help-38

[hollow tangle]

5x^2 - 10x + 3x - 6
5x(x - 2) + 3(x - 2)
(x - 2)(5x + 3)

get it?

[eternal vapor]

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[hollow tangle]

if x^2 coeff in not 1 then do this

[eternal vapor]

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[hollow tangle]

k

[eternal vapor]

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[hollow tangle]

do all the terms have something in common?

[eternal vapor]

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[hollow tangle]

get it?

yes

then?

[eternal vapor]

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[hollow tangle]

.

[eternal vapor]

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[hollow tangle]

basically all brackets should only have x

[eternal vapor]

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[hollow tangle]

no x^2 or anything

no

[eternal vapor]

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[hollow tangle]

no

[dapper swift]

also 5x^2 - 7x - 6 is not (x + 10)(x - 3)

[hollow tangle]

hey higher

[dapper swift]

grrr

[hollow tangle]

.

thnx tho

[eternal vapor]

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[eternal vapor]

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[hollow tangle]

oh wait your his brother

no wait you arent even related to him

mb

[eternal vapor]

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[dapper swift]

I'm higher's secret twin brother

[eternal vapor]

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[hollow tangle]

not secret anymore

so you arent his secret twin brother

eng teachers would make us write an essay on this "paradox"

[eternal vapor]

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[hollow tangle]

yess

now?

[dapper swift]

wait that's still not right sorry

[eternal vapor]

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[dapper swift]

it's (x - 2)(5x + 3)

[eternal vapor]

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[hollow tangle]

thats on me lmao

[dapper swift]

everything else is good tho

[eternal vapor]

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[dapper swift]

try (9x + .....)(x + ....)

[hollow tangle]

fixed it

use the same method

[dapper swift]

so 2 and 1 go in some order in here to make the middle term 19x

[hollow tangle]

write sum and prod

[eternal vapor]

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[eternal vapor]

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[dapper swift]

which part don't you get?

oh yeah we are using different methods

2 and 1 go in some order in the .....

[eternal vapor]

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[eternal vapor]

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[hollow tangle]

is possible

[dapper swift]

well if we do (9x + 2)(x + 1), the middle term is 9x * 1 + 2 * x = 11x, so that's not it

[dapper swift]

no

[dapper swift]

now try (9x + 1)(x + 2)

see what the middle term is now

[eternal vapor]

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[dapper swift]

yeah

[eternal vapor]

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[dapper swift]

nearly

you should factor 81x^2 + 18x + 1 as well

hint: it's a perfect square

[eternal vapor]

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[dapper swift]

oh wait no, 81x^2 + 18x + 1

[eternal vapor]

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[dapper swift]

perfect

[eternal vapor]

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[eternal vapor]

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[dapper swift]

factor 81x^2 + 9x

[eternal vapor]

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[dapper swift]

common factor

[eternal vapor]

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[dapper swift]

difference of squares is a^2 - b^2

[dapper swift]

yeah, 9x times what?

[eternal vapor]

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[dapper swift]

cool

[dapper swift]

so this becomes 9x(9x + 1) + (9x + 1)

write this as 9x (9x + 1) + 1 * (9x + 1)

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[dapper swift]

yeah!

and lastly 2x + 6 = 2(x + 3)

now a lot of things should cancel on your paper

[eternal vapor]

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[dapper swift]

,w (5x^2 - 7x - 6)/(2x^2 - 8) * (9x^2 + 19x + 2)/(x^2 + 7x + 12) * (2x + 6)/(81x^2 + 18x + 1) simplify

[solid kiln]

Results provided by WolframAlpha

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

Requested by sooouth.

[eternal vapor]

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[dapper swift]

hopefully you get this after cancelling

[eternal vapor]

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[dapper swift]

yeah

flip cause you're dividing, to turn it into multiplying

[eternal vapor]

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[dapper swift]

oh well I mean I guess tests make you learn
it's not a pleasant feeling I know

[dapper swift]

oh yeah I wanted to address the last part of the question, 'state the restrictions'

basically, when is this defined?

[eternal vapor]

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[dapper swift]

so do you remember, what number can you never divide by?

[eternal vapor]

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[dapper swift]

yep!

[eternal vapor]

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[dapper swift]

so if you look at the first result here

we don't want (x - 2)(x + 4)(9x + 1) to be 0

oh wait

nah I swear I typed it correctly

[eternal vapor]

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[dapper swift]

yeah

if (x - 2)(x + 4)(9x + 1) = 0, we are dividing by 0 !

[eternal vapor]

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[dapper swift]

x - 2, x + 2, 9x + 1 after you flip

oh and x + 3 after you flip

[eternal vapor]

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[dapper swift]

idk exactly

[dapper swift]

but basically you are cancelling 9x + 1 with (9x + 1)(9x + 1)

so there is one 9x + 1 left on the bottom

[eternal vapor]

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[dapper swift]

so just erase the 2

cause it's (9x - 1)^1 = 9x - 1

[eternal vapor]

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[dapper swift]

then combine everything that isn't cancelled

,w (5x^2 - 7x - 6)/(2x^2 - 8) * (9x^2 + 19x + 2)/(x^2 + 7x + 12) * (2x + 6)/(81x^2 + 18x + 1) simplify

sorry yes there was a typo in the Wolfram Alpha

[solid kiln]

Results provided by WolframAlpha

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

Requested by sooouth.

[dapper swift]

wooohoooo

[eternal vapor]

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[dapper swift]

yes cause 2 and 2 cancels

[eternal vapor]

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[dapper swift]

no leave it like that

but 2 and 2 cancels

[eternal vapor]

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[dapper swift]

also it's 9x + 1 not 9x - 1

[eternal vapor]

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[dapper swift]

it was (9x + 1)^2 the entire time

[eternal vapor]

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[dapper swift]

yeah

[eternal vapor]

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[dapper swift]

how what?

[eternal vapor]

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[dapper swift]

(9x + 1) cancels once with (9x + 1)(9x + 1)

so you get 1 on top and 9x + 1 on the bottom

[eternal vapor]

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[dapper swift]

yeah

my Wolfram output had a typo after all

[eternal vapor]

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[dapper swift]

remember the thing you said about not dividing by 0

[dapper swift]

that means that (9x + 1)(x + 4) can't be 0

[eternal vapor]

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[dapper swift]

oh nearly

yeah so x can't be -4, cause x + 4 would be 0

so how about 9x + 1 = 0 then

[eternal vapor]

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[dapper swift]

yeah you can just solve for x yk

[eternal vapor]

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[dapper swift]

this first

subtract both sides by 1

[eternal vapor]

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[dapper swift]

yeah so x = -1/9

that means that $x \ne -1/9, -4$

there you go

[solid kiln]

higher's secret twin brother

[eternal vapor]

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[dapper swift]

if you recall a^2 - b^2 = (a + b)(a - b)

you just have (4x - 7 + y)(4x - 7 - y)

[eternal vapor]

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[dapper swift]

yes

[eternal vapor]

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[dapper swift]

ok pls leave me alone

if you have another question, just close this one and open a new help channel

[eternal vapor]

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[slow horizon]

Help

[slow horizon]

Is this true or false ?
If it true explain those steps

[trail ingot]

[slow horizon]

[trail ingot]

these are exactly the same, right?

[slow horizon]

Yes

[trail ingot]

so if you throw this on one of them...

[slow horizon]

So?

[strong zinc]

Then they’re not the same? Bc one is $\sum{nk+k^2}$ and the other is is $\sum{nk+k^2+k}$

[solid kiln]

Pixelius

[trail ingot]

yes

[cosmic meadow]

Unless my dyslexia is doing his job, these are not the same.

[slow horizon]

Hmmm

[strong zinc]

Oh I thought this was laylas question lol

[slow horizon]

[slow horizon]

Lol

@strong zinc in fact i want just to understand those two steps

[trail ingot]

if you are asking whether these are equal, then no definitely not

[strong zinc]

Distributing $k(n+k)$ and combining the sums via the linearity of summation?

[solid kiln]

Pixelius

[slow horizon]

Those two steps is a part the solution of this exercise

[slow horizon]

@strong zinc

[trail ingot]

[strong zinc]

As Layla said, line 3 does not equal line 2. That proof must be wrong

[strong zinc]

https://math.stackexchange.com/questions/534714/evaluate-sum-k-0nkn-k
Not identical but should help you

[slow horizon]

Hmm

I used recurrence solving @strong zinc

I will show you my demonstration @strong zinc

[strong zinc]

Let’s see it

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[unkempt wyvern]

Anyone can help?

[torpid lantern]

4/2

2 mins per hour

2/6

1/3 mins per hour

u there?

[trim jolt]

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

[unkempt wyvern]

hey eyah

i'm here

i'm not really sure how to write the equations to find the time with these

i tried converting everything to minutes and then i got these 2 equations but i don't trust them

[mellow tendon]

You can start with writing the velocities of the 2 watches down

Angular velocities
Then put the angular distance as a function of time

[unkempt wyvern]

so john's watch goes 4 mins per 2hours that means 2 minutes per hour = 12degrees/h

and peters watch 2 mins per 6 hours or 1 minute per 3 hours or 2degrees per hour?

but how can i use that to find when the watches will show the same time again

[mellow tendon]

Hmm
J = 24 minutes per 12hour
P = 4 minute per 12 hours
12 hours = 720 minutes
24 * t = 4 * t (mod 720)
20 t=0(mod 720)
20 t = 720
t = 36 * 12 hour = 432 hours

[trim jolt]

@unkempt wyvern Has your question been resolved?

[unkempt wyvern]

so you matched both by hours

what do you mean by mod?

[mellow tendon]

Mod is something used to limit the numbers to a certain set of numbers
It is usually used when dealing with clocks due to the fact that when analog clock passes the 12 hour mark it resets back to 00:00

You can search it
Modular arithmetics

[unkempt wyvern]

Hm ok

Can you give me a thought process to solve these types of problems?

Like what you did was find a common time between both people

Convert all times to minutes

But i don’t get what you did after that

[mellow tendon]

Ok
First I made both velocities with the same unit which is minutes per 12 hours
Then because as said later after 12 hours or 720 minutes the time resets to 0

For the 2 clocks to be equal after time t they need to have the same value in mod 720
So 24 * t = 4 * t in mod 720
Then I subtracted 4*t from both sides to get
20 t = 0 in mod 720
This equation here has an infinite number of solutions
The reason relays on the fact that in modular arithmetics 0 = 720 = 1440 = etc
And it kinda makes sense because the clocks would keep getting matched over and over and over for an infinite amount of times assuming they work indefinitely

But what we are considering here is the first match which why we choosed the value 720 because it is where the first match (after setting) will happen

Solving that gives t = 36
But remember that the unit we used was minutes per 12hours
Meaning that the t here is in the units of 12hours
So to turn it to units of hours we just multiply by 12 to get 432 hours

[unkempt wyvern]

Im gonna understand the modulo thing more because i don’t really get it, but its basically that after that amount of time the clock basically rests in the meaning that 1am and 1pm are the same

Thank you very much for this explanation though really cleared it up

[mellow tendon]

I recommend you to learn about it.
It is a cool and easy field of mathematics

I am thinking of other ways to solve it without moduler arithmetics

[unkempt wyvern]

I know chatgpt is not a reliable source but it did this

[zinc ginkgo]

don't use chatgpt for math

also don't waste helpers time reviewing its nonsense

[unkempt wyvern]

My bad

[mellow tendon]

Ok it was a bit helpful till the point of difference in Tim

So there is a difference velocity 40 minutes

For the 2 clocks to get to the same time
The difference needs to reach 12 hours

So they are basically the same on an analog clock

12 hours = 720 minutes

720 minutes/ 40 minutes= 18

Meaning they will get in time after 18 days

18days * 24hour = 432 hours

That's another valid way to solve the problem @unkempt wyvern

[unkempt wyvern]

Okay i think i get it

The 18 refers to days because both 720 and 40 are per 12 hours?

[mellow tendon]

It refers to days because the difference between 2 clocks increases by 40 minutes per days

So for the difference to reach 12 hours or 720 minutes
It needs 720 minutes over (40 minutes per day) = 18 day

[unkempt wyvern]

Ahh okok i see

Thank you very much for your time

Hope you have a great evening

.close

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[tribal fractal]

can someone check if i did this right? for part (a) i got 5, for (b) i got no zero rows, and for (c) i got 2

[astral mortar]

Maybe you should post your working

[tribal fractal]

ok

so for (a), i believe that there are 2 free variables, and there are seven columns, so the rank is 7-2=5

for (b) no zero rows since the rank = # of rows

for (c) its just the number of free variables

which i found from the number of vectors in the basis for the null space

<@&286206848099549185>

anyone?

please

<@&286206848099549185>

anyone please

.close

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[tropic timber]

Can anyone check this proof for me? I think I did it right but I’m not sure if I used the absolute value property right

[trim jolt]

@tropic timber Has your question been resolved?

[tropic timber]

<@&286206848099549185>

[oak idol]

@tropic timber Hello

[tropic timber]

hi!

does my proof look okay :')

[sleek canopy]

are you still there ?

[tropic timber]

I am..?

[sleek canopy]

you delta should not depend on x

but only on epsilon

[tropic timber]

oh 😭

wait what do you mean

ah

[sleek canopy]

like to prove that f(x)=x is continuous in x=0 you can chose $\delta=\epsilon$

[solid kiln]

everg

[tropic timber]

wait but then how would I get delta=epsilon?

I don't really see a way of getting rid of the x

[sleek canopy]

but you cannot chose $\delta=x\epsilon$

[solid kiln]

everg

[tropic timber]

hm I think I get that

[tropic timber]

so I'd still be trying to find a way to manipulate x-1 so that it's equal to 1/(x^2-2)+1?

would you say I was on the right track with (x+1)(x-1)/(x^2-2) 😭

cuz that's the only way I'm seeing to get an x-1 out of it

[sleek canopy]

something like that

[floral pilot]

i need hel with this guys

[tropic timber]

I'm a bit confused here. It looks like you just added 1 to x and delta, but why did the zero turn into 1-delta instead of just 1?

and is it alright to separate the 1 from the x since they were both in that absolute value?

[floral pilot]

oh okay ty

let me see if it works

i was confused on it

[tropic timber]

I don't think you're in the right place rn 😭 but f(x) would just be the y value, so on a graph, you look for all the values of x (interval on the x axis) where the parabola is above y=0

[floral pilot]

where should i go then?

[tropic timber]

well do you know where y=0 is on the graph?

[sleek canopy]

the only step that is not entirely clear is the definition of M ...actually it exists cause the g function in continous in that interval ...but at the same time you are not giving any value ...but I hope this could let you think about how to deal with delta ... and maybe you can find a proper number to put in place of just M

[floral pilot]

no

[tropic timber]

I'm kind of confused about where you got M and g from 😭

[tropic timber]

hm well do you know what the x and y axis are?

[sleek canopy]

g is defined as that ratio in deltas... it just came out from the computation

[tropic timber]

see how the vertical line is labeled y and the horizontal one is labeled x? they're axes that represent the x and y values of a function. so when you look for y=0, you'll look at the place in the graph where the y axis intersects with the x axis

[tropic timber]

my professor never used those terms before though so idk if that's the right way to go about this 😭

are you saying M and g are variables you've come up with to represent numbers?

[sleek canopy]

try to find a proper number in place of M... so you need to find a number such that (like 1000) such that tha fraction is smaller

now I have to go ...hope this has helped

[tropic timber]

oh I'm not sure i understand 😭

thank you though

[sleek canopy]

g is a function that you have to bound

[tropic timber]

I'll try to figure it out :')

[sleek canopy]

M is a constant ...and you have to find a specific value in place of M such that g(\delta)< that number

[floral pilot]

yeah idk im really confused with this

i miss 1 day and look what happens

it was a new unit too

[tropic timber]

I could help with your thing haha

we can start with the question wording

the graph is of a function, which is basically a relationship between x and y

x is your independent variable, or input, and y is the dependent variable, or output

you'll also see y written as f(x)

just know that y and f(x) are essentially the same thing

for example, let's take the function y=x+1

then if x=1, then y=1+1, so y=2. And if x=2, then y=2+1, so y=3. This holds true for every number, and when we plot those numbers down, we'll get a graph

for example, this is the graph of the example just now, y=x+1

(which can also be written as f(x)=x+1)

on a graph, the y axis, or that thick vertical line, shows the y values

the x axis, or the horizontal line, shows the x values

notice this point? that means when x=0, y=1

[tropic timber]

now your question asks to find all values of x for which f(x)>0

basically, to find the segments of the horizontal axis when the line graphed has a y value greater than 0

[tropic timber]

in the example I just gave, f(x)=x+1, we see that from x=-1 and beyond, the line stays above y=0

now your job is to look at this and look for when this line graphed is above the y level 0

[tropic timber]

does that make sense?

[tropic timber]

I still don't really get it at all 😭
How would I go about finding that number? is there something I'm supposed to base it off of? was I on the right track at all before? your example worked off my previous work, but I don't see how I could use the (x+1)(x-1)/(x^2-2) now that the assumption doesn't include the x-1

[tropic timber]

<@&286206848099549185>

[limpid dawn]

Oh I don't think delta is supposed to depend on x too, only on epsilon

[solid kiln]

bacc the sigma😔🤞

[limpid dawn]

You could use that to estimate (x+1) up again

[tropic timber]

but how would I avoid having to rely on x?

[limpid dawn]

Well we need to somehow use the assumption we are given

[tropic timber]

so to try to get the x+1 to turn into a 1/(x^2-2)+1?

could I square the inequality..?

[limpid dawn]

yea technically

although hmm

if the left side is negative

[tropic timber]

I'm stuck at x^2+2x+1

I don't know how I'd use the x in the 2x 🥲

[limpid dawn]

,, \abs{x-1} \cdot \abs{x+1} \cdot \frac{1}{\abs{x^2-2}} < \delta \cdot \abs{x+1} \cdot \frac{1}{\abs{x^2-2}}

this is where we at

[tropic timber]

wait how did we get there

oh is that what I had before

I thought that was wrong

[limpid dawn]

hmm why

this literally the assumption

[tropic timber]

well you said delta couldn't depend on the x right?

[limpid dawn]

ye

we need to work it out further

[solid kiln]

bacc the sigma😔🤞

[tropic timber]

I don' t get it 😭

[solid kiln]

bacc the sigma😔🤞

[tropic timber]

what am I trying to do with that right now..?

[limpid dawn]

basically we are trying to get rid of the x's and make everything delta

and define that as our epsilon

[tropic timber]

oh wait I thought we were trying to turn the x-1 into 1/(x^2-2)+1

since that's <epsilon so I though twe would then equate delta to epsilon that way

how would we get rid of the xs entirely??

[limpid dawn]

I think we could try to bound |x-1| by considering |x-1| < 1

basically a small distance

that would get us then

-1 < x-1 < 1

[tropic timber]

wait what do you mean "bound"

[limpid dawn]

like the distance from x = 1 should be smaller than 1

[tropic timber]

can we just set x equal to 1 like that?

[limpid dawn]

we are basically shrinking it, considering a small neighbourhood

[limpid dawn]

not equal

but we need something to work with, so we could choose something reasonably

[tropic timber]

what do you mean to work with

like a temporary substitute?

[limpid dawn]

to continue this proof

[tropic timber]

why are we doing that though?

[limpid dawn]

we have the other terms too that we need to take care of, so if we consider a small neighborhood around x=1 we could manage this

[tropic timber]

I'm still not sure I get it..

so we're trying to make everything delta, but why does making x smaller help?

[limpid dawn]

i mean |x-1| < 1000000 how would that help

you are now considering the limit from a very big distance

[tropic timber]

but do we have to set x-1 within something?

[limpid dawn]

no

[tropic timber]

I thought x-1 was just between 0 and delta

how could we change that?

[limpid dawn]

yea and delta

can vary

"there exists a delta ..."

greater 0

[tropic timber]

so we're basically setting delta to 1?

I'm so sorry we can continue and I'll see if I get it as we go 😭

[limpid dawn]

we will see

|x-1| < 1

-1 < x-1 < 1

0 < x < 2

0 < x² < 4

-2 < x² - 2 < 2

|x²-2| < 2

shit

total opposite

1/|x²-2| > 1/2 doesnt help 😭

[tropic timber]

wait huh 😭

did it not work?

[limpid dawn]

yea i have no idea

[tropic timber]

oh-

thanks for trying then 🙏

think I'm gonna skip this problem for now 😭

.close

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Channel closed

Closed by @tropic timber

Use .reopen if this was a mistake.

[limpid dawn]

i cant believe i failed you twice

[tropic timber]

omg no you're fine

would #proofs-and-logic be the place to ask instead?

[limpid dawn]

yes

[tropic timber]

got it 👍

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[rapid ridge]

Dont undetstand this proof

[rapid ridge]

especially on a macro level

[empty wyvern]

Ahh u beat me

[rapid ridge]

:)

[twin yacht]

:)

either way idfk looks like chicken scratch to me 😅

[rapid ridge]

same 😔

[rapid ridge]

context:

@river willow maybe 🥺

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Please do not ping individual helpers unprompted.

[rapid ridge]

😔

pokedance is friend

not foe

[trim jolt]

@rapid ridge Has your question been resolved?

[trim jolt]

@rapid ridge Has your question been resolved?

[trim jolt]

@rapid ridge Has your question been resolved?

[river willow]

hahaha

[river willow]

how i understood it is we used induction and showed using the isomorphism theorems that we can break down the two series into shorter series. and its true on the shorter series due to induction

http://abstract.ups.edu/aata/struct-section-solvable-groups.html

i used this textbook and it could be clearer? give it a look

[rapid ridge]

https://tenor.com/view/cat-gif-7107264540211859015

[rapid ridge]

this is cooked

let me lok though

[river willow]

u good buddy?

i just went over the proof again lol

[trim jolt]

@rapid ridge Has your question been resolved?

[river willow]

say something TnT

[river willow]

@rapid ridge do u need help still

[rapid ridge]

💀💀💀

I kms

[river willow]

can i explain

u understand why the second one is just a series of groups that are all normal in the previous one right

because they are all normal subgroups intersect with a normal subgroup

given normal subgroups A, B of G. an element in A int B is some c.
for all a in A, b in B,
aca^-1=c, as c in A and A is normal in G, a in G (in A)
bcb^-1=c same reason replace A with B

@rapid ridge

r u still following so far

[rapid ridge]

I

Think I'll just

Take the L tbh

😔

[river willow]

nooooo

😭

[rapid ridge]

Thanks as always though poke

[river willow]

did i say something hard

[rapid ridge]

😢😢😢😢

[river willow]

oki well atb

[rapid ridge]

https://tenor.com/view/if-you-dance-ill-dance-and-if-you-dont-ill-dance-anyway-dance-gif-4063821110722853759

[river willow]

where did u know me from anyway

i dont recall having interacted much?

[rapid ridge]

https://tenor.com/view/if-you-dance-ill-dance-and-if-you-dont-ill-dance-anyway-dance-gif-4063821110722853759

Poke fogor me

After sending all the youtube thingya

.close

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[rapid ridge]

Mayb

https://tenor.com/view/el1te-gif-26472339

[river willow]

LOL

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[lusty shuttle]

Honestly I don't even know where to start

[river willow]

what do u not know

[lusty shuttle]

yes I understand the task

but idk how to start solving it

[river willow]

do u know of completing the square

[lusty shuttle]

i know that x vertex is -b/2a

so just -b/2 in our case

[river willow]

just start with

(x-a)^2 +b = 0

this form should be easier to work with

what would the solutions be ?

[lusty shuttle]

how did you get this?

[river willow]

if u expand it u get a general monic quadratic

its completing the square or vertex form

the resulting equation

[lusty shuttle]

ah

makes sense

[river willow]

but since they didnt give u numbers u can start with this equation

[lusty shuttle]

so (x-1)^2 = -b

[river willow]

how did u get a=1?

[lusty shuttle]

because it's monic

[river willow]

no

the general equation is

a(x-b)^2+c

i set that original a to 1 already

and renamed the variables

[lusty shuttle]

ahh

[lusty shuttle]

ah okay that makes more sense now

ok so

x = a+-sqrt(-b)

[river willow]

yea

and whats the point of the vertex

[lusty shuttle]

-b/2a

[lusty shuttle]

so -a/2 with these variables

[river willow]

wait nvm u r right lol

[bright quarry]

damn

[lusty shuttle]

so these are where y(x) intersects x-axis

[river willow]

stop

[bright quarry]

foxfil

dope name

[bright quarry]

what?

[lusty shuttle]

x=b+-sqrt(c)

[bright quarry]

i haven’t even read the question

[river willow]

wait no its a/2

not -a/2

[lusty shuttle]

thanks lol

[lusty shuttle]

why

x vertex is -b/2a

a=1

so -b/2

[river willow]

i got it from this

wait nvm

the vertex is just a

[lusty shuttle]

huh

[river willow]

nvm let me

work it out

[lunar galleon]

I wouldn't use b and c in this

makes it confusing

a(x-h)² + k

[bright quarry]

h and k

[lunar galleon]

yea

[bright quarry]

THE BOYS

[lunar galleon]

😛

[river willow]

yea soo

the vertex is inbetween the two solutions

at least its x value is

[lusty shuttle]

is it like this?

[river willow]

so we take the average of both solutions, and we get a

[dapper swift]

yep

[bright quarry]

yo wait when did you get green?

[lusty shuttle]

what do i do next idk

[dapper swift]

so actually a = sqrt(D)

if you take the difference of the 2 roots

[lusty shuttle]

ohh

yeah

or wait

isnt it 2*sqrt(D)/2

[dapper swift]

oh wait

yeah

[lusty shuttle]

so a = sqrt(D)

= sqrt(b^2-4c)

[river willow]

im knot

[bright quarry]

you’re knot

🪢

[lunar galleon]

this question is trippy

[bright quarry]

my eyes mislead me

[lusty shuttle]

what do i do next idk

[bright quarry]

dope bio

[lunar galleon]

ignore that

I can't remove it

[lusty shuttle]

i did this @lunar galleon

[lunar galleon]

👻

[bright quarry]

interesting

[lusty shuttle]

and then this

[lunar galleon]

isn't it -√D

[lusty shuttle]

no

so if we know that x vertex is -b/2 we can put it in y(x) to get y vertex = -b^2/4

[lunar galleon]

oh I did it the other way around

[lunar galleon]

r we imagining this parabola to be on the left side of the y axis or the right side

[lusty shuttle]

idk

I drew it on the left side

[lunar galleon]

@dapper swift come back 😔

[lunar galleon]

then it would be the other way around

a = -√D

[lusty shuttle]

why

[dapper swift]

no, side lengths are always positive or zero

[lunar galleon]

oh yeah

so the negative don't matter

I'm clueless

[dapper swift]

probably then find the distance between $(\frac{-b + \sqrt{b^2 - 4c}}{2}, 0)$ and $(-\frac{b}{2}, -\frac{b^2}{4} + c)$

[solid kiln]

higher's secret twin brother

[dapper swift]

and set that equal to a

[lusty shuttle]

hmm

how do I find that distance

[dapper swift]

distance formula, so $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

[solid kiln]

higher's secret twin brother

[dapper swift]

jesus that's not a nice answer

honestly I don't understand what the question wants

[lusty shuttle]

D

find b^2-4c

[dapper swift]

from this, isn't it just a^2 = D = b^2 - 4c

so the answer is a^2

which is the side length squared

[lusty shuttle]

but the answer should be a number

thats the problem

[lunar galleon]

😭

[dapper swift]

https://www.wolframalpha.com/input?i=(sqrt(b^2-4c)%2F2)^2+%2B+(-b^2%2F4+%2B+c)^2+%3D+(b^2+-+4c)

the algebra will be long

[lusty shuttle]

it shouldnt be that hard

but idk

[dapper swift]

I think a better way is to let the line of symmetry of the parabola be $x = 0$

so then you can write the parabola as $y = k(x - r)(x + r)$

[solid kiln]

higher's secret twin brother

[dapper swift]

it doesn't matter where the parabola is, like where it's translated

only the shape of the parabola matters

[dapper swift]

so the distance between the roots is $2r$, $a = 2r$

[solid kiln]

higher's secret twin brother

[dapper swift]

and when $x = 0, y = -kr^2$

[solid kiln]

higher's secret twin brother

[lusty shuttle]

okay so b=2 and c=-2 seems to a be correct example

so D=4-(-8)=12

and its a correct answer

[dapper swift]

yes

so from the WA output, you don't want c = b^2/4

cause then x^2 + bx + b^2 / 4 = (x + b/2)^2, so the roots will be both the same (two of the points are the same)

you want c = 1/4 * (b^2 - 12)

then any value of b works

[lusty shuttle]

makes sense

[dapper swift]

algebraically

$b^2 - 4c = b^2 - (b^2 - 12) = 12$ for whatever $b$

[solid kiln]

higher's secret twin brother

[lusty shuttle]

yeah

cool

thanks

[dapper swift]

actually then you can do $\frac{kr^2}{2r} = \frac{3}{2}$

if the base of an equilateral triangle is 2, its height must be sqrt(3)

[solid kiln]

higher's secret twin brother

[dapper swift]

ah so the function is y = k(x^2 - r^2) and the discriminant is just -4c = 4kr^2

we're close

wait it's monic so k = 1

whatever

.close

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[lusty shuttle]

lol

thanks

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[lethal ivy]

I get everything excepts my teachers work for 3c. Why do we care about the minute hand?

wouldnt it be corrct to do 1/12 times 2pi

i dont get it

in that aspect

[sly pewter]

what does 1/12 times 2pi look like

2pi is a full rotation so 1/12 of a full rotation is how much time on a clock @lethal ivy

[lethal ivy]

yes i realized my mistake

but im still confused

a bit

[sly pewter]

https://tenor.com/view/cat-edit-mewing-chains-gif-2436767122158080536

[lethal ivy]

one sec im tryna work it out agabn

again

[sly pewter]

What is confusing you

Okay

[lethal ivy]

im trying to understand what multiplying by the 1/6 rlly means

[sly pewter]

Think of what 1/12 times 2pi means

[lethal ivy]

that means like how much the hour hand moves in 1 hour right

[sly pewter]

yes

So 1 hour = 2pi/12

So if I knew how much time passed in hours, I can translate it into radians right?

[lethal ivy]

wdym

i think i am being confused cause we are looking at 2 different hands when question is looking for 1

[sly pewter]

Not really

[lethal ivy]

but i understand that 1/12 x 2pi is hour hand movemnet in 1 hour

[sly pewter]

yes so

1 hour is 60 minutes

But just because I say minutes doesn’t mean that I’m talking about the minutes hand

I don’t know why the teacher mentioned minute hand but

12 hours = 2pi radians

Divide both sides by 12

1 hour = 2pi/12 radians

how would I find the radians for 10 minute then

[lethal ivy]

divide by 60?

[sly pewter]

The left hand side becomes 1hour/60

Or 1 minute

[lethal ivy]

so thas what u do?

[sly pewter]

Ok you understand this

[lethal ivy]

yes

[sly pewter]

How much

Is 10 minutes in terms of hours

[lethal ivy]

1/6

[sly pewter]

Do you see it now?

[lethal ivy]

wait so divding by 60 was correct?

or no

[sly pewter]

No you neeeded to divide by 6 right

[sly pewter]

60 minutes = 2pi/12 radians

If I divide by 60

1 minute = 2pi/(12*60)

you can multiply by 10 now though

10 minutes = 2pi/(12*6)

[lethal ivy]

so u divide both sides by 6?

1/6?

[sly pewter]

You divide both sides by 6 to get 10 minutes on the left side and the radians on the right

[sly pewter]

This method doesn’t use algebra like I did, but it has similar thinking

[lethal ivy]

ya so i dont get the way my teacher did it

and i am supposed to understand it

i get ur way

[sly pewter]

ok you shouldn’t try to understand the “minute hand” part

But

1/12 * 10/60 * 2pi

Should make sense to you

[lethal ivy]

u did this cause u wanted to get to 10 mins so u thought 60 divdided by what is 10?

do i understand it

[sly pewter]

Yes

[lethal ivy]

it dont for sm rzn

[sly pewter]

Ok so

[lethal ivy]

can i think of it as like units cancelling

[sly pewter]

2 pi is a full rotation

I want 10 minutes and I know that there are 12 hours in a full rotation

[lethal ivy]

that would definetly help

[sly pewter]

I’m not sure what you mean but yes

[lethal ivy]

like unit conversion

styly

style

like making units cancel to get wanted answer

[sly pewter]

I’m not seeing where the units cancel

[lethal ivy]

ok then nvm

i just dont understand my teachers way tho still

[sly pewter]

What Your teacher probably wants to say is

there are 60 minutes in an hour and 12 hours in a full rotation for the hour hand

if I wanted to calculate the rotation of the hour hand after 10 minutes

I need to calculate the fraction of 10 minutes out of 12 hours

does that make sense?

[lethal ivy]

kinda but still dont understabd

[sly pewter]

Hmmm

It all comes from the fact that 12 hours make a full rotation

[lethal ivy]

wait i think i get it

nvm

i worked it out some more and understand

i gtg now but thx for the help

[sly pewter]

Okk cool

[lethal ivy]

appreciate you

.close

[sly pewter]

https://media.discordapp.net/attachments/793333636417847337/956359093512208424/4313-cat-spin.gif

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[serene storm]

When it comes to calculating speed, some people use the symbol ‘s’ to represent speed e.g. s=d/t and some people use the symbol ‘v’ to represent speed e.g v=d/t

[serene storm]

So which one is ‘more correct’

[wary basin]

it depends on what you are trying to say because both are different

s is speed and v is velocity

[serene storm]

This is what I found

[wary basin]

the main difference being s is scalar while v is a vector, meaning s is a singular value while v (or velocity) indicates a speed and direction

in the case of representing one number value you would use s but usually in physics we often use v and indicate direction

at least within my course

[serene storm]

wait so you use s instead of v right

[austere cedar]

Not very often you're working with speed, instead of velocity

[wary basin]

if we are talking singular unit yes use s

because for velocity you should indicate direction

[austere cedar]

In that case, |v| would be pretty clear imo

[serene storm]

i don’t like that bc some may confuse it with absolute value

[wary basin]

however especially in higher level courses you will just use velocity

[serene storm]

but i know that is one way to address it

[serene storm]

well i’m still in high school so we use both speed and velocity

[wary basin]

yeah ofc

basically it depends on the question

[serene storm]

just to clarify, i would write velocity just as a v right?

[wary basin]

but you mostly use velocity

[serene storm]

would i have to write v with an arrow on top to indicate it’s a vector?

like v arrow = displacement over time

[wary basin]

I have seen it notated that way but my course didn't require it

I would ask your teacher for preference

[serene storm]

alright

yea sure

for the formula: 2as = v^2 - u^2

s in this case is distance?

or displacement

[wary basin]

displacement

[serene storm]

alright

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[vagrant prism]

for some reason i am not sure how to prove this

[vagrant prism]

using epsilon definition

[night patio]

i think u can do this graphically/geometry involved

[vagrant prism]

i need to do epsilon stuff

i started by doing \
$\forall \epsilon > 0 \exists N: \forall n > N, |a_n - L| < \epsilon$\
$\forall \epsilon > 0 \exists N: \forall n > N, |b_n - M| < \epsilon$

[solid kiln]

artemetra

[vagrant prism]

my goal is showing $L \leq M$

[solid kiln]

artemetra

[vagrant prism]

maybe if i show $M - L \geq 0$ it will be enough?

[solid kiln]

artemetra

[stark bison]

Assuming the limits exist, yeah, it's easier to first show that a nonnegative sequence has a nonnegative limit and then use that

[vagrant prism]

yeah we assume the limits exist

[vagrant prism]

hm

so are you saying i should consider $b_n - a_n$ as a sequence and show that it goes to $M-L$?

[solid kiln]

artemetra

[stark bison]

Yeah

Anyway rewrite |a_n - L| < eps as -eps < a_n - L < eps and try solving for L

And see if you get anything that implies L to be nonnegative given a_n >= 0

[marble wharf]

meanwhile I would have assumed L > M, then there is a gap between them and then abuse that

[vagrant prism]

but what if there isn't

cuz it's L >= M

oh nvm

you did it by contradiction

yeah fair i'll try that too

[night patio]

oh yeah thats smart

u can consider the ||midpoint of L and M and find epsilon||

epic spoiler fail

[marble wharf]

||

[night patio]

🤦

[vagrant prism]

.close

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[limpid mirage]

f), i got stuck at this point

[limpid mirage]

i don't know how to multiply the power and root

how do i do it further?

That's my all calculations

alr, got it after many tries

.close

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[dull thorn]

Hello
I need help in understanding statistics vocabulary

[dull thorn]

:3

[vagrant prism]

like what?

[dull thorn]

I'll post a list of what I'm having struggle with

What's the difference between a modality and a statistical variable

Uhh

Just this actually

I figured it out

.close

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[past blaze]

can someone help me with this

[dapper swift]

what happens if x = 4?

[past blaze]

x is greater than 3

so 4 but what is y

[dapper swift]

3 : 4 = y : 4

[past blaze]

oh so y is 4

[wraith dragon]

You can write it as 3/4 = y/4

[past blaze]

is that what : is

okay that makes so much more sense

thank u so much

i got it now

.close

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[eager summit]

idk how to proceed

[eager summit]

also asked in #help-39 message

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@eager summit Has your question been resolved?

[eager summit]

@dreamy oar

wait sorry 😭

<@&286206848099549185>

[hollow tangle]

Is your doubt done?

Let's say f(x,y) = ax²+2hxy+by²

How do you know that d f(x, y) / dx = 0?

@eager summit

[eager summit]

because its given f(x,y) = 0

[hollow tangle]

f(x, y) might be 0

But the slope doesn't have to be

[eager summit]

huh

[hollow tangle]

In this case f(x, y) are 2 straight lines

[eager summit]

i was taught to think of it as taking the derivative on both sides

and dc/dx = 0

how does that fail

[hollow tangle]

Give me a minute

I don't know bro try pinging again

[eager summit]

alr, ill just wait for someone else to come

[modern storm]

I was thinking if it was possible to extract any information from the statement to figure out the form of y, but it seems like it doesn't lead to any answer

[modern storm]

This here seems to be correct, I did it and reached the same thing 🤔

[eager summit]

actually i got it

ax^2 + hxy = -hxy - by^2 from the original

=> x(ax +hy) = y(-hx -by)

=> -(ax + hy)/hx + by = y/x

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[cyan blade]

are these both correct?

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@cyan blade Has your question been resolved?

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[proper kernel]

if youre not sure, create a truth table for each of them to make sure that theyre true when they should be

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[astral spade]

I think it should just be this

-j plus with circle m

So not J xor m

I don't know what the notation your teacher wants to use for exclusive or is

[astral spade]

I think it's what I wrote, but I could be wrong

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[quick knoll]

It can generally be said that: \exists x P(x)<==> \neg \forall P(x) and \neg \exists P(x) <==> \forall P(x)?

[nova spire]

uh let's just be clear

I think I see what you mean

but "exists" and "forall" is hardly a statement by itself

"exists x, P(x)" for example

or "forall x, P(x)"

now with this in mind

[quick knoll]

?

yeah u can add P(x) if it do more sense

[nova spire]

well yeah you need it to make sense

otherwise you kinda wrote 'there exists [????] such that'

[quick knoll]

why maths is so exactly

xD

[nova spire]

not saying what exists

and whatever is after the "such that" is cut short

so

take this statement for example

$\exists x(P(x))$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

the negation of that statement is indeed

$\forall x(\neg P(x))$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

and the negation of $\forall x( P(x))$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

is $\exists x(\neg P(x))$

[solid kiln]

rafilou is not not born in 2003

[quick knoll]

can u edit it complete that I can save it in my notes

ah ok perfect