help-28

[honest lark]

thanks

@lofty vine thank u so much bro u taught me the radians mode thing

[lofty vine]

no prob

[honest lark]

it worked

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[plush mortar]

two solutions

[lofty vine]

(0,0) and (sqrt(27), sqrt(27))

[plush mortar]

yes

they are two semiricles

intersection

[lofty vine]

by symmetry of equations u can assume x = y then it becomes a quadratic

[lofty vine]

they look like ellipses

[plush mortar]

its not sqrt 27

acc to graph

[lofty vine]

ah

[plush mortar]

it is (0,0) and ( 3, -3) acc to desmos

[pliant dagger]

There are four solutions if you solve algebraically. I need to learn how to solve algrbraically because of exams

[lofty vine]

yeh u right i made a mistake

[plush mortar]

there are three intersections

[pliant dagger]

(0,0) (0,sqrt(27/2)), (sqrt(27/2), 0) and (3,3)

[plush mortar]

(0,0) and (3,3) are right

[lofty vine]

move the radicals to one side and square the equation

[plush mortar]

how did you got 4 solutions there are only 3

sry

4 are there

[lofty vine]

there are 4 on the graph

[plush mortar]

my mistake

[lofty vine]

np i made mistake as well

[pliant dagger]

x=0 solves the first equation

similarly y=0 solves the second

that's how you get (0,0)

[plush mortar]

ohh

yeah

i find smthng sussy that in second equation x is replaced by y with tespect to first eqn

is there something to it

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[magic glade]

Define the language SAT-2-CLAUSES:

SAT-2-CLAUSES = { ⟨𝜙⟩ | 𝜙 is a CNF formula with exactly two clauses, which has a satisfying assignment }.

Prove that SAT-2-CLAUSES ∈ P.

I get that the complexity is O(n*2^n) (n is the amount of unique variables). This is exponential, and thus not in P. However, my friend says it's O(2n) and thus is in P. What's the correct solution here?

[deep bolt]

why?

[magic glade]

wdym why

[deep bolt]

why should I define the language?

tell me why?

[tacit siren]

ain't nothing but a heartbreak

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@magic glade Has your question been resolved?

[lucid flower]

you don’t need to check all the different arrangements for assignments of truth values to the variables

and even if you literally did the naive thing it still wouldn’t be n*2^n

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[hollow python]

hi

[hollow python]

i need help

hello?

[hybrid zephyr]

Just ask

[hollow python]

ok so we have a number lets say we have N = 111...1 (n times)

or 22..2 or anything n times

[hybrid zephyr]

Mhm

[hybrid zephyr]

!da2a

[hollow python]

ok for the 1st example i found out that 1111...1 = 1+10^2+10^3+......+10^n

[full forum]

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/

[hybrid zephyr]

Ok

[hollow python]

so is there any explanation for that like a general rule that explains this

[hybrid zephyr]

Wdym general rule?

[hollow python]

like what if we have 222222...2 or 333...3

[hybrid zephyr]

Take 2(1111111.....)

[hollow python]

something we can apply for all cases or is it for only one

[torn jolt]

!original

[full forum]

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.

[hybrid zephyr]

So 2* that thing

[hollow python]

hmmmmmm noice noice

[hollow python]

soo is what i said true right?

[hybrid zephyr]

Yes

[hollow python]

lets say we have 11111...1 2010 times

so it will start from 1 to 10^2009

[hybrid zephyr]

Yes

[hollow python]

r u sure

[hybrid zephyr]

Yes

[bronze scaffold]

where ai is the ith digit

[hollow python]

btw the end is n-1 not n

[hybrid zephyr]

Yes

[hollow python]

what

[hybrid zephyr]

11 is also 10 + 1

[hollow python]

do u even understand what i say

[hybrid zephyr]

Yes

[hollow python]

bruh

like frfr

[hybrid zephyr]

n-1 only

[hollow python]

hmm ok ok

thx

for replying to my messages

have a nice day

[hybrid zephyr]

Np

[hollow python]

good luck

!close

uhhh ok ill go

[hybrid zephyr]

.close

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[little owl]

if general form is when everything =0

why come i cant just take away y from both sides to get -1/2x+6-y=0

(i know you cant have fractions in general form is that why?)

@viral jasper how do i make it not a fraction then?

what do i do to get rid of the fration

[viral jasper]

Multiply everything by 2

[little owl]

so i want to turn the fraction into 1?

in this case -1

[little owl]

so 2y=-1x+6

0=-1x+6+2y (final answer?)

[dull seal]

No

[little owl]

what did i do wrong

[dull seal]

6 * 2 is not 6

[little owl]

0=-1x+12+2y

[dull seal]

Better

But

Still a mistake

Sign error

[little owl]

any hints?

[dull seal]

Equation is y=-1/2x+6

Multiply by 2

2y=-x+12

[little owl]

oh just the 1?

[dull seal]

Now we subtract 2y from both sides

-x-2y+12=0

[little owl]

would it be fine if i answered like "0=-x+12-2y"

or does the order mater

[dull seal]

general form is Ax+By+C=0

So i would write like i said

[little owl]

how do i know what a and b is

[dull seal]

You compare directly

[little owl]

ohh based on what x and y is

[dull seal]

A=-1 here because -x is -1 * x

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[wheat garden]

if you multiply each length of a triangle by a positive number, will the new numbers still make a triangle?

[native stratus]

yes, since all you need is the triangle inequality

[umbral dome]

that will create a similar triangle

[viral jasper]

the same or different positive numbers?

[wheat garden]

same

[native stratus]

if c<b<a form a triangle then a<b+c, so kc<kb<ka also form a triangle since ka < k(b+c)

[wheat garden]

if it was a right angle, would the new lengths also make a right angle?

[native stratus]

yes

[viral jasper]

you'd get a similar triangle every time

[wheat garden]

oh ok

ty

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[hollow summit]

when i have abs(5x-40) < 1

[hollow summit]

i can put it to -39/5 < x < 41/5 right?

or 39/5 < x < 41/5

[torn jolt]

yeah

[hollow summit]

https://tenor.com/view/skeleton-gif-5530475079020271091

[twilit leaf]

explain your thinking

[sacred yarrow]

no

[torn jolt]

uh nvm

[hollow summit]

whenevr i have absvaluex < 1

i just put it to -1<x<1

[twilit leaf]

ok

keep going

[ocean cypress]

https://tenor.com/view/rippuwallettu-pepe-wallet-wallet-empty-rip-wallet-gif-23844969

[hollow summit]

thats it

and then i +40 each side and /5

[twilit leaf]

ok, what do you get on the left side

[hollow summit]

39/5

[twilit leaf]

yes

not -39/5

[hollow summit]

why is the left border open

[sacred yarrow]

It’s ok, it happens

[ocean cypress]

#help-28 message

[torn jolt]

👍

[hollow summit]

nvm i forgot to multiply 8 by 5

[devout valley]

Remember you need to set x = 39/5 and x = 41/5 in the original series, and see what happens there - you've only found the radius of convergence so far!

[umbral dome]

yeah the root or ratio test is always inconclusive at the end points

[hollow summit]

yup i just did 5(x-8) and i forgot to multoiply 8

oh but for the absolute value

abs(5x-40) < 1

ik i can do it by putting it as 1< (5x-40) < 1

but im not too sure why

[devout valley]

-1 as the left

[hollow summit]

ye

s

[devout valley]

And e.g. remember that when x is positive, |x| = x, so |x| < 1 is saying x < 1, but when x is negative, |x| = -x, so |x| < 1 is saying -x < 1, equivalently, x > -1, but you need both of those statements true

[devout valley]

(also the testing endpoints is pretty important - you know you definitely converge when |5(x - 8)| < 1, and you definitely diverge when |5(x - 8)| > 1, but you have absolutely no idea if you converge or diverge when |5(x - 8)| = 1 - see here, which is the reason they include 7.8 but not 8.2, after all!)

[hollow summit]

ahhhtysm!

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[ruby raft]

Can someone please explain this statement to me? "For two matrices A and B, each column of AB is a linear combination of the columns of A using weights from the corresponding column of B"

[ruby raft]

How I'm thinking it actually should be is each column of AB is a linear combination of the columns of B using weights from the corresponding row of A.. isn't that what we're doing?

[umbral dome]

\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} e & f \ g & h \end{bmatrix} = \begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix} = \begin{bmatrix} a \begin{pmatrix} e \ g \end{pmatrix} + b \begin{pmatrix} f \ h \end{pmatrix} & c \begin{pmatrix} e \ g \end{pmatrix} + d \begin{pmatrix} f \ h \end{pmatrix} \end{bmatrix}

[glossy valve]

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

[umbral dome]

messed that one up a bit, sorry

[rapid rain]

you mean $\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} e & f \ g & h \end{bmatrix} = \begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix} = \begin{bmatrix} e \begin{pmatrix} a \ c \end{pmatrix} + g \begin{pmatrix} b \ d \end{pmatrix} & f \begin{pmatrix} a \ c \end{pmatrix} + h \begin{pmatrix} b \ d \end{pmatrix} \end{bmatrix}$

[glossy valve]

rafilou2003

[umbral dome]

yeah

[rapid rain]

yep, columns of AB are weighted sums of columns of A

the weights are given by the corresponding column of B

[ruby raft]

okay when you list it out like this it makes sense ig i'm just having trouble gaining an intuitive understanding of this

[umbral dome]

if you multiply Ax where x is a column vector, then the resulting vector is a linear combination of the columns of a with the entries of x as weights:

then AB simply extends that to more columns, since each column of AB is the result of multiplying A by one of the column vectors of B

[ruby raft]

this makes more sense

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[dusty kayak]

What am I missin

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[viral onyx]

Can someone help me figure out what I should do here? Not sure what cot(A) should be, confused about the angle terminating in the 4th quadrant part.

[umbral dome]

the quadrant tells you what the sign of tan(A) should be

[viral onyx]

Quadrant 4 is positive cos and negative sin, so tan(a) is negative right?

[umbral dome]

yes

[viral onyx]

But then isn't cot(A)=-3/sqrt(7)?

That's not an option up there

Oh wait...the 1st one is equivalent to that

K, ty!

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[deep bolt]

Since, sec A is positive thus, A is in the 4th quadrant

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[drifting bronze]

What is the best formula for aproximating the circumference of an elipse? I need precision for my project

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[trail salmon]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[trail salmon]

what is wrong with khan academy

17/7 = 2.42

but it said 2.42 is wrong lol

unbelivable man

[dull seal]

No

17/7 is not 2.42

[trail salmon]

yea it is

[dull seal]

No

That is an approximation

[trail salmon]

2.42857...

[zenith thistle]

2.42857142857 this is the exact value

[trail salmon]

lol

[dull seal]

Is not the same

[trail salmon]

well you know its an acceptable answer

to give an aswer to three decimal places

cause its obvious

[dull seal]

Acceptable depends on the context

[trail salmon]

thats so dumb man

ur not being helfpul

🤦‍♂️

im just gonna use thewiseai and cheat now

[dull seal]

If the question asks you to lake an approximation then it is ok

[vagrant yoke]

can you write 17/7 as answer?

[dull seal]

If they want the exact solution

You let it as 17/7

You need to understand what = means

[trail salmon]

i love cheating with thewiseai

[zenith thistle]

yknow just typing 17/7 is easier than calculating the decimals

[trail salmon]

ik

[vagrant yoke]

so did you do it?

[zenith thistle]

then why waste time?

[hallow walrus]

thewiseai?

[trail salmon]

i need help

[hallow walrus]

are you talking about chatgpt?

[trail salmon]

with geometry

its chatgpt for math

its very accurate

[dull seal]

$= \neq \approx$

[glossy valve]

Samuel

[trail salmon]

until you get to real analysis

cause sometimes it spouts nonsense

[hallow walrus]

not at all

[trail salmon]

nope

its accurate up to like calculus probably

[hallow walrus]

no, it isn't.

[trail salmon]

yea it is

lol

[hallow walrus]

I've seen chatgpt literally type out 1+1=3 like it was nothing

!nogpt

[full forum]

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).

[dull seal]

What is your point?

[trail salmon]

so you are saying its not accurate yet u dont even know what it is

yet

im the one who uses it and has experience with it

is it your word against mine

the person who has used it before vs the person who does not know what it is

[zenith thistle]

ai is not trustworthy

[trail salmon]

nope its pretty accurate

u should give it a try

[dull seal]

This is not related with the question anymore

[trail salmon]

give it any precalculus calculus question it will probably help you out

[hallow walrus]

what makes you think I haven't used it?

[trail salmon]

cause you said thewiseai?

with a question mark

[dull seal]

For discuss this go to #discussion or #chill

[trail salmon]

indicating you dont know what it is

[hallow walrus]

because you wrote that...

[trail salmon]

why did u put the question mark at the end

[hallow walrus]

close this channel if there's no question

[low echo]

,calc 17/7

[glossy valve]

Result:

2.4285714285714

[zenith thistle]

the literal website says that its not very accurate

[trail salmon]

well

i never said chatgpt

i said thewiseai

[low echo]

hey after how long inactivity channel closes automatically ?

[trail salmon]

chatgpt is probably only good for like

english

GEOMETRY IS TOO HARD

[hallow walrus]

any AI that's not made specifically for math shouldn't be used for it

[trail salmon]

i see

[vagrant yoke]

show us this geometry problem

[hallow walrus]

yeah, and stop refusing the things people are saying to you on a server that's literally dedicated for math.

[trail salmon]

well

the reading comprehension is hard

[hallow walrus]

nobody will help you with that attitude lmao

[trail salmon]

its like rotation to -270 degrees reflected over point B translated up 5 units on the reflection x perpendicular to line segment ab to ac

its like reading chinese i dont understand any of it

and i dont really understand anything about rotation or how the shape is suppose to move

[zenith thistle]

what topic are u studying

in geometery

[trail salmon]

ok

sorry kheeri

high school geometry in khan academy

[hallow walrus]

!show

[full forum]

Show your work, and if possible, explain where you are stuck.

[hallow walrus]

first show the question^^

[trail salmon]

i dont understand the definitions

like the reading comphrension

it confuses me

i dont understand anything i cant visualize it

[drifting bronze]

Try drawing it

[trail salmon]

ok

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[twin wolf]

chat how do i solve for x

[grave elm]

is this a joke?

[twin wolf]

no

[grave elm]

holy fuck (25°)^3

[hybrid zephyr]

Its pretty simple ngl

[grave elm]

am i reading it correctly?

[hybrid zephyr]

It just looks complex

[twin wolf]

yes

[grave elm]

Then what the - is this supposed to mean

[twin wolf]

i think its +

[grave elm]

Anyway, square both sides twice, (120/5!)^256 = 1

[hybrid zephyr]

In first one

The numerator is 1+3/4

So 7/4

Denominator is 7/256x

7 cancel

[twin wolf]

is this under the double square root

[hybrid zephyr]

So 256x/4

[hybrid zephyr]

Yes

[torn jolt]

AYYYYY

You know what

[grave elm]

And this is just fancy (4x^2+256)^(1/4)

[torn jolt]

I discovered it just today

[hybrid zephyr]

Yes lol

[hybrid zephyr]

What

[grave elm]

why are we even solving roblox problems

[twin wolf]

wait

so i got

[torn jolt]

It is a robloxmhame

[twin wolf]

double square root

256x/4

is that right

for the left

ok it is

what do i do for the right

[grave elm]

5! = 120

so 120/5! = 1

1^256 = 1

sqrt(1) = 1

[twin wolf]

oh true ture

[grave elm]

(4x^2+256)/1 = 4x^2 + 256

so it's just (4x^2+256)^(1/4)

[twin wolf]

wt about the double square root

[hybrid zephyr]

Double square on both sides

[torn jolt]

(The quadratic equations were a pain in the ass)

[twin wolf]

$\sqrt{\sqrt{\frac{256x}{4}}}=\sqrt{\sqrt{\frac{4x^{2}+256}{1}}}$

[glossy valve]

water beam

[twin wolf]

is that it

[raw dirge]

answers -8 mate

[grave elm]

How was this even solved?

[hybrid zephyr]

Play roblox
Solve math
Tf us this solve roblox

[raw dirge]

cant believe i did this bs

[hybrid zephyr]

Nop x cant be negative

[grave elm]

What's the (25°)^3?

[hybrid zephyr]

25^0 lol

[raw dirge]

oh wait yeah im tweaking

[hybrid zephyr]

Ie 1

[grave elm]

oh it's 25^0

not 25 degrees

[hybrid zephyr]

Yea

No

[twin wolf]

LMAO

[grave elm]

25 degrees would be cursed

[twin wolf]

25 degrees

[hybrid zephyr]

What

[twin wolf]

ok chat i got the right answer which was x = 8

thanks gang

[hybrid zephyr]

U just cant add angles to numbers

[torn jolt]

That ain't degrees

[hybrid zephyr]

Its illegal

[torn jolt]

It is 25^0

[hybrid zephyr]

Ye

[twin wolf]

its roblox anything is possible 🔥

[hybrid zephyr]

Lol

[grave elm]

I can't believe people are solving math problems just because of roblox

[hybrid zephyr]

Lol

[paper acorn]

at least it is not calculus

[twin wolf]

on god i was bored and saw this game called math problems or something

so i was like

aight

[hybrid zephyr]

Math is meth

It's fun

[twin wolf]

anyways gg chat

[torn jolt]

What out for the matrix questions

[grave elm]

highly addictive and dangerous

[torn jolt]

Watch*

[hybrid zephyr]

Lol

[twin wolf]

there are matrices???

[torn jolt]

2nd and 3rd from the last

[twin wolf]

oh my god

😭

[torn jolt]

In extreme level

[twin wolf]

i havent learnt matrices yet

[torn jolt]

Very lengthy

[twin wolf]

is there calculus pls say there is calc

[grave elm]

use ,w

[twin wolf]

OH YEAH

facts

i dont know how to do a matrix calculation on wolfram tho what do i write

[grave elm]

no idea

[paper acorn]

imagine putting the unsolvable problem in the game just to see how kids solve the problem. I think it is time to learn how to make roblox map

[grave elm]

Lol

[paper acorn]

and get a free $1M

[grave elm]

cheapest work force

[paper acorn]

we can says spend $100 for their robux reward

[grave elm]

and you can even make them pay for hints and materials for studying

[paper acorn]

ez money

[twin wolf]

most of the players are like 8th 9th grade so matrices are just rude

[torn jolt]

No

[twin wolf]

what

😭

wheres the fun in that

[paper acorn]

you say matrix is rude but not calculus?

[torn jolt]

As method has said

Questions are just fancy

Except for the quadratic equations

True pain

[twin wolf]

I know the right term is just 1

but what do i do with log base x

is that a change of base

im guessing it is

[torn jolt]

No

What

[grave elm]

log(1000)/log(x)

[twin wolf]

oh

[grave elm]

3/log(x)

[torn jolt]

Base of a log can't be 1

[twin wolf]

@devout valley LMAO

[grave elm]

Wait is this also roblox or no?

[paper acorn]

and log(x^2) = 2log(x)

[grave elm]

yep

[twin wolf]

yes 💀

[grave elm]

it looks quadratic

[paper acorn]

first ask roblox question then put the real homework question in. that's actually clever way to get the solution lol

[twin wolf]

nah this is in game

[paper acorn]

I'm just joking lol. but imagine tho

[twin wolf]

what do i do with $\frac{\log\left(1000\right)}{\log x}-2\log\left(x\right)+1=0$

[glossy valve]

water beam

[twin wolf]

should i get a common denominator

[grave elm]

log(1000) = 3

[hybrid zephyr]

Wait someone asked this question yesterday lol

[twin wolf]

oh

[grave elm]

and then multiply by log x

[twin wolf]

ok

[grave elm]

and it will be quadratic

then do some magic and youll get the roots

maybe sub u = logx if you dont see how it's quadratic

[twin wolf]

oh real

is it not 31.623?

[grave elm]

i think it is

[twin wolf]

why is the game not registering it correct

maybe round up?

[grave elm]

wait no hmm

[twin wolf]

im getting jipped

[torn jolt]

Ah

Sometimes you have to leave it fractions

[twin wolf]

ohh

[grave elm]

Oh it is 31.623, sqrt(1000)

[grave elm]

its not fraction

its root 1000

[torn jolt]

No I am just referring to the questions

[twin wolf]

sqrt1000 isnt working either that sweird

[torn jolt]

Sometimes it's in decimals sometimes in fractions

[grave elm]

3/2 is the solution of quadratic, so 10^3/2 is the solution

is it decadic logarithm?

log(x^2)

or is it base e

[torn jolt]

you probably have to use the other solution

iirc

[twin wolf]

apparently it was 0.1

[torn jolt]

i beat this game n sqrt1000 didnt work

[twin wolf]

im on extreme level now

[grave elm]

What the fuck

That's the 2nd solution..

log(x) = -1

so x = 10^(-1)

[twin wolf]

yep gg

[grave elm]

why they dont accept both

what they mean

its an equation

[twin wolf]

idk it was like that for another quadratic question before they only accepted negative solution for some reason

[grave elm]

not expression

[paper acorn]

poor map creator

[grave elm]

Yeah

We should make something better

[paper acorn]

like webwork?

[grave elm]

and then point people to the roblox map instead of khan academy

[twin wolf]

am i supposed to solve for x

[twin wolf]

exactly

[grave elm]

I have no i dea

the question is stupid

[hybrid zephyr]

Khan academy is kinda slow ngl

[torn jolt]

looks fine

[torn jolt]

Take 4 combiinations

[grave elm]

Try -0.8

[torn jolt]

theres only 3

[twin wolf]

oh i think it wants me to add both solutions from the negative and positive cuz its an absolute value

[grave elm]

it has nothing to do with the fact that its absolute value

[torn jolt]

x<-5/3, -5/3<=x<3/2 , x >= 3/2

[grave elm]

at least i hope so

[paper acorn]

yeah, you can write absolute value as |x| = -x, x for different value of x

[torn jolt]

There are 4

no

[grave elm]

it depends on how you are solving it

cheesus

[torn jolt]

nah

1st term can be negative and positive
Same with the second

[paper acorn]

one condition is collapsable

[grave elm]

yeah

[torn jolt]

nah

I already solved it :/

ok bro

[grave elm]

Well then I claim there are only 2 combinations

[torn jolt]

It has 4 combinations

[paper acorn]

like, it should be obvious that if 3x + 5 < 0, you can also infers to 2x - 3

therefore, you can reduce it down to 3 conditons for example

[torn jolt]

2 because for the other 2 the x value doesn't satisfy

The conditions taken

[grave elm]

There are only 2 combinations because it should be obvious that there are no solutions where -5/3 < x < 3/2

[paper acorn]

yeah, you can keep reduce it down

[grave elm]

Actually, there is just 1 combination

[torn jolt]

that doesnt mean -5/3<x<3/2 is nonexistent

[twin wolf]

this game is cooked

[grave elm]

because its obvious that if x>3/2 then the only solution is x = 4.6

Wait no, there is 0 combinations

because if x<-5/3, then its straightforward to see that x = -5.4 is the only solution

Wait no, there are negative 1 combinations

[twin wolf]

💀

[paper acorn]

another simple way to solve this is to solve
$\sqrt{(2x-3)^2} + \sqrt{(3x+5)^2} = 25$ and you should get 2 equations: for positive x and negative x

[glossy valve]

print("NAME")

[twin wolf]

@torn jolt u said matrix question was very last?

[twin wolf]

oh true

[torn jolt]

2nd and 3rd from last

[twin wolf]

oh god

[torn jolt]

The last one is logs ig

[twin wolf]

💀

i see it

wtf

i dont even do linear algebra

[paper acorn]

it is quite easy tbh because $\text{det}(aX) = a^nX$ for $X\in\mathbb{R}^{n\times n}$

[glossy valve]

print("NAME")

[twin wolf]

i dont know what any of that means 🤣

@devout valley maybe i should take u up on that offer

[torn jolt]

how you gonna solve that?

[paper acorn]

well, I can say that inside is positive and remove the square root

[devout valley]

Told ya

[paper acorn]

I can also say that inside is negative and remove the square root

[torn jolt]

thats the same thing as abs values

sqrt(x²) = |x|

[twin wolf]

teach me the ways 🙏

[paper acorn]

That's the point for solving the equation

[torn jolt]

then whats the point of changing it to sqrt(x²) form

[paper acorn]

because now if I know that x is positive, then sqrt(x^2) = x

[grave elm]

What's the point of changing x + 1 = 2 to x = 1 when they are essentially the same thing?

[paper acorn]

no absolute value

[torn jolt]

??

the same applies to absolute values

|x| = x when x>0

[paper acorn]

yes...

[torn jolt]

|x| = -x when x<0

[paper acorn]

but now you have 2 equations instead of consider for each cases

[torn jolt]

BTW you get a nerd face after completing extreme level

how

its 2x-3 and 3x+5

[paper acorn]

I think it is trivial to expand that no? you will have 2 linear equations

sum it up

done

[torn jolt]

huh

[paper acorn]

because again, if you know one is positive for example, you also know the bound for another term

says if x is positive, I can say rightaway that 5x = 23

and if it is negative, I can say rightaway that 5x = -27

[torn jolt]

?

[paper acorn]

that's the point for writing it this way

[torn jolt]

what if the positive number is less than 3/2

[twin wolf]

im cooked i cant do LA

but im leeching off of this other guy

[paper acorn]

then what?

[torn jolt]

LA?

[paper acorn]

linear algebra

[torn jolt]

Oh

[topaz valley]

,w 25(3x - 24)^-1 (20 + 6) = 5

[glossy valve]

Results provided by WolframAlpha

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

Requested by tearfox

[torn jolt]

you cant just say if x is positive then 2x-3 >0

[paper acorn]

but first I say that (2x-3) > 0 then I can say that x > 0 for example...

not exactly but you see my point

like x > something, 3/2 here for example

[torn jolt]

then u gotta solve for -5/3<=x<3/2

n x<-5/3

[paper acorn]

dude. I see no point argue about this. If x > 0 then
|3-2x| + 3x = 20, agree?

and |2x-3| + 3x + 5 = 25 agree?

(same thing anyways)

[torn jolt]

then you still have to solve for more than 2 cases??

[paper acorn]

why do you fixate on how many case you need to solve?

like, does that matter?

[torn jolt]

??

[paper acorn]

as I note, those two equations are equivalent

so, I end up with 2. for x > 0 and x < 0

[torn jolt]

you have to solve for 0<x<3/2 and x>=3/2 there + you have to consider when x<0

[paper acorn]

so what?

[torn jolt]

so more than 2 cases

[paper acorn]

my point is 2 equations

I just don't understand why you want to win this argument so much

[torn jolt]

u still have to consider the absolute value

in those 2 equations

[paper acorn]

I always say 2 equations. not 2 cases

[torn jolt]

its 4 then

[paper acorn]

ok you win

:)

[torn jolt]

aight thats what i thought

[paper acorn]

can we now talk about roblox?

[twin wolf]

this is what i get for beating the game

[paper acorn]

yeah, reportable image

[twin wolf]

gg guys

that was an emotional journey

thank u for all ur contributions

[torn jolt]

You got past the matrices ?

[twin wolf]

i leeched off of some other guy to do them 🤣

[paper acorn]

matrix question is essentially linear equation

[torn jolt]

Oh

[twin wolf]

there was a final summation question which was so dumb

i put it in my calculator and it froze it 💀

[torn jolt]

Oh yes

Forgor

Thought it was logs

First I thought i was sqrt(-1)

[paper acorn]

det(5A^(-1)B') = 5^2(1/det(A))det(B), det(A) = 3x - 24 and det(B) = 26.

just plug it in, solve for x, done.

[topaz valley]

.

[twin wolf]

👑 wolfram

[paper acorn]

yeah, that equation is essentially it

[grave elm]

You finished it already?

[twin wolf]

yes it was a combination of getting the more intelligent players to do it for us and then also my calculator

[grave elm]

lol

[paper acorn]

seems too easy. let's make something harder

[twin wolf]

for sure

and add calculus

[grave elm]

"Jump in intervals, such that you type out proof of the fermat's last theorem in morse code"

[twin wolf]

charrrt lurking

[paper acorn]

well, I can think of a question like $\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} ; dx$ or other PDF kernel to make life of people who never learn probability at college level before suffer

[glossy valve]

print("NAME")

[twin wolf]

we

[paper acorn]

it is still calculus. but hell you don't want to do by-part or things like that. Like, imagine beta density

[twin wolf]

we'll get them to integrate e^x^2 from -inf to inf using polar coordinates

[paper acorn]

I mean, why bother? We know that $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} ; dx = 1$

[glossy valve]

print("NAME")

[paper acorn]

then you can just multiply by $\sqrt{2\pi}$ and done, get the answer

[glossy valve]

print("NAME")

[twin wolf]

we'll say thats cheating and they have to pretend they dont know that

[paper acorn]

and that is what make this evil

[twin wolf]

🔥

anwyas

going eep

goodnight

.close

[full forum]

Channel closed

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[hot elm]

dk how to approach it at all new to these comp math type questions

[full forum]

@hot elm Has your question been resolved?

[hot elm]

<@&286206848099549185>

[clever sluice]

Hello

[hot elm]

hi

[clever sluice]

Is the question talking about there being two different digits in both?

[hot elm]

im assuming it just gotta add up to 999

[clever sluice]

Idk if that's what it's asking?

But that is a condition

[green sluice]

I think we gotta consider both of the distances

cuz both of them together consist of a road sign

[hot elm]

aight

[green sluice]

soo, there are a total of 90 possible combination of digits.

so if we consider a general case, except 0. calculate the possibilities and check the cases for 0 separately.

not sure tho, I'm suggesting these just from the top of my head. I haven't thought much of it

[hot elm]

alright

[hot elm]

would this be combinatorics?

[green sluice]

yupp

this def is a problem on combinatorics

[hot elm]

ok ty

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[vapid night]

How can i prove that the length of the skewed line is 5,2m

This should be a better pic i think

<@&286206848099549185>

[wise delta]

pythagorean theorem

[warm kite]

@vapid night

it will help you visualize

[vapid night]

Oh

[vapid night]

Yes i got it thank you so much

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[granite elk]

i made everything sin

but the solution uses tan

and i got different values

for my answers

i wanna know if i will still get marks in my exam

[torn jolt]

U r supposed to get different values?

I dont see why u shouldnt get marks

Wait, lemme see

[granite elk]

i think i may have done my method wrong

lemme retry

[neat aspen]

wdym made everything sin

[granite elk]

5sin(x) - 2cos(x) = 3sin(x)
5sin(x) - 2(1-sin(x)) = 3sin(x)
5sin(x) - 2 + 2sin (x) = 3sinx
3sin(x) = 2

then solve from there

[neat aspen]

i zee

sin(x) + cos(x) = 1 is not the identity you're thinking of

[torn jolt]

Hey... Cot(x) = 4.. It cant happen?

[neat aspen]

it's sin^2(x) + cos^2(x) = 1

[granite elk]

ohhh ok

[torn jolt]

Nvm, i stupid

Im*

[granite elk]

i haven't learnt that yet lol

[torn jolt]

Nah, i was wrong, cot(x)=1

[granite elk]

wait but if u square root both sides does it not just make the identity sin(x) + cos(x) = 1

why doesnt that work?

[torn jolt]

If im not wrong, theta=45°,225°

[granite elk]

yeah

it is

[torn jolt]

Great, i still have my sanity

I thought i lost my meth ability after the social science exam

[granite elk]

i have my end of year exams soon so i'm tryna just perfect everything rn

[torn jolt]

Cool

[torn jolt]

The solution btw

[granite elk]

is it not just simpler to just use tan?

[torn jolt]

Ah.. How?

I found it as the simplest path

[torn jolt]

O

Well, both r kinda the same tbh

[granite elk]

yeah i sometimes forget how u can just pull out stuff for the fun of it

[torn jolt]

So, ur problem solved?

[granite elk]

yep now

how do i close again ?

[torn jolt]

.close

[granite elk]

.close

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Use .reopen if this was a mistake.

[neat aspen]

that is not what you get when you square root both sides is why

[woeful spruce]

(a+b)^2 is not equal to a^2 + b^2

root (a+b)^2 = a+b

root (a^2+b^2) is sometehing else

[granite elk]

oh ok i see

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[torpid perch]

The digits of a four-digit positive integer add up to 14. The sum of the two middle digits is nine, and the thousands digit minus the units digit is one. If the integer is divisible by 11, what is the integer?

[somber niche]

You on fire today my guy

so suppose our number is abcd

then a + b + c + d = 14

now write the other conditions like this

[torpid perch]

four digit positive integer

the sum of the two middle digits is 9

b + c = 9

[somber niche]

Yup

what else?

[torpid perch]

the thousand digit minus the unit digit is 1

that is complex

. . .

[somber niche]

nah

that's just a - d = 1

no?

[minor crater]

yes

[torpid perch]

yeah

my bad

[somber niche]

the last bit might be a bit tricky though, but it's also really simple

[torpid perch]

abcd | 11

[somber niche]

Well rewrite abcd in a more convenient way

[torpid perch]

(1000a + 100b+ 10c + 1d )| 11

[somber niche]

More specifically 1000a + 100b + 10c + d = 11 * k

where k is some positive integer

[torpid perch]

okay

[somber niche]

a + b + c + d = 14
b + c = 9
a - d = 1
1000a + 100b + 10c + d = 11 * k

so you got this

[minor crater]

there's actually a divisibility rule for 11 that won't involve k but i personally have never heard of it

[somber niche]

yeah there is but not sure if op is familiar with it

I'm fairly sure we can solve this without it

[grand hatch]

Not true

There are divisibility rules for all primes

But after 11 it gets quite crappy

[somber niche]

One way to solve this now is to solve the system with k as a paremeter

as in, get a, b, c and d in function of k

and then vary k and test what works

though that may be quite tedious

You'd only have to vary k for values that make a in the range of 1 to 9

so you'd probably just have to check 9 values of k or so

[minor crater]

you can solve for a and d easily from these and only then start involving k for b and c

[grand hatch]

You can modify $1000a+100b+10c+d=11k$ to:
$\newline (11-1)^3a+(11-1)^2b+(11-1)c+d=11k \newline $. Expanding gives:
$-a+b-c+d=11(\text{some integer})=11k$

[somber niche]

That's some galaxy brain moves right there

[glossy valve]

quickdoom

[minor crater]

and also since b+c is a constant and they are digits it will be very easy to just brute force

[minor crater]

very nice

[grand hatch]

Its actually just deriving the divisibility rule for 11

[somber niche]

yeah basically

[grand hatch]

And since a,b,c,d come from 0 to 9, you only need to test k=0,1 and -1

[somber niche]

Proof by exhaustion, my favourite

[grand hatch]

I believe this should greatly reduce the working required and atleast make it manageable

[torpid perch]

I think this problem is way above my level

I might surrender

sorry guys

[grand hatch]

Its not

Ypu have 4 linear equations in 4 variables

[somber niche]

Plug in the second equation into the first one

That simplifies things a lot

Or rather, solves half the problem

cause then the first and third equation can be solved for a and d

[torpid perch]

a + b + c + d = 14
b + c = 9
a - d = 1
1000a + 100b + 10c + d = 11 * k

which one

[somber niche]

and then you substitute b from the second one and plug all that into the fourth to solve for c

or b

look at these two

a + b + c + d = 14
b + c = 9

by plugging in the second into the first we get

a + 9 + d = 14

a + d = 5

[granite torrent]

this is not the way

[somber niche]

is it not?

[granite torrent]

this i mean is not* what ur saying is fine

[somber niche]

Show me da wae

[granite torrent]

first thing to this problem is realize b + c = 9 implies a + d = 5 which u did

[somber niche]

Anyway, by observing the new equation
a + d = 5

with the third one
a - d = 1

[granite torrent]

ye that

[somber niche]

we can solve for a and d

we get a = 3 and d = 2

From there on we have :
b + c = 9
3000 + 100b + 10c + 2 = 11 k

now just solve for b and c and then vary k a bit to check around things

[granite torrent]

👀

[somber niche]

Hm?

[grand hatch]

Im not sure how solving -3+b-c+2 =11k is not better than solving 3000+100b+10c+2=11k

[granite torrent]

sorry maybe i misread

[grand hatch]

I was just modifying what system they had arrived at to aid in the calculation

[somber niche]

so anyway, b = 9 - c

3000 + 100 * (9-c) + 10c + 2 = 11 k

3000 + 900 - 100 c + 10 c + 2 = 11k

3902 - 90 c = 11k

90 c = 3902 - 11k

c = (3902 - 11k) / 90

We know c can only be from 1 to 9 so knowing that we vary k

to see what fits

Alternatively

express k

[grand hatch]

Again, you can save a lot of mess(and make the solution more ovvious) by using -3+b-c+2=11k, giving 8-2c=11k => (8-11k)/2=c, and the choice for k here should be pretty obvious (|| k=0 => c=4 and b=5 ||)

[somber niche]

as (3902 - 90c) / 11

and vary c = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

and see for which ones you get a whole number k

and then use that k to get b

[somber niche]

Ye, but that would involve the galaxy brain move, I wanted to show op how they could do it without it

[torpid perch]

,w Solve[{a + b + c + d == 14, b + c == 9, a - d == 1, 1000 a + 100 b + 10 c + d == 11 k}, {a, b, c, d, k}]

[glossy valve]

Results provided by WolframAlpha

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Requested by __melomaniac

[torpid perch]

how do I check my answer?

[grand hatch]

Giving our desired 4 digit number ||3542||, and indeed it does satisfy all conditions

[somber niche]

Yeah

c = 4 is the only one that gives k as a whole number

c = 4 therefore must be the solution for c

and then b = 5

so the number is 3542

[torpid perch]

k, thanks

.close

[full forum]

Channel closed

Closed by @torpid perch

Use .reopen if this was a mistake.

[grand hatch]

You need to be able to interpret the result. It is saying a=2 and c=9-b and d=2 and k= (90b+3092)/11. From here onward the solution has been continued here: #help-28 message

[full forum]

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[foggy glade]

anyone can help me w this derivative pls

[foggy glade]

i m at ln (x^2) + x * 2x/x^2 + 1/x^2

but then ıdk

[brittle steeple]

Seems right

Now just simplify the fraction and probably also the ln(x^2)

[foggy glade]

the fraction x*2x/x^2 does it make 2x/x OR what

it feels weird

[brittle steeple]

Well yes but it simplifies even further

[runic snow]

f(x)=2x.lnx-1/x

[foggy glade]

yeah but how

is my problem

[brittle steeple]

correct but not helpful

[foggy glade]

help pls....

[brittle steeple]

2x/x = 2

[runic snow]

ln[x^a] can be written as aln[x]

[brittle steeple]

except when x=0 but that's usually not considered important enough to worry about for these problems

[foggy glade]

but it s 2x/x^2 no?

so how

[foggy glade]

yes

[brittle steeple]

No, it's x*2x/x^2

2x/x as you said

[brittle steeple]

^

[foggy glade]

and then u simplify again 2x/x to 2???

[brittle steeple]

Yes

[foggy glade]

that s so weird

can ı even do that

[brittle steeple]

Cancel out an x from both sides

Why not

[foggy glade]

no but on 2x/x

like this is what makes it 2 right

not smth else

[brittle steeple]

?

2x/x is 2

[foggy glade]

okay slay

and this do ı have to simplify it

or is it not necessary ln[x^a] can be written as aln[x]

[brittle steeple]

I would

[foggy glade]

will it be easier to derive again ?

for the second derivative

or harder

[brittle steeple]

Easier

Probably

[foggy glade]

oki thanks man

^^^

[brittle steeple]

Np

[foggy glade]

.close

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Channel closed

Closed by @foggy glade

Use .reopen if this was a mistake.

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• Show your work, and if possible, explain where you are stuck.
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[full forum]

Please don't occupy multiple help channels.

[cedar dove]

,rotate

[brittle steeple]

triple lhopitals moment

[cedar dove]

Without lhopitals

Its not allowed

[brittle steeple]

Taylor expansion?

[main ginkgo]

$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

[glossy valve]

Combustion

[main ginkgo]

$\frac{\tan\left(x\right)-\sin\left(x\right)}{x^{3}}=\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}-\sin\left(x\right)}{x^{3}}=\frac{\sin\left(x\right)-\cos\left(x\right)\sin\left(x\right)}{x^{3}\cos\left(x\right)}$

[glossy valve]

Combustion

[main ginkgo]

$\frac{\sin\left(x\right)-\cos\left(x\right)\sin\left(x\right)}{x^{3}\cos\left(x\right)}=\frac{\sin\left(x\right)\left(1-\cos\left(x\right)\right)}{x^{3}\cos\left(x\right)}=\frac{\sin\left(x\right)\left(1-\cos\left(x\right)\right)\left(1+\cos\left(x\right)\right)}{x^{3}\cos\left(x\right)\left(1+\cos\left(x\right)\right)}$

[glossy valve]

Combustion

[main ginkgo]

$\frac{\sin\left(x\right)\left(1-\cos\left(x\right)\right)\left(1+\cos\left(x\right)\right)}{x^{3}\cos\left(x\right)\left(1+\cos\left(x\right)\right)}=\frac{\sin\left(x\right)\left(\sin^{2}\left(x\right)\right)}{x^{3}\cos\left(x\right)\left(1+\cos\left(x\right)\right)}$

[glossy valve]

Combustion

[main ginkgo]

$\left(\frac{\sin\left(x\right)}{x}\right)^{3}\left(\frac{1}{\cos\left(x\right)\left(1+\cos\left(x\right)\right)}\right)$