk]

.

[alpine sable]

Scalar dot Product

[grand girder]

these are infinitesimal quantities so you can change the order or do whatever you want with each of the 4 operations not breaking the tie.

However, if it was something like $\overrightarrow{\mathbf{v}}/dt \cdot \overrightarrow{\mathbf{r}}$ and $\overrightarrow{\mathbf{r}}/dt \cdot \overrightarrow{\mathbf{v}}$, where$\overrightarrow{\mathbf{v}}$ and$\overrightarrow{\mathbf{r}}$ are vectors that vary with time, then the order could indeed matter

[ocean seal]

adzetto

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[alpine sable]

Thx

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[lone heart]

Please don't occupy multiple help channels.

[wary stream]

You already have #help-13 don't open multiple channels

.close

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[last rune]

how to factorise this

[winged shard]

You can use the fact that (a^5 - b^5)=(a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4). Then you agroup (a+b+c)^5 with -1a^5 and you agroup b^5 with -1c^5

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@last rune Has your question been resolved?

[last rune]

i didn't quite understand

elaborate? if you will

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@last rune Has your question been resolved?

[winged shard]

Sorry for the long timing
Can you be specific with what part you didnt undertstand pls?

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[vapid steppe]

is it A or E

[winged shard]

If that graph is the graph of f', what does it mean if f' is 0?

[vapid steppe]

candidate for extrema

[winged shard]

Okay, so how many candidates for extremes do we have?

[vapid steppe]

do the endpoints count

[winged shard]

Dont think so

[vapid steppe]

3

but one of them doesn’t count

so basically 2 extrema

if the interval was closed it would be E?

[winged shard]

So if we have 2 candidates for extremes it is A or B right?

[vapid steppe]

it’s A

but if the interval was closed would it be E?

[winged shard]

Yes
But did you understand?

[oblique trail]

Here!

[vapid steppe]

bro

😂

[winged shard]

xddddddddd

[winged shard]

Why are you saying that?

[vapid steppe]

yes but i have a question still

[vapid steppe]

because since f’ is positive at x=-6 that means the curve increases from x=-6

implying there is a relative min

[winged shard]

Yes, that's the right thought

[vapid steppe]

same thought for x=8 but backwards

so it would be E if it was closed interval?

[winged shard]

Yes it would

[vapid steppe]

#❓how-to-get-help

[winged shard]

apply distributive
Recognize what A*(conjungate of A) means and you are done

[vapid steppe]

.close

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Please don't occupy multiple help channels.

[outer halo]

I agree

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@hasty brook Has your question been resolved?

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[little glen]

hello- just wanted to know if I got these answers correctly! we've just started on directed numbers

[wary stream]

You can use a calculator to check your answer

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[dapper lynx]

Hello math friends!

[dapper lynx]

My understanding of all of this is still a bit shaky

And when I see a recursive solution like this, well, it makes me a bit suspicious that I've done it right

[neon scarab]

For (iii), DI table may help simplfy the process a bit.

[dapper lynx]

XD

It did get a bit ugly, didn't it?

Does the math check out, though?

.close

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[cerulean herald]

How do I know if it passes through -3,4

ik it does because i looked at the answer key but how else can i know

[vapid shuttle]

plug in the points into the equation

do they satisfy the equation?

(-3)^2+(4)^2=?

[limpid turret]

Have you ever heard of the 3-4-5 triangle?

[cerulean herald]

no

[cerulean herald]

9+16?

25

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@cerulean herald Has your question been resolved?

[cerulean herald]

what does this mean

[rapid scroll]

what is on the other side of the equation?

oh

its a circle on a cartisan plane

there is a right triangle

[cerulean herald]

its a cricle

What does tangent to the x axis mean

[rapid scroll]

can you clarify

tangent to the x axis of what?

wait

here is a picture of what it think it means

the center of the circle would be on coordinate 4, -3

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[cerulean herald]

How do I find the foci

[scenic badger]

Touches the x-axis

Not intersects tho

[cerulean herald]

that was my last ticket

lol

[scenic badger]

Yeah but I looked at the diagram

[cerulean herald]

oh

[scenic badger]

The circle in there does not touch the x axis 😬

[scenic badger]

B^2 + C^2 = A^2
So we know B^2 = 9 and A^2 = 25, then find C
The foci are going to be in the vertical direction C units up and C units down from the center

[cerulean herald]

25+9

=34

[scenic badger]

What

No

9 + C^2 = 25

C^2 = 16

[cerulean herald]

whattt

no

[scenic badger]

C = 4

[cerulean herald]

its c = a^2 - b^2

so 4

[scenic badger]

Yea so move the b^2 over to the left side

Yes

[cerulean herald]

yeah

[scenic badger]

Lmao

[cerulean herald]

lol

so than

(h,k+c)

[scenic badger]

Yea

And (h, k-c)

[cerulean herald]

wait

so

5+4?

so the foci is 9?

[scenic badger]

hUh

[cerulean herald]

Wia tso whats the foci

[scenic badger]

No (h,k) is the center which is (0,0)

[cerulean herald]

ye

wait

[scenic badger]

So it would be (0,4) and (0,-4) right

[cerulean herald]

hm

wait

what does the foci even do

on an ellipses

ohh

[scenic badger]

When you beam light from a foci no matter what angle it hits the ellipse it always goes to the other foci

There’s a formal definition of the ellipse using foci but I forgot

[scenic badger]

Yea basically

[cerulean herald]

so its y=4 and y=-4

[scenic badger]

Write it as coordinates

[cerulean herald]

oh

ok

[scenic badger]

This

[cerulean herald]

ty

[scenic badger]

Np

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@cerulean herald Has your question been resolved?

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[floral vault]

I am having trouble solving this limit my professor said It needs to be telescopic

[vapid shuttle]

well it isn't a limit (why did I get "X" this question is about series convergence)

[vale wigeon]

well first off you understand that you should be testing the absolute-value series $\sum_{n=1}^{\infty} \ln\paren{1+\frac1n}$ for convergence, right?

[ocean seal]

Ann

[buoyant storm]

Thing=$ln(\prod_{n=1}^{\infty}(1+\frac{1}{n}))$

[ocean seal]

orthogonal_1

[tawny condor]

@slow hound stop this

[slow hound]

NOW

[tawny condor]

NOW

[slow hound]

who is mao

@alpine sable

[buoyant storm]

Stick to your own channel

[slow hound]

is that a threat

[buoyant storm]

no

it is a request

[buoyant storm]

Oh what did you mean by this

[tawny condor]

nothing, go on

[slow hound]

[buoyant storm]

you made me think what he sent was his own question and he was spamming here

@slow hound sorry

[slow hound]

dw ur fine

[tawny condor]

well its true that its unrelated

[slow hound]

its good you took the side of Sir Eggnote the Redstone

[buoyant storm]

How?

[buoyant storm]

This is correct and related

[tawny condor]

i mean, my message

[floral vault]

Im confused

[slow hound]

just look at this

say your prayers to Lord Redstone, and expand a few terms of the sum

you should see what happens

[floral vault]

how does this tell absolute convergance

[slow hound]

it doesnt

thats the key

[floral vault]

I need to use telescoping

[slow hound]

yeah that is telescoping

[floral vault]

huh

[slow hound]

https://www.khanacademy.org/math/integral-calculus/ic-series/ic-telescoping-series/v/divergence-telescoping-series

i think you may need to review

[floral vault]

we learned all of series in like 4 days prof trying to kill us

how would I find if it convenverges absolutely

[slow hound]

mate did you watch the video

[floral vault]

I have, so since we increase the number of n to infinity the value of the sequence increases. Therefore it diverges

it also occilates with positive and negative

?

I have this other quesiton

I see that we cant use ast

so not sure what to use next

[grand girder]

https://en.wikipedia.org/wiki/Alternating_series_test

[floral vault]

For the given equation I dont see it it follows both the ast rules. it doesnt go to zero and it decreases

[rigid prawn]

well then it isnt...

[grand girder]

Sorry for the confusion, my eyes saw sin instead of ln. You can use limit test.

[floral vault]

my prof said if ast deosnt to try divergance how would I apply that here?

[rigid prawn]

for Σan to converge you need lim n to inf of an=0

this isnt the case here so it has to diverge

[floral vault]

yes and for this example I get ln(0). so it does diverge I see thanks

can someone help me with this question my prof wants me to use telescopiong but not sure how to apply it here

[lone heart]

@floral vault Has your question been resolved?

[rigid prawn]

ln1/n=-ln(n)

[lone heart]

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[warped topaz]

https://i.imgur.com/s0VTb7G.png

[warped topaz]

So x(t) = 8t and y(t) = -8t + 7/8

I dont really know how to get them as numbers though

To plug into ay+bx=c

Btw c needs to be 1, right? Or am I mixing that up with something else...

[rigid prawn]

so -x=-8t....

[warped topaz]

Oh

I see what you mean

[rigid prawn]

the goal is to combine x and y so yeah

[warped topaz]

Ok so we got lucky with this one I think

[rigid prawn]

if it wasnt that easy just solve for t.... t=x/8

and plug in to the other equation

it can get a lot harder i guess though but there are other ways to combine them

[warped topaz]

okok thats easy enough though

[rigid prawn]

for example x=sin^2t y=cos^2t you get x+y=1

[warped topaz]

Ah yeye I know that trig rule

Btw c needs to be 1, right?

Or can I leave it as 7/8

okok apparently it doesnt haha

Thanks!

❤️

.close

[lone heart]

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[warped topaz]

https://i.imgur.com/p2DXoqv.png

[warped topaz]

https://i.imgur.com/6y3jlMA.png

It says 'is a point', does that mean it could ever be a point other than (0,0)?

[vale wigeon]

no

it just is not important for the purposes of whatever they're doing that the point is specifically the origin.

[warped topaz]

okok, makes sense

Thank you!

❤️

.close

[lone heart]

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[wooden epoch]

for this question in my exercise on using the discriminant for quadratics, i can't seem to work out how to solve for a. I've tried equating the 2 equations. ax+1 = -x^2 -x-8 and then transposing all to one side to make it one quadratic. x^2 +(a-1)x -7 = 0. Afterwards i tried use discriminant formula, if the 2 graphs only intersect once then the discriminant should be 0. b^2 - 4ac, (a-1)^2 - (4 x1 x-7) = 0. but i still can't seem to find a.

i continued on: (a-1)^2 = (4 x 1 x -7), (a-1)^2 = -28, which can't be true because if i can't square root both sides because of the -28.

[young finch]

well, you dont need the discriminant

you just know that the line has to go through the parabolas vertex

so find the quadratics vertex then plug those points into y=ax+1 to find a

[wooden epoch]

what's quadratic vertext

[young finch]

its maximum or minimum point

,w graph x^2

[ocean seal]

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[young finch]

this parabolas vertex is at 0,0

,w graph -x^2-x-8

[ocean seal]

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[young finch]

youll have to use something else to find the exact coordinates but you see that this parabolas vertex will be something like -0.5,-7

[wooden epoch]

idk i think they want me to use discriminant because it's in that exercise, i used discriminant for finding intersections of 2 quadratics in other questions but for this question, one of the equasions aren't a quadratic so idk what to do.

oh wait nvm it doesn't matter if they are not both quadratics

[young finch]

discriminant just tells you how many solutions there are to a quadratic

[wooden epoch]

right and they want the value of a to allow there to be only one intersection

this is the worked example they included for these kinds of problems

[young finch]

yeah but you cant do this problem that way

it never asks if there is a solution

of course you could set up an inequality and use the discriminant that way

but thats a waste of time

especially if you do it the way i wrote above

[fiery salmon]

just put y=ax+1 in the eqn of parabola

and put D=0

[wooden epoch]

in my CAS?

[fiery salmon]

cas?

[wooden epoch]

i'm confused how do i do that

[keen rampart]

sup

[wooden epoch]

hey

[fiery salmon]

ax+1=-x^2-x-8

[young finch]

whats x1

[wooden epoch]

ohh yeah equate them

[fiery salmon]

x^2+x(a+1)+9=0

[wooden epoch]

i did that before

[nimble fern]

sup

[fiery salmon]

do D=0

[nimble fern]

note that its
x²+(a+1)x+9=0
instead of
x²+(a-1)x-7=0

[wooden epoch]

oh

yeahhhh thats the mistake i made

lets gooooo

ahahha

i'm going to try again!

YESS it worked

thanks guys so much!!!

[young finch]

alternate method:
$vertex_x=\frac{-b}{2a}=-\frac{1}{2}\$
$vertex_y=-(-\frac{1}{2})^2-(-\frac{1}{2})-8\$
$vertex=(-0.5,-7.35)\$
$-7.25=a(-0.5)-8\$

frick

[wooden epoch]

oop that looks scary

[nimble fern]

😆

[young finch]

if i didnt do a calculation error for second line then a should equal -1.5

[wooden epoch]

enjoy your day/night yall 😁

a = -7 or 5

i got 5

.close

[lone heart]

Channel closed

Closed by @wooden epoch

Use .reopen if this was a mistake.

[young finch]

i messed something up

[nimble fern]

a+1=±6

[wooden epoch]

ohhh

[young finch]

method still stands

[wooden epoch]

right

[nimble fern]

you need both, just saying

[wooden epoch]

yeah i was wondering why i didn't get the other one

i forgot the plus-minus

hehehe

[lone heart]

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[noble elk]

<@&286206848099549185> suppose an equation with the function f(x)=ln(10x) all tangent lines at any point could be written as y=mx+b and if so, does m and b has a functional relation if they do what is the function and if not prove it

[lone heart]

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

[young finch]

its just asking what the derivative of ln(10x) is if im not mistaken

can you copy and paste that to a different channel

and preferably delete it here

[young finch]

prob not actually

[wintry rover]

Oh, ok, sure, i thought that this chann gave you a seperate channel when you posted a question in here

[young finch]

no somebody just nagged it

[noble elk]

no it si not i am asking if there is a relationship between the slope and the x intercept at each tangent line to the graph.

[muted flicker]

lmk wat channel u go to i know how to help

[wintry rover]

Which channel?

[young finch]

#help-7｜zen1thxyz #help-8 #help-22

[wintry rover]

Ok thanks

[noble elk]

could you please help me

[young finch]

,w differentiate ln(10x)

[ocean seal]

Results provided by WolframAlpha

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[young finch]

,w graph 1/x

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Arctic#1647

[wintry rover]

I went to #help-7｜zen1thxyz

[young finch]

1/x is the slope of y=mx+b in question

[noble elk]

1/x is the slope but b is different

[wintry rover]

#help-7｜zen1thxyz

[young finch]

oh wait

@noble elkthey cant because m and x cancel out

since x/x = 1

y-1=b

b has no relation with m

[noble elk]

it is doesnot make a sense if it helps you i will attach my trials after few minutes

[young finch]

alright

[noble elk]

here it is

<@&286206848099549185> help

[muted flicker]

no

[noble elk]

i've sent it

<@&286206848099549185> hellllllllllllllllllllllllllllllllllllllllp

<@&286206848099549185> hellllllllllllllllllllllllllllllllllllllllp

[simple blade]

W

[simple blade]

Shut up

[near apex]

With finding a relationship between m and b ?

ln(10/m)-b = 1

This is what i got.

[near apex]

=> ln(10/m) -1 = b

[muted flicker]

bro relax

ok first what you can do

f(x) = ln(10x)

what is the derivative of this?

[noble elk]

could the ordered pair i mentioned:(0.556,1.86) fit in this equation?

[muted flicker]

bro was begging for help and doesn't even listen

[noble elk]

sorry, it is 1/x

[muted flicker]

ok

now lets take an arbitrary point x_1

what is the gradient of the tangent at the point x_1?

[noble elk]

1/x_1

[muted flicker]

ok

now

what is the y-value at that point?

remember f(x) = ln(10x)

[noble elk]

ln(10x_1

[muted flicker]

yep

ok now we can form an equation in the form

y = mx + b

[near apex]

It should.

[muted flicker]

so we know our gradient m = 1/x_1

and our x = x_1 and y = ln(10x_1)

so then

imm just write it out

we know that

[m = \frac{1}{x_1}]

[ocean seal]

._pentium

[muted flicker]

and we also know that

[b = ln(10x_1) - 1]

[ocean seal]

._pentium

[muted flicker]

right?

[noble elk]

yes

[muted flicker]

rearranging this we get

[x_1 = \frac{1}{m}]

[ocean seal]

._pentium

[muted flicker]

right?

now sub that into the equation with b and you get

[b = ln \left( \frac{10}{m} \right ) - 1]

[ocean seal]

._pentium

[muted flicker]

do you get it?

[noble elk]

yes, thanks bro

[muted flicker]

no worries

next time pls don't spam ping anyone

[noble elk]

sorry

.close

[lone heart]

Channel closed

Closed by @noble elk

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[muted flicker]

its ok someone will come and help you dw

[lone heart]

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[warped topaz]

https://i.imgur.com/ffzkoH6.png

[warped topaz]

I understand the calculations but I dont understand the logic

I thought the direction vector was just the a bit in at+b
But for this you sub in the value for 2 and get the answer of at+b
So its the a bit multiplied by 2 + b

Like arent the velocity vector and the direction vector different? Or no?

[vale wigeon]

overthinking it + overfocusing on details + this is not a straight line

[warped topaz]

https://i.imgur.com/SFj6Ywi.png

This is what I thought direction vectors were

Like x(t) = 4t-3 y(t) = -6t+3
its in the form at+b therefore direction vector is <4,-6>

[warped topaz]

• ratio? 😅

[vale wigeon]

not feeling that disgruntled rn so no

[warped topaz]

I am just confused why the direction vector is different to how I did it before

[vale wigeon]

non straight lines don't HAVE direction vectors at all

[warped topaz]

okok

But the direction vector is the same as the velocity vector right? Or no theyre 2 different things?

[near apex]

Don't they ?
Isn't direction vector of the tangent at that point considered the direction vector at that point?
Although, yes, unlike a line, curves don't have same direction vector all the time.

[near apex]

It is. Since it's rate of change of position, it's equal to velocity. Although don't scale it.

[warped topaz]

Ah ok, makes sense

I have to close this to ask another question, sorry 🙃

And thx everyone!

[near apex]

Sure.

[warped topaz]

❤️

.close

[lone heart]

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[young bramble]

im not really sure what to do here, i've just recently learnt limits and the first principle

[warm flume]

if you substitute x=0, what happens?

[young bramble]

uh

[warm flume]

upstairs 0*sin(0) is 0

[young bramble]

mhm

[warm flume]

and downstairs e^0 - 0 - 1 is 0

have you learnt about anything to do when the limit is of the form "0/0"

[young bramble]

no, not really

[warm flume]

well then

there is a rule called the l'hopitals rule

that the limit f(x)/g(x) is the same as the limit f'(x)/g'(x)

when the limit is of the form "0/0"

do you know how to take the derivatives of the functions in your question?

[young bramble]

i think so

uh

maybe not for the demoinator

[warm flume]

you have a hint in the question about the e^x part

[young bramble]

do i use the first principle on both of them

[young bramble]

ohh

[warm flume]

but yeah look up that lhopitals rule in your book or online, its really handy for a lot of limits

[young bramble]

alright

i think i know what to do now

thank you 🙏

[warm flume]

npnpp

[young bramble]

.close

[lone heart]

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[wide berry]

I am sorry for not using latex,

[tawny condor]

dont be

[vapid shuttle]

do be

[tawny condor]

ok what's your question$.$

[ocean seal]

redstoneplayz09

[tawny condor]

$???$

[ocean seal]

redstoneplayz09

[wide berry]

Its group theory regarding cauchy's theorem, I just want to ask is my logic correct, if prime number p divides o(G) then there exists an element of order p, my proof is that if o(G) is itself a prime then it is itself a cyclic group, and if its not they it may be even/order, if its even then we know that there exists a element of order 2 ( prime), if its odd, then we also know there is a element say 'x' such that x^o(G) = e (identity element) then we can prime factorize o(G) such that o(G) = p1*p2 then (x^p1)^p2 =e is it correct?

[ocean seal]

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

[wide berry]

I mean this

is someone there?

[lone heart]

@wide berry Has your question been resolved?

[wide berry]

<@&286206848099549185>

[lone heart]

@wide berry Has your question been resolved?

[lone heart]

@wide berry Has your question been resolved?

[wide berry]

<@&286206848099549185>

[lament forge]

this isn't correct

firstly, the group containing an element of order 2 doesn't tell you anything about whether it contains an element of order p

...and is also something you would have to prove

secondly, if o(G) is odd that does not necessarily imply that there is some x such that x^o(G) = e, that's only true about cyclic groups and a lot of groups with odd orders aren't cyclic (for instance take the product of two copies of the cyclic group of order 3, that has order 9 and all of its non-identity elements are order 3)

[lone heart]

@wide berry Has your question been resolved?

[lone heart]

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[lone heart]

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[meager bone]

I need help with my pre calc stuff

[meager bone]

im literally so lost

I have an exam and i dont know how to do anything

factoring is hard

[tawny condor]

ok send it

[meager bone]

and im stuck on rational equations and idk what to do

thank you

[tawny condor]

always just send the question when u open a channel

[meager bone]

Ignore what I wrote

That’s old stuff and I’m retaking the class I want to learn instead of just copying the answers

[meager bone]

Got it

[tawny condor]

$\frac{1}{x^2 - x - 6} + \frac{3}{x + 2} = \frac{-4}{x - 3}$

[ocean seal]

redstoneplayz09

[tawny condor]

this is ur question?

[meager bone]

Yes

[tawny condor]

ok

can u try factoring $x^2 - x - 6$

[ocean seal]

redstoneplayz09

[meager bone]

I’m not sure how

Let me try

[tawny condor]

so u know it's gonna be something like

$(x\ \ \ \ )(x\ \ \ \ )$

right

[ocean seal]

redstoneplayz09

[tawny condor]

and u add/subtract from both of these x's

we know its just x and x since you're gonna get x^2 when multiplying it out

[meager bone]

Is it this?

[tawny condor]

is that a 1

[meager bone]

No 7

-7

[tawny condor]

that's not correct anyways

[meager bone]

Oh

How do I factor?

I always struggle

[tawny condor]

the reason that u cant do it is probably because u don't know how to expand

ok if I gave u

[meager bone]

I think I do

[tawny condor]

$(x - 4)(x + 5)$

[meager bone]

Let me try it really quick

[ocean seal]

redstoneplayz09

[tawny condor]

can u expand that for me

[meager bone]

Oh wait hold on let me try

[tawny condor]

you should be at a level where it is easy for you

preferably

[meager bone]

Yea but I haven’t done math in 2 years sadly due to unfortunate circumstances

I’ll try my best

Thanks for ur time btw

[tawny condor]

np

[meager bone]

I don’t know how to do it smh

I only got the X^2

If that’s even what I’m supposed to do

[tawny condor]

ok hmm

so

let's do it step by step

do you know about the distributive law

[meager bone]

Yea

[tawny condor]

$a(b + c) = ab + ac$

[ocean seal]

redstoneplayz09

[meager bone]

Yea

[tawny condor]

$(b + c) \cdot a = ba + ca$

[ocean seal]

redstoneplayz09

[tawny condor]

ok

so

$(x-4)(x+5)$

[ocean seal]

redstoneplayz09

[tawny condor]

let's look at the (x-4) as some number

and we multiply it by parenthesis, which have x and 5

so according to the distributive law:

[meager bone]

Ohhh wait I think I kinda get it

[tawny condor]

$(x-4) \cdot x + (x-4) \cdot 5$

[ocean seal]

redstoneplayz09

[meager bone]

But how do you get 2 X-4’s if they aren’t being squared?

OHHH WAIT

you broke them up so you can distribute each variable?

[tawny condor]

yes

now u distribute again

[meager bone]

Let me try it on paper

[tawny condor]

$(x - 4) \cdot x = x \cdot x - 4 \cdot x$

[ocean seal]

redstoneplayz09

[tawny condor]

this is for the first parenthesis

yes try it

[meager bone]

Wrong pic

Sorry

So is it like that?

[tawny condor]

yes

and now u have -4x + 5x

which is?

[meager bone]

X

?

Because -4+5 is 1

And so you just put X

[tawny condor]

yeah u can think of it like this:

you have -4 of something called x

and u add 5 of that x

so u get 1 of it in the end

[meager bone]

So it gives me X^2+X-20?

[tawny condor]

yes

[meager bone]

Perfect

[tawny condor]

notice what we did

$(x - 4)(x + 5)$

[ocean seal]

redstoneplayz09

[tawny condor]

we just multiplied the x by x,

x by 5

-4 by x

and -4 by 5

[meager bone]

And distribution

[tawny condor]

and added it all up

[meager bone]

Then combined like terms

[tawny condor]

so u can just do it quickly in your head instead of distributing twice

[meager bone]

OHH THATS SMART AF

I gotta get used to mental math so that’s a smart way of thinking of it

Okay

How do I apply that to my current question?

Wait I think I can do some of it

Let me try

So I find something that equals X and something that multiplies to -6

Which is -3•2

[tawny condor]

yes that works

ill show u a good way of thinking about it later

[meager bone]

Okay

[tawny condor]

for now lets continue

[meager bone]

Sure

Thanks

[tawny condor]

$\frac{1}{(x-3)(x+2)} + \frac{3}{x+2} = \frac{-4}{x-3}$

[ocean seal]

redstoneplayz09

[tawny condor]

now, can you make a common denominator

[meager bone]

So I replace the first denominator with our factor first right?

Instead of the X^2-x-6 I put what I got when I factored

That’s just my understanding lmk if I’m thinking wrong

[tawny condor]

yes

[meager bone]

Okay

[tawny condor]

thats what I did

[meager bone]

Got it

[tawny condor]

i just replaced x^2 - x - 6 with (x-3)(x+2)

they are equal

[meager bone]

Yup

Got it

Do I multiply all sides by the LCD?

Because that way I can cancel out the other denominators

Wait I just realised the LCD was also the denom of the 2 other fractions

[tawny condor]

u want to make the denominators the same

notice how u just have to multiply two of them by something

make all of them equal (x-3)(x+2)

[meager bone]

Does that something need to be the same number?

Because I’m thinking multiplying them by -2 and 3 to make them both X+6

[tawny condor]

start with $\frac{3}{x+2}$

[ocean seal]

redstoneplayz09

[tawny condor]

how do you make it have a denominator of (x-3)(x+2)

[meager bone]

Maybe multiply the top by -X?

I’m not rly sure

[tawny condor]

hmm

how do I give a hint without giving it away lol

[meager bone]

Lol

[tawny condor]

you want to multiply the numerator and denominator by the same thing so that it doesn't change the value of the fraction

for example

$\frac{1}{2}$

[ocean seal]

redstoneplayz09

[tawny condor]

if i wanted to make this have a denominator of 6

i would multiply the top and bottom by 3

so u get

$\frac{1 \cdot 3}{2 \cdot 3} = \frac{3}{6}$

[ocean seal]

redstoneplayz09

[tawny condor]

and its still the same fraction

just written with a different denominator

[meager bone]

I’m trying to like think of that while looking at my notebook and I’m lost

Give me a second to figure it out

[tawny condor]

you want $\frac{3}{x+2}$ to have $(x+2) \cdot (x-3)$ in the denominator

[ocean seal]

redstoneplayz09

[tawny condor]

so what do you multiply the top and bottom by?

[meager bone]

Is it supposed to be a single number or is it supposed to be in the form of (X +- ?)

[tawny condor]

something with x

don't be mean to x

u can include him

u dont always need things to be numbers lol

[meager bone]

X+1?

[tawny condor]

no

[tawny condor]

look at this

u already have the (x+2) part

so what else do you need

[meager bone]

This is unnecessarily difficult but my guess is X+3?

I’m so bad with numbers omgggg🤦‍♂️😭

[tawny condor]

close

[tawny condor]

do u understand this

[meager bone]

Yea that’s easy but I don’t get it when there’s like the X and stuff

[tawny condor]

x is just a number

(x+2) is also just a number

and so is (x-3)

[meager bone]

I’m stumped tbh

[tawny condor]

damn idk how to spoonfeed this better

take $\frac{3}{x+2}$

[ocean seal]

redstoneplayz09

[tawny condor]

multiply top and bottom by (x-3)

and tell me what you get

write it down on paper

(dont expand the parenthesis btw)

[meager bone]

What do you mean by Expand? Do you mean like I shouldn’t put them side by side? I’m supposed to distribute right?

Did I do the first part right?

Like the format it’s supposed to be

Now I split both fractions?

[tawny condor]

wtf is

$x^26$

[ocean seal]

redstoneplayz09

[tawny condor]

what is that

i told u not to distribute but forget that

u did it wrong anyways

the denominator is just (x+2)(x-3)

[meager bone]

But isn’t that expanding?

Or is expanding the one I did?

[tawny condor]

u expanded it wrong

$\frac{3}{x+2} \cdot \frac{x-3}{x-3}$

[ocean seal]

redstoneplayz09

[tawny condor]

$= \frac{3 \cdot (x-3)}{(x+2) \cdot (x-3)}$

[ocean seal]

redstoneplayz09

[meager bone]

So just leave it like that? Or is there more I need to do?

[tawny condor]

well leave the denominator like that

u can expand the numerator if u want

[meager bone]

But distribute the top?

[tawny condor]

but do u understand what we did

[meager bone]

Oh okay

[tawny condor]

now we have a denominator of (x+2)(x-3) just like we wanted

[meager bone]

So because we don’t expand the denom we literally just put the number we want which was the X-3 and expand the top?

That’s what I got from that

[tawny condor]

wdym "put the number we want"

we multiply top and bottom by the same thing so that the denominator changes, but the value of the whole fraction doesn't

[meager bone]

Yea the same thing is what I meant

And we don’t expand the denom

Only the numerator

Because we only wanted X+2 and X-3 in the denom

So like that then

[tawny condor]

just because it doesn't matter

easier to look at it when its (x+2)(x-3)

[meager bone]

Got it

[tawny condor]

yes its 3x - 9

nice

good

[meager bone]

So do i replace this new Fraction instead of the old one?

[tawny condor]

$\frac{1}{(x+2)(x-3)} + \frac{3x - 9}{(x+2)(x-3)} = \frac{-4}{x-3}$

[ocean seal]

redstoneplayz09

[meager bone]

Yea got it

[tawny condor]

now do the same for the last fraction

[meager bone]

So would it look like this?

And also do I just make the X-3 squared?

[tawny condor]

well u didnt make any mistakes

but thats not the denominator we want

we want (x+2)(x-3)

u did (x-3)^2

not very helpful

[meager bone]

Ohh because that’s our LCD right?

Like this?

[tawny condor]

yes

[meager bone]

And now I have to cancel out all the demons right?

[tawny condor]

yes

[meager bone]

Denom^

So I multiply

?

[tawny condor]

its the same thing as multiplying both sides of the equations by (x-3)(x+2)

[meager bone]

Got it

[tawny condor]

it will make all the denominators go kaboom

[meager bone]

Let me try

That’s what I got

[tawny condor]

yeah

continue

[meager bone]

I factor now right?

[tawny condor]

nope

look again at what u have

[meager bone]

Wait I think I know what to do

Let me try and lmk if I’m wrong again

[tawny condor]

btw

[tawny condor]

$1 + 3x - 9 = -4x - 8$

[meager bone]

That’s what came to mind

[ocean seal]

redstoneplayz09

[tawny condor]

instead of starting to move stuff already

I would first combine like terms

so you have 1 - 9 which is -8

so u get

[meager bone]

Even though they’re on different sides of the equations?

[tawny condor]

$3x - 8 = -4x - 8$

[meager bone]

I can do that?

[ocean seal]

redstoneplayz09