help-23

[plucky elk]

the limit of the first derivatives of numerator/denominator is also 0/0 so you can use l'hopital again

[mossy gyro]

so l'hopital this again

why does this work tho

😭

[plucky elk]

no i was mistaken

[mossy gyro]

oh crap

[twilit spindle]

careful, l'hop only applies to a 0/0 siutation

[mossy gyro]

yea

[plucky elk]

f''(x) / (x-4) -> ? as x goes to 4

[mossy gyro]

the trick question was that i cant l'hopital after the second

i shouldve stopped and said it didnt exist

thanks for pointing me in the right direction

.close

[safe radish]

Channel closed

Closed by @mossy gyro

Use .reopen if this was a mistake.

[twilit spindle]

[\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f(x)-0}{g(x)-0}=\lim_{x\to c}\frac{f(x)-f(c)}{g(x)-g(c)}=\lim_{x\to c}\frac{\frac{f(x)-f(c)}{x-c}}{\frac{g(x)-g(c)}{x-c}}=\lim_{x\to c}\frac{f'(x)}{g'(x)}]

[flat frigate]

ΠαϳαμαΜαμαΛλαμα

[safe radish]

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[pallid magnet]

another day someone else asked this question:

if I have a set of egyptian fractions that sums to 1 with maximum denominator m, is it guaranteed that there is a way to sum these fractions without an intermediate step involving a denominator larger than m?
it is not true, so the op asked the next obvious question, do u have a counterexample, can anyone here help? the original question is solved obv but i still wanna find a counterexample

[pallid magnet]

does anyone here have an counterexample?

[twilit spindle]

the original question is already proven (to be true?) but you still want to find a counterexample?

or have I just misunderstood you

[safe radish]

@pallid magnet Has your question been resolved?

[pallid magnet]

proven to be false, but i want to find a counterexample

so there is a counterexample somewhere

i've coded a program to find one for me but i think ima die of old age before it finds one, so i'm looking for a math way of making one

[safe radish]

@pallid magnet Has your question been resolved?

[pallid magnet]

<@&286206848099549185>

[bold river]

1/2 + 1/3 + 1/10 + 1/15 = 1, The maximum denominator is m = 15

[pallid magnet]

equal is not larger, therefore it's allowed

i've just got more info, for a max denominator of 500 or less, and a max length (say 1/2 + 1/3 + 1/6, it has a length of 3) of 8 or less, there's no counterexample found within range

tried everything

[bold river]

hmmm

[bold river]

In that case theres no counterexample within the tested range

[pallid magnet]

yes that's exactly what i said

so it gotta be bigger

while my code is brute forcing it, i'm looking for a math way of constructing it

[austere goblet]

the original question also had a counterexample posted.

if I recall correctly it's to do with finding a CE to 1/2 first, if you wanna search it up. forgot which help it is though

[pallid magnet]

nope unless there a different op

none worked which is why i wanna try finding one

[pallid magnet]

nope there's a way

idk how exactly but ik because my program checks for all possible way of adding a set so there's a way

lemme check my code tho

[pallid magnet]

hmm you know what

nvm

.close

[safe radish]

Channel closed

Closed by @pallid magnet

Use .reopen if this was a mistake.

[median sand]

An easy counterexample is that let m=6.

Since you want to show for any fractions a/b + c/d = bd/(gcd(b,d))

[pallid magnet]

bruh read the question carefully

[safe radish]

Available help channel!

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[sinful osprey]

Can someone help me to sketch

[sinful osprey]

A graph

Its y=|2x-a|+b

Where
a and b are positive constants

The only thing that I know is that on the y-axis
x=0
y=a+b <--

Im kinda confused on how to sketch on the x-axis

|2x-a|+b=0
2x-a=±b
Either
2x-a=b or 2x-a=-b
x=(b-a)/2 or x= (a-b)/2

So like

Isn't b>=0

What do I do

How do I sketch ts

[small sandal]

do you know the graph of y=|x|

[sinful osprey]

Yes

V

[small sandal]

mhm

and y=|x-a|?

[sinful osprey]

No

[small sandal]

thats ok

[stoic torrent]

.pin

[sinful osprey]

I know that it wont change tho

[small sandal]

do you remember how to sketch y=(x+1)^2 -3 for example

[sinful osprey]

It would just slide on the x-axis

[sinful osprey]

Yes

[small sandal]

same idea applies here

[sinful osprey]

So

X=a/b
y=b

(a/b, b)?

[small sandal]

where did you get those from? Also could you show the original question if possible

[sinful osprey]

Sure

Afk

[small sandal]

for example it should look like this i believe

and to be more general it could also look like this

[sinful osprey]

I got this

[small sandal]

mhm

[sinful osprey]

But in terms of a and b

Should I find the numbers

Or find it in terms of a and b

[small sandal]

well in part B were given some conditions

see what you can get given those conditions

x=0 is a solution and at x=c we have another solution

[sinful osprey]

Okay wait

Ama do it in a while

.close

[safe radish]

Channel closed

Closed by @sinful osprey

Use .reopen if this was a mistake.

[sinful osprey]

.reopen

[safe radish]

✅ Original question: #help-23 message

[sinful osprey]

Yeah

[sinful osprey]

I got a-b=8

When x=0

[small sandal]

we dont req. values for a and b, the question only asks for the relationship between c and a

[small sandal]

however this is a great start

[sinful osprey]

c= 2a-2b+16

🥹

Can I just substitute that

Into that

[sinful osprey]

This

[small sandal]

just one thing to note, its not a-b=8 it should be a+b=8

[sinful osprey]

Wait nk

No

[sinful osprey]

What the fuck

Okay wajt

Re do

[sinful osprey]

Oh wait I got it

I found

c=(2a+2b-16)/7

[small sandal]

sub b=8-a

[sinful osprey]

Oka

[sinful osprey]

I got 0

[small sandal]

thats one solution

thats the one we are given

[sinful osprey]

So what now😭🥹

[small sandal]

hmm lemme type out what we should have rq.

[sinful osprey]

Okie

[sinful osprey]

Oh wait now i understand why you said substitute b=8-a

[small sandal]

For our case when x=0 we get: $|2x-a| +b = \frac{3}{2} x +8 \implies |-a| +b = \frac{3}{2} (0) + 8$

$\implies a+b=8 \implies b=8-a$

Sub in b and simplify: $|2x-a| = \frac{3}{2} x+a$

now solve the absolute value equation case by case.

[flat frigate]

KB

[small sandal]

so your solving:

1. $2x-a=\frac{3}{2} x+a$

2. $-(2x-a)=\frac{3}{2} x +a$

[flat frigate]

KB

[sinful osprey]

Wheres b

[small sandal]

we subbed b=8-a and simplified

since we want c in terms of a

we want to get rid of b

[sinful osprey]

No wheres b in that

Y= |2x-a|+b = 3/2x + 8

[small sandal]

mhm

sub b=8-a

[sinful osprey]

Y= |2x-a| + 8-a = 3/2x + 8

Y= |2x-a| = 3/2x + a

Oh

OH

x=4a

So like

x=c

So c=4a?

Wait i didnt find the - one

[small sandal]

the - one is our c=0 solution

[sinful osprey]

I got!!!

Thank you so much!!!

.close

[safe radish]

Channel closed

Closed by @sinful osprey

Use .reopen if this was a mistake.

[small sandal]

no problem

[safe radish]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[fickle mantle]

Can someone pls help me

[median sand]

Which question?

[fickle mantle]

a1 and a2, b2, c2 c3 c4, d and e2and 3

[median sand]

I do see not a1,a2 here.

[wraith forge]

d) is so easy

[median sand]

Well, suppose we start with cos330sin60, it ask us to calculate in radian?

[median sand]

Is this sentence necessary?

[opaque fern]

Ok maybe its better to focus on one of them at a time...

[fickle mantle]

Oki

[ionic blaze]

is that a . between 4 and 2 on (c)

or a multiplication

[median sand]

Let's check if sin Beta is in which quadrant?

[fickle mantle]

Huh

[fickle mantle]

Which questions pls

[median sand]

1 a

[median sand]

I am solving question 1 a. Is this the one are you asking for help?

[fickle mantle]

1st and 2nd

[fickle mantle]

Erm there are a lot but someone said we should do one at a time

[median sand]

Right, now note that Beta is in which quadrant?

[fickle mantle]

2nd?

[median sand]

Not quite, check that Beta is between 90 and 270.

[fickle mantle]

I’m not sure

[median sand]

Read the conditions of the question.

[fickle mantle]

It’s either the first or second

[median sand]

Yes, it must be in the second quadrant.

[fickle mantle]

Why

[median sand]

Since sin(Beta) is positive.

Sin only positive in first and second quadrant, we are also given that the angle Beta is in quadrant 2 and 3. So there is an overlap in quadrant 2.

[fickle mantle]

No I mean using the / (90,270)

[median sand]

^

[fickle mantle]

Ok

[median sand]

Now, we have this illustration.

Sin Beta is opposite / hypotenuse.

Right?

[fickle mantle]

Yes

[median sand]

Now, what does sin Beta equal to?

[fickle mantle]

2 root 6 over 5

Can I go for my phone pls

[median sand]

Then, if $\frac{2\sqrt{6}}{5} =\frac{y}{r}$

[flat frigate]

Mercury.

[fickle mantle]

So I can write on my iPad and read from my phone

[median sand]

Yes, feel free to do so!

[fickle mantle]

Ty!

[median sand]

You are welcome!

[fickle mantle]

Back

[median sand]

Hi!

[median sand]

So, let's get back to here.

[fickle mantle]

Ok

[median sand]

So, what is y and r?

[fickle mantle]

Wait why is it in Yh second what’s again

How did I do that

[median sand]

Alright, for which quadrant is Sine positive?

[fickle mantle]

First and 2nd

[median sand]

Correct.

Now, what is the range of Beta?

[fickle mantle]

90,270

How do i do the range part

[median sand]

Which are 2nd and 3rd quadrant, right?

[fickle mantle]

Yes

[median sand]

It is given to you from the beginning already, in the question?

[fickle mantle]

Wait what

[median sand]

Now, since there is an overlap in quadrant 2, we can say Beta is between 90 to 180 degree.

Here,

[fickle mantle]

Where does 90 start from

[median sand]

From the condition given by the question.

[fickle mantle]

No I mean on the Cartesian plane

[median sand]

90 start from the positive side of y-axis.

[fickle mantle]

So from the 90 line to the 270 line

[median sand]

As you can observed from this illustration I got from the Internet.

Correct.

[fickle mantle]

Oki tu

Ty

[median sand]

No worries, so are you understood in this part?

[fickle mantle]

Yes pls

[median sand]

So, recall this sine-cosine-tan formula.

We are given that $\sin{\beta} = \frac{2\sqrt{6}}{5}$

[flat frigate]

Mercury.

[median sand]

By using the sine formula, can you identify which is y, which is r?

[fickle mantle]

Yes pls

[median sand]

Can you identify?

[fickle mantle]

Y is 2 root 6

R is 5

[median sand]

Right.

Now, let's find x.

[fickle mantle]

Ok

[median sand]

You can use Pythagorean formula.

[fickle mantle]

I got 1

[median sand]

Correct, but since Beta is in second quadrant, so x = -1

[fickle mantle]

Oh

Bc x is cos?

[median sand]

Which means, you reflect this triangle

Right.

Reflect it through the y-axis, since the shown image is the triangle in the first quadrant.

[fickle mantle]

Wdym reflect

Also late reply, my dads making me do smth

[median sand]

Something like this.

[fickle mantle]

Oh

[median sand]

No problem!

[fickle mantle]

But why pls

[median sand]

Because we found out Beta is in the 2nd quadrant, right?

The triangle we just solved is in 1st quadrant.

[fickle mantle]

I don’t understand

[median sand]

Oh right, how about this image?

[fickle mantle]

Yes

[median sand]

You got the angle Beta on the 2nd quadrant, right?

Ignore that theta angle.

[fickle mantle]

Yes pls

[median sand]

Now, you still have y and r.

[fickle mantle]

Yes

[median sand]

But x, it's on the negative side of the x-axis.

That's why it is -1 instead of 1, if you see.

[fickle mantle]

Yes!

[median sand]

Thanks!

[median sand]

Right, now back to the question, what is tan formula?

[median sand]

Recall this.

[fickle mantle]

Oop over adj

Y over x

[median sand]

Correct!

Now tan Beta = 2sqrt 6 / -1

There we go, Q.E.D.!

[fickle mantle]

OH

[fickle mantle]

what’s Tht mean

[median sand]

Quod Et Demonstrandum - As desired.

[fickle mantle]

Thats a big word

[errant ravine]

-# as a little nudge this might be usefull

https://tenor.com/view/matematica-gif-7322246

[median sand]

It is Latin

Right, can you do question 1b now, OP?

[fickle mantle]

Yes

[median sand]

Ping me if need help!

[fickle mantle]

Ty

[median sand]

No worries!

[fickle mantle]

Is this how it’s done?

OH

NVM

ITW COS

[median sand]

Yeah, sin(90 + beta) is just cos beta

[fickle mantle]

Tysm

[median sand]

No worries! You did all the hardwork!

[fickle mantle]

Erm what about qiestion b2)

[median sand]

b2?

[round egret]

what's question b2

[median sand]

Check the pin question.

[round egret]

no I meant, there's only b

[errant ravine]

-# maybe its the pdf?

[median sand]

This one.

Can you pin it?

[median sand]

Thanks!

[fickle mantle]

I got question c2 wrong again

[round egret]

tan(60) should be posistive I think

because tan(120) = -tan(60) (Second Quadrant)

and you have -tan(180-60)

[fickle mantle]

Your right

Tysm

What about this pls

[round egret]

third line is wrong

should be cos^2(50) being posistive

[fickle mantle]

Oo your right

Tysm

But can you help me with d

[round egret]

whats d

ok wait

i just pinned it :/

this?

[fickle mantle]

Yes

[fickle mantle]

Sorry

[round egret]

Oh no need to be sorry :p

[round egret]

you can rewrite tan as $\tan \beta = \frac{\sin \beta}{\cos \beta}$

[flat frigate]

MxRgD

[round egret]

now try to mainpulate LHS to look like RHS

[fickle mantle]

I get stuck herem

This is what I did previously

[round egret]

ok so $LHS = \frac{\sin \beta + 1}{\cos \beta (1 + \sin \beta)}$

[flat frigate]

MxRgD

[round egret]

you can factor out the cos

and you can cancel it

[fickle mantle]

But don’t we want it to stay

[round egret]

you can mainpulate RHS to be 1/cos

[fickle mantle]

Oh

I’ll do tht

[round egret]

or if you want to

do $LHS = \frac{1}{\cos \beta} \cdot \frac{\cos \beta}{\cos \beta}$

bleh

[flat frigate]

MxRgD

[round egret]

okay there we go

[fickle mantle]

Why multiply tho?

[round egret]

well you get $LHS = \frac{\cos \beta}{\cos^2 \beta}$ right?

[flat frigate]

MxRgD

[round egret]

and $1 - \sin^2 \beta = \cos^2 \beta$

[flat frigate]

MxRgD

[round egret]

which is what RHS is

[fickle mantle]

Can I solve it again and reach here before pls

[round egret]

yeah sure

[fickle mantle]

Ok I done

[round egret]

did you get it?

[fickle mantle]

tysm

I struggled a lot w this one

[round egret]

np

[fickle mantle]

Sometimes I forget we can change the other side💔

[round egret]

yeah i recommend working backwards if you're stuck

[fickle mantle]

What’s that mean pls

[round egret]

basically what you said, changing the other side a bit

if you're stuck

[fickle mantle]

Ohh k ok

Oh can u also help me with Judy question e1

Like the cos(540 + theta)

[crisp hull]

Can u mark the question

[fickle mantle]

How would u reduce the 540 while still keeping the theta

[round egret]

okay wait let me see what the question is

[fickle mantle]

.

[crisp hull]

Do yk abt allied angles?

[fickle mantle]

Maybe, can u show an example

I don’t rlly learn math terms name

[round egret]

you can do $\cos(540^\circ + \theta) = \cos(180^\circ + \theta)$ instead

[flat frigate]

MxRgD

[fickle mantle]

I don’t get it

[round egret]

since 540 is the same as 360 + 180. Since circles repeat every 360, you can just ignore the 360

[fickle mantle]

Why 180

[crisp hull]

Well listen allied angles r like multiples of 90 with smt being added or subtracted

For eg cos (180 + theta) can be written as cos (2x90+theta)

Now as you see here

There is a even multiple of 90

Now look at quadrants

2x90 means u moved 2 quandrants to 180

U r at 180 now

Now look if smt is being added or subtracted

As in this case theta is being added so u r in 3rd quad

[round egret]

think about what cos(180) is

[crisp hull]

So for even multiple u dknt change the function
So u write as cos theta but negative with it as it is in 3rd quad

[fickle mantle]

Like the reduced for ?

Form

[round egret]

sorry i meant cos(540)

$\cos(540^{\circ} + \theta) = \cos(360^{\circ} + 180^{\circ} + \theta)$

[flat frigate]

MxRgD

[fickle mantle]

Oo

[round egret]

cos(180) is -1 right?

what would cos(560) be

[fickle mantle]

Wdym?

[round egret]

like cos(180) = -1

it's a standard trig value

[fickle mantle]

360+20p

200

[round egret]

sorry I meant 540, idk why I can't read all of a sudden

[fickle mantle]

It’s ok

[round egret]

but yeah what's cos(540)

[fickle mantle]

360+180

[round egret]

mhm

but cos(x) is symmetric right?

if you look at the graph

[fickle mantle]

My father wa ts me to charge his phone

I’ll be back rq

[round egret]

okkk

[round egret]

I'll have to go now as well but let me graph what I mean

[fickle mantle]

Ty for helping me out tho

[fickle mantle]

I’m not sure, I’ve never used the graph to solve questions

[round egret]

i have some time left, ignore the pi's as (pi = 180 degrees and 3pi = 540 degrees)

now you know the distance between 540 and 180 degrees is 360 degrees right?

[fickle mantle]

Yes

[round egret]

they have the same value as you can see from the symmetry of the graph

when you add 360

[fickle mantle]

Yes pls

[round egret]

so what do you think cos(900) is

[fickle mantle]

4

[fickle mantle]

4

[round egret]

How did you get 4?

cos(180) = -1 right

and cos(540) = -1 as well

so cos(900) is?

[fickle mantle]

-1?

[round egret]

yep

because we're +360 each time

[thin bridge]

540

[round egret]

idk why I keep confusing 520 and 540 ;-;

but yes 540

[fickle mantle]

Tysm everyone

[crisp hull]

Try solving that q now

[fickle mantle]

Oki

Would it become tan180

Or tan theta

[crisp hull]

Well you have two options here

You either use formulas

Or allied angles

If you dk abt formulas i think allied angles would be better to go with

[fickle mantle]

Mmh

[crisp hull]

Well i would choose allied angles want me to explajn them to you?

[fickle mantle]

Sure

[crisp hull]

So do you know abt quadrants?

[fickle mantle]

Yes pls

[crisp hull]

Alr do you know what functions r positive and negative in them

The ASTC rule

[fickle mantle]

Yes pls

[crisp hull]

Yk abt it?

[fickle mantle]

Yes

[crisp hull]

So allied angles r multiples of 90 with smt being added or shbtracted

For example

Cos 540+ theta

Can be written as cos (6x90+theta)

[crisp hull]

@fickle mantle

Do u understand this

[fickle mantle]

Kind of

I’ve never seen this b4

[crisp hull]

As 540 divided by 90 is 6

So 6x90=540

We r jst writing it in terms of 90

[crisp hull]

Try yourself divide 540 by 90

[fickle mantle]

Yes 6

[crisp hull]

So 6 x 90 equals 540

U can check this too

If u want

Alr one step is done

[fickle mantle]

But won’t 90 change the sign

[crisp hull]

No

[fickle mantle]

Oh ok

Bc it’s multiplying?

[crisp hull]

So it changes like this

,rccw

[crisp hull]

Now look at what is being multiplied with 90?

[fickle mantle]

6

[crisp hull]

Now u tell me

Make a quadrant of 4

6 means move around 6 times

[fickle mantle]

Wdym

[fickle mantle]

Ohh

[crisp hull]

Like

Wait

[fickle mantle]

From the 360 line?

[fickle mantle]

2nd quard

[crisp hull]

,rccw

[crisp hull]

Visualize like this

From one line to 2nd line

Its 90 degree

So its 1 move

I did 4 moves in this pic

U gotta do 6 moves

So u end up where

?

Like on which line u end up after moving 6

[fickle mantle]

Uh the 2nd line,

2nd, 3rd

[crisp hull]

3rd

The one at which angle is 180

U end up at it

After 6 moves

[median sand]

,rcw

[fickle mantle]

Oh

[fickle mantle]

So the third box?

[crisp hull]

No not the box i mean the lines

[fickle mantle]

Oh ok

[crisp hull]

We gotta look for quadrant now

As we r on 3rd line

So look at cos (6x90+theta)

Is smt being added or subtracted

?

[fickle mantle]

Add

[crisp hull]

To multiples of 90

So yes if we added smt we make it big so move forward right?

[fickle mantle]

Yes pls

[crisp hull]

So if we move from 3rd line froward whats the quadrant

[fickle mantle]

3rd quard

[crisp hull]

Ye right

[fickle mantle]

I think?

[crisp hull]

So u can write it as cos of theta

[fickle mantle]

Ohh

[crisp hull]

But as in 3rd quadrant cos is negative

So u gotta add a negative sign

It becomes -cos theta

[fickle mantle]

Yes

[crisp hull]

So it is solved

[fickle mantle]

I wanted help with the tan part tho

[crisp hull]

Now try doing cos 720 plus theta

[crisp hull]

Ye we move tk that next

[crisp hull]

First show me how you solve this

[fickle mantle]

Oki doki

[fickle mantle]

Cos(90x8

[crisp hull]

Plus theta

8x90+theta

[fickle mantle]

Why plus tho

[crisp hull]

As theta is bejng added in q

Cos (720+theta)

We jst wrote 720 as 8x90

You dont need to worry abt signs here

Then what do u do after writing it in terms of multiples of 90

[fickle mantle]

Count 8 lines

[crisp hull]

Ye

Where u end up

[fickle mantle]

4th quard?

[crisp hull]

No no i told u to look at lines

Not quadrant

[fickle mantle]

Oh line

[crisp hull]

Which line

[fickle mantle]

0 line

[crisp hull]

Ye the line which is at 0 degree

Now look smt is being added or subtracted

And tell me quadrant

[fickle mantle]

1st

[crisp hull]

Ye

Right

As 1st qaudrant has everything positive so

Cos theta is your answer

[fickle mantle]

Yay

[crisp hull]

,rccw

[crisp hull]

On your work u will write like this

So we were dealing with even multiples of 90

Now there is another thing odd multiples of 90

Which has jst 1 step more

Cos 630 +theta

Can be written as?

[fickle mantle]

Cos(7x90+theta)

[crisp hull]

Now look its 7 its an odd no

But no need to worry do same thing move 7 lines

[full orbit]

Four couples are to be seated around a circular table. All the couples are to sit together,except for Jeremy and Candace,in such a way that males and females alternate.How many seating arrangements are possible?

[crisp hull]

Tell me which ljne u end up with

[fickle mantle]

270li e

Line

[crisp hull]

Go to free help channel

[crisp hull]

Yes now which quadrant

[fickle mantle]

4th

Cos +

[crisp hull]

Yes

Now listen

[fickle mantle]

I’m gonna use the washroom

Rq

[crisp hull]

Alr

So if u have a odd multiple of 90 like 7x90 9x90 etc then the function at end changes to its co function like cosine into sine

[fickle mantle]

Back

[fickle mantle]

Ohh ok ok

[crisp hull]

So answer is?

[fickle mantle]

-sin

[crisp hull]

Ye minus sin theta

Keep in mind the sign is always according to the function in your q

Its psoitive theta

Sin theta

Cos tells us the sign

[fickle mantle]

Oh

Wait I don’t get you

[crisp hull]

Wait lemme write

[fickle mantle]

Oki

[crisp hull]

,rccw

[crisp hull]

So cos is psoitive in 4th quad

[fickle mantle]

Yes yes, I understand now

[crisp hull]

So lets solve your q

[fickle mantle]

Ur handwriting is very fancy

Oki

[crisp hull]

Xd

[crisp hull]

.

Do the q step by step using allied angles

Cos 540 plus theta reduces to?

[fickle mantle]

90x6

[crisp hull]

Solve it fully

[fickle mantle]

Cos(90x6 +thetha)

[crisp hull]

Then

It reduces to

R u there

@fickle mantle

[fickle mantle]

Yes

[fickle mantle]

Cos theta

[crisp hull]

Negative cos theta

Cos is negative in 3rd quadrant

Well try for tan (minus theta -180)

It reduces to tan (-(180+theta))

Then- tan (180 +theta)

Now solve this

[fickle mantle]

So I switch the theta and 180?

[crisp hull]

Wdym

1+2 and 2+1 are same thing

U can change the place

Jst making it easier for you to write in that form

[fickle mantle]

Oh ok ok

Ty

[crisp hull]

So what becomes of tan(180+theta)

[fickle mantle]

Tan theta

[crisp hull]

Ye right but we took negative common right?

So it ends up as - tan theta

Lets move to sin(180+theta)

[crisp hull]

@fickle mantle

[fickle mantle]

Ohh yes

[crisp hull]

What becomes of it

[fickle mantle]

-sintheta

[crisp hull]

Ye right one

Now what the question reduces to

Put the values

@fickle mantle

[fickle mantle]

-costheta -tantheta over -sintheta

[crisp hull]

Right look at this now

This is what we jst solved

[fickle mantle]

Yes

[crisp hull]

Now we gotta simplify it

[fickle mantle]

Can we reduce the tan

By the identity?

[crisp hull]

Yes

[fickle mantle]

Oki

[crisp hull]

Tan is sin over cos

Write it

[fickle mantle]

Yes sir or maa’m

1?

[crisp hull]

Minus 1

U got it tho jst be careful abt signs

[fickle mantle]

Wait for the numerator did u make it positive

Bc -x-

[crisp hull]

Above minus multiply minus gives positive

But another minus is below it

So the result is minus 1

[fickle mantle]

Oh ok ok

Tysm

[crisp hull]

Nws

[fickle mantle]

Byee

[crisp hull]

Do u have more questions or jst this

[fickle mantle]

Nope not rn

[crisp hull]

Btw

Another thing

[fickle mantle]

I’ll take a break and continue later

[crisp hull]

When i told u that u move forward

[fickle mantle]

Probably

[crisp hull]

When smt is being added

But when smt is being subtracted

[fickle mantle]

Uhuh?

[crisp hull]

What happens in that case

Liek cos (180-theta)

[crisp hull]

Solve this one then u can go

[fickle mantle]

-cos

[crisp hull]

Yes

Negative cos theta

Quadrant is 2nd btw

In this case

As smt is being subtracted making it smaller so u mkve backwards from the line

I hope you got it gl !

[fickle mantle]

Oki doki

Tysm

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[zinc mural]

ive only really done highschool level math up to the point of understanding some basic calculus, proofs, and linear algebra and im struggling to find a way to progress from there?
i know this sounds ambitious but my goal in maybe a years time or more is to understand differential geometry at an intuitive level..
is this possible?
if so.. how would i even get there?

[timid ridge]

do more calculus then diff eq

[naive dragon]

@earnest nacelle

[zinc mural]

as in multivariable calc?

also should i learn topology as well?
i have no idea what these terms mean ive just heard they're necessary

[mint plume]

if youve done basic calculus then do a thorough treatment of linear algebra

[zinc mural]

any resources?

[mint plume]

ladr by axler

[zinc mural]

alright

tysm

and then from there should i do

multi variable calc

[mint plume]

u could do real analysis

[zinc mural]

ok

is there a speciifc book i should use

[mint plume]

i like charles pugh analysis

[zinc mural]

alright tysm

ok

.close

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[safe radish]

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[brazen nimbus]

can someone help

[errant ravine]

so ur triying to find the highest point on the graph right?

[brazen nimbus]

what do the 1,2 mean

ik what the highest point is

but the context

1,2 means then

120% or what

of its nominal range

[errant ravine]

where is 1.2?

[brazen nimbus]

on the y axis

the highest point

[errant ravine]

r(x) =y

[brazen nimbus]

yeah

[errant ravine]

so y is the qoutient of the actual range

[brazen nimbus]

yes

[errant ravine]

so y = range of electricle vechile / nominal range

[brazen nimbus]

yeah

or nominal range/actual rang

?

[errant ravine]

it is worded as

[brazen nimbus]

oh

[errant ravine]

so i think this implies actual range / nominal range

but not sure i am bad at translating word problems correctly

[brazen nimbus]

the actual range is affected by the outside temperature

right

[errant ravine]

yup^

[brazen nimbus]

okay so uhh

[errant ravine]

reember nominal range is a fixed value

[brazen nimbus]

yep

[errant ravine]

i think u can get the answer from here

lmk if u still need anything!!

[brazen nimbus]

wait so 22 C would be the point where the car has the most actual range?

or

[errant ravine]

x is tempreature

y is the amount of range

[brazen nimbus]

yes

[errant ravine]

so gives that nominal range is a fixed cosntant anyway

[errant ravine]

that means that this is true or false?

[brazen nimbus]

wdym?

[errant ravine]

what u just said do u think its correct or not?

[brazen nimbus]

i mean i think it is i cant think of anything else 😭

[errant ravine]

yes it is correct!

dw!

[brazen nimbus]

but its kinda weird that the actual range is 1,2 * nominal range

like wdym it got more range than it actually can have wtf

[errant ravine]

okay so here is the important thing to know

so maybe nominal range is an average of these condotions so nominal range happens at idk when the temprature is between 4 and 30 or smthing

we dont really know but a good thing to reember is

nominal range probably refers to a realstic range the vechile has

optimal range would be the point hieghest on the graph

[errant ravine]

does that make sense?

[brazen nimbus]

yeah

ty

[errant ravine]

no problem

also dont forget to close the channel when ur done!!

[brazen nimbus]

.close

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[viscid marsh]

Hi could someone explain why u do this?

[radiant ice]

do what exactly

[viscid marsh]

Write -16x as +20x-36x

[radiant ice]

it is to factor

[viscid marsh]

Oh wait

I get it

My bad

[radiant ice]

ur good

[viscid marsh]

.close

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[fallen fiber]

how the fuck do i solve this sum😢

[fallen fiber]

my teacher used this formula but i dont Get the steps after the ratio part

[errant ravine]

the reason for the ratio is because trisections split the line into 3 parts

as so usally midpoint would be the point between two ppoints so the ratio n each side is 1: 1

but here if u get a line and put two points on it

the ratio would be 2:1 instead

[errant ravine]

does that make sense?

[safe radish]

@fallen fiber Has your question been resolved?

[safe radish]

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[icy wadi]

could i have some help on how to approach this problem and how to set up its equation/s?

[stoic torrent]

right well, can you first identify the variable eqns you know?

[drowsy path]

to set up the equation, let’s do it based on the ratios
20 pieces of cherry gum
30 pieces of grape gum
x pieces per pack

so in one scenario, chewbacca loses on pack of cherry gum.
the new amount is 20 - x.

The amount of grape gum remains as 30

so the ratio would be 20 - x over 30

in the second scenario, chewbacca finds five packs of grape gum.
the amount of cherry gum is still twenty, whilst the new amount of grape gum is 30 + 5x

the ratio would be 20 over 30 + 5x

since both of the ratios are equal, you must set them against each other

first, you simplify the denominator on the right side by factoring five

and then you cross multiply

next you subtract 120 from both sides

and then factor the equation

[icy wadi]

i understand what youve said, but how do these equations match with what youve been saying? wouldnt one of the equations be 20-x/30+5x ?

[drowsy path]

20 - x/ 30+5x wouldn’t fit. if we used the equation, you would be describing a third, different universe where both of the events occured at the same time (so he lost cherry gum and gained grape gum)

[icy wadi]

alright, that makes sense

[drowsy path]

okk

[icy wadi]

so would x simply be 14?

[drowsy path]

yes.

[icy wadi]

alrighty!! tysm you really cleared my confusion on this problem i rlly appreciate your help <3

[drowsy path]

ofcc 😉

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[icy wadi]

.close

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[cerulean cloud]

i want help with converting these from triogenometric to rectangular forms, i am also unsure if im correct with the solutions
the very first equation was z²-4z*cosθ+4=0

[gentle oxide]

i can't really read your handwriting

[cerulean cloud]

oh? 😭

z1= -2( zcosθ-√(z²cos²θ+i) )

is this any help

the only difference is the plus

[gentle oxide]

$$z_1 = -2(z\cos(\theta) - \sqrt{z^2 \cos^2(\theta) + i})$$

[flat frigate]

W*

[gentle oxide]

is this correct?

[cerulean cloud]

z2= -2( zcosθ+√(z²cos²θ+i) )

[cerulean cloud]

ok yes thank u

[simple galleon]

parenthesis goes at end

oh it is at the end

nvm then

[cerulean cloud]

it's there dw

[gentle oxide]

and your question is to convert z_1 and z_2 into rectangular equations? or what

i don't particularly understand

!occupied

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

[cerulean cloud]

yes

that is it

[outer pollen]

<@&268886789983436800> potential trolling?

[gentle oxide]

$$z_1 = -2(z\cos(\theta) - \sqrt{z^2 \cos^2(\theta) + i})$$
$$z_2 = -2(z\cos(\theta) + \sqrt{z^2 \cos^2(\theta) + i})$$

[flat frigate]

W*

[simple galleon]

use \bigl and \bigr to make the parens bigger

[gentle oxide]

oh wait i see

and yes im aware im too lazy to care

[simple galleon]

fair enough

[gentle oxide]

@cerulean cloud you are trying to solve $z^2 - 4z \cos(\theta) + 4 = 0$ by converting to rectangular?

[flat frigate]

W*

[gentle oxide]

you should've led with that

[cerulean cloud]

uh no

im asked to solve that

and then turn the solutions into rectangular

[gentle oxide]

right ok

[cerulean cloud]

im unsure about the solutions though

[gentle oxide]

i think you did something wrong yeah

when i did it on scrap it was a lot simpler

[cerulean cloud]

like does z= smth*z make sense

[gentle oxide]

do you mind restarting with me?

[cerulean cloud]

ofcc

[gentle oxide]

okay so

$$z^2 - (4\cos(\theta))z + 4 = 0$$

[cerulean cloud]

Im prob confused on things

[flat frigate]

W*

[gentle oxide]

so i suspect you used the quadratic formula, which is a good choice

[cerulean cloud]

we've got other choices?

[gentle oxide]

probably not lol

anyways this gets you $z = \frac{4\cos\theta \pm \sqrt{16\cos^2\theta - 16}}{2}$

[flat frigate]

W*

[gentle oxide]

can you verify that for me?

[cerulean cloud]

ohhhh

oh my godd

[gentle oxide]

i don't want to bore you with the algebra because i think you can do it yourself

[cerulean cloud]

i added the bare z to the quadratic formula

[gentle oxide]

but please verify

[cerulean cloud]

😭

[gentle oxide]

hahaha

[cerulean cloud]

yess

[gentle oxide]

right i was confused why you had a z in there

[cerulean cloud]

it's true

i didnt notice

[gentle oxide]

anyways

[cerulean cloud]

i had it written as

4zcos(theta)

[gentle oxide]

$\sqrt{16\cos^2(\theta) - 16}$

[flat frigate]

W*

[cerulean cloud]

that would be

[gentle oxide]

what do you think is ideal to simplify this

[cerulean cloud]

16(cos²(theta)+i)

well

(cos²(theta)-1)*16

[gentle oxide]

+i?

[cerulean cloud]

yes

replacing the -1

[gentle oxide]

hmm let's slow down

[cerulean cloud]

sure

[gentle oxide]

$\sqrt{16(\cos^2(\theta) - 1)}$

[flat frigate]

W*

[gentle oxide]

you factored 16 like this i see

why did the 1 turn into an i?

[cerulean cloud]

+(-1)

i=-1

+i

[gentle oxide]

hm?

[cerulean cloud]

is that

not right

[gentle oxide]

i don't think so

[cerulean cloud]

ohhh

it should be

i²

[gentle oxide]

why are you turning the 1 into an i?

[cerulean cloud]

😭

[gentle oxide]

right yeah

i^2 = -1

[cerulean cloud]

well im turning the -1 into an i

i switched them up

[gentle oxide]

there's no reason to here

[cerulean cloud]

oh

OH

it's gonna be sin²θ anyway

[gentle oxide]

$z = \frac{4\cos\theta \pm \sqrt{16(\cos^2\theta - 1)}}{2}$

[flat frigate]

W*

[gentle oxide]

what do you think we should do now

[cerulean cloud]

we can turn the cos²θ-1 into a

-sin²θ

[gentle oxide]

yes

$$z = \frac{4\cos\theta \pm \sqrt{-16\sin^2(\theta)}}{2}$$

[cerulean cloud]

and then

[flat frigate]

W*

[cerulean cloud]

we got

4√-sin²θ

[gentle oxide]

$$z = \frac{4\cos\theta \pm 4\sqrt{-\sin^2(\theta)}}{2}$$

[flat frigate]

W*

[cerulean cloud]

-1=√i

[gentle oxide]

$$z = \frac{4\cos\theta \pm 4i\sqrt{\sin^2(\theta)}}{2}$$

[flat frigate]

W*

[cerulean cloud]

4√(sin²θ*(√i))

[gentle oxide]

where did you get sqaure root of i

[cerulean cloud]

ohh

[gentle oxide]

did you mean square root of -1

[cerulean cloud]

omg yes

no i meant

[gentle oxide]

by the way you shouldn't split it like that technically

[cerulean cloud]

the square of 8

i*

[gentle oxide]

how?

[cerulean cloud]

im trying ok 🥹

[gentle oxide]

you agree with this right

[cerulean cloud]

yes yes i got it

it's √i²*sin²θ

[gentle oxide]

right

$$z = \frac{4\cos\theta \pm 4\sqrt{i^2 \sin^2(\theta)}}{2}$$

[flat frigate]

W*

[cerulean cloud]

we can say it's

4i*sinθ

[gentle oxide]

$$z = \frac{4\cos\theta \pm 4i \sin(\theta)}{2}$$

[cerulean cloud]

ohh it makes sense now

[flat frigate]

W*

[cerulean cloud]

we got 2(cosθ±isinθ)