help-49

[olive herald]

how does the vertex

Get us to that

Oh

Nvm

[lavish venture]

the vertex form is just a form of writing the parabola

[olive herald]

yep

Realized

that

alright

Now I got it

this is just the parabole formula alr

Ah ok

I got it now

[lavish venture]

so once we have the equation of the parabola we can have the derivative

[olive herald]

yep and the angle was that of the derivative

Now we just make any triangle for the trigonometric functions

And calculate the angle

[lavish venture]

so consider the tangent line at the point near (73 or whatever,0)

we can use tan(alpha) = deltay/delta x

using the tangent line

but deltay/delta x ≈ the derivative

[olive herald]

Yep

[lavish venture]

so tan(alpha) ≈ f’(73ish)

[proper kernel]

Hey knief

[lavish venture]

🐐🐐🐐

[proper kernel]

Wait is this still the bridge problem

[lavish venture]

how’s khan academy

[lavish venture]

yea i left him before

[proper kernel]

I don’t study off khan academy

[lavish venture]

was busy

[lavish venture]

not anymore?

whatre you studying

[proper kernel]

I found the official syllabus from my school district

And they just have a full online course

[lavish venture]

the school does??

[proper kernel]

Ye

I think it was for Covid or something

[lavish venture]

did you talk to your counselor

[proper kernel]

They have their own website

Nah

[lavish venture]

you should

tell em what you want to do

[proper kernel]

I will in a few months

I still want to consider whether or not it’s a good idea

[lavish venture]

true

[proper kernel]

So how’s real analysis

[lavish venture]

what was your plan though

other than just getting ahead in math

[proper kernel]

?

[lavish venture]

meh

[proper kernel]

Wdym

[lavish venture]

like did you want to graduate early

[proper kernel]

Nah

So I really like nuclear physics

[lavish venture]

do you?

[proper kernel]

And I know I need multivariable calculus to study it

[lavish venture]

yep

have you consider doing competition physics?

[proper kernel]

And I don’t want to wait another 4-5 years to study it

[lavish venture]

like Fma

[proper kernel]

They have that?

[lavish venture]

yes

[proper kernel]

I don’t know

I’ll think about it

[lavish venture]

F=ma

physics olympiad

[proper kernel]

Did u do it?

[lavish venture]

there’s a bunch of competitions

[lavish venture]

no

my friend and i were looking into it last year

[proper kernel]

I see

[lavish venture]

we wanted to do i think was like physics brawl or something

but we didn’t have enough people

[proper kernel]

Oh

[lavish venture]

i wish i discovered those things earlier

https://discord.gg/phods

they’d know more

if you want to get into it

[proper kernel]

Thx

[lavish venture]

those people are crazy at physics

[proper kernel]

U play any video games in college?

Or do u not have enough time

[lavish venture]

nah lol i quit video games a long time ago

[proper kernel]

Dang

[lavish venture]

i play clash of clans i guess

but that’s not as time consuming

log on like once a day

[proper kernel]

Wait u go to Cornell right

[lavish venture]

for a few minutes

[lavish venture]

no

[proper kernel]

Oh dang

I was hoping u would know my brother :/

He’s a cs major

[lavish venture]

i quit in 9th grade

oh he goes to cornell

[proper kernel]

Ye

[lavish venture]

i know someone that goes there

[proper kernel]

Nice

[lavish venture]

neuroscience guy

[proper kernel]

Dang that’s cool

How far are u planning to get in math

Like there’s a ton of subjects right

[lavish venture]

i’ll see, if i’m not interested enough in research i’ll just go for a masters in like financial engineering or something

[proper kernel]

Oh okay

[lavish venture]

i might also go for a masters in stats

[proper kernel]

Cool

[lavish venture]

i don’t plan on spending my life in academia

[proper kernel]

Do you plan to do research or what

[lavish venture]

if i did i wouldn’t want to do it for my entire life

[proper kernel]

Hmm makes sense

[olive herald]

so just about number 1. You said value and derivative has to be the same for it to be a tangent but like the exact same function wouldn't be a tangent of itself, no?

[lavish venture]

if a problem interested me then i’d do it

[proper kernel]

Hold up Martie needs help

I’m not gonna take up any more of ur time

Cya later 👋

[lavish venture]

see ya

[lavish venture]

well, a function isn’t tangent to itself

[olive herald]

yeah

So like there has to be a third requirement

[lavish venture]

that the functions aren’t equal i suppose

[midnight plank]

@olive herald Has your question been resolved?

[olive herald]

oh btw

Whats the derivative of

16^-x

I'd assume just 16^-x

[midnight plank]

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[late patio]

Did I do this correctly?

[sly rapids]

Hey Ahmed. Can you explain to me what you did to get the second fraction in step two?

[late patio]

I flip the second fraction because i know a/b divided by c/d is equal to a/b • d/c

basically flipped to reciprocal

[sly rapids]

Ah, makes sense. Let me keep reading.

What is after x != 0?
+/- x/3 - y/2?

[late patio]

Just setting the properties to zero so I state the correct restrictions

[sly rapids]

And what do you say the restrictions are? I'm seeing the text above and four circled non-equalities.

[late patio]

x cannot equal

0, 1y/3, -1y/3, -1y/2

My handwriting does need to improve more for sure

[sly rapids]

It's better than mine.
You got -1y/3 from 3x/3 != -y/3. Can you tell me why it is "3x/3 != -y/3"?

[sly rapids]

But you're not coming from 3x+y=0.

[late patio]

I’m setting it to zero to find the restriction/what causes the function to be undefined

[sly rapids]

What are you setting to zero?

[late patio]

3x+y

[sly rapids]

Where is "3x+y" in your simplified equation?

[sly rapids]

Ah, I see!

[late patio]

Apologies on my part as well ahaha

[sly rapids]

No, I forgot you cancelled it. I think you got the problem right.

[late patio]

Just a question am I only stating restrictions for x or for both x and y

[sly rapids]

Just x should be fine. The restrictions use the same equations but solve for the other variable. x != y/3 -> y != 3x

[late patio]

Thank you, good Sir/Madame(?) for your help.

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[twilit field]

We find that a basis of null(T) has dimension 2 , thus by the rank-nullity theorm, a basis of the image will have dim 2 too. But all basis of dim $2$ span $F^2$, thus T is surjective

[grand pond]

A dense set

[twilit field]

We first prove that assuming that there exists as $S$ such that $ST$ is the identity operator, then $T$ is injective.
\
We know that $ST$ being the identity operator is injective. This means that $S(T(v))$ is injective. This means that if $S(T(v_1)) = S(T(v_2)$ ,Then,$T(v_1)=T(v_1)$.

Right so far?

I feel I'm messing up somewhere

We first prove that assuming that there exists as $S$ such that $ST$ is the identity operator, then $T$ is injective.
\
S(T(v)) being the identity mapping , is injective.$S(T(v))$ by definition is a mapping from $V \to V$, thus when we say it's injective , we're mapping distinct elements form $V$ to elements in $V$. Thus, we can conclude that $T(v_1) = T(v_1) \implies v_1=v_2$ as $V$ is the domain, and $v_1,v_2$ are elements in $V$.
\
\
We now prove if $T$ is injective, then there exists $S$, such that $ST$ is the identity operator.
\
As $T$ is injective $T(v_1) = T(V_2) \implies v_1=v_2$. We now define $T(\alpha_1 e_1 +\alpha_2e_2 + \dots + \alpha_ne_n) = \alpha_1 f_1 + \alpha_2 f_2+ \dots \alpha_n f_n+ \alpha_1f_{n+1}+ \dots \alpha 1 f_m$.
\
We then define $S: W \to V$ as follows $S(\alpha_1f_1+\alpha_2f_2+ \dots +\alpha_nf_n+ + \alpha_1f{n+1}+ \dots \alpha _1 f_m) = \alpha_1e_1+ \dots + \alpha_ne_n$
\
where $e_i$ forms a basis of $V$ and $f_i$ froms a basis of $W$.
\
\
From this it's evident that $T(S(v))=v$
\
There thus does exists $S$, such that $ST$ is the identity operator.
\

[grand pond]

A dense set

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

oops, pinged too early by mistake

sorry

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[crimson abyss]

could you help me understand how i could use synthetic division and the interactions with sqrt of 2 in sythentic divisoin

[midnight plank]

@crimson abyss Has your question been resolved?

[royal grove]

this means that x-sqrt(2) is a factor for that polynomial and dividing synthetically will give you coefficient of lower degree polynomial which you should be able to factorize further

[crimson abyss]

ok simpler then i thought

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[elfin zodiac]

How do u figure these problems out with minimal information?

[elfin zodiac]

15,16

[lavish venture]

15 is just f(x) = 2.2

find where that happens

then find f(x) = -6

and divide that x value by 2

[elfin zodiac]

How is the answer A?

I get D

[lavish venture]

0.6/2

[lavish venture]

look where f(x) = -6

it’s at x = 0.6

divide that answer by 2

since we want 2x = 0.6

[elfin zodiac]

Got it

.close

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[grand canopy]

|z| = 2 and 0 < arg(z) < π/2
Prove that arg(z - 2i) = arg(z)/2 - π/4

[grand canopy]

<@&286206848099549185>

[midnight plank]

@grand canopy Has your question been resolved?

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@grand canopy Has your question been resolved?

[grand canopy]

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[strange ravine]

Just confused where it's stopping

[strange ravine]

5pi/4?

...nevermind I'm stupid I'm finding blue not red

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[craggy merlin]

hey could i get help with this

[craggy merlin]

how does my teacher get this?

the root 36x^2

[smoky walrus]

seems like they made a mistake

[fallen galleon]

6 * 6x² = 36x²

and sqrt(a) * sqrt(b) = sqrt(ab)

[smoky walrus]

it's 4√(6x^2) though

[fallen galleon]

oh yeah true

just missing a 4 there

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@craggy merlin Has your question been resolved?

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[half goblet]

can someone help me turn this into a trig problem? i just need something to calculate for a plane landing miniature im gonna make (not to scale)

[half goblet]

something like this just angles needed

[half goblet]

numbers can be changed

[midnight plank]

@half goblet Has your question been resolved?

[half goblet]

<@&286206848099549185>

[restive wadi]

here

[half goblet]

HI

[restive wadi]

hi

we can start

[half goblet]

ok so basically

okay

ive been trying to figure out since earlier what i can do for making a problem 🥲

[last slate]

Is this the question?

[restive wadi]

ok

i will try

[half goblet]

ah no im asking if someone could help me make a problem

[restive wadi]

we can try

[half goblet]

okay

[last slate]

Thats quite a problematic ask

https://tenor.com/view/ba-dum-tsss-drum-band-gif-7320811

[half goblet]

oh

OOPS

okay i thoguht u were serious 😓

[last slate]

No obviously not

Go ahead

What help do you require

[restive wadi]

well making question is a very creative task

[half goblet]

yea 🥲

[half goblet]

just making my own problem and then solving it too afterwards 😭

[last slate]

You wish to create a question using trignometry right?

[half goblet]

yes

[last slate]

Okay

Which grade if I may ask?

[half goblet]

im in 11th

[last slate]

Okay then

Lemme see if I have something interesting

[restive wadi]

should it be 11th grade question?

[half goblet]

yes

[restive wadi]

or higher or lower

[restive wadi]

ok

[half goblet]

just 11th

i just need something related to plane landing

i dont know if im allowed to send links

but theres this one thing im trying to base off

but i cant understand it

actually ill just send pics from it

basically i need something like this

[last slate]

Okay if a ladder is sliding off from its anchor point at a rate of +x metres per second, at time = t, what rate is the distance between the ladders base point and the wall increasing

The angle between the ladder and the base is theta

Express your answer in terms of t, x and theta

[half goblet]

i dont need anything with time 😓 just angles and related to plane landings

[restive wadi]

i gues i got one

[restive wadi]

pilot plans to descend from a cruising altitude of 32,000 feet to a runway at 5,000 feet. Using the 3:1 rule, the descent should begin 96 nautical miles from the runway. However, if the descent path is obstructed by a mountain peak that rises to 15,000 feet and is located 60 nautical miles from the runway:

Q-1:At what altitude should the plane be when it reaches the mountain peak to ensure a safe descent path that clears the peak while following the 3:1 rule?
Q-2:What would be the revised descent angle to clear the mountain, if the descent begins at the same point?

im sorry if its too hard or too easy

[last slate]

Hmm looks nice

I like this question

[restive wadi]

okay then
whoever gets to solve first gets fake 1 million dollars

[half goblet]

hold on im so sorry guys im just asked to do something so i apologize if i reply later to the question 😭

[restive wadi]

np!

[half goblet]

SO SORRY IM BACK

[half goblet]

im gonna use question 1

but i need something with the angle of descent

😭😭

[deep terrace]

Yoo need help

[half goblet]

helloo

[half goblet]

.

[deep terrace]

Need help

[half goblet]

basically i need to formulate a problem and have a solution for it as well

related about angle of descent of a plane landing

just angle

[deep terrace]

Ok show the Qustion

[half goblet]

theres no question 🥲

[deep terrace]

Oo ok

[half goblet]

i need to make my own question and solution

[deep terrace]

Ooo

Ok good luck

[half goblet]

NEVERMIND IM OKAY

[half goblet]

i figured something out with this

thank you all!!

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[boreal ocean]

$2^{n+1}-2 = 2(2^n-1)$

[grand pond]

cristorenzo99

[twilit field]

We attempt to prove this via induction
\
Base case : $2^{2}-2=2$
\
We now suppose $\sum_{i=1}^{n} 2^ i =2^{n+1}$
\
We wish to prove from this that $\sum_{i=1}^{n+1} 2^ i =2^{n+2}$
\
$\sum_{i=1}^{n+1} 2^ i =2^{n+2}= \sum_{i=1}^{n} 2^{i}+2^{n+1}=. 2\cdot 2^{n+1}=2^{n+2}$
\
\
QED

typos, typos

[grand pond]

A dense set

[gaunt imp]

How did you get to the last sum(from i to n+1) being equal to 2^(n+2)?

[viral dagger]

i belive its supposed to be
$$\sum^{n}_{i=1}2^i=2^{n+1}-2$$

[grand pond]

Skill_Issue

[boreal ocean]

I would have done this for the inductive step:
$\sum_{i=1}^{n+1}2^i =\sum_{i=1}^{n}2^i + 2^{n+1}= 2^{n+1}-2+2^{n+1}=2^{n+2}-2 $

[grand pond]

Skill_Issue

[viral dagger]

why did it not work for you D:

[hard umbra]

its like none of you can see this

[dawn dagger]

Skill issue saw it

[viral dagger]

hey i saw that

[boreal ocean]

Now it should work:
$\sum_{i=1}^{n+1}2^i =\sum_{i=1}^{n}2^i + 2^{n+1}= 2^{n+1}-2+2^{n+1}=2^{n+2}-2$

[dawn dagger]

for blakes sake

[grand pond]

cristorenzo99

[boreal ocean]

But I believe you could use also that:

$2^{n+1}-2 = 2(2^n-1) = 2(2^n-1^n) = 2(1+2+2^2+2^3+\dots+2^{n-1}) = 2 + 2^2 + 2^3+\dots + 2^n = \sum_{i=1}^n 2^i$

[grand pond]

cristorenzo99

[twilit field]

oops

[boreal ocean]

Using: $a^n-b^n = (a-b)\Big(\sum_{k=0}^{n-1}a^k b^{n-1-k}\Big)$

[twilit field]

We attempt to prove this via induction
\
Base case : $2^{2}-2=2$
\
We now suppose $\sum_{i=1}^{n} 2^ i =2^{n+1}-2$
\
We wish to prove from this that $\sum_{i=1}^{n+1} 2^ i =2^{n+2}-2$
\
$\sum_{i=1}^{n+1} 2^ i =2^{n+2}-2= \sum_{i=1}^{n} 2^{i}+2^{n+1}-2=. 2\cdot 2^{n+1}-2=2^{n+2}-2$
\
\
QED

[grand pond]

cristorenzo99

A dense set

[boreal ocean]

Start from $\sum_{i=1}^{n+1} 2^i$ or from $2^{n+2}-2$ and find the other one! You can't use that $\sum_{i=1}^{n+1} 2^i = 2^{n+2}-2$ in the first equality because that's what you have to prove!

[grand pond]

cristorenzo99

[twilit field]

I can use the fact that $\sum{i=1}^{n} 2^ i =2^{n+1}-2$

[grand pond]

A dense set

[twilit field]

Whoops

$\sum_{i=1}^{n} 2^ i =2^{n+1}-2$

[grand pond]

A dense set

[sinful topaz]

Yes you can use that

[twilit field]

I thinnk I have

thanks

We attempt to prove this via induction
\
Base case : $\sum_{i=1}^{1} (8i-5) = 3$
\
We now assume that $\sum_{i=1}^{n} (8i-5)=4n^2-n $
\
\
From this we wish to prove that $\sum_{i=1}^{n+1} = 4(n+1)^2-(n+1)$
\
\
$\sum_{i=1}^{n+1} = 4(n+1)^2-n = \sum_{i=1}^{n} (8i-5) + 8(n+1) -5 = 4n^2-n+8n-5= 4(n+1)^2-(n+1)$

[sinful topaz]

you want to prove its 4(n+1)^2 - (n+1)

[twilit field]

oppps

[sinful topaz]

and you cant assume what youre proving

[grand pond]

A dense set

[midnight plank]

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[manic bison]

how to evaluate such large powers

[sage helm]

Have you heard of Newton's identity?

[manic bison]

not yet

[sage helm]

If you have a polynomial $f(x) = ax^2 + bx + c$, with roots $\alpha$ and $\beta$, define $S_n = \alpha^n \pm \beta^n$, then $aS_n + bS_{n-1} + cS_{n-2} = 0$.

[manic bison]

i see

how do we apply this here though

[grand pond]

neon

[sage helm]

Notice that S_n = sqrt(5)T(n)

work from there

[manic bison]

oh alr

any other method?

also how do we do c) and d)

[midnight plank]

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[manic bison]

i need help with polynomials

[daring bison]

send question

[manic bison]

.close

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[midnight plank]

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[rugged bay]

I need help with partial fraction decomposition

[rugged bay]

Integration

No clue how to proceed

[slender walrus]

missing a few () there

[rugged bay]

Where at?

Oh ik what u mean

But even then idk how to

Do this

[slender walrus]

then expand and equate coefficients

[rugged bay]

Who tf are you

[midnight plank]

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

[rugged bay]

I got this

[sinful trout]

and please be respectful ooferm

[rugged bay]

Sorry

[rugged bay]

How do I make it

Here

[slender walrus]

group terms based on the power of x

[rugged bay]

Arent they already grouped

[slender walrus]

not the way you want

[sinful trout]

gather like terms

e.g. Ax^3 and Cx^3

[rugged bay]

Ooohhhh wait I kept thinking u can’t

[sinful trout]

because they have the same exponent on x

why wouldn't you be able to lol

[rugged bay]

How would that look if u combined them?

[sinful trout]

try it yourself

[rugged bay]

(A+C)(x^3)

?

[slender walrus]

something * x^3 + (something else) * x^2 + ...

[rugged bay]

Is this it

For the first one

I’ll get the idea

[sinful trout]

yeah

[rugged bay]

Oohhh

[sinful trout]

yeah

[rugged bay]

Then x = 0

[sinful trout]

uhmmmm

no

[rugged bay]

Oh

[grand pond]

ren

[sinful trout]

now just set the individual coeffs equal and solve

[slender walrus]

plugging in x = 0
will give you an equation about some of your unknowns

[rugged bay]

Why the 1?

[sinful trout]

e.g.

A+C = 1
B+D = 1
etc

[sinful trout]

?

[slender walrus]

essentially the same result from equating the constant term

[rugged bay]

I’m slow okay I see

[rugged bay]

Yes yes

Wait

A and C can be anything no?

Like 3 and -2 to equal 1

Wait wait wait

No

Wait

I know

Stop

Uh I got A is 0

That doesn’t sound right

If c is 1 then A has to be 0

[rugged bay]

What am I doing wrong

[slender walrus]

why do you think it would be an issue if A=0?

[rugged bay]

Oh

Word

I’ll finish this later I got a class rn but Ty so much

.close

[midnight plank]

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[round venture]

Feel like this is such a simple question but theres a few ways to interpret them..
for now im just doing a b and c
a) I think is 8 because 2 possibilities per 4 days = 8
b) I went with 4 cos it works every day and shes checking for 4
c) I think its 5? 3 working days = 3 plus the two different possibilities you can get with the 4th day(working or not working)

[midnight plank]

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[round venture]

someone pls tell me im right😂

[round venture]

<@&286206848099549185>

.close

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[round venture]

.reopen

[midnight plank]

✅

[round venture]

<@&286206848099549185> ok a is actually 16 cos 2^4
would b and c still be correct?
im confused cos i know that binomials are a bit different

[midnight plank]

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[round venture]

.close

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[craggy merlin]

How do i rationalize this

[craggy merlin]

Please help me im gonna tweak out

[loud hemlock]

uhh first idea would be to write 6=2*3=2*sqrt(3)*sqrt(3) and than pull one sqrt(3) out

dont know if that works

[craggy merlin]

What?

Im confused

Sorry

[loud hemlock]

wait a sec i get pen and paper

[neat canyon]

huh

multiply and divide by sqrt(3)

[loud hemlock]

ohh yeah that too xD

[neat canyon]

ig u can simplify before doing that

take common factor 2 out of the num

[craggy merlin]

Yeah but i checked symbolab and it said its incorrect

Oh

Let me try

[short sundial]

my answer is coming (3-root3)/3

[neat canyon]

then multiply and divide by sqrt(3)

[neat canyon]

ye

[neat canyon]

.

[craggy merlin]

Would the bottom square root disappear

?

[neat canyon]

i mean the square roots cross out

but the 3 doesnt disappear

[neat canyon]

why did you just multiply the 3 in the num

u gotta distribute it to the whole term

[craggy merlin]

Oh

Okay

Like this?

[neat canyon]

yes

but the denominator is wrong

what's sqrt(3) * sqrt(3)

its not 1

[craggy merlin]

Uh

Wdym

Like

The 9?

[neat canyon]

no

denominator

the bottom

[craggy merlin]

Thr 2

The 2?

[craggy merlin]

Since i multiplied it by root 3 i thought it cancels

[neat canyon]

it cancels but it doesn't disappear

just the square root goes away

[craggy merlin]

Oh

[neat canyon]

sqrt(3) * sqrt(3) = 3

[craggy merlin]

So it becomes 3 and mulyiplies into 6

[neat canyon]

ye

can you factorize the numerator

[craggy merlin]

Uh

Likr 3?

[neat canyon]

common factor 3 yes

[neat canyon]

its wrong

[loud hemlock]

write everything as a factor of 3

[short sundial]

solve them in 2 parts

it will be much easier

6root3/4root3 -6/4root3

[loud hemlock]

9=3*x and 6 =3*y
then you can devide by 3

[craggy merlin]

Im really confused

[neat canyon]

this is fine

take common factor 3 out of the num

[loud hemlock]

you have that and now want to get out the common factor of 3

[loud hemlock]

noice

[short sundial]

that's what i said

[loud hemlock]

2!=3

we're not engineers

[craggy merlin]

Thank you

[short sundial]

oh no

thats not what i said

[loud hemlock]

that was my solution btw

another way

if it helps

[midnight plank]

@craggy merlin Has your question been resolved?

[midnight plank]

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[dreamy lichen]

. @hard shard this channel can be yours

[fathom apex]

im confused af

[fathom apex]

<@&286206848099549185>

[karmic cloak]

احم السلام عليكم
ولاتهتم شفتك عربي جيت أسلم عليك استنا حد يفهم فهاي الاشياء

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fathom apex]

wa alaykum al salam habibi

i barely understood what u wrote bro im white washed 😭

[last slate]

I can help you, I think I can solve it.

I think it's the last answer, for the first image. Question 5.

I can prove it, but I need to take my ruler, and draw it.

But by looking at it, I think that it is the last one.

[fathom apex]

im getting both A and D

and for question 4 is it D aswell?

[last slate]

I think so, yeah. It should be D by looking at it.

Question 4 is C.

[last slate]

That's for question 5

[midnight plank]

@fathom apex Has your question been resolved?

[midnight plank]

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[midnight plank]

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[shadow knot]

I believe i sorted and found the mean / median correctly. to answer the last part, "What measure do you think best represents the set and why?"
Would the value be something between 13750 to 17950.
or would i pick one of the two values?

Is there a formula to find out the measure that best represents the data set?

[visual mist]

I think it might be that 60000 is considered an outlier, so the median would be better?

[chrome sail]

It’s going to be median bc of the outlier

[visual mist]

great minds think alike

[shadow knot]

Dont we consider outliers when we are plotting graphs, not when we are finding a representive value?

[visual mist]

since the mean is considering the 60000, it is too high

[chrome sail]

well the 60000 throws the mean off by a pretty decent amount so the median “represents” it better

[visual mist]

outliers shouldnt be considered as they dont represent the data properly

[shadow knot]

Ahh

I get you both now

[midnight plank]

@shadow knot Has your question been resolved?

[shadow knot]

<@&286206848099549185>

[proper hedge]

What is the problem bro

[shadow knot]

Is all my working out, and final answer correct?

just asking to confirm before i submit it

[midnight plank]

@shadow knot Has your question been resolved?

[shadow knot]

.close

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[midnight plank]

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[lusty cave]

Hi, I've got this from subbing values and then calculated after the subsition 9 * 2 , 500 and I got S 10 = 472,500 can someone check for me im hesitant i may have done the formula incorrectly.

!status

[midnight plank]

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

[lusty cave]

4

[dusk spear]

yeah, it's right

[lusty cave]

tysm

.close

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[sharp mesa]

I need help setting up the integral

also for sketch, do I just graph x^2 +y^2 + z^2 = 1 and x^2 + y^2 + z^2 = 16?
and focus on the 2nd octant?

[midnight plank]

@sharp mesa Has your question been resolved?

[midnight plank]

@sharp mesa Has your question been resolved?

[midnight plank]

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[upbeat plinth]

One message removed from a suspended account.

[upbeat plinth]

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[polar socket]

oh

[last arch]

derivative of ln x is 1/x

[polar socket]

yeah

[upbeat plinth]

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[pearl hull]

Chain rule

[polar socket]

well whatever is inside the parentheses

[upbeat plinth]

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[polar socket]

u just find the reciprocol

so like

if its ln (2x)

it would be 1/2x

[upbeat plinth]

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[polar socket]

also you also find derivative of 2x but lets ignore that for rn

[upbeat plinth]

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[polar socket]

bc chain rule

[polar socket]

awmmm

no

ln (lnx)

the lnx is in the parentheses right

[upbeat plinth]

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[polar socket]

so like

whats reciprocol of ln x

[upbeat plinth]

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[last arch]

Sub u = ln x

[polar socket]

1/lnx bestie

[upbeat plinth]

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[polar socket]

am i making it more complicated or like

[upbeat plinth]

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[polar socket]

alr

[upbeat plinth]

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[polar socket]

yas

[polar socket]

oh

chain rule remember

[upbeat plinth]

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[polar socket]

yeah

derivative of lnx

and derivative of ln (lnx)

[upbeat plinth]

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[polar socket]

YEAHHHH

[upbeat plinth]

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[polar socket]

its okayyy

[upbeat plinth]

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[polar socket]

yw !!

[upbeat plinth]

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[midnight plank]

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[midnight plank]

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[median ridge]

I’m confused because I thought the mass was supposed to be the hypotenuse

[sharp coral]

the weight always points straight down

[midnight plank]

@median ridge Has your question been resolved?

[sharp coral]

it would be the hypotenuse if you were splitting the weight into components along the coordinate axes of the incline. but that's not what we're doing here

[midnight plank]

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[kindred lagoon]

why can a function and its inverse have intersections on y = -x instead of just on y = x?

because I'm thinking that x and y swap for an inverse, hence the intersection satisfies

(x, y) = (y, x)

which implies y = x because coordinates are ordered.

but this is clearly not true. Gets contradicted easily with the case of y = -x³.

is it false that the coordinates swap when we have an inverse?

[sharp coral]

how are you getting that y = -x³ intersects its inverse in a place not on the line y = x?

[kindred lagoon]

well my brain isn't working right now to do algebra, but it works graphically

intersects at (-1, 1) and (1, -1)

and also (0,0) actually

[sharp coral]

oh i see the problem. y = -x^3 has both (-1, 1) and (1,-1) on its graph, so the corresponding coordinates on the inverse graph are (1, -1) and (-1,1). so although they do not intersect their "corresponding points" they do intersect the opposite point

and that just comes down to the fact that y = -x^3 passes through two points which are symmetric when mirrored across the line y = x

[kindred lagoon]

but the (x, y) = (y, x) doesn't really stand, it feels more like (|y|, |x|) = (|x|, |y|) except that's only sometimes true

yes (-1, 1) = (-1, 1) but that's if the form (x, y) = (x, y) where both are on either function, not of the form (y, x) = (x, y)

[sharp coral]

the "coordinate swap" is a sufficient but not necessary condition for a function to intersect its inverse

[kindred lagoon]

right that makes sense. Is the rule inverted to y = -x for y-reflected functions?

because -x = -x³ is faster to solve than -x^(1/3) = -x³, so might be useful

[sharp coral]

basically the issue arises if you have two points which satisfy $(x_1,y_1) = (y_2,x_2)$ on the graph of the original function (not its inverse), then the graph of the inverse will intersect that function at those points, although they aren't on the line y = x

[grand pond]

cloud

[kindred lagoon]

okay

can it be outside of |y| = |x| though?

let me try to prove it for strictly decreasing actually

no that failed, seems to only work for strictly increasing functions

should I only substitute y = x for strictly increasing functions, and brute force it (f(y) = f(x)) for strictly decreasing?

[midnight plank]

@kindred lagoon Has your question been resolved?

[midnight plank]

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[livid widget]

I need help from question 7-10

[midnight plank]

@livid widget Has your question been resolved?

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[twilit field]

Prove by induction

[sudden yacht]

Have you done the base case?

[twilit field]

I'm typing out my proof rn

would like it checked

[sudden yacht]

Uh sorry

[olive matrix]

you really gotta stop opening channels like this, you know better. Work out your formatting issues in #latex-testing if you need to, or at the very least specifiy the entire context in the opening message.

[twilit field]

We wish to proceed by induction.
\
Base case $1 - \frac{1}{2!} =\frac{1}{2}$
\
We now assume that $\sum_{i=1}^{n} \frac{n}{(n+1)!} = 1- \frac{1}{(n+1)!}$ is true , and from that wish to prove $\sum_{i=1}^{n+1} \frac{n}{(n+1)!} = 1- \frac{1}{(n+1\2)!}$
\
We denote the some of the first $i$ terms by $S_i$
\
We thus have $S_{n+1}= S_n + \frac{n+1}{(n+2)!} = 1 -\frac{1}{(n+1)!} + \frac{n+1}{(n+2)!} = 1- \frac{1}{(n+1)!}( 1 - \frac{n+1}{n+2}) = 1 -\frac{n+1}{(n+2)!}$
\

[grand pond]

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

[twilit field]

Got it. Sorry

[sinful trout]

formatting when

[twilit field]

We wish to proceed by induction.
\
Base case $1 - \frac{1}{2!} =\frac{1}{2}$
\
We now assume that $\sum_{i=1}^{n} \frac{n}{(n+1)!} = 1- \frac{1}{(n+1)!}$ is true , and from that wish to prove $\sum_{i=1}^{n+1} \frac{n}{(n+1)!} = 1- \frac{1}{(n+2)!}$
\
We denote the some of the first $i$ terms by $S_i$
\
We thus have $S_{n+1}= S_n + \frac{n+1}{(n+2)!} = 1 -\frac{1}{(n+1)!} + \frac{n+1}{(n+2)!} = 1- \frac{1}{(n+1)!}( 1 - \frac{n+1}{n+2}) = 1 -\frac{n+1}{(n+2)!}$
\

There we go

[grand pond]

A dense set

[tribal temple]

That last equality

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

What's wrong with it

[tribal temple]

For one, it's not what you're looking for, and for second, you may want to justify how you even got there

Are you saying that $1 - \frac{n + 1}{n + 2} = \frac{n + 1}{n + 2}$?

[grand pond]

@tribal temple

[twilit field]

no

or wait

it's 1/n+2

We wish to proceed by induction.
\
Base case $1 - \frac{1}{2!} =\frac{1}{2}$
\
We now assume that $\sum_{i=1}^{n} \frac{n}{(n+1)!} = 1- \frac{1}{(n+1)!}$ is true , and from that wish to prove $\sum_{i=1}^{n+1} \frac{n}{(n+1)!} = 1- \frac{1}{(n+2)!}$
\
We denote the some of the first $i$ terms by $S_i$
\
We thus have $S_{n+1}= S_n + \frac{n+1}{(n+2)!} = 1 -\frac{1}{(n+1)!} + \frac{n+1}{(n+2)!} = 1- \frac{1}{(n+1)!}( 1 - \frac{n+1}{n+2}) = 1 -\frac{1}{(n+2)!}$
\

[grand pond]

A dense set

[twilit field]

Thanks!

.close

[midnight plank]

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[modest sphinx]

i dont even understand the question 😭

[median gust]

i think they want area VQR

[lethal path]

yeah

[median gust]

start off by labelling

[modest sphinx]

is P to V 90degrees?

[median gust]

find VQ using pythagoras

[median gust]

ah yes

[median gust]

the word "vertically" indicates this

[modest sphinx]

what how

[modest sphinx]

i noticed that but i was rather confused at first

[median gust]

c is your VQ

[modest sphinx]

i know hte formula but i dont really understand the shape 😭

where is the right angled triangle

[lethal path]

PQ is perpendicular to PV
PV is perpendicular to PR

[median gust]

because you are drawin a 3D shape in 2D, the right angles dont look like right angles

[modest sphinx]

gimme a min comprehending

[median gust]

here take this

[modest sphinx]

ookay ill take some times. I think i might have to draw multiple triangles to show each surface for the sake of my own vision XD

[median gust]

ah yes good idea

[modest sphinx]

how am i suppose to

get VQR with this

[modest sphinx]

is angle QPR 90 degrees?

i dont even know if i need that

[median gust]

we do not know

you cannot assume that

[modest sphinx]

oh

what to do

[median gust]

ok so here is the annoying bit

[median gust]

there may be a better way to do this, but i will tell you how i would do it

[modest sphinx]

aight

[median gust]

are you familiar with the sine rule

[modest sphinx]

a/sin a=b/sin b

?

[median gust]

yep

[modest sphinx]

alr yea

[median gust]

so rn you have this

[modest sphinx]

where did u get 2 root 29

[median gust]

they are one and the same

[modest sphinx]

oh i swear i didnt see that

😭

[median gust]

ye that is one problem of breaking up the shape

[median gust]

but we have this

[modest sphinx]

alrrr

[median gust]

you would need to know 2 things:

1. sine rule
2. area of triangle

are you also familiar with this formula

[modest sphinx]

yess

[median gust]

okok good

so, step 1: find this blue angle

using the sine rule

next, find this other blue angle, using 180 - 80 - (the angle you just found)

[modest sphinx]

angle VRQ is 45

[median gust]

step 3: apply this formula with VQ, VR and the angle to get the area

[modest sphinx]

angle QVR is 55

[median gust]

mhm

[modest sphinx]

oh my

[median gust]

there might be a faster way to do this, but this is what i came up with

[modest sphinx]

ay appreciate it

[modest sphinx]

ay

i just tried

using heron's formula

it worked but slightly annoying bcs before i finished it i was not supposed to round the number like i had to take the whole decimals for accuracy 😭

but in the end i could round it to 4 significant figures and it matches the answer sheet

thanksss

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[queen torrent]

hey can someone explain the process on how to do this question to me i'm not too sure where to start

[queen torrent]

i'm gonna be afk ping me pls

[rapid marsh]

@queen torrent u tryna find the value of the variables?

[queen torrent]

yeah

[rapid marsh]

so lets start with 4a

do u see that they are the same triangles but flipped around?

[queen torrent]

yes

OH

[rapid marsh]

you see how they label the same lines with like 1 slash

and the other line with 2 slashes?

[queen torrent]

wait so is y = 5.2

and x = 7.3

[rapid marsh]

yes exactly

yea

need help with the rest?

[queen torrent]

hold up im gonna try to do them first

[rapid marsh]

alr no problem

js ping me

[queen torrent]

aight thank you

it's all good now this was a call for me to go to sleep

that was really easy after you explained it

thanks for that

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[noble scaffold]

I'm a little scared that I'm not doing this right, just want to confirm.
I have to calculate the Circumference of the shape

[grave moth]

Why 1/4

[noble scaffold]

Wait you're right

wait

No

[grave moth]

Try it again

[noble scaffold]

That smaller semicircle, right, the radius is given, if I divide the semicircles radius by 2, i get the diameter of the smaller circle

so i need the smaller circle's radius

so I divided 18, a radius, which acts as a diameter for each smaller circle, by 2 for that radius

OH SH

wait

i see

I mislead myself

I see, the smaller semicircle is still a semicircle, so I needed to divide that by 2 aswell

and divide the 36 by 4 for that semicircle's radius

@grave mothDid I get it right?

[grave moth]

Yes!

113.097 to be more precise

[noble scaffold]

I'm moving!!! Thank you so much!

[noble scaffold]

Yeah, but my school prefers simplicity

So they like it when we round off alot

[grave moth]

Oki

[noble scaffold]

Thank you!

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[radiant jacinth]

ANY HINTS

[dreamy lichen]

differencce of squares

dont factor 82896, factor 17^4 - 5^4

[vivid haven]

17^4 is (17^2)^2

same for 15

5*

[radiant jacinth]

still have no clue lol

wait so step by step

[dreamy lichen]

ill show you an example

[radiant jacinth]

you get (17^2)^2 - (5^2)^2

right?

or am i stupid

[dreamy lichen]

yes, exactly

now apply the difference of squares formula

[grand pond]

MæthIsAlwaysRight

[radiant jacinth]

(17^2+5^2)(17^2-5^2)

[dreamy lichen]

now simplify this further

the first factor cant really be simplfiied

but the second one can

[radiant jacinth]

(17^2+25)(17^2-25)

[dreamy lichen]

(17^2-5^2)

does this look familiar?

you can apply the formula once again

and factorize it even further

(17^2+25) and this one will have to be computed manually, we dont have a sum of squares formula unfortunately

[radiant jacinth]

wait so do we expand (17^2+25) (17+5) (17-5)

[radiant jacinth]

what next

[dreamy lichen]

now just compute it

17 + 5 is?

17 - 5 is?

and prime factorize every factor

until you get whole factorization

[radiant jacinth]

22 and 12

[dreamy lichen]

alright

,calc 17^2 + 5^2

[grand pond]

Result:

314

[dreamy lichen]

so its

314 * 22 * 12

just prime factorize each of these now

[radiant jacinth]

ah i see

[radiant jacinth]

157 * 3 * 11 * 2^4

thanks

but what does the mark scheme do>

[dreamy lichen]

,w (a+b)^4 - (a-b)^4

[grand pond]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[dreamy lichen]

,w expand (a+b)^4 - (a-b)^4

[grand pond]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[dreamy lichen]

yeah, so apparently
(a+b)^4 - (a-b)^4 = 8ab(a^2 + b^2)

you dont really need this tho

[radiant jacinth]

ah ok

thanks

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[jovial hound]

Can someone lmk if my njmbers are right i think the nimbee is way too high but idk what i did wrong 😭😭

[jovial hound]

I used uhhh

Washer method

I forgot my x^6 whoops but i think that would just makw the number bigger

No im slow

😭😭😭

Yeah no i dont get it

😫😫

[last slate]

could you

post

the original question

[jovial hound]

<@&286206848099549185>

[midnight plank]

@jovial hound Has your question been resolved?

[midnight plank]

@jovial hound Has your question been resolved?

[jovial hound]

<@&286206848099549185>

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@jovial hound Has your question been resolved?

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[short path]

how to calculate determinant here ? ( easy method)
$$
\left|\begin{array}{cccc}
\alpha & \theta & \theta & \beta \
\theta & \alpha & \beta & \theta \
\theta & \beta & \alpha & \theta \
\beta & \theta & \theta & \alpha
\end{array}\right|
$$

[grand pond]

Uruma

[midnight plank]

@short path Has your question been resolved?

[short path]

<@&286206848099549185>

[raven nebula]

just try laplace, no?

maybe fixing the second line

[midnight plank]

@short path Has your question been resolved?

[short path]

it's fine thanks

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[prisma otter]

can anyone show me what the sketch may look like

[midnight plank]

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[prisma otter]

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[glossy compass]

If I find the second derivative

Wouldnt i get -2??

I'm looking at my other examples and idk how I got the second derivative..

[distant vigil]

A little confused what you're doing

[glossy compass]

Optimization

[distant vigil]

Do you have original question?

I think it would help for context

[distant vigil]

Yes your second derivative would just be -2

Working looks fine

[glossy compass]

Ohh so we don't have to plus anything in

*plug

[distant vigil]

Also note that 55=110/2

Which makes sense because 55*55 will be larger than any combination

[distant vigil]

I mean apart from p'(x) = 0 to find stat point

-ve concavity then indicates it's a max

[glossy compass]

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[magic orchid]

using the formula tn = t1r^n-1, theres a question asking what height does the ball rise after 10th bounce

[magic orchid]

i plug in my t1 + r value

25(4/5)^10-1

cuz its asking for the 10th

but it doesnt work

cuz somehow its 11-1 as n = 11

BUT THEN-----

in the next question it asks how many days in april whatever 30 DAYS right and then asks how much this person gets in the last day of april so tn=30

i plug in 0.01(3)^30-1

i get it right

like what

i dont get it

someone help me

pls

when do i know when it adds another +1

and when it doesnt

[ancient plank]

Can you share the problem statement?

[ancient plank]

First bounce is considered at 20m, then second bounce at 16m, and so on...