help-17

[short kraken]

wait

sin(2 times0) / (2 times 0)

[vast shale]

lim x---> 0 sinθ/θ = 1

[short kraken]

okay thats why

[vast shale]

got it?

[short kraken]

yep

[vast shale]

see, 0/0s are nice

[short kraken]

as x approaches 0 sinx/x approachs 1

from left and right

[vast shale]

yes!

[short kraken]

is that what lim x--> 0 sinx/x is communicating?

[vast shale]

yep

[short kraken]

thanks guys

[vast shale]

you are welcome

i hope we helped

[short kraken]

you did

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[scenic latch]

What's up guys, kinda dumb silly question but I wanna double check, my problem is
7y + 9y² - 7y² - 6y
I have to simplify the problem but I'm kinda confused because the y²

[lost salmon]

2y^2+2y

[scenic latch]

Uh what 😭

[lost salmon]

y(2y+2)

[gaunt sparrow]

That answer is wrong and you shouldn't be giving out answers in the first place

[scenic latch]

So is y² like a whole different thing than y

[lost salmon]

yh

y*y=y^2

[gaunt sparrow]

Group like terms. y^2 is different that y, but you can group the y^2 together and do the same for the y's

[scenic latch]

Okay thanks

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[pearl moth]

how do I get the domain

f(x)=ln(5−3x)

[pearl moth]

pretty sure i do 5-3x > 0

then when I solve that out I get x < 5/3

[queen root]

yes

[pearl moth]

but how do I put it into like a ordered pair

[queen root]

u mean interval?

[pearl moth]

ye

[queen root]

one way is to

represent ur inequality

[queen root]

this one

on a real number line

then read it out

and write out

it out*

[pearl moth]

(-infinity, 5/3)

cuz it goes infinitly left

[queen root]

bravo

[pearl moth]

or negetive

oo

ty

[queen root]

np

[pearl moth]

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[edgy totem]

im confused on how to do this can i get some assistance? i need to describe how i got answer idk how to do it though

[ornate ember]

in order to be continuous, there can't be a break in the function. So, with that logic, you need to find values where 5bx-6a meets -3b - 4ax at x = -2 as well as -3b - 4ax meets 5x-1 at x = - 2. So how do you think we can do this?

[edgy totem]

set them equal and solve?

i got b= -2a
so a = -11/14
then u sbusititute it back into the equation

so b is 11/7?

[ornate ember]

One moment

(sorry wife is calling me for something, I'll return shortly. If someone else answers then cool, if not i'll hit ti back)

[edgy totem]

alright no problem

🙏

anyhone please

[ornate ember]

Hey Im so sorry, emergency with pet

let's wrap this up

okay so you had the right idea but there might've been a miscalculation somewhere

So when we plug in x = -2 into the top equation, we get -6a -10b, middle equation is 8a - 3b, and finally the bottom equation is -11.

Since they all have to equal one another, then we have the following:
-6a - 10b = 8a - 3b = -11. We can turn this into a system of equations.

-6a - 10b = -11
8a - 3b = -11

So now we can solve the system of equations

want to give it a try real quick?

\begin{align*}
-6a -10b &= ,-11 \
8a - 3b &= ,-11
\end{align*}

[twin meteor]

MellowDramaLlama

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[fleet vault]

how do i find the angle the answer says its pi/3 idk how to get that

[rugged orchid]

which angle are you trying to find?

[fleet vault]

the gradient

so like the one next to the x axis

[rugged orchid]

can you draw the angle you're looking for?

perhaps there are some (right angled) triangles that could help you out

[fleet vault]

oh wait

so the triangle below 2 (y axis)

would have angles 60, 90 and 30?

[rugged orchid]

yes

[fleet vault]

ok thank u i got it i needed someone to prompt me

thank u so much !!

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[compact fable]

Surely there has to be a more efficient way to solve this problem than using Pascal's triangle right? How am I supposed to determine the coeffcients?

[oblique cargo]

do you know coefficients

[compact fable]

I tried searching on googlee

[oblique cargo]

like do you have a computer there

[compact fable]

but i cant find a triangle that goes up to row 20

[oblique cargo]

do you have a computer there??

[compact fable]

I do have a computer

[oblique cargo]

what type??

[compact fable]

Im not sure, its a PC, but I dont think its designed for like math/pascals triangle lol

Like the computer I have is not related to the school work

[oblique cargo]

I thought it was sth like Casio fx-580

[compact fable]

Not sure, I am sorry

Do you mean a calculator?

[oblique cargo]

yeah

[compact fable]

https://www.mathway.com/Algebra

[oblique cargo]

i kinda dumb at English

[compact fable]

Lol no worries

How about mathway does that work?

ti 30x*

apologies

[oblique cargo]

maybe calculate by hands will be find

fine

[compact fable]

I can try that

[oblique cargo]

$C^k_n = (n!) / (k! * (n - k!))$

[compact fable]

Is that a symbol on the calculator?

[oblique cargo]

wait

yes

[twin meteor]

Nguyễn Tuấn Minh

[compact fable]

prb?

then it asks me to choose nPr or nCr

or !

[oblique cargo]

nCr (maybe)

[compact fable]

it then says "Ans nCr __"

I would assume I put 20

but then I get 0 when I press enter again which seems wrong

[oblique cargo]

wait

lemme find a simulator

[compact fable]

Ok

[oblique cargo]

all i found is trash

[compact fable]

Lol no worries

Its fine if U couldnt find the answer

its Just one homework problem of many

[oblique cargo]

$C^k_n = (n!) / (k! * (n - k!))$

[twin meteor]

Nguyễn Tuấn Minh

[compact fable]

Is that a formula I can use?

[oblique cargo]

maybe calculate by hands maybe find

[compact fable]

Yes I can try that

[oblique cargo]

so n in your case is 20

[compact fable]

Row 20

yes

[oblique cargo]

try k from 0 -> 2

that's the coefficient

[oblique cargo]

btw 2 must be square in the last term

[compact fable]

Ohh I forgot u do that with the exponent thanks

[oblique cargo]

i'm just too bored doing the crazy assignments my teacher assign me to

[compact fable]

Lol

Ok thank you for your help I am going to close now!

.close

[oblique cargo]

ok

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[brazen thistle]

I need to factor this trinomial, but I'm stuck on how to do it, could someone help me?

[rare swift]

Split 5x into 6x - x and then factor

[brazen thistle]

sorry, what do you mean by split 5x into 6x - x?

[oblique cargo]

5 = 6 - 1

so 5*x = (6 - 1)*x = 6x - x

[brazen thistle]

ooooh alright, i understand now

thank you

[oblique cargo]

nothing

[brazen thistle]

.close

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[vocal sleet]

@fossil pollen Has your question been resolved?

[inner osprey]

use this

[fossil pollen]

But its asking for trace

And its A³adj(A²)

dedge

[inner osprey]

so?

[rugged orchid]

notice with this you can sub stuff in

[inner osprey]

^ when you "use" a formula you want to figure out way to make it work

[rugged orchid]

we can think about the trace part later, figure out what's inside first

[fossil pollen]

So do i write it like tr(A× A²adj(A²)?

so tr(A× |A|²I)?

and then do i just find the matrix or is there a shorter trick

OHHHH thanks i got it

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[stone oyster]

Help please

[stone oyster]

<@&286206848099549185>

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[idle robin]

.close

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[kindred parcel]

hey!!!!!! i need help in trignometry question......

[vocal sleet]

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.

[twin horizon]

[rare swift]

It's such a hard question that i am not even able to see it

[kindred parcel]

so saddd

[rare swift]

That's a trigonometry que?
Btw take lcm and try the first derivative test

[kindred parcel]

oh i'm sorry i actually had doubt in this question too from maxima and minima

[twin horizon]

It is

Sadly

[kindred parcel]

still not able to do it...

[rare swift]

It's a increasing function atleast for x > 0 if my calc are correct

[twin horizon]

Absolute value of x = +√(x²)

I guess this'll help

[rare swift]

Nvm my calculations arnt correct

[kindred parcel]

yeah i got it

thankss.

🙂

[twin horizon]

Oh ok

[rare swift]

What did u do

[rare swift]

That's some crucial info

[harsh pelican]

did u find it?

a true

b false cause it supposed to be 5/6

[rare swift]

Function is decreasing for x > 0 so minima should be at x = 1

I hope this time my calculations are correct

[rare swift]

What is supposed to be 5/6 minima maxima inflection point

[harsh pelican]

we want the min value of f(x) so we have to make the max denaminator

put -1 instead of x

[rare swift]

Ohh ryt

[harsh pelican]

than you find 1/2 + 1/3

btw sorry if i cant explain what i want cuz im not so good at english

but im tryin

[rare swift]

Fck my calculations are again wrong
Some one find ther derivative of the function and tell wheater it's increasing or decreasing or having a local minima or maxima

[twin horizon]

Yes

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@kindred parcel Has your question been resolved?

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[bleak prawn]

This is 1 of the steps in the solution to a question

How did they integrate sqrt(x) - ln(x) into 2x(sqrt(x))/3 - x(ln(x) - 1)?

I got 2(x^3/2)/3 - xln(x) - x

[gentle thicket]

x^{3/2} = (sqrtx)^3

[bleak prawn]

hmm ok... but for their answer I got 4/3 - 4ln(x) and they got something different

So where did I make my mistake

Wait

I see it

Its supposed to be +4 and +1

Ok my bad

Thank you!

❤️

.close

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[hollow plinth]

Hey

[hollow plinth]

I have almost done this problem

I just need some form of confirmation that i did this problem right ✅️

[vast shale]

How did you do it?

[hollow plinth]

I asked chat gpt to check my work and see if i did it right but i am sure it made a mistake lol

[vocal sleet]

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).

[hollow plinth]

Huh what is your problem if i use ai in my private life. I don't trust ai and this is why i came here lol

I only make use in certain situations

[vast shale]

For matrix computations maybe using something like GNU octave (free analogue of MATLAB) is more reliable

[hollow plinth]

Ohh thx for the tip

So did i do it right

?

[vast shale]

lol

obviously it’s fine to use otherwise but it’s proven to be highly inaccurate at times for math specifically

[hollow plinth]

Yes i know lol

This is why i came here

I solved the question myself and then i ask ai to check it

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[bleak prawn]

So first I got 2x^2 + 9x + p

And I got the discriminant of that and set it > 0

b^2 - 4ac > 0

81 - 4(2)(p) > 0

p < 81/8

But I dont really know what to do with that cause its not at y = 1

[civic otter]

Why are you considering the numerator?

[bleak prawn]

Im not really sure what im doing, I was just guessing 🙃

[civic otter]

Read the text carefully

You need to read more deeply

What exactly has 2 solutions?

[bleak prawn]

y = 1

[bleak prawn]

Thats what I was saying here

Its not that I didnt read I just dk how to do it lol

[civic otter]

So ...

[bleak prawn]

So I dont know 💀

[civic otter]

What is y also?

[bleak prawn]

intersecting f_p(x) twice?

[civic otter]

This doesn't make sense 😬

[bleak prawn]

Ok then I dont know what is y also

[civic otter]

Well, it's f_p(x)

Have you ever seen y = f(x)?

[bleak prawn]

Yes

But in that case I have to sub in x = 0, right?

[civic otter]

So if both 1 and f_p(x) are equal to y, it follows that...

[civic otter]

This also doesn't make sense 😅

[bleak prawn]

I think I will take a break and come back to this question

.close

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[civic otter]

Yeah I think it's better

[bleak prawn]

What does that mean

[civic otter]

That your brain should recover a little bit, so that you can somehow reset all the wrong stuff you've thought of doing

I often see that after a break the brain is clean for new reasoning

And sometimes it turns out it gives the correct way of solving a problem

[bleak prawn]

Ok sorry if I kept saying the wrong stuff im just trying to learn

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[civic otter]

Don't worry, as you said it's something normal, there's also the saying "you learn by your mistakes"

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[vast shale]

Calculate BM in terms of a and c if BM = 2MN and AP = AC

[fervent wasp]

isn’t it resolved?

I saw you post it yesterday

[vast shale]

No one answered

Just someone saying bisector theorem I see it coming, or sm like that

But no

it's not

[dense cedar]

The angle bisector splits the base in ratio of its sides

[vast shale]

M yep, I don't how to use that here

tho

[dense cedar]

Can you draw a perpendicular from B upon AC?

[vast shale]

mm I think yes

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@vast shale Has your question been resolved?

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[distant perch]

( a shuttle number is a number that : the first digit is greater than the second. The second digit is less that the third digit. The third digit is greater than the fourty. The fourty digit is less than the fifty digit... )

How many 7-digit numbers formed by the digits 1, 2, 3, 4, 5, 6 and 7 are shuttle numbers? For example, 4253617 is one of these numbers. But 5372146 is not (2 is greater than 1 and 1 is less than 4) and 2163457 is not either (4 is less than 5).

[thin vale]

@loud walrus

[rare swift]

Is repetition allowed ?

[distant perch]

no, there is only a number

like, there is no 737

[rare swift]

? So it's not allowed

[distant perch]

it is not allowed

[loud walrus]

[rare swift]

What have u tried

[distant perch]

i am struggling to start actually, idk how can i combine this

[rare swift]

Then don't solve this problem solve a smaller case of this problem ie solve for 3 digit numbers and try to make cases

[distant perch]

alright, i will try to do that

[vocal sleet]

@distant perch Has your question been resolved?

[rare swift]

Any progress

[distant perch]

i am thinking yet, give me just a moment please

[distant perch]

that is what i found, 1 means the first digit

so the first one means that are 15 combinations if the first digit is 7

adding it all up i found 70 combinations

[rare swift]

Could you identify some pattern?

[distant perch]

yes, mainly seeing here with 4 digits but i can't prove anything yet, one moment

[distant perch]

the most obvious thing of course is that when the first number is 5 or less we start to remove the lower numbers, so when first digit is 5, we remove 1, when first digit is 4 we remove 2 and so on

[rare swift]

Hmm so ur using brute force to do it
Ok think like this u have 10 digits right and u select 3 out of the 10 digits and arrange them in such a way that a1 > a2 and a2< a3
You can select 3 digits out of ten in 10C3 ways
Let the digits be abc can i argue that there will be one digit out of those 3 which will be the smallest let's say b that will be your middle term a2 in this case and the rest a and c can be either a1 Or a3 so a and c will have 2 choices to be arranged into and b our smallest digit only has one choice which is to be the middle term
So total arrangements possible will be 10C3 × 2 × 1

So do you get my argument?

[distant perch]

i am trying to

[rare swift]

Ok

[distant perch]

sorry, i have to go but i'll consider everything that you said to me, thank you and have a great day

.close

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[royal turret]

simple question but whats the 3.7 inches mean?

[urban edge]

this is a triangular prism

the 3.7in is the height of the triangle from the right side

[royal turret]

thank you but then whats that line with the square

[urban edge]

its a right angle

that means the 3.7in is the height and not some other random line

[royal turret]

thank you

[vocal sleet]

@royal turret Has your question been resolved?

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[brittle vessel]

i need help with all of thsi

<@&286206848099549185>

[shut osprey]

That's ncert dude

[calm light]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shut osprey]

I'll just rapid fire answer all of them dw

[shut osprey]

Ques 11 or ques 12?

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[short kraken]

how does this work

[short kraken]

they pulled out 4/5 to get sinx/x

so (4/5) * 1

but there's no x to plug numbers into

[harsh canopy]

Do you not understand how they pulled 4/5 out?

[short kraken]

they just did..?

[peak shard]

multiplied by 1

4/4

[short kraken]

yes sin4x/4x = sinx/x

[peak shard]

no

[harsh canopy]

also if you expand out that second bit, you will realise you get back to the first expression

[harsh canopy]

You can’t divide the 4 in the sin expression like that

[peak shard]

$\sin(4x) \neq 4 \cdot \sin(x)$

[twin meteor]

tobi

[harsh canopy]

Sin4x/4x = 1/4 * sin4x/x

[short kraken]

im soooo confused

they just pulled out 4/5 and replaced it with 4/4 by multiplying by 1?

[peak shard]

$\frac{\sin(4x)}{5x} = \frac{4}{4} \cdot \frac{\sin(4x)}{5x} = \frac{4}{5} \cdot \frac{\sin(4x)}{4x}$

[twin meteor]

tobi

[short kraken]

yea i feel like theres more steps in between that

i dont get how the swap occurred

[peak shard]

$\frac{a}{b} \cdot \frac{c}{d} = \frac{c}{d} \cdot \frac{a}{b}$

[twin meteor]

tobi

[peak shard]

you can always swap

$a \cdot b = b \cdot a$

[twin meteor]

tobi

[peak shard]

maybe it gets simpler if I do this once

[short kraken]

i mean. duh

but 4x/4x is within the sin expression

[peak shard]

$\frac{\sin(4x)}{5x} = \frac{4}{4} \cdot \frac{\sin(4x)}{5x} = \frac{4 \cdot \sin(4x)}{4 \cdot 5x} = \frac{4 \cdot \sin(4x)}{20x} = \frac{4}{5} \cdot \frac{\sin(4x)}{4x}$

[short kraken]

its not 4/4 * 4sinx/5x

[twin meteor]

tobi

[short kraken]

k but can you show what happens between here

like are they canceling the 4/20

[peak shard]

sure

$\frac{4 \cdot \sin(4x)}{20x} = \frac{4 \cdot \sin(4x)}{5 \cdot 4 \cdot x} = \frac{4}{5} \cdot \frac{\sin(4x)}{4 \cdot x}$

[twin meteor]

tobi

[peak shard]

maybe this here is unclear to you

[short kraken]

oh

thats too convenient

[peak shard]

$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$

[twin meteor]

tobi

[peak shard]

this is always true when b and d are not zero

[short kraken]

okay so why are we doing that in the first place

just to get sin4x/4x = 1?

[peak shard]

okay thats the next question but you agree with me that we didnt change anything right?

[short kraken]

right

theyre still equivalent

[peak shard]

you should also know that constants can be pulled outside of the limit

[short kraken]

right

[peak shard]

$\lim_{x \to 0}} \frac{4}{5} \cdot \frac{\sin(4x)}{4x}$

[twin meteor]

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

[short kraken]

okay

so rn there is an x value to plug 0 into

but sin4x/4x has to become 1 first

[peak shard]

no

[short kraken]

then there is no x value to plug into

[peak shard]

limits never work like this

you never plug in the value

[short kraken]

yea then im lost

[short kraken]

okay thats just untrue

i have definitely plugged in the value into equations

[peak shard]

only for continues functions

[short kraken]

ah

[peak shard]

but limit just means what happens with my function as x approaches that value

it can be the same as plugging in the value, but thats only for continues functions

$\lim_{x \to 0} \frac{4}{5} \cdot \frac{\sin(4x)}{4x} = \frac{4}{5} \cdot \lim_{x \to 0} \cdot \frac{\sin(4x)}{4x}$

[harsh canopy]

Yeah in this case you can basically just do what everyone does and say sin4x/4x tends to 1

[twin meteor]

tobi

[harsh canopy]

Or if you really want you could change variables to something like u = 4x and then you have the limit as u tends to 0 of sin(u)/u (since this is typically the standard result) but that’s not really needed

[short kraken]

thats seems cheaty

just putting it outside the function

[harsh canopy]

You can take constants out of limits

[short kraken]

so if i had 2/x^2

i could take out the 2

[peak shard]

yeah this is not so difficult to show using the definition of the limit

[short kraken]

and just do 2 * limx--> 0 (1/x^2)

[harsh canopy]

Because a limit of a product is the product of the limits

[peak shard]

well this expression does not make sense in the first place but yes

[peak shard]

no

[short kraken]

it doesnt? 😭

it was an example i came up with to check my understanding

[peak shard]

only if both exist

[harsh canopy]

Well yes

[short kraken]

okay so we can take constants out but what do we do with our sinx/x

or sin4x/4x whatever

[harsh canopy]

If you really want to think of it like this you could, but again you don’t really need to:

[peak shard]

$\lim_{x \to 0} \frac{x}{x} \neq \lim_{x \to 0} \frac{1}{x} \cdot \lim_{x \to 0} x$

[twin meteor]

tobi

[short kraken]

no i meant x * lim as x goes to 0 for x

so other way around

[peak shard]

$\lim_{u \to 0} \frac{\sin(u)}{u} = 1$

[twin meteor]

tobi

[short kraken]

are we doing f(0) tho

[peak shard]

you usually prove this geometrically, there are some great videos for this limit

[short kraken]

sin*0/0 is just undefined

[peak shard]

yes thats what I mean with we never really plug in the value

[short kraken]

oh right

specifically for this limit

[harsh canopy]

$\lim_{x \to 0} \frac{4}{5} \cdot \frac{\sin(4x)}{4x} = \lim_{x \to 0} \frac{4}{5} \cdot \lim_{x \to 0} \frac{\sin(4x)}{4x} = \frac{4}{5} \cdot \lim_{x \to 0} \frac{\sin(4x)}{4x}$

[short kraken]

sinx/x

[peak shard]

we just look at the value we get when we approach 0

[short kraken]

as x goes to 0

its always 1

[harsh canopy]

[short kraken]

brah what am i looking at

[peak shard]

sin(x) behaves like x for really small values of x

[harsh canopy]

Yeah gimme a sec

[peak shard]

so for very small x we can cancel x and x to get 1 (physics trick) but it gives some intuition why the equation is true

graph of sin and x

[twin meteor]

eugene_krabs_has_cake

[harsh canopy]

^

[peak shard]

you need a definition of limits for that tho

[short kraken]

okay i get it now

its literally just the identity

[harsh canopy]

I’m not trying to be THAT rigorous, this should do fine

[short kraken]

i know why we can just have 4/5 not how squeeze theorem works unfortunately

so i dont understand how trig functions work here

heres another problem

we turn it into an equivalent form

so we have

[peak shard]

well first of all

[short kraken]

1/(2xcosx^3)

[peak shard]

maybe write this in terms of sine and cosine

yeah

[short kraken]

do i just plug in 0

[peak shard]

no

[harsh canopy]

1/2x isn’t continuous at 0

[peak shard]

you never plug in that value

[short kraken]

what do i do with these type of problems then

[harsh canopy]

Not even considering the numerator

[short kraken]

so im not done yet

[harsh canopy]

You would need both to be continuous at 0, here the denominator isn’t

[short kraken]

right, so more factoring?

[peak shard]

$\lim_{x \to 0^-} \frac{1}{2x \cdot \cos(x^3)}$

[short kraken]

cos of 0 is just 1

[peak shard]

yes

[short kraken]

so i do plug in 0

[twin meteor]

tobi

[short kraken]

i mean id get 1/0

[peak shard]

thats bad

[short kraken]

right

so what do we do

[peak shard]

we dont divide by zero

but thats the correct idea

you can think about it like this

what happens for values very close to 0

[short kraken]

but i want to approach 0 from the left so i can plug values in up to 0 that wont give me 1/0

[peak shard]

that are always negative

what happens for cos(x^3) when x is almost 0

you already said it, I just want you to think differently about limits

[short kraken]

it approaches 1?

[peak shard]

yeah

and what happens for small negative values with 1/x, as x goes to 0 from the left

[short kraken]

negative infinity

[peak shard]

so for very small values you mulitply 1 with negative infinity

which is just negative infinity

you can (again) ignore the factor of 1/2

[short kraken]

because this expression is essentially 1/x?

[peak shard]

$\lim_{x \to 0^-} \frac{1}{2x \cdot \cos(x^3)} = \frac{1}{2} \cdot \lim_{x \to 0^-} \frac{1}{x \cdot \cos(x^3)}$

[twin meteor]

tobi

[peak shard]

constants can be pulled outside of the limit

[short kraken]

oh thats what u meant

so the idea is that as x approaches 0 from the left f(x) approachs neg inf here

[peak shard]

yeah

what would happen if x goes to 0+?

[short kraken]

pos inf?

[peak shard]

yeah

the cosine term here doesnt matter because cos(0) = 1 and cosine is a continues function

I can recommend trying to understand the epsilon delta definition of a limit

[short kraken]

yea its a lot

[peak shard]

this helped me with intuition

[short kraken]

i need to get to it

so we can say our limit is neg inf

[peak shard]

maybe try to proof:
$\lim_{x \to 4} (5x - 7) = 13$

[twin meteor]

tobi

[short kraken]

well as x gets closer and closer to 4 we get closer and closer to 13 as our solution

idk how to do proofs yet

[peak shard]

[
\lim_{{x \to c}} f(x) = L
]

means that for every (\epsilon > 0), there exists a (\delta > 0) such that for every (x),

[
0 < |x - c| < \delta \implies |f(x) - L| < \epsilon
]

[twin meteor]

tobi

[peak shard]

thats the definition of "lim"

and this definition shows that you never plug in the value c here

but maybe thats to much information if you have never seen this before

[short kraken]

i have seen it before i just dont understand whats it trying to communicate to me and why i should care about it

epsilon is just the y value right

this i get

[peak shard]

yes and if your function is not continues you couldnt shrink the red border as much as you want

if the green gets very close

if its close to 1

[short kraken]

why is that problematic

[peak shard]

if thats the case the limit does not exist

[short kraken]

it never shrinking means we wont reach our limit?

right

[peak shard]

we get very close to c (x axis)

and look at the y axis and look at what value we approach

in this case we cant approach anything because the function is not continues

https://youtu.be/kfF40MiS7zA?si=SOFJCIry-H_1vhYm&t=501

[short kraken]

hospitals rule 💔 '

[peak shard]

the timestamp is good

[short kraken]

ok so ill watch the video but i just have some questions

[peak shard]

sure

[short kraken]

so for problems like these

i never actually plug in 0

i just want to know whats happening around 0

[peak shard]

yes and with around 0 you mean very very close to 0

[short kraken]

yep

from whatever direction they tell me

[peak shard]

yeah

[short kraken]

BUT

if i can plug in 0 without getting 0/0 or #/0

is that my limit

?

[peak shard]

well sometimes

for polynomials yes

ln(0) is also undefinied for example

[short kraken]

well here i just get 12

e^0 is 1

[peak shard]

you are missing something here

what happens with -2/x as x goes to 0 from the left

[peak shard]

yes thats correct but thats not what the limit is equal to here

[short kraken]

it gets bigger

[peak shard]

$\lim_{x \to 0^-} 12 e^{-2/x} = 12 \lim_{x \to 0^-} e^{-2/x} = 12 \exp(\lim_{x \to 0^-} -2/x)$

[short kraken]

nvm

smaller

[twin meteor]

tobi

[peak shard]

-2/x goes to +inf as x goes to 0 from the left

[peak shard]

do you know what I did here?

[short kraken]

you isolated it because its the value that actually matters?

[peak shard]

on the first step I pulled 12 outside of the limit

[short kraken]

right

because we want to know what -2/x is doing and the 12 doesnt matter

[peak shard]

yes and know

we want to know what $e^{-2/x}$ is doing with x around 0

[twin meteor]

tobi

[peak shard]

the exponential function is continues tho, so we can interchange limit and function

thats what I did in the 2nd step

[short kraken]

got it

[short kraken]

so it approaches positive infinity

[peak shard]

yes nice

[short kraken]

so thats my solution?

[peak shard]

yeah

[short kraken]

so when in doubt plug in values around 0 to see whats happening

in other words, make a table

[peak shard]

for not very crazy functions this will work yes

[short kraken]

awesome thanks

[peak shard]

you could have functions that oscillate infinitely often next to the value that you are looking at

[short kraken]

then i would have to plug in values from the unit circle?

[peak shard]

this is what I mean with oscillate infinitely often

[short kraken]

ah

[peak shard]

or this function

[vocal sleet]

@short kraken Has your question been resolved?

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[sharp laurel]

i need help with a math question

[vast shale]

Ask

[sharp laurel]

i know you need a calculator to do this but i dont know the step

[vocal sleet]

@sharp laurel Has your question been resolved?

[sharp laurel]

no

[calm light]

which step specifically?

[sharp laurel]

well i need to find the linear regression

[calm light]

alright

[sharp laurel]

using that table but

i am suppose to type this table in the calculator to find the regression but i dont know how

[calm light]

what calculator are you using?

[sharp laurel]

Texas Instruments TI-36X Pro

[calm light]

idk if that calculator can even do regression

isnt it a scientific calculator?

[sharp laurel]

yeh thats the only calculator i have

i didnt know you need a specific one

[calm light]

can you use desmos?

[sharp laurel]

yea sure

[calm light]

so in desmos you can type "table"

and it'll make a table

you can input all your data in that table

and then type y1 ~ mx1 + b

and it'll give you the values for m and b

as well as the correlation coefficient

[sharp laurel]

ok ok

oh wait

i got it

thanks

i found the decimals

now i basically have to round them up

by putting it into an equation first

thanks bro your a big help

[calm light]

np little bro

!done

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[sharp laurel]

https://tenor.com/view/شكرًا-لك-gif-15712474265662603866

.close

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[opal violet]

Linear Algebra. Doesnt vector AB mean adding Vectors A and B, why are they subtracting it for vector V.

[opal violet]

Also I don't really get the question either

If it says that Vector V is Vector AB, wouldn't they be the same thing

Or am I just dumb

[heavy yoke]

vector AB points from point A to point B. vector v points from the origin to (2,5). they are mathematically equal because they have the same magnitude and direction

you can find vector AB = B - A

[vocal sleet]

@opal violet Has your question been resolved?

[opal violet]

Why can't u just do A + B?

Instead of B - A?

[vocal sleet]

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[amber badger]

How am i not getting this right? I swear i got it i can’t understand why its wrong

[heavy yoke]

pay attention to which divide bar is larger

[amber badger]

oh my god i’m so fucking stupid

😭😭

thank you

!close

how do i close

.close

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[short kraken]

this is just mean

What am I supposed to plug into s(a)

[copper crypt]

The formula for s(t) but with a instead of t

[short kraken]

(-4.9a^2+250) - (-4.9t^2+250)/a-t

(-4.9a^2 + 4.9t^2) /(a-t)

[copper crypt]

So far so good

[short kraken]

4.9(-a^2 + t^2)/(a-t)

Or
4.9(t^2 -a^2)/(a-t)

4.9(t-a)(t+a)/(a-t)

-4.9(t-a)(t+a)/(t-a)

Cancel (t-a)

-4.9(t+a)

-4.9t - 4.9a = 0

-4.9t = 4.9a

t = -a

??? Is this making sense

sorry I'm on mobile lmao

[lone linden]

This is fine so far

[lone linden]

Why are you setting it equal to zero

[short kraken]

Because I dont know what else to do

[short kraken]

I dont know what "the limit of the function as t approaches a" means

I dont know what a is

So I have no idea what they're asking me to find

[lone linden]

It represents the value the function approaches as t approaches a

So you can do direct substitution now

Substitute t=a

[short kraken]

-4.9(2a)?

so -9.8a

[lone linden]

t approaches a, not a approaches t

[short kraken]

-9.8t?

Hm now what

I sub in -9.8t?

[lone linden]

What does “hit the ground” tell you

[short kraken]

Nvm

[short kraken]

Does that mean "approaching"

[lone linden]

It’s talking about the specific instant where the object reaches the ground

[short kraken]

right

what's that telling me to do?

[lone linden]

When you hit the ground, what’s the height?

[short kraken]

oh 250

[lone linden]

…

What’s the height at the ground?

[short kraken]

0?

[lone linden]

Yeah

[short kraken]

so I need s(t) to equal 0

a t value that will make it 0 I mean

[lone linden]

Yeah

[short kraken]

So sqrt51?

[lone linden]

,w 250/4.9

[twin meteor]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[lone linden]

Don’t round during intermediate steps

But you have the right idea

[short kraken]

So t = sqrt(250/4.9)

So that's my answer for the first question

[short kraken]

Oops forgot this

I plug t into here right

[lone linden]

Yeah that’s for the velocity

[short kraken]

ohhh

so this is the first answer

The then we do -9.8*(sqrt259/4.9) for the velocity

[lone linden]

Typo but yeah

[short kraken]

thank you

[lone linden]

Anything else?

[short kraken]

Im good for now

.close

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[vocal sleet]

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[naive oak]

how would I set up an infinite sum to solve this

(ik its been like 3 minutes but I really gotta go soon 😭 so <@&286206848099549185> )

hope yall dont mind 😶‍🌫️

[gentle thicket]

what have you tried?

[naive oak]

i think its more just I have no clue how to approach it

the procedure they show for how money ccumulates over time its a weird process, one which depends on the previous outcome after a combination of operations

so its like a nested sort of summation

and I just have no clue how to make it something which can easily be calculated out to infinity

(ie a geometric sum)

[gentle thicket]

i don't think it forms a nice geometric series

[naive oak]

ye exactly

so idk what to do 💀

[gentle thicket]

could be potential AGP or some combination of something else

[naive oak]

agp meaning?

[gentle thicket]

arithmetico-geometric series

[naive oak]

gotchu

what wld u think wld be the best course of action then

wait, could it be I could set it up like a growth rate function

like e^kt

[gentle thicket]

I think you should start by calculating some of the amounts given to charity
then observe the pattern and try to generate its general term

[naive oak]

gotchu

ill do that then

[gentle thicket]

this always helps in any sequence and series question

[naive oak]

yeye

wait, is there a common ratio of 0.945?

[gentle thicket]

okay I thinK I'm not smart enough

it forms a geometric series

[naive oak]

LUL

yea cause I just realized the addition can be interpreted as a 1.05 multiplication

and then multiply that by 0.9

for the amount withdrawn

and u get that 0.945 ratio

aight tyyy

.close

[vocal sleet]

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[vocal sleet]

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[vast coral]

Hi, can anyone explain how 4 and 5 are done?

<@&286206848099549185>

[simple terrace]

yo @vast coral

for the 5th one you could use product rule

[vast coral]

The uv’ + vu’? Idk how to apply it 😦

[simple terrace]

uv' + vu' is for the derivatives

for the antiderivative the uv rule is different

[vast coral]

Ohh

[simple terrace]

it is the (first function multiplied by the integral of the second function) minus {Integral of[(dervative of first function)*(integral of second function)]}

sounds complicated

in terms of u and v

[vast coral]

Wait so u = e^2x? And v 2xe^2x?

[simple terrace]

it's u ∫v - ∫ (u' ∫v)

[simple terrace]

wait

they asked the antiderivative of smtg else though, check

they asked the antiderivative of

x * e^2x

[vast coral]

Ye

[simple terrace]

Yep

do you know abt ILATES

[vast coral]

No 😭

[simple terrace]

Aight listen

what're the two terms here?

[vast coral]

V and U?

[simple terrace]

yes but from this , there are two terms in the product

what are those

[vast coral]

X * e^2x and 2xe^2x?

[simple terrace]

no no bud

see it's like

the question is trying to confuse you i think

the first half of the question is basically useless

[vast coral]

omgg do u mean X and e^2x

[simple terrace]

just to confuse you i think

[simple terrace]

Yasss boi you get it

[vast coral]

[simple terrace]

now, we must choose u and v right?

[vast coral]

OHH

[simple terrace]

for uv rule in differentiation

[vast coral]

Ye

[simple terrace]

the order deosn't matter

but for antiderivative

[vast coral]

So u=x and v=e^2x?

[simple terrace]

there is an order

[simple terrace]

yess but you guessed it

[vast coral]

ohh what is the rule for order?

[simple terrace]

if the question was ∫ e^2x * x

would you take e^2x as u?

no bcuz of the order

[vast coral]

Prob not

[simple terrace]

I - inverse trig
L- Logarithmic
A - Algebraic
T - Trigonometric
E - Exponential

so which ever comes first, you take THAT as u

you get it?

[vast coral]

Ohhh yep

E is at end so v

[simple terrace]

so here ∫ e^2x * x

x is algebraic and e^2x is exponential

[vast coral]

Ahh I seee

[simple terrace]

so whihc becomes u?

[vast coral]

X

[simple terrace]

YEP

got it my boi

now just simply substitute

[vast coral]

Nice teaching

[simple terrace]

u ∫v - ∫ (u' ∫v)

u=x and v=e^2x

[vast coral]

Ye

[simple terrace]

integral of e^2x is?

[vast coral]

1/2 e ^2x?

[simple terrace]

derivative of x is?

[simple terrace]

YAS

[vast coral]

1?

[simple terrace]

perfect

[vast coral]

Nicee

[simple terrace]

so it becomes ?

[vast coral]

Just 1/2e*2x

[simple terrace]

(1/2 e ^2x )- ∫ (1* 1/2 e ^2x) dx

[vast coral]

Yeppp

[simple terrace]

or simply

(1/2 e ^2x )- ∫ (1/2 e ^2x) dx

∫ (1/2 e ^2x) dx is ?

1/2 ∫ ( e ^2x) dx

[vast coral]

1/2e^2x x 1/2

So 1.4?

1/4*

[simple terrace]

YEP

[vast coral]

oooo

[simple terrace]

1/4 ( e ^2x)

the final would be

(1/2 e ^2x ) - 1/4 ( e ^2x)

[vast coral]

Ohhh that makes sm sense now

Ty

[simple terrace]

what's the e option

[simple terrace]

[vast coral]

Just that

[simple terrace]

yayy??

LESGO

[vast coral]

:DD

[simple terrace]

cheers 🍷

[vast coral]

Cheers, ty!

[vocal sleet]

@vast coral Has your question been resolved?

[vocal sleet]

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[shut osprey]

.close

[vocal sleet]

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[steep inlet]

Do i just leave it at "65" or should i go for "65.009" (shown in the calculator) if yes, how?

[oak magnet]

Are you supposed to go with decimals or not ?

[steep inlet]

Must be decimals

Ig

[oak magnet]

Then you can continue till you get that 9

[steep inlet]

So i add both 0 to my answer and the 5?

But if i do that, I'll get .009 first than >503

A number greater than 503

[oak magnet]

Well, ig you can stop at 6.5 tho

[steep inlet]

65

[oak magnet]

65 ?

[steep inlet]

Yeah, i got 65

[oak magnet]

65*50.3 ?

[steep inlet]

It says 65.009

[oak magnet]

Ive only read that you write 327 divided by 50.3 in green

[steep inlet]

Oh

[oak magnet]

Then if it is 3270 indeed

[steep inlet]

Thanks for noticing that😭

It's 6.5

Thankss

.close 🫶

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[kind light]

m^2-n^2=2023^2 where m,n is a positive integer, how do i find the smallest m

[kind light]

like ik its (m+n)(m-n) but im not sure how to pick factors of 2023^2 such that m is the smallest

[inner osprey]

by "smallest" do you mean "closest to 0" or "most negative"

[rare swift]

Wish it was smallest n

[kind light]

fixed it >:(

[rare swift]

Should be Closest to 0

[inner osprey]

ty

[sly sierra]

did you find the prime factorization of 2023?

[thin vale]

hello sir

[kind light]

7×17^2

confused it with 2024 :(

[atomic pine]

you should factor (m^2-n^2)

[inner osprey]

so you'll have to solve m + n = A and m - n = B where AB = 2023^2

[kind light]

(m-n)(m+n)

[inner osprey]

notice that solving this, you get that m = (A + B)/2 and n = (A - B)/2

good so far?

[kind light]

yep

[sly sierra]

@thin vale AB sighted

[inner osprey]

so true

[thin vale]

💿

[inner osprey]

WLOG i will say m > n and A > B

so to minimize m, we have that A + B is as small as possible

can you think of when this would be true?

[kind light]

sr

when A and B are the closest but not equal?