help-38

[covert spindle]

u got this

[whole coral]

(Hint: what type of function is sin?)

[covert spindle]

^^

[radiant portal]

cos(a)sin(2a)

[covert spindle]

yes!

so now give me your P_a and P_pi-a

[radiant portal]

(cos(a)sin(2a), cos(a))
(cos(a)sin(2a), cos(a)sin(2a))

Which is wrong?

The 1st or second

[whole coral]

[radiant portal]

Ohhhh

(cos(a)sin(2a), cos(a))
(cos(a)sin(2a), -cos(a))

[covert spindle]

yup

now tell me

what does this say?

[radiant portal]

I think I get it

The x coordinates are the same

Which means its vertical, right?

[covert spindle]

yep!

[radiant portal]

Awesome!!

[covert spindle]

see sometimes just do something

and hope that it makes sense at the end

[radiant portal]

Thats fine advice for homework, but idk about that in an exam 😝

Which is what im actually worried about

[covert spindle]

its better to try something than nothing 🙂

[radiant portal]

Oh btw

The x coordinates are the same
Which means its vertical, right?
How would I write at the end this bit

Like in an exam

You need to write; thus, this concludes blah blah blah

Or whatever

Whats like the official way to write it

[covert spindle]

probably what u said would be allowed

just say the x coordinates are always the same

only the y values differ so it can only be a vertical line

[radiant portal]

Dont you need to write it in like fancier language or something'

In an exam

[covert spindle]

not really I think

oh there is one case where it is not a line

[radiant portal]

Are you suuuuuuuuuuuure

[covert spindle]

can u think of one?

[radiant portal]

The origin

Or the vertical tangent

[covert spindle]

well keep in mind the vertical line (the length) depends on our y values

and our y values depend on a

(0,0) is indeed a point where this happens

aaah it is the only point

u can see it in the graph

but uhm (x,0) (x,-0) = (x,0) (x,0) so if cos(a) = 0 , a= pi/2 +pi*k

[radiant portal]

Wait what

[covert spindle]

so in short

the line depends on the y-values right?

so lets ignore the x values here

[radiant portal]

ahhh yeye wait

[covert spindle]

y=-y can only be equal to each other if y=0

[radiant portal]

I get it

[covert spindle]

and y=0 means cos(a) = 0

alright I got to focus on my own homework again

good luck you got this 🙂

[radiant portal]

Thank you for your help!!

[covert spindle]

I felt honored!

[radiant portal]

And Chartbit too

Thanks to both of you

❤️

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[ashen minnow]

10th question

[real rose]

Can you visualize it

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@ashen minnow Has your question been resolved?

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[ember ivy]

Hi

[plush vigil]

split the fraction

t-3/t^2 = 1/t - 3/t^2

[trim lichen]

(t-3)/t^2.

don't drop the brackets there.

good point otherwise.

[plush vigil]

i really need to learn latex

[real rose]

Yes

[ember ivy]

greaat

Can you solve this with this:

[plush vigil]

$(t-3)/t^2 = 1/t - 3/t^2$

[solid kiln]

cerosi

[plush vigil]

i did that first try

[hidden dew]

$\frac{t-3}{t^2}=\frac{1}{t}-\frac{3}{t^2}$

[solid kiln]

yoboiqimmah

[real rose]

https://en.wikibooks.org/wiki/LaTeX

[plush vigil]

that looks much better

[ember ivy]

How to do it this way

[real rose]

This is a repeating linear factor

So this does not work unfortunately

[ember ivy]

ah

okay

because we have two repeating factors?

[real rose]

You either pain yourself with a u-sub of x^2+6x+9 or do what you did

[real rose]

Yes

x=A(x+3)+B(x+3)

See how this becomes a problem

[ember ivy]

not really

Why is it a problem

But in my calc. I get "No solution"

[real rose]

A=-B

$\frac{A}{x+3}-\frac{A}{x+3}=0$

[solid kiln]

mathisfun

[real rose]

SO uh...

Self explanatory

[ember ivy]

cool

thank you

.solved

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[frail hound]

Need to check my answer.

[frail hound]

70ab was my answer

[real rose]

For which question

[frail hound]

1.3

And hello again

[frail hound]

Second one, by the way

[real rose]

I kinda can't see it

😔

It says a^ b^ +abc+b^2=0?

[frail hound]

Third. 🤦

Not second

[frail hound]

Sorry about that

[real rose]

I can't see it in general man

I'm squinting my eyes

You have the question too zoomed out

[spring current]

Click the image and use this button

If youre on phone idk how well it works, its pretty good for pc though

[real rose]

Danke

[spring current]

Bitte

[frail hound]

This better?

[real rose]

Show work

[real rose]

Yep

[frail hound]

Thank you

. close

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Please don't occupy multiple help channels.

[red mountain]

<@&268886789983436800>

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[drowsy trench]

"A regular triangle is inscribed in a circle with a radius of 18 cm. Calculate the length of the side of the triangle."

[drowsy trench]

i think its called Equilateral triangle

all corners 60

degrees

[jagged sparrow]

you can use cosine rule pretty sure

if AB is 18 then AC is also 18

AB bisects A so 60/2 is 30

then you have an isoceles triangle ABC with the angles 30,30,120

then just use cosine rule

[jagged sparrow]

AC^2 = 18^2 + 18^2 - 2(18x18xcos(120))

[drowsy trench]

so the side is 18 sqrt of 3?

[jagged sparrow]

ye

is that the answer

[drowsy trench]

i got that yup

thanks

[jagged sparrow]

nice

[drowsy trench]

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[frail hound]

Hello, just wanted to ask if you could check my answer

[trim lichen]

show the problem & your work

[frail hound]

1.3.1

[trim lichen]

this one?

[frail hound]

Yep

[trim lichen]

i don't understand what you've written here

what is this symbol here for example

[frail hound]

That was supposed to be brackets. Kind of rushed to show my working

[trim lichen]

uh, don't rush maybe?

sit down, write your work out properly.

also what is 43.4 doing there? isn't the cover cost 43.75...?

definitely redo this entire work! and do it slowly, carefully and properly.

[frail hound]

Done

So...

[trim lichen]

this pic is blurry and i can't make out the first line

$3500 = x \cdot \frac{(x-4) \cdot 43.75}{4}$

[solid kiln]

Ann

[trim lichen]

is that it?

[frail hound]

Yeah

[trim lichen]

can you explain how you arrived at this equation

because i am not quite seeing it

and i am a little bit suspicious of that division by 4

[frail hound]

Well, since 4 people have not payed and the rest have to cover the cost. I thought that since the remaining have to cover the cost, why don't I make the equation give me the total all 4 owe. With this I can get the the cost of one person by dividing the total the four owe by 4

[trim lichen]

uhh

i am confused ngl

make the equation give me the total all 4 owe.
ok i guess you can argue the amount that the 4 broke people owe to their other friends is 43.75(x-4)

but... this inter-friend debt has no bearing on the total cost of the taxi so i do not see whyyou would equate it to the taxi price

so you're off the mark here i am afraid

[ashen forge]

okay so

if x people paid

3500/x

[trim lichen]

!nosols

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[trim lichen]

preemptively

also i was gonna explain this anyway

or at least try to take OP through it

[ashen forge]

okay?

[trim lichen]

actually if you want you can take over instead

[ashen forge]

no no go ahead

[trim lichen]

(but don't dump the entire solution at him)

ok

[ashen forge]

you seem to be better at words

[topaz heart]

Wrods

[trim lichen]

@frail hound you still here?

[frail hound]

Yeah

[trim lichen]

also btw does R stand for rand?

[frail hound]

Yep

[trim lichen]

ok

right so here is how i would have reasoned through this problem

[ashen forge]

rand?

which country

[trim lichen]

south african currency

[ashen forge]

ah

[trim lichen]

the taxi costs 3500 rands and this would be shared between x people.
so if all was according to plan, each person's share would be 3500/x

however 4 people are broke

so the other x-4 people instead have to split the cost, paying 3500/(x-4) each

@frail hound does this make sense thus far

[frail hound]

Yeah

[ashen forge]

wait

the equation he wrote is correct i think

[trim lichen]

is it...?

i am about to write down an equation that i am certain is correct.

[ashen forge]

yeah well thats what im getting after simplifying

[trim lichen]

if it turns out equivalent to OP's then power to him i guess, but i'm doubtful.

what i was gonna say now

is that the paying friends have to pay 43.75 rand above the previously agreed upon amount

thus $\frac{3500}{x-4} = \frac{3500}{x} + 43.75$

[solid kiln]

Ann

[trim lichen]

is the equation we will get

@ashen forge is this what you started with

[ashen forge]

ha ya

that simplifies to his equation obv

[trim lichen]

obv?

i do not see how it is so obvious, can you enlighten me?

[ashen forge]

wait am i tripping

[topaz heart]

Probably

[ashen forge]

just do the algebra

oh obv 😭 im sorry i meant to type "ig" i have no idea how i mistyped to obv

[trim lichen]

oh ok yeah

[ashen forge]

its like 2 am idk what im typing

[trim lichen]

after some algebraic BS it does simplify

[topaz heart]

Please go to sleep

[trim lichen]

to... i guess 3500*4 = 43.75x(x-4)

[ashen forge]

yes, but im confused how he obtained it directly

[trim lichen]

me too

[ashen forge]

xenobla

[frail hound]

Still here

[ashen forge]

im fine bbg

[topaz heart]

Please bb

[ashen forge]

how did you get the simplified form direclty?

[ashen forge]

or did you get this too

[trim lichen]

,calc 3500/16 - 3500/20

[solid kiln]

Result:

43.75

[trim lichen]

checks out

[ashen forge]

its a holiday ill stay up

[topaz heart]

Hmm okayy 😒

[ashen forge]

hye it could also be -16 friends

[frail hound]

No, just got it directly

[ashen forge]

yeah how come

explain your logic

[ashen forge]

dw im fine, ill sleep if im proper tired

[topaz heart]

No you won't don't lie to me

[ashen forge]

ill stay up as long as you do

[frail hound]

It would be better if I wrote it down

[topaz heart]

Hmm fair enough

[ashen forge]

okay yes

please do

[topaz heart]

Yeah maybe write a little next to each step

[frail hound]

Ignore 4= total amount

Anyway thank you guys.

. close

[ashen forge]

why cost for one of them is total amount/4?

i dont think that makes sense

[topaz heart]

What do you do with the 3500

@frail hound

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[hot verge]

Is this the correct way to cretae a binary search tree

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@hot verge Has your question been resolved?

[hot verge]

<@&286206848099549185>

[hot verge]

.close

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[torn socket]

can somebody explain to me what im supposed to do with this

[torn socket]

i cant find a calculator to explain the steps

[dusty sleet]

You have to use the fundamental theorem of calculus

[torn socket]

wait i literally just understood it i think

yep okay i got it nevermind sorry

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[dusty sleet]

https://youtu.be/3G5SbkBHqlQ?feature=shared

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[timid imp]

You can write that in vertex form pretty easily

And then expand the polynomial

[rotund basin]

Do I not just use the substitution method?

[timid imp]

Idk what that would be

[rotund basin]

Can you elaborate?

I don't understand sorry

[timid imp]

You can write this in the form $a(x-h)^2 + k$

[solid kiln]

NahhFam

[timid imp]

Where h is the shift left/right and k is the shift up/down

[rotund basin]

Write what in that form?

[timid imp]

That parabola

So it’s shifted to the right by 1

And shifted up by 8

And then the value of a is going to be negative since the parabola opens down

[rotund basin]

Is this a known formula?

[timid imp]

Yes

It’s the vertex form of a quadratic

[rotund basin]

I'll try

[rotund basin]

Is this equal to y or?

[timid imp]

Yep

[rotund basin]

And I just compare two vertices?

[timid imp]

Wym

There will only be one vertex

[rotund basin]

What is the shifted to the right value that you're talking about

And the up and down one

[timid imp]

Tell me the coordinates of the vertex

The values of h and k in that formula are (h,k) where (h,k) are the coordinates of the vertex

[rotund basin]

(1, 8)

[timid imp]

Yea

[rotund basin]

oh

[timid imp]

So plug that into the formula

[rotund basin]

but x is 1 then?

a(x-1)^2+8

[timid imp]

No

x stays as x

$a(x-1)^2 + 8$

[solid kiln]

NahhFam

[timid imp]

After you plug in your values for h and k

[rotund basin]

And then solve the polynomial?

Does y stay as y too?

[timid imp]

Yea you can find a by inspection if you know how to do it

If not you can expand that polynomial with a included

And use the zeroes to solve for a

[rotund basin]

Inspection?

[timid imp]

Like just looking at the graph

[rotund basin]

Why do we need to solve a?

[timid imp]

You need a value for a still

For the vertex form

You have h and k but not a

[timid imp]

For this

[rotund basin]

What does y=a(x-h)^2+k even solve

[rotund basin]

or x or y

[timid imp]

It’s an alternate form for $ax^2 + bx + c$

[solid kiln]

NahhFam

[rotund basin]

But there's no b or c

[timid imp]

Yea

After you find a you can expand the polynomial

Expanding $a(x-h)^2 + k$ gives $ax^2 + bx + c$

[solid kiln]

NahhFam

[rotund basin]

$y=ax^2-2ax+a+8$ @timid imp

[solid kiln]

Elliot Pixel

[rotund basin]

That's what I got

[timid imp]

Yea I think thats right

[rotund basin]

But how would I see a from inspection of the graph?

[timid imp]

a streches or compresses the graph

Thats the function of a

[rotund basin]

Yeah but how would i notice its value by looking at the graph?

[timid imp]

If the graph is compressed in the horizontal direction, $|a| > 1$ and if it’s streched in the horizontal direction, &0 < |a| < 1$

[solid kiln]

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

[timid imp]

And in this case it’s being compressed

So $|a| > 1$

[solid kiln]

NahhFam

[rotund basin]

But what do I do with $y=ax^2-2ax+a+8$?

[solid kiln]

Elliot Pixel

[timid imp]

You can plug in the zeroes of the function

And solve for a

So you know f(-1) = 0

So $0 = a(1^2) - 2a(1) + (1) + 8$

[solid kiln]

NahhFam

[timid imp]

And you can solve for a

[rotund basin]

$x_1=-1,x_2=3$

[solid kiln]

Elliot Pixel

[timid imp]

Meant -1 whereever I put a 1 btw

My mistake

[rotund basin]

You mean +a+8 not +(1)+8 right?

[timid imp]

I do

My bad

[rotund basin]

Do I use these or do you mean in the y intercept?

[timid imp]

Wym

[rotund basin]

For when x is 0 or when y is 0?

[timid imp]

y

Lol

When y is = to 0

[rotund basin]

But then a will have two values

Cause $x_1=-1,x_2=3$

[solid kiln]

Elliot Pixel

[timid imp]

It wont

They give the same thing

[rotund basin]

Oh

[timid imp]

You can verify it urself if u want

But all you gotta do is solve for a

And plug it into the quadratic

[rotund basin]

a=-11/3

[timid imp]

Not what I got

[rotund basin]

I'll show what I did

[timid imp]

You plugged in 3 for a

Fourth line from bottom

U did what I did by mistake xd

[rotund basin]

wdym 3 for a

[timid imp]

It should be $0 = a(3)^2 -2a(3) + a + 8$

[solid kiln]

NahhFam

[rotund basin]

oh shi

lmao

[timid imp]

Not $0 = a(3)^2 -2a(3) + 3 + 8$

[solid kiln]

NahhFam

[rotund basin]

How come we both did it lol

I'll blame it on the time being 12:59 am

@timid imp I got a = -2

Damn straight

[timid imp]

I gotta leave now, good luck with solving anything else

[rotund basin]

thanks

for your help

.close

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[acoustic sphinx]

can someone please help me it says i need to find the solutions to the equation

[acoustic sphinx]

it wants the general solution

in the book it has a different answer to what I have and I'm really confused

[restive vortex]

cot = 1/tan

1/tan(2x-pi/6) = sqrt(3)

then

[acoustic sphinx]

tan(2x-pi/6)=1/sqrt 3

[restive vortex]

yeah

rationalize it

[acoustic sphinx]

sqrt 3 /3

[restive vortex]

suppose tan (a) = sqrt3/3

find the value of a

[acoustic sphinx]

reference angle is pi/6

[restive vortex]

a = 2x-pi/6

i mean

tan is periodic function

so is there a limit

because if there is not there will be infinitely many answer

[acoustic sphinx]

yeah it wants teh general solutions

[restive vortex]

we can assume that the range is 0 <= x <= 2pi

[acoustic sphinx]

no

it wants the general solution

[restive vortex]

alr

i understand

for what value does a such that tan a = sqrt3/3

[acoustic sphinx]

pi/6

[restive vortex]

so 2x -pi/6 = pi/6

[acoustic sphinx]

isn't it 2x -pi/6 = n*pi + pi/6?

right?

[restive vortex]

oh yeah i forgot

[acoustic sphinx]

k so it then become 2x=n*pi + pi/3

[restive vortex]

yes

👍

[acoustic sphinx]

x=(n*pi)/2 + pi/6

[restive vortex]

yeah

[acoustic sphinx]

x=((3n*pi)+pi)/6

x=((3n+1)pi)/6

[restive vortex]

(3n+1)*pi/6

[acoustic sphinx]

idk i think it is somehow wrong

[restive vortex]

how so

[acoustic sphinx]

if n>1 it is the wrong solution

[restive vortex]

we literally write the same thing

[acoustic sphinx]

yeah but i'm not sure if it is correct

[restive vortex]

try checking for n > 1

meaning n = 2, 3,4....

tan 7pi/6 = 1/sqrt(3)

[acoustic sphinx]

my bad

[acoustic sphinx]

we're right I forgot that you have to put it back into the initial thing

[acoustic sphinx]

thanks

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[glad grotto]

Can someome please help me understand and complete question 1 properly and then I can do question 2 myself

[glad grotto]

The more straight forward you are with ur explaining the better I understand

Any help is appreciated

Wrong question mb

It's this one

I just need to know how many flags there should be

<@&286206848099549185>

@ me if u can help

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@glad grotto Has your question been resolved?

[glad grotto]

<@&286206848099549185>

[stable flower]

Wait a sec

This will maybe help but im not sure

You there?

[glad grotto]

Yo

Yea ik that formula

[glad grotto]

I need help with this question

[trim jolt]

@glad grotto Has your question been resolved?

[glad grotto]

Screw this

.close

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[lofty depot]

is my work clear here? my sketch is kinda shit and i skipped a few steps.. my teacher is fine with skipping easy steps but he still wants my steps to be clear enough to understand

[trim jolt]

@lofty depot Has your question been resolved?

[elder linden]

its absolutely clear.
your teacher is a prick

[lofty depot]

LMAO 😭 no hes chill but he just wants to make sure I understand everything so he doesnt like it when i skip stuff

[elder linden]

kidding...

but yeah
everything looks good

[lofty depot]

also this one... kinda weird but when i use symmetry and integrate between 0 and root 3 I get 128pisqrt(3)/5 but when i integrate from - root 3 to root 3 I get 28pi root3. im assuming its the latter because im actually integrating what i need to integrate

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@lofty depot Has your question been resolved?

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@lofty depot Has your question been resolved?

[north pond]

the solid is a cylinder with a height equal to the diameter of its base then, right?

[lofty depot]

yes I would think so

[north pond]

wait shouldn't the volume of the solid be 2.pi units then?
height*area = (diameter).(pi(diameter/2)^2) = (2).(pi.1^2) = 2.pi

since the base is a unit circle

16/3 isn't close to 2.pi

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[wraith hinge]

if red for a>0 and blue for a<0

[eternal nimbus]

what's the question

[wraith hinge]

is red for a>0

and blue for a<0

[eternal nimbus]

your image is a bit hard to make out

[vagrant prism]

yes to both

[wraith hinge]

?

[vagrant prism]

red for a>0 – yes
blue for a<0 – also yes

[wraith hinge]

@eternal nimbus

[dusty sleet]

!noping

[trim jolt]

Please do not ping individual helpers unprompted.

[wraith hinge]

then f(0) is c right

[wraith hinge]

my bad my friend i sent him the image clearly this time as he asked for it

[dusty sleet]

Oh sorry

I didn't realise that

[wraith hinge]

like if i had an eqn:
ax^2 + bx + c=0
f(0) = c.
therefore:
a(f(0))>0 then c>0 or a<0 and c<0

[dusty sleet]

Why are you calculating a(f(0))?

[wraith hinge]

see image my friend

[dusty sleet]

(whatever that notation means)

[wraith hinge]

wdym u dont know functions?

[dusty sleet]

I know them

I asked you because a(f(0)) sounds like a function a(x) evaluated at x = f(0)

[wraith hinge]

see img xd

[dusty sleet]

Therefore I didn't understand it

Yeah let me see it

[eternal nimbus]

mathematicans should never have been allowed to design the notation of math

[wraith hinge]

ok pls ans q .-.

o7

[dusty sleet]

I don't understand what these conditions are for

[wraith hinge]

for making sure that

[eternal nimbus]

I think he means multiplying by a, because a seems to be a constant in this context

[dusty sleet]

Like, can you send the part before?

[wraith hinge]

this is the

first part

of the concept

OH

[dusty sleet]

But which concept?

[wraith hinge]

Myy bad my friend

give me a second

here you go @dusty sleet

can someone

explain now

plssss

[cedar flower]

I can try if everyone else has gone but I'm not sure what you need explaining

[cobalt karma]

That's reasonable logic. But I could be wrong. You're evaluating the product of the constant c and the leading coefficient, and saying that both have to either be positive or negative, for the value to be greater than zero—did I interpret your inference correctly?

[wraith hinge]

yes

[cobalt karma]

Yes, then I agree with your statement.

[wraith hinge]

can u help me with a sum too?

[cobalt karma]

Sure, just send it over.

[wraith hinge]

@cobalt karma this is the soln but

how is D>=0

we werent taught functions too

[trim jolt]

@wraith hinge Has your question been resolved?

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[cobalt karma]

Great question; had to think about it for a bit. Because, recall the definition of x being "real." There is a qualification that the discriminant must meet, in order for this to be the case.

[cobalt karma]

Looks like it closed right before I answered 😆 Reopen this post, or create a new one if you still have questions!

.close

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[stuck sapphire]

Hello!

Corollary Let $S$ and $T$ be finite subsets of a vector space $V$ with $S\subseteq T$. If $S$ spans $V$, then $T$ also spans $V$.

I'm confused by the wording "spans". Does it mean span $S = V$ or $V \subseteq $ span $S$? Tenks!

[solid kiln]

明suhi

[marble wharf]

both in this case

span S subset V is trivial cause V is a vector space and S subset V

but in general when you say that some set A spans some set B, then that means span A = B

[stuck sapphire]

icic. Thanks! Then the corollary actually says "then span $S = V = $ span $T$" right

[solid kiln]

明suhi

[marble wharf]

yes

[stuck sapphire]

Owki. Thanks!

Thanks @marble wharf

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[wraith hinge]

yea

[wraith hinge]

.reopen

@cobalt karma ig i reopened it??

[dusty sleet]

Yep, as soon as you wrote has been automatically reopened

[wraith hinge]

oh

ok

[tame sorrel]

@wraith hinge could you send what your question is again?

[wraith hinge]

quadratic eqns and sure one sec

[wraith hinge]

after this soln there is a soln for it but i dont understand it

[tame sorrel]

do you know how to find the range of a function

[wraith hinge]

idk functions properly but yea ik range and stuff

can u tell me the diff between a function and a quadratic equation

[tame sorrel]

well a quadratic equation focuses on finding specific solutions.
like a function is a relationship between variables

if you have y = ax^2+bx+c here y is a function of x
now if you want to find what value of x gives u specific y iit gives u an equation

[wraith hinge]

yea but then

if i bring

y to the other side

it becomes a

quadratic how

[tame sorrel]

it does not

because y is not a constant it is dependent on x

[wraith hinge]

see the soln

[tame sorrel]

in the solution what they have done is assume y to be any real constant and show that the equation you get from that is true for all real x given that condition

[wraith hinge]

how can it be constant

[gray sun]

nowhere in the quadratic formula derivation do you require that the coefficients have to be a constant

u may treat it as such for visualization but yeah… it doesn’t need to be constant

[wraith hinge]

heh

[tame sorrel]

they must be independent of x though

right?

[storm peak]

ye

[storm peak]

what do u not understand

[wraith hinge]

functions.

and

quadratic eqn difference

cuz idk how D<=0 and D>=0 comes.

[storm peak]

first time u sue d greater than

zero cuz the eq has real roots

[wraith hinge]

u dont know that

how do u even say

[tame sorrel]

what youre trying to show is that y can be any real value for x belongs to R right?

you can prove that by showing that for any k belongs to R y=k gives you a real value of x

[wraith hinge]

dude

pls teach me functions properl

y

like the definition

idk functions properly thats why im

not able to properly figure out..

[storm peak]

like

range in this question is y

range is ntg but values of y

[wraith hinge]

yes.

yes.

[storm peak]

for whihc x is defiend

[wraith hinge]

ik that

[storm peak]

that means real

[wraith hinge]

ddude

how does

[tame sorrel]

in a simple way you can say a function is a relationship between an input and an output
x is input and y is output
you get only one y for one x

[wraith hinge]

x-a*x-c - y)(x-b) =0 a equation then

[wraith hinge]

Yes ik that

and stuff

[storm peak]

continue

[wraith hinge]

its a function..

[tame sorrel]

did u understand this

[wraith hinge]

like a straight line function

[tame sorrel]

no its an equation

[wraith hinge]

EXACTLY!

TELL ME THE DIFFERENCE

sorry

[storm peak]

honestly like udk even need all this for ur question

like if u want to range x shld be defiend

that means real

so u do

D>=0

[tame sorrel]

that is an equation in 2 variables

if you wrote one variable in terms of the other only using the equation
like y in terms of x only (y is a function of x)
or x in terms of y only (x is a function y)

[wraith hinge]

i think im getting irritated you mind if i come after 1 hr or smth? (its been 6 days and i have still been working on this q.)

[storm peak]

💀

no offence

[wraith hinge]

im shit as fuck thanks ik.

[storm peak]

but i dont understand what ur even asking

[wraith hinge]

just give me some time.

[tame sorrel]

okay

[tame sorrel]

this is the simplest explanation i can give

[wraith hinge]

ill just refresh.

[wraith hinge]

yes this makes kind of sense

[wraith hinge]

no offence, but you rush to give an answer. aeria tries to understand my doubt and is answering..

[storm peak]

i mean i still dont understand wht ur doubt even is

[tame sorrel]

the equation is the relationship between x and y.
Now you are trying to use this to write x as a function of y

[trim jolt]

@wraith hinge Has your question been resolved?

[unique sonnet]

This solution looks wrong. It starts by assuming the answer, then deduces something true and says that's enough.

I have frequently found over the years that students worry that they're missing something important when it comes to functions. But there's nothing deep about functions. It's just a rule that gives an output when you give it an input.

The only requirement is that if you give it the same input multiple times, it always gives the same input. So the function cannot flip a coin in its calculation, for example.

Everything else you want to know about functions is actually just studying specific types of functions. In this case: quadratics.

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[past cargo]

is y^2 + (z - 6)y + (7 - z^2 - 6z) = 0 a quadratic in y

[past cargo]

given that z = z(y)

well technically it's a quadratic form right

[wraith hinge]

I think so

[tame sorrel]

no its not, for example take z = y^2

[past cargo]

yeah i thought of that but well

consider finding the range of a rational function whose denominator is of degree 2 and numerator is of degree 1

you transform it into a quadratic in x

in that case y is clearly a function of x

but people still consider the discriminant for the range

[tame sorrel]

there you consider y to be a parameter you are varying it independent of x and trying to find if a possible x can you that y

[random oar]

can anyone explain the equation, as it is quite difficlt of me?????

[past cargo]

what

[past cargo]

you consider z to be a parameter you are varying it independent of y

how does that exact reasoning not work for the above

[past cargo]

just for the record, this is a quadratic in y btw

[wraith hinge]

What is the question now then

[past cargo]

idk what the confusion is

my question is up there

😭

[past cargo]

^

[wraith hinge]

Didn’t u answer urself lol

[past cargo]

i want an explanation not just an answer

I made the question myself

Why would i care about the answer

[limpid dawn]

nope,look up the definition

[tame sorrel]

my bad yes here you can treat it as a quadratic to find conditions on z

[past cargo]

not really worried about quadratic forms but more worried about just regular quadratics

[wraith hinge]

yes.

[past cargo]

also okay let me look at the definition of quadratic forms

[wraith hinge]

its a quadratic in y.

u can make it a quadratic in z too.

[past cargo]

according to the definition, it is indeed a quadratic form tho

[wraith hinge]

yep this also

is right.

[past cargo]

y^2 -> deg 2
(z - 6)y -> deg 2
(z^2 + O(z)) -> deg 2

[limpid dawn]

x and y are independent, z is not

[wraith hinge]

why u say

z is dependent..

functions is smth ig me and mmm7 dont know properly

[limpid dawn]

because it was given

[past cargo]

where does that come into the definition

[past cargo]

?

[wraith hinge]

oh 💀

[wraith hinge]

well that makes me

[limpid dawn]

x and y can be anything in that expression as opposed to z

[past cargo]

yeah but the definition is just:

a quadratic form is a polynomial with terms all of degree two

[limpid dawn]

If z is not in the form of ay²+by+c then it's over

[limpid dawn]

Thats an example on wikipedia

[past cargo]

it's not an example tho 😭

it's the definition on line 1

the example is when they say "for example.."

[wraith hinge]

But I don’t think it holds here because z is a function of y

It doesn’t really make sense if they are dependent

[limpid dawn]

?

[past cargo]

check the first sentence

that's the definition, no?

[limpid dawn]

Unless stated otherwise, x and y are independent

[limpid dawn]

Yes and now pick z = e^y

Then you have a problem

[past cargo]

idgi tbh 😭

[limpid dawn]

abcdegh

[tame sorrel]

check polynomial definition

[past cargo]

?????

[past cargo]

anyway fine we

can say that we care about independence between variables bc that sounds about right

[tame sorrel]

the variables in the polynomial must be independent

[limpid dawn]

is that a question?

[past cargo]

no i mean your point makes sense

but how do you answer this now

not related to quadratic forms btw^

[limpid dawn]

i am just reading a set of instructions? what am i supposed to do?

[past cargo]

also it's not a quadratic form because what i wrote here is wrong

so i guess it didn't matter that z and y were independent or not

[past cargo]

y = y(x) then why are we considering the discriminant

if it's technically not a quadratic in x

,, y(x) = \frac{P_1 (x)}{P_2 (x)}

[solid kiln]

mmmm7

[past cargo]

where P_1 is a first order polynomial and P_2 is a second order polynomial

with some algebra you can "construct" a quadratic in x

but then y = y(x) so like that suggest you can't

[limpid dawn]

ok hold on

[past cargo]

okay here let's give numbers

,, y = \frac{x - 3}{x^2 - 2x + 5}

[solid kiln]

mmmm7

[past cargo]

this when simplifies gets you to $x^2(y) - x(-2y - 1) + (5y - 3) = 0$

[solid kiln]

mmmm7

[past cargo]

hopefully the math is correct there

but that's not the big idea

[past cargo]

people use discriminant here to then uncover the range

[limpid dawn]

ok thats new to me to do that

[past cargo]

but then it's not a quadratic

cuz y = y(x)

[limpid dawn]

i think you treat y as a constant but i cannot really make sense of it

[past cargo]

yeah that's why i was asking

Lol

btw as proof

that this works

,w range of (x - 1)/((x-3)(x-2))

[solid kiln]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by mmmm70177

[limpid dawn]

is there a source to this

[past cargo]

,, y = \frac{x-1}{x^2 - 5x + 6} \implies x^2 y - x(5y +1) + (6y + 1) = 0

[solid kiln]

mmmm7

[past cargo]

the discriminant is (5y + 1)^2 - 4(y)(6y + 1) right?

we want it to be >= 0 to have real solutions for x for a given y

now watch this

[limpid dawn]

ah okay

[past cargo]

,w (5y + 1)^2 - 4(y)(6y + 1) >= 0

[solid kiln]

Results provided by WolframAlpha

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

Requested by mmmm70177

[past cargo]

remember that^

,w range of (x - 1)/((x-3)(x-2))

[solid kiln]

Results provided by WolframAlpha

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

Requested by mmmm70177

[past cargo]

see

[limpid dawn]

yea

[wraith hinge]

That’s cool

[limpid dawn]

you are essentially treating y as a constant

thats what it seems to me

[past cargo]

no then in other cases you could also do the same thing

[limpid dawn]

and then by that rationale it is a quadratic in x

[past cargo]

yeah that was my reasoning thus far, but like hmm

not convinced

[past cargo]

also no

i think it's common enough?

not something novel

[limpid dawn]

ok but like

imagine you replaced y

say t

just a stupid variable

not a function

then you do the same process

and you just figured all possible values for t

so yea you are treating y as a constant and not something that depends on something else anymore

[past cargo]

yeah in that sense y = y(x) doesn't matter

but going back to my very first question then

[limpid dawn]

the thing is when you said z = z(y)

[past cargo]

u can extend this notion to practically my question above

[limpid dawn]

i thought z(y) could be anything

[past cargo]

well

i read in a stackexchange answer

i don't remember

but they said it is indeed a quadratic in y

same question

can't find the link tho

[limpid dawn]

yes cause you treat z as a constant

[past cargo]

yeah but then we're just neglecting z = z(y)

😭

pick a quintic and then make it a quadratic 😭

and treat it as a constant?

[limpid dawn]

well you dont

in the situation you are you dont pick it

you get 0 = 0

[limpid dawn]

When you plug what y was prior to your transformation then you just get 0 = 0

you cant suddenly say y can be a quintic

in that sense it's basically given, or constant

[past cargo]

no i'm saying if you have the liberty to pick when u can start making things constant and when u can't

then technically everything can be quadratic in that sense

[past cargo]

the answer to my question to this was then it is indeed a quadratic in y

but the fact that we just decide to treat z as a constant is a bit weird

i mean i used to do that all this time but like

z = z(y)

it's far from being constant

[limpid dawn]

well

[limpid dawn]

y remains forever (x-1)/[(x-3)(x-2)]

it's not an arbitrary function of x

so in that sense it's constant

[past cargo]

what 😭

in that case all functions are constants

[past cargo]

dk what this means

[limpid dawn]

you cant pick y it was given to you

[limpid dawn]

thats not what i meant

I think the best would be to do the following

y is a function of x bla bla

now regarding the range

[past cargo]

no the range example i just manufactured myself

i can make sense of the range example

bc i thought of it

basically just imagine y = 3 for example

then you want x to have real values for that

now keep changing y

in each case you want x to have only real values

so you can treat it as a quadratic in x

because for each fixed a, we consider the case where x assumes real values

[limpid dawn]

you set or declare y to be an arbitrary but constant number now say t. What you are doing is basically equating every possible number of the codomain to (x-1)/[(x-3)(x-2)] and then you arrive at some quadratic equation. Since you want a solution for x to exist you consider the discrimimant of it to be non negative

the possible solutions for t give the range values because for those values of the codomain there exist a solution of x

[past cargo]

so i was right about my above thing indeed being a quadratic?

[past cargo]

^this?

structurally it is a quadratic

[limpid dawn]

if you say given z is now a constant then i would say yes

[wraith hinge]

I would say that traditionally it isn’t but u can treat it as one to get certain stuff

[limpid dawn]

y²+ye^z+1 is still a quadratic in y given z is a constant

y being a constant now feels only weird because you are not introducing a new variable thats all

in the end you arrive at the same

[wraith hinge]

This makes much more sense now that i think about it when we rearrange the function into a quadratic it has to be equal to 0 and so we check for which values of y there is a solution because there has to be a solution and so by that we get the range of y because the range of y is the range for which the equation holds

[limpid dawn]

yea

[wraith hinge]

So ye treating it as a quadratic might not be traditional but it can help us get certain things

[limpid dawn]

Imagine you did the process with actual numbers

y = 1

y = 2

and so on

everytime you get that quadratic and you see if an x exists

well you would never finish

so the idea is you pick an arbitrary constant

y = t

[past cargo]

btw i think we only care about the structure

[limpid dawn]

and do that process once

[past cargo]

like if you have a cubic in x

and then a quadratic in pi

[limpid dawn]

and well y = y you arrive at the same it's just mathematical laziness

[past cargo]

you can solve the resulting quadratic in pi

Lol

consider:

,, x^3 + \pi x^2 + (3 \pi - 2) x + \pi^2 - 4 = 0

[solid kiln]

mmmm7

[past cargo]

this is a quadratic in pi right?

[wraith hinge]

Yea

[limpid dawn]

if x is constant yea

[past cargo]

no it apparently doesn't matter

[wraith hinge]

X doesn’t have to be a constant now

[limpid dawn]

i mean saying in pi is basically saying x is constant

[past cargo]

normally u give this to a regular person

he'll just say it's a cubic in x

over x in R

no?

[limpid dawn]

maybe try to read again what i wrote and why it's constant

[past cargo]

no our reasoning coincides

i agree with that

[past cargo]

but here you can do the same right?

just treat x as a constant

\pi is the variable

[limpid dawn]

choose x = e^pi

if you say it doesnt matter

[past cargo]

😭 no if you treat it as a quadratic in pi

u get x = 2 - pi

[wraith hinge]

But u don’t have to plug it

[past cargo]

i just tried it

which is indeed one of the solutions

[wraith hinge]

We only care about the image

[past cargo]

i think

,w x^3 + \pi x^2 + (3 \pi - 2) x + \pi^2 - 4 = 0

[limpid dawn]

ok yea

[solid kiln]

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[past cargo]

yeah see

[limpid dawn]

yea

[past cargo]

so okay what we're getting at here is

you can just start to treat anything as constants

at one point

[wraith hinge]

Yes

[past cargo]

actually wtf moment

[wraith hinge]

Because we are only looking at certain properties

Lol

[limpid dawn]

you do that out of laziness

the rigorous way would be to introduce a constant

let me try one last time

[wraith hinge]

But this laziness makes sense

Why wouldn’t we be able to

[limpid dawn]

you are able to but you start confusing things

[wraith hinge]

We are only looking for the value of the “constant” which is by equaling it to 0 the image of our function constant

[limpid dawn]

I think if you used y = f(x)

you wouldn't be as confused

because f(x) refers to the function depending on x and y is basically the value you get for some x, making it basically a constant value

[wraith hinge]

I get what you mean calling it constant is stupid but it’s not because I think it is a constant it’s just the way we are treating it is the same as we would a constant

[past imp]

[limpid dawn]

I will help better next time relax

[grim sparrow]

[merry gyro]

why wormin dawg

what I do

[wraith hinge]

Lol

[grim sparrow]

your fault for making the role

[unreal surge]

What is going on here?

[umbral hazel]

Solo raid

[unreal surge]

What did he do?

[umbral hazel]

Which he

I already told you what Moyai is doing

[unreal surge]

you mean he was spamming?

[umbral hazel]

Spam-pinging to be exact

[unreal surge]

yeah report him

[trim jolt]

@past cargo Has your question been resolved?

[past cargo]

haha thanks guys

[trim jolt]

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[tame tusk]

I have no idea where to even start with this problem?

[elder linden]

it's in the form dy/dx + Px = Q, where P and Q are constants or functions purely in x

these have a standard solution. but its a multi step process
you'll need to find the integrating factor first