help-33

[magic hare]

Meanwhile 1^+ is something a tiny bit bigger than 1

But its still 1

It acts as 1 in most cases

But not all the time

So lets try to solve it together

We plug in 1^+

[stable axle]

i'm sorry, but how would I write that into a calculator? 😅

[magic hare]

You really cant

[stable axle]

oh

[magic hare]

Its like 1

But it doesnt act like 1 when you substract it by itself

[magic hare]

For example here in the denominator

(1^+)-1

A number a bit bigger than 1

Minus 1

Doesnt that give us something a bit bigger than 0?

[stable axle]

yes, should be

[magic hare]

So thats 0^+

Now lets repeat the same thing

When we devide

1/0.1

What do we get?

1/0.00000000001

We are getting closer to 0

Yes it was 10

If you try something even closer to 0

1/0.00000000000000001 you get an even bigger value

So the infinitely tiny bit bigger than 0 thing

When its 1/ by that number

[stable axle]

is the limit?

[magic hare]

Not the limit

But

When its 1/ (that number)

You should get something extremely large right?

[stable axle]

yes

[magic hare]

Which is infinity

Because 0^+ is something infinitely a little bit bigger than 0

Limit is really just all about approaching and stuff about infinity

1/(0^+)=inf

But as you saw

From the right we got -inf

And from the left we got +inf

[stable axle]

yes

[magic hare]

Yeah but we originally got -1/(0^+) which is just -[1/(0^+)]

So its -[inf] which is just -inf

So from the right its -inf

And from the left its +inf

They are not equal amd therefor there is no limit

It diverges

One side is plus inf and the other is negative inf

Again, to find the limit the left and right side must be equal to eachother

You will understand it better when we look at the next question

The answer to the first question is: diverges

Diverges means that it takes infinitely multiple values

And not one

[stable axle]

is it wrong to call this infinity or is it the same?

[magic hare]

Its wrong because infinity is just positive infinity, but the thing is when we get closer to x=1 its going to negative infinity and positive infinity at the same time

Lets try the second question

Maybe that could clear the fog

What is the function again?

[stable axle]

For the second function:

Should I simplify this one, like we tried for the first one?

[magic hare]

Sorry had to go flr s bit

For a bit*

[magic hare]

Here you can clearly see that you cant simplify it so you dont have to

We can skip that part and instantly look for right and left limit

Because if we plug 1 we get 1/0

Same problem

So lets start with taking the left limit

Can you do it?

What was (1^-)-1 again?

[stable axle]

Sorry, i'm slow

[magic hare]

If 1^- is something a little bit smaller than 1

And we substract 1 from it

What do we get?

[stable axle]

negative number

[magic hare]

No no not at all its fine, its not an easy concept to grasp

Yes we get a negative number that is a little bit smaller than 0

So 0^- right?

[stable axle]

yes

[magic hare]

And if we square that

What does it turn into?

From negative

[stable axle]

to positive

[magic hare]

Exactly

But its still 0

Because 0² is 0

But this time its a little bit bigger than 0

Because we squared it

It became positive

So at the end we are left with 1/(0^+)

Imagine

(-0.01)²

You get 0.00001 right?

[stable axle]

yes

[magic hare]

Yeah its similar to this

It was something a bit smaller than 0 which is negative

When we squared it it became something a bit bigger than 0 which is positive

So finally we have 1/(0^+)

And what was the answer to that again?

1 divided by something really really really close to zero

And its positive

[stable axle]

I would have thought of 0.99999..., but that was for the other one close to 1

[magic hare]

Mhm so its not that

Its more like

0.000000000000000000000000... 1

Infinitely close to 0

But its still not 0

So 1/ 0.000000000000000000000000000000... 1

And what is the answer to that?

Something really big right?

[stable axle]

yes

[magic hare]

Which is infinity

We accept it as infinity

Becaus again there are infinite amount of 0s before it gets to the 1 so its 0.00000(infinite times)1

Yeah so what is the answer to the left limit overall?

1/(0^+)=?

[stable axle]

isn't it still infinity?

[magic hare]

Yes

That is true

So the left limit is infinity

Now lets check the right limit

Basically similar but this time we have something a bit bigger than 1 instead of something a bit smaller than 1

Which means

What is the answer for

[stable axle]

that was 0.999999999...

right?

[magic hare]

1.0000000000(infinite times)1 - 1

[magic hare]

(1^-) is that

(1^+) is 1.0000(infinitely)1

So what do we get if we do (1^+)-1

(1.0000...1)-1=?

What is the answer to that?

[stable axle]

I would have thought of something like infinity, but I guess we are not done yet so we can't conclude that yet?😅

[magic hare]

It is infinity

Because we get the same thing

1/(0^+)²= 1(0^+)=inf

So the right limit is also infinity

And the left limit is also infinity

And when both right and left limits are the same value

That is the answer to our main limit

So

What is the answer for the second question now?

[stable axle]

infinity

[magic hare]

Exactly

Now onto our last function

[stable axle]

is it possible that those three functions doesn't have a limit through 1?

[magic hare]

The first one diverges

The second one is infinity

These are the limits

Of these functions

What we calculated were the limits of the functions

[stable axle]

If the limit is infinity, wouldn't it "not be" a limit?

[magic hare]

It countsas a limit

Counts*

As a limit

[stable axle]

ok

[magic hare]

Sometimes negative infinity

Sometimes positive infinity

Could be an answer to a limit

But it cant be something like 1/0 0/0 1^inf because these are what we call an indeterminate form

Or undefined stuff

Thats where we have to check for left and right limits

0^0

These are all indeterminate forms

When you plug the number and you get one of these values you either have to simplify(mostly the case) or you have to check the right and left limit

If you just get infinity or negative infinity then these are valid answers

And you dont need to simplify

But you usually get those when you check the right or left limit

Anyways

Lets check the last function

I will give you another example if you need to understand the topic better

[stable axle]

thank you. As I understand it right now from what we have done, you just plug in the the number from x → a as we did here (your screenshot)

[magic hare]

Yes

We just plug that number in and try it

Thats the first thing we do

If we get 0/0 or 1/0 then we try to simplify or check the right and left limit

What was the last function?

[magic hare]

This one should be pretty simple

[stable axle]

i tried to plug in 1, it gave me 0

[magic hare]

So thats the answer

🙌

You dont need to check right and left limit if it just works

[stable axle]

ok, thank you

[magic hare]

No problem, ask anything you want

You just need to grasp what approaching from the left and from the right means

[stable axle]

if you have time, can you give me a brief explanation on how to check left and right limits?

[magic hare]

And you will be good

Yeah sure

Let me see

You might need this

These are conclusions

So when you plug in the value you might get these

And these are the answers to them

I dont like memorizing

So you shouldnt memorize it

You should try to understand it

So let me provide an example

Try this

[stable axle]

simplifying first?

[magic hare]

We plug in the number first

[stable axle]

ah, ok

[magic hare]

What do we get?

[stable axle]

i would get 11/0.
Isn't that the same problem as before?

[magic hare]

Yup

Its undefined so what do we try now?

[stable axle]

left and right limits

[magic hare]

Yess

Now lets start with the right limit this time

We try to plug in first

[stable axle]

i'm sorry but was there a formula for it or did we just plug it in like this (from your screenshot):

[magic hare]

No formula

[stable axle]

yes, i thought of that

[magic hare]

We just plug it in

We do the same thing

Its a normal limit after all

Just different number (2^+) in this case

[magic hare]

What is the result for this?

[stable axle]

one moment

[magic hare]

Take ur time

[stable axle]

was the + bigger than two?

[magic hare]

Yes

So something like 2.000000000....1

[stable axle]

so it's like that for every x→ n, just with the different number.

[magic hare]

Yes

Thats how you know

What value its gonna get

When x is aproaching n

We try that first

So what do we get?

[stable axle]

yes

[magic hare]

What is 11/(0^+)?

[stable axle]

would it be something like 11/0.00000000.....1?

[magic hare]

Yeah and what is that?

[stable axle]

infinity?

[magic hare]

Yes.

So the right limit is infinity

Now lets check the left limit

[stable axle]

that was the 2^-, right?

[magic hare]

Yupp

First step of taking limits?

[stable axle]

plug in the n from x→n

[magic hare]

Yes

Now what do we get when we do that?

X-->(2^-)

What do we get when we plug that?

[stable axle]

is the method correct?

[magic hare]

Yes

That is correct!

Now what is the last conclusion?

Like whats the value of that?

Something a bit smaller than 8+3 is something a bit smaller than 11 so 11^-

But what about the denominator?

[stable axle]

the value of this

would that be:
wouldn't that be 1.999999....?

[magic hare]

No

Its

Not

[stable axle]

oh yeah, 0.0000000....1?

[magic hare]

Yes but negative

Cuz its a bit smaller than 0

So its -0.00000...1

[stable axle]

ah, yes

[magic hare]

Now whats the answer for that conclusion?

[stable axle]

so it would become -infinity

[magic hare]

Exactly

Great job

Now

Left limit is -inf

Right limit is inf

So is there a limit?

[stable axle]

as you said before, it would count as a limit

[magic hare]

Oh did i?

[stable axle]

yes

[magic hare]

My bad i meant that only infinity was a limit

[stable axle]

ah, ok

[magic hare]

It doesnt count as a limit when the right and left limit are not equal

Which is the case here

One is left limit the other is right limit

[stable axle]

Then i guess that there is no limit here...

[magic hare]

So the limit doesnt exist it diverges

Doesnt coverge

Converge*

Its the same thing really

No limit

Limit does not exist

or DNE

I cant tell the difference myself

But it is certain that when the right limit doesnt match the left limit then there is no main limit

[stable axle]

ok, so they both have to be infinity

[magic hare]

No they both have to be the same value

They can both be 0,2,3,inf, -inf

But they both MUST be equal for a limit to exist

So here if they were both inf

Then the answer would be inf

If they both were -inf then the answer is -inf

Inf they both were 0 then the answer would be 0

[stable axle]

so about this, wouldn't it be the last function,
sqrt (x-1)?

[magic hare]

The second also has a limit value

Because remember? The right and the left limit were equal to eachother

They were both inf

So there is a limit

Value

[stable axle]

so it's both, right?

[magic hare]

1/(x-1)² and sqrt(x-1) yes

[stable axle]

to understand correctly, do they both have to be inf?
Does it still count if the right and left was -inf and inf?

or does it have to be exactly equal?

[magic hare]

Yes they both have to be the same value

Yes

[stable axle]

ok

I think that was it.

[magic hare]

Gg

[stable axle]

Thank you really much

[magic hare]

No worries, thanks for listening :))

[stable axle]

I really appreciate that time you used and help.

[magic hare]

Dont mention it

Good luck

And have a nice rest of your day

[stable axle]

and you too

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[marsh citrus]

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[naive rock]

is this just asking me to find the solution to the defferential equation

[naive rock]

also what does exp mean in E

[fervent rampart]

,, \exp(x) = e^x

[elfin berry]

cloud

[naive rock]

ok

but what is the question asking us to solve

[fervent rampart]

each function is a solution to one of the differential equations. you can figure out which one is which either by solving the DE (if you know how) or by plugging in the given solutions and seeing if it matches

[naive rock]

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[naive rock]

is number 1 correct

they dont eqaul eachother

[eager tinsel]

$6exp(7x)=6e^{7x}$

[elfin berry]

77²

[naive rock]

ohhhhhhhhhhh

thx

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[eternal sequoia]

How can I expand this out so its moreover the comparison of two sums?

[eternal sequoia]

Rather

it is in this form

Where U(Si) can be substituted into E_i but is there a way to write a comparison with one sum on one side of a inequality, and the other sums for the other energies on the other side?

I would expect it would look something like this but I have no clue if this comparison is true about the two sums

[marsh citrus]

@eternal sequoia Has your question been resolved?

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[tame ore]

anyone got a way to start?

[marsh citrus]

@tame ore Has your question been resolved?

[potent swan]

Hey @tame ore . Gladly!

[tame ore]

thanks 🙂

[potent swan]

Key points to consider:

When do they each start?
Where are they going?
George: Alton -> Chawton -> Basing -> Alton
Jim: Alton -> Basing -> Chawton -> Alton

What we would have to do is find their km/h, then divide that with their total distance

I can't give you direct calculations/answers, as that is agains the guideline, but let me know if you need any more help!

(feel free to ping me! :)

[tame ore]

i think i shld form some simultaenous equations but idk what the s d t is for them invidiually

so

George takes 2 hours in total to travel 12km at 6km/hr

Jim takes 1 hour 30 minutes to travel 12km at 18km/hr but starts 30 minutes later

<@&286206848099549185>

i asked a q

[potent swan]

Yes! Correct!

Now that you know this, you may divide the distance between them, with their total sum of km/h's!

[tame ore]

wdym

as in like

[potent swan]

as in: 5/24 km per hour

you then get the minutes

[tame ore]

whys it 5?

sorry im a bit slow

[potent swan]

sorry, that's my bad! The 30 minutes early start, gives him 3 km

From 12:00 to 12:30, George runs for 30 minutes (or 0.5 hours) before Jim even starts.
In 0.5 hours at 6 km/h, George covers 6×0.5=3 km.

does that make sense? :)

[tame ore]

yeah got the 3km

can i j model it as a straight line

so easier to see

like two people moving directly to eachother

[tame ore]

yeah that makes sense

so then the distance between them is 9km

[potent swan]

The 5 km is the total distance between Chawton and Basing, which is where George and Jim are heading towards each other

[tame ore]

can i set up some simultaenous equations?

[tame ore]

yeah

is there a way i can set up two equations

and find exactly where they meet

[potent swan]

Sure!

[tame ore]

is there a way i can fill this out

[potent swan]

Yes!

[tame ore]

ik george has alr travelled 3km

and travels for 30 mins more than jim

[potent swan]

alright, I'll try to explain it with 2 equations instead

[tame ore]

thanks

im j confused what the respective distances and times wld be

and then we can solve to find a distance and see how for it is from Basing

[potent swan]

first, find the time to meet, which is the distance they have to move dividided by the relative speed (Relative speed being both of their's km/h combined)

Let me know when you have got that

and feel free to ping me

[tame ore]

yh im not sure how to get the distance

@potent swan how do i find the distance?

ik theres 9km distance between them

?

idek

[potent swan]

distance is just speed x time
Relative Speed = Speed of Object 1 + Speed of Object 2 (when they are moving towards each other)

Time to Meet = Distance/Relative speed
​
George's speed: 6 km/h
jims speedies: 18 km/h
Relative speed, those two + each other (you got that)
The total distance between Chawton and Basing is 5 km.

I think you got it :)

Let me know if you need help (feel free to ping)

[tame ore]

yeah im not sure

can u tell me the answer so i can work back

im still stuck

[potent swan]

I'm not allowed to tell direct answers/calculations

ah, right @tame ore

the decimal number it gives ishann, would be the hours

you can turn those hours into minutes by * with 60! (why 60 you may ask? Well because there are 60 minutes in 1 hour!)

Let me know when you have got to that! :) (feel free to ping)

@tame ore

[marsh citrus]

@tame ore Has your question been resolved?

[tame ore]

oh yeah i got it now

thanks

i j watched a quick vid on realtive speed

its the same as saying one is stationary and the other object is moving at the speed of boith combined

thats the time at which they meet and then i sub it back into one for speed x time to get the distance

[potent swan]

Perfect!

Do you need me to help you further? :)

[tame ore]

nah its ok i got thanks

.close

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[celest folio]

alr so i have (6x^6/x)/(x^2) * (1-ln(x))

how do i simplify the first term

can i do 6x^6/x * x^-2

[hard gull]

Is this correct?

Nvm it's correct :d

[hard gull]

Actually u can do that and use law of indices

[celest folio]

can i do exponent rules

what is law of indices

i only know exponent rules

[hard gull]

Exponent rules, same

[celest folio]

so a^m * a^n = a^m-n

well, +

[celest folio]

but we have a -2

so its gonna be minus

[hard gull]

Ye

[celest folio]

ok ty

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[fluid umbra]

Ten red cards and ten black cards are placed in a bag. You choose one card and then another without replacing the first card. What is the probability that the first card will be red and the second will be black?

[unreal geode]

!status

[marsh citrus]

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

[rugged spindle]

20 cards may be chosen for the first one, and 19 may be chosen for the second one (since you remove one). what's the probability of picking a red card for just the first card pick?

[marsh citrus]

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[marsh citrus]

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[undone bear]

if anyone knows how to do this i would appreciate it, im working on other problems and will keep discord open

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[frail lodge]

Can someone show that
[
\frac{(n+1)(3n^2+1)}{n(3(n+1)^2+1)}
]

is at most 1?

[elfin berry]

Kakaka

[frail lodge]

Chatgpt is giving me this gem

[final skiff]

Is n any natural number, integer, real, etc?

[frail lodge]

integer n≥1

Ah, I suppose you could work backwards by assuming >1, then showing contradiction, but I want to it get algebraically directly

Here is their solution

I don't get how they got the first inequality

[main idol]

they took the (n+1)st term and divided by the nth term

assuming $a_n = \frac{(n+1)(3n^2+1)}{n(3(n+1)^2+1)}$

[elfin berry]

riemann

[frail lodge]

Don't think so?

here is the full context

It's from a Real analysis class

[main idol]

oh then yes you're right that's not a_n

[main idol]

are you trying to understand their first inequality?

[frail lodge]

yeah

[main idol]

or some other method?

this specifically?

[frail lodge]

yes

Hollup

[main idol]

to see the inequality holds, try working backwards by assuming it's true. then cancel the 3n^2 + 1 term, and divide the top and bottom by (n+1)

[frail lodge]

yes, but that's like

completely missing the forwards thought process

like how would i have thought of making an upper bound looking like this in the first place

also

I ultimately want to show that $a_{n+1}\leq a_n$ for monotontically decreasing right?

[elfin berry]

Kakaka

[frail lodge]

suppose on the contrary that $a_{n+1}\geq a_n$

[elfin berry]

Kakaka

[main idol]

real analysis is largely about memorizing tricks

[frail lodge]

Then $\frac{n+1}{3(n+1)^2+1}>\frac n{3n^2+1}$ for $n\geq 1$

aint this true?

[elfin berry]

Kakaka

[frail lodge]

wait wait

[marsh citrus]

@frail lodge Has your question been resolved?

[marsh citrus]

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[robust fractal]

If an augmented matrix in reduced row echelon form has 2 rows and 3 columns (to the left of the vertical bar), then the corresponding linear system has infinitely many solutions.
the answer is false but i dont know why

[marsh citrus]

@robust fractal Has your question been resolved?

[static quarry]

consider:

1 1 1 | 0
1 1 1 | 1

how many solutions does that have?

[robust fractal]

no solutions?

[marsh citrus]

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[frosty gyro]

in intersection we multiply while in union we add right

[frosty gyro]

.close

[marsh citrus]

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[slate galleon]

can someone help me with this?

as how i understand it, to prove that limit does not exist we need to show that the limit has different values for different values of a variable like lets say m, like we have to choose a path for example y = mx and then show that the limit is dependent on only m

[leaden monolith]

We don’t need to

But we can

[slate galleon]

then how?

[leaden monolith]

This only considers straight lines through the origin

For a limit to exist we need every possible path through the origin to be the same

[slate galleon]

ok

[leaden monolith]

Straight lines are not the only path

Hence I say we can use straight lines to prove the limit does not exist

But we may choose non-straight lines to do so as well

[leaden monolith]

This is an example of a non-straight line

[slate galleon]

yeah i understand that, but even after putting in y = -sin(x) i am unable to simplify it.....

maybe i am doing something wrong

[leaden monolith]

Uh looks weird

The limit becomes just x-> 0

Perhaps you can use lhopital’s rule?

[slate galleon]

i tried

[leaden monolith]

Where do you get stuck with that

[marsh citrus]

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[dense hemlock]

yo

[dense hemlock]

the second derivative is d^2y/dx^2

why isn't the d in the denominator squared

i asked my teacher and she said it's because the square is suppsed to cover dx (as in (dx)^2) but parebthesis isnt written for whatever reason

is this correct?

[leaden monolith]

This is just notation

Just like how we write y*y as y²

[dense hemlock]

yes but

(d/dx)(dy/dx)

makes sense that it's d^2 on top

but dx * dx is (dx)^2

[leaden monolith]

Nope the command hates me

$\frac{d^2}{dxdx}y$

[elfin berry]

frosst

[leaden monolith]

Because when you differentiate with respect to different variables it looks like that

$\pdv[2]{f}{x}{y}$

[elfin berry]

frosst

[dense hemlock]

yeah makes sense

[leaden monolith]

Here the order is important

[dense hemlock]

but how is (dx)(dx)=dx^2 and not =(dx)^2

[leaden monolith]

It’s just for simplicity

[dense hemlock]

mk

[lucid zenith]

you're not actually doing dx * dx

it's just notation

derivatives are not fractions

[fervent rampart]

because everyone knows what it means and doesn't want to use more pen strokes

[marsh citrus]

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[magic olive]

that's what i got so far

[marsh citrus]

@magic olive Has your question been resolved?

[magic olive]

<@&286206848099549185>

[marsh citrus]

@magic olive Has your question been resolved?

[marsh citrus]

@magic olive Has your question been resolved?

[magic olive]

Nah not yet

[round furnace]

You have found alpha and beta using eq 1 and 2

Now use eq 3

Just plug those values for alpha and beta in

[round furnace]

@magic olive ?

[magic olive]

do you know how to do part b?

[round furnace]

Do you know how a rhombus looks like?

[magic olive]

yeah

[round furnace]

Well I’ll just give my idea because i honestly dont know how to hint this without giving the (what i think is the) answer

Add $\vec{AB}$ and $\vec{BC}$

[elfin berry]

thijs2725

[round furnace]

If you draw it in 2d with 3 random points abc youll see what i mean

[magic olive]

hmm

[marsh citrus]

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[marsh citrus]

@velvet pike Has your question been resolved?

[marsh citrus]

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[flint vault]

That’s what I have so far

[limber hearth]

2n+2=2(n+1)

[flint vault]

Yup

Cancel 1 power from bottom?

[limber hearth]

Become 2n+1 indeed

[flint vault]

should i split the bottom bracket

[limber hearth]

Like (n+1)*(n+1)^2n?

[flint vault]

yeah

not sure where to go from there lol

[limber hearth]

(n+1)^2n = (n(1+1/n))^2n

[flint vault]

woahhh

eeeeee

[limber hearth]

:)

[flint vault]

didnt think of that

that gets rid of n^2n

Am I missing something

Does it just diverge

[limber hearth]

For me it does

[flint vault]

Let me try putting that in

Apparently not

It’s supposed to be less than 1 which I’m confused on

[rocky flicker]

i might not be much help here but isnt (1+1/n)^n=e

or i mean

it goes into e

[flint vault]

Yes but the n on top

That goes to infinity

Infinity/e is infinity

[rocky flicker]

yeah i see

[flint vault]

Oh wait

I completely forgot one of the brackets lol

There’s still (n+1)

[rocky flicker]

wait you just completely ignored the n+1?

[flint vault]

Forgot**

[rocky flicker]

ok so its 2(2n+1)/e^2(n+1) right?

[flint vault]

Yes

[rocky flicker]

well there you have it you take out 2/e^2 now and you have same level on both sides

it should be 2

[flint vault]

So 4/e^2

[rocky flicker]

yeah

[flint vault]

Alright that worked

Thanks for the help

[rocky flicker]

no problem

[flint vault]

.close

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[iron meadow]

think about what could you multiply both sides by to cancel out some square roots?

try to multiply both sides by $\sqrt{\f{x_2}{x_1}}$

[elfin berry]

∫oosh

[iron meadow]

on one side, it will simpliffy to x_2 / x_1, then on the other it'll simplify to 1, see if you can solve from there, should be pretty straightforward

yes on the left

[magic hare]

You isolate x_2

[iron meadow]

so after we multiply both sides by the factor we said, it becomes:

$\f{1}{2p_1}\f{x_2}{x_1}=\f{1}{2(p_2+t)}$

[elfin berry]

∫oosh

[iron meadow]

does that make sense so far how the square roots simplify?

now just isolate x_2 on one side

and should also probably include in the solution that x_2 and x_1 can't be 0 by the original equation

[magic hare]

Lets hope its included

If not mention it so that you get extra credits 😎

[iron meadow]

that doesn't seem right

check your arithmetic

yes, it would be clearer to write it as (p_2+t)x_2 / (p_1) = x_1 to make sure you arent implying x_2 is in the denominator there

[marsh citrus]

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[severe owl]

let's aaoron's age be x

Then form relations

[iron meadow]

or better yet use A for aaron's age and R for ron's age

[severe owl]

One variable ?

[iron meadow]

now see if you can use every piece of information to write an equation

so "Aaron is 5 years younger than Ron"... how would you write this using the variables R and A

@strong trench ?

I guess the OP left...

[marsh citrus]

@strong trench Has your question been resolved?

[marsh citrus]

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[dapper robin]

ok silly doubt does $\sqrt{x^2} = \pm{x}$ or $\sqrt{x^2} = |x|$

[elfin berry]

PianoDolphin

[late geode]

the latter

[dapper robin]

ok

[thorny bison]

|x|

[lucid turret]

root of x^2 is just x

-x will satisfy tho it is not approved

[thorny bison]

Think about it for a min what u just said

[lucid turret]

x should be positive

[dapper robin]

so when I have $\sqrt{x^2} = 3$ then does $x = \pm{3}$ or just 3

[elfin berry]

PianoDolphin

[late geode]

x is a variable, it isn't inherently positive/negative

[dapper robin]

wait sorry

[thorny bison]

wat

[dapper robin]

I meant

[weary karma]

+-3

[lucid turret]

so the best is to use:
root (x^2) = abs(x)

[thorny bison]

Yes

[lucid turret]

Yes, +-3

because it will satisfy the equation

but doing normal calculations like root (3^2),
you can't give the answer as -3

[wispy ingot]

Sin x = log x
Solve for x

[dapper robin]

so if I have equation then I generally take +- because there we are finding values that satisfy that equatoin

[thorny bison]

x = |3|
x = +3 , -3

[glossy flint]

!occupied

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[lucid turret]

Prof. Alberto Z. 🙏

[dapper robin]

right?

so I will take x to be both +- in this case

so when will I use mod then?

[glossy flint]

I'm just a student, lol

[lucid turret]

yes, because we are finding values which would satisy the equation given.

[dapper robin]

exactly

[thorny bison]

2.219 smthin

[dapper robin]

but when then do I ues mod

[lucid turret]

when needed

[thorny bison]

Everytime

[lucid turret]

like in root(x^2) = |x|

[late geode]

it'd be clearer if you write the intermediate step
$$|x| = 3$$
and the solutions to that would be $x=-3,3$

[elfin berry]

ℝαμOmeganato5

[lucid turret]

yeah

you can also show it that way

[dapper robin]

alright got it

[thorny bison]

I did

For him

[dapper robin]

you should have put another equals to in the midle

middle

[lucid turret]

Prof. Alberto Z. has all the knowledge of Maths and its required expertise.

He's not a student.

A 256 year old Prof.

[late geode]

issues arise when you try to shortcut it to $\pm$, take
$$\sqrt{x^2} = -3$$
if you try that, you'd get $\pm x = -3$ \
and reach the wrong conclusion that $x=-3,3$ for that equation

[elfin berry]

ℝαμOmeganato5

[dapper robin]

there are no solutions to this eq in the real numbers right?

[lucid turret]

wait

[late geode]

yes

[lucid turret]

|x| can't be negative

so we're done basically

[dapper robin]

yep

[late geode]

whether
sqrt(x^2) is x or -x
depends on the sign of x

[lucid turret]

in calculations

[late geode]

so its poor to try to represent that with a lone +-

[lucid turret]

Ok i will use |x| from now onwards

[thorny bison]

bro has all the knowledge still can't solve the Millenium problems 😔😔

[thorny bison]

We have no chance

[stoic slate]

What is this? I guess it is a typo

[lucid turret]

use logic and discretion where necessary

to avoid errors

[thorny bison]

It's english (yea it's a typo and I'm lazy to remove it)

[stoic slate]

Do not use this channel for that

No, you have a channel for that

It is not up to me to close this channel

[late geode]

Use #bots for stuff like that. Don't do it in other people's channels.
Please don't pressure other people into closing their channel.

[stoic slate]

#bots here you can

[lucid turret]

sorry

[marsh citrus]

@dapper robin Has your question been resolved?

[marsh citrus]

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[lyric bay]

Hello

[lyric bay]

Why is this the normal vector for this plane?

[graceful mountain]

The plane equation is derived from the dot product between the normal vector and a given vector
And as such if N is a normal vector you would get N_1 * x_1 + N_2 * x_2 and so forth.
And that gives you your expression, where you can just "read" the values of the normal vector.

[lyric bay]

Ohh ic

Thanks!

.close

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[alpine tendon]

do i do integration by parts

[alpine tendon]

this problems overwhelming me i feel like so much is going on

[late geode]

consider integration by substitution

[alpine tendon]

can i distribute the square

[late geode]

yes

[alpine tendon]

ok

so 2x/100 e^(-x^2/100) dx

u=x^2/100 du=2x/100 dx

[late geode]

yeh

[alpine tendon]

so then its just e^-u du

what do i do abt the t and 0

[late geode]

update the bounds
when x=0, u = ?
when x=t, u = ?

[alpine tendon]

x=0, u = 0
x=t, u = t^2/100

[late geode]

yeh

[alpine tendon]

so ans is -e^((-t^2)/100) +1

thanks

.close

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[covert scarab]

how to calculate a 3D regression vector

[low ruin]

!original

[marsh citrus]

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.

@covert scarab Has your question been resolved?

[covert scarab]

Its a question I have, I wasnt given it. If I can calculate a 2d regression line how can I go about calculating a 3d regression vector given a set of points

[low ruin]

What do you mean by 3D regression vector.

do you want to have 3 independent variables

[covert scarab]

2 independant

to have an x and z variables be independant and the y be the dependant

[low ruin]

You can use the following

https://en.wikipedia.org/wiki/Ordinary_least_squares#Assumptions

The part you want to refer to is matrix/vector formulation

Particularly the formula for beta-hat

[covert scarab]

.close

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[marsh citrus]

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[regal dagger]

can anyone help me w inverse trigo

[marsh citrus]

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[fossil pebble]

why does this work?

[fossil pebble]

I know how it’s done but i do better when i understand why it works as opposed to just “this works”

[glossy flint]

Because the second row is the square of the first row

[fossil pebble]

wdym?

[glossy flint]

What's 1²?

[fossil pebble]

1

oh are you talking about the inputs and outputs?

I was asking about why doing the little carrot thing works in finding out the degress of the polynomial associated

[glossy flint]

Yeah sorry

[fossil pebble]

all good

[glossy flint]

Which carrot thing? 🤔

[fossil pebble]

Essentially it says that the degree of the polynomial can be find by evaluating successive differences in the output values. The number of diffrences needed to make a constant value (in this instance 2) determines the degree of the polynomial

which once again, in this instance would be 2

[glossy flint]

Uhm I've never seen this thing, I believe there's a special theorem about those kind of sequences, but I'm not aware of it

[marsh citrus]

@fossil pebble Has your question been resolved?

[fossil pebble]

<@&286206848099549185>

[eager tinsel]

what's exactly the problem you're facing?

[fossil pebble]

understanding why it works

i don’t actually understand it

that’s just what my teacher told me to do

[eager tinsel]

see the The degree of polynomial = The Order of difference, at which the difference becomes constant

[eager tinsel]

like here we got 2nd order difference to be constant

[fossil pebble]

which means a second degree polynomial

[eager tinsel]

yes

[fossil pebble]

is there a reason why it works? Or is that just how it is

[eager tinsel]

you can prove it

[fossil pebble]

how would i go about doing this?

or finding a proof

[eager tinsel]

okay nvm, idk the general proof that'll be accepted formally

but you'll get the idea of how to prove it generally

so let some function be f(n)

and decide it's degree to whatever you want

for example take degree=2

so it will be $f(n)=an^{2}+bn+c$

[elfin berry]

77²

[eager tinsel]

now you find the f(n-1)

[eager tinsel]

are you here btw?

[fossil pebble]

mhm

[eager tinsel]

okok

[eager tinsel]

.

[fossil pebble]

so i just shove in (n-1)?

[eager tinsel]

yes

then you do f(n)-f(n-1)

so like the first order difference

[fossil pebble]

Ohhh im seeing

[eager tinsel]

once you solve it, you'll get some linear thing

call that thing g(n)

then find g(n-1)

and g(n)-g(n-1) will be constant

which also will be the second order difference

and since we assumed f to be 2nd degree

so you can get the idea of proving this thing for general order

[fossil pebble]

yeah

thank you

[eager tinsel]

np

[fossil pebble]

.close

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[round furnace]

Anyone that has an idea how to solve this one?

[night mica]

no idea

[round furnace]

I would also be happy with someone explaining what happened in this part of the solution

I am completely lost from the point where he defined the contour $\Gamma$

[elfin berry]

thijs2725

[round furnace]

Nah scratch that I am also already lost at the part where he defines the function f(z)

[spark otter]

it's like a variable change

"z = x^b"

so you look at the function you're integrating "after" the variable change

[round furnace]

O yeah I see it now woops this was obvious

[spark otter]

my best guess as to what Gamma looks like

it starts at R

does a loop gamma

(I drew a clockwise loop but I think it's counterclockwise)

then goes towards r

then does a clockwise loop Phi

and goes back to R to finish

the goal is to make the countour encompass more and more of the positive real number line

so r-> 0, R -> infinity

[round furnace]

Ooooh like that ok that makes it a lot more clear

Do you have an idea why some of the integrals for the contour are in terms of x and others in terms of z?

Is it just because we are integrating along the real number line there?

[marsh citrus]

@round furnace Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[stable garnet]

I have a question on number 39 and here my work but I don’t know if I did it right

[cunning fiber]

this is fine

I have no idea what step 3 is supposed to mean

[stable garnet]

To separate the x from 6

So it’ would be just x=

[cunning fiber]

so $x=\frac{4\sqrt 3}{6}$

[elfin berry]

Civil Service Pigeon

[cunning fiber]

ah then that's fine

it looked like the 4 was an index of the radical

[cunning fiber]

just simplify it and you're done

[stable garnet]

To simplify it do I have to root the 6 or smt?

[cunning fiber]

4/6 = ?

[stable garnet]

oh thank you so it would be

[cunning fiber]

✅

[stable garnet]

Thank you

[cunning fiber]

!done

[marsh citrus]

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

[stable garnet]

.close

[marsh citrus]

Channel closed

Closed by @stable garnet

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

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Remember:
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[stray raven]

how to find the compliment angle of this expression?

[stray raven]

does it turn into this

67° 25' 52"

[marsh citrus]

@stray raven Has your question been resolved?

[stray raven]

.close

[marsh citrus]

Channel closed

Closed by @stray raven

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Available help channel!

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Remember:
• 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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[idle bough]

This is what i got so far

What do i do next..

[unreal oxide]

Okay so uve found f'(x)

U just need to find f'(pi/8)

[idle bough]

Oh .. hmm

[unreal oxide]

Can u do it?

[idle bough]

Imma try

[idle bough]

I don't think i understand

[unreal oxide]

Why did u write a 4 in the 3rd line?

In the numerator

Aftrr sec

U r correct upto the second line

[idle bough]

Oh.. cause its sec^2.2(pi/8)?

[unreal oxide]

Its sec²2(pi/8)

sec²x for any x stands for (secx)²

Evaluate secx first, then square the result

[idle bough]

Ohh

[idle bough]

Is the answer 4?

Uhm hi?

[unreal oxide]

Yes this is correct

[marsh citrus]

@idle bough Has your question been resolved?

[idle bough]

Thankyou so much

[marsh citrus]

Channel closed

Closed by @idle bough

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[marsh citrus]

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Channel closed

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Available help channel!

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Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[merry crane]

i graphed it

but I can't tell

[merry crane]

ignore the table and green dots

btw this is the first graph

[peak hearth]

hello! could anyone possibly help me with this problem and explain how you recieved the answer? I'm aware im doing this very wrong lol

I apologize if i intruded in the conversation :(((

[cloud field]

cos(θ)cos(θ)?

[merry crane]

hey this channel is occupied, it would be better to go to the avaialble math help channels which is under "MATH HELP(AVAILABLE)"

[merry crane]

what's with it?

[peak hearth]

oh i apologize! this is my first time using the server

[cloud field]

the question is csc cos

you typed cos cos

[merry crane]

#❓how-to-get-help message

it's fine it can be confusing

[merry crane]

oh my bad

that just makes the 2 graphs the same

so is none of the studetnts wrong

plz help?

<@&286206848099549185>

[marsh citrus]

@merry crane Has your question been resolved?

[vernal mantle]

This is sneaky

Both answers say the same thing

[merry crane]

so are they both correct

but now how would I answer question B

can I write (1/sin(theta))*cos(theta)+1/cos(theta)

[vernal mantle]

I mean just fiddle around with the definitions ig