help-10

[royal karma]

is it true?

[rigid lintel]

yes

but you could verify that yourself

with pictures or a more formal proof if you like

[royal karma]

yes i would like to get formal proof

can you help me with that?

im kinda stuck

[rigid lintel]

how far along are you

[royal karma]

idk how to continute

[rigid lintel]

write the definition of $B\Delta A$

[warm shale]

Gijs

[royal karma]

i did

no?

[rigid lintel]

oh yeah

you just have to show that if A subset of B and x is in B \ A then x must be in B \delta A

and vice versa

[royal karma]

it needs to be the opposite

like this

cause its BdeltaA
not AdeltaB

right?

[rigid lintel]

same thing

[royal karma]

how can i show that?

[royal karma]

can you write it down for me please?

[rigid lintel]

its really quite trivial

just write the definitions of A delta B and B delta A

[royal karma]

which one will be more helpful

1st or 2nd

[rigid lintel]

i would say the top is easiest

in this case

[royal karma]

what now?

@rigid lintel

[rigid lintel]

,rotate

[royal karma]

yes sorry

[rigid lintel]

sorry i cant read half of this

[royal karma]

it's basically

[rigid lintel]

what is this

[royal karma]

exactly what u wrote

the thing before the : is

by definition of symmetric

[royal karma]

this

[rigid lintel]

this is what you have to do

[royal karma]

im sorry, i cant understand this 😦
is it possible for you to do a formal proof, in this way i'll understand better
im dumb in set theory..

[rigid lintel]

suppose A subset B then let x be an element of B \ A then x is in B but x is not in A hence ... therefore x is in the symmetric difference of A and B

something like that

dont overcomplicate it

[royal karma]

ok

Thank you so much! @rigid lintel

[rigid lintel]

you did it?

nice!

[royal karma]

I think so 🙂

i need to submit the assignment in 7 minutes

so i hope it's right

i don't have enough time to be sure

[rigid lintel]

good luck

[royal karma]

Thanks buddy!

@rigid lintel Appreciate that.
I need to practice more...

I'm under pressure of assignments unfortunately

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[glossy yacht]

?

[twin sapphire]

take your shirt as a whip and try to hunt it down

[wild swallow]

[tardy epoch]

#chill for nonsense

.close

[obtuse pebble]

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[zenith raft]

i know what do in that situation when it's 10 pm but can't help if it's 12 pm

[wild swallow]

[zenith raft]

i wish 😭

[obtuse pebble]

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[proper dragon]

Hi, sorry this might be a dumb question, but how did the n exponentials just disappear here ?

[kind hawk]

dw, not a dumb question

whats the full context?

but it looks like an error

[proper dragon]

Well, the first expression you see is a series. I have to prove whether or not it converges or not using the comparison limit test
Thing is I'm stuck with what expression I'm supposed to compare it to, so I asked ChatGPT and apparently you can rewrite the equation like that

[winter ether]

Yeah it goes from an equality to an expression doesn’t make sense

[wild swallow]

[winter ether]

Ask wolfram alpha not chatgpt 💀

[wild swallow]

chatgpt is wrong

[proper dragon]

It's only giving me the answer on if it converges or not, but with no proof

[winter ether]

Ah

[proper dragon]

Though I think I was using symbolab, not wolfram

[winter ether]

Try wa, it usually has step by step that includes conv tests

Not sure tho

[kind hawk]

chatgpt tells a lot of useless garbage

[winter ether]

Chatgpt is a language model not a mathematical model so you have no way of knowing whether or not what it says is actually correct

[proper dragon]

Haha it's been useful in programming, so I figured it might help me in math

[kind hawk]

it sounds mathematically correct but it does not actually know math, it only knows stuff that sounds like it

[proper dragon]

Well, now I know. Thanks for the heads up ! ^^

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[fading herald]

How is this answer right? I'm get 90% confidence level not 80%, I basically took the ME/SE to get the critical value and looked at the z table

[obtuse pebble]

@fading herald Has your question been resolved?

[obtuse pebble]

@fading herald Has your question been resolved?

[obtuse pebble]

@fading herald Has your question been resolved?

[obtuse pebble]

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[magic cairn]

hi, how did my instructor tell that f(x) and g(x) are both > 0? I understand how he got f'(x) and g'(x) > 0 and f''(x) and g''(x) < 0 tho

[neon eagle]

he couldn't i think he made a mistake

[magic cairn]

so it was purely an assumption...

ok thank you :)

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[obtuse pebble]

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[timid silo]

f(x)= -csc(1/2(x + pi/6))

[timid silo]

I just need help graphing

desmos not helping

[hybrid gull]

What happens when you try to graph it on Desmos?

[timid silo]

is like this

im just confuesed where to plot the points on the hgraph

and where the asymtotes are

[hybrid gull]

You also have cos x in there as well

[timid silo]

yeah

just as a refernce as the sec

[brave bramble]

Is that not the graph right there?

[timid silo]

yeah but the idk where the aymptotes are locted

[brave bramble]

They're located where
cos(1/2(x + π/6)) = 0
1/2(x + π/6) = kπ + π/2

For any integer k

[timid silo]

This what I have so far

[brave bramble]

Nice, you can add on what I told you

[timid silo]

so Just add the points π/2 , π, 3π/2 and 2π?

and the aymoties will be at every pi/6?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[vital walrus]

how to find the domain and range of y=√(1-x^2)

[trail cloak]

Ooo

You know how if the square root gets any negative numbers, it gives you math error?

[vital walrus]

ye

[trail cloak]

You try to find the x values that avoid that problem

That means

[vital walrus]

1-x^2 bigger than or equal to 0

[trail cloak]

Yeppp

[vital walrus]

1 bigger than or eual to x^2

[trail cloak]

Exactly

[vital walrus]

what to do now

[trail cloak]

Solve for x

Also keep in mind

[vital walrus]

x smaller than or equal to 1, -1

[trail cloak]

Yeah

This means that x should be between 1 and -1

Inclusive

[vital walrus]

does the smaller one (-1) go at the back

[trail cloak]

Yepp

[vital walrus]

when u get the 2 answers of 1 and -1 is it just the smaller one that goes behind

and the bigger one goes in front

[trail cloak]

Yep

[vital walrus]

aight

[trail cloak]

That's if the number is bigger than x^2

[vital walrus]

-1 < x < 1 but with the equal to as well

[trail cloak]

Yeah

[vital walrus]

wdym

[trail cloak]

You know how the absolute value function works right?

[vital walrus]

ye the | |

[trail cloak]

Yep

[vital walrus]

makes it positive

[trail cloak]

Mhm

It's a similar case here

So that x^2 stays less than or equal to 1

X itself has to be between -1 and positive 1

[vital walrus]

where does the absolute value come in

[trail cloak]

$\sqrt{x^2} = |x|$

[warm shale]

VulcanOne

[vital walrus]

so x^2=1 |x|=1

[trail cloak]

Wait that's not the same

[vital walrus]

x=|1|?

[trail cloak]

When you want to find x in the expression x^2 = 1, you take the square root right?

[vital walrus]

ye

[trail cloak]

Taking the square root of x^2 turns it into |x|

And 1 stays 1

But when you wanna know what's x, you look at the x values in the absolute value function

Like

|-1| = 1, and |1| = 1

But

X^2 is not the same as |x|

[vital walrus]

ye but when u sqaure root it it becomes |x|?

[trail cloak]

Yeah, in that case, that's true

[vital walrus]

would it be true in every case

[trail cloak]

Nope

$\sqrt{x^2} = |x|$

[warm shale]

VulcanOne

[trail cloak]

Lol

Anyways

[vital walrus]

idk how to find the range tho lol

[trail cloak]

Well the range is the range of the square root

[vital walrus]

y=√(1-x^2)

uh square both sides?

[trail cloak]

Nope

You see the absolute minimum

And absolute Maximum

Absolute minimum is when the square root gives 0 right?

[vital walrus]

uh ye

[trail cloak]

Now since x is bounded, you can't get to infinity

So you have to see the absolute maximum

[vital walrus]

for range do u just solve for y

[trail cloak]

Nope

[vital walrus]

i got to x= √(1-y^2)

[trail cloak]

Now solving this will give you -1<=y<=1

Which makes no sense

Because x is bounded between-1 and 1 already

This is a new function btw

[vital walrus]

ok

[trail cloak]

Finding the range can be either through graphing or through trying your function at various points

[vital walrus]

can u not find range algebraially

[trail cloak]

Hmm

You can but you'll have to eliminate some y-values

[vital walrus]

i asked a different question before y=1/ √(16-x^2) and u could do it algebraically

√-1/y^2+16=x^2

√16-1/y^2=x

16-1/y^2 >=0

16>=1/y^2

y^2<=1/16

y=+-1/4

[trail cloak]

Wait

[vital walrus]

i think

[trail cloak]

The greater than or equal sign is wrong in this case

We want the fraction to stay above 1/0

[vital walrus]

for which one

[trail cloak]

Or else it will be undefined

The 1/sqrt(16-x^2)

[vital walrus]

16-x^2>=0?

@trail cloak u still there?

[trail cloak]

Yeah

Sorry for that

[vital walrus]

lol math so confusing

[trail cloak]

Sometimes yeah

If you don't understand it

Anyways

Let's get back to our original question

[vital walrus]

aight

[trail cloak]

$\sqrt{1-x^2}$

[warm shale]

VulcanOne

[trail cloak]

You found the domain of this function right?

[vital walrus]

yeah

-1<=x<=1

[trail cloak]

$-1 \le x \le 1$

[warm shale]

VulcanOne

[trail cloak]

And you found an inequality for y right?

[vital walrus]

no

[trail cloak]

Let's find it

$y = \sqrt{1-x^2}$

[warm shale]

VulcanOne

[trail cloak]

$y^2 = 1-x^2$

[warm shale]

VulcanOne

[vital walrus]

yep

[trail cloak]

$x^2 = 1-y^2$

[warm shale]

VulcanOne

[vital walrus]

yep

[trail cloak]

$x = \sqrt{1-y^2}$

[warm shale]

VulcanOne

[vital walrus]

yeo

[trail cloak]

Looks the same as our original

[vital walrus]

1-y^2>=0?

[trail cloak]

So $-1 \le y \le 1$

[warm shale]

VulcanOne

[trail cloak]

Yeah

[vital walrus]

yeah

but its wrong lol

[trail cloak]

Yeah here's why

[vital walrus]

do u also have to get the 0

from the orginal

[trail cloak]

Wdym?

As in like

See the minimum y from the original?

[vital walrus]

idk something like that

cos answer says 0<=y<=1

[trail cloak]

Yeah the minimum y is when sqrt = 0

$\sqrt{1-1} = 0$

[warm shale]

VulcanOne

[trail cloak]

That's your minimum y from the original

That's what I meant by eliminating y values

We found that y can go to 1 though

And it cannot cross 0

From the original

So the range is $0\le y\le 1$

[warm shale]

VulcanOne

[vital walrus]

hmm ye ig that makes sense

[trail cloak]

Yeah

You have to check both your original and the x function

[vital walrus]

so if the origninal supercedes it the range changes

[trail cloak]

Yeah

Like

The x function can go from -infinity to positive infinity but the original can only go from 0 to infinity then you give priority to the original

[vital walrus]

alright

also when for example y^2=a would it be -√a<y<√a

like when its squared it would have 2 answers and it would be between them

[trail cloak]

Yep

But the original is for example y = sqrt(a)

[vital walrus]

a wouldnt be negtaive

[trail cloak]

Yep

[vital walrus]

so just one answer of y<=√a

[trail cloak]

Yepp

[vital walrus]

alraight thanks

have a good rest of ur day

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[obtuse pebble]

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[timid silo]

what am i doing wrong im pretty positive im right

[nocturne minnow]

Check your last row

[timid silo]

should zero come first?

because this is correct

[nocturne minnow]

The columns represent x, y, and z

[nocturne minnow]

So you saying 3, 1, 0 meant 3x, 1y, and 0z, was that what the equation had?

[timid silo]

ah

thank you

silly me

.close

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[keen heart]

how do I do this

[keen heart]

like my thinking is that I do the derivative of 3sin(x) on the derivative of x then just do 3cos(0)/4

giving me 3/4

or is that thinking incorrect?

[gilded needle]

seems fine to me

[keen heart]

ok ty

[obtuse pebble]

@keen heart Has your question been resolved?

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[gilded needle]

correct for what?

[tardy epoch]

Can you show the original question

[zenith raft]

post the qn haha

[fierce lagoon]

11/100 is correct if 11/100 - x = 0

Okay then we cannot assess

You just gave us a random@number

How are we supposed to know if it's correct lol

[tardy epoch]

Come back when you figure out the question

[fierce lagoon]

I was joking with that one lol

Ain't no way

[zenith raft]

what the heck

[gilded needle]

is this math for artists or something? 😁

[zenith raft]

everyone here is :c

[gilded needle]

sure, why not?

[fierce lagoon]

We legit have no context because like

[gilded needle]

you can even have i/100

[fierce lagoon]

LOL

i/100 squares

Ah yes

[zenith raft]

.close if you mean that

[obtuse pebble]

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[zenith raft]

oh

[obtuse pebble]

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[rugged wind]

help

[rugged wind]

wtf

wtf

wtf

bro

<@&268886789983436800>

PLS

WTF

WHY DID I HAVE TO WITNESS THAT

<@&268886789983436800>

<@&268886789983436800>

<@&268886789983436800>

<@&268886789983436800>

a

a

a

a

aa

a

a

aa

aa

a

as

aa

[sterile latch]

Got it

[unreal lodge]

[rugged wind]

holy fk

[timid silo]

i am bamboozled

[unreal lodge]

Happened to me too lol

[wispy patio]

lmaoooo

[green epoch]

What happened here

[rugged wind]

whats the fastest way

how do i do the question

[green epoch]

Just write down the expansion till x²

[rugged wind]

its to n power

[green epoch]

You'll have 3 x² coefficients one from the x-x multiplication and one from 1-x multiplication and the x² coefficient

[green epoch]

Yea write it in terms of n

[timid silo]

1 X coefficient of x^2 in (1-x/2)^n + coefficient of x in (1-x/2)^n

[rugged wind]

wait how

im getting nC2

[rugged wind]

im getting nC2

how to calculate

[timid silo]

(1-x/2)^n = nC0 - nC1 (x/2) + nC2 (x/2)^2 ...

[rugged wind]

yes

nC2 causing problems

[timid silo]

it has to be nC2/4

[rugged wind]

yes

[timid silo]

make up your mind

[rugged wind]

i show you pxiture

@timid silo @green epoch

[timid silo]

yes so add -n/2 and nC2/4 now

[rugged wind]

how to add?

-n/2 + nC2/4?

[timid silo]

nC2 = n!/((n-2)!2!) = n(n-1)(n-2)!/((n-2)!2!) = n(n-1)/2

nC2/4 = n(n-1)/8

[rugged wind]

this good so far? @timid silo

[timid silo]

yes and cancel the factorial parts

[rugged wind]

how?

[timid silo]

n!= n(n-1)(n-2)!

[rugged wind]

is that formula

[timid silo]

no

do you not know what factorial is

[rugged wind]

repeated multiplication until 1

[timid silo]

yes so
n!= n(n-1)(n-2)(n-3).....2.1
(n-2)!= (n-2)(n-3)....2.1
n!= n(n-1)(n-2)!

[rugged wind]

ah

i see whatchu did now

thx a bunch bro

[obtuse pebble]

@rugged wind Has your question been resolved?

[obtuse pebble]

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[delicate umbra]

You live in the country of “Memoryland”, it is your first day of work in the bar “Precios en la
head” and you have to charge 5 full lunches and 3 coffees. You ask the price to your
colleague and tells you: for 3 lunches and 6 coffees we charge €12.15 and for 2 lunches and 5
coffees €9.05.
a. How much does a coffee cost? And a lunch? (2 points)
b. How much do you have to charge your customers? (0.5 points)

[delicate umbra]

Good morning ^^. I just did a subtraction

And got X + Y = 3,10

How can I know the X and Y valor?

I don't want the answer, just help with the process

[narrow vault]

they gave you 2 different meal combos for two different prices. can you represent them as equations?

[delicate umbra]

3x + 6y = 12,15
2x + 5y = 9,05

[narrow vault]

looks good

[delicate umbra]

So I subtract it

x + y = 3,10

[narrow vault]

now find a way to get rid of x or y

[delicate umbra]

uhm

[narrow vault]

putting your equations next to each other:

3x + 6y = 12,15
2x + 5y = 9,05
x + y = 3,10

[delicate umbra]

x + 6y = 12,15 / 3?

[narrow vault]

no, that's not correct, because you only multiplied 3x by 1/3 instead of 3x+6y

[delicate umbra]

oof

6y = 12,15 - 3x

duh, but that doesn't do anything

[whole dock]

Why dont you use substitution here

Its way easier

[delicate umbra]

What you mean?

[whole dock]

For the first equation solve for x and substitute that value of x in the second equation

[delicate umbra]

Is what I'm trying to do

_>

<_<

3x = 12,16 - 6y

[whole dock]

Yes

[delicate umbra]

But I don't know how tu just "ignore" y

or X

[whole dock]

Now divide both sides by 3

[delicate umbra]

x = 4,05 - 3y

Right?

[whole dock]

Yes

[delicate umbra]

oh

so now

[whole dock]

No actually

[delicate umbra]

?

No?

[whole dock]

Its 6y/3 is 2y

[delicate umbra]

Ops

x = 4,05 - 2y

Sorry

[whole dock]

Yes

[delicate umbra]

So the next part now is

[whole dock]

in the second equation

[delicate umbra]

2x + 5y = 9,05

[whole dock]

Yes put the value of x in it

[delicate umbra]

8,10-4y + 5y = 9,05

[whole dock]

Now calculate y

[delicate umbra]

y = 9,05 - 8,10

y = 0,95

:DDDDDDDDDDD

[whole dock]

Yes

Good

Now put that value of y in first equation to get x

[delicate umbra]

X + 0.95 = 3,10

[whole dock]

Since you have one known value the other can be easily calculated for

[delicate umbra]

Ok

[whole dock]

And what do you get

[delicate umbra]

The sad thing is that mooore or less I was reaching this number

[whole dock]

2.15

[delicate umbra]

But then I was wait, full lunch can't be 2,15

To cheap

[whole dock]

XD

[delicate umbra]

But man, it's just math, not real life

[whole dock]

That's true

But math has built the life we live in

The questions may be a bit unrelatable but the theory is very much applicable irl

[frail sierra]

Om bro so ineed help so i get A in a test i need a p

[delicate umbra]

?

I think you can't ask help for tests

I think

[whole dock]

Yeah

[whole dock]

Was that all, do you have any questions?

[delicate umbra]

No, thanks ^^

[whole dock]

Np

[wise talon]

You cant cheat on tests

[delicate umbra]

I've an exam 3 of January, I'm doing all the past year tests as practice

[wise talon]

If you need like revision help thats fine

[delicate umbra]

Thanks ColdTee, I'll ask for more help sooner than later :P

[whole dock]

Good luck

[whole dock]

Glad to help

[obtuse pebble]

@delicate umbra Has your question been resolved?

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[sturdy ibex]

What is derivative of y^3 with respect to x?

[sturdy ibex]

Context:

That little dy/dx after 3y^2 is confusing me

[lost tree]

its the chain rule

treat y as a function of x

[sturdy ibex]

How so

In that example, in particular

[lost tree]

well y is a function of x right

like depending on different values u put for x

u get diff values for y

[sturdy ibex]

It's implicit functions

[lost tree]

so y depends on x

yeah

so y^3 then means (a function of x)^3

[sturdy ibex]

How is y a fucntion of x?

[lost tree]

?

so x^3 + y^3 = 3axy

if u put x=1, then 1 + y^3 = 3a*1*y

u can solve this for y

[sturdy ibex]

Then can we say that x is also a function of y? So for implicit functions, both variables can be expressed as each others' functions?

[lost tree]

u can also say that yes

[sturdy ibex]

Ah ok that clears it a bit

[lost tree]

but u treat it as a function of the variable u want to differentiate it wrt to

[sturdy ibex]

Yes true

[lost tree]

if u wanted to find dx/dy

then treat x as a functino of y

[sturdy ibex]

Yea

So back to the problem, why is dy/dx there?

[lost tree]

coz

[lost tree]

y^3 means (a function of x)^3

so derivative of this is

[sturdy ibex]

Yes

[lost tree]

3*(function of x)^2 * (derivative of the function inside wrt x)

right

[sturdy ibex]

I'm not getting that

[lost tree]

for example derivative of (sinx)^2

wrt x

is

2*(sinx) * derivative of sinx wrt x

so 2sinxcosx

[sturdy ibex]

Yes

[lost tree]

same thing here?

ur function of x here is y

[sturdy ibex]

Ok i see

But why is the resultant dy/dx tho

[lost tree]

so (derivative of the function inside wrt x) = derivative of y wrt x

coz function inside = y

derivative of y wrt x = dy/dx

does that make sense

[sturdy ibex]

I can't

Understand

Sorry

[lost tree]

😦

what is confusing u

[sturdy ibex]

just a second

what will be the function y w.r.t x here?

if thats even possible to claculate

[lost tree]

u want to find y as a function of x?

[sturdy ibex]

yes, i think

ig it would clear it a bit

[lost tree]

well that is difficult to do

u can find y as a function of x

by subjecting y

from x^3 + y^3 = 3axy

but its a cubic

[sturdy ibex]

true

yea its hard

im going to think about the question and ping you when i am done and ask whats actually confusing me, will that be alright?

[serene kiln]

3

[lost tree]

okay

[sturdy ibex]

@lost tree ok so y is a function of x, since it is asking to differentiate in w.r.t x. now, how is the chain rule applied here?

[lost tree]

so in ur equation u got y^3 right

[sturdy ibex]

yes

[lost tree]

and cubing itself is a function right

[sturdy ibex]

yes

is it?

[lost tree]

yes

[sturdy ibex]

ig yea

[lost tree]

i can define g(y) = y^3

[sturdy ibex]

ah ok

[lost tree]

but now y is again a function of x

maybe well write y = f(x)

[sturdy ibex]

ok

[lost tree]

so y^3 is now g(y) = g(f(x))

[sturdy ibex]

yes

[lost tree]

so y^3 is a function of a function

and to differentiate it u apply the chain rule

[sturdy ibex]

ok i see

[lost tree]

the derivative of g(f(x)) wrt x, is g'(f(x)) * f'(x) right

[sturdy ibex]

yes

[lost tree]

so g' is the function which squares its input and multiplies it by 3

right

[sturdy ibex]

uhhh what

ahh

yea

[lost tree]

u get it?

okay

[sturdy ibex]

yes

[lost tree]

so g'(f(x)) = 3*[ f(x) ]^2

[sturdy ibex]

yes

[lost tree]

but f(x) = y right

[sturdy ibex]

that gives 3y^2

[lost tree]

yes

then u multiply this by f'(x)

f'(x) means the derivative of f wrt x

[sturdy ibex]

yes

[lost tree]

so d/dx f(x)

but f(x) = y

d/dx f(x) = d/dx y

[sturdy ibex]

yep

[lost tree]

so do u get it

[sturdy ibex]

that y=f(x) is what helped

yes i do now

[lost tree]

oh

[sturdy ibex]

thank you

[lost tree]

👍

[sturdy ibex]

alright

.close

[obtuse pebble]

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[cursive arrow]

Hello! Can someone please help me solving for x in this problem?

[rigid lintel]

,w factor 2187

[warm shale]

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Requested by Gijs#0201

[cursive arrow]

anyone?

the farthest I've got till is x/3 - 21x^2 - 1/× = -64

but I'm having trouble solving after that

[wise talon]

$7\cdot{(-1 + -8 + 3x^3)} = x/3 + 1/3 + -1/x$

[warm shale]

DerpZ

[wise talon]

then solve

[cursive arrow]

im sorry, I'm having a bit of trouble understanding what you did here. [im quite dumb]

[wise talon]

from here

[wise talon]

then $3^{x/3 + 1/3 + -1/x} = \left(3^7\right)^{-1+-8+3x^3}$

[warm shale]

DerpZ

[cursive arrow]

then you just equate the exponents right?

[wise talon]

yup

[cursive arrow]

okay tysm!

.close

[wise talon]

interesting lol

.close

[obtuse pebble]

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[hoary moon]

x^(3cosx) derivative right ?

<@&286206848099549185>

[rigid lintel]

you cant use lhopital if you dont have a fraction

but you can make one

you already have the answer thoughg

[hoary moon]

how can i make

logaritmize?

@rigid lintel

[calm quail]

Rewrite in the base of e

[hoary moon]

with ln

what is equilivaent

[calm quail]

When you rewrite it,you can use L’Hopital rule in the part of power

[hoary moon]

e^x^(ln3sinx)

is it correct ?

[calm quail]

Because exponential is the continuous function,so after you rewrite it,you can put the limit before the power

No it’s e^[ln (x^3sin(x)]

[hoary moon]

before the which power

[calm quail]

Before ln

[hoary moon]

can you solve it to paper please

[hoary moon]

thank you so muchh

i cant solve it

[calm quail]

ln(a^b)=blna
Then you rewrite in the form as fraction,finally use L'Hopital rule

[hoary moon]

i found 3sinx*lnx=lny

lny=(3cosx)/x

[calm quail]

I don’t understand what you wrote
Rewrite in the form as the fraction means
ab=a/(1/b) or b/(1/a)
Then it is the form of indeterminate(in this problem),which is the case you can use L'Hopital rule

indeterminate form ：0/0 or ±∞/∞

This is the meaning of rewrite in the form as fraction

To check if it is indeterminate

[obtuse pebble]

@hoary moon Has your question been resolved?

[obtuse pebble]

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[timid silo]

ok

[warped agate]

bruh

[trail musk]

Lol

[rigid pine]

Is there two cheaters now? Same question as the other cheater. <@&268886789983436800>

[obtuse pebble]

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[fierce lagoon]

.wait what

[rigid pine]

See help 11.

[rigid pine]

.close

[obtuse pebble]

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Use .reopen if this was a mistake.

[rigid pine]

Same test as the other person.

[fierce lagoon]

@long plinth this guy

[oblique glacier]

got it

[fierce lagoon]

"No life kid" damn he got the original insults ong

[obtuse pebble]

Available help channel!

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[warped fulcrum]

Bit confused. What are the equilibrium points of a system of non linear differential equations and what are the stationary points?

[warped fulcrum]

Say the system of differential equations is (x', y')^T = (f2(t, x, y), f1(t, x, y))^T then are the stationary points just the points for which (x', y')^T = 0?

[obtuse pebble]

@warped fulcrum Has your question been resolved?

[warped fulcrum]

@proven zephyr

Oh

<@&286206848099549185>

[proven zephyr]

?

oh wrong ping nvm

[warped fulcrum]

Yeah sorry about that

.close

[obtuse pebble]

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[narrow vault]

it's past midnight and my brain is shriveling and struggling with basic limits

show $\lim_{n \rightarrow \infty} \frac{n^{\frac{1}{k}}}{[ \ln(n)]^k} = \infty \forall k>0$

[narrow vault]

$\forall L \in \mathbb{R}, \exists \varepsilon >0, N \in \mathbb{N}_0, st.$

$$\begin{align}
|\frac{n^{\frac{1}{k}}}{[\ln(n)]^k} - L| &> \varepsilon\
n^{\frac{1}{k}} &> (\varepsilon + L) [\ln(n)]^k\
n &> (\varepsilon+L)^k [\ln(n)]^{k^2} > (\varepsilon+L)^k\
\end{align}$$

for $n > e$, so we take $N = \max{3, (\varepsilon+L)^k}$

my eyes don't see anything wrong with it (and the result is supposed to be immediate) but my brain is rejecting this proof

[warm shale]

IV

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

[unreal musk]

Wait hang on, confused about your above?

Thought $a_n$ goes to $\infty$ if for any $H >0$, there’s an $n_{H}$ such that for any $n \geq n_{H}$, $a_n > H$?

[warm shale]

chartbit

[narrow vault]

wait holy shit im showing divergence when i should be showing increasing to infinity

[unreal musk]

Mind you I think that’s the definition of going to infinity, that’s from memory so might be wrong on that one 🤔

Actually I am right, look at me

[narrow vault]

yeah ok it'd help if i was proving what was being asked
it's as simple as it's supposed to be just did the completely wrong thing

.close

[obtuse pebble]

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[obtuse pebble]

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[wary kettle]

How do I solve these types of limits?

[rigid lintel]

find the terms which dominate

so that you can setup inequalities

[wary kettle]

x+1?

[rigid lintel]

well not really

[alpine raven]

x+1 is an exponent, not a term

[wary kettle]

4^x+5^x?

[alpine raven]

on the numerator, which terms grow faster than the other when x goes to infty

[wary kettle]

the 5^x?

[alpine raven]

yes, now factor by 5^x

[wary kettle]

sorry, my english is not that good, so you mean 5^x(4^x/5^x+1)?

[alpine raven]

$5^x \left(\left(\frac{4}{5}\right)^x + 1 \right)$

[warm shale]

Ｈｅｒｅｌｓ

[wary kettle]

oh, okey, tha makes sense, so what now?

[alpine raven]

you do the same at the denominator

[wary kettle]

$5^{x+1}((\frac{4}{5})^{x+1}-1)$

[warm shale]

marejak023

[alpine raven]

yes now you have :
$$\frac{1}{5} \frac{\left(\frac{4}{5} \right)^x + 1}{\left(\frac{4}{5} \right)^x -1}$$

[warm shale]

Ｈｅｒｅｌｓ

[alpine raven]

and thats easier to solve now

[wary kettle]

wait, where did the 1/5 come from and where is x+1?

[alpine raven]

$\frac{5^x}{5^{x+1}}$

[warm shale]

Ｈｅｒｅｌｓ

[alpine raven]

://

bro

#❓how-to-get-help

[novel gull]

oh

[wary kettle]

um, what does that mean, i really suck at these basic algebra things :d

[alpine raven]

$\frac{5^x}{5^x \times 5} = \frac{1}{5}$

[warm shale]

Ｈｅｒｅｌｓ

[wary kettle]

so $n^{x+1} = n^x+n^1$?

[warm shale]

marejak023

[alpine raven]

[wary kettle]

oh fuck

im dumb

i see it now lmao

yeah, that makes sense

so the answer is -1/5

[alpine raven]

yes but u need to prove it

why do you think (4/5)^x goes to 0

[wary kettle]

n^x as x goes to infinity goes to zero for n<1

[spiral knot]

-1<n<1

[wary kettle]

oh yeah, of course

[alpine raven]

[wary kettle]

.close

[obtuse pebble]

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[strange fiber]

We have the following information about the linear mapping f : R^2 → R^2:
$$f\circ f = f, f\binom{1}{1} = \binom{1}{2}$$

Determine the image of the vector (x, y)^T when f is displayed (depending on x and y).

[warm shale]

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

[royal basin]

google translate at it again, huh?

also

what stage 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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@strange fiber

[obtuse pebble]

@strange fiber Has your question been resolved?

[strange fiber]

1

For the linear map f: R^2 -> R^2, we have the following information:

Determine the image of the vector (x, y)^T under the map f (depending on x and y).

[royal basin]

okay so

first off

can you say what f(1, 2) is?

knowing that f(1,1) = (1,2) and f^2 = f, you should be able to find the value of f(1,2).

[strange fiber]

If for all (x,y), f(x, y) = (x, 2x), then f² = f and f(1, 1) = (1, 2)

[strange fiber]

image

[royal basin]

hm. well then, i guess you have jumped ahead and landed on the answer, haven't you?

[strange fiber]

i dont even know

if it is correct

if it makes sense

[royal basin]

well, it does happen to be correct. the map (x,y) |-> (x, 2x) is idempotent and sends (1,1) to (1,2).

[strange fiber]

what about

(x,2y)

@royal basin

[royal basin]

... do you mean "I am unable to verify whether this mapping satisfies f^2 = f and would like somebody else to do it for me"?

[strange fiber]

it also

corrects it

it also makes it

[royal basin]

no.

[strange fiber]

1,1 to 1,2

[royal basin]

sure but (x,y) |-> (x,2y) is not idempotent anymore

[strange fiber]

never heard of

[royal basin]

applying it twice gives (x, 4y)

[strange fiber]

idempotent

[royal basin]

idempotent means equal to its own square

the map (x,y) |-> (x,2y) does not satisfy f^2 = f.

[grim eagle]

To verify whether the mapping f(x,y) = (x, 2x) satisfies the condition f^2 = f, we can simply substitute this mapping into the equation f^2 = f and see if it holds.

Substituting, we get:
$$f(f(x,y)) = f(x, 2x)$$
$$f(x, 2x) = (x, 2x)$$

Since this equation holds for all values of x and y, we can conclude that the mapping f(x,y) = (x, 2x) satisfies the condition f^2 = f.

[warm shale]

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

[strange fiber]

i dont understand

@royal basinanyways i cant just write on papes (x,2x),i just looked at it and saw it but i dont really know what steps do i have to write

[digital moat]

The intensely smart person is here again tonight

[royal basin]

... well i tried to guide you step by step but you went and jumped way ahead of me.

[strange fiber]

okay, so back to start

[strange fiber]

Is it Image?

[strange fiber]

huh?

[royal basin]

no... "f(1,2) = image" makes no sense at all

at best it's like being asked "what is 2+2?" and saying "number"...

[strange fiber]

hahah well its not incorrect

tbh i dont know what it is

or maybe i know but im thinking of something way more difficult than what ure asking

@royal basin

[royal basin]

that, or there is a language barrier

[strange fiber]

yeh maybe

so could u tell me what was the answer u were looking for?

maybe I really dont know and it will be something new

[royal basin]

i was looking for f(1,2) = (1,2)

because f(1,2) = f(f(1,1)) = f(1,1) = (1,2)

[strange fiber]

That makes sense, i wasn't thinkin ab it but it is clear to me

hold up so ill tell u

my algebra professor is mentally ill and expects me to explain everything in the biggest detail possible

so first i have to do some overview of the task

1. We are given that f ∘ f = f, which means that applying f twice to any vector is the same as applying f once. This condition is called idempotence.

2. We are also given that f(1, 1) = (1, 2). This means that the image of the vector (1, 1) under the linear map f is (1, 2).

@royal basin are my 2 points correct?

[royal basin]

is that what your professor expects you to write?

[strange fiber]

i have to explain everything in order to get full points from the homework

[royal basin]

disgusting

[strange fiber]

yeh well so u can just say 1) correct 2) correct

or both incorrect

i will be writing it and sending it here

and u will be just saying correct or incorrect

is that ok?

@royal basin

[royal basin]

... i would really rather not!

[strange fiber]

oh why

:(

@royal basin I wrote it down completely, do you think i can dm you it?

[royal basin]

no, i would rather you didn't.

[strange fiber]

Fine

[strange fiber]

From f(1, 1) = (1, 2) you get a + b = 1. From f(1, 2) = (1, 2) you get a + 2b = 1, hence b = 0 and a = 1.

does this make sense?

@royal basin

[royal basin]

no, and i do not want to be pinged about this anymore.

[strange fiber]

lol

[obtuse pebble]

@strange fiber Has your question been resolved?

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[obtuse pebble]

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[meager hull]

Finding the inflection points of the original function from the derivative

[meager hull]

Ignore the writing but like it's safe to say when x=0 and x=2 are inflection points right?

[spiral knot]

yeah

[meager hull]

Epic

But like one more thing, so inflection points are where the concavity changes

[spiral knot]

so inflection points are when f ''(x) = 0

[meager hull]

So like what would the og graph look like sorta for the derivative to come out like that and still have x=0 as an inflection

[meager hull]

Ah

Yk that's like finding a puzzle piece missing

Ahhhh

[spiral knot]

f ''(x) < 0 when f '(x) decreasing
f ''(x) > 0 when f '(x) increasing

[meager hull]

Ok

[spiral knot]

so it's clear that (0,-2) and (2,0) are the point of inflection

[spiral knot]

yeah , i know haha

[meager hull]

Alr tyyy

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[obtuse pebble]

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[limber wharf]

Hello, can someone explain me what does this mean?
$a_m \equiv 1 (\mod p)$

[warm shale]

Frogieder

[royal basin]

do you know modular arithmetic

[limber wharf]

Yes

I know what does modulus mean

[royal basin]

this means a_m is congruent to 1 modulo p

we can't tell you what a_m is because it could be just about anything. but it is probably the m'th term of some sequence called (a_1, a_2, ...)

[limber wharf]

Yes, it is

Oh, so does that mean that p * n + 1 = a_m for some integer n?

[haughty coyote]

By definition of congruence, yes

[limber wharf]

Alright, thank you

For some reason, this specific expression got me bamboozled. Now I kinda know how to handle it

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[frozen jolt]

how do i factor this with only the use of square?

[frozen jolt]

well

factor it with grouping

[daring rock]

try rearranging the terms first, from highest degree to lowest

@frozen jolt

[frozen jolt]

2z^5-2z^4-z^3+z^2

[daring rock]

Now you should be able to factor in groups

Or are you confused about how grouping works in general?

[frozen jolt]

no

it's just

if you look

above the submit button

i can only used ^2 power

and idk how to factor it with just ^2

cuz i also need ^3

[daring rock]

I don't think you should get a power of 3

just 2

[frozen jolt]

so

[daring rock]

oh maybe it's not factorable farther

wait let me actually write it out lol

[frozen jolt]

ok

[daring rock]

yeah 2 should be the only power you need

Can you show what you did?

[frozen jolt]

i think you can't factor it

but apparently you can

wait

i should just group it with z^2

[daring rock]

I'm not exactly sure what you mean

but yes it is factorable

[frozen jolt]

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[vital falcon]

Hello could please someone help me?

I already found the point P(2|3) but I dont know how to get the equation for the tangent now.

[tardy epoch]

have you learned derivatives yet

[vital falcon]

yes

[high lily]

don't really need derivatives for this

note that the tangent at P will be perpendicular to the radius/diameter through P

[vital falcon]

but what does it help me

[high lily]

you have coordinates of points on that diameter which could be used to determine its slope
and hence the slope of a line perpendicular to that

(which will be the slope of your tangent line)

[vital falcon]

I have already calculated that the gradient of AOP is 2

[high lily]

and do you know the relation between slopes of perpendicular lines?

[vital falcon]

no

[high lily]

it's something you should know if you're already up to derivatives,
a quick search could be done to refresh your memory

[vital falcon]

slopes of perpendicular lines are opposite reciprocals

So the gradient of the tangent is -1/2

[high lily]

opposite is bad, you should use negative

you now have the slope and a point on your desired line,
you should be able to get the equation frombhere

[vital falcon]

yes now i get it y=-1/2x+4

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[marble thicket]

I need help

[marble thicket]

How do you find the area of a triangle when given three vertices

[patent jetty]

Take half of determinant

Modulus of that

x1 y1 1
x2 y2 1
x3 y3 1

I think u also include z if it's 3d stuff

[patent jetty]

ig*

[marble thicket]

I did not understand a single thing you said because we had a different method in my class which I forgot

My problem looks like this

Nvm is the answer-9

So how do I end this?

[tardy epoch]

.close

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[tardy epoch]

I did it for you. But you can type what I did to do the same thing

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[desert river]

i just made my first math paper. can some one take a quick look and tell me about any vital flaws?

[warm canopy]

Go ahead

[rigid lintel]

you might want to blur your name

[desert river]

ehh

here is new one

[dark stirrup]

abstract is nice and succinct

perhaps mentioning your methods would be good to include

That's personal preference though.

In my experience, an abstract transforms each section into a sentence. But just preference. Nothing wrong with your abstract imo

"Newton’s third law of motion states that force is the
product of mass acceleration"
I think you need an and between mass acceleration

In your intro, maybe mention how differentiation is a measure of the instantaneous change of a function. Just another opinion