help-36

[tranquil pine]

sqrt18

[fathom walrus]

so whats the area of the inner square

[tranquil pine]

so 18 is the area of the square

which is

half of the 1st one

so ts js halfing

no?

[fathom walrus]

let me ask my sir

@magic sparrow

do you concur?

[tranquil pine]

😭

[magic sparrow]

Let me ask my sir

@tiny gorge

[fathom walrus]

LOL

[tranquil pine]

😭

[fathom walrus]

wait if Bungo asks his sir...itll become an infinite geometric series

1/1-r

[tiny gorge]

what is your doubt?

[fathom walrus]

hilbert can now rest in peace

[magic sparrow]

Stephen had a doubt about Alana’s doubt

[tiny gorge]

ohh recursive doubts

[fathom walrus]

thats correct, alana proposed the idea that the area of the squares are halving each time

[fathom walrus]

yes exactly

[magic sparrow]

And we thought it be best we ask our sir

[tiny gorge]

indeed i too must ask my sir

@barren hound

[midnight crest]

Since y'all have sirs can one of you become my sir?

[fathom walrus]

Bungo's sir is a mod

[tiny gorge]

i tried to choose a sir with a sense of humor 😃

[magic sparrow]

I have doubts about this…

[fathom walrus]

@tranquil pine

we're currently asking our sirs

please wait patiently

[tranquil pine]

😔

[fathom walrus]

this is why you should never hire a consultant

[tranquil pine]

lesson learned

[magic sparrow]

That is what my sir would tell me

[midnight crest]

I shall become your disciple!

[fathom walrus]

lol

[magic sparrow]

Amazing we shall embark on our first problem together in #discussion shortly

[fathom walrus]

have u progressed with the problem @tranquil pine

[midnight crest]

but basically we all agree we got no idea how to solve it

[tranquil pine]

no, im trying to figure out what the formula would be

[fathom walrus]

u have area of square 1

u have the ratio

its a geometric series

just go from there

[autumn geode]

^^

[fathom walrus]

$\frac {$area of first square}${1-r}$

[soft zealot]

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

[magic sparrow]

Bro

\text{}

[fathom walrus]

idk advanced latex

we should ask sir snow

or sir mehdi

[magic sparrow]

@final saddle

[fathom walrus]

or maybe sir modmail

[final saddle]

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[opaque rune]

Can someone please help me with this

[magic sparrow]

What did you scratch out

[final saddle]

@opaque rune Has your question been resolved?

[magic sparrow]

What did you blur out?

[opaque rune]

It’s nothing it just says proficient

Now can you please help me

[magic sparrow]

Why did you scratch it out then?

[opaque rune]

Because it’s not used I didn’t want anyone to get confused

@magic sparrow can you please help me with the question

[tiny nova]

I think you might have to work with Derivatives

The thing is since it is a cube, you need to take a derivative of x^3

I think

I dont know why they were talking about a birthday

date

Wait

No

x^3 is volume

Not surface area

6x^2 is surface area

So take derivative of that I believe

Then plug in 24

So when the surface area is 24, the derivative of that when x is 24 would be the rate of change at that instant

[final saddle]

@opaque rune Has your question been resolved?

[opaque rune]

@tiny nova

You just use a random birthday

[tiny nova]

What do you mean

Oh

So you take your own birthday

Right?

[opaque rune]

Yeah just use any

I think it says just use the day number

Just use 20

So 2-0 = 2

@tiny nova

[tiny nova]

Oh

Okay

So what date should we pick

Like 17th?

that would be negative 8

So they have to add up to 24?

[opaque rune]

Can we just pick 2

@tiny nova

What do I do with it though, where do I plug it in

[tiny nova]

I really dont get what the birthday is supposed to mean

[opaque rune]

Just use a number

[tiny nova]

Oh

Okay

[opaque rune]

Let’s use 2-0 = 2

[tiny nova]

Okay

So do we plug in 2?

I dont think so

[opaque rune]

What formulas do we use

[tiny nova]

Because that would tell us what the rate of change is when the time is 2

I think we have to figure out at what instance(date) the rate of change is 24

so plug in number into the derivative of 6x^2

the derivative of 6x^2 is 12x

Wait

to find when it is equal to 24

We can set it equal to 24

So its 12x=24

Then just solve for x

[opaque rune]

2

[tiny nova]

Yeah

You were right

Now we have to find the derivative of x^3

since at the end they ask for the rate of change at that instance of the volume

so to do rate of change take the derivative

of x^3

[opaque rune]

Where did you get 6x^2 from before

[tiny nova]

Thats is the formula for surface area, since there are 6 square sides, and the formula for area is x^2, and there are 6 of them

Hence 6x^2

But then we were using the surface area to find the correct time that we needed

since we could set it equal to 24 and find the right time

Right now we are applying that instance to the volume of the cube

To find the correct rate of change

So to get the rate of change in terms of x we need to take the derivative

Then we can plug in the 2 we got before

So, calculate the derivative of x^3

[opaque rune]

3x^2

[tiny nova]

Nice

So plug in the 2 we got before into that

[opaque rune]

12

[tiny nova]

So I think, I might need to double check but that is the correct answer

[opaque rune]

What are the variables for this

[tiny nova]

Yeah I think it is right

[opaque rune]

Is it dA/dt or something

[tiny nova]

d/dx

wait

Actually

Let me think

We are only working with differientiating in terms of the x value, which is the time, so it could be d/dx or d/dt, depending on your preference

[opaque rune]

So d/dx= 12

[tiny nova]

Well no, it would be,

d/dx (x^3) = 3x^2, when x=2 it equals 12

Wait

They ask for the answer in cubic meters be hour

So its, 12 cubic meters/per hour

Just in case

[opaque rune]

Like this?

The bottom of it says d/dx = 12 cubic meters/ per hr

@tiny nova

[tiny nova]

I don't think you can use d/dx like that

d/dx = 12, is like saying sin() = 12

you have to have somehting next to it

[opaque rune]

So it just = 12

[tiny nova]

Well yeah

[opaque rune]

So what do I write with it

[tiny nova]

Or if you want the d/dx you have to write, d/dx (x^3) = 12 {x=2}

The curly brackets say that x=2

not multiplication

The paranthesis around the x^3, is there for clarity purposes but usually its not written

[opaque rune]

Ok so the answer is 12

[tiny nova]

I believe so

[opaque rune]

Alright thanks

[tiny nova]

Also, in the last step something is wrong

[opaque rune]

What

[tiny nova]

You wroter d/dx of (2)^3 = 3(2)^2, but since you replaced x with 2, you need to write,

d/d2 (2^3)

Not d/dx

[opaque rune]

Ok thanks

[tiny nova]

Yup

Good luck

[final saddle]

@opaque rune Has your question been resolved?

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[tranquil pine]

how solve

[warm python]

!status

[final saddle]

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

[tranquil pine]

1

[warm python]

what do you get if you add the two matrices on the left

[tranquil pine]

what is a matrices

which numebrs

[warm python]

,w define matrix

[soft zealot]

Results provided by WolframAlpha

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

Requested by physicsrocks

[warm python]

see the first defn

[barren hound]

matrices is the plural of matrix. do you know what a matrix is?

[tranquil pine]

i'm not good with definitions

just tell me the numbers that are matrix or whatever

[vital crag]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:matrices/x9e81a4f98389efdf:mat-intro/v/introduction-to-the-matrix

[tranquil pine]

ok

so how do i solve this

what's the first thing i have to do

[barren hound]

you have to watch that video Riemann posted

[tranquil pine]

ok

ok do matrix are numbers in a box that are columns and rows

so

what's the first step to solving

uhh

where did y'all go

[barren hound]

sorry

uh did that video not talk about adding matrices?

[tranquil pine]

no

[barren hound]

maybe you'll need to look at some of the continuing ones

i don't really want to recapitulate your entire class, which would have covered this

[tranquil pine]

no i'm studying for the act

which is tomorrow

i never learnt this before

[barren hound]

sigh ok

matrix addition you just add corresponding entries

so like add top left + top left to get top left

[tranquil pine]

so the first y is 4

or do i gotta do something with the x

[barren hound]

not sure where you got 4

[tranquil pine]

3+(1)x

[barren hound]

err

yes y = 3 + x

[tranquil pine]

ok

x+3

[barren hound]

x+3 = y, yes

what else can you ascertain?

[tranquil pine]

what does that word mean

[barren hound]

figure out

[tranquil pine]

the other y is

x/2+5

[barren hound]

the other y?

they're both y

[tranquil pine]

true

[barren hound]

but yes, you also have y = x/2 + 5

now can you use that to solve for x and y?

[tranquil pine]

uhh

maybe

ok so i have no idea what that means

[barren hound]

y = x + 3
y = x/2 + 5

so since both of those things equal y, they equal each other, so
x + 3 = x/2 + 5

[tranquil pine]

makes sense

[barren hound]

is there something you can do with this equation?

[tranquil pine]

you can maybe simplify it a bit possibly

[barren hound]

you can do more than that

[tranquil pine]

uhh

[barren hound]

it's an equation with one variable (x)

[tranquil pine]

yea idk

[barren hound]

can you solve it for x?

[tranquil pine]

idk what that means

[barren hound]

then in terms of reviewing for the act you aren't at all qualified for the matrix stuff and should be reviewing algebra 1 concepts

it means manipulating the equation to the point that your have something like x = 382

[tranquil pine]

i see

is it like

[barren hound]

have you taken an algebra class?

[tranquil pine]

2=4

wait

[tranquil pine]

yes

i'm taking math honors

pre calc next year

[barren hound]

lungo if you're unable to go from 2x + 3 = 13 to x = 5 then you're going to have a really awful time in pre calc

[tranquil pine]

i just need a good teacher

i'll be good

[tranquil pine]

wait i thought it was x+3=x/2+5

[barren hound]

i made up a different, slightly easier equation just to demonstrate

[tranquil pine]

i got

2x+4=x

but idk what to do after this

if i divide x it'll be gone

so x won't equal anything

[barren hound]

no, baby, it won't; but you didn't perform the algebra correctly and that's not what you would do in that situation anyway

i see you multiplied both sides by 2 and tried to subtract 6, but you seem to have gotten confused at some point

I'm telling you with love that you will learn nothing in pre calc if you don't have a rock solid foundation in algebra

[tranquil pine]

ok but

i subtract 3 first

so only x is on the left side

wait no

i subtract 5

[barren hound]

okay, either of those will work.

[tranquil pine]

ok

imma subtract 3 then

then i uhh

[barren hound]

write out the current state. don't try to hold it all in your head

[tranquil pine]

what do i do with x=x/2+2

that's the main thing that's confusing me

[barren hound]

you have a few options

it seems like last time you multiplied both sides by 2

and that would work

[tranquil pine]

2x=x+4?

what do i do now

the two x's are messing me up

you can't cancel them because x won't equal anything if it's gone

it'll just be numbers

[barren hound]

try subtracting x instead

but hang on

if you were to divide both sides by x

the x wouldn't be entirely gone, since you'd have 4/x

it's not a fruitful path. but x wouldn't entirely go away.

[tranquil pine]

if i subtract x what would happen

[barren hound]

what would happen?

[tranquil pine]

2=4?

that doesn't make sense

[barren hound]

what?

i don't understand how you got that

[tranquil pine]

ok so

you subtract x from the right

then you subtract it from the left

the x on the 2 cancels

[barren hound]

... no it doesn't. lungo you desperately need to review and drill algebra and possibly pre algebra

[tranquil pine]

it doesn't cancel? so what does it do

[barren hound]

you have 2x - x

you can think about this as having two apples and taking away one

what are you left with?

[tranquil pine]

1

[barren hound]

like the abstract concept of 1?

[tranquil pine]

idk what that means

[barren hound]

uh

if you had two xylophones (here represented by the letter x) and you lost one of them, what would you have?

[tranquil pine]

so it's 4

i see it now

[barren hound]

what's 4?

[tranquil pine]

x=4

so if x equals 4

then what

[barren hound]

sigh then I think at some point we were trying to find x and y

what do we know about y?

[tranquil pine]

y is

x+3 and x/2+5

[barren hound]

ok

can you use what you just learned?

[tranquil pine]

to what

the only think i learned was to add the things and that's what y is and to solve for x

if x is equal to 4 does that mean the coordinate for x is 4

but how do you get 7

solve for y

do i just put a y= next to the things

no

that's dumb

nvm

solve for y to get y

how to i solve for y

what equation to solve

but we already know y

y=x+3 and x/2+5

[barren hound]

if you have x=4

then whenever you see x you can replace it with 4

[tranquil pine]

oh i remember

7

ok

that was easy

[barren hound]

what about 7

that was not easy for you.

[tranquil pine]

it's easy now

[barren hound]

please, listen to me when I say that you will have a significantly better time if you do not go into pre calc next year, unless you do a lot of review and practice of these things

[tranquil pine]

probably

wish i didn't spend so much time just solving that one problem

i have a whole act to study for

[barren hound]

eg you should do one of these worksheets a day, or one from an earlier bank
https://math-drills.com/algebra/alg_linear_axplusbeqcvar_001.php

[tranquil pine]

i just needed to be reminded

[barren hound]

you needed the entire problem walked through.

[tranquil pine]

i did

it's a new concept

ight i gtg bro

i shouldn't have spent so much time here bro

.close

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[uneven light]

idk how to start

[rain compass]

you should try literally anything

[uneven light]

UGHHHHH

[rain compass]

it worked last time ;)

[uneven light]

u substitution dint work

what now

[rain compass]

what u sub did you try

[uneven light]

2x + 1

[rain compass]

why didnt that work

(it should)

[uneven light]

there is a leftover x

[rain compass]

i beg to differ

[uneven light]

wut

[rain compass]

try taking the derivative of 2x + 1 again

[uneven light]

Oh

maybe i am a little slow

[rain compass]

no, but you seem determined to not try anything when you see an integral

even though you clearly have good intuition for what to do

[uneven light]

i am intimidated by these integrals because it's a new chapter and i was expecting to have to do another method to do the integral

😭

[rain compass]

yeah, i was suggesting you be more confident in your abilities

the worst thing you can do is not try

[uneven light]

that is true

did i do everything right

[rain compass]

2tan^-1(u)

but i think it is fine otherwise

[uneven light]

okok

[uneven light]

whats inverse tan of 1

[rain compass]

unit circle

[uneven light]

is inverse tan just the same as cot

idk

[rain compass]

nah

tan^-1(1) is an angle such that
tan(theta) = 1

so find a theta with a tan of 1
specifically between -pi/2 and pi/2

[uneven light]

ohhh

gotcha

yippie

thank you for the help

.close

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[weak ridge]

I’m confused on this

[final saddle]

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

[unreal anvil]

do you recall what is "standard form of quadratic function"?

[weak ridge]

Yeah finding the quadratic function

Then turning it into a standard form

It’s usually given

Wait nvm I figured it out

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[dusky kiln]

.close

.reopen

i have a question

[dusky kiln]

i'm learning quadratic functions and graphing them

so i understand how to graph them and everything but now i'm getting confused with vertex form and intercept form

can't i just convert both to standard and solve?

[rain compass]

you can

[dusky kiln]

and then i can solve for the roots with the quadratic formula?

and find the vertex form too

thank you for your time have a nice day

.close

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[jagged flare]

If $$A=\frac{1+2}{3}+\frac{1+2+3}{9}+\frac{1+2+3+4}{27}+.....$$ then what is $8A$

[soft zealot]

Skill_Issue

[jagged flare]

idk where to start

[nocturne spire]

Start symplifying

[jagged flare]

how

[nocturne spire]

A= 1+6/9+10/27

[jagged flare]

but theres infinite og them

[nocturne spire]

A=2

If rounded to the nearest tenth

So 8A would be 16

16 final answer

[jagged flare]

dawg you cant do that 😭

[muted wraith]

LMAO

[jagged flare]

i need a precise answer

[inland kettle]

@jagged flare tried summations??

try setting one up

[jagged flare]

ifk how to, ans whats summations?

[inland kettle]

notice that the top is just 0.5(n^2 + n), for the last number

[jagged flare]

ok

then what

[vital crag]

find the pattern for denominator since ren gave you the numerator

[jagged flare]

3^(n-1)

wait

since iys asking for 8a, whst if i multiply by 9 and subract a

[vital crag]

yea try that

[jagged flare]

3(1+2)+(1+2+3)+(1+2+3+4)/3+...
(1+2)/3+(1+2+3)/9+(1+2+3+4)/9

if i subtract its
3(1+2)+(1+2+3)+(3+4)/3+(4+5)/9+(5+6)/27

eh i dont see a continuation

ok what if $$\frac{3}{2}(\sum^{\infty}_{n=2}\frac{n^2+n}{3^n})$$, would there be any merit to that

whats the infinity symbol?

[rain compass]

\infty

[inland kettle]

$\frac{1}{6}\sum_{n=2}^{\infty}\frac{n^{2}+n}{3^{n}}$ isn't this the summation

[soft zealot]

ren

[jagged flare]

dunno

[jagged flare]

its this right for the bottom

[inland kettle]

well, the series starts from 0.5(n^2 + n)/3^(n-1) for n =2

ytes

[inland kettle]

then this turns into what i have above

[jagged flare]

then if we multiply the denominator and numerator by 3 you get 3/2(n^2+n)/3^n

[inland kettle]

hang on

um ok i see

[soft zealot]

Skill_Issue

[inland kettle]

but that can't be right, since our summations are different.

[jagged flare]

why?

[inland kettle]

because the coefficients aren't the same...?

[jagged flare]

oh yeah

[inland kettle]

OH MY BAD

SORRY

I WAS WRONG

ur summation is right

i accidentally used n+1 instead of n-1, ur good

[jagged flare]

ok, but how do we continue from here?

[soft zealot]

ren

[inland kettle]

thinking

continues to think

ummm

this is a weird one, to be honest.

i haven't really gotten to the point of evaluating summations like this

just a question, what course/grade this

[jagged flare]

olympiad grade 8

[inland kettle]

what?

no way

then there's gotta b smth simpler/

1+1/6+10/27+5/27+7/81...

[jagged flare]

probably, olympiad questiins normally have simple answers that needs creativity

[inland kettle]

grade 8?

[jagged flare]

i mean my friends from grade 8 got it in our olympiad class

[inland kettle]

i feel 16 is correct

[jagged flare]

and i was normally in their class aswell but i wasnt at their class at that time

[inland kettle]

well

[jagged flare]

but how do you get it?

[inland kettle]

i evaluated it using desmos

and i knwo the answr

it's not 2

[inland kettle]

intuition lmao

[vital crag]

do you know calculus

[inland kettle]

@vital crag apparently this is a grade 8 olympiad question

[jagged flare]

only the very basics

[inland kettle]

;-;

im dying rn
my shame knows no bounds

i can't solve a freaking grade 8 question

[vital crag]

it's close to the second derivative of the function 1/(1-x) evaluated at x=3.

with some shift and multiplicative constants

[jagged flare]

wha, what is a second derivative

[inland kettle]

LMAOOOOO

[vital crag]

yea nevermind then

[jagged flare]

i kinda only know what a regular derivative is

[inland kettle]

riemann, what do you think we should do

without calculus

[jagged flare]

i mean i can maybe ask the teacher if youd like

[inland kettle]

second derivative is just the derivative of the derivative, or how fast the derivative changes

[jagged flare]

oh

[inland kettle]

i mean if u want sure, but this is tricky. i wanna do it myself

it's a good challenge

thinks

geometric implications, maybe?

riemann continues to type

[vital crag]

you can handwave and re-arrange the sum to group all the terms with 1 in the numerator together to get one geometric series of r^n with r=3. and then similarly for 2/3 + 2/9 + ... to get twice that.
3/9 + 3/27 + ... is another geometric series

[inland kettle]

MY GOD

RIEMANN ARE YOU A WIZARD

HOW GOATED IS THIS MAN

[jagged flare]

what

[inland kettle]

,w expansion of 1.5 sum from n=2 to 9 for ((n^2+n)/3^n)

[soft zealot]

Results provided by WolframAlpha

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

Requested by justarenegade

[inland kettle]

omfg

aaaaaaaaaa

[jagged flare]

ok the teacher said "there is 2 steps until you can get an infinite geometric summation"

[inland kettle]

ok look i gtg smwh pretty soon

[jagged flare]

"multiply both sides by 1/3 then subtract, then do it again"

[inland kettle]

can we try to wrap this up

[inland kettle]

wat?

hang on sorry i gtg

DM me ok

[jagged flare]

$$A=\frac{1+2}{3}+\frac{1+2+3}{9}+\frac{1+2+3+4}{27}+.....$$
$$A/3=\frac{1+2}{9}+\frac{1+2+3}{27}+\frac{1+2+3+4}{81}+.....$$
$$2A/3=\frac{3}{3}+\frac{3}{9}+\frac{4}{27}+.....$$

[soft zealot]

Skill_Issue

[inland kettle]

@jagged flare i gtg
DM me

[jagged flare]

$$A=\frac{1+2}{3}+\frac{1+2+3}{9}+\frac{1+2+3+4}{27}+.....$$
$$\frac{A}{3}=\frac{1+2}{9}+\frac{1+2+3}{27}+\frac{1+2+3+4}{81}+.....$$
$$\frac{2A}{3}=\frac{3}{3}+\frac{3}{9}+\frac{4}{27}+.....$$
$$\frac{2A}{9}=\frac{3}{9}+\frac{3}{27}+\frac{4}{81}+....$$
$$\frac{4A}{9}=\frac{3}{3}+\frac{1}{27}+\frac{1}{81}+....$$
$$\frac{4A}{9}=1+\frac{\frac{1}{27}}{\frac{2}{3}}$$
$$\frac{4A}{9}=1+\frac{2}{81}=\frac{83}{81}$$
$$8A=\frac{166}{9}$$

[soft zealot]

Skill_Issue

[nocturne spire]

Idk what that means

[jagged flare]

close

[final saddle]

@jagged flare Has your question been resolved?

[jagged flare]

.close

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[wanton yarrow]

I've been working on this proof using PMI, and I'm just a bit stuck on the process

[signal vector]

Use the fact that 3>1

And 3>2

[wanton yarrow]

I'm confused

How can I just compare the numbers

[soft zealot]

smidgin

[wanton yarrow]

How did you get 1 + 2^n+1 + 2 + 2^n

[signal vector]

Expand the brackets, and write 3=2+1

3+2^(n+1)+2^(n)=1+2+2^(n+1)+2^n

[trail mango]

easier way to look at it might just be 3 + 3*2^n > 1 + 2*2^n because 3 > 1 and 3 > 2

[signal vector]

I said that, but she wasn't happy with it

[trail mango]

[wanton yarrow]

wdym wasn't happy i was just confused

[signal vector]

If this helps:
3>2 => 3•2^n>2^(n+1) [multiplying both sides by 2^n] add this inequality with 3>1 to get 3+3•2^(n+1)>1+2^(n+1)

[wanton yarrow]

.close

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[steel kite]

okay

how would I go about solving this

give me hints

I wanna try figure it out

[abstract summit]

$\frac{d y}{dx}=x^2\text{ and }y(1)=-2$

[soft zealot]

nico

[abstract summit]

did you reach calculus yet?

[steel kite]

uh

no

we havent done calculus yet

so uh that formula you just showed me

is really unfamiliar with me

[neon slate]

may wanna double check that derivative anyway 😄

[abstract summit]

it's 3am for me 😄

i'm tired ok...

[steel kite]

yeah thats okay ofc

but uh

how would I go about solving this

[honest ruin]

pixel didnt we just do this lmao

[neon slate]

so without calculus

[steel kite]

this was really easy

this is like

[honest ruin]

its literally the question we just did

[steel kite]

really different

idk

[honest ruin]

but its worded differently

[steel kite]

seems different to me

[neon slate]

how do you think of a tangent line? it's a line that touches the graph at just one point right?

[steel kite]

yeah

[honest ruin]

youre trying to do the same thing here

[steel kite]

the other questions, I had two equations that were equal to each other

[honest ruin]

firstly figure out the equation for the straight line

[steel kite]

I just substituted then used discriminant

[honest ruin]

you know straight means y= x

so figure that eq out first

it gives you the points

[neon slate]

so the equation of the line, knowing the point it goes through but not the slope is:
y + 2 = m(x-1)

[honest ruin]

^

soosh got u

[neon slate]

we'll write in terms of y: y = m(x-1) - 2
and we have the equation of the parabola: y = x^2

[steel kite]

okay

wait

slow down a bit

[neon slate]

we have the equation of the line, without knowing m and the equation of the parabola thats all so far

you're familiar with point-slope form for a line right? then the line one should be clear

[steel kite]

oh thats what you did

i haven't used that in ages 💀

yeah thats what I was confused about

alright

[neon slate]

well line stuff comes up in all math so keep that handy in your toolbox of tricks

[honest ruin]

always remember that algebra builds on itself 💀

alg2 is just alg1 but building on it

[steel kite]

y = m(x-1) + 2

where are we going with this now

[neon slate]

we have both equations in terms of y, now just set the y's equal to each other, we know the tangent touches at one point, so for some x both the y's and x's will match up, i.e.:
m(x-1) - 2 = x^2

[steel kite]

oh

right

$x^2 + m(x-1) - 2 = 0$

[soft zealot]

pixel

[steel kite]

wait let me write down what we have so far in my book

[neon slate]

that's...not right you didn't take care of the negatives correctly

$x^2-m(x-1)+2=0$

[soft zealot]

Soosh

[steel kite]

oooh right

[neon slate]

so we have x^2-mx + (m+2) = 0 aka a quadratic equation

now we need to think a bit about the geometry of a tangent line, it touches a parabola in only one point right

[steel kite]

yes

[neon slate]

(the bottom line)

[steel kite]

thats right

[neon slate]

but we can have a generic line that touches two points like the top line, that's not a tangent

[steel kite]

mhmm

[neon slate]

so the sweet spot since m can vary is when that quadratic equation has only one solution

[steel kite]

yep

so how would we only find one solution?

[neon slate]

so looking at our good friend the quadratic formula, remember that we get the usual two solutions from when we do + for one and - the other

[steel kite]

yep

[neon slate]

but...if that term after the +/i is ZERO then there's only one solution

so that's our sweet spot that we need, basically we need b^2 - 4ac = 0

[steel kite]

so when discriminant (b^2 -4ac) is equal to 0, theres one solution

so we use the discriminant

now I plug in numbers?

[neon slate]

x^2-mx + (m+2) = 0
a = 1
so b = -m
c = m + 2

[steel kite]

yep

$m^2-4\cdot 1 \cdot (m + 2)$

[soft zealot]

pixel

[steel kite]

$m^2 -4m + 8 = 0$

quadratic

right?

[neon slate]

you need to

[soft zealot]

pixel

[neon slate]

pay attention to negative signs more

[steel kite]

oh

wait hold on let me try find my mistake

only negative is b = -m

(-m)^2

is just m^2

no?

[neon slate]

-4(2) = ?

[steel kite]

-8

[neon slate]

right so - 8

instead of +8

[steel kite]

is the negative not included as value of 'b'?

ok so say b = -4

it would be (-4)^2

b = -m

would be (-m)^2

am i missing something here

OH

💀 omg

i was arguing about the WRONG THING

💀

i apologise

[neon slate]

you were at this step ^ then youre just doing a simple distribution: -4(m+2)

[steel kite]

$m^2-4m-8=0$

[soft zealot]

pixel

[neon slate]

yeah

[steel kite]

bro im so sorry 😭

thank you for being patient

[neon slate]

so new quadratic in m, doesn't seem to factor i think so quad formula again 🙂

this time need to actually solve for m

[steel kite]

ok one sec

ima write down what we just did

and then solve for m

$\frac{4\pm\sqrt{48}}{2}$

[soft zealot]

pixel

[steel kite]

thats what I reached

when solving for m

from equation: $m^2-4m-8=0$

[soft zealot]

pixel

[neon slate]

ya, you can simplfy that to $2\pm2\sqrt{3}$

[soft zealot]

Soosh

[steel kite]

alright

now what

sub back into $y=m(x-1)-2$?

let me reread question

[neon slate]

MINUS 2 😠

these minus signs are going to be hurt you on tests if you don't pay attention 😅
y: y = m(x-1) - 2

[steel kite]

so we find equation of every straight line that passes through (1, -2) and is tangent to y=x^2

[soft zealot]

pixel

[steel kite]

😭 lord

yes yes

my bad

bro i wrote it wrong on my book too

💀

i was reading off my book

[neon slate]

yeah now we have: $y=(2\pm\sqrt{3})(x-1)-2$

[soft zealot]

Soosh

[steel kite]

alr

do we uh

expand

[neon slate]

makes sense that thered be two tangents since the parabola is symmetric

[steel kite]

yeah

[neon slate]

so 1 tangent on each side meet at that whatever point

now thats the answer, graph the lines to double check your work

along w the parabola and confirm they seem tangent

[steel kite]

yep graphed

confirmed they are tangent

I was confused at first, I was like wheres the second equation

but i forgot it was a quadratic

so

$\pm$

[soft zealot]

pixel

[steel kite]

$y=(2\pm 2\sqrt{3})(x-1)-2$

[soft zealot]

pixel

[steel kite]

u missed a 2 before

sqrt

[neon slate]

ya

[steel kite]

thank you

i got 2 more questions here but i think i can answer them on my own

ill post question and answer in a bit to check if im correct

so ill keep channel open

C = -13

mhmmmm

$-2x^2-6x+2=mx+6 \ -2x^2-6x-mx-4 = 0 \ -2x^2-x(6+m)-4=0 \ \Delta b^2-4ac = 0 \ (m+6)^2-4\cdot -2 \cdot -4 = 0 \ m^2+12m+36-32 = 0 \ m^2+12m+4 = 0 \ \ $then use quadration formula: $\ \frac{-12\pm\sqrt{128}}{2} \ -6\pm 4\sqrt{2} = 0$

[soft zealot]

pixel

[steel kite]

what now

😭

idk if ive done it correctly

.close

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[floral shadow]

Idk how to do 13 c)

[final saddle]

@floral shadow Has your question been resolved?

[final saddle]

@floral shadow Has your question been resolved?

[craggy plank]

Set ax^2+bx+c=0 and plug the given points into the set function to solve a b and c

[floral shadow]

so like a simultanous equation?

[floral shadow]

then i solve?

[craggy plank]

yeah

[floral shadow]

ight

lemme try

[craggy plank]

Ping me when you’re done

[floral shadow]

i got a=3 b=2 c=-4

i think its wrong tho

@craggy plank

cus it doesnt wrok when i sub it in

oh wait i got it

.close

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[craggy plank]

Sorry, I was solving questions for others XD

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[fleet zinc]

im new to limits and im not sure if i can solve this limit like this:

[void valley]

What have you tried

[fleet zinc]

and say that its infinity

[void valley]

You can split the limits like that if both of them exist

[fleet zinc]

so its ok to do this in this case?

[void valley]

In this case it doesn't seem so

[fleet zinc]

ah

well my attempts resulted in this and in getting 0*infinity

so im not sure

[void valley]

Yes

You can solve it simply "plugging" inf in

[fleet zinc]

wait but isnt the parentheses equal to 0?

[green ibex]

yea its not an indeterminate form so you can solve by 'plugging in'

[fleet zinc]

oh wait im dumb

it was supposed to be x *( that sqrt - 1)

let me send the right pic, sorry

[void valley]

That explains a lot huh

[fleet zinc]

yep sorry, my bad

[void valley]

Okay, maybe try multiplying by conjugate?

Then simplify

[fleet zinc]

oh right i havent thought about that

yeah i get 1/2 now which is most likely right

thanks!

.close

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[hollow cliff]

hello, do my answers and solutions make sense?

im looking for the maximum and minimum the product of 2 numbers that sum up to 56. (x, y ∈ Z+)

[hollow cliff]

first pic is solving for max and the remaining 3 is for minimum

[west berry]

Seems right

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[desert dagger]

Precalc “Is there one triangle, two triangle or no triangle?”

[desert dagger]

,rotate

[desert dagger]

55 degrees ^

I just took a test and I wanna know if I got it right

[vapid knot]

one triangle i believe

[desert dagger]

BROO

[vapid knot]

what do u think

[desert dagger]

that’s what I had at first but he kept talking about no triangles the past few days

[inland kettle]

hmm

[desert dagger]

should’ve gone with my logic instincts

[inland kettle]

use cosine rule

[vapid knot]

yup

[inland kettle]

$c = \sqrt{a^2+b^2-2ab\cos{55}}$

[soft zealot]

ren

[inland kettle]

,w cos 55 deg

[soft zealot]

Results provided by WolframAlpha

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Requested by justarenegade

[inland kettle]

,w sqrt (10^2 + 12^2 - 240cos55deg)

[desert dagger]

ahh ok

[soft zealot]

Results provided by WolframAlpha

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Requested by justarenegade

[inland kettle]

yes triangle

[desert dagger]

.cloze

[inland kettle]

@desert dagger

.close not .cloze

xD

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[tranquil pine]

Best I got is,

2sinxcosx = sin2x

but i dont know how go go from there lol

sin2x = sinP + sin Q = sin(P+Q)

2x = P+q x = P+Q /2

well

sin (P+Q)

well either way

no

[dapper jolt]

try thinking about the ways in which you can get the right hand side form

like which existing trignometric addition/subtraction formulas give that form of 2sin and cos something

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[delicate ivy]

can someone precisely state what this highlighted notation means

[junior obsidian]

https://www.youtube.com/watch?v=TQjEhCGG0yQ&t=11s

[delicate ivy]

<@&286206848099549185>

[scarlet sequoia]

limit of (f(a+tv)-f(a))/t when t-> 0 i presume

it's the directional derivative of f at a, with direction v

[delicate ivy]

v is any arbitrary vector, is a just the point?

[scarlet sequoia]

that's the gist

[delicate ivy]

aight, thanks again

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[chrome hemlock]

excuse me

[chrome hemlock]

how do i do this

idk where to start

question A

[vapid knot]

for this u can make 6 spaces

now make 2 cases

case 1: last letter is a
case 2: last letter is i

now u can use the multiplication principle to proceed

[chrome hemlock]

so XXXXXA

and XXXXXI

[vapid knot]

yup

[chrome hemlock]

but it says begin and end?

[vapid knot]

oh sorry i read that wrong mb

so IXXXXA and AXXXXI

[chrome hemlock]

AXXXXI and IXXXXA

[vapid knot]

yup

[chrome hemlock]

uh so

how do i use the multiplcation principle?

[vapid knot]

oh so basically here assume the 1st X is one of the letters of the word DARWIN except E and A

so we have 4 choices

now for the 2nd X we have 3 choices as the 1st X has been fixed

[chrome hemlock]

wait i dont understand the multiplication principle

[vapid knot]

so we proceed till we have 4!

[chrome hemlock]

can we pls use an easier question like this one first

[vapid knot]

sure

[vapid knot]

okay so to go from Sydney to Adelaide there are 6 possible routes

[chrome hemlock]

oh ya

[vapid knot]

similarly 5 to go from Adelaide to Perth

[chrome hemlock]

6x5?

or is that wrong

[vapid knot]

so assuming I fix 1 route from Sydney to Adelaide, how many choices do i have to go from Adelaide to Perth? 5 right? Now there are 6 initial routes I can chose
So total ways = 5 + 5 + 5 + 5 + 5+ 5 = 6x5 = 30

[vapid knot]

yup

[chrome hemlock]

like my instinct is to just multiply them

but i dont understand why

[vapid knot]

does this provide a helpful intuition?

[chrome hemlock]

hm

[vapid knot]

if not i can explain it another way

[chrome hemlock]

yes please :D

sry if im kinda slow its 12:30 am

[vapid knot]

alright gimme a minute

[chrome hemlock]

spent the whole day studying physics forgot about marh

[vapid knot]

it's fine man
i spent hours trying to understand this when i started learning this

https://math.stackexchange.com/questions/4660011/is-there-a-relation-between-principle-of-addition-principle-of-multiplication-a/4664671#4664671

check out the answer to this

[heady delta]

Wrong channel

[vapid knot]

it might help

[chrome hemlock]

really? that makes me feel better lol

i thought i was really slow

[vapid knot]

nah math can be hard sometimes xD
take ur time trying to figure stuff out

[chrome hemlock]

ohh we start seeing these diagrams later in the textbook too

[nova iron]

hi, i can try to explain if you wish

if u still don't understand

[chrome hemlock]

sure have a go :D

[nova iron]

Okay so these are the three places: S = Sydney, A = Adelaide, and P = Perth

From the question you have four flights and two trains from sydney to adelaide so simplifying this: 6 ways to get from sydney to adelaide

because we don't really care if he travels via flight/train

[chrome hemlock]

yes

[nova iron]

I added 6 pathways

now in a similar manner, we have two flights and three trains from adelaide to perth so basically 5 modes of transportation from adelaide to perth

good so far?

[chrome hemlock]

yes

[nova iron]

now let's say i take the very first pathway from sydney to adelaide

i have 5 pathways to choose from to go from adelaide to perth right?

[chrome hemlock]

yes

[nova iron]

do you see where we're getting at?

so each of those one blue pathways = 5 choices

there are 6 blue pathways = 6 * 5 choices

[chrome hemlock]

oh yes

[nova iron]

and then so on yeah

[chrome hemlock]

ohhh i see

ok thank u :D

6 lines from syd to adl 5 from adl to perth to get all ways 6 x 5

makes sense

thx

[nova iron]

sure np

[chrome hemlock]

.close

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[quasi plinth]

,tex Find all the eigenvalues and eigenvectors of ( A = \begin{pmatrix} 1 & 3 \ 1 & -1 \end{pmatrix} ). If ( f: \mathbb{R}^2 \to \mathbb{R}^2 ) is the linear transformation defined by ( f(\mathbf{x}) = A\mathbf{x} ), decide if there exists a basis ( B ) of ( \mathbb{R}^2 ) such that ( M_B(f) ) is diagonal.

[soft zealot]

renato

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[tiny gorge]

i guess you're supposed to assume that the value increases at a constant rate annually since it was produced?

[soft zealot]

nox💫

[tiny gorge]

the value of the coin in 1909 would surely have been 0.01, no?

it is after all a penny

[soft zealot]

nox💫

[tiny gorge]

i think the rate itself would be constant (unless stated otherwise)

i would set it up something like:
$0.01(1+r)^n = 3200$

[soft zealot]

Bungo

[tiny gorge]

where $n$ is the number of years elapsed from 1909 to 2015

[soft zealot]

Bungo