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[alpine sable]

$xe^{x+x^x}=2$

[ocean seal]

little kitty

[wind cloak]

,w solve xe^(x + x^x) = 2

[ocean seal]

Results provided by WolframAlpha

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Requested by NEONPerseus#2128

[alpine sable]

and the explanation????????

😰

[wind cloak]

Leave that to God

[alpine sable]

😒

.close

[lone heart]

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[wind cloak]

That's a really nice relation there

And it is indeed true

[alpine sable]

ookay....

[wind cloak]

What's your question

[alpine sable]

^

[humble heath]

good luck bro

[wind cloak]

What's the question

[limpid turret]

There is information missing. What are x and y?

[alpine sable]

this is all it give me

[soft pulsar]

That is the form of the solution.

We cannot solve anything because we have no information on how it travels

[alpine sable]

oh

.close

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[alpine sable]

Determine the equation of a line if it's passing trough points A(5, -2) and B(5, 2).

[limpid spade]

https://www.cuemath.com/geometry/two-point-form/ yw

[alpine sable]

Can't solve it due to it being y-y1 =4/0 (x-x1)

Can't divide by zero

I know the formula already

[limpid spade]

Draw the dots and line

What will you see

[willow patio]

No, he is right. He can't draw the line because it is a vertical line and has an undefined slope

This looks like a problem with no solution, right @alpine sable?

[alpine sable]

Yes

[willow patio]

Yea, I agree with you

[alpine sable]

I mean i can solve it by drawing it and i can say y=2 ez but im not sure thats gonna give me points on the exam

[alpine sable]

So does it not have a solution?

Do i just say that?

[willow patio]

Yup, 'no solution'

[gray isle]

there's definitely a solution

not all equations of lines have y in them

[willow patio]

Eh? why?

plenty of problems have no solution

[alpine sable]

How

[limpid spade]

x=5 is a line

[gray isle]

did you attempt to plot the points and draw the line through them?

[alpine sable]

Yes

[gray isle]

show what you drew

[alpine sable]

I got a vertical line

[gray isle]

and best describe that line

[alpine sable]

Passing trough point x=5

[gray isle]

as simply as possible

x=5 is not a point

notice that all points on that line have an x-coord of 5

[alpine sable]

5, any y

[gray isle]

that line is simply x=5

[alpine sable]

So how do i get to that without drawing?

[gray isle]

by noticing that both points have an x-coord of 5

[willow patio]

oh, duh. Yea he is right

[limpid spade]

''The equation of a vertical line passing through (a, b) is of the form (x = a).
This is an exceptional case where the two-point form cannot be used.''

[willow patio]

Sry, mb

x=5 is the line

[alpine sable]

Right so for any other case i would just write whatever number is the same in both points?

For example A(a,b) and (c,b)
Line would be y=b?

Right?

[willow patio]

ya

[alpine sable]

Thank you

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[raven girder]

Which rules do I need to apply here to get the derivative?

I was thinking chain rule (english?) first but then idk what to multiply the outer derivative with

[soft pulsar]

What is the full problem?

Are you integrating or taking the derivative?

[raven girder]

I need to take the derivative of the above function

[marsh rapids]

Looked pretty clear to me

Derivatives are just a bunch of rules

[raven girder]

chain rule

[marsh rapids]

This one is most easily considered as a fraction

[raven girder]

quotient rule

so just quotient rule then?

I was confused because of the e^x^2

thought that's another function so need chain rule somehow

[marsh rapids]

Quotient rule and then see what that gives

And then apply the chain rule if you need to

[raven girder]

ohh I thought chain rule must always come first

nevermind then

[marsh rapids]

There's no rules, only conditions for when properties apply

[marsh rapids]

You can see this as a fraction

But you can also see it as a chain rule first

[raven girder]

you mean I could apply the chain rule in the numerator and again in the denominator first, then do quotient rule?

[marsh rapids]

let f(x) = (x - 1)/(x+1)
g(x) = e^x^2
Then you want (f o g)'(x)

So long as it makes sense

[marsh rapids]

Would that make sense ?

[raven girder]

it doesn't make sense in my head

l0l

but if I apply quotient rule and then after that chain rule, did I not calculate f''(x) instead of f'(x) ?

that confuses me

[marsh rapids]

Take this step by step and you'll see

[raven girder]

k sec

why does wolframalpha say this

I thought e^x doesn't change

[marsh rapids]

It doesn't

[raven girder]

I guess this is the chain rule?

[marsh rapids]

But here it's (t -> e^t)(x^2)

[raven girder]

oh I think now I get it

[marsh rapids]

Sorry gonna have to go

[raven girder]

thank you

I got the correct result

.close

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[raven girder]

Can someone explain this to me?

For example if this was 2x then the derivative would be 2, no? So the x disappears

why is there a + 1?

do you need to apply chain rule here?

oh product rule

nvm

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[wet lark]

💀

[lone heart]

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[alpine sable]

Determine the unknown coordinate of point A so that the points A(2, y), B(5, 2/3) and C(-4, 6) are collinear

[alpine sable]

I've already determined that the explicit equation of a line gotten from points B and C is y=-16/27x + 98/27

Now im not sure how should I use that to actually get the y coordinate of point A

[gray isle]

sub in x=2

you have the equation of the line
sub in x=2, to find the y-coord when x=2

[alpine sable]

sub?

[gray isle]

substitute

[alpine sable]

what

[gray isle]

what what

[alpine sable]

substitute what

[gray isle]

x=2

into the equation of the line you just found

[alpine sable]

and thats gonna give me y of the point A?

[gray isle]

yes...

[alpine sable]

This is magic

.close

[lone heart]

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[gray isle]

you've actually taken the longer approach

finding the equation of the line isn't actually needed

[alpine sable]

is there a better one?

[gray isle]

the slope between any pair of points on a straight line will be the same

so you could do something like
slope(AB) = slope(BC)

[alpine sable]

idk this looks way more complicated

to solve

[gray isle]

try before making any comments

[alpine sable]

i dont even know where to begin

[gray isle]

slope(AB) = slope(BC)

[alpine sable]

still

[gray isle]

like to get your equation of the line you would've applied stuff like slope formula...

[alpine sable]

yeah

[gray isle]

so apply that for the points A,B as well

[alpine sable]

how?

oh

[gray isle]

its the same formula

[alpine sable]

yeah but if i use your equation i have to apply formula 2 times

this way i just apply once no?

and then substitute?

[lone heart]

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[gray isle]

well you also had to apply stuff to get the equation of the line in addition to finding the slope

[gray isle]

like actually try it

instead of making assumptions or speculating about whether its going to be better/worse

[alpine sable]

yeah okay

well thank you

[gray isle]

.close

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[stable acorn]

is what I’ve done right?

[long axle]

$(a-b)^2 \neq a^2 - b^2$

[ocean seal]

Stephen

[long axle]

Line 3 into line 4

[stable acorn]

ahhh I see thank you

Let me redo and get back

ok I think I’ve overcomplicated this and there’s an easier way

[stable acorn]

how would you go about this?

[long axle]

Hmm

$5\sin x - 5\cos x = 2 \newline (5\sin x)^2 = (2 + 5\cos x)^2$

[ocean seal]

Stephen

[long axle]

See if u can make something of that

[stable acorn]

Think I already see it

Thank you

Once again will get back haha

[stable acorn]

Hiya

Sorry for taking so long but one of my answers is wrong

Instead of 151.4 it shud be 208.6

[long axle]

Lemme see

[stable acorn]

it’s to be solved for 0<theta<360

[long axle]

Ok I put it in mathway, and I got 61, 298, 151, 208

[stable acorn]

yes how would I get 208

[long axle]

Now I’d just plug those into the original problem

[long axle]

u did arccos?

[stable acorn]

Yep but I only got one answer

I’ll show

I’m sure there’s a way on this calc to get more than one answer

I just don’t know how to use it lol

[long axle]

Cosine is positive in quadrant 1 and 4

(Sqrt(46) - 2)/10 is positive

Therefore there will be an answer in quad 1 and quad 4

[stable acorn]

yeah I’ve just realised I’m going to need to use a cast diagram

By chance would you know if I can do that thru this graphical calc

[long axle]

No clue

[stable acorn]

Ah np but thank you so much for all the help ur way was 100x easier haha

[long axle]

No problem

Once u do find out how to get the other solutions, make sure to plug each of them into the original problem to see if they work

Especially since we squared both sides at the start

[stable acorn]

yep will do thank u!

[lone heart]

@stable acorn Has your question been resolved?

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[grand cape]

New parabolas are created by applying the operations in the figure to the parabola f(x) = x^2 + 1 defined in real numbers.

What is the area, in square units, of the triangle formed by joining the apex of these parabolas?

[grand cape]

My solution way is started like that;

A1 is 10/6 but 2nd area idk how can I find

<@&286206848099549185>

[lone heart]

@grand cape Has your question been resolved?

[grand cape]

2nd part of triangle is under the x line

[royal socket]

What's the second image?

[royal socket]

This one

[grand cape]

1st

[grand cape]

2nd

[grand cape]

3rd

Of my solution ways

And im sure that it’s correct

But just the last of triangle area I need to find under triangle

[royal socket]

is the third just

a redraw

of second image

but not to scale

[grand cape]

This one?

[royal socket]

Are the two triangles you draw the same

drew*

[grand cape]

Yeah I know

I writing step by step so

[royal socket]

Have u ever tried finding area of triangle from coordinates

[grand cape]

Before exams I can remember what I made

[grand cape]

How can I make it?

My geometry is so bad

Kinda remember somethings from 10 years ago lol

[royal socket]

Draw this rectangle

This u should see it

[grand cape]

You meant find rectangle area and minus the empty areas for find triangle?

[royal socket]

Yes

Exactly that

[royal socket]

Then*

[grand cape]

Imma trying rn

[grand cape]

I found 3

But YouTube teacher says 5

KA is Red Area and SA is Black area here and TA is total area

[royal socket]

A1 + A2 + A3 = 7

not 9

[grand cape]

Oh sheesh yeah mb

[royal socket]

Ye A3 = 2*4/2

[grand cape]

I corrected yup now everything is fine

Thanks a lot man

I appreciated

[royal socket]

Yw

[royal socket]

a bit of a weird way this lol

[grand cape]

Well Turkish education system is so bad

[royal socket]

😅

[grand cape]

Imagine I’m 28 and engineering student

And wanna return education life preparing exam but this guys and questions are weird af

[royal socket]

That's still impressive

[grand cape]

Using university method for some questions

[royal socket]

And good luck for your exams

[grand cape]

Tysm ❤️

[royal socket]

(Also you can dm me if you have any more questions)

[grand cape]

Added you and probably I will have soon haha

[royal socket]

I might not always be free lol but I'll check then whenever I am

them*

[royal socket]

Ooo Lagrange interpolation 😳

[grand cape]

Okay sounds so good 🐱

[grand cape]

Yeah from engineering

[royal socket]

Yes

[grand cape]

This guy using weird method for this problems

And idk how the high school students get it

[royal socket]

ik that looks a bit weird lol

[grand cape]

Yeah lol

[royal socket]

Does this require Lagrange interpolation?

[grand cape]

Ofc it is

[royal socket]

Because tbh P(5) you could set it to anything you want

[royal socket]

(by extending Lagrange to include P(5))

[grand cape]

Well I solved a lot of different questions with Lagrange but with this guys method just tried to get an hour

[royal socket]

does the question say use Lagrange though?

[grand cape]

We have a good sentence in turkey, the best way is what you know haha

[royal socket]

haha

[grand cape]

Nah you free to use anything

[royal socket]

oh...

[grand cape]

Cause exam is test

[royal socket]

I mean I would have just done P(5) = 5/6

[grand cape]

But it’s 2/3

[royal socket]

from P(0) to P(4) you can just guess P(n) = n/(n+1)

[royal socket]

that's from lagrange

[grand cape]

Also this guy solution and test answer also 2/3

And ai solutions

[royal socket]

Oh I guess the test is on Lagrange then? 😂

[grand cape]

Nah it’s just high school test

For joining university

But well they asking kinda weird stuffs

This question only about basic polynomial functions

Wait I will show you his solution

[royal socket]

Ohhh P(x) is a polynomial

[royal socket]

I didn't know what it said lol

yes then you would use Lagrange

[royal socket]

I didn't think it had to be a polynomial

(should've guessed it from P but oh well)

[grand cape]

P(x) is a 4th degree polynomial and
It gave values of p(0) - p(1) ….etc
Asking what is P(5)

[royal socket]

Ah

Yes then Lagrange makes perfect sense

[grand cape]

Wait I show how high school teachers making solutions

[grand cape]

Look at this

[royal socket]

😭😭😭😭

I mean I guess that's a way of doing it

[grand cape]

I listened what he told about solution in my language but well I’m dumb or he is

Idk hahaha

[royal socket]

Actually that's

Quite a smart way of doing it haha..

[grand cape]

How can I learn one of the math unit idk

[royal socket]

This bit makes sure it's a deg 4 polynomial

[grand cape]

On university also I failed with a unit

[royal socket]

practice is the best thing you can do

[grand cape]

permutation, combination, binomial, probability

[royal socket]

but I wouldn't know... I'm not in uni yet

[royal socket]

practice questions are very good for those

[grand cape]

Doesn’t matter I really can not get the sense of it haha

[royal socket]

😅 Then try to make sense of them

[grand cape]

4 years failed on it lol

[royal socket]

watch YouTube videos on nPr and nCr

or just search it up

[grand cape]

Thanks for tips

[royal socket]

Yw

[grand cape]

I’m returning to parabolas sadly

Haha

Have a nice day

[royal socket]

U too!

[lone heart]

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[cerulean blaze]

.open

[stable pecan]

translation pls?

[cerulean blaze]

Oh

a=?

Its parallelogram

[stable pecan]

what have you tried?

[cerulean blaze]

Umm

[stable pecan]

when you are stuck, its a good idea to review all the properties you have learnt and try to apply them

[cerulean blaze]

I found some angles but

I couldnt get it end

[cerulean blaze]

I think i know all of em but

[stable pecan]

can you show what you found?

[cerulean blaze]

Kk

[stable pecan]

also, another idea is try to work backwards, i.e, think which angle do you need to find so you can find a

[cerulean blaze]

@stable pecan

x+y=90

[stable pecan]

shouldn't DEF be 75?

[cerulean blaze]

Yes my bad

[stable pecan]

adding to this, since opposite angle of a parallelogram add to 180, if you find ADC, you can find a

you only need to find y to find ADC

[cerulean blaze]

I know it but

Idk actually how can i find adc

[stable pecan]

I think y can be found if you make one more equation involving x and y

alternatively, you can find x and since ACB = 45, you know that a = 135-x due to triangle sum property

[cerulean blaze]

If y equals 45, x needs to be 45 too

[stable pecan]

for this, try using the triangles DEF and FEC

[cerulean blaze]

But if x=45

[royal socket]

(x is not 45)

[cerulean blaze]

But i found that x+y=90

[royal socket]

you need to find y

[royal socket]

that's not enough

[cerulean blaze]

My brain doesn't work rn

[stable pecan]

give it a restart

[cerulean blaze]

Im working math for 3 hours

[stable pecan]

wash you face, have some water, review all the properties, then take a fresh look at this question

[cerulean blaze]

Lemme try

Thank you for help @stable pecan and @royal socket

And sorry for my English lol

Okok i solved it

X is 15

Y is 75

This question took 20 mins

[royal socket]

👍

[acoustic olive]

you dont need equation to solve

its actually really easy

[royal socket]

"easy" is different to different people

[acoustic olive]

sorry thats not what i meant

but you dont need equation

since triangle DEC is a Right Triangle and EF is a median then FC=EF

from that <FEC=<FCE

[lone heart]

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[summer pollen]

is there a geometric object that is represented by the integral of the area of a circle (1/3 * pi * radius^3)?

[vapid shuttle]

Do you know the formula for the volume of a sphere?

[summer pollen]

yeah it's four time that amount

[vapid shuttle]

So, perhaps that could represent the volume of a quarter sphere?

[summer pollen]

yeah I guess but I hoped there was something else

.close

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[meager lark]

I have been stuck on this problem for a while, can someone help me solve part b and c?

[lone heart]

@meager lark Has your question been resolved?

[lone heart]

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[meager lark]

.reopen

[lone heart]

✅

[meager lark]

No

<@&286206848099549185>

[foggy current]

I'll try

[meager lark]

Ok

[foggy current]

Are we trying to maximize the volume or no?

[meager lark]

My professor wasnt specific about that but I would assume yes

[foggy current]

Ok this will be much easier

[meager lark]

Do you think it would be possible for us to join a vc so I can understand the problem better?

[foggy current]

I'm doing other stuff at the same time so I can't do that

[meager lark]

Ok I understand

[foggy current]

Is this a calculus question?

[meager lark]

Yes

[foggy current]

ok

In order to maximize the volume we'll just assume that 4x+y=108 (because 4x+y<=108 from the question)

and solving for x we get x=(108-y)/4

the rest should be straightforward

[covert solar]

.

[meager lark]

So the formula for volume would be, V= h * ((108-y)/4)?

[foggy current]

Yeah, probably

[meager lark]

"probably"😭 ?

[meager lark]

You are not sure?

[foggy current]

I'm not 100% sure because I'm a human

and humans make mistakes

[meager lark]

Ok. I have one last quuestion

[meager lark]

Does this account for both the length and the width or just one of them?

[foggy current]

the length is x and the height is y

[meager lark]

So how would a write the volume as a function of x alone?

[foggy current]

h=x

wait no

h=y

As the problem stated

So V = y * (108 - y) / 4

[meager lark]

Is that "y" on the outside representing the height of the box of the height of the base?

[foggy current]

The y is the height of the box

[meager lark]

Ok. what about the height of the base?

[foggy current]

It is x

[meager lark]

Where is x

[foggy current]

Oh I completely forgot about that

And I have to have dinner and my mom doesn't allow me using the computer while eating

[meager lark]

Ok lets finish up then

[lone heart]

@meager lark Has your question been resolved?

[foggy current]

I'm back

I also proved that 4x+y=108 must be true to maximize the volume while eating

So from that, we get y=108-4x

and V=x^2y=x^2(108-4x)

[meager lark]

I dont think volume should have another equal sign within its own definition

[foggy current]

Which part are you talking about?

[meager lark]

When you say "=x^2(108-4x)

[foggy current]

In other words, I'm saying that $V=x^2(108-4x)$

[ocean seal]

Math Is Fun

[meager lark]

Ohhhhh

I get it now

Thank you!

[lone heart]

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[round valley]

can someone check my answer?

[willow patio]

Um, there is definitely discontinuity at x = 1 right?

The denominator would be 0 and therefore undefined

@round valley ?

[round valley]

what

wait

[round valley]

no

thats asymptote

which is a point that the graph is infinitly approching but never touches

[willow patio]

ah, and I assume everything stays to the right of the asymptote then?

[alpine sable]

Hi I'm new and I need help with my math question

[round valley]

bruh

[alpine sable]

yo?

[round valley]

go type ur question in one of the help channels without username

as you can see, this channel is occupied

[round valley]

uh

depends

[alpine sable]

It doesn't look occupied to me

[round valley]

bro...

there are rules

[willow patio]

Hey, please read #❓how-to-get-help

[round valley]

anyway

uh

discountinuity is like

the factors of the polynomial

cancel out

so like if you factor the top polynomial and bottom there are 2 of the same factors

[willow patio]

Yea, I see what you mean.

[round valley]

ye

i just don't know if my math checks out

cuz i got

((x+3)(x+2)(x+1))/((x-2)(x-1)(x-3))

[willow patio]

naw, you're right.

I thought you meant if the polynomial was continuous or not.

[round valley]

oh

mb lol

[willow patio]

Np, you are good!

It was my fault really for not reading properly!

[round valley]

alr

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[gaunt moat]

deer

[gaunt moat]

lord i need help i might actually be dumb

[long axle]

Why?!

[gaunt moat]

a house is 43 feet long 40 ft wide and 8 feet tall what is the cubic volume of the house in meters

i swear i just do 43 * 40 * 8

13760ft to cubic

[long axle]

Wdym by this

[gaunt moat]

that is 43 * 40 * 8

[long axle]

“To cubic”

[gaunt moat]

yes

[long axle]

It’s alrdy in cubic feet

[gaunt moat]

needs to be in cubic meters tho

my bad

[long axle]

Oh

[gaunt moat]

13760 ( 1/3.281^3)

i swear this is how you do it

but my number is off by a multiple of 10 and idk why

[long axle]

Lemme try it

[gaunt moat]

im getting 389.58

[long axle]

That’s not the right answer?

[gaunt moat]

and its telling me its 3.89

apparently not

[long axle]

Send a ss

[gaunt moat]

welp nvm now its accepting it but on the pratice problem it also was not

[long axle]

Ok

[long axle]

Now I want to go to ihop 💯

Maybe Perkins idk

[gaunt moat]

there are no diffrences in these problems right ?

this is stress inducing for zero reason

[long axle]

Look at the units

It says x 10^2 m^3

Which means the answer u put in the box times 100

[gaunt moat]

thats bs

[long axle]

Yea fr

Idk why they’d do that

[gaunt moat]

im gona kms

[long axle]

Webassign be garbage sometimes

[gaunt moat]

not really but thats dumb

anyways thx

.close

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[long axle]

Np

[lone heart]

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[modest vine]

How would you do this?

[lone heart]

@modest vine Has your question been resolved?

[lone heart]

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[fervent briar]

sat math help

[fervent briar]

can someone explain this

explanation doesn't make sense to me

i understand that there 360 degrees in a real circle

but how does that have to do anything with area

the question originally asked for "the fraction of the area of the shaded region to the entire circle"

so it should be (the area of the shaded region) / pi r^2

but we dont know its radius lol.

[modest vine]

No

[long axle]

More degrees means more area!!!

Sector angle is directly proportional to both circumference and area

[sour dove]

So basically the idea with this is that stuff will cancel out.

The area of a sector is $\frac{\theta}{360^\circ} \pi r^2$. The area of a full circle is $\pi r^2$.

So when it's asking for the fraction of the area, we divide the sector area by the full area, like so:
$\\\frac{\frac{\theta}{360^\circ} \pi r^2}{\pi r^2}$. As you can see the areas (including the radius) is cancelled out, and we're left with $\frac{\theta}{360^\circ}$

[ocean seal]

MellowDramaLlama

[sour dove]

so then we get 100/360, which is about .278, or about 27.8%

[lone heart]

@fervent briar Has your question been resolved?

[fervent briar]

Oh my god

thank you so much

i totally forgot about the area of sector formula

forgot i have to memorize that

the khan academy explanation totally neglected that factor

thank you so much

.close

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[drowsy sedge]

If I am given a plane curve, lets say r(t) = 7ln(t)i - (4/t - 3)j + 0k for some domain lets say 0 <= t <= 4

And a point, lets say P(1,2) on r(t) (I have not actually checked if thats on r(t))

How on earth can I determine the radius of the osculating circle of r(t) at P?

This is a homework question, but I basically re-made the question. I can discuss, just dont dump the answer pls

[drowsy sedge]

<@&286206848099549185>

[lone heart]

@drowsy sedge Has your question been resolved?

[tacit arch]

r(t) isn't a circle

[drowsy sedge]

well

im trying to make a circle

to find the curvature

at the point p

ye?

[tacit arch]

Equations in Figure 11.1.5
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/11%3A_Parametric_Equations_and_Polar_Coordinates/11.01%3A_Parametric_Equations

[drowsy sedge]

hmm, i dont think thats exactly what im looking for

I know the equation Kappa = DT/ds

and p = 1/kappa

[drowsy sedge]

.close

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[nova void]

.help

[lone heart]

Commands:
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[nova void]

Hey can anyone answer this math question?

[solid crane]

what?

[lone heart]

@nova void Has your question been resolved?

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[bitter crest]

.help

[lone heart]

Commands:
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consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[bitter crest]

.help

[lone heart]

Commands:
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consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[bitter crest]

.ask

[bitter crest]

I gotta go to training but I have the first part the values of a and b both being 1

However I do not no how to approach part b1 and b2

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[vapid shuttle]

We define a function $f$:$\mathbb{Q} \xrightarrow{} \mathbb{Z}$ s.t.\
\begin{equation}
f(x) =
\left{
\begin{array}{lr}
0, & \text{if } q=0\

    1 & \text{if } q=1\\
    p_1^{2r_1}\ldots p_n^{2r_n} & \text{if} q>0, q\in \mathbb{N} \setminus \{1\}\\
    p_1^{2r_1}\ldots p_n^{2r_n} q_1^{2s_1-1} \ldots q_m^{2s_m-1} & \text{if} q>0, q\in \mathbb{Q} \setminus \mathbb{N} \\
    -1 & \text{if } q=-1\\
    -p_1^{2r_1}\ldots p_n^{2r_n} & \text{if} q<0, -q\in \mathbb{N} \setminus \{1\}\\
    -p_1^{2r_1}\ldots p_n^{2r_n} q_1^{2s_1-1} \ldots q_m^{2s_m-1} & \text{if} q<0, -q\in \mathbb{Q} \setminus \mathbb{N}
    
\end{array}

\right}
\end{equation}
Where any listing of p's/q's is the corresponding prime factorization of $|q|$.

[ocean seal]

AustinU

[vapid shuttle]

I am trying to prove that f(x) is surjective

My idea is that I should split the integers into cases of =0 =1 =-1, and then >1 and <-1

since I have already shown that the first 3 cases are included by my function

then all I would have left to show is that

for any integer z>1, or y<-1, there is are x's such that f(x) = z and f(x)=y

does that seem like the right approach?

If so, I'm not really sure how to do that, so I'd appreciate some help getting on the right track.

[mortal trellis]

get the prime fact of z and then find the relevant fraction

[keen plinth]

also you can collapse your cases down

[vapid shuttle]

and if z<-1 then the prime factorization of |z| right?

so like, say we are looking for the x s.t f(x)=-7

[mortal trellis]

same thing really

[vapid shuttle]

p.f is 7

times negative 1

oh well

[keen plinth]

[
f : \Q \to \Z, \quad f(q) = \sign(q) p_1^{2r_1} \dotsm p_n^{2r_n} q_1^{2s_1 - 1} \dotsm q_m^{2s_m - 1}
]
where
[
q = \f {p_1^{r_n} \dotsm p_n^{r_n}} {q_1^{s_1} \dotsm q_m^{s_m}}
]

[mortal trellis]

the positive and negative cases are just a minus away from each other. so if you solve the positive one then the negative one is solved aswell

[vapid shuttle]

agreed

[vapid shuttle]

what about the cases of q=0, q=1, q=-1

do I still need to include those?

[keen plinth]

sign(0) = 0

when q = 1 or -1, you just have nothing in the prime factorisation

[mortal trellis]

ehh depends on def

[keen plinth]

so all the exponents are 0

[vapid shuttle]

I used that to prove injectivity denascite

it was easier I thought

[keen plinth]

typo in case 2?

[vapid shuttle]

Would this just be, -1/7 then?

[vapid shuttle]

oh forgot the equal sign and second half

yes

typo

[vapid shuttle]

So, I can do this for specific cases, but I'm not sure how to transform that into a proof for all cases

Focusing on first showing the z<-1 has x s.t. f(x)=z

these z are just the negative versions of all n in N right? So, we let x=-1/z, and our f(x) will then give us z

z=-8

let x=-1/8

[keen plinth]

integer z<-1?

[vapid shuttle]

yeah

[vapid shuttle]

then f(x)=-8

so it works?

[keen plinth]

you would do whatever you do for integer z>1, adding a negative sign

[mortal trellis]

f(-1/8) = -2^5

[vapid shuttle]

oh right

I am forgetting to p.f

that is only the case for integers < -1 where they are prime

isn't it

where their p.f is themself

Okay, I'll do the case with integers z>1 then instead, since like snow said it would be more clear to start here

integers z>1, if z is prime then let x=1/z, then f(x)=z ?

If z is not prime, then I need to figure that out now

[mortal trellis]

take some random number, lets say $2^2 3^5 7^3 11^4$ and try to construct the relevant fraction

[ocean seal]

Denascite

[mortal trellis]

if you can do that, you can do the general case

[vapid shuttle]

let $x=\frac{2\cdot 11^{2}}{7^{2}\cdot 3^{3}}$

[ocean seal]

AustinU

[vapid shuttle]

right?

then f(x)=2^2 3^5 7^3 11^4

[keen plinth]

then do it in general

[mortal trellis]

yes

[vapid shuttle]

Would it be like this

For any $z \in \mathbb{Z}, z>1$ Construct its p.f $z=p_1^{r_1} p_2^{r_2} \ldots p_n^{r_n}$ and then for any $r_i$ odd, place $p_i^{\frac{r_i-1}{2}}$ in the denominator of $x$ and for any $r_j$ even, place $p_j^{\frac{r_j}{2}}$ in the numerator of x. Then $f(x)=z$.

[ocean seal]

AustinU

[mortal trellis]

yes

[vapid shuttle]

https://tenor.com/view/suits-dance-dancing-jam-music-gif-5174541

[mortal trellis]

negative z is solved with that aswell

[vapid shuttle]

TY Denascite and Snow!!

mhm

[mortal trellis]

and z=1 you could just take as the special case of "empty product"

which by def is 1

[vapid shuttle]

yeah, for z=1 the x that produces it is x=1

since I defined it that way

same for z=-1 and z=0

would be x=-1 and x=0

Alright, I think I am good to go then, unless you have any other pointers?

Thanks again for the help!

.close

[lone heart]

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[vapid shuttle]

uh oh, a contradiction

[lone heart]

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[errant dagger]

How would I do this?

if it was in R^3 I would just do cross product and check the normal vectors if they are parralel

[vale wigeon]

solve the linear system (first param) = (second param)

[errant dagger]

hm?

first / second params of which

ah wait wait wait

in order for the planes to be parralel

they must not intersect right?

[vale wigeon]

exactly

[errant dagger]

unless they intersect everywhere

[errant dagger]

so could I just equate the planes and do gaussian elimination?

and if there are infinite solutions theyre the same plane

wait

hm

how would I know if they are the same plane>?

[vale wigeon]

it matters whether the infinite solutions form a one parameter family or a two parameter family

[errant dagger]

ah wait it would be same plane if soln had 2 parameters

would intersect across a line if 1 parameter

cant intersect at a single point

and if no solution then the planes are parralel and not equal

is that right>?

[vale wigeon]

yes, except the planes can in fact intersect at a single point

this is R^4

[errant dagger]

😳

i got one more question

i dont even reallyt know what that is

[golden canyon]

Det means determinant

[errant dagger]

yee

i think we've only done like

yeah im not really sure tbh

what this means or how to do it

I used determinants to calculate cross product

[golden canyon]

Notice that if f(x, y, z) = 0 is a plane then translations will also give a plane

Then I think calculating the determinant is the easiest way

https://www.mathsisfun.com/algebra/matrix-determinant.html

This approach works nicely

[ripe oriole]

this channel is occupied you can open a new one

@thorny sluice

send your question to one of these

no worries

[gentle creek]

x=5

[cedar sail]

I can help you

[lone heart]

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[keen glade]

"Let a be a positive number. then the expression (a^-3 : a^-2)^-1/2 is equal to:.." i tried to solve it and got a^1/2 would this be correct?

[solid crane]

is that a ratio symabol?

if it is then your answer is correct

[keen glade]

yes it's a division ":"

[keen glade]

great, thank you!

[mortal trellis]

division in text is written as /

[keen glade]

oh sorry

[mortal trellis]

dw, now you know it for next time

[keen glade]

😁

.close

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[nimble fern]

as Ann said, and what you did,
we found d(arcsin(x)+arccos(x))/dx=0

so, the anti-dervative should be equal to some constant

which with some steps, you can find it equal to pi/2

[heady void]

uuuuh so we just take the antiderivative for arcsin(x)+ arccos(x)?

[nimble fern]

nah

[heady void]

?

pls explain further

[nimble fern]

we take the antiderivative of
d(arcsin(x)+arccos(x))/dx

[heady void]

but isn't that just arcsin(x)+arccos(x)?

[nimble fern]

exactly

[heady void]

what what's that gonna help with that?

isn;t that juat saying ?

[heady void]

I still don;t get it

.close

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[quasi sonnet]

What did I do wrong

[stable pecan]

5th line I think

[quasi sonnet]

Where the heart is?

[stable pecan]

The line above it

sorry, but its possible I may be misreading due to your handwriting

[quasi sonnet]

Well idk how to solve it

I just keep getting a wrong answer

[torn elk]

this is wrong

-32x not -36x

[quasi sonnet]

How?

[torn elk]

16*2 = 32, not 36

[quasi sonnet]

Oh

[quasi sonnet]

I did 2*8 as 18

Aha

Thx

.close

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[lucid phoenix]

please help me

this one is like a torture for me so i would greatly appreciate some help rn

[wind cloak]

Rewrite the term under the surd as a square

[lucid phoenix]

ok

whats next?

[lucid phoenix]

hello?

[vale wigeon]

show what you've got.

[lucid phoenix]

2+√2=a+b√2

is it correct?

[vale wigeon]

,w (2 + sqrt(2))^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by Ann#0413

[vale wigeon]

sure is

2+√2=a+b√2
from this, can you say what a and b are individually?

[lucid phoenix]

no

can you show me please?

[vale wigeon]

not even knowing that a and b are integers?

this is literally pattern matching

[lucid phoenix]

ummmmmm

a=2+√2-b√2?

[vale wigeon]

no, overthinking it.

${\color{red}2} + {\color{blue}1}\sqrt{2} = {\color{red}a} + {\color{blue}b}\sqrt{2}$

[ocean seal]

Ann (glomed)

[lucid phoenix]

oh

i get it now

tks

[lone heart]

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[dense helm]

how to solve z^2 + iz̄=1

[unborn verge]

Rewrite z = a+ib and solve it

It'll end up as two equations of two variables

[lone heart]

@dense helm Has your question been resolved?

[dense helm]

and then?

i got
x^2 - y^2 + 2ixy + ix + y = 1

[unborn verge]

Then x(2y+ 1) = 0 and x²-y² + y = 1

So either x = 0 and you use the second equation to find y

Or y = -0.5 ans you use the second equation to find x

[dense helm]

x(2y+ 1) = 0 why you removed the imaginary parts

i m not getting it

[unborn verge]

Well the 1 has an imaginary part of 0

So the imaginary part of the left side must be equal to 0

I just wrote 1 = 1 +0i and identified the real part and the imaginary part

[dense helm]

and then what should i do

[unborn verge]

... well you then have the two solutions for your equation

[dense helm]

no 😭

[unborn verge]

Which part is unclear

[dense helm]

wait give me a minute

[dense helm]

the solutions should be
z1 = (-i/2 + √7) / 2
z2 = (-i/2 - √7 / 2

[unborn verge]

Looks good to me ig

Since the condition x=0 leads to northing

[dense helm]

so conditions x = 0

[dense helm]

are + - ( √7 / 2) for real part

and the y are -1/2 imaginary part

[unborn verge]

Yep

[dense helm]

idk I feel like I just did the exercise but got nothing

I didn't understand well the system part

[unborn verge]

The basic idea was to explicit the real/imaginary parts

Then identify

Then solve

[dense helm]

oh ok i got it

but why the exercise count only the solutions for x=0 ?

[lone heart]

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[hollow thicket]

how would i solve for c

[ocean hawk]

you'll have to integrate, looks like

[hollow thicket]

oh wiat oops posted the wrong question lol

c for this question

[ocean hawk]

what have you tried/what is your intuition

[hollow thicket]

basically what i did was solve the derivative

and i got dy/dx = sin(x/3)

x=a so dy/dx=sin(a/3)

normal of the curve is -1/sin(a/3)

and then I subbed in the point (9,0) in the equation y= -1/sin(a/3) x + c

which was 0 = -9/sin(a/3) = c

and c = 9/sin(a/3)

that wasn

t the correct answer for the y-intercept though :(

is it necessary for me to use the the other line formula y - y1 = m(x - x1)

[ocean hawk]

I was just thinking it might be

[granite badger]

so you wanna find the gradient at x=a and because it meets at right angles, you wanna find the perpendicular gradient (m1m2 = -1)

[hollow thicket]

oh why is that?

[hollow thicket]

was that the normal -1/sin(a/3)?

[granite badger]

out of curiosity, is this the methods textbook?

[ocean hawk]

just a thought. doesn't look like it helps though

[hollow thicket]

yeee

[granite badger]

which page/chapter

[hollow thicket]

question 12 of 11J Integration

[granite badger]

oh haha my class isn’t up to integration yet, do you have the worked solutions with you? i can send it

[ocean hawk]

what book?

[granite badger]

it’s a common book for the australian math curriculum

[hollow thicket]

thank you lol though i alraedy have the worked solutions and i still can't understand it

[granite badger]

lemme look at it and try and explain it

as you said dy/dx = sin(x/3)

and the question asks find x=a

remember the form y - y1 = m(x - x1)

[hollow thicket]

oh yeah

[granite badger]

we know that the gradient is perdendicular and the dy/dx is the gradient

[hollow thicket]

so i can't use y=mx+c in this scenario?

[granite badger]

no

[hollow thicket]

mhm

[granite badger]

always use

[hollow thicket]

ohh

why is that/

?

[granite badger]

y = mx + c is the general form of a line, and requires a gradient and the y-int, y - y1 = m(x - x1) requires a point and a gradient

most of the time you will be given a point

i don’t recall seeing a question giving the y-int

because with the point, you can find the gradient given the x value of the point then the tangent (you would’ve learnt this in differentiation)

@hollow thicket does it make sense?

[hollow thicket]

ohhhhh

[hollow thicket]

yeah sounds good

[granite badger]

ok let’s get back then, because dy/dx gives us the gradient, then -1/(dy/dx) should give the perpendicular gradient

= -1/(sin(x/3))

this is our m value

remember the question said x = a, so m = -1/(sin(a/3))

now we can use our formula y - y1 =
m(x - x1)

the question gives us y = 3 - 3cos(a/3) and x = a

these are y1 and x1

[hollow thicket]

sorry wait so going to this formula y = mx + c

[granite badger]

yes eventually

[hollow thicket]

can't we still use it?

[granite badger]

no, we don’t have a y-int

[hollow thicket]

can't you solve with it with just any point on the line though

[granite badger]

i would say so but the question asks for the equation of the normal where x = a

we do not know the coordinates in (x, y) form

this is part ci

part cii gives you a point to substitute

[hollow thicket]

ohhhh

okay so for the first part we just ignore that point okay understandable

[granite badger]

so for part ci we get
y - (3 - 3cos(a/3)) = 1/(sin(a/3)) * (x - a)

now for part cii, we can sub in (9,0)

and solve for a

you need to use cas

if there’s smth you don’t understand ping me

[hollow thicket]

i understand it now appreciate it

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[lone heart]

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[fresh arch]

I know that gamma = -(alpha+beta) fixes the sum of root problem but i can't figure out how to prove that c = alpha * beta ( assuming that's what we have to do)

[vale wigeon]

**-**c = αβ

[fresh arch]

From which equation did you get that

<@&286206848099549185>

[stark lake]

From the fact that α and β are the roots of x^2 -γx-c=0

[fresh arch]

But that is the thing that we need to prove

[stark lake]

Oh sorry

[alpine sable]

@fresh arch

Do you know what a root is

[fresh arch]

Yes

[alpine sable]

Then using the roots, rewrite both equations into something you can work with

(factor)

[fresh arch]

.close

[lone heart]

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