help-4

[worn sparrow]

but you can use axioms to prove that a rule of inference in a natural deduction system is a valid procedure in a different, axiomatic system, for example

[vale dock]

@supple terrace Has your question been resolved?

[supple terrace]

perfect

thank you so much for the help

appreciated

[vale dock]

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[eternal trench]

Hello, I am stuck on this problem and I need to calculate the quantity conjugated ( I THINK ) in order to solve the problem and find the inverse of a reel ( the whole thing is in French so if you don’t understand tell me so I can translate it properly )

Also I know that square root n+1 + square root n = 1 divided by square root n+1 - square root n

[marble arch]

So question 1?

[eternal trench]

Yes !

[marble arch]

Alright, can you please translate it

[eternal trench]

Show that for any natural number n, the number
Square root {n+1} - square root{n}
has as its reciprocal (inverse) the number
Square root {n+1} + square root {n}.

[marble arch]

Ahhh well apologies but I'm not very good with proofs

[eternal trench]

that’s totally fine! Thank you for answering !!

[marble arch]

I can help you with the second part though

[eternal trench]

I’d love to ! If that’s okay

Would you need me to translate it

[marble arch]

Ahhhh yes

Just to be sure

[eternal trench]

yes ?

[marble arch]

I meant I kinda get what it's asking, but just to be sure can you translate it

[eternal trench]

Yepp I’m actually just copy pasting from my friend who translated it a while before so it should be good

We consider the function f defined on {R} by
f(x) = x^2 - 12x + 20,
and we denote by{C} its representative curve in a Cartesian coordinate system.

A polynomial function is a function made up of sums of positive integer powers of x. The degree of a polynomial function is the highest power of x.
1. Determine the degree of the function f. Deduce the general shape of its representative curve.
2. a. Show that for all real x, f(x) = (x - 6)^2 - 16.
b. Deduce the minimum value of f on {R}.
c. Then sketch the graph of the curve {C}_f, using a scale of your choice.
3. For which value(s) of x does the curve {C}_f intersect the x-axis? Justify your answer with a calculation.

[marble arch]

Ahhh that's probably the second exercise

[eternal trench]

my bad did you mean the second part of the ex1

[marble arch]

For now, yes

[hardy coral]

(1.1) is showing that $\sqrt{n+1} + \sqrt{n} = \frac{1}{\sqrt{n+1} - \sqrt{n}}$ for all $n \in \mathbb{N}$

[rocky lotus]

Nel

[eternal trench]

Thank you !!

[marble arch]

We can solve this too later

[hardy coral]

Do you know what a conjugate is?

[eternal trench]

2. Deduce the value of
   A = \frac{1}{\sqrt{2} - \sqrt{1}} - \frac{1}{\sqrt{3} - \sqrt{2}} + \frac{1}{\sqrt{4} - \sqrt{3}}.

[marble arch]

Alright

[eternal trench]

I looked it up but it was a little confusing since my teacher didn’t explain it to us at all

[marble arch]

Okay for now let's solve (1.1) through Nel's help

[hardy coral]

Do you know the identity (a+b)(a-b) ?

[eternal trench]

yep !

[hardy coral]

Can you make that work on the right-hand side?

[eternal trench]

You mean using the identities to solve it ?

[hardy coral]

Well on either side actually, both work

Yes

[eternal trench]

I’m not really great at it but I understand the concept and I can normally apply it depending on how hard it is

[hardy coral]

The point is to multiply either side by some quantity x/x

[eternal trench]

Right

[hardy coral]

Can you figure out that x?

[eternal trench]

To find it out we need to basically isolate it on one side using the identities or the other method

[hardy coral]

No that's not what I'm talking about

[eternal trench]

oh I got confused so sorry

[hardy coral]

$\sqrt{n+1} + \sqrt{n} = (\sqrt{n+1} + \sqrt{n}) \cdot \frac{X}{X}$ for some quantity $X$

[rocky lotus]

Nel

[hardy coral]

Up to you to figure out what X should be to lead to the original equality

[eternal trench]

ohhh alright thank youuu

And I need to use the conjugate for that or just the identity

[hardy coral]

Use whatever you think works

[eternal trench]

okay ill try that thank you

[marble arch]

Can you ping me once we can start with the other parts

[eternal trench]

alright !! Thank you

[hardy coral]

So did you get it?

[vale dock]

@eternal trench Has your question been resolved?

[marble arch]

Welp, looks like she left

[vale dock]

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[eternal trench]

I just did !

[eternal trench]

Sorry it took some time I had to do an errand I’m so sorry

[weary pilot]

,av aicha

[rocky lotus]

aaicha's Avatar

Click here to view the image.

[weary pilot]

okay

[weary pilot]

hmm

is it solved

do you have other questions

you just started a help channel

on accident ye

[eternal trench]

yes the others questions in the exercice and I’m so sorry abt that I didn’t think it’d close already

[weary pilot]

okayyy

is it the same image

[eternal trench]

yes!

[weary pilot]

what question?

[hardy coral]

It closes if you don't react to the bot

[weary pilot]

ALWAYS ANSWER THE BOT.

damnit you ruined the joke

[eternal trench]

Right I apologize

Loll

[hardy coral]

[eternal trench]

Super messy handwriting I’m sorry

[weary pilot]

,rccw

[weary pilot]

what question hm

[eternal trench]

But this is basically what I came up with using the identity

[eternal trench]

That

One

[weary pilot]

,rccw

[hardy coral]

Sure that works but it would be neater if you went from one side to the other directly

[eternal trench]

It’s in French but the other questions were translated so I could put them back in that one

[weary pilot]

'un petit exercice' bien sur

[eternal trench]

I don’t really understand what you mean sorry

[eternal trench]

RIGHT LOL

[viscid spade]

"Évaluation non surveillée"

[hardy coral]

$\sqrt{n+1} + \sqrt{n} = \ldots = \ldots = \frac{1}{\sqrt{n+1} - \sqrt{n}}$

[rocky lotus]

Nel

[hardy coral]

(or the other way around)

[weary pilot]

yeah who the fuck wrote this

[weary pilot]

dw ill prolly get it

and for the rest we can always kidnap @viscid spade

[viscid spade]

Sure

[weary pilot]

le residentiel francais

[eternal trench]

Oh so basically I should put (n+1)-n in between

[weary pilot]

yeah

[viscid spade]

The text has some issues in the second exercice kek

[weary pilot]

ooh

lemem see

wtf is that

$\mathcal{C}$?

[rocky lotus]

Περσυ

[hardy coral]

Basically yes

[weary pilot]

no

[eternal trench]

In the actual exercice ?

[viscid spade]

Next one

[eternal trench]

ohh and would that make it easier or just more correct

But THNK YOU

[hardy coral]

It just shows directly what you are proving

[eternal trench]

righttt I understand

[hardy coral]

You can write the identity (a+b)(a-b) = a^2-b^2 on the side if you want, to show that you're using it

[viscid spade]

(On dit parfois "multiplier par le conjugué", cela fait apparaitre l'identité que Nel vient de dire)

[eternal trench]

Ohhhh right I’ll do that and I finally understand thank youuu

Bc my brain was like stuck for a hot min when I saw the hw

[weary pilot]

her insistence on speaking english hahaha

[eternal trench]

And then for the second exercice it’s telling me that I’m supposed to know the value of A with what we just did

[eternal trench]

AH

[weary pilot]

also you have nice english well done

[eternal trench]

Merciiiii

[weary pilot]

(its exercise in english btw)

[eternal trench]

ah oui j’ai pas remarqué pardon !! et merci

[hardy coral]

Yes, you can transform each term using the equality you just proved

[eternal trench]

Ah si basically doing what I did for each one of them and then just substacting then adding

[hardy coral]

You don't need to repeat the proof, just use the result

[hardy coral]

$\mathscr{C}$

[rocky lotus]

Nel

[eternal trench]

okay so for example A= sqr2+1-1

wait no since n is supposed to be the same thing

[weary pilot]

ah

[eternal trench]

(0,4+1)- 0,4 =1

Wait but then A= 1

It’s all equal to 1 then

[hardy coral]

No

[eternal trench]

I did something wrong then

I replace (n+1)-n

By

The two square roots subtracted

Which is probably the problem

I think I should have done

1 divided by square root n +1 + sir n divided by the same thing except substrates before sqr n and then idk

No that’s not it

[hardy coral]

$A = \frac{1}{\sqrt{2}-\sqrt{1}} - \frac{1}{\sqrt{3}-\sqrt{2}} + \frac{1}{\sqrt{4}-\sqrt{3}}$

[rocky lotus]

Nel

[hardy coral]

And you know $\sqrt{n+1} + \sqrt{n} = \frac{1}{\sqrt{n+1} - \sqrt{n}}$

[rocky lotus]

Nel

[eternal trench]

right

[hardy coral]

Each of these terms look a lot like the right-hand side, don't you think?

[eternal trench]

ohhh yeahhh

So we’ll divide it

On both side or

Nvm

[hardy coral]

$\frac{1}{\sqrt{2}-\sqrt{1}} = ?$

[rocky lotus]

Nel

[eternal trench]

So it’s sqr 2 + sqr 1 which equals to that

Since sqr 2 is

Sqr n + 1

[hardy coral]

Right

So replace it

Same for the other two terms

Do you see it or not

[eternal trench]

Sorry I’m back

And when I replace them all

I do the calculations like normal

[hardy coral]

Show what you get after replacing all the terms

[eternal trench]

Alright!

Also not well written

[hardy coral]

,rcw

[hardy coral]

Ok that's not how you write these things

[hardy coral]

Write this on one line, then on the next line write the terms after replacement

[eternal trench]

Oh so just the replacement

[hardy coral]

Yes

[eternal trench]

RIGHT I UNDERSTAND

[hardy coral]

(a = b) - (c = d) doesn't mean anything

[eternal trench]

okay ill dit that sorry my brain doesn’t click like it normally does when I’m working on maths

[hardy coral]

Not quite right

[eternal trench]

Is it equivalence instead of equal

<=>

[hardy coral]

No, equal is fine

If you have something like P = Q+R, then -P = -(Q+R) = -Q-R, not -Q+R

[eternal trench]

ohhh so I should remove the + in between and add them before I add a ()

Parenthesis

Ses

[hardy coral]

Not sure what you mean

[eternal trench]

So like (sqr2+sqr1) -(sqr3+sqr2)+(sqr4+sqr3)

[hardy coral]

Yes ok

Now try to simplify

[eternal trench]

Okay

I tried something but I think I’m not allowed to

[hardy coral]

That's definitely not right

You are overthinking it

[eternal trench]

Yeah that’s what I thought but I feel like I just can’t seem to get the logical answer

[hardy coral]

What is -(sqrt(3) + sqrt(2)) ?

[eternal trench]

-0,3

Not sure

[hardy coral]

Don't calculate it

Just simplify

[eternal trench]

To simplify I should just get rid of the square roots of them

[hardy coral]

No

Again you are overthinking it

-(sqrt(3) + sqrt(2)) = ?

[eternal trench]

-(1,7)+(1,4) if it’s not that I don’t think I’m smart enough to find out by myself

[hardy coral]

I said don't calculate it

-(A+B) = ?

[eternal trench]

-AB

[hardy coral]

No

[eternal trench]

I don’t think I can get it by myself

[hardy coral]

Is -(5+7) equal to -35 ?

[eternal trench]

nope

-12

[hardy coral]

Okay, how about -(4+x) ?

[eternal trench]

but to do 4+x it would be 4x no ?

[hardy coral]

How

Why would adding be the same as multiplying

[eternal trench]

it wouldn’t but I don’t know how I can do 4+x unless I did like

[hardy coral]

Ok let's try something else
(6+x) - (1+x) = ?

[eternal trench]

Without calculating

[hardy coral]

Show me step by step how to simplify this

[eternal trench]

okay

6+1 6+x

X +1

X+x

[hardy coral]

Are you trying to do (6+x)(1+x) ??

[eternal trench]

I confused it with factorization but I don’t think I know how to do it with -

[hardy coral]

Sorry but I'm honestly not sure how to teach you basic algebra

It's like parentheses are scaring you for some reason

[eternal trench]

it’s fine thank you so much for your help you’ve been so patient with me

[eternal trench]

LOL

maybe

[hardy coral]

@viscid spade @weary pilot I hope you don't mind stepping in, you might understand her better than I can

[viscid spade]

La deuxième ligne est correcte

La troisième non, reprends à partir de la deuxième

[eternal trench]

D’accord mais le problème c’est que je suis perdu sur ce que je dois faire

Je pense que c’est parce que j’ai pas eu de prof de math pendant 2/3 ans donc j’ai un gros écart

[viscid spade]

Fais en sorte de ne plus avoir de parentheses et après tu peux additionner ce qui doit aller ensemble

[eternal trench]

d’accord

Donc je peut juste faire racine carré de 2 + racine carré de 1 afin de ne plus avoir de parenthèse

[viscid spade]

Quand tu as un + devant une parenthèse tu peux directement les enlever, par contre si il y a un - il faut distribuer le -

Donc -(a+b) = - a - b

[eternal trench]

Ah donc racine carré de 3 devient - racine carré de 3

[viscid spade]

... de 3 tu veux dire

[eternal trench]

oui miss click désolé

[viscid spade]

Oui

Et pareil pour l'autre

En faite ca change le signe de ce qu'il y a dans la parenthèse

[eternal trench]

Er les autres puis ce que c’est des + sa change pas

Ok sa a du sens et c’est super logique

[viscid spade]

Oublies pas le racine de 2 aussi

[eternal trench]

Oui ducoup pour racine carré de 2 et de 2 c’est +, pour celui de et et 2 c’est - et pour 4et 3 c’est +

[viscid spade]

Et ça donne quoi comme résultat ?

[eternal trench]

Donc la je calcule

La racine carré

Et la différence c’est que se sera négatif

Pour les racine carré de 3 et de 2 seulement car le reste est positif

[vale dock]

@eternal trench Has your question been resolved?

[vale dock]

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[open coral]

Based on my work, I know I need to check to see if the second derivative is less than 0. But , the professor has the second derivative of f(mu), but of which function?

[zealous pendant]

Nitpicking here, but when proving something you want to end at it, not start with it and reach equality, the steps you use are reversible though so it doesn’t matter (for i)

[open coral]

This is a past assignment. I am trying to restudy for my comp exam in December.
Unfortunately, I did not show the work. I didn’t prove is in my past work & that’s why my professor wrote the orange part on my last sheet.

I want to go through and prove it now. But I don’t know which function I need to derive.

Do I derive the original pdf? Or do I derive the 1/σsqrt(2pi) ?

[zealous pendant]

To find f’’(x) you take the derivative of/derive the function f’(x) then put in x=mu

(Stick to x until you need to put it in)

[open coral]

OH

So take this derivative & then plug mu in for x

[zealous pendant]

Yeah, otherwise you get f’’(x)=0 which won’t be the answer

[open coral]

Okk, I am going to try to solve for this to make sure I get some function. Then to plug mu in for x.
Thank you so much! That is what I was stuck on

[zealous pendant]

The way I think about it is you’re trying to approach x so if you put in a value for x you cannot approach it anymore

[open coral]

Thank you! I am going to do that & make sure I get what the professor has

[vale dock]

@open coral Has your question been resolved?

[vale dock]

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[gentle meadow]

so when factoring p(x)=(x+2)(4x^2+11x-3), i got p(x)= (x+2)(x+3)(x-(1/4)). But the answer was actually (x+2)(x+3)(4x-1). I was wondering if they were the same

[hardy coral]

No, you are missing a factor of 4

[gentle meadow]

oh i factored 4x^2+11x-3 and got (x+3) and (x-1/4)

i dont know how they got 4x-1

[hardy coral]

You are still missing a factor of 4

(x+3)(x-1/4) can't produce 4x^2

[zealous pendant]

But the quadratic has the first term with a coefficient of 4 and yours doesn’t

[gentle meadow]

oh

[zealous pendant]

Just because you have the roots of a quadratic doesn’t mean you have factored it

You also need the coefficient which will not affect the roots location but changes the quadratic

[hardy coral]

Just because -3 and 1/4 are roots doesn't mean you can factor like (x+3)(x-1/4) directly

[gentle meadow]

so how can i turn it into 4x-1

[hardy coral]

By multiplying by that missing factor of 4

[gentle meadow]

when divided the coefficient of 4x^2 by 1 do i multiply 4 by x

other way around

mb

[zealous pendant]

When you factor you want to make sure that when you expand you get the original thing, basically factorising is expansion in reverse

[gentle meadow]

oh ok

so i got the roots (x+3) and (x-1/4), but i have to multiply 4 by x so i am not missing the factor of 4

[hardy coral]

Solving 4x^2 + 11x - 3 = 0 and x^2 + 11/4 x - 3/4 = 0 will get you the same solutions

[gentle meadow]

ohh alright

[hardy coral]

You need to make sure that the leading coefficient (the one on x^2) is correct

[zealous pendant]

Multiply (x+3)(x-1/4) by 4

[gentle meadow]

this helped me a lot

ty

[zealous pendant]

So the answer is 4(x-3)(x-1/4) but you can move the 4 into the 2nd bracket to avoid fractions

[gentle meadow]

.close

[vale dock]

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[vale dock]

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[oak crane]

which calculator is easier to use? casio or sharp?

[oak crane]

I need to calculate 2/2E03 on casio

the E he did here is exponent

do i just use ^ for exponent?

[spring jackal]

well, E is x 10^, not just ^

but there should be a button called either EXP or x 10^x on a casio

[oak crane]

yeah i have x10^x

how do i use it

[spring jackal]

that would be the same as the E

[oak crane]

so 2/2 x10^x then 3?

[spring jackal]

do 2 / ( 2 x 10^x 3 ) =

[oak crane]

oki lemme try it

[spring jackal]

you may not need the brackets, that might be something to experiment with

[oak crane]

i got 1m

[spring jackal]

._.

[oak crane]

😭

sorry im stupid

i have the engineering notation turned on

lemme send the image wait

[spring jackal]

not at all! calculators enjoy gaslighting you i swear

[oak crane]

this is what I have

[oak crane]

real

sooo did i do this correct

[oak crane]

i should get this output

[spring jackal]

oh, just do 2103, it handles the multiplication for you

[oak crane]

i got 2000

that gives 2k

what 😭

what did i do wrong

[spring jackal]

and the minute i need my own calculator, shes unfindable

try with no brackets?

[oak crane]

I did

still same

[spring jackal]

idt it matters but

try going into menu and picking 2 - Sci

[quick sparrow]

this result is correct

nvm

ignore me

[oak crane]

Holy shit

that worked

thanks a lot

you're the best

.close

[vale dock]

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[vale dock]

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[low trench]

Hi there!!! Sorry if this is a rude way to do this but I’ve got a math test tommorw and its completey killing me here, I’ve got no clue how to figure out a consistent formula for these and in general just don’t see anyway to figure it out💔

[cunning sky]

there isnt

a consistent

formula

😇 learn em one by one

[low trench]

Ah, I see then

Do you have any idea on them then? I’m seriously lost on this paper 😭

[cunning sky]

geometric area
percentage
velocity/speed
delta (for q16 and 17)

[low trench]

Ohh I see!

Would you recommend anyway to learn those/ examples to where I could get practice working them out?

[vale dock]

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[mystic anchor]

If I have $a^2d^2+b^2c^2=3b^2d^2$ is there a way to show that this is a contradiction because of both sides' parity without expanding each variable to $2k+1$ or $2k$

[hazy pivot]

Well you can cancel one of the terms

That being said

!original please cuz that would help us understand what you're trying to do

[vale dock]

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.

[wintry oxide]

Are a b c d integers ?

[mystic anchor]

This is the original question

[normal pond]

Set n=ad, m=bc, then n^2 = 2m^2 and you end up with the same deal as in the proof that sqrt(2) is irrational.

[mystic anchor]

x = a/b, y = c/d
b, d != 0

Then i plugged in and got rid of the fractions

Im trying to prove by contradiction

oops i meant = 3c^2d^2 not cb

[hazy pivot]

Yeah I was just working that out in my head lol

I think you mean b²d² though

On the right

Cuz it would have the denominators

[mystic anchor]

oh yeah

im high ash

[rocky lotus]

trmcburger

[mystic anchor]

is the best way to show the contradiction to expand all the variables into 2k+1 or 2k depending on the case or is there a faster way

[wintry oxide]

Ok so they’re integers in your definition

[mystic anchor]

yeah

[wintry oxide]

Work modulo 3

[hazy pivot]

Yeah that makes it much easier right

[wintry oxide]

Yup

Also you can always manage to have the same denominator for a pair of rationals

So just simplify it to : a^2 + b^2 = 3c^2

And check possibilities modulo 3

[normal pond]

To save a slight amount of work: consider what values a square number can have (mod 3) first.

[vale dock]

@mystic anchor Has your question been resolved?

[vale dock]

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[limpid stirrup]

how might I approach this question? do I change into a surd mabye?

[spring jackal]

can you substitute something to turn this into a quadratic?

[limpid stirrup]

im not sure this is all the information given and I don't see anything to substitue

[spring jackal]

what i mean is, set x = (something new) that you came up with. solve for that (something new), then turn it "back into" x

[limpid stirrup]

okay thank you ill try that

[limpid stirrup]

I decided to do y=sqrtx can I just get a confirmation if im on the right track?

[spring jackal]

yep, thats the correct substitution :)

[limpid stirrup]

okay thank you 🙏

.close

[vale dock]

Channel closed

Closed by @limpid stirrup

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[vale dock]

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[agile zodiac]

I don’t know what I am doing wrong

[vale dock]

Please don't occupy multiple help channels.

[normal hollow]

Which part do u think is wrong?

[agile zodiac]

I have no idea

I’m completely lost

[ashen prawn]

solve for f(g(x)) and g(f(x)) seperately for 2 and 3

then place values

[agile zodiac]

I’m confused because that’s what I thought I did

[wraith heart]

closed your other channel

[normal hollow]

Why is g(3)=0

[agile zodiac]

Wait that’s three not zero rigjt

[normal hollow]

It's supposed to be 3 and not 0,yes

And f(-1)? How is it -5?

[agile zodiac]

Sorry my phone died but I think I got it now

I honestly think I got it wrong the first time I tried and then just panicked 💀

[normal hollow]

Js calculation mistake prolly,ig u understood how this works

[agile zodiac]

Yea I think I need to just stop studying tonight and give myself a break

[vale dock]

@agile zodiac Has your question been resolved?

[vale dock]

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[craggy girder]

!status

[vale dock]

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

[misty skiff]

i dont know where to begin or atleast attempted to but got stuck

[craggy girder]

Do you know calculus, because we will be using it

[misty skiff]

like how far are we talking

[ashen prawn]

as the CNC machine moves with constant acceleration, do try to apply the equations of motion alongthe determinate direction once

[craggy girder]

Basic integration

acceleration depends on velocity?

[ashen prawn]

might be but still quite necessary

[misty skiff]

I am in first year and we're still on derivative but I've taken a look at integration before

[craggy girder]

Oh

Let's start with writing our first eqn

[misty skiff]

actually ive been using the power rule for integration alot recently so im familiar with that

[craggy girder]

Ok that's great

I guess that is all we need

[misty skiff]

sounds good

[craggy girder]

So a = -0.3•v

Do you know that a = dv/dt

[misty skiff]

yes i do

[craggy girder]

Try using that to find v in terms of t

[misty skiff]

so do i just integrate both side to get v(t)?

[craggy girder]

Yup

[misty skiff]

ok thats what i did

[craggy girder]

I recommend pen/paper
Then sending a pic

[misty skiff]

1 sec

[craggy girder]

Incorrect

[misty skiff]

ok what did i do wrong

[craggy girder]

Write a as dv/dt

Now take dt to the other side and v to the other side

Then integrate

[misty skiff]

what difference does that make?

[craggy girder]

v is not independent of t

You can't say that $\int v dt = vt $

You can't say that $\int v dt = vt$

[misty skiff]

i thought it was just a notation

[rocky lotus]

Wumpus Man

[craggy girder]

Oh it's not

$\int v dt = v \int dt = vt$

[rocky lotus]

Wumpus Man

[craggy girder]

This is what you are doing but that means v is independent of t with it is not

So it is incorrect

[misty skiff]

so like this?

[misty wing]

you want to keep all the v stuff on one side and all the t stuff on the other side

are you familiar with separation of variables?

[misty skiff]

so id have to move v to the dv side?

[craggy girder]

Yess

[misty skiff]

ok but then how'd i integrate that

[craggy girder]

You have not done integration of dv/v?

[misty skiff]

not yet atleast

[craggy girder]

I don't think you should attempt this question then

[misty skiff]

it's part of my assignment for dynamics class

[misty wing]

how much calculus do you know?

[misty skiff]

ive no idea why he puts this in tbh

[craggy girder]

.

[misty skiff]

i've done subsitution and integration by parts before but thats just out of curiosity

[misty wing]

are you familiar with $\int \frac 1x dx$

[rocky lotus]

LocalLunatic

[misty skiff]

in school we're just on derivative

[misty skiff]

ln |x|?

[misty wing]

yup

[craggy girder]

Yes

[misty skiff]

so you're saying dv/v is basically 1/x?

[craggy girder]

Nope

dv/v is lnv

[misty skiff]

but it yields the same result no?

wait actually nvm i get it now

its just chain rule lmao

[craggy girder]

Nice

[misty skiff]

and im assuming im in the right direction

[craggy girder]

you are right

but you missed the +c

lol

[misty skiff]

yea i realized

lmao

[craggy girder]

ok nice

send pic of corrected oone

[craggy girder]

That is wrong

you get
lnv = -0.3t + c
Taking exponent

$v = e ^ {-0.3t + c} = e ^ {-0.3t} * e ^ c = v_0*e ^ {-0.3t}$

[misty skiff]

ohhh lmao wait

its cuz i got lazy and put + C in the wrong line

[rocky lotus]

Wumpus Man

[craggy girder]

lol happens

[misty skiff]

wait i just have a question regarding plus c

[craggy girder]

ask

[misty skiff]

because what i've known so far is that + C is jsut an initial start or a shift to the graph

like when taking the integral of acceleration of gravity for example. i'd get -9.81(t) + C

and i can just do a direct substitution to this C if its given

[misty skiff]

in this case what is C?

[craggy girder]

c is the initial start

if you draw a graph of lnv vs t then it is the shift in the graph

[misty skiff]

okay but is it good enough to say that e^c = v0?

or do i have to solve for it

[craggy girder]

yes it is ok to say that

[misty skiff]

ok i think i get it now

[craggy girder]

nice

[misty skiff]

thanks!

[craggy girder]

btw we add the +c because of the following reason

lets say you have
f(x) = 2x + 3
g(x) = 2x + 5

[misty skiff]

oh ya that one ik

[craggy girder]

ohk nice

[misty skiff]

ok wait 1 more question actually

so you're saying to integrate i'd have to make a relation like a = dv/dt then multiply by dt on both side?

[craggy girder]

yes

[misty skiff]

ok

[craggy girder]

we define d/dt as an "operator" but we also treat it as a fraction

[misty skiff]

but how does it work if i just have a function lets say y = 3x^2

[craggy girder]

even i dont know why/how

[misty skiff]

ya but i've always tohught the same with the integral

that the ∫ always comes with dx

like an operator

∫ f(x) dx

[craggy girder]

there is this thing called differential equations

these are eqns in terms of x, y, dy,/dx d^2y/dx^2 ...

they dont have the int sign in them, you have to add it yourself

ill give you an example
y = 3x^2
dy/dx = 6x

this is a diff eqn

it also have no int sign

[misty skiff]

ok

wdym by that?

[craggy girder]

which line?

[misty skiff]

.

[craggy girder]

$\frac{dy}{dx} = 6x$

[rocky lotus]

Wumpus Man

[craggy girder]

see now integration sign

but now i can take dx to the other side

$\dy = 6x\dx$

[misty skiff]

okay

then now u can take the integral?

[craggy girder]

now as this relation is true for all x

(this is very important)

we can take the integral on both sides

[misty skiff]

if you use y' though how would that work

can we just alternate between these notations just like that

[craggy girder]

we can

y' is used because dy/dx is pretty lengthy to write

[misty skiff]

ok it's because my highschool calc teacher told us to not treat it like a fraction

[craggy girder]

ohh well afaik we can

tho it might be a good idea to just be careful while doing that

[misty skiff]

alr alr sounds good

i'll just ask my prof once we get to integration lol

tysm btw

ima go to sleep now

.close

[vale dock]

Channel closed

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

Hi, i'm having trouble with a complex numbers exam question. I need help with (c) (i), which ive attached the worked solutions to. How do you know that ZX is [z-(3+4i)]? I'm having trouble visualising how the solution works

[pure pulsar]

!status

[vale dock]

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

[vocal tusk]

if ur having troubel visualizing, draw it

[sly bone]

c (i). im confused about this step (from the worked solutions). Why is ZX equal to [z-(3+4i)]

[vocal tusk]

do u know how to add vectors graphically

[sly bone]

I just want to understand the solutions cos this is a question i got wrong

[pure pulsar]

Use the triangle inequality

Whats the answer for (a)you got?

[sly bone]

(a) is |z|<=2

(a) and (b) I get but I have trouble with (c)

[pure pulsar]

Thats the hint for c

Using triangle inequality, you can prove it ig

|z - (3+4i)| <= |z + (3+4i)|

[vocal tusk]

bruh

[pure pulsar]

@sly bone back tracking is a good way for proving

[sly bone]

what do you mean back tracking

[pure pulsar]

When you see the 7 try to see if the equation you have can be converted to get 7

[pure pulsar]

If we put z = 2 in this, we get the answer for c @sly bone

I do by trying different things that might help me to make the problem simpler

[sly bone]

ok thanks

[pure pulsar]

Also if you directly put z=2 in the given question of c, you will get the answer

It will be 3 ig

[sly bone]

Thanks for the help @pure pulsar @vocal tusk I figured it out

[sly bone]

I was tweaking

.close

[vale dock]

Channel closed

Closed by @sly bone

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[west schooner]

Need help understanding how this math modeling problem was solved

[shut mirage]

what do you understand about the solution thus far?

[west schooner]

I don't understand where 8x + 8x + 12y = 3600 is from

[shut mirage]

sure

do you have a diagram of what you think the situation looks like?

[west schooner]

nope

[shut mirage]

this is a rough sketch of the situation

[west schooner]

ohh

so x is the 2 ends

and y is the parallel one

but where is A = xy from

[shut mirage]

what's the area of a rectangle?

[west schooner]

I mean like is A = xy the formula for any shape or just rectangles

[shut mirage]

in this form it is only applicable to rectangles and squares (where all of the sides are perpendicular to each other)

[west schooner]

Ah i see

[shut mirage]

is that the only confusion you have?

[west schooner]

why was y isolated

[bleak kiln]

hi

im here for pre university math

[shut mirage]

because y itself depends on x

[late creek]

Helllo pre university math, Im isitho

[stark wedge]

hello and welcome to the server

this channel is currently occupied

[west schooner]

can you elaborate?

[shut mirage]

hi, and welcome to the server! this is an ongoing help channel, though, so please head to #discussion or #chill to chat!

[shut mirage]

ok, see

you have a finite amount of fence

if you spend more money for the fence on the two $8 sides

do you then agree you would have less fence for the parallel side?

[late creek]

Because the function is supposed to be in terms of the length of the end or x

[shut mirage]

or rather, less money to spend on the fence for the parallel side

[west schooner]

ohh since 3600 is the budget

[shut mirage]

correct, your budget is fixed

[late creek]

A = xy, the question asks for function representing area entirely in terms of x.

[west schooner]

what if the given on b is the length of the parallel fence instead

[shut mirage]

on your journey to derive the function for a), you would have come across an expression for the parallel side in terms of the length of the two ends (x)

use it to find the length of either end of the fence (the x sides)

[west schooner]

am I going the right direction?

[shut mirage]

what are you trying to do in this case?

are you answering part a)?

[west schooner]

im tryna do this

[shut mirage]

so you're trying to find x in terms of y

[west schooner]

I think?

[late creek]

Just substitute x then...

[shut mirage]

yes, looks correct

[west schooner]

did I do it right?

[shut mirage]

this is for your changed problem?

,w (-3/4)(100^2) + 225(100)

[rocky lotus]

Results provided by WolframAlpha

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

Requested by gentle_flower11

[shut mirage]

yes

[west schooner]

how about this problem

from what I understand I need to use the area formula for both rectangle and circle?

[shut mirage]

half a circle, but yes

then add them together

[west schooner]

A=πr^2 for circle right?

[shut mirage]

pi*r**^2**

but otherwise yes

[west schooner]

is perimeter just another way of saying area?

[shut mirage]

no

perimeter is the length of the enclosure

or the outside of the shape

area is the amount of surface covered by the object

[west schooner]

so perimeter is x?

[shut mirage]

no

[west schooner]

can you give an example?

[shut mirage]

x is just the base. obviously the base alone won't surround the whole frame

well, what's the perimeter of a square?

(with side length of, say, a)

[west schooner]

no idea

is perimeter just adding the length of all sides?

[shut mirage]

yes

[west schooner]

so 4a?

[shut mirage]

correct

[west schooner]

im confused what to do next

how do I make the mathematical model

[shut mirage]

as with the previous question, find y in terms of x

hint: first find the circumference (perimeter) of the semicircle

[west schooner]

perimeter for circle is 2πr right?

would that mean a semicircle would be 2πr/2 since its half?

[shut mirage]

exactly so

[shut mirage]

you could have cancelled the 2 from 2pi and the 2 in the denominator to make it simpler, but that works

[west schooner]

like this?

[shut mirage]

correct

[west schooner]

would the rectangle part be x + 2y since the top part is the semi circle?

[shut mirage]

absolutely yes

well spotted

[shut mirage]

yes

[west schooner]

how do I divide the -π x/2 part?

[shut mirage]

divide the x/2 part. it would suffice

[west schooner]

what's next

[shut mirage]

the area of the whole frame

[west schooner]

that's the final answer?

[shut mirage]

no

you're asked for the area of the whole frame. now is the time to start work on it

[shut mirage]

reminder that this only accounts for the rectangle

but yes, good start

[west schooner]

ohh so I have to use it on this?

[shut mirage]

you're on the right track with this

you just need to remember the area of the semicircle

[shut mirage]

you want to add the area of the semicircle, not multiply

[shut mirage]

bracket the (x/2) and leave the ^2 outside the bracket

like so: $\left(\frac{x}{2}\right)^2$

[rocky lotus]

Céline

[shut mirage]

looks good

[west schooner]

what's next

do I distribute x?

[shut mirage]

nothing else, you're done

[west schooner]

oh

[shut mirage]

you can if you want to, but nothing changes here

[west schooner]

where did 9 - 2x come from?

oh wait I understood it just now lol

nvm

[west schooner]

what does feasible domain mean?

[vale dock]

@west schooner Has your question been resolved?

[pallid stone]

hi

i am new here

[runic aspen]

yo there are so many help channels how do ik which to use

[vale dock]

@west schooner Has your question been resolved?

[lyric sundial]

Pick one among the available ones

Or if you want to ask something about a specific topic, choose the suitable channel having that topic as its name

[vale dock]

@west schooner Has your question been resolved?

[shut mirage]

oh not done yet?

[shut mirage]

because you are working with lengths, some values of x don't make sense

such as negative lengths

[west schooner]

ah i see

so I just transpose?

[shut mirage]

transpose has a rather specific meaning in my head in math.
I would need you to explain what you mean by that word, because chances are it's not what I'm thinking about

[west schooner]

im talking about the 9 - 2x > 0

[shut mirage]

uhuh

[west schooner]

I understand it now

thanks 😄

[shut mirage]

oh ok, nice

[west schooner]

how do I close this?

.close

[vale dock]

Channel closed

Closed by @west schooner

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[viscid harness]

What is the (4 * sin(A) * sin(A) + 2 * cos(90 - A)) / (tan(B) * tan(B)) when the answer of 4 * x² -2 * (1 + √3) * x + √3 == 0 is cos(A) and cos(B) and 0 < A && A < B && B < 90 && A != B?

[pure pulsar]

!status

[vale dock]

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

[pure pulsar]

!original

[vale dock]

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.

[viscid harness]

1

[pure pulsar]

.

[river shale]

,rccw

[pure pulsar]

Wait let me translate this first

[turbid valve]

It's been correctly translated

[pure pulsar]

Oh yeah

Thanks

[turbid valve]

"If the roots of the quadratic [] are cosA and cosB, what is the value of [fraction]? [constraints on A and B]"

[stark wedge]

the wording looks awkward in English (i assume not in Korean)

[turbid valve]

It's kinda fine?

[stark wedge]

i wonder why they chose not to write sin^2(A) and tan^2(B)

[turbid valve]

It's written in a programming style tbh

[viscid harness]

I tried using the formula of a * x² + b * x + c == 0 (a != 0) where a = 4, b = -2 - 2 * √3 , c = √3

[turbid valve]

No I mean the English translation

[turbid valve]

I mean yes, that would be a route

[pure pulsar]

Finding sum of the roots and product of the roots first might help you to make it easier ig

Can clearly see the roots when we find sum and product

[viscid harness]

To find the sum, don't you have to find the roots first?

[pure pulsar]

Nope

[stark wedge]

hang on

[pure pulsar]

Sum of the roots = -b/a

[stark wedge]

the expression isnt symmetric in A and B

i think we actually have to find the roots here

[pure pulsar]

Thats what we are doing ann

[stark wedge]

w/o simply using QF?

[pure pulsar]

Product of the roots = c/a

[pure pulsar]

@viscid harness

We have formulas bro

[turbid valve]

[I presume by finding "two roots whose sum is ... and whose product is..."]

...I'd still just QF it

[stark wedge]

dont call me "bro" please.

(or OP, who also has she/her pronouns...)

[viscid harness]

2 / √3 + 2 ?

[pure pulsar]

1 + root3/2 is sum of the roots

And product of the roots is c/a

[pure pulsar]

I dont check pronouns mate i just check the problems, my bad

[stark wedge]

there are role diamonds

[pure pulsar]

Product is root3/4

[stark wedge]

you're missing brackets here.

you meant (1 + sqrt(3))/2

[pure pulsar]

Can you write it using texit?

[stark wedge]

$\frac{1+\sqrt{3}}{2}$ is the sum of the roots.

[rocky lotus]

Ann

Ann

[pure pulsar]

Also root3/4 too

[stark wedge]

\frac{}{} makes fractions; \sqrt{} makes square roots, dollars begin and end math formulas.

[pure pulsar]

I will eventually forget so pls if you can

[stark wedge]

....

nah you can figure out the LaTeX yourself m8

[pure pulsar]

I dont need to, just wanna help the person understand it better

Or is this my exam?😭

[viscid harness]

Can the sum and product can always find root directly without a * x² + b * x + c == 0(a != 0) formula?

[pure pulsar]

Nope ig

But here you dont need to i think

Because its clearly visible, what the two roots will be

[pure pulsar]

@viscid harness This is sum of the roots and root3/4 is product of the roots, so what 2 numbers will make their product as root3/4

Or if you find Quadratic formula easy, you can do it that way

[viscid harness]

Maybe the root is (1 + √3) / 4 + (1 / 2) * √(1 + √3 / 2) and (1 + √3) / 4 - (1 / 2) * √(1 + √3 / 2)?

[pure pulsar]

Remember, it has to be an angle value aswell

We have sin and cos ahead waiting for us

[viscid harness]

It has chance to be incorrect

[pure pulsar]

From what i see, the roots are root3/2 and 1/2

@viscid harness

[viscid harness]

But isn't √3 / 8 != √3 / 4 ?

[pure pulsar]

Product is root3/4 right

So root3/2 * 1/2 is same right?

[viscid harness]

It simplifies to √(3 / 8)

[pure pulsar]

How? Can you show your work?

[viscid harness]

√(3 / 2) / 2 == √(3 / (2 * 2²))

Hold on if it was 1 / 2 and √3 / 2 it is true

And the A and B would be 60 and 30

[pure pulsar]

No A is smaller so A is 30 and B is 60

As cos is decreasing on (0,90)

[pure pulsar]

Also, please take 2-3 similar examples and find roots by whichever method you find easy, just so you get a hold on it

[viscid harness]

I think I solved it

[pure pulsar]

Pls share your final answer

[viscid harness]

2 / 3

[pure pulsar]

Yep, great job mate

If you think your problem is solved, pls close the ticket and have a great day

[vale dock]

@viscid harness Has your question been resolved?

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[tawny sundial]

please explain the solution

[tawny sundial]

!helpers

[vale dock]

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

[tawny sundial]

!Helpers

[vale dock]

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

[marble arch]

You gotta @ them, anyways there's a 15 minute rule here

!15m

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[tawny sundial]

oh, thanks

[marble arch]

Alright, so where's the issue?

[tawny sundial]

why do we have to find y1 and y2

and how do I trace the curve?

[marble arch]

Alright, well basically the thing is that the circle is present in both + and - y-axis

[tawny sundial]

okay

[marble arch]

And its equation requires you to square root the right function, which rightfully yields two answers, i.e. +f(x) and -f(x)

So that is why you get two equations for y, which are just the same equation only one is positive and the other is negative

[tawny sundial]

understood, thanks

[marble arch]

That's why you use Y1 and Y2

[tawny sundial]

now what about the curve

how do I know it's a circle?

[marble arch]

Let me see

Well it's related to the equation of a circle, let me confirm it first though

Wait a second

Dude that's not even a circle in the first place

Check it out on Desmos, and let me know once you have

Honestly looks more like a guitar pick

[tawny sundial]

Yeah I checked

Yes

[marble arch]

Well, the equation for a circle is (x-h)^2 + (y - g)^2 = r^2

[tawny sundial]

But the figure doesn't have anything to with the solution so I guess it doesn't matter

[pure pulsar]

Its a semi circle if you put the value ig

[tawny sundial]

thanks for your help

[pure pulsar]

When its x-3

[tawny sundial]

more like a distorted semi circle

like somebody pulled it from the side

[marble arch]

Also I'm not sure how they expect you to trace that figure from this equation

[tawny sundial]

how do I close this channel?

[marble arch]

I mean you can still find the area without the figure

[tawny sundial]

yeah

[pure pulsar]

.close

[marble arch]

Yup so my network was terrible and I didn't see the messages

But yeah, also one more thing, the reason you have +y and -y is because the positive indicates the area of the region above the x-axis, and the negative indicates the area of the region below the x-axis

[tawny sundial]

got it, thanks

.close

[vale dock]

Channel closed

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[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[soft geode]

Okay so i need help with top one

Im a bit too tired today to comprehend some of it, i believe i have to bring b^-2 to the front

I can just shot my work

Show

Im getting nowhere with this

[shut mirage]

,rccw

[soft geode]

I guess i could remove negative power

But thats all i can come up with

The questions asks to simplify fraction by the way

[blazing raft]

try substituting a^1/2 and b^-1

[soft geode]

What does that word mean

[blazing raft]

replacing

[umbral isle]

Seems you can also reverse the sign of the denominator and take the a^(1/2) out to reduce it more

[soft geode]

Oh yea i thought of it yeah

[soft geode]

This i mean

[blazing raft]

ok ya it works

[soft geode]

Ok i just realized im stupid

I thought i was gonna get 1-b or smth

Ok one sec

[blazing raft]

replace a^1/2 with A and b^-1 with B

then it becomes a lot simpler

[soft geode]

What

Ok i just have what i have now

Ill just keep sending my work

So now i have this

Do i add - in front

To get -a and +b?

[blazing raft]

wdym

[soft geode]

Uhh

Just add minus in the bottom and change all pluses to minuses

And minuses to pluses

No idea how to explain 🙁

Ill just send what i would have done

And then simplify

[blazing raft]

gtg

[soft geode]

Wait hold up i dont know if there should be - infront of a^1/2

Aight if someone can help let me know before i go crazy

[umbral isle]

Yeah there shouldnt

Only 1 negative sign is needed

[soft geode]

Ok now what

Or is this all

[umbral isle]

I think thats all

[soft geode]

Photomath says i have to get 1 on top but i cant

I can get this but im not sure how to get rid of b at the top

[umbral isle]

Wdym you cant, just divide both by b^(-1)

[soft geode]

I only know how to get rid of it in the bottom

[soft geode]

Yea but i only know how to do it for top here

Like b^-2 / b^-1 = b^-2-1

No?

Or does that work too

b^-3 on top

[umbral isle]

No it becomes b^-1

[soft geode]

Oh right

My bad

So b^-1 on top is fine?

[umbral isle]

Just change b^-1 to 1/b

That probably looks better

[soft geode]

Yea

Wait

Can i do 0.25^1/2

Or is that just 0.5

[umbral isle]

Uhhhhh just do 0.5 i guess

[soft geode]

Yea i was asking if you can even calculate that

But realized you can

Okay i think thats all appereantly it wasnt that bad thanks

[umbral isle]

Np👍

[soft geode]

.close

[vale dock]

Channel closed

Closed by @soft geode

Use .reopen if this was a mistake.

[vale dock]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lost oracle]

hey i think the general attack is to show sup A >= lim inf s_n and sup A <= lim inf s_n but im not sure how to go about this, any advice?

[pure pulsar]

I think you should first try what you think and when you get stuck, you can ask for help