help-0

[tardy robin]

how do i goo about the 2nd part

of question 10

yup

[pale kestrel]

X ~ (1, s)
P(X > 1.05) = 0.01

So

200 rods

probability 4

are over

hm hm hm

So this is just binomial

you see why?

[tardy robin]

yea

but RIGHT

[pale kestrel]

The normal distribution is irrelevant

[tardy robin]

Ya but but

u see it can alsoo be

[pale kestrel]

maybe its normal binomial

[tardy robin]

JUST RIGght fit

[pale kestrel]

huh?

[tardy robin]

okay so it can be more than

less than

and equal

[pale kestrel]

uh...

I think the only thing thats relevant is 1%

Your probability is 0.01

[tardy robin]

yea

[pale kestrel]

you have 200 trials

and 4 or more oversized

cumulative binomial?

this sounds like.

[tardy robin]

okay

so my n= 200

and p is 0.01

but then i need to find 4 of them

so do i just multiply the probability by 4?

[pale kestrel]

no no no

Tell me, what are the parameters to binomial

[tardy robin]

no of trials

[pale kestrel]

X ~ B(???)

[tardy robin]

and probability

n,p

right

[pale kestrel]

X ~ B(n, p)

probably

[tardy robin]

ye

[pale kestrel]

And so we want

X ~ B(200, 0.01)

And then we want

P(X >= 4)

[tardy robin]

ohh

[pale kestrel]

yes?

[tardy robin]

okay

so this just gives me my answer

got it

[pale kestrel]

Im curious what the answer is 🤔

[tardy robin]

wait lemme press the calculoator

okay i will need to do

P(X<3)

which is 0.85803

[pale kestrel]

huh, didnt expect the probability to be so big

ig it makes sense

[tardy robin]

okay i will try the other part on my own

if i dont get it would it be fine if i pmed u regrading it?

[pale kestrel]

i might be afk

you can

[tardy robin]

okay thank you

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[shrewd topaz]

I'm kind of lost with NFA's can someone help?
Image
for example if this was a dfa it would be easy
but as NFA i don't really get it what should and shouldnt be

[shrewd topaz]

for example is this acceptable

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@shrewd topaz Has your question been resolved?

[shrewd topaz]

<@&286206848099549185>

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@shrewd topaz Has your question been resolved?

[rapid nova]

no

it accepts 01

this'll work

[lone heart]

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[rose rivet]

Please can you help with this

[echo socket]

Have you tried setting (2n + 47)/(n + 1) equal to 5?

[rapid nova]

just solve (2n+47)/(n+1) = 5 for n

for the second one divide both numerator and denominator by n

[rose rivet]

yes

42/3?

14

what about the second part

<@&286206848099549185>

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[flint ravine]

f'(0) when f(x) = 2tanx+5x^2sin(1/x) (x>0,x<0) f(x)=0 (x=0)

[arctic gust]

do you mean 2tanx + 5(x)^(2sin(1/x))?

[echo socket]

I think they meant 5(x^2)sin(1/x)

For this problem you can use the limit definition of the derivative

[echo socket]

Basically f'(0) is the same thing as the limit of (f(x) - f(0))/(x - 0) as x approaches 0

[flint ravine]

Yup

[echo socket]

In this case f(0) = 0, so you need to solve the limit (2tanx + 5x^2sin(1/x))/x

as x approaches 0

[flint ravine]

So just lHopital on that?

[echo socket]

Obviously not

Because then you'll need to differentiate 2tanx + 5x^2sin(1/x) at 0

But you are solving for the derivative at 0 right now

[flint ravine]

Okok

I get it now

Tysm

[echo socket]

So you can solve the limit or do you want me to help you there as well?

[flint ravine]

Its fine

Thank you for the help

.close

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[alpine sable]

Hello. I'm solving a big O notation problem. I got C >= log2(e) + 1. But it's not true at all for n = 2 (n>=k where k=2). Have I done some mistake here?

[alpine sable]

The graph on desmos

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

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@alpine sable Has your question been resolved?

[tacit arch]

I think setting k=2 seems unjustified here. Start back with |1+log (n)| <= Clog(n) and solve for n

It can be as crude as you want because big O notation is so forgiving

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[proper venture]

can someone help me with this problem (Prob 2 parts c and d):

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[scenic flame]

Find an equation in slope-intercept form (where possible) for the line.

[tacit arch]

do you know the general equation for slope-intercept form?

[scenic flame]

Not really

And I have 2 questions that are very much alike

I would possibly like help with 3 and to maybe try 4 by myself

Once I understand

[scenic flame]

sorry for the ping ^

[long iris]

Well maybe you can try to use perpendicular lines slopes multiplation is 1

I think that should work.

Slope of 2x-7y=61 is 2/7

m*(2/7)=1

[scenic flame]

Idk how to show the work….or solve the problem

[void niche]

Slope intercept form is $y=mx+b$. The first step is to make the given equation look like that.
Then, use the following fact: If you are trying to find the line perpendicular to $y=m_1x+b_1$, the new lines slope should be $m_2=-\frac{1}{m_1}$, the negative reciprocal.
You should then have a formula that looks like $y=m_2x+b_2$ for the new line, with $b_2$ unknown. This is where you plug in the given point in the question for $x$ and $y$. Then solve for $b_2$.

[ocean seal]

Migillope

[long iris]

Ye

[void niche]

Does this make sense? I can give an example, if that would help

[long iris]

I think it would be better

[scenic flame]

This is all too confusing for me

[void niche]

ok, lets work through the first one

#3, that is

[long iris]

Then try going by step by step

[void niche]

Step 1: Make it look like y=mx +b. So, you need to get y on one side of your equation, and everything else on the other side

Your work should look like this:
$\ 2x-7y = 61 \ -7y = 61-2x \ y = \frac{61-2x}{-7}$

[ocean seal]

Migillope

[void niche]

Which you can simplify to:

$y=\frac{2}{7}x-\frac{61}{7}$

[ocean seal]

Migillope

[void niche]

Does that make sense so far?

[scenic flame]

I guess…

[void niche]

Does it or does it not? If it doesn't tell me where I lost you, so I can clarify

[scenic flame]

This

[void niche]

Which step?

1st 2nd or 3rd equation

[scenic flame]

2nd

[ocean seal]

Migillope

Migillope

[void niche]

If it helps, my students typically like to write it out like this

[tacit arch]

but hopefully better handwriting

🔥

[proper venture]

@void niche

if you can help me with my linear algebra problem, it would mean the world to me loll

[scenic flame]

That makes more sense

Now what about the third step?

[long iris]

But he made a mistake here

[void niche]

on the third step, you are dividing both sides by -7

[long iris]

It must be 7y=2x-61

If im not wrong

[void niche]

That's the same thing, you just multiplied everything by -1

[long iris]

Wouldnt it be -7y=61-2x then?

[void niche]

that is what i wrote

[scenic flame]

This is confusing for me 😅

[proper venture]

@void niche once you're done helping the other people ofc. Sorry, I didn't know I was interrupting a session but it would help me out sooo much if someone can help me with lin algebra

[long iris]

This made me confused

[void niche]

Oh, I thought you meant in the latex!

yeah, I forgot the negative sign there, in front of the 7

[void niche]

I understand! Which part are you confused on? Still the third line?

[void niche]

I maybe can, I haven't done linear algebra in a while though. it also looks like riemann is helping you

[scenic flame]

So do I just write a big like under this and divide by -7?

[void niche]

should be -7=61-2x, I forgot the negative sign when I wrote it by hand

[inner fable]

Yes

[solemn grove]

tell me what you know about gradients of lines

and perpendicular lines

[proper venture]

@void niche oh no, I think he/she is helping with another problem. But once you're done, could i show you my problem?

[solemn grove]

what is your q

[void niche]

but yes, now you divide both sides by -7 (thats what the big line means: division.) -7y divided by -7 is just y, hence why in the latex the -7 "disappeared" from the left side of the equation

[scenic flame]

Where?

[solemn grove]

oi oi oi, let them do it their way

[long iris]

#help-1

[scenic flame]

I’m so confused there’s so many people typing at once

[void niche]

(can yall move this conversation to a different channel)

[proper venture]

@solemn grove okay, let's move this to help-1

[void niche]

I'll rewrite this

[proper venture]

@long iris ill move it to help-1

[scenic flame]

Thanks! It helps seeing it in writing I think for me

[void niche]

I apologize in advance for handwriting

[scenic flame]

Wait

I did -2x

Instead of /-7

[void niche]

That was the previous step, I didn't rewrite that

[scenic flame]

Oh ok

[void niche]

So from the beginning,

[void niche]

yes! looks great

ok, now our end goal at the moment is to get the equation to look like $y=mx+b$

[ocean seal]

Migillope

[void niche]

so the only remaining step is to separate the terms on the right hand side. I'll write it down, then explain

[void niche]

oop, wait, should be 61-2x in the numerator in the last line. Sorry, I'm writing a bit hastily

anyway: here is the next step

a general rule is that whenever you have something being subtracted in a numerator over a single thing in the denominator, you can "separate" the fraction, like I did here

The only thing to be careful of is that you keep track of the negative signs. The first term, $-\frac{61}{7}$ is negative because in the line above, $61$ was positive and $-7$ is negative. A positive and a negative make a negative. In the second term, the $2x$ was negative, and the $7$ negative as well, and two negatives make a positive

[ocean seal]

Migillope

[void niche]

make sense?

I know that was maybe a lot to follow, so let me know if you would like me to elaborate on anything there

[scenic flame]

Yeah I think it makes sense

[void niche]

great! So can you show what work you have so far?

[void niche]

Great. Now, just to be clear, you understand that the x is not in the denominator in the last line, correct?

[scenic flame]

Oh I thought it was 2 over 7x

[scenic flame]

I saw this and thought it was 7x

[void niche]

nope! It's 2/7 times x

It can be tricky writing fractions in front of variables, its an easy mistake to make

[scenic flame]

2

• X
  7

Like that?

[void niche]

correct

[scenic flame]

Got it

[void niche]

ok, now we are almost done, we just want to swap the positions of the terms to make it looks like this

we're allowed to do this for the same reason that 1+2=2+1: commutativity

now it looks like y = mx+b where m is 2/7

so far so good?

[scenic flame]

Yeah

I’m gonna try to make it all fit in the box

[void niche]

that will be a tribulation, but I have faith in you

ok: so this is where we use the fact I mentioned earlier

[scenic flame]

Does that mean it may not fit?

[void niche]

It will if you write small and start at the top again

If you have a line with the equation $y=m_1x+b_1$, then the line perpendicular to it has the form $y=-\frac{1}{m_1}x+b_2$

[ocean seal]

Migillope

[void niche]

This looks scary, but it isn't bad

In our case, $m_1=\frac{2}{7}$ (do you see why?). Then we just need to flip the fraction and slap a negative sign in front to obtain $-\frac{7}{2}$

[ocean seal]

Migillope

[scenic flame]

Sigh this is gonna be a very long night I have 27 problems >-<

[void niche]

you can do it!

[scenic flame]

So all I do is flip the 7 and 2?

[void niche]

and put a negative sign in front

that gives you the equation for the second line which looks like this

now you only have one step left: you need to figure out what $b_2$ is

[ocean seal]

Migillope

[scenic flame]

Wait

Where did b come out of?

[void niche]

This is a new line/equation, not the same as the original

we know the slope (m=-7/2), but we dont know when it crosses the y-axis. That is what the b_2 represents

You can essentially just write a question mark there if you prefer, we will fill it in with a number after the next step (at which point we will be done)

let me draw a picture

Now, do you remember what it means to be "perpendicular"?

[scenic flame]

Like this?

[scenic flame]

Yeah like across from eachother but not exactly the same?

Different from parallel

[void niche]

Yeah, so a line being perpendicular to another one means that they meet at a 90 degree angle

In this case, there are a LOT of lines which satisfy this requirement:

for example, I'll draw a few

but! Something you can notice here is that they all cross the y-axis at different points

Do you see that?

[scenic flame]

Yeah

[void niche]

ok, so we know there is only one solution, though. We need to figure out which line is the one that we want

that's what the b_2 represents: it is the point where the line intersects with the y-axis. As you can see in the picture, if we know where the line intersects the y-axis, there is only one possible line!

do you follow?

[scenic flame]

Yeah

[void niche]

awesome, so we have this

awe why did it turn sideways

(its the same picture from before)

[scenic flame]

I see

[void niche]

We need to solve for b_2. Luckily the problem gave us a point that is on the line, (-1,9)

so you can just plug in -1 for x and 9 for y, then solve

Do you know how to solve for b_2 from here?

woops hold on

[long iris]

Which question are you in 3 or 4?

[void niche]

3

[scenic flame]

No-

[void niche]

Do you know how to add/subtract fractions?

[scenic flame]

Not rlly

Kinda

Also

When my friend did it he just did this

[void niche]

ok, so you need to get the denominators to be the same, then add/subtract the numerators. See: https://www.kristakingmath.com/blog/adding-and-subtracting-fractions

[long iris]

Migillope i just tried another thing and i found -7x+11=2y is it true?

[void niche]

We're doing all the same steps, he just didn't write them down. Also, he lost points because me made an arithmetic error somewhere

yes

[scenic flame]

Can we take a little pause if that’s ok? My mother insists I do something In the middle of homework-

I’m so sorry

[void niche]

no problem

[scenic flame]

I can ping you when I’m back

[void niche]

👍

[proper venture]

@void niche any chance you're free now? Im getting help in help-1 but im still a bit confused

[long iris]

I miss what you did there but i did it with slope and it was way shorter

[void niche]

@proper venture it looks like riemann is leading you down the right path, just try to answer his questions and youll get there

[proper venture]

okay, can i ping you later if there's a point we both get stuck?

[void niche]

im not sure what you mean that you did it with slope, but this is pretty much as quick as it goes. I'm just elaborating on each step and giving explanations

[void niche]

yeah but dont message in this channel pls

[long iris]

I did it like if a equation is ax+b=cy+d c/a is the slope

And for the first equation slope was 2/7

They are two perpendicular lines their slopes multiplation must be -1

[void niche]

I don't think it is a great idea to use equations without meaning. Can you prove that formula is true? Do you know why it is true? If so, then it is a fine shortcut, but it only really saves 2 lines of work anyway

[long iris]

Oh ok

[scenic flame]

Sorry quick question

My friend wants to know where he went wrong in solving it so he doesnt make the same mistake again

[void niche]

this is incorrect

[scenic flame]

@void niche this channel won’t close itself right? Cause this thing may take a bit

I’m so sorry

[void niche]

np, it closes after some period of inactivity

[scenic flame]

Oh…

[void niche]

you can just send a dot every now and then

the period is quite long

[scenic flame]

Alrighty

Tysm

.

[lone heart]

@scenic flame Has your question been resolved?

[lone heart]

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[wise basin]

.rotate

This is what I did but I don’t think it’s right lol

Answer doesn’t make sense so it definitely is not

[placid zinc]

You've correctly identified that:
w•h = x² + 7x + 8

Because you started with lhw, and divided out l

[wise basin]

Yeah

Then I used synthetic division

[placid zinc]

You don't want wh, you want w and h.

Still a step away.

[wise basin]

Ah crap

Now I’m confused

Nvm

I tried factoring it but I can’t

[placid zinc]

Well, quadratic formula always works

[wise basin]

I’m dumb

For some reason I dropped the negative

It should have been x^2 + 7x - 8

Because -36 + 28 is not +8

[placid zinc]

Haha. Well, that factors a lot better now!

[wise basin]

Yes the other one was not possible to factor into whole numbers

[jade sleet]

Hello

What do you need help with

[wise basin]

Figured it out

.clpse

.close

[lone heart]

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[lone heart]

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[rocky ledge]

hello I just need a little bit of help. so I understand everything except for why the power (2/3) becomes a negative

[glass lichen]

$a^{-n}=\frac{1}{a^n}$

[ocean seal]

Mosh

[glass lichen]

just an exponent law

[rocky ledge]

so its reverse of that image?

wait

i dont get it

[gray isle]

which part of that law don't you understand

[lone heart]

@rocky ledge Has your question been resolved?

[rocky ledge]

i get that in order to get a negative index number u need to put a 1 over it but how does it work for the other way?

[gray isle]

the same principle applies

consider that
$$a^{-n}\cdot{a^n} = 1$$

[ocean seal]

ℝamonov

[gray isle]

dividing both sides by $a^n$ to get what mosh posted,
you can also divide both sides by $a^{-n}$ to get
$$a^n = \frac{1}{a^{-n}}$$

[ocean seal]

ℝamonov

[gray isle]

or use various other exponent laws to justify it

[alpine sable]

Alr thx

[rocky ledge]

?

[lone heart]

@rocky ledge Has your question been resolved?

[lone heart]

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[gaunt herald]

Formula of compund interest?

[alpine sable]

ye?

[gaunt herald]

150000x(1-2.4/100)^4?

[alpine sable]

is the answer wrong or smthing

[gaunt herald]

Yeah its coming wrong..

[alpine sable]

hm

let me solve

[gaunt herald]

Okayy

Or it might be my calculator? Idk

[alpine sable]

no

its

interest

the amount of interest

not the final answer

[gaunt herald]

Hmm

Ohhh

150000 - ans?

[alpine sable]

yep

whats the answer

you've been fgiven?

[gaunt herald]

13889.8

The answer on the yt video is

[alpine sable]

alright

can you gimme a min

i think i did something wrong

[gaunt herald]

14,926.74

Take ur time man

[woven spade]

150 000*(1.024)^4-150 000

@gaunt herald

[gaunt herald]

Yeah

[woven spade]

plug that into wolfram

[alpine sable]

= w 150 000*(1.024)^4-150 000

[woven spade]

, 150000*(1.024)^4-150000

,calc 150 000*(1.024)^4-150 000

[ocean seal]

The following error occured while calculating:
Error: Unexpected part "000" (char 5)

[woven spade]

...

[alpine sable]

na we already did that

the answers wrong

[woven spade]

really

aight

[alpine sable]

wait

is it

[gaunt herald]

Wait how did u get 1.024

[woven spade]

2.4 percent

[gaunt herald]

Okkayy but then itsss

[alpine sable]

,calc 150000*(1.024)^4-150000

[ocean seal]

Result:

14926.7441664

[alpine sable]

EY

ye i think i made a mistake

in calc

ulation

[woven spade]

look

[alpine sable]

yep bremperor's right

[woven spade]

if you have 5 dollars

[gaunt herald]

1-2.4/100?

[woven spade]

and you have 10 percent interest

[gaunt herald]

Yesss

[woven spade]

wouldn't it be 5*1.10?

[gaunt herald]

I dont know tbh

[woven spade]

it is

[alpine sable]

LOL

[gaunt herald]

From where did we get the 1

🥲

[woven spade]

the 10 comes from the fact that percentages are just fractions out of 100

the 1 comes from the fact that you keep the original

now of course

[gaunt herald]

Waitwait

OHHH YEAHHH U RIGHT

But man

I just did a dumb mistake lookk

[woven spade]

?

[gaunt herald]

Instead of putting minus it was supposed to be +

[woven spade]

don't worry

happens to the best of us

[gaunt herald]

As well as i get what you mean thank youu

[woven spade]

welcome :)

always glad 2 help

close?

[gaunt herald]

Nono not yet

[woven spade]

ok

[gaunt herald]

Why did we do 4-150000 tho?

[woven spade]

we don't

[gaunt herald]

4 minus or divide?

[woven spade]

it's the whole jazz before

minus 150000

[gaunt herald]

OHhHhh

[gaunt herald]

Like here

[woven spade]

(150000*(1.024)^4)-150000

think of it that way

[gaunt herald]

OhHhh

[woven spade]

:)

[gaunt herald]

Yyesss

I GET UUU

THANK YOU SO MUCH AGAIN

[woven spade]

glad 2 help

[gaunt herald]

@alpine sable THANK YOU TOO MANN really appreicate it

[alpine sable]

yes i did everything

[gaunt herald]

Im glad so many kind ppl exist and i hope they increase like the prices is increasing💀

Take care yall

[woven spade]

least i could do, i receive so much help from the server

give back what i owe i guess

take care dude

[gaunt herald]

U too man and and for that im also trying my best but the more i see rooms the more complicated stuff i see

And like im just taking igcses this year o..

So.*

Thanks againnn

[woven spade]

welcome man

[gaunt herald]

.close

[lone heart]

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[sour peak]

The product of two numbers is 50. Assume one number is x. Express the sum
of the two numbers in terms of x.

[rancid rivet]

x+50/x

?

[woven spade]

i think that's it

ye

that's your answer

[sour peak]

waht how

[woven spade]

look

one number is x

the other is 50/x because x times that number is 50

you want the sum of the numbers

which is x +50/x

[sour peak]

ohh okeoke

[woven spade]

.close

[lone heart]

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[celest herald]

How do you do this?

[lone heart]

@celest herald Has your question been resolved?

[celest herald]

<@&286206848099549185>

[alpine sable]

aight im solving it

dont close

[celest herald]

Ok

[alpine sable]

Cubopia im still here btw

this is a complicated question

what grade u in?

yo @split oriole

sorry for ping

but could u help out lol

[split oriole]

Wait i am helping someone

[alpine sable]

bet

[celest herald]

its for a competition

apparently a grade 7-9 question

@alpine sable

if you want to ask multiple questions do you use the same channel or make different channels for each question?

[alpine sable]

no fucking clue

[lone heart]

@celest herald Has your question been resolved?

[lone heart]

@celest herald Has your question been resolved?

[alpine sable]

yo @split oriole

if ur not free its cool

but if ur free

could u help pls lol

[split oriole]

I will try ,but can't promise

[alpine sable]

ty ily

[celest herald]

Is it possible?

[split oriole]

Dm

Me

[alpine sable]

can u show me solution aswell 🙂

[split oriole]

I haven't solved this

May be question have some issue

Can't say

[alpine sable]

7 -9

yep

the question def has issue

[celest herald]

Mhm

[alpine sable]

Im in grade 7

wait

do u habe answr

[celest herald]

No

[alpine sable]

hm

[grizzled path]

the answer is 45 deg.

but i dont have the steps to get to that

i just drew it in desmos and measured the angle

[alpine sable]

LOL

[alpine sable]

wai

did u measure it with a device

or did u figure it out

cause i laughed thinking u measured it with some kinda website or smthig

[grizzled path]

desmos has a geometry mode

and it allows you to do things like that

i did measure it with desmos

[lone heart]

@celest herald Has your question been resolved?

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[rocky atlas]

What is tangent of the angle?

[split oriole]

Tan theta

[rocky atlas]

I see

Why is tan theta= (m2-m1)/(1+m1m2) suddenly converted (a2b1-a1b2)/(a1a2+b1b2)

Nvm got it

.close

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[fallen rain]

I think is 3, but i'm note entirely sure

[sage violet]

yep u are right supposedly

[fallen rain]

How many people enjoy action only?

I'm not sure about this question?

[sage violet]

ooh ok

so you have all of the rest of them filled ryt?

and you have 120 total people

so just subtract all of those numbers

and

your arithmetic may be wrong

[fallen rain]

I got 41

[fallen rain]

I didn't add 7 to total intersection for that answer

[sage violet]

how about the one that doesnt like any

[fallen rain]

o yeah

41-1=40

thank you @sage violet

.close

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[fading torrent]

'make r the subject of the formula A=πr^2'

[fading torrent]

would it be r = √A/√π

[glass lichen]

it'd help if there was r in this

[fading torrent]

there.

[glass lichen]

then yes, assuming r>0

[fading torrent]

ok

and

'make c the subject of the formula a = P/b+c'

[small trellis]

Yeah, but you can just write the fraction in one square root

[fading torrent]

okay

[glass lichen]

unclear typing

a=P/b+c or a=P/(b+c)

[fading torrent]

second one

[small trellis]

How do you get a negative radius?

How could you, I mean

[glass lichen]

You couldn't however a priori you don't know if r is positive.

[fading torrent]

'make c the subject of the formula a=P/(b+c)'

[lone heart]

@fading torrent Has your question been resolved?

[fading torrent]

<@&286206848099549185>

[crimson carbon]

Get the reciprocal get b+c and yeah that's about it

[fading torrent]

what

can u explain step by step please @crimson carbon

[crimson carbon]

So what I'm tryna do here is get b+c as the numerator to do that you'll have to get the reciprocal (multiplicative inverse).

The reciprocal of a number x is denoted as 1/x, or x^(-1); a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. So it just swaps the numerator and the denominator. Of course whatever you do to either side of the equation you'll have to do to the other.

So that would get me 1/a=(b+c)/P

Now of course get rid of P

That'll be P*1/a which is P/a=b+c

subtract b

P/a-b=c

Transposition yeah

[fading torrent]

ok

i didnt understand that at all

but thx

[crimson carbon]

ok what don't you get

[fading torrent]

everythign

we havent learnt that yet

[crimson carbon]

ok let's simplify everything i said

Well let's say there's 2 ways we could do this one way which i've shown

We'll do the second one then i'll get back to simplifying the first method

oh no wait

Have you learnt how to factor?

factorise?

[fading torrent]

no

i might have actually

just do the factorising one then i might recognise it

[crimson carbon]

actually i'm being stupid bruh

ignore me no need for the factorising okay

So let me just do it

and hopefully you see what goes on

a=P/(b+c)
a(b+c)=P
b+c=P/a
c=P/a-b

[fading torrent]

yes i understand that

thank you v much

[crimson carbon]

okay I want to explain the first too

So what i done was I got the reciprocal which basically swaps the numerator and the denominator

as said the reciprocal of a fraction a/b is b/a

So when i get the reciprocal of the fraction in the RHS (Right Hand Side of the equation, LHS (LH, LS) = RHS (RH, RS)) you have to do it on the LHS too

Because whatever you do on one side you gotta do it to the other

So I have a=P/(b+c)

Get the reciprical

1/a=(b+c)/P

Then we multiply P

you get P*1/a which is just P/a because a1=a, and a(b/c)=(ab)/c

[fading torrent]

yes

[fading torrent]

do u mean c=P/(a-b)?

[crimson carbon]

so yeah subtract b

no

We use () for grouping and clarity because y'kno writing P/a-b is ambigious cause people interpret it as

[fading torrent]

so like this

right?

[crimson carbon]

yes

[fading torrent]

ok thxc

[crimson carbon]

But do you understand though?

[fading torrent]

yes

[crimson carbon]

are you sure?

[fading torrent]

ive been working on this hw for 3 days

[crimson carbon]

[fading torrent]

so im glad to get it finished

:D

[crimson carbon]

okay so when it comes to algebraic manipulation

Learn some properties and methods

and when solving for a variable you leave whatever is acting on it last

and work the order of operation backwards

idk how to explain it properly

so for example

ab+c=d solve for b
ab=d-c
b=(d-c)/a

[fading torrent]

yes

[crimson carbon]

okay i hope my explanations weren't so convoluted

[fading torrent]

any resources u would recommend for improving at maths

[crimson carbon]

Dunno honestly

Just watch videos

or buy books

I'd usually watch Khan Academy or The Organic Chemistry Tutor

on YT

[open folio]

^^^

[crimson carbon]

books are a lovely source of learning

[fading torrent]

ok i will look into books then

[crimson carbon]

pirate them or buy them 🙂

Go read Spivak's Calculus

[fading torrent]

lol

i am not picking further maths so i dont have to do calculus B)

[crimson carbon]

mk

But don't you learn calc as a maths student tho

like a non-further maths student

I mean that's if you're taking maths for further education

[fading torrent]

no here im doing just normal maths gcse

[crimson carbon]

What year you in

[fading torrent]

in england it would be 10

[crimson carbon]

mk

[fading torrent]

wbu

[crimson carbon]

one above you

year 11

Close the channel once you're done with .close

I believe she's done

.close

[lone heart]

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@tawny dawn Has your question been resolved?

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[jaunty bridge]

so far, i know that BCD is an isoceles and BC is sqrt 6 too

[crimson carbon]

Look at some triangle proportionality theorems

and it is given that segment BD is a bisector of angle ABC

so you could use the angle bisector theorem

what am i saying

well it could apply but you don't have necessary things

I need to read questions more often

So since you know that BD is a bisector of angle ABC

then necessarily <CBD=<ABD

Have you been taught to use the sine rule?

[jaunty bridge]

yea but kinda forgotten how to

[crimson carbon]

ok no worries

[crimson carbon]

a,b,c are the sides of the triangle, alpha, beta, gamma are their respective angles

the opposite angles

i mean quite evident

is my discord dying

test message

ok i'm on the web browser

because my discord app is dying

[jaunty bridge]

ooo okok

[crimson carbon]

okay and R is the radius of the triangle's circumcircle

brb

god damn my mum needs me to go lift some things

i'll be back in like

idk 10 mins

okay nevermind

where was i

okay yeah sometimes the law is expressed using the multiplicative inverse (reciprocal)

Because why not

Usually whatever you're solving for (angle, or side) you'd have them on the numerator

i.e. if you were solving for angles you'd have sin alpha/a = sin beta/b yada yada

Okay given that BD is root 6, and you have angle BAC = 45 degs and knowing that angle ABD = 30 degs you can solve for DA

just sub them in

[jaunty bridge]

(sqrt 6)/ 45 = AC / 60

?

[crimson carbon]

no

[jaunty bridge]

hm

[crimson carbon]

Where'd ac come from and where'd the sins go

[jaunty bridge]

oh

[crimson carbon]

$\frac{\overline{DA}}{\sin(30^\circ)}=\frac{\sqrt{6}}{\sin(45^\circ)}$

[crimson carbon]

Yeah

Should be obv

[jaunty bridge]

but the q wants to find AC

[crimson carbon]

Yes i know that

But we have to find AD and DC separately

AD+DC=AC

okay what aren't you getting

[jaunty bridge]

why cant find ac straigh away hm

also does having sin or sqrt in the eqn mean i have to use calculayor?

[crimson carbon]

I mean it's not a choice really

you'd have to

Unless you are a human calculator

[jaunty bridge]

ah

this paper cant use calculator tho hmm

[crimson carbon]

oh wtf

Okay then

Are you familiar with the unit circle?

and your trivial identities?

[jaunty bridge]

uhhh no

ahhhhhhhhhhh

[crimson carbon]

Okay since you have sin(30), and sin(45) just look at the unit circle

and find the values

But why can't you use a calculator

[jaunty bridge]

this math olympiad comp

cant use calc

[jaunty bridge]

ah okok

[crimson carbon]

oh wtf olympiad

okay

Just remember the unit circle

did you find them yet?

they're adjacent to each other

[jaunty bridge]

yeap

[crimson carbon]

The angles expressed as pi is radians

which you probably know

another unit of measure for angles

you got degrees, radians, gradians

so sin(30 degs) and sin(45) is 1/2 and sqrt(2)/2 respectively

Which means

$\overline{DA}=\frac{.5\sqrt{6}}{\frac{\sqrt{2}}{2}}$

[crimson carbon]

Apply a fractional rule

then factor

[jaunty bridge]

hmm why cant find AC straight away?

[crimson carbon]

then cancel

Because we don't have the necessary information to calculate it directly

Wait let me have a good look at the information provided and see what i could do with it

I don't think we could calculate it directly

Yeah i don't think we can

We have to find CD and DA

you there?

[lone heart]

@jaunty bridge Has your question been resolved?

[lone heart]

Channel closed

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[jaunty bridge]

.reopen

[lone heart]

✅

[jaunty bridge]

@crimson carbon i think i found a way

theres something when u draw a perpendicular line from C to BA

u get a 30-60-90 triangle

which gives a ratio of 2a : a: sqrt 3a

sqrt 6 = 2a

and then CEA is an isoceles right angle

then pythogaros walaaaaaaaaaaaaaaaaa

AC is 3

tqsm

[crimson carbon]

wait what's 'a'

nvm

aight close it if you're done

i'll check again later

gonna have to leave now

[lone heart]

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Please don't occupy multiple help channels.

[ocean seal]

Kepe

Kepe

[lone heart]

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[brave remnant]

I need help with python. Can someone help me in voice chat? It is urgent :/

[brave remnant]

<@&286206848099549185>

[bleak ridge]

You need help with like syntax or something?

Or math

[brave remnant]

no. I am using this framework called pandas

and I am trying to plot a figure

but I want to minimize the ammount of data being used from the CSV

I can't

I have read the documentation

and stackoverflow + other sources

nothing works

still have the same issue

[bleak ridge]

Why is it urgent?

[brave remnant]

assignments is due very soon :p 😦

[bleak ridge]

So youre just trying to optimize something that already works?

[brave remnant]

yes

when you run it now it almost explodes my pc

because it is reading like 2 gb of data

[bleak ridge]

What kinda data is it

Like a x y plot?

[brave remnant]

a CSV

yes

exactly

[bleak ridge]

Hows it formatted about

x,y,x,y,x,y or smthn?

[brave remnant]

do you have competence in this?

[bleak ridge]

Ive never used pandas but I have cs experience

[brave remnant]

can you join my channel

much easier

[bleak ridge]

Atm not really

[lone heart]

@brave remnant Has your question been resolved?

[tacit arch]

There's a python discord server that's pretty big. There's also a data science server that has a couple people familiar with pandas

[lone heart]

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[rose rivet]

Is this correct?

[rose rivet]

.close

[lone heart]

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[rose rivet]

.close

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[wide raven]

help required

I tried using the parametric form of a circle and plugging in the form in the first equation but it just looks like it ends nowhere

[alpine sable]

question 1 is 5a plus 3a

[wary stream]

Are you asking a question or giving out an answer to the question asked above?

[alpine sable]

im asking a question

[wary stream]

Because if you're asking a question, don't ask in an occupied channel
Read #❓how-to-get-help

[alpine sable]

ok thank u so much

[wary stream]

You were linked that like 3 times before

[wide raven]

thanks bud!

[alpine sable]

ok sorry

bye

[lone heart]

@wide raven Has your question been resolved?

[lone heart]

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[alpine sable]

In an examination 35% of the candidates failed in Mathematics, 40% failed in Science and 15% failed in both. If the total number of candidates be 500, find the number of candidates who passed in both the subjects.

[alpine sable]

Can someone help pls

i think i can take a stab

give me a moment to draw something

[glass lichen]

what have you tried?

[alpine sable]

so we have 4 distinct groups of students

were looking for the number that passed both, the bottom right box

[glass lichen]

Try waiting for them to actually respond

[alpine sable]

yeah

@alpine sable you still there?

welp im going to assume you got busy if you end up needing help with this again give me an @

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[alpine sable]

Hey guys

[alpine sable]

Can someone help me out

I’m not sure how to find the average of these numbers first 0.59, 0.59, 0.60

[undone hinge]

what is the average of 1, 2, 3?

its 2 so how did you find that mathematically?

[alpine sable]

I know I have to add the three numbers and divide the numbers by how many there are

However it gives me a strange number after I do so

[crimson carbon]

which is .593333333...

yeah?

nothing wrong with that

the 3 is just recurring

[alpine sable]

It gives 1.38

[crimson carbon]

how'd you get that

[alpine sable]

Just did 0.59+0.59+0.60 divided by 3

[crimson carbon]

and you got 1.38?

[alpine sable]

Yes

[crimson carbon]

[alpine sable]

Might be something wrong with my calculator or something

[crimson carbon]

$\frac{2\cdot.59+.6}{3}$

[crimson carbon]

are you sure you exactly typed this

[alpine sable]

Oh no I didn’t type that in fraction form

I just did it normally that’s probably why

When I do it in brackets it says 0.5933 though

[crimson carbon]

tmw fractions are abnormal

[crimson carbon]

yes that's probably why

(2*.59+.6)/3 is not the same as 2*.59+.6/3

[alpine sable]

Alright so just confirming 0.60, 0.60, 0,59 would be 0.5966

Just to check if I’m doing the formula right

[crimson carbon]

yeah .5966666667

[alpine sable]

Alright thanks for helping me out

[crimson carbon]

The AM is also called the Arithmetic Average

or simply mean or average when the context is sufficiently clear

I just expressed it using summation notation

otherwise it's just

[crimson carbon]

@alpine sable if you're done close the channel using .close

[alpine sable]

.close

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[alpine sable]

Why is mass negative?

Apparently it's F = -80 * 1.5 according to the mark scheme

[wary stream]

I don't think it's necessarily saying mass is negative but acceleration due to gravity it negative

Down is negative

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@alpine sable Has your question been resolved?

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[worldly igloo]

I am being asked to prove that among 1,002 positive integers, there are always two integers whose sum or difference is a multiple of 2,000. I'm not really sure where to start, but will I end up having to prove that some sum divides 2,000?

[warm orchid]

I think yes, it look like an application of congru modulo or whatever it's called in your country

(the equal with 3 bars)

[worldly igloo]

Hmm interesting

[warm orchid]

But no idea on how to do it

[worldly igloo]

Oh, okay

.close

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