help-17

[paper shoal]

<@&286206848099549185>

[stone flower]

Use by parts after 2 by parts this equation will be repeated

Ln x goes to 1/x2 and 1/x2 goes to ln x

Ln x is u

1/x2 is dv

[stone flower]

Then u take it to other side

@paper shoal did u get it?

[paper shoal]

ohhh I see

so it's a 2 cycle

I've seen it with trig but i didn't recognize it here cuz it looked weird

ty

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[dusk bear]

Hi

[dusk bear]

Help 😭

[tough valve]

!status

[vocal sleet]

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

[dusk bear]

Uh

2

Like

How can u solve it without knowing the other angles or the length of any sides

[thin vale]

SOHCAHTOA

[dusk bear]

But how

[thin vale]

draw the triangle

[dusk bear]

Yup

[thin vale]

okay

now try finding sinT

SOHACHTOA tells you it should be what

[dusk bear]

Wait how

Idk the length of any of the sides

[thin vale]

it doesn't matter

draw the triangle

label the sides

[dusk bear]

Ok

Uh

Wait im so confused

[thin vale]

have you drawn the triangle?

[dusk bear]

Yeah

[thin vale]

Have you labeled the sides?

[dusk bear]

Do u just write opposite over hypotenuse for the answer

[thin vale]

No

[dusk bear]

Oh

[thin vale]

what is the opposite side?

specifically

[dusk bear]

Of T?

[thin vale]

the opposite side from the angle T

[dusk bear]

A

[thin vale]

A isn't a side

it's a point

[dusk bear]

RA

[thin vale]

yes

that is the opposite side length

RA

and the hypotenuse is what

[dusk bear]

TA?

Wait so its just RA over TA?

[thin vale]

yes

yes

[dusk bear]

😭

[thin vale]

for sin(T)

[dusk bear]

Oh okk

Ty!

[thin vale]

that is why the problem says, in terms of RA, TA,...

no problem

[dusk bear]

Oh oops 😭

.close

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[waxen hawk]

Why, in “S.H.M”, omega = sqrt(k/m)

[pale perch]

w=2pi f

[waxen hawk]

why

[pale perch]

F=-kx and F=ma
-kx=ma -x/a=m/k
a=-w^2 x in shm

so -x/(-w^2 x)=m/k
1/w^2=m/k
w^2=k/m

is one way

[vocal sleet]

@waxen hawk Has your question been resolved?

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[waxen hawk]

I see, thank you so much!

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[hasty pilot]

hello i have a math question and i wanted to know how to solve it so, i am very lost so can you show me step by step please

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@hasty pilot Has your question been resolved?

[hasty pilot]

No

No

<@&286206848099549185>

[hasty pilot]

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[inner osprey]

what have you tried,

[finite maple]

well i did

a + b = 0

a+b+g+d = -b/a aka 5/6

= g+d=5/6

idk what to do from there

[viral copper]

if you assume the roots to be $\alpha$ and $-\alpha$

[twin meteor]

NEON

[finite maple]

cant i js do alpha beta gamma and delta?

[viral copper]

then you can assume a general factorized form of this equation as $6(x^2 - \alpha^2)(x^2 + px + q) = 0$

[viral copper]

sure but since their sum is 0 this is convenient

[finite maple]

cant i js do alpha + beta = 0

[viral copper]

$\alpha = -\beta$

[twin meteor]

NEON

[viral copper]

so now the two roots are $\alpha$ and $-\alpha$

[twin meteor]

NEON

[viral copper]

same thing :/

[finite maple]

yeah alr

[twin meteor]

NEON

[finite maple]

huh

howd u get that factor

cant i do the question using vieta formula?

[viral copper]

maybe

[finite maple]

cuz im pretty sure im meant to

[viral copper]

the two roots are alpha and - alpha

so multiply them in factored form

[viral copper]

alright then just list out all of them and

try to solve for all unknowns

[inner osprey]

u can

... is certainly valid approach

[inner osprey]

apply vieta's relations (use other coefficients)

use ur roots as a, b, c, -c (or whatever variables u want) (also imo it helps to immed use "-c" instead of "d" or smth... makes the equations more clear)

(other helper showed u this but i was referencing ur original workings)

[vocal sleet]

@finite maple Has your question been resolved?

[finite maple]

uhh sorry for the delayed response but

ok well i tried to solve it

so far got 2 roots

a and -a

im looking at my teachers working out but dont understand this

it says sub 1 and 5 simultaneously

but how do u sub bg=-1 into b+g = -5/6

[inner osprey]

this is common type of quadratic

xy = something
x + y = something

here what ur teacher did is g = -5/6 - b
then, bg = -1 becomes b(-5/6 - b) = -1

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[finite maple]

nvm yeah lol

got it ty

.close

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[vocal sleet]

✅

[finite maple]

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[frail sigil]

can someone explain to me why anyone would use a natural log instead of just using the percent to decimal conversion

[frail sigil]

this homeworks answers are so bad

[livid horizon]

,w ln(1.00289)

[twin meteor]

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Requested by masked.sniper

[frail sigil]

how am i supposed to know

that its supposed to be aln

instead of just using what they gave me

[livid horizon]

If you solve the differential equation, you'll get ln on both sides eventually, so you it'll be easier to take antiln

[livid horizon]

Without a calc, you shouldn't know it
But even if you don't convert it at the start you'll end up converting it after solving eventually

[frail sigil]

but its not like they instructed me to find k

they gave me k in the first place

so i shouldn't have to go back and find it again

[livid horizon]

If you solve it both ways you get the same ans
The only difference is that they found the antilog at the start itself, but in the other method you'll be finding it in the end

[frail sigil]

yea so the homework should hvae different answers that are acceptable

[livid horizon]

Yeah ig

[frail sigil]

this site always has one way

and if u dont find it u get a garb gradfe

grade

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[pulsar chasm]

,w cx+x^2 / 2^x

[twin meteor]

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Requested by shuxx_1

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@pulsar chasm Has your question been resolved?

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[compact elk]

Hello, I was under the impression that $\int{\csc \theta d\theta} = -ln \abs{\csc \theta + \cot \theta}$

[twin meteor]

dp

[compact elk]

but I saw this on a video

ignore the 1/6

Are both expressions correct or have i made a mistake copying

[compact elk]

Thanks, is there any property of logarithms I'm missing?

I can't go from one side to the other

or is it something regarding trigonometry

[strong grove]

https://www.cuemath.com/calculus/integral-of-cosec-x/ they have also derived it

[compact elk]

can't thank you enough

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[vast shale]

algebra question

[vast shale]

binomial theorem

[steel snow]

Send the problem

[vast shale]

just a sec

.close

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[vast shale]

,reopen

.reopen

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✅

[vast shale]

the 9th question

binomial theorem

can somebody try out this

.close

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[frozen shuttle]

help

[frozen shuttle]

the first one is 4

but i dont get the second one

[compact elk]

what is the correct answer?

3?

[frozen shuttle]

there's no markscheme for it

[compact elk]

oh ok

it most likely means inverse function

saying $f^{-1}(x)$ is like saying "Hello, I got -1; what number did I pluck into my function"

[frozen shuttle]

oke so its 3.5?

[twin meteor]

dp

[frozen shuttle]

hmmm

[compact elk]

When does the value of f(x) = 1

[frozen shuttle]

at 3

[compact elk]

that's your answer

[frozen shuttle]

so if it was f^-2 then the answer would be 4?

wait

no

[compact elk]

no

[frozen shuttle]

if it was f^-1(2)

[compact elk]

when does f(x) = 2?

[frozen shuttle]

then it would be 3 still

i think im getting the hang of it

the ni have another question

[compact elk]

no no

[frozen shuttle]

if u have time

[compact elk]

I have time

[compact elk]

but $f^{-1}(2)$ is not 3

[twin meteor]

dp

[vast shale]

1

[frozen shuttle]

oh okey

[vast shale]

fof(0)=1

[compact elk]

please, I'm on it

[vast shale]

oh soory

[compact elk]

and that is not correct

when does f(x) = 2?

[vast shale]

no

[frozen shuttle]

2.5?

[compact elk]

if we look at the function, we see that f(x) is = 2 more or less between 2 and 3

[frozen shuttle]

whut

[compact elk]

I'd say 2.5 aswell

yes

[vast shale]

that is not how composite functions work

[compact elk]

we're still on question b

now, let's do the one you sent after

Find the value of $(f\circ f)(0)$

[twin meteor]

dp

[compact elk]

do you know what that means

[vast shale]

u tell me

[compact elk]

I'm talking to @frozen shuttle , please stay on your channel

[frozen shuttle]

is it not f(f(x))

[compact elk]

it is, yes

so we would first calculate f(x) and then put that value again in f(x)

[frozen shuttle]

ok ok

wait

so f(0)=3

then f(3)=3

[compact elk]

$f(f(0)) = f(3)$

[twin meteor]

dp

[compact elk]

but f(3) is not 3

[frozen shuttle]

im a stupid or are the y lines between 0 and 2 one too many?

2/3?

[compact elk]

Hmm

no

[frozen shuttle]

waiot

[compact elk]

There is a line for y = 0, y = 1, y = 2...

[frozen shuttle]

my eyes are messing up

[compact elk]

but only even lines are labeled

[frozen shuttle]

its 1

[compact elk]

it is one :)

[frozen shuttle]

the lines on my screen are varying in thickness so it was har to tell

the nthere is c

oh man

[compact elk]

woo

let's first find the domain

What is the domain for f(x)?

[frozen shuttle]

-3 to 6?

[compact elk]

Ok

[frozen shuttle]

waaait

yes

[compact elk]

Are you familiar with interval notation?

[frozen shuttle]

not sure

[compact elk]

[] and ()

like [-3,6] for instance

[frozen shuttle]

oh yeah

[compact elk]

ok

so we know that $D f(x) = [-3,6]$

[twin meteor]

dp

[compact elk]

so, where would we have to look to find the domain of g(x)?

[frozen shuttle]

how do you mean

[compact elk]

if $g(x) = f(x+5)-5$, where is the "problematic" part

[twin meteor]

dp

[compact elk]

the one that will cause us trouble

[frozen shuttle]

f(x+5)-5?

[compact elk]

yes, but more precisely

is it f(x+5) or -5

the thing that limits the domain

[frozen shuttle]

x+5?

[compact elk]

f(x+5), yeah

so, if f(x)'s domain is [-3,6], in this case, x+5 must be in the [-3,6] interval

right?

because, imagine x = 3

g(x) would be trying to do f(x+5) = f(8) which doesnt exist

[frozen shuttle]

so x+5 would have to be less than 6?

[compact elk]

and bigger than -3

Do you understand why?

[frozen shuttle]

would it be [-8,1]

yes i think i do

[compact elk]

yes, that would be the domain of f(x)

let's look at the range now

$g(x) = f(x+5)-5$

[twin meteor]

dp

[compact elk]

that's basically saying

[frozen shuttle]

no it sthe -5?

[compact elk]

??

[frozen shuttle]

wait no keep explaining

[compact elk]

ok

That's like saying

hello

I'm g(x) and I'm taking f(x+5)'s value and substracting 5

so if f(x+5) is 10 i'll give out 5

[frozen shuttle]

riight

[compact elk]

what would be the range of g(x) then

[frozen shuttle]

i should be getting it but im too tired to understand it

[compact elk]

no worries

let's think this together

if $f(x+5) = -3$, what would $g(x) = f(x+5)-5$ be

[twin meteor]

dp

[frozen shuttle]

-8?

[compact elk]

yes

[frozen shuttle]

right

[compact elk]

so g(x) is just removing 5 from every number of the function

so what happens to the range if we remove 5 from each number

[frozen shuttle]

it geos down 5 steps on the graph?

[compact elk]

yes

so first, what is the range of f(x)

[frozen shuttle]

[5,-4]

or would it be written the other way around?

[compact elk]

normally you would write the smaller number first

but yeah, your answer is correct

so if the range of f(x) is [-4,5]

what is the range of g(x)

[frozen shuttle]

[-9,0]

[compact elk]

that's correct :)

[frozen shuttle]

and the x+5part is just gone

[compact elk]

it's not gone

It just doesn't matter for the range

[frozen shuttle]

but its the domain?

[compact elk]

the range is only vertica

[frozen shuttle]

yes alright

[compact elk]

f(x+5) just moves the function 5 steps to the left

but doesn't change the range

so we can ignore it

the range of $f(x)-5 = f(x+10)-5 = f(x+99999999)-5$

[twin meteor]

dp

[compact elk]

the important bit is the -5

[frozen shuttle]

ok ok

thank you

[compact elk]

np

[frozen shuttle]

now i can fail my mocks tomorrow atleast knowing this:)

thank you

[compact elk]

good luck :)

[frozen shuttle]

you too

[compact elk]

IB?

[frozen shuttle]

yes

sadly

[compact elk]

haha

nice

good luck :)

[frozen shuttle]

oh well

life goes on

anyways thank you

[compact elk]

np

[frozen shuttle]

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[karmic adder]

#help-25 message

[compact elk]

.state

no

I meant

where are you at

[compact elk]

do you know where to start?

[karmic adder]

I used substitution but I’m stuck

[compact elk]

ok

I assume you got

[sly sierra]

fyi for future reference, it's !status

[compact elk]

yeah, just found it

kinda new to the bot

thanks :)

I assume you got $\frac{\sqrt{36\sec^2(\theta)-36}}{6\sec(\theta)}$

[karmic adder]

This is what I got so far

[compact elk]

Firstly, are you able to factor something out?

maybe seeing this way helps

[twin meteor]

dp

[karmic adder]

Yes there is a common factor

[compact elk]

which one?

[karmic adder]

36 is common

In the numerator

[compact elk]

$\sqrt{36} = 6$

[twin meteor]

dp

[compact elk]

right?

so if we divide by 6 on both sides we get

[karmic adder]

Yes that’s right

[compact elk]

$\frac{\sqrt{\sec^2(\theta)-1}}{\sec(\theta)}$

[twin meteor]

dp

[compact elk]

If you have the list of trig identities next to you, you will find the one to use obvious

if not, I'll help you through it

[karmic adder]

I have the list with me. sec x = 1/cos(x)

[compact elk]

In this case, since they are squared, there might be a trig squared identity to help you

like $\sin^2(\theta) + \cos^2(\theta) = 1$

is there one of them that applies here?

[twin meteor]

dp

[karmic adder]

Yes. Tan^2 (x) + 1 = sec^2 (x)

[compact elk]

right!

what do we have now?

[compact elk]

yeah, but these ones cancel out

and if we cancel the 1s

we can also cancel the square root

I wouldn't substitute the denominator though, since $1+tan^2(\theta)$ is more complex than $sec^2(x)$

[twin meteor]

dp

[compact elk]

we want the simpler expression

[karmic adder]

I think this is simplified? Or not yet?

[compact elk]

there is a way to simplify it further

if we put $\sec^2{\theta}$ instead of $1+tan^2{\theta}$

[twin meteor]

dp

[karmic adder]

Oh so it becomes 1/cos^2 (theta)?

[compact elk]

you would get

$\frac{\frac{sin\theta}{cos\theta}}{\frac{1}{cos^2\theta}}$

[twin meteor]

dp

[compact elk]

which can be simplified further

[compact elk]

yes

that would be the most simplified form

[cosmic kayak]

It was sec(x) not sec^2(x)

[compact elk]

?

[cosmic kayak]

It simplifies to just sin(x)

[compact elk]

true true

we made a mistake along the way

$\frac{\frac{sin\theta}{cos\theta}}{\frac{1}{cos\theta}}$

[twin meteor]

dp

[compact elk]

It's just this

thanks :)

I don't know when the numerator changed from sec to sec^2

[karmic adder]

Ah thanks for pointing that out:)

Got it correct. Thanks!

.close

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[queen verge]

help please

[queen verge]

Im only looking for the answer if thats ok

Because im in a rush

[main knot]

check rules bro

[queen verge]

Fine ill take help

[main knot]

hopefully someone can assist you with solving it yourself

[queen verge]

But it will have to be quick

[main knot]

I'd be on it but I'm not sure how to solve it, it's either something I've not learned, or something I learned like 10 years ago

[queen verge]

rip lol

[main knot]

I'll look on google incase nobody shows up

[queen verge]

oh ok thanks 👍

[main knot]

right so a hexagon has all sides equal according to its definition

and the p and q are vectors

like a measure of distance travelled in some direction

and so if BC= 6p, and all sides are equal then it should tell you AO

[queen verge]

oh ok thanks

i understand it now

how do i close this channel?

/close

[main knot]

with a period

[queen verge]

.close

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[mighty iris]

a very simple problem, however, i keep getting the wrong answer ( f(x) = x^4+px^2+qx+r passes through the point (-1;16). and at this point, f''(x)= -f'(x)=16. Find the values of q,r, p )

[mighty iris]

p=2, q=-8 ( however, the book says q=24, so i am puzzled)

[steel snow]

Can you show your work

[mighty iris]

-f'(x)=16 ergo f'(x)= (-16), f'(x) = 4x^3+2px+q

f'(-1) = -4-4+q

[steel snow]

-f’(-1)=16 btw

[mighty iris]

yeah but why

nvm

[steel snow]

It says “at this point”

[mighty iris]

i get that part, but thats not the issue

[steel snow]

What part of your work is the issue

[mighty iris]

'- ( -8+q ) = 16 right?

[steel snow]

Uhh

No

because q=24

[mighty iris]

i mean yeah it should be 24

[steel snow]

you mustve done something wrong with a negative then

[mighty iris]

'-(4* -1^3+ 2 '' 2*-1 +q)

-(-4-4+q) = 16, did i make a mistake

f'(x)=4x^3+4x+q

cuz i dont see it, but ik its probably somewhere

' - (-8+q) = 16

maybe i am just misunderstanding the problem

but at the same time it says -f'(x)=16

[steel snow]

Let me check

Sorry I’m having lunch rn

[mighty iris]

its cool man

[steel snow]

You’re missing p

[mighty iris]

i have replaced it with 2

[steel snow]

Was that correct

[mighty iris]

yeah , i am postive

[steel snow]

Could it have been -2

[mighty iris]

cuz f''(x) = 12x^2+2p =16

so f''(-1) = 12+2p = 16

p =2

so without a doubt

[steel snow]

I don’t see an error I guess

hmmm

[mighty iris]

here's what i have noticed if you write 4 * -1^3 +22-1 + q = 16, you get 24

[steel snow]

I don’t know why there is -f(x)

[mighty iris]

however it kinda violates the given terms of this problem -f'(x)=16

[steel snow]

it is suspicious

ok idk someone else check this work because I think it’s right <@&286206848099549185>

[mighty iris]

maybe its just a mistake in the book

the worst part is that its not a difficult problem, i just cant calculate properly, or at least, so i think

<@&286206848099549185> just aid to understand whether my point makes sense or not, cuz man i am full of doubt but it makes sense to me

https://tenor.com/view/virtual-insanity-gif-20466967

Whatever it is, let it be a mistake in the book

<@&286206848099549185>, just in case

[vast shale]

I think that the minus sign was a misprint

otherwise if we do assume their answers are legit

Then f'(x) would have to be = 16

Not -16

[mighty iris]

yeah probably a misprint

aight thanks yall, for helping me and giving me some faith in my initial thoughts

[vocal sleet]

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[vocal sleet]

@floral jay Has your question been resolved?

[floral jay]

<@&286206848099549185>

[vast shale]

What

[floral jay]

what

[vast shale]

You pinged

[floral jay]

your not a helper young blud

on blud this shit ass

[vast shale]

Do you still need help my pookie bear

[lyric fossil]

what have you tried

[floral jay]

on blud I do

I tried pluging every number into desmos and that shit dont work

LMAO

[vast shale]

Ok to find the bottom values

It's just x intercepts

So like points that are (___, 0)

[floral jay]

ight so bottum would be +4 and -1

[vast shale]

Do you still need help

[floral jay]

nah I had to call people to ask

[vast shale]

?

Why would you open a channel then

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

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[lunar geyser]

https://media.discordapp.net/attachments/1142576082835492864/1214001023744745572/Screenshot_2024-03-03_at_4.06.35_PM.png?ex=65f7858f&is=65e5108f&hm=a3f149779f7b53587fcc19df4224a2905d38b363d40986fb41cdffb4160092c4&=&format=webp&quality=lossless&width=2160&height=456

[lunar geyser]

related rates problem

<@&286206848099549185>

[lunar geyser]

mb

<@&286206848099549185>

.clos3e

/close

.close

[vocal sleet]

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[radiant tree]

posting question

[radiant tree]

i dont exactly understand how converting time units with standard deviations works

let's say

im given the passage .. Shipping the
new phone to the warehouses takes 1 month. The standard deviation of the shipping time is 1 week.
Please note that a year has 12 months, and each month has 4 weeks. Further, a normal distribution for
the distribution of the lead time is assumed.

and i want the average lead time and std lead time in months

like is the std lead time 1/4 or 1* sqr root (4)

in months?

[heavy yoke]

unit conversions don't change the standard deviation,so for example standard deviation of 1 hour would be the same as a standard deviation of 60 minutes

[radiant tree]

so there is no scaling factor?

[heavy yoke]

you use the same scaling factor as when converting units normally, but since scaling factors are considered "exact values" they just multiply directly with no additional steps

[radiant tree]

could you give an example in the context of months and weeks?

[heavy yoke]

for example, if you have an average of 3 months with standard deviation 0.5 months, then that corresponds to an average of 3 months × 4 weeks/month = 12 weeks with standard deviation 0.5 months × 2 weeks/month = 2 weeks.
any time you multiply by an exact (defined rather than measured) value (for example unit conversions), the uncertainty is simply multiplied by that value as well

(of course, a month isn't exactly 4 weeks, but we seem to be ignoring that for this question)

[radiant tree]

nevermind - had to think..

thanks

.close

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[azure moat]

Help

[azure moat]

I need help with this geometry thing

I don’t know how to do any of these

<@&286206848099549185>

Got a big quiz tmrw im lost

😭😭 pls someone

[vocal sleet]

@azure moat Has your question been resolved?

[azure moat]

😑

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@azure moat Has your question been resolved?

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[leaden crane]

If I have this graph

[leaden crane]

Is the first one -infinity, second one is 1, third one doesn't exist, and fourth one is -3?

.close

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[vocal sleet]

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[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[wicked tundra]

Hiii

[wicked tundra]

Need a quick recap on floor function

and solving the question

highest power of 12 in 100!

I solved it earlier but I need some revision on how to solve it agian

[brisk moss]

you are trying to use legendre's formula i imagine?

[wicked tundra]

Yep

[brisk moss]

ok, any progress?

[wicked tundra]

I dont remember the way of solving it

I mean I have slight idea how we use 2 and 3 as the primes

and found the powers

but dont exactly remember how

[brisk moss]

for each factor of 12, there needs to be two factors of 2 and a factor of 3

[civic drift]

https://youtu.be/sOf_G3gQvf8?si=CnEC722vuWWKhb57

This one helped me last time

He explains well

[wicked tundra]

k thanx I'll check it out

can I keep it open for 20 mins?

[civic drift]

I'm not sure how long it will be open but in case this closes open a new one, I'LL try to reopen it incase it closes.

[wicked tundra]

thanx

tho I got it

Ima close it

.close

[vocal sleet]

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[vocal sleet]

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[warm kindle]

find the linear function with the properties f(-4)=0 and the slope is -3

[warm kindle]

i dont know what to do

beyond the obvious slope

[vast shale]

help me for number 20 b

[warm kindle]

.close

[vocal sleet]

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[crude arrow]

why did you close

now I can't help you 😦

[raven owl]

That wasnt even the owner

[vast shale]

wait

how can i opened it

[raven owl]

You need to claim ur own channel

Go to a channel under math help available

[vocal sleet]

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[sullen condor]

where did I go wrong?

[civic drift]

10x^2 × x = 10x^3

[sullen condor]

what about the 2

doesn't that also get multiplied in

[civic drift]

10x^2 × 2 = 20x^2

[sullen condor]

10x^2 * x * 2 = 20x^3

[civic drift]

Is that a new property you discovered cause Im confused

Plus it's wrong

You have applied the property wrong

Can you explain why you wrote that?

[sullen condor]

okay i got it

thanks

.close

[vocal sleet]

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[civic drift]

https://www.khanacademy.org/math/4th-grade-foundations-engageny/4th-m3-engage-ny-foundations/4th-m3-te-foundations/a/distributive-property-explained

[sullen condor]

yeah i got it

[vocal sleet]

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[wind torrent]

Can someone tell me how to change this unit

L should be litre

[jagged cargo]

convert litre to cubic meter first

[signal pendant]

1l = 0.001 cubic meter

[jagged cargo]

[wind torrent]

Oh thank out. @jagged cargo @signal pendant

How about this one? Which part was wrong

[jagged cargo]

!1q

[vocal sleet]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

@wind torrent Has your question been resolved?

[silk osprey]

.close

[vocal sleet]

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[amber forum]

Can I have help please

[silk osprey]

!da2a

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[fervent wasp]

whats your problem?

[amber forum]

I need to put this in this kind of form

Ik how to solve it once it’s in the form I just don’t know how to get it in that form

[near tartan]

Sum of all internal angles of a triangle is 180°

[fervent wasp]

what's the sum of the three angles of a traingle?

...

[amber forum]

180

But

How do I put it into an equation?

[fervent wasp]

add everyting up

they're equal to 180

[amber forum]

Can you write it out for me the equation

So I can solve it

[fervent wasp]

6x+14+5x+15+3x=180

it's literally place them together XD

[amber forum]

And then I solve left to right?

[fervent wasp]

yeah

[amber forum]

Like that

?

I got one more question

[fervent wasp]

sure

[amber forum]

That’s correct tho?

[fervent wasp]

it is

[amber forum]

Ya!

[amber forum]

And what about for this one?

[wise pumice]

Can you help

Me

[amber forum]

What…

This is my ticket…

@fervent wasp sorry to bother you there is just one last question I’m kinda stumped with

[amber forum]

Can anyone help me with this quickly

<@&286206848099549185>

@amber forum when you can help me please

[vocal sleet]

@amber forum Has your question been resolved?

[amber forum]

It has not lol

[vocal sleet]

@amber forum Has your question been resolved?

[vocal sleet]

@amber forum Has your question been resolved?

[short drum]

think about how you can express three arbitrary consecutive integers and then write down what it says there

[vocal sleet]

@amber forum Has your question been resolved?

[amber forum]

<@&286206848099549185>

can someone please help me with this]

i have been asking for a min

[somber harbor]

@amber forum

You on big man?

@amber forum

Bruh moment

[amber forum]

yep

[somber harbor]

So here we have three consecutive integers.

Lets call them a, b, c respectively.

How would you express b in terms of a?

[vocal sleet]

@amber forum Has your question been resolved?

[vocal sleet]

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[vocal sleet]

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[whole narwhal]

How do I find the solution set for (a,b,c) such that a+b = c and b-a = (c-2)/2?

[zinc quail]

solve for any var

then you can declare one of the vars from the reduced equation to be arbitrary

[whole narwhal]

I tried that I think but I got infinite answers. Is it possible?

[zinc quail]

well ofc

you have two equations

and thee vars

if I give you the following set of equations:

A+B = 1

then it has infinitely many solutions for pairs (A,B)

e.g. (2,-1), (3,-2), (4,-3), ...

@whole narwhal is it clear how you get the set of solutions?

or at which step are you stuck

[whole narwhal]

I've found that y = 3x-2 and z = 4x-2, but i am stuck here

[zinc quail]

if these are true then you are done

we can declare x to be the arbitrary variable

then your set of solutions is (x, 3x-2, 4x-2) for all x e R

@whole narwhal

[vocal sleet]

@whole narwhal Has your question been resolved?

[zinc quail]

hope it clarified itself :3

[vocal sleet]

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[silk kindle]

can someone help me with this

[fervent wasp]

!status

[vocal sleet]

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

[silk kindle]

i don't knwo where to begin

[vocal sleet]

@silk kindle Has your question been resolved?

[silk kindle]

<@&286206848099549185>

[gentle elk]

you just have to come up with a system of equations that satisfy what the question tells you

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[fallen agate]

pls help me prove this

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[ashen fractal]

Bart Simpson decides to jump Springfield Gorge on a skateboard. Springfield Gorge is 50m wide and Bart uses a 1.0m high ramp inclined at 30. What speed does Bart need to get to the other side?

[pallid zenith]

whatd you try

[ashen fractal]

im trying to send a screenshot

idk why i have a really bad time understanding this concept, because i was really good at 1-dimentional motion

[pallid zenith]

well, if you feel really good about 1-d motion, you can turn this into that kind of problem

sorry im having a hard time totally parsing your work its been a hot minute since i took phys

[pallid zenith]

idk that we need tan here

heres my thought

do you believe that like, you can split this problem into two dimensions?

i mean, you can look at vertical motion and horizontal motion separately

[ashen fractal]

i think i can but idk how

because i think the y(x) = ... is a cruical part

[pallid zenith]

sorry im pushing through it

[vocal sleet]

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[vocal sleet]

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[royal vapor]

Need help with 1.2.3

[royal vapor]

Do I add all values of x that I got from previous questions

[vast shale]

cant see shşt

[royal vapor]

Just click on it

[vast shale]

i did

still nuin

[vast shale]

u think im stupid?

[royal vapor]

🤨

[royal vapor]

Dude chill I am just surprised

It looks clear on my phone

[vast shale]

https://media.discordapp.net/attachments/1124005516922863637/1130395646935961600/attachment.gif

[royal vapor]

This better?

Can you please answer

[edgy sapphire]

you have to combine the information from the last 2 questions yea

hopefully there's only one whole number that works with your answer in 1.2.1 and 1.2.2

[vivid cedar]

@royal vapor from the first condition u should get x is in [-5,8] and from the second one that x is in (5/4; infinity)

so u make the intersection between them and the natural numbers. Thats the answer

[vocal sleet]

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[hazy sparrow]

If $f$ is a continuous function $f: \mathbb R \to \mathbb R$ with limits $a$ and $b$ at respectively $x\to \infty$ and $x\to -\infty$, then how would I go about showing that it is uniformly continuous on the entire domain?

[twin meteor]

SORCERER💉🐴

[hazy sparrow]

I thought I could divide the domain into 3 areas $(-\infty, K]$, $[-K, K]$ and $[K, \infty)$, and go from there possibly

[twin meteor]

SORCERER💉🐴

[hybrid flicker]

that's exactly it

for a given epsilon, choose K such that it works

[hazy sparrow]

Would you mind giving a hint or just getting me started, because I'm having a little trouble seeing how to use the limits

[hybrid flicker]

$\lim_{x\to\infty}f(x) = a$ means that for $\varepsilon >0$, ...

[twin meteor]

rafilou2003

[hazy sparrow]

Right, if I unfold the definition then I can use the triangle equality to get a $K$ such that $|f(x) - f(y)| < \epsilon$ for $x,y \geq K$

[twin meteor]

SORCERER💉🐴

[hazy sparrow]

And likewise on the negative region

[hybrid flicker]

yes

well more like 2epsilon rather than 1

but doesn't change the result

and then on [-K,K] you use uniform continuity as well

[hazy sparrow]

I get [-K, K] for free I believe

[hybrid flicker]

and then for x,y not in the same

you go through f(-K) and f(K)

[hazy sparrow]

Wdym?

[hybrid flicker]

like if x and y are not in the same region

you write f(x) - f(y) = f(x) - f(K) + f(K) - f(y) or something depending on the situation

[hazy sparrow]

Hmm

[hazy sparrow]

It's because there are different cases

I get it

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@hazy sparrow Has your question been resolved?

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[vast shale]

k\

[vast shale]

.close

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[sullen coyote]

Please help, i have to solve the different values of a polynom but when I have to draw it the values don't add up

[noble spindle]

You are missing one zero

How did you get zeros?

I mean maybe you are not

[sullen coyote]

i used horner's algorythm

[noble spindle]

I didnt check one is a double

[sullen coyote]

to get them

[sullen coyote]

oh?

idk like i think i got them but when i try to draw it it doesnt add up

[noble spindle]

Ok you are not missing zeros but like most of your zeros are incorrect

[sullen coyote]

ah

[noble spindle]

This is a biquadratic

$$a(x^2)^2 + b(x^2) + c = 0$$

[twin meteor]

casework

[noble spindle]

You just solve for x^2

[sullen coyote]

lemme just write that down

[noble spindle]

I mean you can also factor the eq very easily

[sullen coyote]

eq?

[noble spindle]

Equation

[sullen coyote]

in what way

[noble spindle]

$$-2x^4 + 4x^2 - 2 = -2(x^4 - 2x^2 + 1)$$

[twin meteor]

casework

[noble spindle]

And you can hopefully now see what is in the parantheses

[sullen coyote]

ohhhh i see

can i try to solve before u close?

[noble spindle]

I dont close the channel you do

[sullen coyote]

thank god

[noble spindle]

I mean there are some people in the server that can close other ppl's channels but its normally for some good reason

[sullen coyote]

so do i apply the horner's algorhythm to the parantheses numbers?

[noble spindle]

And the channel closes auto in few mins of inactivity (like 20/30 or something like that)

[noble spindle]

You can just factor it

$$(a - 1)^2 = a^2 - 2a + 1$$

[twin meteor]

casework

[noble spindle]

Looks similar?

[sullen coyote]

holld up

so its

-2(x2 - 1)2 ?

[noble spindle]

Yes

Now you can factor it again

[sullen coyote]

-2(x^2 - 1)^2*

[noble spindle]

As $x^2 - 1 = (x - 1)(x + 1)$

[twin meteor]

casework

[sullen coyote]

so then u get

-2((x+1)(x-1))^2

[noble spindle]

Yeah i mean both are squared

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

[twin meteor]

casework

[noble spindle]

Now you can easily see zeros

[sullen coyote]

so now x1,2 = -1 and x3,4 = 1 ?

[noble spindle]

Yes

[sullen coyote]

omg

[noble spindle]

Its a double zero so i hope you know how to draw that

[sullen coyote]

this is it

yes i do

thank you so much

.close

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[worthy citrus]

a screenshot of the question would help

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@lyric sable Has your question been resolved?

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[weak hedge]

A​ DC-9 aircraft leaves an airport from a runway whose bearing is N 42 degrees E. After flying for one-half
​mile, the pilot requests permission to turn90 degreess and head toward the southeast. The permission is granted. After the airplane goes 1 mile in this​ direction, what bearing should the control tower use to locate the​ aircraft?

[weak hedge]

.close

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@fresh mantle Has your question been resolved?

[edgy sapphire]

sure, are you concered about why it breaks the theorem?

[fresh mantle]

Basically, the question before that is about an ellipse. We are being told to look up what the four vertex theorem is and comment on whether it agrees with the plot of the curvature of the ellipse. In that case, i believe it does, reason being there is 2 max and 2 minimas as well, and hence the sign of the curvature changes at least 4 times at 4 different vertices (the curvature for the ellipse is double U shaped which im assuming is because of the period going from 0 to 2 pi)

On the other hand, regarding the question i posted, we are also asked to say if we think the theorem agrees with the plot of the curve. In my opinion it doesnt, but im not sure about the proper explanations regarding both questions

For the question in the picture, the plot of the curve looks somewhat like a cochleoid-which is similar to a cardioid

[edgy sapphire]

I believe it breaks one of the preconditions to the theorem, so there's no guarantee there's four vertices

[fresh mantle]

Does it HAVE to be a simple curve? Because obviously a cochleoid isnt a simple one since it intersects its own

But i can show u the plots for both questions if it helps, would help me better probably if you could explain it while looking at my plots

[edgy sapphire]

yea I can see the plot on wolframalpha

[fresh mantle]

Ye i used mathematica

[edgy sapphire]

I think it's just that it's not a simple curve yea, all the sources I see mention that except one part of wikipedia O_o

[fresh mantle]

Exactly. Most mention it, but 1 source doesnt

So its confusing

Which is why im trying to look for another reason

And i was thinking the amount of vertices, change in sign when it comes to its curvature could be a reason. If any <4, it means it doesnt agree with the theorem. But is it less than 4?

[edgy sapphire]

do you have the curvature plot for this one?

[fresh mantle]

its this, ye

[edgy sapphire]

yea looks like 2 vertices

[fresh mantle]

care to elaborate so i can see it more clearly? Ive been doing maths for 20 hours, im not too bright atm

Is it because the highest point of the curve isnt on any axis

[edgy sapphire]

like for max/min curvatures

[fresh mantle]

Hmmm, what do u think about the other one then? Do the plots make sense to you? And if so, since im guessing the answer is yes for the other one, how would you say its different compared to this one?

[edgy sapphire]

I count 4 for the ellipse so it's fine

[fresh mantle]

I mean, since its 2 U shaped plots for the curvature thats basically 2x2=4 vertices, sign changing at all 4 points of the ellipse

Ye thats my idea too

What about the one at the left? Is that not being counted because we have to choose 1 of the points as the starting one?

[edgy sapphire]

I still beleive it's just that the second one intersects itself

I don't count the left because it's periodic on (0,2pi] and you have to choose on endpoint and not both

[fresh mantle]

Makes complete sense

Second plot is also on same period, so its the same way-then it would make sense that it has 2 changes of sign and 2 vertices if you look at both the curve and curvature plot

So it could be both the number of vertices+the intersection related thing

Im just gonna write both down, i wanted to understand it better or see how someone who knows even more than me thinks about it

[edgy sapphire]

what do you mean by both? altogether it's that there's only 2 vertices but that doesn't disagree with the theorem because of the simple loop precondition

[fresh mantle]

Wait, wdym because of the simple loop precondition? If its 2 vertices and the theorem states that it has to be at least 4, doesnt that kinda "solve" it?

By both i meant i wanted to give the 2<4 reasoning as well as the fact that it intersects itself and its not a simple curve

[edgy sapphire]

uhh that seems like a logic issue

like 2<4 means it disagrees with the theorem, but you can't really disagree with a theorem haha

[fresh mantle]

No but

What i mean is

We are being asked if the statement of the theorem agrees with our plot, and since our plot has 2 vertices and theorem requires 4+, it just disagrees with it

[edgy sapphire]

I guess I'm reading "agree" in a weird way, but I feel you'd be required to explain why it disagrees

idk I think I'm overcomplicating it lol

[fresh mantle]

I get what u mean but u are ye, basically what we are asked is just to say if its valid for our plot or not, we havent even studied this, in the coursework we just got told to look the theorem up and relate it to our plots for no reason

And the explanation for it would just be what we talked about regarding the number of vertices+the intersection thing and it not being a simple curve i guess-as an answer to my coursework's question

Thanks a lot tho, imma go sleep and write it up tomorrow, gn