help-33

[rapid spire]

Did u get 2 equations?

[thin bobcat]

Yea

[rapid spire]

Now subtract one from the other using elimination

[thin bobcat]

Do you mean collect the like terms ?

[rapid spire]

Wait

[thin bobcat]

8a + b - a - b ?

[rapid spire]

U got number 1 and number 2 right?

So now subtract 1 from 2

That would give u 7a

[thin bobcat]

I got one question why are we subtracting it ?

[rapid spire]

So that we can find a and b

Show me your equations

[thin bobcat]

Ok got it

[rapid spire]

It should be 7a=42

7a left hand side

[thin bobcat]

Ok

[rapid spire]

You can find b now by substituting the value of a in the 1st equation

[thin bobcat]

Ok bro makes sense, we sub for a (6)and sub the x for 1?

Either equation will work right ?

[rapid spire]

yh

[rapid spire]

6 for a

[thin bobcat]

So I can replace the x either 1 or 2 ?

[rapid spire]

yh u would get the same answer

[thin bobcat]

6 for a ?

[rapid spire]

i meant for a

[thin bobcat]

Ok

Thank you

[rapid spire]

you're welcome

[thin bobcat]

👍

[elfin stone]

@thin bobcat Do you have anymore questions?

[thin bobcat]

Yes

[calm harbor]

.solved

[marsh citrus]

Channel closed

Closed by @calm harbor

Use .reopen if this was a mistake.

[calm harbor]

.reopen

[marsh citrus]

✅

[thin bobcat]

How do you do 11

I don’t get it

What does this general rule mean ?

[calm harbor]

Do you still remember remainder theorem?

[thin bobcat]

Yes

[calm harbor]

What is the context of it?

[thin bobcat]

To find the remainder without doing the long division?

[calm harbor]

yes

So for any polynomial f(x) divided by (x-k), f(k) will be its remainder, right?

[thin bobcat]

Right

[calm harbor]

Right, so how do we determine whether (x-a) or (x+a) are the factors of $x^n - a^n$ or $x^n + a^n$

[elfin berry]

🌙 ЅκψΑиdΝιɡħτ

[calm harbor]

This is the first question

@thin bobcat

[thin bobcat]

Through the remainder Therom?

[calm harbor]

Yes, what should you do specifically?

[thin bobcat]

I don’t get it

[calm harbor]

What does remainder theorem says?

Pretend that I'm a friend of yours and I've never heard of this term before, how would you explain this technique to me?

[thin bobcat]

Oh ok, you must get one of your factor (x+a), x=-a then multiply with the coefficient and add/subtract

[calm harbor]

okay, but what does the result represent after we do all these work?

[thin bobcat]

We can find the other factors and the remainder

[calm harbor]

We can find the other factors and **the remainder **

Fantastic, now we have the common ground

[thin bobcat]

Ok

[calm harbor]

So basically you plug a value in and check if it has a result, if no then the divisor is the factor of the polynomial

[thin bobcat]

So my sender would be we would use the remainder therom the factors and remainder

[calm harbor]

For example Let's take $x^n - a^n$ and $x-a$ for example. When x = a is plugged into the polynomial, it is obvious that the result would be zero, meaning that the latter is the factor of the former

[thin bobcat]

Oh ok

[elfin berry]

🌙 ЅκψΑиdΝιɡħτ

[calm harbor]

Do you agree with that?

Remainder = 0 >> factor

[thin bobcat]

I understand it, it’s the very first step before doing the remainder therom

You must find the 0

[calm harbor]

Alright, so can you tell me in what condition will (x+a) be the factor of x^n - a^n?

x+a, mb

[thin bobcat]

=0

[calm harbor]

no, I'm asking about the condition of n

[thin bobcat]

The n must be positive integer

[calm harbor]

odd or even?

[thin bobcat]

Odd

[calm harbor]

yes

why?

[thin bobcat]

When you subsitute x = -a in to your equation , you can get the result of 0 needed for the factored theorem ?

[marsh citrus]

@thin bobcat Has your question been resolved?

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[quick oyster]

Can someone please give me a starting point
Someone told me to consider triangle EOB but idk how to (O is the center of the circle)

[quick oyster]

Sorry it hasn’t been 15 mins but I have to go soon

<@&286206848099549185>

Where are you helpers

[wispy stump]

Start with symmetry observations

[quick oyster]

Yes I know that part

[wispy stump]

E has to be uniquely determined

[quick oyster]

But I need to use PoP

[wispy stump]

Whats the current status

[wispy stump]

Yep

[quick oyster]

Nothing

But idk how

[wispy stump]

You know that point b lies outside w

[quick oyster]

Yes I know that

[calm harbor]

I drew a cute diagram for that if anyone needs it

[wispy stump]

@quick oyster

[quick oyster]

YES I HAVE THE VASICS

I HAVE A DIAGRAM

[vital oracle]

there’s a meme for this

[calm harbor]

what's PoP?

[wispy stump]

Power of a point

[quick oyster]

I gotta go sleep

If you guys have hints please Dm

.close

[marsh citrus]

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[marsh citrus]

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[night relic]

its a very cute diagram :D

[night relic]

oops

.close

[marsh citrus]

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[solemn minnow]

can anyone help me?

[calm harbor]

[solemn minnow]

2018K = 2018 (mod 10000)

1009K = 1009 (mod 5000)

and then?

what should i do

[solemn minnow]

K= 1 (mod 5000)

[indigo nest]

...no

You wouldn't divide the base of the mod

[solemn minnow]

but

[indigo nest]

You didn't do it here

[solemn minnow]

GCD (10000,2) = 2

am i wrong?

[indigo nest]

Wait I'm stupid

[solemn minnow]

wwww

[indigo nest]

,calc 2018 * 5001

[elfin berry]

Result:

1.0092018e+7

[indigo nest]

Yeah I'm stupid I forgot how modular division works

[solemn minnow]

its okayy

[indigo nest]

Anyway I don't think you can do any better

You want the number so you want k = 1 mod 5000 such that 2018k is less than 10^9

[solemn minnow]

yes

[bleak meadow]

i don't think your initial system of congruences is correct, or at least you're missing one congruence

[solemn minnow]

hmmm

[bleak meadow]

unless you baked it into this one, which I don't think it is

consider that you also want the answer to be a multiple of 2018

[solemn minnow]

oh yea

you have solution?

[bleak meadow]

i think i have, but of course it would be better for you to work it out

(i'm not sure i'm right anyway)

[solemn minnow]

just show me your solution

[bleak meadow]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[solemn minnow]

lemme check

damnn

tell me the steps

[bleak meadow]

after forming your two congruences, you should probably find out how often the solutions repeat (by finding the distance between each repeat).
then, find out the smallest such solution, and figure out how many repeats you can fit within your size constraint

as for how often the solutions repeat, you're gonna have to do a little finicking with your congruences and perhaps use a property common to both mods (maybe the smallest of them...?)

[solemn minnow]

so i have to do that over and over again until i find the smallest solution?

[bleak meadow]

the smallest solution should be obvious

[solemn minnow]

what modulo should i use

[bleak meadow]

now that's the fun part

[bleak meadow]

^

[solemn minnow]

between 2018 and 10⁹?

[bleak meadow]

if you mean the new mod, yes

it's definitely between those two numbers

but you'll need to use the two original mods to find the new mod

[solemn minnow]

damnn

[bleak meadow]

I will say this: one of the original mods you should use is 10000, not 5000

[solemn minnow]

and then 100000?

[bleak meadow]

nope

the other mod is a 4-digit number

it's also given in the question statement and is the other mod I mentioned that you did not consider

[solemn minnow]

Is there a pattern in the congruence results?

[bleak meadow]

well, yes I suppose?

your new mod is exactly the distance between each solution

[solemn minnow]

waitt

so first, 2018K=2018 (mod 10000) am i right?

[bleak meadow]

I would rather say $N \equiv 2018 \pmod{10000}$

[elfin berry]

Céline

[bleak meadow]

and ofc i screw my LaTeX up lol

[solemn minnow]

wwww

[bleak meadow]

also, I have been informed there are more solutions than at first I thought

so let me recheck

[solemn minnow]

okii

how if N<10⁹/2018

[bleak meadow]

ok nvm I stick to my guns

[solemn minnow]

hmm

[bleak meadow]

we'll deal with that size constraint later

first, identify the second congruence

[bleak meadow]

this is the first

[solemn minnow]

first N=12018 right?

[bleak meadow]

nonono, this is already the first congruence

[solemn minnow]

second congruence, i should multiply by2?

idk how to find second congruence

[bleak meadow]

look at the question again

[solemn minnow]

N=2018+10000K (mod10000)

[bleak meadow]

what is the other condition that a number should have?

[bleak meadow]

you have only dealt with the last four digits

you have not dealt with one other condition

[solemn minnow]

first condition is 4-last digit 2018

N=2018K+2018(mod10000)?

[bleak meadow]

no

the mod itself should be different

[solemn minnow]

hmmm

10001?

[bleak meadow]

the mod is a 4-digit number

[solemn minnow]

3000

?

[bleak meadow]

how do you write the congruence for the multiple of a number?

[bleak meadow]

and no, not the other congruence here

first, do you know what a congruence is?

[solemn minnow]

a=b+kn

a= b (mod n)

[bleak meadow]

in other words, a number x is congruent to another number y, mod some N, if x divided by N leaves the same remainder as y divided by N

agreed?

[solemn minnow]

yeah agree

so, we use CRT?

[bleak meadow]

not yet

we need the other congruence, then we can CRT our way to victory

but first

what does being a multiple of another number mean?

if I take a number A that is a multiple of some other number B, what is the remainder of A/B?

[solemn minnow]

0?

[bleak meadow]

so in other words, $A \equiv 0 \pmod B$

[elfin berry]

Céline

[solemn minnow]

okayy

[bleak meadow]

see where I'm getting at now?

[solemn minnow]

yea i see

[bleak meadow]

so what's the second congruence?

[solemn minnow]

is it a multiple of 2018

[bleak meadow]

yeah, but how would you write it as a congruence like this?

[solemn minnow]

2018N= 0 (mod 2018)

?

[bleak meadow]

funny you should do that, because 2018 is ofc going to be 0 mod 2018, so you could have skipped writing it

[solemn minnow]

2018N+K = K (mod 2018)

[bleak meadow]

well, I said to skip writing something, and you added to it

the second congruence is $N \equiv 0 \pmod{2018}$

[elfin berry]

Céline

[bleak meadow]

that's it

[solemn minnow]

what should I do actually, is look for another modulo that has four digits?

[bleak meadow]

does 2018 not have four digits?

[solemn minnow]

yes

[bleak meadow]

exactly.

[solemn minnow]

from this we can conclude N multiple by 2018

[bleak meadow]

this is our second condition

now, we need to know our common modulus

and here you can bring out your CRT. find the common mod of 2018 and 10000

(the new mod is 8 digits long.)

[solemn minnow]

ohhh okay

so from N=2018(mod 10000) and N=0 (mod 2018) we use CRT right?

[bleak meadow]

you can use the CRT directly on the mods themselves.

[solemn minnow]

it is mod10090000?

[bleak meadow]

yes

[solemn minnow]

whats next?

[bleak meadow]

now, what's the smallest solution to the problem?

[solemn minnow]

so N=2018(mod10090000)?

[bleak meadow]

yes, sure

but we're interested in the smallest solution

[solemn minnow]

5000?

ehh

[bleak meadow]

5000 neither ends in 2018 nor is a multiple of 2018

[solemn minnow]

just 100090000-2018?

[bleak meadow]

no

what's the smallest multiple of 2018?

[solemn minnow]

ohhh

[bleak meadow]

🤔

,w 7018 mod 2018

[elfin berry]

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[bleak meadow]

(super direct hint: you're already looking at the smallest solution. in fact, you have seen it said multiple times throughout this conversation)

[solemn minnow]

umm

2018?

[bleak meadow]

finally!

2018 is a multiple of 2018, exactly (1 times 2018), and for obvious reasons it ends in 2018

so it fits the bill as our smallest solution ✅

[solemn minnow]

i forgot if 2018=2018(mod10090000)

[bleak meadow]

a number is always its own multiple, ma'am

[solemn minnow]

i see

[solemn minnow]

yeah youre right

[bleak meadow]

anyway

we now know our smallest solution

and we know the distance between solutions

[solemn minnow]

yes

[bleak meadow]

so we can express solutions in the form 2018 + 10090000k

and now, we will finally bring in our size constraint

[solemn minnow]

because N=2018(mod10090000) right?

[bleak meadow]

we know that we want whatever solution to be less than 10^9

[solemn minnow]

agree

yea

[bleak meadow]

and we know that we have a solution for every integer value of k

so now, we need to find the applicable range of values of k

set up an inequality to solve for k

[solemn minnow]

so K=9?

max of K

[bleak meadow]

too small

there are way more solutions than that

[solemn minnow]

ehh wait

2018+10090000K<10⁹?

[bleak meadow]

mhm

solve for k

[solemn minnow]

99?

10⁹-2018

and then divided by 10090000

[bleak meadow]

could have just put those two parts in the same sentence

but yes

[solemn minnow]

tysm @bleak meadow

[bleak meadow]

wait

[solemn minnow]

lmaoo im sorry

[bleak meadow]

so what's your final answer?

[solemn minnow]

my first solution was very far

hehe

[bleak meadow]

I meant your final answer

[solemn minnow]

ohh

[bleak meadow]

^

[solemn minnow]

99

[bleak meadow]

wrong

did you remember that 2018 is also a solution (corresponding to k = 0)?

[solemn minnow]

damnn

i forget

so all solution is 100?

[bleak meadow]

correct

curious, what grade is this question?

[solemn minnow]

idk

[bleak meadow]

what grade are you then, if I may ask?

[solemn minnow]

i think math olympiad

in my country

[solemn minnow]

im grade 11

[bleak meadow]

I see.

okay, so is there anything else you would like to ask?

[solemn minnow]

Is this a common question or do you think it needs further study?

[bleak meadow]

wdym?

[solemn minnow]

i mean

Do you think this is a normal problem or do you think it hard problem?

[bleak meadow]

can't it be both?

so basically you want to know how hard I think the problem is

[solemn minnow]

yes

hehe

[bleak meadow]

if you understand the constraints, and the CRT, then you can kill this in 10-15 minutes if you're meticulous and used my method

while guiding you through I was notified of another solution that could kill this faster, but it jumped quite a few hoops

the original problem you had with the question was that you formed the wrong congruences

(or straight up missed one)

so I would say, it's hard until you know exactly how to translate the words into math and apply the math you know

I mean, you seem to know the math well enough

except for the whole "number is a multiple of itself" thing, which is worth remembering

[solemn minnow]

ohh okay

[bleak meadow]

so overall, I would say, if you said this was an olympiad question, I would say this would be on the easier side

if this was homework though, then this was indeed quite a good mental workout

[solemn minnow]

umm okayy

i have last question, when should we look for another modulo?

[bleak meadow]

wdym

[solemn minnow]

like from this question

first i use mod10000

and then we should find the smallest modulo?

[bleak meadow]

no

we translated the constraints to a system of congruences

we aren't deliberately looking for another modulo (well, other than the CRT -> new mod part)

it all depends on the question

[solemn minnow]

i see now

So we need a lot of practice so that we know which modulo to use in certain questions?

[bleak meadow]

that, and know what multiples/divisors mean

[solemn minnow]

ahh okayy

tysm @bleak meadow

sorry if i look stupid solving this problem

.close

[marsh citrus]

Channel closed

Closed by @solemn minnow

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[marsh citrus]

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[dreamy hedge]

Hi, can anyone check my work on this exercise pls.

[marsh citrus]

Please don't occupy multiple help channels.

[dreamy hedge]

.end

.close

[marsh citrus]

Channel closed

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[wispy stump]

Correct

[marsh citrus]

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[full oyster]

Dont know where to start limits and derivatives

[bleak meadow]

do you have a question you're stuck with?

[full oyster]

i cant solve a basic equation which has limit

but to be particular heres the ques

[runic temple]

https://www.khanacademy.org

[rancid geode]

ds gives 0/0

consider factoring

[full oyster]

ya i did till there

wait

lemme factor

im getting x^2/x-3

[rancid geode]

good

so whats the answer

[full oyster]

-4 ?

[rancid geode]

cool

[full oyster]

damn

thats it ?

what

[rancid geode]

ye

,w lim as x-> 2 of (x^3 - 2x^2)/(x^2 - 5x + 6)

[full oyster]

it was easy

[elfin berry]

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[rancid geode]

boom

[full oyster]

wow

idk why does it look so hard

thanks a ton

.close

[marsh citrus]

Channel closed

Closed by @full oyster

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[marsh citrus]

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Send your question here to claim the channel.

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[bronze sparrow]

can someone help me check if my answer is correct

[indigo nest]

Sure

[blazing dew]

Send it here

[bronze sparrow]

it mention that i shouldn't use quadratic formula

[bleak meadow]

think you slipped up here

[bronze sparrow]

what the issue?

[wanton sphinx]

plus minus should be on the square root

not the 1

[bronze sparrow]

is there a difference?

[bleak meadow]

yes

big difference

[bronze sparrow]

damn

when i write i was like " whatever, it should be the same"

[bleak meadow]

you have no justification to \pm the 1 here

[glossy flint]

How can it be the same??

[bronze sparrow]

now think about it

yeah it is not multiplication

it is difference

so the answer is x=2.32 or x=-0.32?

[bleak meadow]

first answer wrong sign

[bronze sparrow]

isnt it 1+1.32?

[bleak meadow]

actually both answers wrong sign i believe

[bleak meadow]

x+1 = 1.32

[indigo nest]

,calc sqrt(1.75)

[elfin berry]

Result:

1.3228756555323

[bronze sparrow]

ohhhhhhhhh

[bleak meadow]

when you shove that 1 over to the right it had better not be still positive

[bronze sparrow]

wait

x+1 no?

[bleak meadow]

uhuh

[bronze sparrow]

ohhhhhhhhh

yeah i got it

0.32 and -2.32 right?

thanks

btw

i got a question

[bronze sparrow]

in this part

can i do the squareroot first then divide by 4?

[bleak meadow]

no

[bronze sparrow]

or i must do the division first

[bleak meadow]

if you do the square root first, you'll also have to sqrt the 4

[bronze sparrow]

ohhhhh

[bleak meadow]

a quick reminder that you cannot selectively apply operations to some terms and not to others

[glossy flint]

You can do the square root and then you can divide by 2 (not 4)

But I don't recommend it, because you'll need the absolute value on x+1

[bronze sparrow]

yeah

thank guy

i think i got it

but i think you will see me again very soon

.close

[marsh citrus]

Channel closed

Closed by @bronze sparrow

Use .reopen if this was a mistake.

[bronze sparrow]

.reopen

[marsh citrus]

✅

[bronze sparrow]

i forgot to ask something

when should i apply the plus and minus?

is it that there is negative?

[bleak meadow]

when taking the square root of both sides

[bronze sparrow]

wdym

when putting square root?

[bleak meadow]

this step right here

[bronze sparrow]

so when i square root another side

[bleak meadow]

specifically

[bronze sparrow]

i have to put minus and plus?

ohhhh

[bleak meadow]

both sides

technically you rooted both sides

[bronze sparrow]

yeah true

thank

.

.close

[marsh citrus]

Channel closed

Closed by @bronze sparrow

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[bronze sparrow]

feel wrong

[indigo nest]

,w 3x^2 + 5x - 7 = 0

Oof

[tawdry rampart]

Blud the boxes are there so you write inside them 🥀

[indigo nest]

Boxes are a suggestion

[elfin berry]

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[bronze sparrow]

wtf is this

question not even that deep

[bronze sparrow]

if so

ohhhhhhhhhhh

[indigo nest]

You didn’t multiply properly

[bronze sparrow]

forgot to multiply

[indigo nest]

Yes

[indigo nest]

Was just me checking dw

[bronze sparrow]

how about now?

lmao

wait

forgot to square root

thank guy

.close

[marsh citrus]

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[bronze sparrow]

should be last question

[raw hawk]

,w 2x²-3x+2=0

[elfin berry]

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[bronze sparrow]

dont understand

what is the i for?

[raw hawk]

Where did the - sign go

Anyways I think you should try it again

[bronze sparrow]

isnt it here

[raw hawk]

Try using the quadratic formula then complete the square

And see if you get the same thing

[bronze sparrow]

it said that i shouldn't be using quadratic formula

[raw hawk]

You can use it to check though?

[bronze sparrow]

true

[raw hawk]

And I don't think it's a good idea to convert the fractions to decimals

Except in your answer

[bronze sparrow]

ohhh

ok

let me just redo it

[raw hawk]

i=√-1 btw

[bronze sparrow]

how do you square root fraction btw?

i still got the same answer

[raw hawk]

,, \sqrt{\frac{x}{y}}= \frac{\sqrt{x}}{\sqrt{y}}

[elfin berry]

David

[bronze sparrow]

ohhh

Still doesn't feel right

[amber birch]

in this step the -3/2 x suddenly disappeared

and you replaced it with -3/4

[bronze sparrow]

because -3/2 should be divide by 2 when got square

which is -3/4

[amber birch]

yes, so because you subtracted (3/4)^2

you need to add back (3/4)^2, so as to not change what you started with

[marsh citrus]

@bronze sparrow Has your question been resolved?

[bronze sparrow]

I'm confused

Do you have a worked example?

[amber birch]

https://www.youtube.com/watch?v=2MKigAgPZMQ
yeah

[marsh citrus]

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[tall pond]

I think the answer is D right?

[tall pond]

S2 has to be necessary and sufficient for S1 surely

because if it had any other factor then suddenly it's not sufficient for S1

like 1/15 or 1/14

yes they have factors of 2 and 5 but they have another factor

so I think I can exclude B, C, E from being solutions for sure

and then for S3 and S1 i just test the contrapositive

wait is S2 not the opposite of S3?

so then the statements are basically S1, S2, Not S2

[devout mauve]

there is a slight problem with numbers like 2/1

[tall pond]

and we know if not S2 then not S1 is true because s2 is necessary for s1

[spark otter]

no...

[devout mauve]

imo S2 is not well formulated

[tall pond]

oh

[spark otter]

imo S3 is also not well formulated, unless voluntary

e.g is 10 a power of 2 or a power of 5?

[devout mauve]

fair

[tall pond]

10 = 10/1 = 10/2^0 = 10/5^0?

[devout mauve]

q=10 has a factor 10 which is not a power of 2 or 5

[spark otter]

i'm talking about S3, so it's the denominator

[tall pond]

thats not the denominator

[devout mauve]

1/10

q=10

[tall pond]

1/10

10 = 5*2

?

[devout mauve]

but 10 itself is not a power of 2 or 5

and it is a factor of q

[spark otter]

and so, S3 is correct

[tall pond]

it has a power of 2 and power of 5 as factors though

i think thats what the question means

[devout mauve]

but thats not what S3 says

[spark otter]

but S2 is correct too

[devout mauve]

all around, throw the question away

you cant do logic with imprecise statements

[tall pond]

oh wait

the question is problematic then

i should probably change the question to say it has no factors that aren't a power of 2 or 5 other than itself?

[devout mauve]

is this your own question?

[glass quiver]

just say prime factors

[spark otter]

let's say, to fix the question, that S2 : "The only possible prime factors of q are 2 and 5"

and we don't change S3

[tall pond]

if im being honest

yeah i tried to make a question

[spark otter]

then S1 ≡ S2 and -S3 => S2

[devout mauve]

no S1 and S2 are not equiv

2/1

[spark otter]

here

[devout mauve]

ok thats how you meant it

vacuous truths my beloved

[ashen fog]

PayPal

[spark otter]

S2 isn't equivalent to "the only possible prime factors...", in your statement you're implying q has exactly those as factors

[tall pond]

2 or 5

mb

[spark otter]

and so 3/2 for example doesn't verify S2 as 5 isn't a factor

[spark otter]

makes even less sense

"the only prime factors of q are 2 or 5"

[tall pond]

?

[spark otter]

and again, 2/1 doesn't verify S2 because it has neither

[devout mauve]

why not just rewrite it properly and say "if z is a prime factor of q, then z is 2 or 5"

aka \forall z dividing q, z prime, z in {2,5}

and S3 is \exists z dividing q, z prime, z not in {2,5}

[spark otter]

(and in that case A is correct)

[devout mauve]

(the benefit with writing it like this is that you can easily negate statements like this by just swapping all quantifiers etc)

no

S3 is fucked up

[tall pond]

pls tell me it isnt fucked this time

[devout mauve]

yes

[marsh citrus]

@tall pond Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[main fulcrum]

is this clear

its x-->infinity

[fickle vessel]

$$\lim_{x \to \infty} \frac{6x - 10}{20 - x}$$

[elfin berry]

@fickle vessel

[main fulcrum]

i dont understand what to do with infinity

[umbral oasis]

Divide the numerator and denominator by x

[fickle vessel]

Have you done any work so far?, any attempt at simplifying the fraction or anything?

[umbral oasis]

Point is that the -10 and 20 in the numerator and denominator don't matter

[main fulcrum]

no

[umbral oasis]

So you can think of it as 6x/(-x)

[main fulcrum]

whats the logic behind it

[umbral oasis]

.

[main fulcrum]

why does -10 and 20 not matter

[umbral oasis]

.

[balmy shadow]

Because the numbers are vanishingly small

[fickle vessel]

As x goes to infinity 6x >>> 10, and 6x - 10 becomes approximately equal to 6x

Think about it this way

[umbral oasis]

Another way to see it would be that 6x and -x grow very large in absolute value compared to -10,20 as x->infty

[wispy stump]

@main fulcrum 1/infinity = ?

[main fulcrum]

so its so small that it wont even make a difference?

[fickle vessel]

If I tell you, your glass of water has 6.022 * 10^60 molecules of water, but when we actually count them, it comes out to be 6.022 * 10^60 + 10, would you care about those extra 10?

[main fulcrum]

nah

[fickle vessel]

Exactly

[main fulcrum]

i see

[fickle vessel]

As numbers approach infinity

the smaller numbers lose significance

[main fulcrum]

so we just divide the coefficients ?

[fickle vessel]

in case of addition and subtraction at least

Yes

Well

[main fulcrum]

so just -6

[fickle vessel]

Yup

[main fulcrum]

ahh

I see

Thanks to everyone for helping

[fickle vessel]

You're welcome

[main fulcrum]

.close

[fickle vessel]

!done

[marsh citrus]

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

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[marsh citrus]

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[full oyster]

when i end up with 0/0 while dealing with limits how do i know what to do next so i dont end w zero

[stoic saddle]

it really really really depends.

[fervent rampart]

that's generally a sign that you need to do some algebra to get something on the numerator and denominator of the fraction to cancel. what sort of algebra is very context dependent

[stoic saddle]

the general idea is to factorize numerator & denominator, locate the factors which are causing the zero, and cancel them out or otherwise get rid of them somehow

(by legal means, which doesnt mean just poofing them out of existence!)

show what limit(s) you're dealing with.

[full oyster]

sometimes im able to see through the equation and i just factorise them (if they can be factorised) but manier times im left clueless about what to do next

[stoic saddle]

show some of the limits that left you clueless

[stoic saddle]

this looks factorizable to me.

[full oyster]

what the indian guy did was made the 81 into 9^2 and made the x^4 in whole square

[stoic saddle]

if you confirmed you got 0/0 then you know for free that (x-3) is a factor of num and denom.

x^4 - 81 can be factorized by applying the difference of squares identity twice

you know the difference of squares identity, yes?

[full oyster]

yes

so like

every equation can be factorised ?

there has to be some

that cannot be factorised

[fickle vessel]

Not every, but this one can be

[stoic saddle]

these are not "equations"

[fickle vessel]

Also true

[full oyster]

what shall i do when they cannot be

[stoic saddle]

anyway, when it comes to specifically rational functions, factorization can p much always help you.

you may encounter shit that got roots in it.

[full oyster]

yes

[stoic saddle]

or shit with trig functions in it.

[full oyster]

roots

yes trigs

what to do during that time

[stoic saddle]

a common trick for dealing with rooty shit is multiplication by the conjugate.

[full oyster]

specially the roots

[full oyster]

thats it ?

just multiply with conjugate of denom

[stoic saddle]

multiply with conjugate of the problematic part

wherever it might be

maybe you could show an example problem with rooty shit that i could explain in detail

[full oyster]

wait

i guess i forgot the timestamp where he solved the root one

ill notify u when such type of rooty ques occur next time

[stoic saddle]

do you have anything else to ask right now

[full oyster]

nope im done

.close

[marsh citrus]

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[marsh citrus]

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[lavish pelican]

how to do this integration

[lavish pelican]

i took sqrt cos theta common

and got it to
(1+t^2)/(sqrt t+1)^8 from 0 to infinity

[noble geyser]

huh what did you do??

[lavish pelican]

tan theta = t

[noble geyser]

mhm

[lavish pelican]

then thats what i got...

[main idol]

did you try u = sqrt(t)

[lavish pelican]

ok that will give 2(u+u^5)/(u+1)^8 from 0 to infinity

after that what to do

[main idol]

think of another substitution

[novel juniper]

Try factoring out $\sqrt{\sin(\theta)}$

[lavish pelican]

btw this is the original question, the msg im replying to is what i got it down to

[elfin berry]

wai

[lavish pelican]

i cant think of anything..u=tan^2 theta? but i feel like thats going in circles

[main idol]

i mean from here

yes don't go in circles

[lavish pelican]

isnt that the same as factoring out cos theta

[lavish pelican]

do i put u as 1/u?

[main idol]

no

[noble geyser]

i still dont understand how you got that

prob because im dumb

[main idol]

you have a polynomial on top and another polynomial at the bottom

[noble geyser]

yea that integration by parts no?

[main idol]

you can do a simple sub to get just powers of u

then you just use power rule to integrate

[lavish pelican]

u+1 = t?

[main idol]

yea try it

[lavish pelican]

bet bet

2[(t-1)+(t-1)^5)]t^8 from 1 to infinity

the t-1^5 still looks kinda tough

[noble geyser]

shouldnt it be t-1

[lavish pelican]

oh ye sorry

[noble geyser]

shouldnt this be integration by parts now

?

oh wait no you can just solve it

[lavish pelican]

yeah nvm i got it from here we can just split and for the second integral u-->1/u

.close

[marsh citrus]

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[elfin badger]

does anyone know how to round these numbers to 1d.p

and keep the table?

[jagged relic]

Like an Excel command?

[fringe nacelle]

do you actually want th enumbers rounded or just have the display rounded

[elfin badger]

i figured it out

[fringe nacelle]

then .close , glad you figurd it out

[elfin badger]

.close

[marsh citrus]

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[vernal mica]

@knotty trellis the other channel closed?

[main idol]

channels timeout. can you just repost your question and progress here

[vernal mica]

this was my progress and i was correcting my errors

[knotty trellis]

Do u also have the corrected version or are u working on that?

[vernal mica]

im nearly done

i forgot to add .[C] near the end i fixed it tho

[knotty trellis]

Now its pretty solid, id just add "Note that projc(a) + projc(b) = projc(a+b)" somewhere and the arrow u drew shpuld really be pointing from projcb, not the |c|

In general, dont be afraid to use words in proofs

Proofs dont have to be only equations, in fact, there is usually more text than equations

But anyway, your proof is pretty solid and i think ur prof should understand it

[vernal mica]

okay i added the note thanks

[knotty trellis]

Np, enjoy ur prom

[vernal mica]

can you check the first problem for me as well. part 1 of the problem gave me a bit of hassle but part 2 was simple enough but itd be good for you to check so i dont make any mistakes

[knotty trellis]

Ill prolly head to sleep now, but if u just post it, someone else will definitely check it soon

[vernal mica]

its quick dw way easier than the second part

[vernal mica]

thank you i will

Problem 1. Consider a particle in one dimension, whose position as function
of time t is given by
x(t) = α t cos(ωt).

1. Give the dimensions of the constants α and ω.
2. Give the velocity v(t) and the acceleration a(t).

can anyone check if its right

[marsh citrus]

@vernal mica Has your question been resolved?

[marsh citrus]

@vernal mica Has your question been resolved?

[cunning fiber]

,w differentiate a t cos(bt) with respect to t

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[cunning fiber]

,w second derivative of a t cos(bt) with respect to t

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[vernal mica]

thank you

[marsh citrus]

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[loud temple]

Is this correct?

[queen sun]

No

Do you know about the rule for a triangle with points on a circle with one side being the diameter?

[loud temple]

I think

[loud temple]

Yh

[queen sun]

Let D be the centre of the circle

All the edges coming out of D are the same length, so all the points A, B, C can be on a circle

[marsh citrus]

@loud temple Has your question been resolved?

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[ancient thunder]

so like i dont understand this at all

[ancient thunder]

like i understand a bit but other than that i dont understand most of it like i feel slow

[static quarry]

can you paste a screenshot of whatever it is?

[ancient thunder]

its like a whole topic but its about the Sampling distribution of the sampling mean

and like Central limit theorem, Z-value etv

etc

[static quarry]

well can you single out a specific question you have?

help channels aren't really for teaching an entire lecture of material

[ancient thunder]

YEAH i know but its like

i literally cannot grasp it no matter how much videos i watch

id like to atleast have someone explain to me like the general idea

and not the entire lecture material if that makes sense

[static quarry]

like maybe tell us the first thing from the lecture or video that you're not sure about

[ancient thunder]

all of itttt...

[marsh citrus]

@ancient thunder Has your question been resolved?

[marsh citrus]

@ancient thunder Has your question been resolved?

[marsh citrus]

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[stable quail]

trying to decipher my teacher's solutions for implicit differentiation- im confused with how he goes from line 3 to line 4? specifically how he went from (1+y') to y' (1/2 (x+y)^-1/2).....

[stable quail]

like its kinda late so maybe i missed smth but huh???

[fringe nacelle]

he distributed it over the left factor

[stable quail]

oh wait. yeah the first term in line 4 is just the 1 from 1+y' ?

[fringe nacelle]

yup

[stable quail]

ok yes

htan kyoyu!!!!!!!

[fringe nacelle]

np

[stable quail]

.close

[marsh citrus]

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[pliant coral]

to find the linear speed of the bicycle I just need to find the linear speed of the back wheel right? and just ignore the front one

[pliant coral]

what I've done so far was use this formula

and got

nevermind

I did it wrong

how do I find displacement when its given in this weird form? because typically you just multiply by 2pi but it doesn't work in this case I believe

[night relic]

right now, you have the RPM of the pedals, and you want the RPM of the back wheel (the front wheel will have the same angular velocity, or RPM)

what can you use to find the RPM of the wheel?

[pliant coral]

mmm

[night relic]

~ microwaves

[pliant coral]

I'm trying to think

we have the RPM of the pedals

but I just don't see how we can use what we have to get the RPM of the wheels

[night relic]

well, what facts do we have relating the pedals and wheel?

[pliant coral]

we have the diameter

which gives us the radius

and the radius of the rear sprocket

and the RPM of the Cyclist

[night relic]

theres an additional piece of information: "Thus every time the cyclist makes one full revolution with the pedals," then how many revolutions does the wheel make?

[pliant coral]

4?

wait that doesn't sound right

somthing to do with this

I just can't figure out what

[night relic]

yeah exactly

what this sentence is saying is: 1 revolution of the pedals = 2 revolutions of the wheel

[pliant coral]

but 4/2 is 2

so isn't it saying 2 = 2 revolutions?

[night relic]

it's saying 1 pedal rev = 2 wheel revs

[pliant coral]

oh

okay

[night relic]

it also gives you the 4 and the 2 with the chain, so you can do it yourself

but then they also gave you the ratio, which is nice :)

anyways, what we have now is $$180 \frac{\text{pedal rotations}}{\text{min}}.$$ We want to multiply this by something to get $\frac{\text{wheel rotations}}{\text{min}}$

[elfin berry]

χασιβ ♥

[pliant coral]

2

[night relic]

then we can get the radians/minute, and use the radius to get linear displacement, like the example u showed

[pliant coral]

so

180*2

=

360

[night relic]

exactly!

so 360 rotations/min is how many radians/min

[pliant coral]

uhhh

I don't know what the conversion factor is

if there is one

oh wait

its 180pi right!?

[night relic]

i was just going off of this lol

assuming that's from your notes or something?

[pliant coral]

thats from our textbook

*free online textbook

okay

so it wouldnt be 180pi

it would be

360*2pi

which is

720pi

and then we take that

and multiply it by 9

and get 6480pi as our final awnser

[night relic]

this is what i got too :D

[pliant coral]

: D

yay

it was correct

and then to get ti miles/hour

is pretty simple

just multiply numorator by 12

and then 5280

and denominator by 60

and get

which simplifies to

6842880pi

and it was wrong : ((

wait I did that wrong and my textbook litteraly explains it : /

[night relic]

you had the right idea, just flipped

[pliant coral]

it said thats wrong too

: C

[night relic]

you are currently at $\frac{6; 480\pi}{63; 360} \frac{\text{miles}}{\text{minute}}$

[elfin berry]

χασιβ ♥

[night relic]

also, they want it to two decimal places, so probably an approximate answer

[pliant coral]

oh

so

I need to divided the denominator by 60

[pliant coral]

they didn't care about that on the first problem

[night relic]

interesting

[night relic]

yep, aka multiply the whole thing by 60

[night relic]

i would recommend writing all of these steps out, tedious as it may be, just to make sure nothing gets missed :)

[pliant coral]

mk

I've kinda done that on paper

[pliant coral]

fingers crossed

this is my last try on this problem

[night relic]

hmm im not getting the same thing

[pliant coral]

hmm

oh

I think I divded by 6

instead of 60

by accident

[night relic]

that would explain it lol, i was off by a factor of 10

[pliant coral]

so it should actually be this

[night relic]

looks good to me 🤞

[pliant coral]

OMG

FINNALLY

2 hours of my life

have been spent on this problem

thank you so much for helping me!

.close

[marsh citrus]

Channel closed

Closed by @pliant coral

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
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[prime scroll]

.claim

?claim

[proud ice]

Just say anything that doesn't start with a period

[prime scroll]

claim

[proud ice]

It's already yours to use

[prime scroll]

oh ok

it never say my name

[proud ice]

discord won't put periods in channel names

[prime scroll]

ok

[prime scroll]

can help me with alegbra

[proud ice]

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[proud ice]

But no, I cannot. Someone else will though. I must head out

[prime scroll]

Does anyone know about algebra?

[proud ice]

again, no need to ask if anyone knows. Just ask your question

[novel juniper]

what kind of algebra

[prime scroll]

uh

like

this is myf irst question

its simple for u i think

but not for me

[novel juniper]

ah, can you make the denominator the same on the left

[prime scroll]

i dont know how to solve it

[prime scroll]

at all

[novel juniper]

The thing is I have a class now, so I can't help, sorry