help-17

[chilly rapids]

I need help on this and a simple explanation please

[edgy gulch]

The second one, mean is just another word for average in maths

or in statistics rather

[chilly rapids]

Oh I see

Thank you

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[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

[vast shale]

1

answers:
highest point: (4, 2, 2)
lowest point: (-4/3, -2/3, -10/3)

I just need to do the work

But idk how

[patent nymph]

do you know how to use the method of lagrange multipliers?

[vast shale]

yeah

[patent nymph]

you can use it here

[regal bane]

The tough part is the intersection curve

I'd solve the second one for z, then sub that into the first one.

[vast shale]

mhm

Well I found lambda

I think

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

:/

.close

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[spiral acorn]

CANSNq

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Please don't occupy multiple help channels.

[patent nymph]

.close

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[thin frigate]

why is this false?

that's the inverse right?

[cerulean gazelle]

Sin^-1 is the reciprocal of sin (arcsin)

[thin frigate]

and?

[paper depot]

inverse

not reciprocal

if you're french it's an understandable mistake but it's a very sore point of mathematical translation between fr and en

what you call réciproque we call inverse, and what you call inverse we call reciprocal

[cerulean gazelle]

Yes I am French and we say reciprocal function

Sorry

[thin frigate]

Bruh

is the dude who translates math mentally challenged

[paper depot]

what "dude" are you talking about

[thin frigate]

I guess it makes sense if he's french

[paper depot]

i mean it sounds like you didnt do what they ask you to do there

they dont ask you to find the inverse of the function, they ask you to find its derivative

[thin frigate]

exactly

rn it's arcsin(x^5)

so it's inverse is sin(x^5)

right?

then the derivative is cos(x^5)(5x^4)

but that's incorrect

[cerulean gazelle]

They're not asking for the derivative of the inverse, but of the function y

[vocal sleet]

@thin frigate Has your question been resolved?

[thin frigate]

Ah, what does that mean exactly

[cerulean gazelle]

You need to find dy so you just use u'(arcsin'(u))

[thin frigate]

what is dy/dx

[cerulean gazelle]

The derivative of y with respect to the variable x, to put it simply

[thin frigate]

I'm confused about that, the difference between dy/dx and dx/dy

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[brittle umbra]

why is $y=u(x)v(x)$

[twin meteor]

TheWhiteShadow

[vocal sleet]

@brittle umbra Has your question been resolved?

[vocal sleet]

@brittle umbra Has your question been resolved?

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@brittle umbra Has your question been resolved?

[vocal sleet]

@brittle umbra Has your question been resolved?

[vast shale]

wew

s

s

s

s

[brittle umbra]

?

[vocal sleet]

@brittle umbra Has your question been resolved?

[solemn cape]

this just mean that y is the product of two functions, u and v

[flat whale]

Because it works as ye solution

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[dire rose]

how do i turn this into vertex form? -> x^2-8x+16=4

[pale perch]

have you tried

also

that doesnt have a vertex

[pale perch]

this is just two straight lines of the form x=a

[soft walrus]

did you mean x^2-8x+16-4=y?

or is what you gave the expression given in the problem?

[dire rose]

yeah thats what it is in the problem

unless i copied it incorrectly one sec

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[wispy wigeon]

how do i do this problem

[hybrid wind]

Oo vectors

[pale perch]

have you found u x v

[vocal sleet]

@wispy wigeon Has your question been resolved?

[wispy wigeon]

yes

(-5a-9)i + (5a+15)j -10k

[pale perch]

yeah alright
what do you know about orthogonal vectors

[wispy wigeon]

they are orthogonal if the dot product is zero?

[pale perch]

nice

[wispy wigeon]

what is i in this case?

in the question

[pale perch]

the vector, i

1i+0j+0k

[wispy wigeon]

oh i see

so i need to determine <1i+0j+0k> * <(-5a-9)i + (5a+15)j -10k> = 0?

[pale perch]

you need to get the a that satisfies that, yeah

[wispy wigeon]

umm... how do i actually do that... 😅

[pale perch]

do the dot product

[wispy wigeon]

that is u * v = |u| |v| cos(90 degrees) right?

[pale perch]

well, yeah in this case
what i meant specifically was to evaluate the LHS in the fashion of
(ai+bj+ck) dot (di+ej+fk)=ad+be+cf (=0 here)

[wispy wigeon]

oh i see

[wispy wigeon]

that is (-5a-9)i?

since the j and k are zero in the i vector

[pale perch]

there would be no i

youd just get -5a-9=0

[wispy wigeon]

OHHHH

and then just algebra

for a?

[pale perch]

yup

[wispy wigeon]

neato

[pale perch]

indeedo

[wispy wigeon]

got -9/5, seems correct!

thank you!!

[pale perch]

no worries

good work

[wispy wigeon]

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[abstract escarp]

I have no idea how to start this

[abstract escarp]

ik you find derivative of arctan(3x) but dont know where to go next

[hasty lark]

Do you have the slope of the line through those two points?

[abstract escarp]

yeah its uh

3/9x^2+1

basically derivative of arctan(3x)

I just dont know where to go from there

[hasty lark]

nope

What is the slope of a line through the points (3,0) and (0,1)?

[abstract escarp]

-1/3

OH WAIT IS THAT IT?

[hasty lark]

That's the slope of the line

Now, what is the slope of a line perpendicular to the line (3,0) and (0,1)

[boreal tundra]

hello

may i pop in

[abstract escarp]

sure

[boreal tundra]

rise/run..

1/3

[hasty lark]

...

.>

[abstract escarp]

;-;

[boreal tundra]

thats bad right

[abstract escarp]

im having brain rot right now

[hasty lark]

, w slope of line through (3,0) and (0,1)

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[boreal tundra]

its -1/3

sorry

forgot negative

[abstract escarp]

yeah

[hasty lark]

Yes, and we already had that

[abstract escarp]

is that litterally it?

[hasty lark]

No

That's just one step

Now we need to know when another line is perpendicular to that line

[abstract escarp]

its not right?

[hasty lark]

The slope is right, but it's just one step

[abstract escarp]

mhm

my brain isnt working right now sorry

[hasty lark]

You might have to wait until it is working to do calculus

[abstract escarp]

you are right

OH WAIT

I got it

it was 0

I had to use desmos

but thanks

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[potent stirrup]

is Pythagoras theorem defined to be true for all points in a unit circle?

[potent stirrup]

because the sense of right triangles will go away when x>90

if I consider the point (cos(100), sin(100)) , i'm unable to construct a right triangle with angle 100

<@&286206848099549185>

[blissful fable]

you can construct a right triangle of angle 80

using the negative x-axis

[blissful fable]

Pythagoras' theorem isnt "defined" by the unit circle

[potent stirrup]

so i should define cosx being the x coordinate and sinx being y coordinate

so Pythagoras theorem holds without the issue of triangles?

[blissful fable]

cosx is already defined as the x-coordinate

you dont have to define it to be that

[potent stirrup]

yes

i mean, if I just think as being x coordinate

instead of triangles and all of that

it would make sense ig

[blissful fable]

wdym

what do you have to find??

[potent stirrup]

sin^2(30) + cos^2(30)= 1

clearly i can draw a triangle and show it to you

can you draw a triangle and show me sin^2(100) + cos^2(100)= 1?

[blissful fable]

it's the same triangle as you would draw for sin^2(80) + cos^2(80) = 1, just that it's flipped around the y-axis

mirror image of the 80 degree triangle

[potent stirrup]

thank you :)

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[vast shale]

question 13

[vast shale]

apply derivative if you can

i mean use it if you can

anybody ?

[vocal sleet]

@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

[vast shale]

please ?

question 13

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@vast shale Has your question been resolved?

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[haughty star]

can some one give a hint as how to approach this integral ? I've tried u sub and by part it looks like a repeating integral but it isnt

[haughty star]

$\int_{0}^{\infty} x^n e^{-x}$

[twin meteor]

xr

[sharp mortar]

,w int_0^infty x^n e^-x

[twin meteor]

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[sharp mortar]

you need to know about the gamma function

[haughty star]

havent touched that yet

[sharp mortar]

(unless you have some constraints on n?)

if n is a natural, do ibp n times

[haughty star]

there no constraints on n

but il have a look on the gamma function

[sharp mortar]

then you need some analysis to do this question

[haughty star]

alright thanks ^^

.close

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[peak axle]

howw

imma try it for a while

[tidal umbra]

divide it

[vast shale]

think of it as units

[peak axle]

yea but how do u divide it if there's like two missing variables

[vast shale]

you can factor :p

[
(3a-2b) \tsx{phones}, \quad (6a^2+5-6b^2) \tss{pesos}, \quad (???)\f{\t{phones}}{\t{pesos}}
]

[peak axle]

alright but how do i get the divisor for synthetic division

like the (3a-2b)

[vast shale]

you dont need to do synthetic division at all

[peak axle]

but we need to use it

[vast shale]

im pretty sure synthetic division is inapplicable here

multiple variables in the dividend doesn't work well with synthetic division

did they explicitly tell u to use synthetic? @peak axle

because i think long division should work

[peak axle]

oooo

[peak axle]

not rlly,i guess

i just dunno what to use now

[vast shale]

long division!

it definitely would work

do you know how to set up the long divison here? @peak axle

[peak axle]

i think i forgot how to do long division

is it like the fraction thing

[vast shale]

its like

this guy for example

[peak axle]

oh wait

is the answer 2a+3b

[peak axle]

broo idunno how to use that, the cancel thing seems easier

[sleek island]

1/(2a+3b)

[peak axle]

isnt 3a-2b the divisor tho

[vast shale]

yes

well you need to decide

do you want to factor or do u want to do long division

[vast shale]

this will be like

2-3 steps to do

u can factor too but might take a bit more to figure out

[peak axle]

i tried factoring it out like trial and error

and just cancelled the two same thing

[marble plover]

its simple div

[vast shale]

ok

what u got then?

if its 2a+3b then you're done

if it isnt you fucked up somewhere

[peak axle]

ye its that

[vast shale]

then thats it ye

also

i recommend to use long division in situations where you have to use long division/synthetic division

long division is just more applicable

[peak axle]

okii, then I'll try reviewing that

tenkss

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[digital totem]

hi

[peak axle]

Something we can help you with?

[digital totem]

I am looking for someone who can explain the full solution to me. If you want to help please try solving the problem first and explain it to me. i have been stuck on this for 3 hours and would like to ONLY see HOW it is solved.

The problem is asking to show that the LHS = RHS

this is as far as i have come so far:

[mild flower]

you almost have it but there's an extremely concerning algebraic error

$\f1{1-\tan²x} \redneq -\tan²x$, you can't cancel like that

[twin meteor]

Haylsune Miku

[digital totem]

this problem is the first time i am seeing tan2x so my bad

can you walk me through the solution ?

[mild flower]

ok i see more algebra issues, you need to review exactly how cancellation works

you also need to be careful not to confuse a / (b/c) with (a/b) / c, as you did on line 3 on the right

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@digital totem Has your question been resolved?

[digital totem]

<@&286206848099549185>

[west bone]

you need to be careful not to confuse a / (b/c) with (a/b) / c, as you did on line 3 on the right

[digital totem]

so how do i solve the problem?

[digital totem]

<@&286206848099549185> i don't have much time left before i have to submit this problem. help please

[marble plover]

?the prove part

[digital totem]

wdym

[marble plover]

i mean we need to prove lhs and rhs?

[digital totem]

yep

when simplified, they should be the exact same expression

[marble plover]

okh

@digital totem

[digital totem]

yeah

[marble plover]

i think i got it

[digital totem]

can you explain pls?

[marble plover]

let me do the rhs part

@digital totem

[digital totem]

yes

[marble plover]

dm

[digital totem]

you?

[marble plover]

yeah

[digital totem]

.close

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[shut matrix]

calculus help

[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/

[shut matrix]

can anyone help me with this

[hybrid flicker]

how do you find extrema of a differentiable function?

[shut matrix]

no max and mix using differentiation

[hybrid flicker]

ok

[shut matrix]

can you help? @hybrid flicker

[hybrid flicker]

is this a condition given by your prof or is it because you don't want to?

[shut matrix]

no condition

[hybrid flicker]

so why not using differentiation?

[shut matrix]

i was absent for this class

[hybrid flicker]

well better catch up, sorry

[shut matrix]

can you tell me the steps

[hybrid flicker]

find the extrema : find the critical points f'(x) = 0

then, having found the extrema, check if they are max, min, or neither

[hybrid flicker]

Usually f''(x) is a good indicator, doesn't work all the time but works many times.
What works in the general case, fixing x an extrema:

• If f is decreasing before x and increasing afterwards, then x is a min.
• If f is increasing before x and decreasing afterwards, then x is a max.
• If neither, then x is neither.

[shut matrix]

is this the way to it

[hybrid flicker]

those are your critical points for f(x) = x^3 - 3x^2 +1

[shut matrix]

no i mean the steps

[hybrid flicker]

this is the first step

[hybrid flicker]

.

[shut matrix]

what is next step

[hybrid flicker]

.

[shut matrix]

bro i understand nothing

[hybrid flicker]

How do you plot a maximum using a curve? You draw a function that goes up and up, and then goes down. The exact point where it starts to go down is your max, right?

Well, a bit before reaching the point, the function is increasing, and a bit after, the function is decreasing

[shut matrix]

understood bro max value is 5 min value is 1

thanks

@hybrid flicker

can you help me with uv rule

6+4 sqr root x cosec x find first derivative

[hybrid flicker]

can you put parentheses here?

[shut matrix]

no there is no parenthesis

[hybrid flicker]

when you write sqr root x cosec x I can read it as sqrt(xcosec(x)) as well as sqrt(x)cosec(x)

parentheses are important

[hybrid wind]

Finding first derivative here?

[shut matrix]

yes

firsst derivative

[hybrid wind]

So we need to essentially find the derivative of sqrt(x) and cscx

But we see that they are multiplied by a factor of 4

And 6 falls away since it's a constant and derivative of constant is zero

It would be 4[use product rule]

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[potent stirrup]

It’s okay

post it again

I want to try FLT

[patent ore]

I have to use Fermat's little theorem to solve mod(19^32, 17)

I thought since 19^(32 mod 16) = 19^0 the answer would be 0, but the correct answer is 1. What did I do wrong?

[potent stirrup]

19^32 mod 17?

[patent ore]

19 to the power of 32 modulo 17

exactly

[potent stirrup]

By F.L.T, $19^{16} \equiv 1 \pmod{17}$ right?

[twin meteor]

Dubleyou

[potent stirrup]

a^{p-1} congruent to 1 mod p

[patent ore]

wow, I mean I don't know if that's correct but since 32 mod 16 is 0 why use 16?

1 is the answer, that's true

aaaaah i see.

[potent stirrup]

I mean, i just used the statement of FLT

[patent ore]

Yeah I don't know why I'm making it so complicated

but tysm this cleared it up for me! Haha

[potent stirrup]

$a^{p-1} \equiv 1 \pmod{p}$

[twin meteor]

Dubleyou

[potent stirrup]

There you go

Can you try my question lel?

[patent ore]

ur question?

oooh lets see

[potent stirrup]

Stupid question tbh

[patent ore]

oh my god I have no clue what im even looking at, I know sin cos tan all that but man

last time I had this type of math was years ago

[potent stirrup]

How do we know both have equal 30 degrees

looks evidently same to eyes, but is there something we call for this?

[patent ore]

hmm well since a full circle is 360 degrees I'm confused as to how 150 + 30 + 30 = 210 degrees is half the circle

but I feel like I'm just completely misunderstanding

they might think I'm seriously helping u but i dont feel like I'll be able to lmao, maybe try closing this channel and re-asking in a new one?

[digital totem]

what are you trying to find

[potent stirrup]

essentially I’m wondering if I write 30 in the right, does that always imply 30 on left?

[digital totem]

can you be more specific? what right? which 30 degree? sin or cos

[potent stirrup]

i’m not asking specifically about trig ratios here

suppose i mark 30 degrees from positive x axis

[potent stirrup]

essentially how would you explain why we wrote 30 in the left triangle?

[digital totem]

they are both 30 degree angles but the left triangle is in the second quadrant, so it has different coordinates

[potent stirrup]

we say they are congruent or something right?

[digital totem]

they have lines in the x and y axis signifying that they are the same length.

if you look at the y axis for the triangles there is 1 stripe or line and the x axis has 2.

[potent stirrup]

oh i see

I’m geometry is pretty basic

I’m trying to work

[digital totem]

if you calculate sin 150 and sin30 they are the exact same value. though the left cos will be negative

[potent stirrup]

sin(x)= sin(180-x) right?

[digital totem]

-150 to get the 30 degrees

[potent stirrup]

I’m actually looking for it’s proof

[digital totem]

i've done a few that look like this actually. proving that the triangle on the right side of the unit circle being the same as the left side

[potent stirrup]

can you show?

[digital totem]

some of the expressions have different variables than in english though. i'll share rn. ask for translation if you dont understand

[potent stirrup]

sure

[digital totem]

the second one doesnt have solution, my bad. here:

[potent stirrup]

Let me see

Thanks for sharing in advance

[digital totem]

np

[vocal sleet]

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[sullen condor]

yo

[sullen condor]

,rotate

[sullen condor]

2348

simplify

no idea what i’m doing

[digital totem]

lmao är du svensk?

[sullen condor]

självklart

[loud pivot]

$\frac{a\cdot(ba^2-25b)}{b\cdot(5b-2a^2)}$

[twin meteor]

USS-Enterprise

[loud pivot]

This it?

[digital totem]

första frågan kan svaras genom att ta bort a från täljaren och nämnaren. sedan dividera -8/8, eftersom -1 * 8 = -8

[loud pivot]

Nevermind I am an idiot

$\frac{10a^3 - 25ab}{5b^2 - 2a^2b}$

[twin meteor]

USS-Enterprise

[sullen condor]

yes

correct

[sullen condor]

uppgift 2348

[loud pivot]

Well you can factor something out of each and you'll end up with something really nice

[sullen condor]

i tried to

[loud pivot]

Well

Is a the most you can factor out of the numerator?

[sullen condor]

nah

5a

[loud pivot]

Why didn't you do that then

[sullen condor]

i’m dumb

[loud pivot]

So what do we get

[digital totem]

is this matematik 1?

[sullen condor]

yes

5a(2a^2 - 5b)

prov imorn 😁🔫

[loud pivot]

$\frac{5a(2a^2-5b)}{b(5b-2a^2)}$

[twin meteor]

USS-Enterprise

[loud pivot]

Now notice something similar about the second factors

[sullen condor]

the denominator snf the numerator look almost identical

[loud pivot]

Almost

What makes it non identical

Rearrange one of the factors

$\frac{5a(-5b + 2a^2)}{b(5b-2a^2)}$

[twin meteor]

USS-Enterprise

[loud pivot]

It goes minus plus and plus minus

If only we could factor something else out again

[sullen condor]

-1

[loud pivot]

Exactly

[sullen condor]

yayayay

i love math sometimes

[loud pivot]

Where do you want to factor it out

numerator or denominator

[sullen condor]

numerator

[loud pivot]

$\frac{5a(-(5b - 2a^2))}{b(5b-2a^2)}$

[twin meteor]

USS-Enterprise

[loud pivot]

$\frac{-5a(5b - 2a^2)}{b(5b-2a^2)}$

[twin meteor]

USS-Enterprise

[loud pivot]

Well now it's over

What do we do

[sullen condor]

,rotate

[sullen condor]

bruh

[loud pivot]

😂

[sullen condor]

,rotate

[loud pivot]

There you go

Now what

[sullen condor]

i’m guessing the -2a^2 + 5b cancels out

[loud pivot]

Of course. Your main operation is multiplication

Think of -2a^2 + 5b as x

We get $\frac{-5a \cdot x}{b \cdot x}$

[twin meteor]

USS-Enterprise

[loud pivot]

x cancel out

[sullen condor]

holy damnn

thank you for telling me that

[loud pivot]

No problem 🙂

[sullen condor]

that made sense

[loud pivot]

It's always good to visualize things

more simply

[sullen condor]

hold on

can you help me with 1 last question

i wanna visualize ot too

[loud pivot]

I'll try

[sullen condor]

,rotate

[sullen condor]

2347

nigel is basically saying that a-8/8-a = -1

and we need to prove that it’s true

[loud pivot]

Yes

[digital totem]

not prove. we have validate if its true

[sullen condor]

yes

🙌

[loud pivot]

Well

What can you think of doing?

[sullen condor]

i guess factoring out -1

[loud pivot]

Great

If you rearrange it again

$\frac{-8 + a}{8-a}$

[twin meteor]

USS-Enterprise

[loud pivot]

We are where we were earlier

minus plus and plus minus

So we factor out a minus somewhere

[sullen condor]

right

im gonna factor the numerator

[loud pivot]

👍

[loud pivot]

There you go

And again, think of 8-a as x, or hug the denominator around with parantheses for, again, better visualization

$\frac{-(8-a)}{(8-a)}$

[twin meteor]

USS-Enterprise

[loud pivot]

Again cancels out

[sullen condor]

yes

[loud pivot]

And we get -1

[sullen condor]

tysm

[loud pivot]

Do you have to consider restrictions?

Because $8-a \neq 0$

[twin meteor]

USS-Enterprise

[loud pivot]

Otherwise you are dividing by 0

Which can't happen

If not, nevermind this

[sullen condor]

the denominator is the same thing as (8-a) x 1

that’s how i visualize it

[loud pivot]

Yes.

What I am asking is though

Whether you have to consider restrictions

Which means "what can't a equal"

Because if a = 8, we get 8-8, which is 0 and we are dividing by 0

[sullen condor]

ohh

[loud pivot]

Which is undefined

But if you don't have to do this it's irrelevant

[sullen condor]

yeah

aleight

tysm for the help

[loud pivot]

No problem 🙃

[sullen condor]

.close

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[pearl coyote]

Idk if I'm just stupid or what but I'm practicing exponential growth and decay for the first time in years and I'm confused how to compound like this (compound 3 times a year for t years)

[pearl coyote]

Is it saying the initial $400 is growing by 9% three times a year? Because I'd think that'd be what I wrote

[vocal sleet]

@pearl coyote Has your question been resolved?

[vocal sleet]

@pearl coyote Has your question been resolved?

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@pearl coyote Has your question been resolved?

[boreal tundra]

you there

i see the problem

its 400(1.09)^3t

idk its basically the same

but its not a very

regular way of saying it

[pearl coyote]

how the heck does that work

so your doing the compounding 3 times a year?

[boreal tundra]

yeah

idk why you got it wrong

that is pretty right

[pearl coyote]

huh

[boreal tundra]

the same

[pearl coyote]

idk maybe the grading thing only accepts your format

[boreal tundra]

yeah

but both is right

[pearl coyote]

thank you for the answer i guess i was right but you more correct according to format

.close

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[craggy hamlet]

$2=2x\sqrt{x}$

[twin meteor]

putridplanet

[craggy hamlet]

im stuck

$1=x\sqrt{x}$

[twin meteor]

putridplanet

[craggy hamlet]

now what

[regal bane]

1 = x^(3/2)

[craggy hamlet]

where did ^3/2 come from

[regal bane]

x^(1)x^(1/2)

= x^(1 + 1/2)
= x^(3/2)

[craggy hamlet]

im solving for x

and the answer sheet says x=1

[regal bane]

Okay well, this is where we're at right now

[craggy hamlet]

oh so we square both two get rid of square

then we take the cubed root of both to et rid of the cubed

and in both cases one stays the same

got it

[vocal sleet]

@craggy hamlet Has your question been resolved?

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[mild raven]

its neither?

a tangent line drawn at c = 50 is the exact line at s(50)

is it not

[sleek island]

Is line above or below the graph

[mild raven]

eventually its above

[sleek island]

Yeah so

[mild raven]

so i guess its asking about the entiretly of the line at c = 50

oh no shit

it says that

explicitly

[sleek island]

Use the approximation you got and compare it to the real result

That should somewhat help ig

[mild raven]

yeah

looks like this

basically

[sleek island]

Yeah so prolly above guess

[mild raven]

boss moves

ur a real one

.close

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[sour wave]

I need help understanding this relation between sums

[vast shale]

didnt u ask that already

[sour wave]

yes but i had another question that wasn't resolved

i understand it better now but i cant see the entire picture yet. it mainly becomes a problem with viewing -a1 and -a2

[magic wasp]

What is it you don't understand?

[sour wave]

i cant grasp my head around what a_n is supposed to be, is it supposed to be -7 or 1

\

[magic wasp]

You can't know, you only know what the sum is

[sour wave]

i know this is wrong but i cant just use the difference from a1 and a2 to get to a3

[magic wasp]

You do not need to find a_3 or any other a_n

[sour wave]

then what do i need to look for

[magic wasp]

You want to know the sum of all a_n from n = 3 to +inf

You have the sum of all a_n from n = 1 to +inf, and also a_1 and a_2

[sour wave]

Trust me when i say i hear you clearly and that makes sense but something just isnt clicking from there

[magic wasp]

Said another way, you want a_3+a_4+..., and you have a_1+a_2+a_3+a_4+... as well as a_1 and a_2

[sour wave]

oh wait

is it the sum from n=3 to inf of an, and you subtract a_1, and a_2

[magic wasp]

Uh no

[sour wave]

darn

[magic wasp]

Maybe I misunderstand what you mean

I can't really make it more obvious

You have a_1, a_2, and (a_1+a_2+R), you want to find R

[sour wave]

is it that a_1 + 1_2 + R sum to 1

a_2

[magic wasp]

Yes

[sour wave]

oh that was much simpler than i thought, sorry about that

thank you so much

[magic wasp]

No worries

[sour wave]

.close

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[thin vale]

I was hoping someone could check my proof for clarity/correctness please! Tag me if you do

[vocal sleet]

@thin vale Has your question been resolved?

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@thin vale Has your question been resolved?

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@thin vale Has your question been resolved?

[proven fulcrum]

lgtm

[sleek island]

Yeah I don't see nothing wrong

[thin vale]

Idk what “igtm” means, but do you have any feedback other than that?

[sleek island]

What do you mean by clarify

[thin vale]

Clarity

That it is clear

Easy to read and understand

Flows

No hiccups

[sleek island]

I mean if a dumbass like me understands, it's easy to read and to understand

[proven fulcrum]

oh sorry it means "looks good to me" lmao

the logic is fine, and it's typesetted properly so good

[thin vale]

Great thank you!

.close

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[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[atomic lance]

Could anyone help me out with this question?

[vocal sleet]

@atomic lance Has your question been resolved?

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[slender salmon]

In this question, give all answers to two decimal places.
Bryan decides to purchase a new car with a price of €14 000, but cannot afford the full
amount. The car dealership offers two options to finance a loan.
Finance option A:
A 6 year loan at a nominal annual interest rate of 14 % compounded quarterly.
No deposit required and repayments are made each quarter.
(a) (i) Find the repayment made each quarter

I tried to first find the final value by using FV = 14000 (1+14/1200)^24 the formula for financial applications adn then dividing it by 24 to find the rapayment for each quarter but the answer does not match with the one i have in the solutions.

[vocal sleet]

@slender salmon Has your question been resolved?

[vocal sleet]

@slender salmon Has your question been resolved?

[vocal sleet]

@slender salmon Has your question been resolved?

[slender salmon]

<@&286206848099549185>

[vocal sleet]

@slender salmon Has your question been resolved?

[outer warren]

there are 4 quarters in a year, not 12

i.e. you should have r/4, not 4/12

also annuities formula should be used as the amount is being paid off during that time

[slender salmon]

what is the annuities formula?

[slender salmon]

.close

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[strange owl]

Can I get a second opinion on my answer here?

[strange owl]

I dont know if that is the answer or if the answer is closer to: "a limit does not exist when the graph/function is continuous with no breaks"

[pale perch]

not enough

a function can have a limit at a point where it is undefined

do you know about left and right limits

[strange owl]

im assuming that is referring to limits regarding the x value aproaching a value from the left or right side?

[pale perch]

yeah

it is defined that a limit exists at a point when the left and right limits are equal

[strange owl]

is there a simple example of this?

[pale perch]

those limits have to exist too obviously (no infinities or - infinities)

[pale perch]

hm f(x) be defined as: 1 for x>=0 and as -1 for x<0

the limit at 0 doesnt exist, from the right its 1, from the left its -1

if a function is continuous youll generally have the limit existing everywhere, issues happen more so when its not, so i get what you were going for

just needs a bit more specific wording is all

[strange owl]

sorry, i am so confused by that explanation. which one of my explainations are you referring to? the first or the second?

("a limit does not exist when the graph/function is continuous with no breaks")

[pale perch]

this one

[strange owl]

this one is incorrect then?

[pale perch]

doesnt make sense, y=1 is continuous with no breaks, its limits exist everywhere

[strange owl]

ok, so to summarize (please correct me if im still no exactly right), "a limit does not exist when the limit's value is undefined"?

[pale perch]

thats a circular definition
what does 'a limits value is undefined' actually mean

you just said a limit doesnt exist when a limit doesnt exist

[strange owl]

For this example, could you explain what you mean?

@pale perch

Perhaps I don't fully understand the question being asked

[pale perch]

i would say the limit as x approaches 0 doesnt exist because the limits on either side approaching dont exist either - they both tend to an infinity

the conditions for a limit to exist are:
the left limit exists
the right limit exists
the right and left limits are equal

here it fails everything

[strange owl]

Can you specify when you say "the left limit exists" or "the right limit exists"?

[pale perch]

well, the left limit here as you can see, will be -infinity, so it doesnt exist
if a limit=infinity or something, then that limit doesnt exist

the right limit tends to +infinity, so it also doesnt exist

[strange owl]

oh?

<@&268886789983436800>

[strange owl]

So.. is it fair to say that "a limit does not exist when the left limit and right limit approach opposite infinities"?

[pale perch]

i mean, no

a limit exists, when the left limit exists, the right limit exists, the left and right limits are equal

them approaching different infinities doesnt matter

neither of them exist

,w graph (1/|x|)

[twin meteor]

Results provided by WolframAlpha

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

Requested by aldrnari_

[pale perch]

they both approach +infinity here, still dont exist

[strange owl]

ok.. i see.

[pale perch]

the point is
the limit exists on both sides
limit on both sides are equal

then the limit exists if the above holds

[strange owl]

my brain hurts, but i think i got it. thanks

[vocal sleet]

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[fluid bough]

Oi

[pale perch]

what

[pale perch]

no

it rounds to 7.36 to 2dp
what is the lowest a number could be to round up to it
whats the highest a number could be so it rounds down to it

[fluid bough]

I don’t get ur

It*

The lowest is 7.40

[pale perch]

7.4 doesnt round to 7.36

what would 7.354 round to

[fluid bough]

How come?

[pale perch]

7.40 just rounds to 7.40 to 2dp (ie it rounds to itself)

do you understand rounding?

[fluid bough]

Oh yeah

Idk why I said that I get it I get it !

I’ll type in and see if it’s correct

[pale perch]

type what in?

[fluid bough]

7.40

I put 7.30…

[pale perch]

i told you that doesnt round to 7.36

[fluid bough]

It’s not correct

[pale perch]

neither does this

[pale perch]

tell me

[fluid bough]

7.35

[pale perch]

good, what does 7.355 round to

[fluid bough]

7.40

[pale perch]

to 1dp yes

we're rounding to 2dp though

remember

[fluid bough]

7.350?

No 7.36

[pale perch]

so 7.355 is our lower bound for L
if you go any lower, youll round to 7.35 rather than 7.36

[fluid bough]

Ok

I’ll put it down and see

Wait do I find the lb of it ?

[pale perch]

you just found the lower bound

you still need to find the upper bound though

[fluid bough]

@pale perch this?

[pale perch]

this is a good visualisation, yeah

what must L be less than for it to round to 7.36

[fluid bough]

Now I don’t know what to put tho

[pale perch]

the lower bound is 7.355 (first box)

[pale perch]

the upper bound (second box) is the answer to this question

[fluid bough]

Ohh!

Okay

But

It’s meant to be 2dps?

[pale perch]

no, L rounds to 7.36 to 2dp
the actual value of L can fall in a range

if L was 7.36444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444.....
it would still round to 7.36

[fluid bough]

So why do I have to put 7.355?

That’s not 2 decimal places?

[pale perch]

it doesnt have to be!

it just is the lowest value L could be such that it WILL round to 7.36 when rounded to 2dp

[fluid bough]

I don’t understand

Ohhh!

I get it

Yess I got it! @pale perch

And btw can u help me with another? If you’re free?

I put 47.13

[pale perch]

mathswatch, what a throwback

[twin meteor]

Result:

110.5682368

[pale perch]

oops

,calc 3.142*(11.2)^2

[twin meteor]

Result:

394.13248

[pale perch]

howd you get this number?

[fluid bough]

So

[fluid bough]

haha lol

11.2 x 3.142 x 2 = pi r ^2

Answer = 47.13

Wait no

70.3808

2dp

7.38

[pale perch]

2pi * r is the circumference

pi r^2 is the area

[fluid bough]

Yeah

That’s what I did

It’s still incorrect

.close

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[narrow shadow]

I have task that goes: if f(x) = arcsin x, show that f(x) < f'(x)x for 0<x<1. I know that f'(x) = 1/sqrt(1-x^2). I've never come across a similar task, so I'm uncertain as to how I'm supposed to approach it.

[narrow shadow]

Initially I was thinking of maybe taking the integral of f(x) and f'(x)x over the interval and see if the absolute value of the integral is larger than the other, I know that f(x) is strictly growing so I guess it would work for that, but I haven't checked if the same applies to f'(x)x

[vocal sleet]

@narrow shadow Has your question been resolved?

[narrow shadow]

I tried the integration method and I did find that the area under f(x) from 0 to 1 is smaller than that of f'(x)x, so I guess my question just becomes, would that count as showing f(x) < f'(x)x for 0<x<1?

[vocal sleet]

@narrow shadow Has your question been resolved?

[narrow shadow]

<@&286206848099549185> I would just like to know if my solution would be valid? if not maybe another direction I can go in to try and solve it?

[vocal sleet]

@narrow shadow Has your question been resolved?

[brittle girder]

This isn't enough, you would need to show this for the integral from 0 to t, for all t in (0,1)

I'd say take the sine of both sides (justify this) and show the inequality holds

[narrow shadow]

I'll try this, but I'm also wondering why wouldn't my method be enough?

[brittle girder]

As a counterwxample, Suppose you have f(x) = 1 and g(x)= x^2, defined on [0,2]. Then integral of g on [0,2] is bigger than that of f(x), but there are points where f(x) > g(x)

[narrow shadow]

Yeah, I see your point

[brittle girder]

The integrals tells you about thr total area, but doesn't tell you if one function is bigger than the other at a single point. That is why you'd have to look at every possible integral in the domain

Loosely speaking,Integrals don't give info about what happens to a function locally

[narrow shadow]

If I use the method you suggested, taking sin of both sides, I obviously get x on one side, but it's unclear to me how I'd use this to prove that for a given interval, that one side should be larger than the other for that interval

[brittle girder]

Well first you need to justify that taking sine of both sides preserves the inequali

Hint:||monotonicity||

[narrow shadow]

Might be wrong here, but wouldn't the inequality be preserved as long as we're not multiplying/dividing by negatives?

[brittle girder]

Hmm, I dont think so?

A < b doesn't imply sin(a) < sin(b)

Eg a=0, b=3pi/4

[narrow shadow]

Well in that case I'm quite uncertain. When x approaches 1, f'(x)x goes to infinity

[brittle girder]

I Don't think that should be an issue

,w sin(x/(1-sqrt(x^2)))

[twin meteor]

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Requested by catGPT#9757

[brittle girder]

Hmm

[narrow shadow]

That would lead to issues with the inequality, as it would shift given an x close enough to 1 or -1

[brittle girder]

Yeah that'll be an issue

Oh wait I'm stupid

I think you can take derivative of both sides of the original

Show f(0) = xf'(x) at 0 (they're both 0)

And show f'(x) < (xf'(x))' for all x in [0,1)

[narrow shadow]

ooooo

you just prove that one function strictly grows faster than the other given the same start point?

[brittle girder]

Yeah

[narrow shadow]

damn

wish I would've realised that sooner 🤦‍♂️

But thank you a lot for the help!

[brittle girder]

[narrow shadow]

.close

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[velvet cove]

can someone explain this

[pale perch]

the ball experiences the 24N force, it doesnt experience the -24N force
the cue is the opposite

24N acts on the ball
-24N acts on the cue

[hidden niche]

you would need to hit the ball again with the cue for the ball to stay stationary

[velvet cove]

why tho\

how do u know which is negative

[pale perch]

you choose which one is negative

pick a direction to be positive

[velvet cove]

oh

then why doesnt the cue go backward

[vocal sleet]

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[upper zodiac]

why sin and cos? couldn't they both just be sin?

[vocal sleet]

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[tall plume]

What did I do incorrectly?

[tall plume]

I’ve rechecked and i don’t get how it’s incorrect

[flat whale]

,rotate

[flat whale]

2x times 1 = 2x, not 3x

[tall plume]

omg

that’s such a dumb mistake

tysm!!

.close

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[timber hemlock]

.reopen

[vocal sleet]

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[clever nova]

Oh mb I thought it was unused

.close

idkwhy it has my name

[pale perch]

already closed, this one was yours though

[clever nova]

oh

[pale perch]

dw about it

[clever nova]

gotcha

[vocal sleet]

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[solemn cape]

What do you now about parallel lines (functions) ?

There’s a thing to now if two graphs are parallel

First, write your function as $y’=ax+b$ where $a,b\in\mathbb{R}$ what can you say about the value of $b$ ?

[twin meteor]

deus ex machina

[solemn cape]

Don’t worry

What do you know about the line you want to graph ?

You know that the line passes trough the point (0;6)

« Line that contains C »

[vocal sleet]

@vast shale Has your question been resolved?

[edgy gulch]

if two linear equations are parallel what must be the same

theres only two things in a linear equation

slope and y intercept

which one do you think needs to be the same for two lines to be parallel

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@vast shale Has your question been resolved?

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[dreamy marsh]

If there's two people on frictionless ice and they both push each other. Person A(55kg) pushes person B(60kg) with 680N of force as person pushes person A with 520N of force. What'd the acceleration of them both?

[dreamy marsh]

I don't understand what force I'd use here. Because if I look at person A, they push with 680N of force but they're also going to experience that force back. Do I do 680-520 or something?

[dark kiln]

my intuition is that there's no back force

like in addition to 520

[dreamy marsh]

What what about every reaction has a equal and opposite reaction? If person A pushes 680N against a wall they'd experience 680N backwards, no?

[dark kiln]

yes, and it makes sense because you don't move as a result

and here it's different

i'm thinking, i don't know yet

i think it would work out if you draw all 4 forces and then sorta treat both people as one thing, single mass and acceleration

but generally i'm completely confused and i give up

no, that's nonsense, you would simply get 0 net force sorry

[vocal sleet]

@dreamy marsh Has your question been resolved?

[dark kiln]

@dreamy marshwhen they ask acceleration of them both, does it mean it's one number, not 2?

or you don't know?

@dreamy marshit would make most sense if you add both numbers

so there's same force, and 2 different accelerations

[dreamy marsh]

No, they ask for acceleration for both seperatly so two different numbers

[dark kiln]

right

the point is that these two forces cannot be all there is, there's is no way they wouldn't be equal

person A is pushed back with 680N and also 520N

then the net force is the same between them

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[grand frigate]

If I have line
y = x+2
and a parabola
-x^2 + 3
How do I find the area enclosed between those two equations and the x axis?

[grand frigate]

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