precalculus

[tawny nacelle]

yeah

[viscid thistle]

[tawny nacelle]

basically what onion said

${100 \choose r}(\frac{2}{x^2})^{r}(3x)^{100-r}$

[viscid thistle]

why?

[tawny nacelle]

@spring thunder yeah isnt it 100 ??

@spring thunder why u gotta confus me

[viscid thistle]

so that means the power of the term must be greater x^3 ?

[obsidian monolith]

soap:

[tawny nacelle]

IT"S r

my bad

[viscid thistle]

but how do I say that the same foes for the other terms

[tawny nacelle]

it is r right ?

[viscid thistle]

what r?

[tawny nacelle]

nvm

[viscid thistle]

the power is 100

[tawny nacelle]

the power in 2/x^2 is r

[viscid thistle]

well the power is 100

[tawny nacelle]

im talking about hte r + 1 th term mate

anyways

simplify that

${100 \choose r}(2)^r x^{-2r}(3)^{100 - r} x^{100 - r}$

[viscid thistle]

?

[obsidian monolith]

soap:

[tawny nacelle]

${100 \choose r}(2)^r(3)^{100 - r} x^{100 - 3r}$

[obsidian monolith]

soap:

[tawny nacelle]

so u want 100 - 3r = 3

[viscid thistle]

shd be sth like this

[tawny nacelle]

,w solve 100 - 3r = 3

[obsidian monolith]

Display results online and refine query

Requested by soap#7503

[tawny nacelle]

[viscid thistle]

help !

[spring thunder]

we're just talking about one of the terms of the expansion

[tawny nacelle]

haaalp

[viscid thistle]

fuuuck

yes

one of the terms

[spring thunder]

you're not gonna write the whole expansion if there's 100 terms

[viscid thistle]

but the power is 100

exactly

so what should I do

I am so effing confused

bruh I am too dumb

[tawny nacelle]

why isnt the thing im doing working !!

[spring thunder]

well it works

[tawny nacelle]

ik

[spring thunder]

didn't you tell us the answer was 0?

[tawny nacelle]

why did i mess up ?!?!

[viscid thistle]

yes i did

but I need to prove it

[spring thunder]

you didn't mess up

it's exactly what you should find

[tawny nacelle]

OH YEAH

[spring thunder]

ie there's no x^3 term

[tawny nacelle]

r isnt an integer

[viscid thistle]

I know intuitively that it is 0 since expanding it and divided by x^2

[tawny nacelle]

so no term with x^3

[viscid thistle]

no terms would be x^3 ish

but how???

[tawny nacelle]

r isnt an integer bruh

[viscid thistle]

should I describe the whole thing

[tawny nacelle]

we just proved it

[viscid thistle]

goddamn

[tawny nacelle]

i literally wrote the whole thing u need

[viscid thistle]

?

wtf

halp

[tawny nacelle]

we just showed r inst a fking ineger

[viscid thistle]

wait what is r?

[tawny nacelle]

oh god

its a variable

[viscid thistle]

I know

lije what does taht refer to

and why is it that r +1 term is what I want @tawny nacelle

[spring thunder]

it just refers to some term of the binomial expansion of (...)^100

[viscid thistle]

oic

no

\

no wait

it shd be

[tawny nacelle]

r + 1 makes life easier

cause then u can just plug r for the r + 1th term

[viscid thistle]

R+2/x^2

so can you write the solution for me again

plz

wtf

[spring thunder]

and fuck we don't have any cat litter

gtg

[slow wharf]

[viscid thistle]

@spring thunder

is this correct

It proves that there is no x^3 tern amiright?

[tawny nacelle]

this doesnt prove it bruh

[viscid thistle]

?

[tawny nacelle]

u say that let the r + 1th term be ......

u write all that that

and then u set the power of x which is 100 - 3r = 3

solve for r

u get r = some non integer

[viscid thistle]

bruh

right?

[tawny nacelle]

yeah bruh

[viscid thistle]

yaya

POGGERS

actually the same thing, i guess i got a bit confused at first

[slow wharf]

What is the correct procedure to check the domain of a function that has a fraction?

For example

$y= \sqrt{\frac{x+2}{5-x}}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I know that the condition is that

$\frac{x+2}{5-x} \ge 0$

[obsidian monolith]

Autistic Hoodie:

[viscid thistle]

Distribute the square root to to and bottom

[serene heath]

ye that works

[viscid thistle]

Then set to greater than or equal to 0

[eternal lotus]

Well first of all x=/5

[serene heath]

and thats easily solveable

also that

[slow wharf]

@eternal lotus Yeah I understand that part

But how do you solve

[eternal lotus]

Considering two cases for that inequality can help u understand to solve it

[slow wharf]

https://cdn.discordapp.com/attachments/363224154469826562/540305765089411072/162305820107735040.png

[eternal lotus]

When num and dem are positive

[serene heath]

multiply both sides by (5-x)^2

[eternal lotus]

or that

[serene heath]

and solve

[slow wharf]

@serene heath Are you saying that

It doesn't matter

that I can just do

$x+2 \ge 0$

[obsidian monolith]

Autistic Hoodie:

[eternal lotus]

U have a (5-x) as well

[serene heath]

both sides

[slow wharf]

I have to do $5-x \ge 0$ aswell

[serene heath]

not just one

[obsidian monolith]

Autistic Hoodie:

[eternal lotus]

$(x+2)(5-x) \ge 0$

[obsidian monolith]

stephen:

[slow wharf]

Oh

[serene heath]

[eternal lotus]

We can times (5-x)^2 and preserve inequality because we know what we're multiplying by is positive btw

[slow wharf]

That's an awesome trick

But wait

I get

$-x^2+3x+10 \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

ANd the answer to that is 5 and -2

[eternal lotus]

I suggest you sketch the graph to solve the inequality

[serene heath]

yea dont expand it out

a quick sketch should help

[slow wharf]

Yeah

[serene heath]

like stephen said

[slow wharf]

Awesome

This makes it so much simpler

[serene heath]

[slow wharf]

I wish I knew this earlier

Before I would do

$x+2 \ge -1$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I mean

I would still get -2 and 5

But I would need to test it to see if which one is the real value

Like

$x+2 \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$x \ge -2$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$5-x \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$5 \ge x \ x \le 5$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Hmm, seem to do the same trick?

Is this a correct way too? Because I don't think I get the correct result sometimes

[eternal lotus]

Mhm but 5-x =/0

U also have to consider

[slow wharf]

Yeah, that too

[viscid thistle]

How do you do this

[eternal lotus]

$x+2 \le 0$ and $5-x<0$

[obsidian monolith]

stephen:

[slow wharf]

@eternal lotus Yeah

But does it always work like that?

[eternal lotus]

Yep

[slow wharf]

Hmm

[eternal lotus]

I dont see why it wouldnt

[kind pier]

can anyone tell me if im doing significant figures correctly?

[viscid thistle]

@kind pier

post

[kind pier]

Wait that didn’t show up..

[viscid thistle]

,rotatoe 270

,rotate 270

[viscid thistle]

4 is correct

3 is correct

4 iscorrect

0.030900 is not 4

0.0005 is not 4

[kind pier]

dam so then its 8 and 5?

[viscid thistle]

no, you're forgetting trailing and leading zeros rule

any huge line of 0's in front of after don't count

with that in mind lets not something

1230 is only 3 sig figs

notice that 0 doesnt count

but why?

was it 1225?

1226?

1228?

[kind pier]

khan academy said because no decimal

lol

[viscid thistle]

well point is

if we got 3 sig figs

that means that 4th one was roudned up

or down really, a lot goes into it

but and string of 0's before or after dont matter

So lets look at

0.030900

COMPLETELY IGNORE and 0's before or after the non-zeros

what digits matter here?

[kind pier]

309 and 5

?

[viscid thistle]

correct

309 matters, because its not 308

its not 310

but 309 for sure

[tawny nacelle]

all numbers MATTER

[viscid thistle]

except in sig figs

dats ur lesson 4 2day

[tawny nacelle]

[kind pier]

oh wait cool i get it... that number was 308(number over 5 then other numbers for next 2 decimal spots)

[viscid thistle]

right the point is

they knew it to that place

if I told you 14+94

you might say 118

but well, we only know the ones and tens places (2 spots)

what if its 14.9 and 94.9?

thats 119.8

trouble is, you can't be too sure really all the time about what matters in numbers

(thats what sig figs tell you)

its sort of the "close enough" or "as good as I can with given data" approach

[kind pier]

that second explanation was great

i get it and why we are using them

thanks

[viscid thistle]

np m8

@kind pier if you want a more worldly approach (or thing to read) about it, heres a bit to read

Also come to think of it

That 0.030900 they could say those trailing 0s matter

So that's up to your professor to determine really

You got any notes?

It is

[kind pier]

we haven't gone over this yet it's tomorrows lec/lab, but I just readd through your thing

[viscid thistle]

It says they are

[kind pier]

and at the top of the pciture i sent it does say that

[viscid thistle]

Its on that page yeah

So count those too

30900

[kind pier]

so if we count those 2 zeros that means we didnt round right?

[viscid thistle]

Well maybe

0.0030901?

[kind pier]

or if we did it was at the last 0

[viscid thistle]

02 03 04?

It could be rounded to that last 0

Could be two added numbers

0.003090

+0.0000001

Adding is tricky too

[deep trail]

hello

Not sure how to do #1

[viscid thistle]

"Spectral Triples for the Variants of the Sierpinski Gasket"

and they make her teach precalculus

shameful

just draw it with all of those properties

here's an example

see if you can come up with another example

[slow wharf]

What would be the simplest way to solve

$3x^4-x^3-9x^2-3x \le 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Without using calculator?

I though of doing something like this

$x^2(3x^2-x-9-\frac{3}{x}) \le 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$3x^2-x-9-\frac{3}{x} \le 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$3x^2-x-9 \le \frac{3}{x}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$x(3x^2-x-9) \le 3$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$3x^2-x-9 \le 3$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

seems like i am not allowed to do that

[eternal lotus]

You may transpose the "x" if x>0, preserving the inequality

[serene heath]

third last pic

you could sketch the 2

and solve it like that

[slow wharf]

@serene heath Whut?

[serene heath]

https://cdn.discordapp.com/attachments/363224154469826562/540484528859250689/162305820107735040.png

this one

you could setch the 2 functions

*sketch

[slow wharf]

How is it two functions?

It's a polynomial

[serene heath]

3/x and the one on the left

[slow wharf]

$3x^2-x-9 \le 0 \ \frac{3}{x} \ge 0$

[obsidian monolith]

Autistic Hoodie:

[eternal lotus]

I just popped $3x^4-x^3-9x^2-3x \le 0$ in the calculator and the results are not exact, I'm not too sure that you can do this question without a calculator

[obsidian monolith]

stephen:

[slow wharf]

The method on the left doesn't always yield the correct result, the one on the right is better?

[odd yarrow]

Maybe but the right is unarguably more tedious from what I'm seeing

butt I think the right

[atomic zodiac]

you did the left incorrectly

[odd yarrow]

but*

[slow wharf]

@atomic zodiac Really?

[odd yarrow]

Is more accurate

[atomic zodiac]

and i've never seen the right before

[slow wharf]

I believe I need to make a table and test for value for the left one?

[atomic zodiac]

you're looking for wherea division is < 0

so either the top or bottom has to be negative

but not both

easiest way is to draw a number line

and label where the numerator is pos/neg

and label where the denominator is pos/neg

then it's easy to see where the region you want is

[slow wharf]

Like this?

[atomic zodiac]

uh

idk what that is

[slow wharf]

._.

[atomic zodiac]

nvm what i said before i thought you were looking for < 0

but it's >0 so you need pos/pos regions or neg/neg

does turning it into a multiplication even work every time

does that make sense @slow wharf

[slow wharf]

I don't understand it

[atomic zodiac]

above number line is where -x-2 is positive and negative

below x=-2 it's positive and above that it's negative

below the number line is where 2x+1 is positive and negative

below x=-1/2 it's negative and above that it's positive

you want where it's >0 so you need the regions where it's positive in the numerator and positive in the denominator

or where it's negative in the numerator and negative in the denominator

for these functions the overlap is just when they're both negative

@slow wharf

[summer siren]

How the heck do I start with this

<@&286206848099549185>

[serene heath]

try not to ping helpers before 15 mins have passed

but for your q, try writing u and v as functions of time then differentiating with respect to time

[summer siren]

Sorry it was quite urgent

[slow wharf]

How would I go about solving

$|z-2| + |z+2| \le 2$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

uhh try to think of it geometrically ig

what would it represent if that was an equal sign instead

waaait a min..

is this the same q you tried a while ago @slow wharf

[slow wharf]

I think I asked it like

Months ago

[serene heath]

where the answer is the empty set

[slow wharf]

Sec

yup

[serene heath]

oh no

the flashbacks

the horror

[slow wharf]

Omg >_>

[serene heath]

algebra didnt work last time right

so ig we can visualize what it represents

prettt sure thatd look like some sort of ellipse

,w plot |z-2|+|z+2|=5

[obsidian monolith]

Display results online and refine query

Requested by lemon catto#7133

[serene heath]

no

how do i do argand diagrams on this thing

[slow wharf]

I have no idea

[serene heath]

,w plot |z-2|+|z+2|=2 in an argand diagram

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by lemon catto#7133

[slow wharf]

Symbolab has an interesting solution

[serene heath]

is it $\emptyset$

[obsidian monolith]

lemon catto:

[slow wharf]

It is No:Solution:for:z\in \mathbb{R}

$No:Solution:for:z\in \mathbb{R}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

But Z is C

[serene heath]

well obvs

ye

we are in C

hmm hang on

[atomic zodiac]

there's no solution in C too

p sure

[serene heath]

ye we should reach that conclusion through algebra

but we dont

clearly doin somethin wrong

[slow wharf]

Maybe I should take another example

That has an answer

[serene heath]

no

dont give up

[slow wharf]

Omg >_>

[serene heath]

dw

[slow wharf]

How about we look at it this way

VIsualize it

|Z| iz the distance from the center to the complex number right

distance is always positive?

And cannot be smaller than 0

No?

Adding two distances will always be positive

I think that's the answer

[atomic zodiac]

,w plot |z+2|+|z-2|

[obsidian monolith]

Display results online and refine query

Requested by Gamedolf#0001

[serene heath]

ok after some algebrs you get this

16x^2+64x+64+16y^2=8x+4

which cant be right since its supposed to be some sort of ellipse

[slow wharf]

I get a whole lot different number when doing is algebricly

[serene heath]

yee i prolly messed up big time

theres prolly a simple and intuitive solution to this we're not seeing

repost q and @ helpers tbh

[slow wharf]

I have a similar one wait

$|z-1-i| + |z-3-i| \ge 4$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

This one is supposed to be solved algebraically

[slow wharf]

@atomic zodiac You have an idea?

<@&286206848099549185> Any idea?

[elfin night]

Is the answer $\left|z-2-i\right|\ge1$ ?

[obsidian monolith]

Colen:

[atomic zodiac]

i was just doing some algebra

[elfin night]

algebra

[atomic zodiac]

didn't really get anywhere cause i think i made a mistake

[serene heath]

algebruh

[slow wharf]

$\frac{(x-2)^2}{4} + \frac{(y-1)^2}{3} \ge 1$

[serene heath]

it should lead to the answer tho

[elfin night]

if thats the answer I know how if not then idk

\frac

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

That's the answer

[serene heath]

ye

its an ellipse

the algebra is just too long

[elfin night]

$\left(x-1\right)^2+\left(x-3\right)^2+2\left(y-1\right)^2\ge4$

[obsidian monolith]

Colen:

[elfin night]

something along these lines

I think

[serene heath]

[elfin night]

[serene heath]

you thonk?

[elfin night]

Idk I remember doing it this way in FP1

[serene heath]

fp1

[elfin night]

but it was never an ellipse

[serene heath]

nice

[elfin night]

f p 1 is bae

it was always like

[serene heath]

it was ok

[elfin night]

single there though

not two abs

[serene heath]

not enough complex stuff

[elfin night]

do FP3

has mor complex stuffs

[serene heath]

does it

[elfin night]

ye has polar form n stuff

[serene heath]

ik it has vectors and hyperbolic stuff

most of which ik

wait fp2 had polar

[elfin night]

[serene heath]

im talkin edexcel

[elfin night]

OCR supremacy tho

https://cdn.discordapp.com/emojis/504810017057275914.png?v=1

[serene heath]

haha

no

[elfin night]

I think all my exams were OCR

[serene heath]

imagine not doin de moivres till fp3 smh

normal or mei

[elfin night]

normie

shush urself

[serene heath]

mei is jusy

[elfin night]

mei is harder no?

[serene heath]

yes

much

[elfin night]

I would've done if they offered but we don't choose

[serene heath]

well its alot less guided i suppose

[elfin night]

[serene heath]

back to the q

[elfin night]

yus

[serene heath]

@slow wharf have u tried goin through the algebra?

[slow wharf]

Doing it algebraically yields that answer?

[serene heath]

it most defo should

if it doesnt youre doin somethin wrong

[slow wharf]

When I do it I get huge and huge numbers

Like

Long ones

[elfin night]

[slow wharf]

With xs

How do I explain

It takes long to get to the answer

And has lots of x^3 and even x^4

[serene heath]

shouldnt be gettin those

[slow wharf]

I will try to solve it using latex here and you find my mistake?

$|z-1-i| + |z-3-i| \ge 4$

[elfin night]

$\left|x+yi-1-i\right|+\left|x+yi-3-i\right| \ \left|\left(x-1\right)+i\left(y-1\right)\right|+\left|\left(x-3\right)+i\left(y-1\right)\right| \ \left(x-1\right)^2+2\left(y-1\right)^2+\left(x-3\right)^2\ge4 $ I thought this was the technique, but no right answer

[obsidian monolith]

Autistic Hoodie:

Colen:

[elfin night]

nice emote

comes out 2 circle

forgot my abs stuffs

@serene heath answer pls ❤

[slow wharf]

$|x+iy-1-i|+|x+iy-3-i| \ge 4 \ |x-1+i(y-1)| + |x-3 + i(y-1)| \ge 4 \ \sqrt{(x-1)^2+(y-1)^2} + \sqrt{(x-3)^2+(y-1)^2} \ge 4$

[elfin night]

ye

[serene heath]

@elfin night ye thats not how abs works

[elfin night]

brain betrayal

[serene heath]

ye good so far

[obsidian monolith]

Autistic Hoodie:

[elfin night]

🤢

[slow wharf]

What am I supposed to do next?

Square it?

Or is there simpler way?

I don't think there is

[serene heath]

move one root

to the other side

then square both sides

then move it again and square again

i think that should work

[elfin night]

:/

[serene heath]

if it doesnt i will

[slow wharf]

$\sqrt{(x-1)^2+(y-1)^2} \ge 4 + \sqrt{(x-3)^2+(y-1)^2} \ (x-1)^2+(y-1)^2 \ge 16 + 8\sqrt{(x-3)^2+(y-1)^2} + (x-3)^2 + (y-1)^2$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$(x-1)^2+(y-1)^2 - (x-3)^2 - (y-1)^2 -16 \ge 8\sqrt{(x-3)^2+(y-1^2)}$

[elfin night]

lmao

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$(x-1)^2-(x-3)^2 - 16 \ge 8\sqrt{(x-3)^2 + (y-1)^2}$

Seems good s ofar

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Seems good so far now

$x^2-2x+1 - (x^2-2x+9) - 16 \ge 8\sqrt{(x-3)^2 + *(y-1)^2}$

[obsidian monolith]

Autistic Hoodie:

[rocky bison]

[slow wharf]

$-3 \ge \sqrt{(x-3)^2+(y-1)^2}$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

oooh

this looks promising

hang on tho

b4 u square

[rocky bison]

How does it

-3 is bigger than the root of the sum of squares

[serene heath]

lmao ye

[slow wharf]

It's complex numbers

No?

[rocky bison]

Given you're using x and y i assume you're in R^2

not C

[serene heath]

i was gon say the inequality should flip when squaring?

[rocky bison]

i'm missing context

but the sum of squares is a bit oof

[slow wharf]

It is this

$|z-1-i| + |z-3-i| \ge 4$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

https://cdn.discordapp.com/attachments/363224154469826562/540550887152549890/162305820107735040.png

[rocky bison]

Ok so we do actually deal with complex numbres

[serene heath]

oh

[rocky bison]

You're trying to find what

z

[slow wharf]

Yes

[serene heath]

or the locus

w/e

[rocky bison]

oke

i leave lemon to hälp then seeming as i might've jumped the gun

[slow wharf]

$9 \ge (x-3)^2 + (y-1)^2 \ 9 \ge x^2 - 2x - 9 + y^2 - 2y + 1 \ 17 \ge x^2 - 2x + y^2 - 2y$

[obsidian monolith]

Autistic Hoodie:

[atomic zodiac]

[slow wharf]

Doesn't seem right

@atomic zodiac You were able to solve it?

[atomic zodiac]

you had equation of a circle

which one we solving

[slow wharf]

$|z-1-i| + |z-3-i| \ge 4$

[atomic zodiac]

https://cdn.discordapp.com/attachments/363224154469826562/540550887152549890/162305820107735040.png

[obsidian monolith]

Autistic Hoodie:

[atomic zodiac]

cool

[atomic zodiac]

squaring is such a pain

[slow wharf]

Yeah...

[rotund musk]

Alright this is starting to hurt my brain area. 9e^3x=100. Can't figure out how to configure it to ln form

[slow wharf]

$x=\frac{ln(\frac{100}{9})}{3}$

[obsidian monolith]

Autistic Hoodie:

[atomic zodiac]

3(x-2)^2+4(y-1)^2=12

?

[slow wharf]

sec

[atomic zodiac]

oh right inequality

uh

3(x-2)^2+4(y-1)^2≥12

[slow wharf]

Yeah

It's correct

[atomic zodiac]

pain

[slow wharf]

._.

[atomic zodiac]

i don't remember doing this in fp1/2/3

like there was no +

[slow wharf]

I don't know what that is but I am the first year of computer science college

[atomic zodiac]

further pure

A level maths

wait

...

isn't this the shit where polar form makes it 1000x easier

[slow wharf]

Is it?

Because I find it stupid that it is so long

[flat turret]

I mean, the fact that it's an ellipse is pretty easy to see

Since you have $|z-a|+|z-b|\ge 4$

[obsidian monolith]

Gonzo17:

[flat turret]

Where a, b are fixed complex numbers (1+i and 3+i)

And the locus of all points with a fixed distance from two loci is an ellipse

and that's just the area outside the ellipse $|z-a|+|z-b|=4$

[obsidian monolith]

Gonzo17:

[flat turret]

Now it would be good to move from C to R^2

So the sum of distances of (x,y) to (1, 1) and (3, 1) is 4

Now we can transform everything by $(x,y)\mapsto (x-2, y-1)$

[obsidian monolith]

Gonzo17:

[flat turret]

To get the more familiar "loci on the x-axis symmetric wrt the y-axis"

And the "new" loci would be (-1,0) and (1,0)

[atomic zodiac]

loci or foci?

[sleek socket]

help

[serene heath]

dont we all need it 😔

[sleek socket]

@viscid thistle

@serene heath please help me

[serene heath]

with what

you haven't even posted a question

[sleek socket]

i can't I lost my phone cant take a picture

[serene heath]

aight np

lemme just read your mind real quick

can you type it tho

?

[sleek socket]

yes. im gonna

[serene heath]

oke

[sleek socket]

2x^2+18x-14 solve for g(3x)

then the same thing but solve for g(1+5m)

[serene heath]

wut

can you type the question exactly as it is

[sleek socket]

i did....

[serene heath]

fr?

what's g tho

[sleek socket]

yes... now do you understand why im confused

[serene heath]

well yea

[sleek socket]

and its probably like f(x)

[serene heath]

question is very incomplete

[sleek socket]

but instead of f its g

[serene heath]

my best guess is that first thing is g(x)

and you're meant to find expressions for g(3x)

and the other one

[sleek socket]

find each function

my bad looked at a different one

[serene heath]

[sleek socket]

man it takes 45 mins to eat

[viscid thistle]

5min

[sleek socket]

for me....

either way I guess Ill get help at school tomorrow

[sharp bay]

What is the greatest number of relative maxima that a polynomial of degree 15 can have ?

I think its 7 but im not sure how to verify it.

[slender river]

first derivative will be a 14th degree polynomial

then it follows from the definition of a maximum that if the polynomial is sufficiently nice then it will utilize all 7 of its maximas

or watever that means

[sharp bay]

oh

[slender river]

oh shit this is precal

sorry

uh

without calculus uhhhh

im sure there's some dubious algebra or some theorem kek

ehhhh idk

[sharp bay]

ty for the verification

i think ima play around with graphing dif functions to test it

[viscid thistle]

No its degree -1

@slender river @sharp bay

A polynomial of degree n has n-1 solutions

[slender river]

that's the property i just said

15-1 is 14

then minimas and maximas are half and half if its a fully fleshed out curve if u know wat i mean

[viscid thistle]

That's that dubious algebra theorem u spek of

[slender river]

so 7

oH

yeah

OHHHH

the solutions are the key

and then u just use like

continuity or something to justify the maximas exist

kek

[viscid thistle]

I think in US school you get it throw at you as a fact not how its a fact

[valid vector]

What's the range of the following thing the "x + 9" being inside a square root?

h(x) = − x + 9

I wrote Range: (-Inf, -9] Why is it wrong?

@odd lichen

[viscid thistle]

$$ y = -\sqrt{x+9} $$

is this what you have?

[valid vector]

Yeah, but with the negative outside the square root. I figured it out I think, the Range should be from (-Inf, 0]

[obsidian monolith]

yami:

[viscid thistle]

so this then

[bright dock]

Yo guys quick refresher, how do I find the range of a rational function?

[viscid thistle]

the range are consists of all the possible values your function can have

range of $ f(x) = x² $ would be $ \bR^+ $

[obsidian monolith]

yami:

[viscid thistle]

including 0

[rocky bison]

@viscid thistle

How much are you ready for

We going the full mile

[viscid thistle]

hell yeah

[rocky bison]

So

[viscid thistle]

[rocky bison]

Do you know what natural log is

[viscid thistle]

loge()

[rocky bison]

ye p much

What about exponentials?

Know roughly how to manipulate them?

and some log rules?

[viscid thistle]

log maybe exponentials uhh

[rocky bison]

If you're not too confident on log and such why you tryna get ur head around the derivs of it

[viscid thistle]

I need a basic understanding of integration for my super speedy robot PID

[rocky bison]

ur question keeps changing????

[viscid thistle]

someone told me I need an understanding of derivatives first

so uhh

here I am

[rocky bison]

it's not something i'd rush tho

get comfortable on the pre-requisites first

like manipulating basic expressions algebraically and understanding exponential/logarithmic rules

otherwise differentials will make no sense and integrals would make even less sense

[viscid thistle]

by exponentiation you just mean equations with exponents right?

[rocky bison]

Yes

And generally manipulating them

with log and such

[viscid thistle]

oh I thought you meant fancy n^x graphs

[rocky bison]

That too

It's all super useful

[viscid thistle]

where do i go for these 'log rules'

the full mile became a marathon

[rocky bison]

https://www.examsolutions.net/as-maths/edexcel/pure-maths-as-tutorials/#log_exponential

[viscid thistle]

aighty

[dire pagoda]

Can anyone explain this unit circle problem to me?

[bleak slate]

idk

[rocky bison]

That's not a problem

[viscid thistle]

hmm, can anyone gimmie a little insight on how to prove that this equation is an identity?

tan(x+pi/4)=(1+tanx)/(1-tanx)

[eternal lotus]

Using the formula, $tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a) \cdot tan(b)}$ should do it.

[obsidian monolith]

stephen:

[bright dock]

I got a D on my midterm

[serene heath]

nice

[slender river]

sad

[slow wharf]

@serene heath Remember the problem with the complex numbers

[serene heath]

ye

[slow wharf]

Well, there's a method to solve it 50 times simpler

I just found it out

[serene heath]

how can I forget

ooh?

[slow wharf]

Yeah...

[serene heath]

show

[slow wharf]

Apparently

$|z_1 + z_2| \le |z_1|+|z_2|$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

oh

[slow wharf]

Inequality of triangle

[serene heath]

triangle

ye

how does that come into play

oh wait

I see

[slow wharf]

$|z-1-i| + |z-3-i| \ge 4 \ |z-1-i+z-3-i| \ge 4$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I believe so

$|2z-4-2i| \ge 4 \ |2x + 2iy -4 -2i| \ge 4 \ |2x - 4 + i(2y - 2)| \ge 4 \ \sqrt{(2x-4)^2+(2y-2)^2} \ge 4 \ (2x-4)^2+(2y-2)^2 \ge 16$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$4x^2 - 16x + 16 +5y^2 - 8y + 4 \ge 16$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

wait

no need to expand

[slow wharf]

Oh

[serene heath]

factor out a 2 from both brackets

[slow wharf]

$2(x-2)^2 + 2(y-1)^2 \ge 16$

[obsidian monolith]

Autistic Hoodie:

[serene heath]

no

its gon be 4

not 2

[slow wharf]

What?

I don't know how to do that ._.

[flat turret]

But this is all wrong tho, you can't just use an inequality like that

[serene heath]

well rip

[flat turret]

You were supposed to find all the points for which
$|z-1-i| + |z-3-i| \ge 4$ ?

[obsidian monolith]

Gonzo17:

[slow wharf]

Yeah

[serene heath]

ye

[flat turret]

Well all the points

With using the triangle inequality you're losing some

[serene heath]

[flat turret]

$|z-1-i+z-3-i|\ge 4$ implies $|z-1-i| + |z-3-i| \ge 4$

[obsidian monolith]

Gonzo17:

[flat turret]

but it's not equivalent

So you're gonna get a smaller region "inside" the actual one

[slow wharf]

The answer i am supposed to get is (x-2)^2/4 + (y-1)^2/3 \ge 1

$\frac{(x-2)^2}{4} + \frac{(y-1)^2}{3} \ge 1$

[serene heath]

why doesn't the algebra work

or maybe it does and I'm just bein dum

[flat turret]

Yeah you just have to know your ways around ellipses

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

But the answer I get using this method is

$\frac{(x-2)^2}{4} + \frac{(y-1)^2}{4} \ge 1$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Is that due to it being inaccurate or did my professor maybe write it wrong?

[flat turret]

Well yeah, as I said, you're losing some points

I just explained it

Using the triangle inequality makes you lose some points

[slow wharf]

Is there another trick like that?

Because it's painful to do it the other way...

[flat turret]

triangle inequality can help you prove that there's no points satisfying something

Well in this case it's some manipulating with ellpises

Basically there's descriptions of ellipses which kinda do the job for you

the region is an ellipse since it's "(distance from z to 1+i)+(distance from z to 3+i)=4"

and that's one of descriptions of an ellipse

a set of points with equal sum of distances to two points in the plane

And then you can use another representation of an ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

[obsidian monolith]

Gonzo17:

[slow wharf]

This all sounds so complicated :/

[slender river]

ellipses r fun tho

[flat turret]

Ya they are

especially if you get all the properties from slicing a cone with a plane

[slender river]

👀

[serene heath]

ellipses bad

[slender river]

im still hung up on proving that an ellipse whose other focus is at infinity is a parabola

“at infintiy”

[flat turret]

Wdym other focus is infinity?

[slender river]

u set focal distance as a constant

set semiminor axis constant

let distance from vertex to centre go to infinity

i know a super handwavy method

but i want a better way

if there even is one

er what i mean by focal distance

is from vertex to focus

[flat turret]

[slender river]

eheh

its weird

um

[flat turret]

How do you define a parabola though?

[slender river]

wat u mean

many ways to do it

theres one definition with directriz

x

[flat turret]

Yeah just wondering which one would be the nicest to work with

[slender river]

i dunno

depends on usage tbh

the definition im trying for is for a proof i know how to do but depends on this definition that i cant defend

ok now on computer yay

[green anvil]

Anyone on atm on here ?

[viscid thistle]

ye

[green anvil]

Im trying to simplify a 1+t-9 / (t-9)(t-8)
for a Difference Quotients

[viscid thistle]

wat u ned

what you gotta do?

you wanna derivative that?

[green anvil]

Cancel the common factors

[viscid thistle]

there be none really unless you copy wrong

[green anvil]

Thats what I Thought but its not wanting to accept it :/

[viscid thistle]

send screen shot of it

[green anvil]

How do I even screenshot this..

[viscid thistle]

PrtScn

print screen then control v

[viscid thistle]

its telling you you can group

1+(t-9)

cancel that out

1/(t-8)

actually

1+1

so 2/(t-8)

[green anvil]

Still dosent want to take it.. This is odd

[viscid thistle]

,w simplify (1+t-9)/((t-9)(t-8)

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[viscid thistle]

what?

[atomic zodiac]

wat

[viscid thistle]

,w simplify (1+t-9)/((t-9)(t-8))

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[atomic zodiac]

1-9 is -8

you have t-8 on top and t-8 on bottom

[viscid thistle]

🤦

[atomic zodiac]

xd

[viscid thistle]

have big dumb

@green anvil

[green anvil]

Soo... Im still lost... lmao

[viscid thistle]

what he say

1+t-9

t+1-9

t+(1-9)

t+(-8)

t-8

cancel that t-8 in top with bottom

[green anvil]

OHHHH
I get it

Gee that was horrible.

The 1 / t-9 is the correct way,
This class is fun....

Thank you!

[hollow plover]

anyone help me

?

[slender river]

no

[slender river]

lol yes

ok

[spring thunder]

oof

[hollow plover]

please

is that answer at the bottom right?

the graph is symmetriic to the y axis and the orgiin?

[slender river]

frick i forgot what to the origin meant

uhhh

yeah y axis is correct

but idk about origin lemme th0nk

[spring thunder]

not to the origin tho

[hollow plover]

ok

why not the origin

if you take -1,4 and make it 1,-4

oh waot

ur right

its not the origin

cause 1,-4 is not on the graph

[spring thunder]

^ _ v

[hollow plover]

pls

am i right

with what i said

so its y-axis

😭

okk

im right haha

[viscid thistle]

How do I do the third one... my teacher never taught log in a summation 😦

[tawny nacelle]

it's just the sum of first log_2 n natural numbers

pretty sure the formula goes like

$\frac{\log_2 n (\log_2 n + 1)}{2}$

[obsidian monolith]

soap:

[tawny nacelle]

cause the formula for the sum of first n natural numbers is

$S_n = \frac{n(n+1)}{2}$

[obsidian monolith]

soap:

[tawny nacelle]

it's just replaced it log_2 n

@viscid thistle

[viscid thistle]

Oh I see. Ty

[tawny nacelle]

[serene heath]

[hollow plover]

anyone please help me with this

i dont know why my answer is wrong

i used the standard formula and made x=0 and y=0

and i got what i put in the box

????

[viscid thistle]

@hollow plover do you have a smartphone?

go download an app called Photomath, it has your answer

its free

[hollow plover]

my smartphone is pretty old so i cant download the app oof

[hollow plover]

i got it i found a calculator my answer was correct its just that i had to write out the +'s and -'s lol

[true talon]

can someone help me with some distance between points questions

i know I need the distance formula but I forgot how to do it

[slow wharf]

Distance between to points is very simple

$d(A,B) = \sqrt{(x_1-x_2)^2+(y_1-y_2)}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Or if you are using Euler's system then it is

$d(A, B) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

[obsidian monolith]

Autistic Hoodie:

[zinc prairie]

if the domain of a log is x > 0, and the domain of a polynomial is all real numbers, why would the domain of logbase4of(1+x+2x^2) be all real numbers?

http://prntscr.com/mg8h4b

[tawny nacelle]

1 + x + 2x^2 gotta be greater than 0

[zinc prairie]

yea and i found the zeroes, theyre: (-1/2, 1)

so im confused

[tawny nacelle]

,w graph 2x^2 + x + 1

[obsidian monolith]

Display results online and refine query

Requested by soap#7503

[tawny nacelle]

how exactly did u find roots again ?

[zinc prairie]

wait

i typo'ed

[tawny nacelle]

[zinc prairie]

its a negative in front of 2x^2

[tawny nacelle]

oh

,w graph 1 + x - 2x^2

[obsidian monolith]

Display results online and refine query

Requested by soap#7503

[tawny nacelle]

do you see when the function is positive ?

[zinc prairie]

kgjwenrgjnk wait i got the right answer

i typed it into calc wrong

but work was right

[tawny nacelle]

nice

[zinc prairie]

should be [0,4/5)U(4/5,1)

I AM SMORT

http://prntscr.com/mg8kzr

how can this be negative infinity if the domain of a even root is x>=0

why is it not [0,6]

[tawny nacelle]

where did u see that the domain is x>0 ??

[zinc prairie]

http://prntscr.com/mg8lgx

[tawny nacelle]

that's for f(x)

not just x

in this case your f(x) is 12 - 2x

[zinc prairie]

im....

so stupid

[queen lion]

Why its rhl is 0?

[tawny nacelle]

cause $0 \leq 0^+ < 1$

[obsidian monolith]

soap: