help-17

[whole forge]

Looking at where the arrows go

[jagged cargo]

i'm surprised you got 1a and 1b wrong, while 2a is correct

they are quite similar

[atomic jasper]

did you look at the correct points?

[whole forge]

Not sure honestly.

[atomic jasper]

theyre all about x=1

leftsided limit, rightsided limit, limit

[whole forge]

Correct me if I am wrong on this one

Answers should be
A. -2
B. 2
C. 2
And D. Is still 2

[atomic jasper]

still wrong

we are looking at this right?

[whole forge]

Yes

[atomic jasper]

so what happens to the left limit?

[whole forge]

It moves up?

[atomic jasper]

this is the left limit

what does this go towards?

[whole forge]

Towards the right

Since I am looking at the arrow as well

[atomic jasper]

i meant, which point

which y value

[whole forge]

1

[atomic jasper]

exactly

[whole forge]

Oh

So let me get this straight

[atomic jasper]

$\lim_{x\to 1^-}f(x)=1$

[twin meteor]

Bonk

[whole forge]

The limits are the y right essentially?

I feel like I am finally recalling my grade 11 stock knowledge

[atomic jasper]

its what the function approaches

[whole forge]

Or maybe I’m wrong

[atomic jasper]

if you were to put in values of x closer and closer to 1

you can see that i slowly goes towards 1

[whole forge]

Which is the other hollow circle on the left?

So you mean to say it just transfers to the left hollow circle since they are the same shape and design?
And approaching from the left so there?

[atomic jasper]

im not sure what you mean by this

[whole forge]

Sorry that was confusing

So let me get the rules straight

[atomic jasper]

but like question 2a and 2b you got correct

[whole forge]

When the limit is negative it is approaching from the left and is going towards the right side of the graph

The positive limits works in an opposite way

[atomic jasper]

dont think of it as positive and negative limits

[atomic jasper]

its a left or right sided limit

or approaching from below or above

[whole forge]

So this is why it is 1?

I am using my visualization

[atomic jasper]

how did you get 2a and 2b?

[whole forge]

Visualization.

Lucky guess you could say in a way

I was just observing the circles and arrows

Ah not lucky since 2a is wrong

[atomic jasper]

no, 2a and 2b are correct

[whole forge]

Oh I see

[atomic jasper]

so im wondering how you got that correct but not 1a and 1b

[whole forge]

Lucky guess

I understand it more now

I watched this video

[atomic jasper]

the left sided limit at x->1 of f(x) is 1

and the right limit is the same

the general limit is also 1

and the function value f(1)=2

[whole forge]

https://youtu.be/rCQALMXhAhI?si=NgwPv4pWB_Pp7SBv

[atomic jasper]

if this helps, great

[whole forge]

So a and b is 1 right?

[atomic jasper]

yup

and c?

[whole forge]

Are they the same cause I just go back and forth or is it a different reason?

[atomic jasper]

they are the same because if you go from the left and right you approach the same point

there is no "jump"

like if you look at -1, there is a jump

[whole forge]

Yes

[atomic jasper]

$\lim_{x\to -1^-}f(x)=3\neq 1\lim_{x\to -1^+}f(x)$

[twin meteor]

Bonk

[whole forge]

Hmm

C is also 1?

Or 2 since there is a marble on top?

[atomic jasper]

$\lim_{x\to x_0}f(x)=L$ if and only if $\lim_{x\to x_0^-}f(x)=\lim_{x\to x_0^+}f(x)=L$

[whole forge]

I don’t get it

[twin meteor]

Bonk

[atomic jasper]

which part

but yes, c=1

because a and b are equal and are 1

[whole forge]

Oh

Ohhh

[whole forge]

This

I am coming back to my grade 11 math’s with knowledge I forgot from grade 11 calculus

[atomic jasper]

😄

[whole forge]

If you ask me why the reason as to why I am trying this

Well it is because tomorrow we have a test in physics and there are bonus grade 11 calculus quesitons and if we get them right

Big exam bonus scores

So I am trying to get all of these right

So I can figure out number 1 and number 2 now

Hmmm

What about number 3 and 4.
You guys didn’t mention that part

Did I get it right?

[atomic jasper]

lemme check

only 3d is wrong

(for question 3)

[whole forge]

For question 4?

[atomic jasper]

4a,b,c are wrong, only 4d is correct

[whole forge]

Ohhh

So 1D is correct right?

D= 2

[whole forge]

Btw am I allowed to like ping you or other helpers here when replying or nah?

[atomic jasper]

yeah idm

but generally i am quite active in the channels

so i will see it either way

[atomic jasper]

yes

[whole forge]

So 2D is = 5

?

[atomic jasper]

yup

[whole forge]

Ohhh

2C is also -6 since it is similar to number 1

[atomic jasper]

wrong point

you are looking at x=-8

but it asks for x=8

i know its a bit confusing

i got ocnfused a few times aswell haha

[whole forge]

This is satisfying

Finally understanding it

So 3 d is = 1.0?

[atomic jasper]

no... its -1.0

notice how the dot below is filled in and the dot above is not

[whole forge]

Why downwards instead of up?

[whole forge]

Ohh

So that’s why

So if it was in opposite position.
The answers would be different?

[atomic jasper]

wdym opposite position?

if the top dot would be filled in then f(0.5)=1.0

is that what youre asking?

[whole forge]

If the top dot was filled and the bottom dot wasn’t

[whole forge]

Would jt be opposite or a same answer?

[atomic jasper]

it would be opposite

since it changes function value

[whole forge]

I see now

So my only mistakes left are 4 a,b and c

Hmmm

Spoil me a bit on the answers on 4 a, b and c.

[atomic jasper]

,rccw

[atomic jasper]

4a,b,c = 4

[whole forge]

Wait

It’s literally just that?

[atomic jasper]

yup

[whole forge]

Damn

[atomic jasper]

there is only a perforation

so the limit exists

and if the limit exists, its equal to the left and right limit

[whole forge]

Any tips or advice on solving this kind of problem?

If I ever need to backtrack quickly in the future.

[atomic jasper]

imagine there is a bead on the line

and you slowly push that bead into the position

*but never actually reaching it!!!

and you do that from the left and from the right

if you get the same answer, the limit is that answer

[whole forge]

I see

So allow me to summarize everything up and verify it all to you

So this is revised and everything I have answered so far

I am much more confident in answering part 2 because I used a calculator

Just had to figure out test 1

@bonk

[whole forge]

So what do ya think so far?

Everything correct?

[atomic jasper]

2c is wrong

[whole forge]

Huh?

You told me 2c is negative 8

[atomic jasper]

the function doesnt even go down to -8

[whole forge]

Ohhh

[atomic jasper]

x->8

not the y

very different

[whole forge]

So it is 2 then?

[atomic jasper]

$\lim_{x\to 8}g(x)=5$

[twin meteor]

Bonk

[whole forge]

Ohhh

[atomic jasper]

where did i say that?

[whole forge]

So like a marble

[whole forge]

Sorry I mistook this

[whole forge]

Well your advice does work

[atomic jasper]

3a is wrong

[whole forge]

I guess the answer to all this is practice

[atomic jasper]

4 is correct

[atomic jasper]

well, thats most of maths tbh

[whole forge]

1.0?

No wait

-1.0?

[atomic jasper]

10 is wrong

[atomic jasper]

yes

[atomic jasper]

is the 4x_4 at 11 supposed to be a - or a +?

[whole forge]

You’re asking limits right?

Or the number itself

If the number itself

Well it is a _

If there are exponents

There are substituents I think?

Since that’s what I remember they are called at in chemistry

[atomic jasper]

11 is correct

i think its a typo

i assumed it to be a - just like you did

[whole forge]

What’s the answer on 10?

[atomic jasper]

,w graph 1/x^3

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by .bonk__

[atomic jasper]

ah wait nvm nvm

youre right

it DNE

i thought it was simply the RH limit

[whole forge]

Aight aight

So no more corrections at this point?

[atomic jasper]

nope

[whole forge]

Okay

[whole forge]

Thank you very much kind person

[atomic jasper]

np

👍

[vocal sleet]

@whole forge Has your question been resolved?

[vocal sleet]

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[delicate nest]

Can someone help me solve this? I calculated it and it gave me the answer if the $9213.04 was the interest and not total

[delicate nest]

Using simple interest

Idk what I’m doing wrong 😭

[grizzled vault]

what answer should you be getting?

[delicate nest]

9000

I got

[grizzled vault]

is this rounded or something?

the 9000

[delicate nest]

I think so to the nearest hundredth

[grizzled vault]

well i don't do this kind of finance math 😭

[delicate nest]

In F there’s the answer

[grizzled vault]

,rotate

[grizzled vault]

,calc 9213.04/1.02367

[twin meteor]

Result:

9000.0097687731

[delicate nest]

But what I don’t get is there is 2 P’s but the answer ditches one

[grizzled vault]

oh i see

i got the same answer

[grizzled vault]

Prt is only the interest

you need to add it to the initial principle

[delicate nest]

I’m so lost

😢

So when I calculate the interest I don’t have to include the p?

[grizzled vault]

it's like if you invest 1000

the "extra" money is the interest you get

let's say that's 100

what's your new "total"

1000 + 100 right?

[delicate nest]

Yes

[grizzled vault]

so I = Prt is the money you get from interest

the "extra money"

they want the new total

so P + Prt

[delicate nest]

But how can I calculate the interest if I don’t know what the principle is?

[grizzled vault]

btw brb

[vocal sleet]

@delicate nest Has your question been resolved?

[grizzled vault]

u can't

but you don't have to calculate it separately, do you?

they want you to calculate the "new total"

new total = old total + interest

old total = P

interest = Prt

new total = P + Prt

you know new total 9213.04, r = 4.8%, and t = 180/365 years

so 9213.04 = P(1 + rt) = P(1 + (4.8/100) * (180/365)) and yeah it's basic algebra to solve for P

[vocal sleet]

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[vocal sleet]

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[sleek birch]

on the cartesian plane we can never form an equilateral triangle with 3 points with integer co ordinates, but can we form an equilteral triangle with 2 integer co ordinates and the 3rd abitrarily close to an integer?

[sleek birch]

like can we form an equilateral triangle w 2 integer co ord vertices and the 3rd infinitely close to an integer?

i was fooling around in desmos and found some examples that are reasonable close to this but is there a way to rigorously prove this?

[grizzled vault]

wdym arbitrarily close to an integer??

[sleek birch]

like

if i give u some very small number

like 0.0000000000000000001

is it possible to find an equilateral triangle w 2 integer co ordinates and a third this close to an integer

i got some examples for 0.1 0.001 by brute forcing

<@&286206848099549185>

[delicate nest]

Is it 1+??

[grizzled vault]

P + Prt = P(1 + rt)

btw best to start a new channel

joemama is using this channel now

[delicate nest]

Can you help me in 11?

[vocal sleet]

@sleek birch Has your question been resolved?

[vocal sleet]

@sleek birch Has your question been resolved?

[vocal sleet]

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[severe stone]

I tried to use LH and the definition of the derative but yet it is wrong

[flat whale]

definition of derivative would be easiest yes

[severe stone]

they say f(x)=sinx²

that is wrong right?

[flat whale]

nope

[severe stone]

HUH

[flat whale]

do you know the limit definition of derivative at a point?

[severe stone]

lim x->a (f(x)-f(a))/(x-a) = f'(a)?

[flat whale]

yes

[flat whale]

what happens when a=0 for this f(x)

oh i see

[severe stone]

f(0)=sin(9)

[flat whale]

yes f(x) should be sin[(3+x)^2]

[severe stone]

aaah okay so the answer would be 2 sin(3)cos(3)?

[hushed pewter]

How did LH not work?

!show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[flat whale]

,w lim x to 0 (sin((x+3)^2) - sin(9))/x

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[flat whale]

,calc 6cos(9)

[twin meteor]

Result:

-5.4667815713081

[severe stone]

damn i got it wrong

[flat whale]

,calc 2sin(3)cos(3)

[twin meteor]

Result:

-0.27941549819893

[flat whale]

messed up chain rule somewhere i think

[severe stone]

omw

[flat whale]

mistake here

write $\sin((x+3)^2) = \sin(g(x))$, $g(x) = (x+3)^2$ and use chain rule to find the derivative

[twin meteor]

riemann

[severe stone]

oh so it would be f'(x)= cos((x+3)²) *2(x+3)

how do i know if it is sin((x+3)²) and when it is (sin(x+3))²?

cuz now i could have guessed it because of the sin(9) which is sin(3²) but what if i couldnt see that

[flat whale]

$\sin^2(3+x) = (\sin(3+x))^2$

[twin meteor]

riemann

[flat whale]

it's quite confusing unfortunately

[severe stone]

ooooh thank you did not know that

[severe stone]

ye true especially with ^-1 and stuff 💀

[silk osprey]

i mean it is definitely true that you can interpret this as f’(0) for f(x) = sin(3+x)^2 but it may be easier to see it as f’(3) for sin(x^2)

[flat whale]

yea that's even worse

[silk osprey]

since it follows directly from the definition of derivative

[flat whale]

ah shit this is easier

[severe stone]

Riemann learning from someone else 👀

[silk osprey]

his way is correct

[flat whale]

instead a=3 would make the original f(x) = sin(x^2) work

[severe stone]

Ye true but it is not the fastest

but wouldnt i need to substitute first

x+3=u ?

[silk osprey]

no

x is really h here

[severe stone]

x=0 -> u=3?

[flat whale]

you still need to do chain rule either way

[silk osprey]

in definition of derivative

you still get 6cos(9)

[severe stone]

oooh with the h definition method

ye true

damn

math so fun 💀

[silk osprey]

2xcos(x^2) vs 2(x+3)cos(x+3)^2

[severe stone]

aaah thats why they used sinx²

[silk osprey]

$f’(\mathcolor{red}{x}) = \lim_{h \to 0} \frac{f(\mathcolor{red}{x}+h) - f(\mathcolor{red}{x})}{h}$

x = 3

that’s how i see it

[severe stone]

aaah ye

smart

One day i will see all that stuff too

[flat whale]

textbf x

[silk osprey]

to put emphasis

[flat whale]

that's quite the dedication

[silk osprey]

no vectors lol

[severe stone]

After this subject i will get vector calculus

i am definitely cooked :(

[silk osprey]

nah

[severe stone]

Matrix wasnt great

[silk osprey]

oh wait i can use colors right

[severe stone]

what is the difference between an absolute maximum and local maximum? Local is at the end of the domain?

[silk osprey]

$\mathcolor{red}{x}$

[twin meteor]

knief

[silk osprey]

ohhh

i’ll do that instead

[severe stone]

aura

[twin meteor]

knief

[silk osprey]

much better

[severe stone]

okay imma close it, thank you guys :)

.close

[vocal sleet]

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[flat whale]

red 👀

i personally like orange

$f’(\mathcolor{orange}{x}) = \lim_{h \to 0} \frac{f(\mathcolor{green}{x}+h) - f(\mathcolor{blue}{x})}{h}$

[twin meteor]

riemann

[flat whale]

that blue is bad

[vocal sleet]

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[halcyon thorn]

Each natural number m < 100 is assigned a natural number F(m), also less than 100. The sequence a_ 1 = 1, a _(k+1) = F(a_k) is constructed. Prove that there is a number n < 100 for which a_n = a_2n.

[halcyon thorn]

so a_k = F(F(......F(1))) k-1 times

.close

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[light geode]

I’m dealing with this question. I need help understanding the concept/ checking if I’m right

What I’ve understood is that with sigma notations you have to take the lower limit and add anything between that and the upper limit( as in if 1 is lower limit and 3 is upper, I’d also have to include 2)

So what I need help understanding is would I go about this question like this

1^1/1 + 2^1/2+ 3^1/3

Or is k is supposed to stay 1 at all times?

[gaunt sparrow]

The k=1 at the bottom of the summation symbol tells you two things.

1. the variable that changes across iterations is k
2. the first iteration starts at 1.

Every iteration, this variable is incremented by 1 (add 1 to it).

Then the number at the top tells you when you stop.

Then it's just a matter of finding out which of the summand (what's after the summation symbol) is a general expression for the sum you're trying to express.

You can usually spot the pattern just looking at what you've written down already, i.e.
1^1/1 + 2^1/2+ 3^1/3

Which of the expressions do you think fits?

[light geode]

I’d assume A since it’s 1 over k which is the same as a root

[light geode]

So whenever we’re dealing with k I can change it until I’ve reached the upper limit

[gaunt sparrow]

Yes

[light geode]

Appreciate the help

.close

[vocal sleet]

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[light geode]

.reopen

[vocal sleet]

✅

[light geode]

Ok I’m completely lost here.

Which equation do I use

[grizzled vault]

anything u like

[hazy spindle]

I'd say 10, just using intuition and basic equations (420 = n(-3+87)/2 gets you n=10). Idk tho, forgot HS maths

I'd love it if somebody could help me with my sht :d

[hazy spindle]

I'm dumb, its not finished

[grizzled vault]

a_n = a_1 + (n - 1) d which gives you 87 = -3 + (n-1)d

S_n = n/2 (-3 + 87) = 420

and then you have a simultaneous to solve

[dark kiln]

84 is double the average
so 42 is average, so there's 10 terms

[light geode]

It’s 7😭😭

Idk how

I’ve tried every single equation

[grizzled vault]

d is 10

and a = -3

-3 + 10 =7

[dark kiln]

so it goes +90 in 9 steps

[light geode]

I’ll try both again

And see what I get

Ty

[grizzled vault]

,w 87 = -3 + (n-1)d and n/2 (-3 + 87) =420

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by mmmm70177

[grizzled vault]

should get that for values of n and d

[twin meteor]

mmmm7

[light geode]

a_n and n are both different?

Ok I got 7

I was doing 87/2 the whole time 😭

[grizzled vault]

ofc

a_n means the nth term

n means the number of terms

[light geode]

Ohhh damn

Well ty for clarifying

[vocal sleet]

@light geode Has your question been resolved?

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[chilly onyx]

for a functional depending on multiple functions such as $ (A_{\nu}) $

[twin meteor]

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

[chilly onyx]

and $ T_nu $, with the lagrangian

i tried to extremalize this directly without taking the variation separately
v

and i said that we find the corresponding euler lagrange equations :

but here :

it says that you have to take the variation separately

i was wondering if i wrote bullshit or it is equivalent

[vocal sleet]

@chilly onyx Has your question been resolved?

[chilly onyx]

<@&286206848099549185>

[vocal sleet]

@chilly onyx Has your question been resolved?

[vocal sleet]

@chilly onyx Has your question been resolved?

[chilly onyx]

Useless

[flat whale]

!done

[vocal sleet]

If you are done with this channel, please mark your problem as solved by typing .close

[vocal sleet]

@chilly onyx Has your question been resolved?

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[zinc blaze]

can someone help me with this?

[sharp lynx]

,rccw

[zinc blaze]

I just dont know how to do this problem

[sharp lynx]

have you been introduced to the pythagorean theorem?

[zinc blaze]

yes

[sharp lynx]

what does it say?

[zinc blaze]

huh?

[sharp lynx]

what is the pythagorean theorem?

[zinc blaze]

oh a^2+b^2=c^2

[sharp lynx]

and a,b,c are what exactly?

i would appreciate fuller sentences

[zinc blaze]

i dont remember exactly

[sharp lynx]

then you should should go back and take a closer look at the theorem

[zinc blaze]

i know what they are i just dont know how to explain it

[sharp lynx]

then you don't know what they are

[zinc blaze]

you're asking where a b and c go right?

[sharp lynx]

i'm not asking where anything goes, i'm asking what they represent

"a^2 + b^2 = c^2" with no further context is just a plain old equation. We need to know what a,b,c actually are for it to have meaning

(and for you to be able to use it for your question)

[zinc blaze]

i never learnt that, i just learnt the equation and how to solve it

[sharp lynx]

how was the pythagorean theorem presented to you?

is it in a textbook? Its own worksheet? Written notes?

[zinc blaze]

written notes and the teacher just saying how it works

[sharp lynx]

please show me these notes

[zinc blaze]

I learnt this a year ago, so i dont have the notes with me, i just memorized the equation

[sharp lynx]

well that's not good

https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-pythagorean-theorem/v/the-pythagorean-theorem watch this before continuing with your worksheet

[zinc blaze]

i never watch math videos because i never pick up on what they teach, sorry

[sharp lynx]

well that's a shame. Good luck then

[crude surge]

[zinc blaze]

good luck in what???

<@&286206848099549185>

<@&286206848099549185>

[sacred hinge]

Oh these ones

So an and b are essentially legs of a 90 degree triangle….

C=is the hypotenuse

Or longest side

[zinc blaze]

oh ok

[sacred hinge]

You can also think of c

As the diagonal to 90 degree

And I will clarify

[sacred hinge]

On this sheet

The variables are wrong

Well not wrong per say

Pythag thereom is essentially just a general thereom

[zinc blaze]

ok

[sacred hinge]

So for the triangle on the left.

A and b are the legs

Which means they sit

Um sort of like this

Notice how they are on opposite sides of the 90 degree angle

Like the 90 degree angle is in between the a and b

[zinc blaze]

yeah i see that

[sacred hinge]

Where a and b can be anything, just not a negative number

C

As you can see is the length of c is diagonal to the 90 degree

Utilizing this previous knowledge

We can just do

[zinc blaze]

ok

[sacred hinge]

A=5.5 b=3.1

So pythag thereom states

5.5^2+3.1^2=c^2

So first we can solve the a^2+b^2 portion

5.5^2+3.1^2=39.86

So we have 39.86=c^2

Honestly a and b can be either or

Just remember that the 90 degree angle must be in between a and b

[zinc blaze]

6.3=c

[sacred hinge]

Yes

Well approximately

So for your problem

the hypotynse is equal to approximately 6.3

So a=6.3

[zinc blaze]

and then for the second triangle you would do c^2-b^2=a^2?

[sacred hinge]

I mean yes that’s just a way of manipulating the function to solve one of the leg value

So for this problem

5.5^2-2.7^2=22.96

22.96=a^2

[zinc blaze]

and then that is

4.8

[sacred hinge]

22.96^1/2

[zinc blaze]

huh?

[sacred hinge]

4.79

Or 4.8

[zinc blaze]

or yeah

[sacred hinge]

same thing

N^1/2 is the same thing as sqrt n

[zinc blaze]

oh ok

[sacred hinge]

Does this clear up any confusion with using the pythag thereom

[zinc blaze]

yes thank you

[sacred hinge]

No problem

[zinc blaze]

i do have 2 more problems

but they're something different i think

[sacred hinge]

What are they?

[zinc blaze]

they're about complimentary angles

[sacred hinge]

Complimentary =90 degrees

[zinc blaze]

ok

[sacred hinge]

so, notice how AD and EG are perpendicular

Which means you can write it as

[sacred hinge]

Boxes represent 90 degrees

[zinc blaze]

ok

[sacred hinge]

I’m gonna rephrase this

Complementary is the sum of angles that equal 90 degrees

I would focus on the upper quadrant though

[zinc blaze]

i see it

[sacred hinge]

So

[zinc blaze]

each quadrant is 90 degrees

and on the frist and second one

you could figure out the other degree

[sacred hinge]

Yes!

Precisely

So to find the other

[zinc blaze]

the 4th one would obviosuly be 90

[sacred hinge]

You would subtract 90 degrees from the known angle

[zinc blaze]

but the 3rd one

[sacred hinge]

They just vertical angles

Or

[zinc blaze]

ohhh

so then the 3rd one would be 40 and 50 degrees as well

?

[sacred hinge]

Mhm

At least what I can tell

[zinc blaze]

oh ok! thank you!

i think i can do the other one on my own

[sacred hinge]

Oh okay!

Good luck bro

Geometry gets fun

[zinc blaze]

thanks once again

[sacred hinge]

You’ll be able to be really creative

[zinc blaze]

i like geomoerty overall

[zinc blaze]

yes!

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[latent marten]

Is arc length the circumference of the small gear?

Ye it is

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[ionic flax]

special triangles

[kind light]

??

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@ionic flax Has your question been resolved?

[delicate olive]

No

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[latent marten]

might be a really dumb question, but how can X = 280 if the domain is 180<x<270

[latent marten]

i checked the key

[lilac pebble]

yeah that doesn't seem right

[latent marten]

Possibly with reference angles?

or 100% its off

this is practice diploma questions so i doubt its wrong, i think their taken straight from real diplomas

[lilac pebble]

i don't see how reference angles would make it right

well it's always possible that i'm missing something

[latent marten]

Idk how to properly solve it, i got -60degrees lol

i checked the key and got confused

CAST rule applys?

if sin is negative, quadrant 3 and 4

@woeful igloo U know how to solve this?

heres my work its kind of shit lol

ignore the bottom half, thats a diff question

[woeful igloo]

Yea, 240deg should be x-40.

I guess what they meant to say was x-40 should be in 3rd Quadrant

Coz otherwise, there is also a solution in 4th quadrant

[latent marten]

Wdym?

[lilac pebble]

there are no solutions as written

[woeful igloo]

Maybe they edited the question later on as an afterthought and forgot to fix the rest of it

[latent marten]

Like the reference angle is in 3rd quadrant only?

Ahh ok

ill forget that one

this one, i got really close answer but C is throwing me off

I got 122, answer is 121

Not sure why it will be 1npi

since if im not wrong, the only NPV is 0,1

sinx = 1, and cosx = 0

actually nv

I think ,if cosx = X value, x = 0 at 90deg and 180deg right

idk trig like this hurts my brain

[latent marten]

Ye i think thats right, im just confused cuz cosx = 0, cos is = 0 at 0,1 and 0,-1, but inverse cos of 0 = 0, 1 only

Just need clarification on my understanding from someone

[vocal sleet]

@latent marten Has your question been resolved?

[stone gazelle]

From cost/(sint - 1) = … I would move sint - 1 over to the other side

[vocal sleet]

@latent marten Has your question been resolved?

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[hot gust]

Prove or disprove: if lim(f(x))²=1 in x0 then lim f(x) = -1 or lim f(x) = 1 in x0?

[hot gust]

I did the following : we get lim(f(x)f(x)) = lim f(x) * lim f(x) = 1 so it has to be either 1 or -1

[left bobcat]

seems ok to me

[hot gust]

So it's true? 🤔

Also, prove of disprove: lim f(x)² = 1 in x0, so there's always a neighborhood around x0 such that for every x in that neighborhood, f(x) is not equal to 0

1/2 🤔

[oak magnet]

Nvm

[hot gust]

It's times haha

[signal pendant]

The limit need not exist at all

[regal bane]

This is assuming lim f(x) exists

I am sniped

[hot gust]

What's a counterexample?

[oak magnet]

You both are messager xd

[signal pendant]

Consider the function that is -1 for all x<=0 and 1 for all x >0. Limit doesnt exist at 0, but lim f(x)^2=1 at 0

[hot gust]

Got it

[hot gust]

So this is also false?

Oh wait

I think it's still true

[oak magnet]

It would be true without the lim for sure imo

[signal pendant]

That is indeed true, since for any eps>0, there is a neighbourhood around x0 where |f(x)^2 -1| < eps for all x in neighbourhood, so f(x) is either in (1-sqrt(eps), 1+sqrt(eps)) or (-1-sqrt(eps),-1+sqrt(eps)) for all x in neughbourhood

Taking eps =1/2 we get such a neighbourhood where f does not become 0 anywhere

[hot gust]

Got it. Thanks a lot

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[honest light]

I need help solving when a balloon becomes empty. Its volume is 5000cm^3 and the speed it decreases at can be solved with the formula (20-0.01x) cm^3/s

[mental egret]

the answer is in terms of "x" ?

nvm lol

[honest light]

its the time they want i think

[mental egret]

!ss

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Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mental egret]

A pic of Q would help :)

[honest light]

no pic just text in a foreign language and i gave the relevant info volume is 5000cm^3 it is leaking according to this formula (20-0.01x) cm^3/s at then the question is how long before the balloon is empty?

[mental egret]

So I'm assuming the time is represented by variable "x"

So that your equation looks something like; $-\dv{V}{x} = (20 - 0.01x)$

[honest light]

I think so since we have a speed which it decreases and a volume

[twin meteor]

Arya

[mental egret]

can you solve it? you just need to integrate both sides, w.r.t. dx

[honest light]

wait so its like -5000x = (20x-0.01* x^2/2

[mental egret]

...

Let's do this step by step pls?

what do you get on integrating this?

0 ≤ x ≤ t

[honest light]

idk how the way you written it

you said both sides but is not only one sided made to an intregraal?

do you mean both sides should be turned into primatives or just one side

[mental egret]

nope, you have to integrate both sides right? since there's an equality sign, if you do any operation on the right, it should also be done on the left

It'd be better to show you what I mean

$- [V(x)]_0^t = \left[20x - 0.01 \cdot \frac{x^2}{2}\right]_0^t$

[twin meteor]

Arya

[mental egret]

did you get it?

[honest light]

so you turn both sides into primative functions within those brackets for intregrals?

[mental egret]

that is how seperable differential equations are solved ✅
not to mention, if you do an operation on one side of the equality, you MUST do it on the other side as well

[honest light]

so the primative of the volume would be 5000x?

or what is this part or is uit just refering to the decrease function as a whole

[mental egret]

this is me writing the "rate of decrease of Volume" [hence, the -ve sign]

this is what we get after integrating on both sides

Now, we put the boundary conditions, at time "t" the volume, as a function of time is V(t), and at time t = 0, the volume initially was 5000 cm^3

So the LHS is basically, -[V(t) - 5000]

and the RHS is 20t - 0.005t²

So your volume function at time "t", would just be V(t) = 0.005t² - 20t + 5000

@honest lightDid you get this?

[honest light]

yeah

[mental egret]

can solve the quadratic to get the time when V(t) = 0

[honest light]

ah I see now why this didn't work. -0.005t^2 +20t-5000=0

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[vital scroll]

help please

[grizzled vault]

yeah you have different options

you can try to invoke $A = \frac 12 ab \sin(C)$

[twin meteor]

mmmm7

[vital scroll]

okay so

which place do I do that for

cos all the options are confusing me

that rule is forrr finding the area i think?

[grizzled vault]

wdym which place?

u could try finding angle Q

[vital scroll]

thats for finding area so first i need to find q then but for finding q won't I need to find p first?

[grizzled vault]

no you can find Q with law of sines

finding Q gives you P for free

[vital scroll]

the sine rule?

[grizzled vault]

yes

[grizzled vault]

and then you have the setup for this

[vital scroll]

okay so uhh

ill do that and come back <3

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[vocal sleet]

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[finite swallow]

<@&268886789983436800>

[mild flower]

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[solemn spear]

this is my reasoning for picking a

why am I worng?

oh wait nvm

[half imp]

o((x+1)^3) doesn't mean (x+1)^3

[solemn spear]

it can tho right?

[half imp]

no

o means strictly asympotically smaller

(x+1)^3 is in O((x+1)^3) but not o((x+1)^3)

in any case the derivative is wrong

you have 3x^2 + 4x + x instead of 3x^2 + 4x + 1

[solemn spear]

ye saw that

[half imp]

if something is o((x+1)^3) as x->-1 then the first 3 derivatives will be 0 at x=-1

[solemn spear]

can you give an example to something that is o((x+1)^3)

[half imp]

(x+1)^4

(x+1)^(10/3)

[solemn spear]

So I can just put any of these and get the right answer?

the exponent has to be bigger than 3 right?

[half imp]

yes

something like sin(x+1)^4 also works

[solemn spear]

oh wait if this is true can't I just ignore it?

[half imp]

sin^2(x+1) - sin(x+1)tan(x+1) works i think?

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

etc

[solemn spear]

.

[half imp]

yes

0 is o((x+1)^3)

but notably

you can obtain the values of the first 3 derivatives by ignoring it specifically

which in this case is sufficient

but note that they could put something in there which is sometimes true and sometimes false depending on what the extra asymptotic function is and in that case you couldn't just take one example

like imagine they put f(0) = 0 as a possibility

this is true when your extra function is 0 but not true if it's something like (x+1)^4

[solemn spear]

I see

anyway I solved it now

thx for the help

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[frail violet]

Is my approach to part a correct

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@frail violet Has your question been resolved?

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@frail violet Has your question been resolved?

[frail violet]

A is the matrix which is given in the defination

or thats what I am calling it atleast

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[buoyant star]

I'm unsure as to how i can solve this problem. I was able to complete another one with a square root in the denominator but this one beats me...

[buoyant star]

direct substituting ofc doesn't work, it leads to 0/0 which is undefined

[gaunt sparrow]

Whenever there are roots mixed with addition/subtraction, think conjugate

[buoyant star]

the x is approaching a -1 btw, it's hard to see in the pic

[buoyant star]

is that like multiplying top and bottom with either? That's what i did for my last question

[gaunt sparrow]

You multiply it by the conjugate of the part with the root

[buoyant star]

Yeah that

[gaunt sparrow]

Here the conjugate of $\sqrt{x^2+8} - 3$ is $\sqrt{x^2+8} + 3$

[twin meteor]

Azyrashacorki

[buoyant star]

Right. So i multiply x+1 with that?

I wrote it down but im not really sure how i'd continue from there

im thinking the conjugate would cancel out with the original so i'd be left with sqrt x^2+8 +3/x+1?

[gaunt sparrow]

You multiply the numerator and denominator by the conjugate

So that you effectively multiply by 1

Then the numerator is a difference of squares

The denominator is a bit messier, but don't expand it yet since it should cancel

[buoyant star]

is this not right

That's what i tried doing

[buoyant star]

did i do this right? ^

<@&286206848099549185>

[neat rose]

,rot

.rotate

[buoyant star]

here i'll just send it rotated

[neat rose]

$= \frac{\sqrt{x^2+8}-3}{x+1}\cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}$ $$= \frac{x^2-1}{x+1} \cdot \frac{1}{\sqrt{x^2+8}+3}$$

[buoyant star]

huh that's different from what i was told to do earlier

[neat rose]

let me simplify it for you

[twin meteor]

Goëtia

[neat rose]

Better @buoyant star ?

[buoyant star]

Sure but now i'm just confused as to how you got that end result. where did x^2-1 come from?

[neat rose]

multiplication of the conjugates

[buoyant star]

i mean yeah but how did you get there? What was multiplied?

[neat rose]

the numerators

[buoyant star]

yes but where did -1 come from? I'd get is -3 and +3 got canceled out but i have no clue what you did

[neat rose]

bro what?

(a-b)(a+b) = ?

[buoyant star]

what? hold on give me a min

wait how do you even multiply when the numbers are in roots?

I'm not getting any of this

[neat rose]

$(\sqrt{x^2+8}-3)\cdot (\sqrt{x^2+8}+3) = \sqrt{x^2+8}\cdot \sqrt{x^2+8} + 3\sqrt{x^2+8}-3\sqrt{x^2+8}-3\cdot 3$

[twin meteor]

Goëtia

[buoyant star]

Okay i see now

[buoyant star]

so then where do we go from here?

[neat rose]

you simplify then thing on the left and plug -1 u get -1/3

[buoyant star]

Oh i think i got it

Thank you

[neat rose]

np

[buoyant star]

.close

[obsidian quest]

Can someone solve 7A I don’t get how they get 1/2

<@&286206848099549185>

[outer warren]

1/2 isn't the solution to 7A

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[halcyon sluice]

Let $C_0(X)$ be the space of all continuous functions that tend to $0$ at infinity. It is claimed this space coincides with the space of all continuous, bounded functions $C_b(X)$ when $X$ is compact. I don't see why.

[twin meteor]

psie

[rough patrol]

So let me understand this

$C$ is a function from a space $X$ to $\mathbb{R}$

[twin meteor]

Arnavutköy

[rough patrol]

what is "infinity" then?

If the domain was the number line this would be obvious

[halcyon sluice]

exactly, this I don't understand either. If X=[0,1], we could have f(x)=x+1. Does it tend to 0 "at infinity"?

[rough patrol]

no, i just don't think this original statement is well-defined

[halcyon sluice]

it's what's written in my book 😔

[rough patrol]

this seems to be analysis. which textbook?

[halcyon sluice]

well, this one

[rough patrol]

i am almost certain that some other properties of X were given beforehand

[cobalt crypt]

is X a subset of R^n?

[halcyon sluice]

well, first it was stated that $X$ is a separable locally compact metric space, for which $f\in C_0(X)$ iff for every $\epsilon>0$, we can find a compact subset $K$ of $X$ such that $|f(x)|<\epsilon$ for every $x\in X\setminus K$. Clearly, $C_0(X)\subset C_b(X)$, and then they claim that if $X$ is compact, then $C_0(X)= C_b(X)$.

[cobalt crypt]

E is X?

[halcyon sluice]

yes, sorry 🙂

will edit

[twin meteor]

psie

[cobalt crypt]

okay so that's what "tend to 0 at infinity" means in this context

[halcyon sluice]

yes

[cobalt crypt]

but you can see that taking K = X here works always when X is compact

[halcyon sluice]

ahh yes, indeed

[cobalt crypt]

essentially compact sets have no points that go off to infinity

so the tending to 0 condition is null

[halcyon sluice]

what does this mean? 🙂

[cobalt crypt]

it's not a very precise statement

but say for example compact subsets of R^n are always bounded

there are no points that are arbitrarily far from others

compact metric spaces are also bounded

but compact is a bit stronger than bounded

[halcyon sluice]

ok 👍

[cobalt crypt]

of course, continuous images of compact things are compact so any function defined on compact X must be bounded

[halcyon sluice]

yeah

[cobalt crypt]

so here, there's no "infinity" in your X in the first place

so you could in fact say that f tends to 0 at infinity

because it is vacuously true

[halcyon sluice]

ah, yes, makes sense

thank you

[cobalt crypt]

👍

[halcyon sluice]

.close

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[thick ocean]

Explain this one please

[gaunt sparrow]

,rccw

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@thick ocean Has your question been resolved?

[thick ocean]

No

[steady plover]

Write the general Equation of a normal and compare it with the given equation

[thick ocean]

Upload image of solution

I don't understand

[steady plover]

do you know the general equation of a normal?

In terms of parametric

[thick ocean]

No

[mental egret]

Slope of tangent to parabola at any point (h, k) is 2a/k
So, for normal, the slope becomes -k/2a. Now y = mx + c is normal => m = -k/2a or k = -2am so h = am²

So the point (am², -2am) satisfies the line y = mx + c

Plug values to get c

[thick ocean]

Slope of tangent to parabola at any point (h,k) is 2a/k. How ?

[mental egret]

y² = 4ax => yy' = 2a => y' = 2a/y

[silver tusk]

@thick ocean did u get the solution

ill solve

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@thick ocean Has your question been resolved?

[silver tusk]

@thick ocean do u want the soln

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[cedar wedge]

yall wgat do i even do

[sudden compass]

what does rate of change mean to you?

[cedar wedge]

erm the rate of change

the ratio

like how much they change when 1 goes 2 to

idk how to put it into words

but

the only part i really need help on is

f(x) = 3^x

and g(x) = 3x

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[calm reef]

guys

[calm reef]

i have a hard time solving this:

i dont know the asnwer to any of these

beside e

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@calm reef Has your question been resolved?

[calm reef]

no

[vocal sleet]

@calm reef Has your question been resolved?

[mild flower]

what have you done so far?

do you know what a limit is?

[calm reef]

i dont understand the negative and positive signs above the numbers in the limit

[mild flower]

ok, have you attempted to learn about those?

those are called left-side and right-side limits

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[sturdy root]

<@&286206848099549185> let's say there are two circles with their centers A1 and A2 and their radii are equal to r1 and r2 ,let n be the distance between the centers of the two circles and n<r1+r2. Is there a way to find how much is the area that is formed when the two circles are overlapping ?

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[modern storm]

I have a cursed way, i got a similar problem a few months ago.

basically:

Calculate the points of intersection between the two circles

Connect those two

Using the points of intersection, create a sector of each circle, considereding the angles for both

From each sector, substract the triangle formed between the points of intersection, and the center of each circle

and the "curved sections" remaining will be the overlap.

Visual Showcase

[sudden compass]

this is the correct way, it isnt cursed

[modern storm]

oh, i sincerely hoped there was a simpler method

good then

originally i used integrals, using the crossing section as the x-axis

bad idea

[sturdy root]

So it is possible to get a formula for this?

[modern storm]

yeah. it wont be pretty tho

[sudden compass]

yes but tbh its better to do it thisw ay

[modern storm]

mostly cause it needs you to calculate the crossing points

which calls for linear (vectorial) algebra-

[sturdy root]

But it can also be done geometrically?

[modern storm]

im trying to come up with some somewhat easy geometric solution for the crossing points

also, if for some reason, one circle "eats" the other, then the area is equal to the smaller circle area

ngl, not findin any lead on that. can we just pretend that we know the intersection points by solving a non-linear system equation?

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[tight shard]

Hello guys, need help with integration. Could anyone help me finding mistake in my solution?

[vast shale]

Post your question!

[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/

[lost barn]

Post the work u did