help-26

[desert pulsar]

And convert it to a fraction

Then convert the remaining

But let’s do something simpler

0.(7)

@neon iron

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[undone flicker]

A theorem says it will be a minor axis which is a^2 =9

<@&286206848099549185>

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[tight rivet]

How’s this wrong

[toxic grove]

(x+5)^2 is always positive non-negative for real x

-6 for example works here

[tight rivet]

So what’s the answer

[sweet shard]

divide both sides by (x+5)^2

then solve for x

[toxic grove]

ah I made a small mistake

I meant non negative there

[tight rivet]

9?

[sweet shard]

you can do this if you rule out x=-5

[sweet shard]

what about 9

can you write in complete sentences

[tight rivet]

X=9 if I divide both sides by (x+5)^2

[sweet shard]

when you divide both sides of an inequality, you still have an inequality remaining

[tight rivet]

9<0

[sweet shard]

e.g. $2x > 4$ implies $x > 2$

[thorny flame]

riemann

[sweet shard]

do you think this is true?

[tight rivet]

Oh yeah

Nvm

I have no idea how to do it then

[sweet shard]

start here

and divide by (x+5)^2 after ruling out out x can't be -5

[tight rivet]

X<9

[sweet shard]

yes

[sweet shard]

do you understand why x can't be -5

[tight rivet]

Yeah

Cus it would be 0

[sweet shard]

so combine this

[sweet shard]

with the mathematical statement of what this means

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[warm grail]

hi

[warm grail]

i looked back at my notes but im really lost

[keen matrix]

are you familiar with the conjugate root theorem?

[warm grail]

is that the same as

fundamental theorem of algebra?

[keen matrix]

similar uses but different idea

do you know what a complex conjugate is?

[warm grail]

we haven’t learned that in class

so no

[keen matrix]

well when a polynomial root has $a+bi$ as a factor, it must have $a-bi$ as a factor if it has integer coefficients

[thorny flame]

MrFancy

[warm grail]

ah yes i have that in my notes

the a + bi

[keen matrix]

so if the polynomial has 2i as one of it's roots what must be the other root?

[warm grail]

-2i?

[keen matrix]

so now we know three roots

but this is all we need, do you know why?

[warm grail]

3 degrees

??

[keen matrix]

very good

ok so now do you know how to write polynomials in terms of their roots? :)

[warm grail]

uhh kinda??

i have it in my notes too

i can try

(x-3) (x - (-2i)) (x - (2i))

💀💀

[keen matrix]

there that's it, that's all it asks you for :) do you think you can do the rest from here?

[warm grail]

OHHH

what if theres a square root?

[keen matrix]

same thing applies, if we want integer coefficients :) put a negative in front of that guy

[warm grail]

ahh

hold on

[warm grail]

do we simplify this more?

[keen matrix]

that's a linear factorization mein freund :)

[warm grail]

wait i can show u my notes

like why is it x^2 + 6x + 10??

[keen matrix]

$x^2+6x+10$ simply has roots $x=3-i,3+i$

[thorny flame]

MrFancy

[warm grail]

ah

so do i have to do that too?

or would my original answer work too

[keen matrix]

x^2+6x+10 is not linear, the questions asks for a linear factorization :)

[warm grail]

oh wait is this linear factorization

Ohhh

i just reread it

ah thanks

so for square roots

we dont simplify those

it just says the same?

!!!!!!

thanks :DD

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[limpid violet]

hello

[limpid violet]

Finding the domain

anybody know why i got it wrong?

I usually start with the bottom one (2), but it had no x, so i wasnt sure how to go about it

[keen matrix]

what's your goal here?

ah I see domain

[limpid violet]

finding the domain

[keen matrix]

mein freund where would this function be undefined? :)

[limpid violet]

if the bottom would equal 0

[keen matrix]

so why is the domain x>=-6?

[limpid violet]

the bottom had no x, only 2

so i didnt know how to go about it

and just did the top 😂

[keen matrix]

so if there's no value that makes this function undefined what is the domain? :)

[limpid violet]

all real values?

[keen matrix]

I don't like that denominator is zero rule of thumb cause things like this happen, it's a good idea but not 100%

[limpid violet]

ahhh i see

[keen matrix]

just find places where the function become undefined this therefore, will not be part of the domain

at x=0 1/x becomes undefined because of division of zero

at x=-1 sqrt(x) becomes undefined because of negatives inside radicals

things like these cause (undefinedness?) :)

[limpid violet]

okay i see

thank you 🙏

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[native loom]

@willow ridge

[finite olive]

solve inside brackets first

make denominators same

multiply 1/5 by 3/3

and 2/3 by 5/5

and add

you get 3-(13/5)

now try do the rest

3 is also 3/1

so make the denominator 15 by multiplying by 1 by 15

[willow ridge]

Ok thank you (:

I've got it now

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[golden zodiac]

the second page it what i tried although i didn’t get very far

i was trying to prove that MN (M and N are the centre of my bottom two circles) and BC are parallel because then the proof would fall out easily after that but it hasn’t really work…

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[lucid junco]

I’m confused, is the solution wrong?

Because when I take arcsin(-sqrt(3)/2 I get -pi/3 as a base angle

But the solution says that the base angle is POSITIVE pi/3

Any ideas

[sterile finch]

The base angle is relative to the reference angle, both pi and 2pi.

From pi, it would be pi + pi/3 and from 2pi it would be 2pi - pi/3.

[lucid junco]

So why would the base angle be positive instead of negative which was I got on my calculator?

[sterile finch]

I'm making a graph to help explain.

[lucid junco]

Sure, thanks

[sterile finch]

You are solving for the values of theta whose sine is equal to -sqrt(3)/2.

There are two possible angles which are both 60 degrees, or pi/3.

The explanation should have further explained that the base angle was relative to the reference angles pi and 2pi.

In this case, 4pi/3 and 5pi/3.

[lucid junco]

Right so Quadrants 1,2 are where sin<0 which is what you showed in the graph.

To get to quadrant 2 wouldn’t I need to perform theta - pi

[sterile finch]

You have the quadrants in the wrong order. Q1 is the upper-right quadrant and subsequent quadrants are in a counterclockwise order.

[lucid junco]

Well I thought I’d draw it like this because we are looking at the negative angles so we go backwards

?

[sterile finch]

No, for the sake of clarity, Q1 is always in the upper-right quadrant.

[lucid junco]

Okay

so to get to Quadrant 3 then we would need pi + theta correct me if I’m wrong

[sterile finch]

Correct.

[lucid junco]

And to get quadrant 4 it is 2pi - theta

[sterile finch]

Correct.

[lucid junco]

and we’ve established that the base angle is -pi/3 so

Q3: pi - (-pi/3)

Q4: 2pi - (-pi/3)

[sterile finch]

And you can generalize the possible angles by noting that the two angles are symmetric around 3pi/2.

[lucid junco]

Q3: 4pi/3

Q4: 7pi/3

[sterile finch]

Which would be 3pi/2 ± pi/6.

[lucid junco]

Hold on so why do we not look at 60 degrees (-pi/3) anymore?

[sterile finch]

This is a "better" way to solve this type of problem in a more general sense.

[lucid junco]

Okay so looking at its coterminal would help us solve it easier?

[sterile finch]

It takes a bit of repetition to understand why.

You can use one angle, 3pi/2, for the coterminal angle, rather than two angles, pi and 2pi.

3pi/2 + (2n)pi ± pi/6

[lucid junco]

Yeah so it’s symmetrical around 270 degrees

[sterile finch]

Correct.

[lucid junco]

Does this mean that we can use that to find other solutions ?

[sterile finch]

Yes, if you are looking for solutions in a given interval.

[lucid junco]

Okay

So if we want to find more solutions we can add 3pi/2 to -pi/6?

[sterile finch]

You can add 2npi to 3pi/2, for all natural values of n; ..., -3, -2, -1, 0, 1, 2, 3, ... .

[lucid junco]

Hm I’ve never really worked with finding general solutions because we haven’t learnt it in class since we only do equations with a restricted interval

Can you explain the 2npi

[sterile finch]

It's a mathematical way to write out all of the coterminal angles for a given angle. It's somewhat implied that n means all Natural numbers, {..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ...}.

Rather than write every negative and positive integers.

[lucid junco]

I see. But wouldn’t adding 3pi/2 + 2npi + (- pi/6) make us overshoot and miss the solution?

[sterile finch]

This is where the given interval comes into play as a constraint on the solutions that you will need to solve for.

[lucid junco]

Okay

So we want 0 to 3

Does that mean we put 3 into n

[sterile finch]

Do you understand what I did here?

[lucid junco]

So you’ve made the arguement of sin equal to the co terminal (2pi/3) + 2npi +- pi/6

But wouldn’t it just be - pi/6 because we’ve established the base angle as -pi/6 so it can’t be positive as well?

[sterile finch]

No, you will need both + and - pi/6 because both angles sine value are equal to -sqrt(3)/2.

So they are both possible values of x that could lie in the interval [0,3].

[lucid junco]

Oh I thought the +- only applied to even functions

hm okay

[sterile finch]

As mentioned earlier, this is just a way of generalizing a solution that has two angles that are symmetric around one angle, 3pi/2.

[lucid junco]

Oh because it’s symmetrical around 270 it will have -+ is that why?

[sterile finch]

Correct.

You could also solve this in terms of pi + pi/3 and 2pi - pi/3, but that is more work.

[lucid junco]

Okay

[sterile finch]

Anywhos, from the image above, you can do some rearranging to solve for x in terms of n by multiplying both sides by 3/2pi.

[lucid junco]

Okay ill try that

[sterile finch]

Any luck?

[lucid junco]

I’m having a little trouble with the rearranging part which is a little embarrassing lol I don’t know what’s going on

[sterile finch]

The pi's all cancel out which is nice.

[lucid junco]

Okay so for x instead of multiplying xby 2pi/3 you divided by 3/2pi which makes sense. Ok I think I see how you did it now

So now for the “n” part what shall we do about that

[sterile finch]

So this is where the generalization comes into play. Set x = 0 and solve for n. This will help you find the lower bound for n.

And likewise, set x = 3 and solve for n. This will help you find the upper bound for n.

[lucid junco]

Ah okay

So how many solutions does this get us?

[sterile finch]

Keep in mind, that you want n to be integers so if you get a fraction, you will either need to round up or down for the integer solution.

You will get the two solutions given in the explanation you posted which should be n = 0 only. Plug that into the equation above and you will get your two solutions.

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[neon iron]

what is going on in this first step? im not sure how he was able to e^ix form (amc 12b q24

[tough cargo]

(e^(ix) - e^(-ix))/(2i) = sin(x)

That said i think they made a typo?

Like they forgot to include the negative on the 2nd exponent in the pi/7 and 2pi/7 term

[neon iron]

oh thank you

i never knew about this identity

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[pseudo prairie]

hyy

[radiant tapir]

did you have a question?

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[bold obsidian]

Is the x-value of the maximum point -1?

[stuck forum]

What's your attempt so far?

[bold obsidian]

I found the vertex which is (-1, 245)

And since it’s saying maximum that must indicate the parabola is going down right

[stuck forum]

You can also determine that from the sign of the quadratic term x^2

But yes

[bold obsidian]

Does that mean I’m right

The x-value of the maximum point is -1?

[stuck forum]

It is

[bold obsidian]

Yay thank you!

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[regal echo]

how do I do this/

[stuck forum]

Is there more to this problem?

[regal echo]

im also confused on how these values of a and b will make g(x) = g^-1(x)

[stuck forum]

Okay

How do you find the inverse of a function?

What's the strategy?

[regal echo]

swap x and y

[stuck forum]

Right, so give me the equation for g inverse

[regal echo]

G-1(x) =(b-a)/(x-1) -a

[regal echo]

But how do I get this?

[stuck forum]

So to find a and b, you need to equate $g(x)$ and $g^{-1}(x)$

[thorny flame]

TooManyCooks

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@regal echo Has your question been resolved?

[stuck forum]

Have you tried equating them yet??

[regal echo]

yeah soz about to

[stuck forum]

You're actually almsot done

finding the inverse is 90% of the work

[regal echo]

it deosn't work on my calculator...

there's two unknowns right?

so don't we need at least two equations to solve that?

[stuck forum]

Actually you don't need to

Let me show you. Give me a sec

$g(x) = 1 + \frac{b-a}{x+a} = -a + \frac{b-a}{x-1} = g^{-1}(x)$

[thorny flame]

TooManyCooks

[stuck forum]

You don't have to solve anything at all

You want this to be valid for all x

However, you can easily see that letting a = -1 does that

For any b

and any x

well, except x = 1

That's it, just let a = -1

Wait did I fuck up my sign

[regal echo]

ahhhhh that makes sense

thank you brother

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[willow scroll]

Help on 3.

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[willow scroll]

<@&286206848099549185>

[willow scroll]

<@&286206848099549185>

bro please

bruh

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[molten lion]

em

Ok Jimmy, considering adding the higher rows with the lower rows to make it simpler first

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[tulip eagle]

b^2 - 4ac >0
b^2 - 4ac = (k-1)^2

[tulip eagle]

when k=1,
wont (k-1)^2 = 0 and not >0

so the line and the curve don't meet for all values of k?

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[fallow heart]

Are you talking about the discriminant Δ of the equation you get after intersecting the two curves?

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[tulip eagle]

yeah

[fallow heart]

Ok nice, yes that (k - 1)² is correct. Since (k - 1)² ≥ 0 for every value of k, it means that there always exist at least one intersection

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[dry warren]

which formula should i use to get this?

<@&286206848099549185>

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[pulsar sun]

[pulsar sun]

At the rings initial position, it is at rest right?

[pulsar sun]

SMH I doubt it's gonna move at a constant speed

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[rapid mantle]

why did they make 1.175 negative?

[rapid mantle]

sd is 18 mean is 96

[vital relic]

the diagram is a bit misleading

can u indicate the area 0.88 refers to

its from -z_t and then to the right all the way (to infinity) right?

So then you need to find what -z_t is, but I presume your reverse cumulative table/calculator

does P(t > X) ?

thats where the minus sign comes from - since the normal distribution is symmetric about the mean

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[rapid mantle]

oh right I see

thanks

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[regal folio]

It's question i

[prisma holly]

Solve the numerator and denominator one at a time

[regal folio]

Am I doing it right?

[prisma holly]

yeah

It's easier to simplify them separately

Rather than like that

[fleet canopy]

ur in grade 9?

[fleet canopy]

i learnt it in a diff way

[regal folio]

Do I cross out 2x^2 since it's both on the denominator and nominator?

[prisma holly]

noo

That's not how it works

[regal folio]

Man I just entered college and lost all my math skills

[fleet canopy]

oh sad

[regal folio]

I'm relearning it ok :(

[prisma holly]

you can only cross them out if they're in multiplication

[prisma holly]

did you solve them together?

[fleet canopy]

no like

wait i will solve and send it

idk hwo to explain

[prisma holly]

alr

[prisma holly]

Split the middle term now

x[2x² - (4-3)x - 6]

[prisma holly]

the numerator ^^

do you know how to solve that)

[fleet canopy]

yeh

[regal folio]

Is this the denominator?

[prisma holly]

no it's the numerator

the denominator is 2x² - (4+3)x + 6

[fleet canopy]

oh wait

i didnt read the q properly

should we rationalize it? or? simplify?

[prisma holly]

@regal folio want me to solve and send it for you?

[prisma holly]

Simplify it

[regal folio]

I wanna try and learn it first

[prisma holly]

alright

Just ask me whenever you're stuck

[fleet canopy]

brb

[regal folio]

Why is there a 4-3 in the numerator?

[prisma holly]

well we look at the sign of the last term when we're splitting the middle term look at the sign of the last term

It's '-' so we have to find a number that equals to the product of the first and the last coefficients when multiplied and the coefficient of the middle term when subtracted

The product of first and last terms is 2 x 6 which is 12 and is also equal to 4 x 3

The coefficient of the middle term is 1 which also equals to 4-3

[fleet canopy]

idk how to simplify it

ik how to rationalize it

tell me how to do it too ;-;

i dont get it

[fleet canopy]

once ur done solving send it

[prisma holly]

alr wait

[fleet canopy]

hm

[regal folio]

I remember learning to do that using a method in class. Something like this

[fleet canopy]

?

what is this

[regal folio]

It's.... nit complete yet

[scenic locust]

so u know long division method

[regal folio]

Yeah but it's kinda hard for me. I remember there being another mehod

But am not sure

[scenic locust]

oh

[fleet canopy]

@prisma holly can u plss solve it and send it

[scenic locust]

uh so

the answer is one single (a+b) bracket thing

[regal folio]

Honestly yeah, I think it would be easier for you to explain the work process of it please

[fleet canopy]

yh..

[scenic locust]

that is solving quadratic eqn method?

are you allowed to use caculator to solve quadratic and cubic eqns?

[regal folio]

Yeah, iirc you could do it on the calculator but I'm using the fs-570ex version know so I dunno

[scenic locust]

um

go to menu

find "equation"

polynomial

and degree (highest power)

[prisma holly]

mb I don't know

[fleet canopy]

uh

[regal folio]

Which?

[scenic locust]

2

[prisma holly]

wait what

[scenic locust]

here ill give u what it is for faster

[prisma holly]

this is used to find the value of x

he isn't asked to find it

[scenic locust]

top = solutions: (x-2) or (3x+2)
bottom solutions: (x-2) or (3x-2)

[scenic locust]

hes simplyfying

u gotta do some cancel shit

do u know the

this method

either equating or comparing coeffs

[regal folio]

Not really

[scenic locust]

ah

[scenic locust]

wtf

just use caculator

[prisma holly]

I solved em individually if you don't understand anything hmu

[fleet canopy]

;-;

[fleet canopy]

u solve it

[scenic locust]

the cubic and quad eqn yes

individually

not the qn tho

[prisma holly]

I'm pretty sure the calculator doesn't show these it just shows the direct value of x

[scenic locust]

u gotta reverse work it

[scenic locust]

.

i did this in 10s

[prisma holly]

yeah ik

But

[fleet canopy]

wait i will try solving it too

[prisma holly]

It's kinda complicated

[scenic locust]

~~i do not know how to do with caculator 💀 ~~

[prisma holly]

And I don't think they understand it completely

[scenic locust]

btw @regal folio im quite sure that if u dont know this method

[scenic locust]

.

u gotta use long div

no other way

[fleet canopy]

what method is that?

[scenic locust]

its called long division ltierally

[regal folio]

Wait so which one is correct?

[regal folio]

Is this used to simplify?

[scenic locust]

no its used to solve the entire qn

[prisma holly]

hmm wdym which one?

[fleet canopy]

what do u do after this?

[prisma holly]

There's only one answer there

[scenic locust]

wait...

[fleet canopy]

is this even right?

[scenic locust]

um

looks right to me

[prisma holly]

don't ask me

[scenic locust]

wait

[prisma holly]

I never understood this shit

[fleet canopy]

mhm

[regal folio]

Oh nvm, understood it

[scenic locust]

oh wait nvm nvm

misread mb

[fleet canopy]

no its 21

what do u do after this?

[scenic locust]

well technically the answer is

[fleet canopy]

hm?

[scenic locust]

wait im very confuse

my caculator tells me theres no remainder

but i seeu didnothing wrong

[fleet canopy]

no its not the reminder

we still gotta solve it

idk how ;-;

[scenic locust]

but u got a remainder

we cannot ignore it

[fleet canopy]

hm

[regal folio]

I'm not sure if this is a bit much too ask but could ya'll monitor me doing another question just in case I mess something up?

[scenic locust]

no prob

[regal folio]

ii.

[scenic locust]

@fleet canopy give me awhile i try this method

[fleet canopy]

okayy

[scenic locust]

wtf

ok i see the problem

for 2x^3-x^2-6x, a possible answer for x is 0

[regal folio]

I think I can do something with the denominators but not sure how to do it

Should I make the denominators same by multiplying them with each other?

[scenic locust]

wait

i check

[fleet canopy]

huh

[scenic locust]

can u turn all the eqns into (ax+b)(cx+d) form first

then u will see something magical

[regal folio]

0_0

[scenic locust]

use ur caculator

it can be done very fast

[regal folio]

On it

[scenic locust]

uhh @fleet canopy sorry i got no idea how to do that i never seen an equation give x=0 value before

[prisma holly]

You solved the first one?

[scenic locust]

no lol

oh wait

[fleet canopy]

lmao-

[scenic locust]

@prisma holly u know how to solve when theres an x=0?

also it cannot be constant in last bracket cuz then there wont be x^3

[prisma holly]

send how you got x = 0

[regal folio]

Yep

[scenic locust]

caculator

[prisma holly]

hmm

Which equation

[prisma holly]

nice

[regal folio]

The one you showed me?

[scenic locust]

$2x^{3}-x^{2}-6x$

[thorny flame]

MrZhongZuChongTu

[prisma holly]

What does it equal to tho?

That's the real question

[scenic locust]

OHHHHHH

[prisma holly]

yes

[scenic locust]

wait nvm

it equal 0

[prisma holly]

In the calc it equals to 0

so you got 0

[scenic locust]

it does = 0 in the question

[prisma holly]

what?

Send the question

Where is it

[scenic locust]

its this one

[scenic locust]

this is the logn div

the cubic eqn literal = (x-2)(3x-2)(x+3)+9x+18

[prisma holly]

hmm

[scenic locust]

how is it going?

[prisma holly]

i don't really understand long division

Lemme put it in my calc

[fleet canopy]

;-;

[scenic locust]

bruh which asshole created math

[fleet canopy]

im so confused rn

[regal folio]

OK up until the right denominator

[scenic locust]

ok give me what u have so far

[scenic locust]

ahh u see

[regal folio]

It only shows x instead of x1 and x2

For the last one

[scenic locust]

that means (x+3)(x+3)

cuz the solution is the same

[regal folio]

Oh ok

[scenic locust]

now u do some cancelling u should be left with something in the form of

[thorny flame]

MrZhongZuChongTu

[regal folio]

Is this right?

[scenic locust]

yep

[regal folio]

Do I now make the denominators the same?

[scenic locust]

yep

do the times

[prisma holly]

When I put it in my calc I got 2 and -3/2

[scenic locust]

x3=0

i think u missed

[prisma holly]

what?

[fleet canopy]

when u guys finally solve it

and get the ans

ping me

and pls explain it once to me als

;-;

thank u

[prisma holly]

ion even know anymore

[scenic locust]

same bruh

what is happening

[prisma holly]

you solved it?

[scenic locust]

no

see x=0

[prisma holly]

bruh

[fleet canopy]

yall ignoring me?

[prisma holly]

noo

[fleet canopy]

nvm i will try to solve it with u guys

together

[scenic locust]

we havent solve yet

[prisma holly]

yeah

[fleet canopy]

yeh

@regal folio

did u figure it out?

[regal folio]

Is this right?

[scenic locust]

potato how u working on the cancelling

ok wow tahst

ok u mad ea mistake

i think u got confuse for times with +

its juts numerator x numterator, denom x denom

[regal folio]

Or yeah

So if it is addition then I do it right?

[scenic locust]

yep

[regal folio]

If it is multiplication there's no need right?

[scenic locust]

yea

[prisma holly]

man

[fleet canopy]

oh god

[scenic locust]

btw for this qn the answer should be

[prisma holly]

that eqn ain't a cubic eqn

☠️

where's the constant

[scenic locust]

i assume = 0

[fleet canopy]

;-;

[regal folio]

Can I just punch the numbers into my calculator?

Or must it be done manually?

[scenic locust]

manually

[fleet canopy]

i mean we need the working

[scenic locust]

i assume u talking ab ii

[regal folio]

Yea

[scenic locust]

$\frac{(x-2)(3x-2)(x+3)+9x+18}{(x-2)(3x-2)}$

[regal folio]

Wait, so if I multiply it, won't I have to multiply (x+3)(x+4) first?

[thorny flame]

MrZhongZuChongTu

[scenic locust]

no dont multiply

leave it in bracket form

u can cancel

show me what u got

[scenic locust]

now cancel

also remove the [ ] its abit confusing (u2y)

[regal folio]

Ok

[prisma holly]

man

did you solve the equation

[scenic locust]

.

this is the furthest i can get

idk how simplify anymore

[scenic locust]

yepp

thats it

[regal folio]

That's the answer?

[scenic locust]

mm

[regal folio]

Wait so is it?

[scenic locust]

is it what?

[regal folio]

The answer

[scenic locust]

yes

thast it

[regal folio]

OK cool

[scenic locust]

i gotta go, good luck with other qns

[regal folio]

Thx

[fleet canopy]

can u explain it to me

[prisma holly]

I'm here if you need me

[fleet canopy]

i dont understand ;-;

someone explain

[prisma holly]

hmm

Which step?

[fleet canopy]

idk everything

uh

[prisma holly]

well the first step he just split the middle term of all 4 equations and wrote the result

Then on the second step he multiplied them and cross out the same terms

[fleet canopy]

mhm

[prisma holly]

And he just repeated it until he got the answer

So you get it now?

@regal folio do you still need help with other questions? If not you should close the channel

[fleet canopy]

got it

thnx

[prisma holly]

yw

[regal folio]

I wanna do one more question if that's fine

[prisma holly]

we should make a gc and solve the questions there fr

[regal folio]

iii

I'll do it first then I'll ask if I need anything

[prisma holly]

alright

@regal folio have you solved it

[regal folio]

Am I doing good so far?

[prisma holly]

yeah

[regal folio]

I can just cross it out from here right?

[prisma holly]

yeah

[regal folio]

I don't need to put them under one line

[prisma holly]

nope no need for that

[regal folio]

Correct?

Well is it right?

[prisma holly]

yeah

you could write it in a shorter form

On the numerator: 4x² - 1

[regal folio]

so i multiply the two brackets

Ok

Thx for helping

See ya later

.close

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[gloomy ivy]

what is the rule for going from step 2 to 3? how does the log dissapear

[half edge]

$a^{\log_a b} = b$

[thorny flame]

Stephen

[gloomy ivy]

thank you!

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[mystic arrow]

is there a better way to have this kind of curve?

[mystic arrow]

preferably it's should be a y =

since I need this for a code

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@mystic arrow Has your question been resolved?

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[lilac furnace]

Idk where to start at all for the 9

[ionic oar]

convert all in terms of one function

[lilac furnace]

Wdym ?

[bold nebula]

doesn't it seem like quadratic to you?

[lilac furnace]

Yeah that’s what I was thinking

[bold nebula]

yes, that's a way

[lilac furnace]

But I didn’t know what do till after

Since it’s cos

Not just x

[bold nebula]

substitute u = cos(x) and use quadratic formula

[lilac furnace]

Oooohh

Yeah

[bold nebula]

solve for u and then for x

[lilac furnace]

Ok

I will

@bold nebula

I solve -1/2=cos x now ?

And 2=cos x?

[bold nebula]

yes

but you can see cos(x) = 2 is impossible

[lilac furnace]

Yeah right

So just -1/2?

[bold nebula]

yup, cos(x) = -1/2

[lilac furnace]

2pi/3 and 4pi/3 ?

[2kpi]

What do I do after that ? @bold nebula

[distant nymph]

You are done. Just write the general solutions.

[topaz sinew]

@lilac furnace Has your question been resolved?

[lilac furnace]

Yeah thanks

[topaz sinew]

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[neon iron]

hello

[neon iron]

how can i prove this using the epsilon delta definition

given that epsilon is > 0

and assuming that 0<|x| < delta

how can i then check that |1/1+x^2 -1| is less than epsilon

[craggy haven]

you get to pick the delta

and it will depend on epsilon

try it for epsilon = 0.1 for instance

[neon iron]

yeah i need to pick delta that will depend on epsilon

in a form like delta = epsilon/2

[craggy haven]

sure

[neon iron]

not with actual numbers

i just dont know how to factor f(x) to be |x| so i can relate the delta to the epsilon

if you get what im saying

like how do i prove this limit being = 1

[hard parcel]

im working on it why is this so hard

[neon iron]

cause it was an exam question

[hard parcel]

fair

[neon iron]

cause usually you can just do some algebra and get |....| < epsilon to be the same as the thing thats |...| < delta

but i have literally no idea how to do that stuff on here

[hard parcel]

yeah im trying minimuns and using triangle ineq but got stuck so now ive decided to redo ;-;

[craggy haven]

try it for examples though

like try it for epsilon = 0.1

what delta do you need?

what about for epsilon = 0.01

[neon iron]

ok i tried this and am here rn
0.01 < 1/1+x^2 < 1.01

i cant multiply with 1+x^2 cause then idk if it is + or -

[craggy haven]

i think you messed up that algebra slightly

[neon iron]

hmmm ok

[craggy haven]

once you fix the algebra error you can invert all sides (take the reciprocal)

[neon iron]

so like x < epsilon = +- sqrt/11/33)

so i make delta +-sqrt(11/33)

im so lost lol

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

nah

@hard parcel you got it or nah

i still dont know how to do it

[hard parcel]

nope ;-;

ive just been staring at this i have no clue on how to get a fraction with x^2 +1 as the denominator

[neon iron]

oh well

thanks regardless

.close

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[bright hawk]

Could somebody help me understand how to prove a variant of the mutilated chessboard problem?

[bright hawk]

If one were to remove two squares of the same color, I can understand that perfectly covering it with 2x1 domino tiles would be impossible

However, in the case that you remove one black and one white square, I can't really find a convincing way to prove that there must be a perfect cover

I've gone so far as to conclude that there could be 31 white and 31 black left and therefore the cover would have 31 dominos in it (a domino has to cover 1 black and 1 white)

but that doesn't seem rigourous enough, where to go from here?

[odd pagoda]

take a covering of the normal chessboard. this then gets broken in two places. can you repair this somehow?

[bright hawk]

broken meaning what?

[odd pagoda]

well 2 places get removed

[bright hawk]

2 squares get removed

no you can't

oh.

wow that was actually much simpler than i thought it would be

if those two squares were next to eachother

that would be like a normal cover of the chessboard with one domino removed so that case still must give a perfect cover

thank you!

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[odd pagoda]

what

no

not that easy

[odd pagoda]

if, sure

but if not, then what

@bright hawk

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[hybrid fog]

Can I take -1 out the limit?

I actually cannot tell where I can do that

Take powers out of the limit

[topaz sinew]

@hybrid fog Has your question been resolved?

[final jungle]

$\left(1+\frac{1}{n}\right)^{-1}\frac{1}{\left(1+\frac{1}{n}\right)}$, and then use the properties of limit.

[thorny flame]

Van21st

[final jungle]

$\left(1+\frac{1}{n}\right)^{-1}=\frac{1}{\left(1+\frac{1}{n}\right)}$

[thorny flame]

Van21st

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[hybrid fog]

Thankyou

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[barren laurel]

hi, does anyone know how to prove that nth degree polynomials divided by (t-1/2)^2 are an affine subspace of nth degree polynomials and find the corresponding vector space ? i know that in order to be a affine subspace for every point of (nth degree polynomials divided by (t-1/2)^2 )set must for each vector of the corresponding vector space the operation between them must give a point in the (nth degree polynomials divided by (t-1/2)^2 )set .

[barren laurel]

a hint is to work for start for n=3 or n=4

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@barren laurel Has your question been resolved?

[barren laurel]

<@&286206848099549185>

can i show that (nth degree polynomials divided by (t-1/2)^2 )set is a vector space ?

[barren laurel]

.close

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[south zodiac]

How can I find the rank of this matrix

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[crude anchor]

Well since it’s an orthogonal matrix it is full rank

Meaning it’s rank is equal to its dimension

Wait I might be mistaken

This may not be orthogonal

[south zodiac]

Yes exactly

We don’t know the inverse

[crude anchor]

If you can manage to compute the eigenvalues then it is the number of distinct eigenvalues of the matrix

[south zodiac]

I didn’t really get it

[crude anchor]

You mean you can’t compute the eigenvalues

[south zodiac]

I didn’t study them yet

[crude anchor]

Ah i see

[south zodiac]

Well all I know is that it’s a symmetric matrix

And I don’t know if we can conclude something about it’s rank

Did you get it or no ?

[crude anchor]

please dont ping people directly, but im still stumped trying to work on it

[south zodiac]

Aa okay im sorry thank you

[crude anchor]

i have a hunch that this matrix has rank n

but im struggling to prove it

i think that if you can manage to show that it has an inverse that might be the best bet

[south zodiac]

Can you please tell me the approaches that you used

[crude anchor]

well i didnt solve it

im jsut sharing my hunch

what i tried though was showing that the kernel of the matrix is the zero vector

but i couldnt complete my proof

[south zodiac]

Yeah I get you

So If I manage to prove that it has an inverse what can I do next

[crude anchor]

that completes the proof

for if a matrix is invertible then it has full rank

i guess it kind of comes from something called the rank nullity theorem

which i dont know if youve heard of

but suffic it to say, if you can show that there exists an inverse then youve shown that the rank of C is n

i guess you could also argue that the determinant is non-zero and so the rank is n

[south zodiac]

I dont really know this theorem

I’ll try to prove that the determinant is different than 0

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[south zodiac]

<@&286206848099549185>

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[agile moth]

when doing this should i make a propositional statement first?

such as P(x) = x takes a job in the idustry, Q(x) = x goes to graduate school

then the negation would be ¬P(x) V ¬Q(x)

?

would that be correct for a)

[hollow shard]

a) is P(x) or Q(x)

so ¬(P(x) v Q(x)) = ¬P(x) ^ ¬Q(x)

[agile moth]

ok

then b) would be P(x) ^ Q(x) = ¬P(x) V ¬Q(x)

correct?

[hollow shard]

sure

which means Yochiko doesnt now java or doesnt know calculus

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[agile moth]

https://cdn.discordapp.com/attachments/1148648248379191328/1149538959114915911/image.png

[agile moth]

how do I solve this with a sequence of equivalences

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<@&286206848099549185>

@urban grove you are good at discrete math right?

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[valid marsh]

please don’t ping specific people

[valid marsh]

You can start by using the definition of implies

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[topaz sinew]

Please don't occupy multiple help channels.

[pastel juniper]

closed the previous channel

ask both the questions here

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[raven thicket]

<@&286206848099549185>

.close

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[burnt mountain]

about these values

I don't know how to calculate them really, I know (a) is sqrt(3) because I learned that It was an important angle

Im making it more complicated every time

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[stuck forum]

Okay. What does 7pi/6 look like?

[burnt mountain]

hhmm

like an angle

150 degrees

[stuck forum]

ok how about this

what does pi/6 look like

[burnt mountain]

like a 30 degree angle

? ?

[stuck forum]

Exactly. pi/6 converts to 30degrees

Now what about 6pi/6

[burnt mountain]

that's equal to pi

or 180degrees

[stuck forum]

Good

so 7pi/6 = 6pi/6 + pi/6

So it's like you start with pi, and then add another pi/6

Does that help you picture it?

[burnt mountain]

Oh I understand

[stuck forum]

Great. So what does that tell you about sin (7pi/6)?

If you know what sin(pi/6) is

[burnt mountain]

its equal to sin [ pi + pi/6 ]

and since its a sum inside the sine

its equal to sin(pi)cos(pi/6)+cos(pi)sin(pi/6)

since sin(pi) is 0 I cancel the first term

and its equal to cos(pi)sin(pi/6)

so (-1)(1/2)

-1/2

🥹🥹🥹🥹🥹🥹🥹🥹

✅

[stuck forum]

Good job

You could do that trick for all special angles: 30, 45, and 60

Just gotta imagine where you are first and see what your angle is with respect to the closest x-axis

Here it's helpful to think of 30 degrees measured from pi

[burnt mountain]

I understand

thanks you again\

but what about the sec(5pi/3) ?

oh its not defined right

nvm it is

[stuck forum]

You good now?

[burnt mountain]

i think im about to get it

1 sec(ond)

I couldn't get it

[stuck forum]

Haha which one?

Let's work on it

[burnt mountain]

nvmmm

[burnt mountain]

it worked on this one too

that deconstruction method

thanks a lot again (again )

.close

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[normal vapor]

can someone explain question 1

[opal vault]

which part are you having trouble with?

[normal vapor]

ok

i need

help with sin 3pi x = sqrt2/2

[opal vault]

yep

[normal vapor]

idk how to get all answers within the limit

[opal vault]

first of all, what angles a do you know that have sin(a) = sqrt(2)/2?

[normal vapor]

pi/4 3pi/4