help-33

[cunning fiber]

it took like 2 seconds dw lol

[marsh citrus]

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[odd crest]

where are you stuck

[marsh citrus]

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

[buoyant jetty]

1

[odd crest]

You have to use the fact that ST and QR are parrallel

That is step 2 ish

step 1 is to understand that you have a relation between the lower bit of S and R

and then by step 2 it means you can find the upper and lower bit of S and the upper bit of T

does that make any sense

i wrote it terribly

The entire angle of S is known and it is?

[buoyant jetty]

ok i see

40 + 180-( 2R + 8) + R = 180

40 - 2R - 8 + R = 0

32 = R

,calc 180-32

[elfin berry]

Result:

148

[buoyant jetty]

str = 148

[buoyant jetty]

what do you think kat

[odd crest]

it seems right

[buoyant jetty]

arigato kat sama

[odd crest]

カッツ‐先生*

:P

[buoyant jetty]

^^

[odd crest]

i probably wrote the wrong katakana for kat

[buoyant jetty]

🤭

.solved

[marsh citrus]

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[worn sluice]

You hop between basic geometry and linear algebra, what are you?

[buoyant jetty]

I need to study for my linear algebraexam but i get bored when no one replies so i take a break doing Aops Alcumus

[marsh citrus]

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[amber birch]

.close

[severe owl]

Why is he spamming!?

[worn sluice]

South spamming close in every channel ban him

[marsh citrus]

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[buoyant jetty]

3. Let $P(x) = x^2 - 2ix - 17$. Find a polynomial $Q \in \mathbb{R}[x]$ of minimal degree such that all the roots of $P$ are roots of $Q$ and such that $Q(1) = -520$.

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[spark siren]

you did such an example already here. The way to go is the same.

• Solve for the roots
• create an new (real) polynom with this roots, ...
• ... and fulfilling the additional condition for Q(1).

[buoyant jetty]

problem is how to find roots of P

[spark siren]

1. then you should say this as statement.
2. there is a formula for the roots of a quadratic polynom.

[buoyant jetty]

a = 1

b = -2i

c = -17

[spark siren]

well, you have everything you need.

[buoyant jetty]

,, x = \frac{(2i) \pm \sqrt{-4 -4\times -17}}{2}

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

,w sqrt(-4 -4*(-17))

[elfin berry]

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[buoyant jetty]

x1 = i + 4

x2 = i - 4

,w expand (x-(i+4))(x - (i-4))

[elfin berry]

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[buoyant jetty]

I am confused

Q(x) = (x-(i+4))(x - (i-4)).R(x)

[spark siren]

you should know from the other example that you need the conjugate complex numbers to get a real polynom based on complex roots.

[buoyant jetty]

oh right, complex conjugate theorem because real coefficients

how do I do that?

wait a second

so its degree 4

?

[spark siren]

look at the other example. you did it already.

[buoyant jetty]

wdym?

[spark siren]

you did the step (expand a poylnom wite conjugate complex numbers to get a polynom with real coeefizients) in the other exaple. you should have learned this already. do it.
dont expect to get everything spooenfeeded.

[buoyant jetty]

the problem is

one root of P is already complex conjugate of the other

[spark siren]

no its not.

[buoyant jetty]

oh right my bad

Roots of P
i + 4
i- 4
Roots of Q
i + 4
4 - i
i- 4
-4 - i

like this?

[spark siren]

yes

[buoyant jetty]

Q(x) is of deg 4

Q(x) = (x - (i+4))(x-(i-4))(x-(4-i))(x-(-4-i)).R(x)

,w (x - (i+4))(x-(i-4))(x-(4-i))(x-(-4-i)) at x = 1

[elfin berry]

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[buoyant jetty]

Q(1) = 260R(x) = -520

,calc -520/260

[elfin berry]

Result:

-2

[buoyant jetty]

Q(x) = -2(x - (i+4))(x-(i-4))(x-(4-i))(x-(-4-i))

Hopefully I didnt messed up

[marsh citrus]

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[buoyant jetty]

.close

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[fierce fulcrum]

given that x^2 -11x + 28 is a factor of x^4 +kx^3 - 67x^2 +394x - 504, evaluate the sum of the four roots of the equation x^4 +kx^3 - 67x^2 +394x - 504 = 0

[pure jacinth]

Do you understand the question? Have you tried anything? Do you know long division?

[fierce fulcrum]

not really, i have tried factoring it so x^2 - 11x + 28)(x^2 +kx-67)

i dont know if i did it right

im just learning this stuff

[pure jacinth]

Multiply it out to see if it's right.

[fierce fulcrum]

no

so 67 is split up

[pure jacinth]

Wait, you shouldn't be getting k in your factor

[fierce fulcrum]

yea i dont no

do i factor the k out

[pure jacinth]

You know it starts with x^2

[fierce fulcrum]

like what possible numbers

yes

[pure jacinth]

And the two constants in the factors have to multiply to 504

[fierce fulcrum]

yes

28 x 18

[pure jacinth]

So now use a new variable for the linear term, multiply it out, and match coefficients.

[fierce fulcrum]

what do you mean

by

new variable for the linear term

@pure jacinth

ohhhh

so (x^2 -11x+28)(x^2 + bx - 18)

so i get x^4 + bx^3 - 18x^2-11x^3-(11xb) x^2 + 198x + 28x^2 + (28xb)x - 504

[pure jacinth]

Gather like terms

[fierce fulcrum]

wait so we know there is 394x and 198x +(28xb)x = 394x which means (28xb) = 196 so b = 7

@pure jacinth

is that right

how do we solve the sum of the four roots then

so k = -4

<@&286206848099549185>

now how do i evaluate the sum of the four roots of the equation

do i do division

or do i factor the fourth degree polynomial

<@&286206848099549185>

[marsh citrus]

@fierce fulcrum Has your question been resolved?

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[dapper field]

I think its 590 but thats not one of the options

nvm i was right the options were wrong

teacher exempted it

.close

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[uneven fjord]

Hey, this isn’t a question about a math exercise, more like an open question. I’m studying a career that’s pretty far from math itself, literally all I do is proportions and basic addition and subtraction. I used to be really passionate about math, Could you recommend a good starting point or resources to get back into math after a long break? Thanks ❤️

[main idol]

Khan academy

[uneven fjord]

Oh, I didn’t know the website, I searched for it and it seems really complete, thanks a lot!!

.close

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[still temple]

Could somebody help me with proving that xn is bounded? I know we'll use induction but I'm not able to prove it.

[regal coral]

do you have an idea for what might work as a bound?

try picking an x_1 and computing a few terms

[still temple]

I think since x_1>1 the lower bound will be 1, for upper bound i used 2 since we're subtracting from 2 in the x_n+1 term

So basically 1<x_n<2 was my idea

[regal coral]

1 should work

2 can fail on the first term

[still temple]

Yes and that's why I'm not able to prove it for n=1 in induction

What do I do?

[regal coral]

that shouldn't interfere with the proof for the lower bound, i don't think

we start with the base case

x_1 > 1 is true because it's given

now what is the induction hypothesis?

[still temple]

Assume that for some k belongs to n,
1 < x_k < 2 is true

[regal coral]

oh you want to prove both bounds at the same time

then we need to pick a correct upper bound after all

[still temple]

Without an upper bound as well, i won't be able to prove its bounded right?

[regal coral]

if u prove it's decreasing then i think you only need the lower bound

i think you can use x_1 as the upper bound

i think each term is going to be smaller than the last

[still temple]

yes

[regal coral]

let's try the induction with 1 and x_1 as the bounds

[still temple]

oh actually that makes sense.

x_1 will be an upper bound since it's a decreasing sequence

[regal coral]

yeah

[still temple]

i think I'll be able to do it now, thank you!

[regal coral]

you're welcome

[still temple]

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[marsh citrus]

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[spark mason]

answer is e^3

[sleek lake]

@tepid houndis it e^3 or you don't know?

[tepid hound]

I dont know

But how?

[sleek lake]

i don't know, i thought you just both have the answer key

[tepid hound]

This can never be e^3. It’s just a mediocre problem

The answer should be an integer or a simple fraction

[sleek lake]

it looks like base 6

so 55+1=100 = 36

yeah

[tepid hound]

Good

[sleek lake]

you basically can't tell if they want 100 or 36

probably 100

[tepid hound]

This is so easy

My frist answer was 79

But it was wrong

Thanks for helping me out!

.close

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[main idol]

Compare S and its transpose

[marsh citrus]

@blissful pawn Has your question been resolved?

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[blissful pawn]

what even is a difference equation HELP

[marsh citrus]

@blissful pawn Has your question been resolved?

[blissful pawn]

<@&286206848099549185>

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@blissful pawn Has your question been resolved?

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<@&286206848099549185>

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[cursive flower]

Can someone help me question 10

[marsh citrus]

@cursive flower Has your question been resolved?

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[zenith needle]

Help

[zenith needle]

Can anyone explain

<@&286206848099549185>

[fathom totem]

Start wit making a diagram of what is said

[zenith needle]

I can't go understand this question

[obsidian sand]

Can anyone help me at help 14

[fathom totem]

Think of it like this

A sector is like a slice of pizza

[zenith needle]

Ye I did that

and it's= 4 radius

Why is there no 4r

[fathom totem]

Now no matter how this the pizza is

[fathom totem]

How much does the perimeter have to be

[zenith needle]

360° ?

[fathom totem]

No no

[zenith needle]

180

[fathom totem]

Make the angle between the 2 edges be 0

So it's pretty much a straight line

[zenith needle]

yea

[fathom totem]

How long is that line

[zenith needle]

but is that related to 4 of radius

[zenith needle]

as the radius ?

[fathom totem]

Leave that part for now

[fathom totem]

Yep

[zenith needle]

Alright

[fathom totem]

So now

[zenith needle]

So I got 2 lines both are which same as radius

[fathom totem]

Wait a sec

Lemme make a diagram

[zenith needle]

Ok

[zenith needle]

Oh thanks a lot

[fathom totem]

I'm in moving car no judge

Ok so the total perimeter is 4r

[zenith needle]

oo np i got the diagram

This what I made ☠️

[fathom totem]

I mean close enough ig

[zenith needle]

Ye and we got 2 r as two lines

[fathom totem]

Yep

Now the curved part is 2r

[zenith needle]

oh yea

[zenith needle]

is that curved one the 2r

[fathom totem]

Now jus put it in the perimeter formula

[fathom totem]

Yes

[zenith needle]

we get 2r+2r ?

[fathom totem]

(Angle/360) x 2x pie x r= perimeter of sector

[zenith needle]

what is sector perimeter formula

I just got theta/360 πr² in sector

[fathom totem]

Wait u don't know that?

[zenith needle]

Nah i just got area of sector , area of arc z area of segment

major arc , major segment that's all

[fathom totem]

A wait a sec

[zenith needle]

ok

[fathom totem]

Ye u gonna need to put this formula to get it

Idk how to do without it

I was thinking put this formula and substitute the value of angle/360

[zenith needle]

Ok tell me the formula

[fathom totem]

Then most of the terms would get canceled

[zenith needle]

I'll learn that

[fathom totem]

It pretty much same thing

But instead of pie x r^2

It 2x pie x r

[zenith needle]

Theta/360 π

[fathom totem]

X2 r

[zenith needle]

theta/360 x² r

Is that the formula

[fathom totem]

Theta/360 x 2 pie r

That x is multiple

[zenith needle]

oo

Oh i was thinking it was x

[fathom totem]

Ye srry

[zenith needle]

Ultimately it's theta/360 2πr

[fathom totem]

Yep

[zenith needle]

is that lengh of arc

[fathom totem]

Yes

[zenith needle]

Yo so just left is to put the formulas now

[fathom totem]

Jus put theta/360 value from this into this formula

Now wait a sec ima go afk for like 2 min

[zenith needle]

ok

[zenith needle]

Ain't both the formula same

[fathom totem]

Almost

It jus that one you have for area and other for length of arc

@zenith needle
Did you get ur answer or do you want to see the whole method

[zenith needle]

Should I put 180/360 on the LHS and 90/360 RHs

[fathom totem]

Wait a sec

Lemme show you

[zenith needle]

I can't get this last thing

about the theta values

[fathom totem]

Wait 1 sec lemme write it down

[zenith needle]

ok

[fathom totem]

1st step we isolated theta/360

[zenith needle]

oo i was doing it wrong me been putting theta/360 2πr = 0/360 2πr which i was getting 0

so is the formula of perimeter of sector is 2r

[fathom totem]

The 1 line's LHS

Is perimeter of arc

Not sector

For sector you also add 2r

[zenith needle]

oh yeah arc*

Yeah it got it thanks a lot dude

[fathom totem]

Yw

[zenith needle]

It was my last one of this chapter 🍺

[fathom totem]

Jus remember this caus Lota ppl mix it up

[zenith needle]

ye the formulas

[fathom totem]

Well if we done here

Ima go do some questions

Gl and gb

[zenith needle]

Your a great guy dude have a great day

[zenith needle]

Alright bye

[fathom totem]

Thx

[zenith needle]

.close

[marsh citrus]

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[vague tendon]

For part b, I need to find P( chi square value w 15 degrees of freedom < 5.4 ) but using the statistical tables it doesnt give me a precise value, how am i expected to find the precise value in an exam where you dont need a calcuator thats able to calculate chi square values

[vague tendon]

for reference this was the question (part b)

<@&286206848099549185>

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[orchid oracle]

How is the given integral split in those 2 written next to it?

[elfin berry]

Pro_Hecker

[orchid oracle]

i see

another identity i have to remember

great

#555245

[marsh citrus]

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[kind lagoon]

hi i already asked it but i cant come over it. I want to show that sin(x^2) is noit evennly continous

[kind lagoon]

and for that i can take an epsilon and x_0 and x where it goes wrong

[spark otter]

evenly continuous?

[knotty trellis]

is that the same thing as uniformly?

[kind lagoon]

but if x_0 is depending on Delta i cant find a way to solve the epsilon

[kind lagoon]

sry my englisch is not that great

[kind lagoon]

but yes thats it

[spark otter]

so

[kind lagoon]

but the prob is if i choose somthing like x_0= Delta/2 + pi and x=Pi im not able to solv3e itz

[spark otter]

uniform continuity means that no matter the max "y" distance

for any two points of "x" distance small enough

their "y" distance won't go above the threshold

so if you wanna disprove it

[spark otter]

instead of reasoning with a specific delta

you can maybe try to find a sequence (x_n,x'_n)

such that their distance goes to 0

(so no matter the delta, after some point you'll be under)

and the distance |f(x_n) - f(x'_n)| stays big enough, maybe constant

[hollow glen]

so, find an $\varepsilon > 0$ such that $\forall \delta > 0, \exists x, y \in \mathbb{R}, |x-y| < \delta $ and $|\sin(x^2) - \sin(y^2)| \geq \epsilon$?

[elfin berry]

rbit

[hollow glen]

,w graph sin(x^2) from -15 to 15

[elfin berry]

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Requested by rbit

[spark otter]

maybe this is more representative

[hollow glen]

one hint is to consider the minima and maxima

[kind lagoon]

so i should find a sequence that disapprove the epsilon but not the delta criteria

[spark otter]

well yes

uniform continuity is

"for any epsilon, there is a delta such that ""delta criterion approved -> epsilon criterion approved""

[kind lagoon]

and because goes to zero and delta>0 it has to counbt

[spark otter]

so trying to disprove A -> B

means you need some instances

where A is true

but B isn't

[kind lagoon]

oh now i see itr

[kind lagoon]

ah

thank yopu you 2 fr

[spark otter]

so yeah as a hint

[spark otter]

you see this oscillates between -1 and 1

find the values of x such that sin(x^2) = -1 and sin(x^2) = 1

and notice that the further you go on the x axis

[kind lagoon]

but one question in the lesson they said that it could be hhelpful that x_0 should depend on delta but here it wouldnt be the case ig

[spark otter]

the "faster" you go from -1 to 1

[hollow glen]

it will depend on delta

[spark otter]

it comes to this indeed

[kind lagoon]

and if delta is getting smaller it cant be uniformily for the whole funtion

[spark otter]

"after some point" depends on which delta you picked

the smaller the delta, the "longer" you'll have to wait

[kind lagoon]

yeah ok understand it

thank you very much really

.close

[marsh citrus]

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[marsh citrus]

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[obtuse plume]

How do I know for x = 0, if it's a zero point, or a extreme point or turning point?

[obtuse plume]

To take as an example f(x) = (1/4)x^4-x^3

[summer grove]

you need to find the first or second derivative

[obtuse plume]

So when you've done that, how do know if it's either one of those points? Does one override another? For example if we take

(1/4)*0^4 - 0^4 = 0

But if we take the derivative

0^3-3 x 0^2 = 0

Both are 0

[summer grove]

you need to take the derivative first, then plug in the value

$f(x)=\frac{x^4}{4}-x^3\f'(x)=x^3-3x^2$

[elfin berry]

Bonk

[summer grove]

$\begin{align*}f(x)&=\frac{x^4}{4}-x^3\f'(x)&=x^3-3x^2\end{align*}$

[obtuse plume]

Yeah exactly, my thought here is that if we plug in 0 for both, we get out 0. Whereas if we plug in 0 for the original function we know thats a zero point, and if we plug in 0 in f', we know that x = 0, could either be a top or low point

[elfin berry]

Bonk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quiet anvil]

To decide between an extremum and a false positive, it has to do with the number of derivatives required to make the function non zero

You called it a turning point I think, so I will add well

A critical point is an extremum if the first derivative is zero and then the next non-zero derivative is an even derivative

It is a turning point if the first derivative is zero and the next non-zero derivative is an odd derivative

[obtuse plume]

Well if we take it a step further (not this function where it will be revelant), but if you end up with that f''(x) = 2x, we have another time where x = 0, which should be the turning point

For example in the function i originally gave, I thought that 0 was a zero point, however it turns out that it is a turning point

My question is why, and how can we check which one it is

[quiet anvil]

So you example f(x) = x^4 / 4 - x^3 evaluated at x = 0 we have the following

f'(x) = x^3 - 3x^2
f'(0) = 0

f''(x) = 3x^2 - 6x
f''(0) = 0

f'''(x) = 6x - 6
f'''(0) = -6

So because the 3rd derivative is non-zero the function has a turning point.

It additionally has a zero point here, because f(0) = 0.

[obtuse plume]

So it is fully possible to have a turning point and zero point to be the same?

As in turning point is where x = 0, but the zero point is also when x = 0

[quiet anvil]

Yes, they are not mutually exclusive

[obtuse plume]

Ah alright, that clears up a lot of confusion, I originally thought that not 2 the same points could have the same x value

[obtuse plume]

So moral of the story, 2 points can have the same x value?

[quiet anvil]

No

A turning point or an extreme point can also be a zero

It's the same point with two different properties

@obtuse plume

[obtuse plume]

But in this case the zero point would be (0,0), and the turning point would be (0,0)?

[quiet anvil]

https://www.desmos.com/calculator/psgwdvacna

Here are a few examples

So at -2 we have a point that is both a zero point and an extreme point, but around -1 we have a point that is just an extreme point

At positive 2 we have a point that is both a zero point and a turning point. And I tried to make "just a turning point" at 0 well, but it failed because I am bad at math whoops lol but it can in principle be done

Hope this helps!

[obtuse plume]

Ah I see! So a point can basically have multiple properties

[quiet anvil]

Yes precisely

[obtuse plume]

Thanks!

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[gaunt gate]

for $n\geq 7$

[elfin berry]

isomorphic to god

[quiet anvil]

sigma(n) is the sum of divisors function? For instance, sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28?

[gaunt gate]

yup

[hollow glen]

consider the harmonic series

[quiet anvil]

Let $n = \prod_i p_i^{k_i}$, each divisor is some unique selection of these terms. If we rewrite these in a clever way we can get the sum of the divisors is $\prod_i \qty(\frac{p_i^{k_i + 1} - 1}{p_i - 1})$.

However, this feels like a blind path, I don't see how this helps. Just writing it down to avoid repetition of work.

[elfin berry]

OmnipotentEntity

[marsh citrus]

@gaunt gate Has your question been resolved?

[gaunt gate]

hm

can i show that $\sigma(n)\leq n (\frac{1}{2} + ... + \frac{1}{n})$?

[quiet anvil]

oh yeah

that works perfectly

[hollow glen]

why 1/2?

[gaunt gate]

i'm using this inequality

follows from it

[quiet anvil]

you can use the fact that H_n + gamma approx log(n)

[elfin berry]

isomorphic to god

[regal coral]

Those are the possible factors of n

A factor f of n is equal to n/k for some 1<=k<=n

[gaunt gate]

missing the n in the RHS tho

[regal coral]

Oh yeah

[gaunt gate]

but yea, the question is done

this eventually becomes bigger

because of the mid terms

hmm

$\sum_{i=1}^{n} \frac{1}{i} - \sum_{d|n} \frac{1}{d} \geq 1$

need to show this i think

[elfin berry]

isomorphic to god

[gaunt gate]

.close

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[lusty crane]

Help please how do I do this?

[stoic slate]

What have you tried?

[lusty crane]

I haven't tried anything I don't know where ti start-

[stoic slate]

Rewrite the denominator of each fraction with prime factors

Can you do this?

[lusty crane]

Yes

[lusty crane]

As in like

Simpler?

[stoic slate]

Like $8 = 2^3$

[elfin berry]

Samuel

[lusty crane]

Yeah okay

[stoic slate]

Show after you do this

Do not give answer

And let me explain the procedure

[full nova]

let you cook bro

[stoic slate]

She has to know the whole procedure for the future with more complicated fractions

[lusty crane]

yes please im studying for an exam

[stoic slate]

2 * 2 can be written as 2^2

Ok now, look at each denominator

First one $2^3$

[elfin berry]

Samuel

[stoic slate]

Second one $2^2*x$

[elfin berry]

Samuel

[stoic slate]

What do they have in common?

[lusty crane]

Both have 2s

2^2 fits in them both

[stoic slate]

Perfect

And now the important question

What is missing in each one from the other?

[lusty crane]

one has an extra x2 the other has an x

[stoic slate]

Extra x2?

I see one has one x and the other does not have x

Maybe you can see clearer like this

$2^2 \cdot 2 \quad \text{vs} \quad 2^2 \cdot x$

[elfin berry]

Samuel

[stoic slate]

The first one does NOT have an x, the second one does NOT have a 2

Can you see?

[lusty crane]

Yes

That makes sense

[stoic slate]

Ok so

Now we want them to be equal

Because with equal denominator you know how to add fractions

Right?

[lusty crane]

Yes

[stoic slate]

But how can we make them equal without changing the number?

Do you have any idea?

[lusty crane]

We add an x to one and a 2 to the other?

[stoic slate]

But that would change the number

What is the number you can multiply with that doesnt change a number?

[lusty crane]

1

[stoic slate]

Good so we multiply by 1

$2^2 \cdot 2 \cdot 1 \quad \text{vs} \quad 2^2 \cdot x \cdot 1$

[elfin berry]

Samuel

[stoic slate]

Like this the expression is the same, agree?

Now we are gonna rewrite that 1 in the way we need

[lusty crane]

Yes

[stoic slate]

Do you agree that any number except 0 divided by itself is =1?

[lusty crane]

Yes

[stoic slate]

For example 2/2 = 1

And x/x=1

[lusty crane]

Yes

[stoic slate]

Like this

$2^2 \cdot 2 \cdot \frac{x}{x} \quad \text{vs} \quad 2^2 \cdot x \cdot \frac{2}{2}$

[elfin berry]

Samuel

[lusty crane]

Yes that makes sense

[lusty crane]

why did you choose x/x tho?

Like why x as the number

[stoic slate]

For the first one we choose x/x because x was missing in the first denominator that the second denominator has

[lusty crane]

Ohh

[stoic slate]

And for the second i chose 2/2 because a 2 was missing from the first denomonator

[lusty crane]

That makes sense now

[stoic slate]

So now we go back to the original problem

And rewrite like this

$\frac{7}{2^2 \cdot 2} \cdot \frac{x}{x} - \frac{x + 3}{2^2 \cdot x} \cdot \frac{2}{2}$

[elfin berry]

Samuel

[stoic slate]

Now, do you know how to multiply fractions?

[lusty crane]

Yes

[stoic slate]

$\frac{7 \cdot x}{2^2 \cdot 2 \cdot x} - \frac{(x + 3) \cdot 2}{2^2 \cdot x \cdot 2}$

[elfin berry]

Samuel

[lusty crane]

Okay

Do i multiply them with each other or subtract them?

[stoic slate]

Yes now you can rewrite denominator in the simple form

$\frac{7x}{8x} - \frac{2x + 6}{8x}$

[elfin berry]

Samuel

[stoic slate]

Now with equal denominators you can subtract the numerators

Remember the negative outside the fraction affects the whole fraction

[lusty crane]

Alright

[stoic slate]

No

You made a mistake

[stoic slate]

Read this

[lusty crane]

I don't get it

[stoic slate]

$\frac{7x}{8x} - \frac{2x + 6}{8x} = \frac{7x - (2x + 6)}{8x} = \frac{7x - 2x - 6}{8x} = \frac{5x - 6}{8x}$

[elfin berry]

Samuel

[lusty crane]

Why does +6 turn to -6?

[stoic slate]

The minus in front of the fraction affects the whole fraction

Not just the first number in the numerator

[summer grove]

$-\frac{a+b}{c}=\frac{-(a+b)}{c}=\frac{-a-b}{c}$

[elfin berry]

Bonk

[lusty crane]

Cos -+6 = -6?

[summer grove]

positive*negative = negative

[lusty crane]

Alright noted

[lusty crane]

So this is the answer?

[stoic slate]

You have to see yourself and try yourself

You also need confidence on what you see and what you do

[lusty crane]

okay noted i made notes on what u said ill practice similar questions with those steps

Thank you very much I appreciate it<3

[stoic slate]

!done

[marsh citrus]

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[lusty crane]

.Close

wait

.close

[marsh citrus]

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[marsh citrus]

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[wraith locust]

Is this right?

[summer grove]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wraith locust]

it’s number 54

<@&286206848099549185>

[meager scaffold]

Yeah, thats right :)

[wraith locust]

Okay ty

[marsh citrus]

@wraith locust Has your question been resolved?

[marsh citrus]

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[rose tiger]

Im doing quadratic equations and im wondering about the c term as most of them are the y intercept yet i have a problem that doesnt interact with the y intercept at all

[main idol]

,rccw

[rose tiger]

Huh?

[proud basin]

what am i looking at

[rose tiger]

Its a table and i did my work on it

I found the equation but im wondering how can the c term be -24 when its not the y ibtercept

[proud basin]

it is the y intercept?

its when x=0

[rose tiger]

Ye and thats not what it says on the table

There is no part in the table where x=0

[proud basin]

that doesent matter

[rose tiger]

So how should i find the c term?

[proud basin]

the c in ax^2+bx+c?

[rose tiger]

I solved it below but the c term is trippin me out

You see where it is -x squared plus 14x-24

The -24 is tripping me out how can that be

[proud basin]

why is it tripping you out?

[rose tiger]

Ill show you why

You see how this table shows x=0 so its -12 for the c term well thats how you check it

The other one doesnt have that

[proud basin]

oh

just plug in a value of x and check if it corresponds with y

any x if you really want to maek sure

[rose tiger]

Except the vertex?

[proud basin]

if you really want to make sure and check yeah

[rose tiger]

And thats another way to check and see if the c term is correct if the table doesnt give you the point where x=0?

[proud basin]

yeah

[rose tiger]

Alr also one last thing my next unit is imaginary numbers on a scale of 1-10 how difficult is it?

[amber birch]

I'm not sure how deep you're going

I suspect you're just doing the four operations, so for example

simplify (2 + 3i)(5 - i)

[rose tiger]

I just started learning so like with i

[proud basin]

its subjective ig

[amber birch]

you just need to apply foil / distributive law for this

whenever you have i * i, that equals -1

by definition

[rose tiger]

I thought -x squared root

[amber birch]

also you can use the quadratic formula to find the non-real roots of a quadratic

it's the same as always just for example sqrt(-64) = sqrt(64) sqrt(-1) = 8i

[amber birch]

equivalent, $i = \sqrt{-1} \implies i^2 = -1$

[elfin berry]

south

[rose tiger]

Wait i dont get the 8i part

[amber birch]

laws of indices, $\sqrt{ab} = \sqrt{a} \sqrt{b}$

[elfin berry]

south

[rose tiger]

I get it now

[lone heart]

The "trick" to remember is "i one, i one, two negatives in the middle" because the powers of i repeats every 4 terms so i, 1, -i, -1 which corresponds to i^1, i^2, i^3, i^4

[rose tiger]

Its hard to get help since the class im in at my school is two years ahead from everyone else

[rose tiger]

Whats the i circle

[amber birch]

basically this, powers of i

but you can see it on the complex plane

[rose tiger]

Alr ye ok i think im good now tysm for your guys help

[amber birch]

no worries!

[marsh citrus]

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[amber birch]

.close

[marsh citrus]

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[rose tiger]

I got a basic question so when you split the 12 square root into simpler terms lets as shown when 4 splits into two 2’s when you wrote it as 2 square root of threes is it because you just pick one of the 2’s from the 4 or is because its simplified from the square root of 4 and square root of 3

[lone heart]

,rotate

[amber birch]

the second thing you said

$\sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$

[elfin berry]

south

[rose tiger]

So its just simplified from the square root of 4 and three

When your splitting the 12 would you take whatever two numbers are the final you can split before one can go on and the other cant?

[amber birch]

yeah you keep splitting, so that means find the prime factorisation first

then you pair up factors

so 2 appears twice so you can pair those up

[rose tiger]

So once you cant split both numbers you choose that one?

[amber birch]

if you had 4 copies of 2 that would also work, or 6 copies etc

[amber birch]

I don't know what you mean by this exactly

[rose tiger]

You see the image above where its 4 and three but the 4 can keep going

[amber birch]

ah yes, exactly

so you need to keep going until you can't split any more

[rose tiger]

Would you pick the four and three since its only 2 below that

[amber birch]

then you will have found the prime factorisation

[amber birch]

yeah if you recognise 4 is a square already you don't need to

[rose tiger]

Ok so if you cant split both numbers anymore you pick those?

[amber birch]

so yeah you just need to reach a point where you can recognise a square number actually

[amber birch]

yeah so as I was saying, if you're still stuck you do this

you get the prime factorisation

then pair up numbers that appear twice

[rose tiger]

Alr it makes sense its just havnt been practicing over winter break so ik refreshing my memory

[amber birch]

if you try 24 for example, factors are 2, 2, 2, 3

pair up 2 and 2

so you're left with sqrt(2 * 2) sqrt(2) sqrt(3)

= 2 sqrt(2) sqrt(3)
= 2 sqrt(6)

[rose tiger]

Ah so even if it can keep going but splits into even more stop at where its only 2?

[amber birch]

I mean you can't split 2, 2, 2, 3 any further

those are all prime numbers

[rose tiger]

Wait hold on

So its square root of 24 so could you do 12 and 2?

[lone heart]

Well you want to prime factorize

[rose tiger]

I suck with vocab

[lone heart]

Break the number into its prime numbers

Do you know what a prime number is?

[rose tiger]

Kind of i know 7 is a prime number but kinda forgot

[lone heart]

A prime number is a number that is the product of 1 and itself

[amber birch]

12 is not a perfect square

[rose tiger]

Ah alr

[amber birch]

you'd need to split it as 6 and 4

now 4 is a perfect square

yeah like a prime number has no other factors other than 1 and itself

[rose tiger]

Ok so you split until you cant

Then what?

[amber birch]

this

[rose tiger]

Wait so we have 3 2’s so 3 square root of 2’s?

Times 1 square root of 3?

[amber birch]

exactly

my point is, you said you have $\sqrt{2} \sqrt{2} \sqrt{2} \sqrt{3}$

[elfin berry]

south

[amber birch]

you can combine $\sqrt{2} \sqrt{2} = 2$

[elfin berry]

south

[lone heart]

https://www.youtube.com/watch?v=u2Z1hoXSrXk&ab_channel=Mario'sMathTutoring
Maybe this will help

[rose tiger]

Alr i get it now tysm

[amber birch]

np!

[amber birch]

.close

[marsh citrus]

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[rose tiger]

This equation i wrote no solution yet now i think i am wrong it is possible am i correct you can solve it? Also i dont use the quadratic formula i make to much mistakes with it

[late geode]

work seems ok so far, try continuing

[rose tiger]

Ik i can solve it its just im trying to figure out why the hell i wrote no solution

[late geode]

forget about what happened before and just solve from here

[rose tiger]

Ok

Omg i figured out why ok were good thx

.close

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[proud basin]

i did casework and split into 2,3,4,5,6, ans hey is 4 where did i go wrong?

[cobalt sedge]

If your probabilities are a/b and c/d per case, then the total probability is (a+c) / (b+d) not a/b + c/d.

[regal coral]

are we assuming the first player picks from {2,3,4,5,6} at random?

[proud basin]

wait what

[frosty thicket]

its (wanted cases)/(total cases)

[proud basin]

its

[frosty thicket]

so if you have (wanted cases1)/(total cases2) and (wanted cases2)/(total cases2), the amount of (net wanted cases)/(net total cases) is the mediant

[cobalt sedge]

You seem to have added up the probabilities? 1/2 + 1/2 + 1/5 + 1/5

[proud basin]

yeah then divided by 1/6 i thought thats what i had to do?

[frosty thicket]

how many numbers can the first player pick?

[cobalt sedge]

The size of sample space for each of the cases is different, so that wont work here

[proud basin]

5 oh uhh

[frosty thicket]

the problem statement isnt very clear

[proud basin]

7/25

[frosty thicket]

because it doesnt state whether p1 is random, playing to win, or playing to lose

[cobalt sedge]

Go from first principles. How big is the sample space (how many wasys can you pick the balls in total)

[regal coral]

is this wrong?

[cobalt sedge]

and how many of them are where you are winning

[frosty thicket]

this problem could be stated so much better

[proud basin]

ans key is if its a/b then a+b=4

im guessinf 1/3?

[regal coral]

i can't see how 7/25 is wrong

which would give 32

[frosty thicket]

im chalking it up to bad problem statement

[proud basin]

maybe

[frosty thicket]

if there were players, they would be playing to win

so they would pick a 1/2 case

that would be my rationale

[regal coral]

yeah this gives 1/3

but i don't like it

because it requires the first player to pick a number from {2,3,4,5,6} with a certain distribution, i think

[amber birch]

yeah

[cobalt sedge]

Its easy, theres only 3 eventualities, you win/you lose/you tie. Makes sense the answer is 1/3

[frosty thicket]

then that would be a poorly written question

[amber birch]

like the assumption is that 2, 3, 4, 5, 6 are picked uniformly

I took one look and skipped cause of this

[frosty thicket]

my assumption is that the players want to win

where is this question from?

[proud basin]

by your logic if its 50/50 of rolling a coin and dice and you had to get either heads or 1, then if you roll 12 times you get 6 that roll coin and 6 that roll dice, so 3 works 1 dont, so 4/12=1/3
so 1/2(1/2+1/6)=1/3
but (1+1)/(2+6)=2/8=1/4?

[regal coral]

that's also a reasonable assumption

[proud basin]

seamo

[frosty thicket]

perhaps its a mistranslation

[proud basin]

its original is in english afaik

[frosty thicket]

oh uh nvm

[proud basin]

alternatively, maybe ans key os just wrong

[frosty thicket]

seamo is southeast asia yeah?

[proud basin]

yeh

ok ill just accept something is wrong here, either me, the question, or the answer key

.close thanks

[marsh citrus]

Channel closed

Closed by @proud basin

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[marsh citrus]

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[faint narwhal]

How to find the angles between n vectors in 3D space, such that the minimum angles for every pair of vector is maximized

[marsh citrus]

@faint narwhal Has your question been resolved?

[knotty trellis]

https://math.stackexchange.com/questions/3778459/maximum-angle-of-separation-between-n-vectors-in-m-dimensions

might be relevant

someone answered that hte general answer is unknown

https://en.wikipedia.org/wiki/Thomson_problem

here is a table

[knotty trellis]

It's actually not the exact same problem, but is similar

[cobalt sedge]

This is very much related to spherical packing, which doesnt have known solutions for most values of n

[knotty trellis]

https://en.wikipedia.org/wiki/Sphere_packing_in_a_sphere

[marsh citrus]

Channel closed

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[marsh citrus]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[hallow plover]

over here @severe owl

i be stuck in the washing machine once again

HELP!!!!

lmao

[hallow plover]

ok so back to the exercise

so, this is what ive tried

which, if you look closely

doesnt work

[severe owl]

Wait

[hallow plover]

mhm

[severe owl]

I don't think the function can be integrated without using complex functions

[hallow plover]

complex me up, dawg

[severe owl]

Give me some more time

[hallow plover]

i aint scared

[hallow plover]

i mean, if uve got an idea

u can tell me so that we both try it at the same time

[severe owl]

Let n be 1

And solve

It gives x⁴/2

That integrates to x⁵/10

Now if you put n= 1 in answers and check you won't get such an answer

[hallow plover]

oh ok

[severe owl]

I mean the function cannot be integrated, we can try an approximation by expanding x^5n-1

Wait let's try wolfram

[hallow plover]

i mean, it seems like ure saying that our integral is a recurrance

[hallow plover]

no no, this has to have a very specific and thorough solution

[severe owl]

,w integrate [x^(5n-1)]/[(1+x^(n-1)]

[elfin berry]

Results provided by WolframAlpha

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Requested by xor_11

[severe owl]

Maybe there's some error in printing !

[hallow plover]

it isnt

i can show you their solution if u want

[severe owl]

Wait

[inland hemlock]

perhaps u need to think about some properties of the function

[hallow plover]

but they pull something out of their asses so like

[inland hemlock]

i don’t think u have to integrate that

[hallow plover]

idk, their solution seems like an abracadabra type thing

[hallow plover]

they didnt either

[inland hemlock]

exactly

nobody in their right mind expect u to integrate that

[hallow plover]

yea, uve gotta make that fraction simpler first

like...wayyyy simpler

nevermind

im making einstein shiver in his grave from fear

[severe owl]

I still say, the question is incorrect

[hallow plover]

let me send u what i got to

[hallow plover]

wrong

watch this:

boom

idk what to do from here

but boom!

[severe owl]

See the question you send

[hallow plover]

oh

[severe owl]

It reads x^(5n-1)/(1+x^(n-1))

[hallow plover]

first of all, the fact that you figured that the question is wrong

is so impressive

but yea you are right,

the actual question is the one written on my paper

[severe owl]

Can I get the correct one ?

[hallow plover]

yea sure, im translating them from another language

thats why chatgpt fucks up sometimes

[severe owl]

[x^(5m-1) + x^(n-1)]/(1+x^6n) right ?

[hallow plover]

here

im very sorry about the fact that i put the wrong question the first time

ill be more careful next time

can you put the last thing i got here into wolfram?

[severe owl]

You used n = 1 ?

[hallow plover]

,w integrate [u^4+1]/[1+u^6]

[hallow plover]

nope

[elfin berry]

Results provided by WolframAlpha

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Requested by sizzywizzy0_0_32713

[hallow plover]

but check my solution

[severe owl]

Yeah it's is correct considering x^3n= u, would have made it easier

But still in your answers

[hallow plover]

i tried that too

some while ago

[hallow plover]

idk, i cant see it in my answers so

[severe owl]

Ok !

[hallow plover]

ill try x^3n = u as well i guess

[severe owl]

Ok

Let me tell you another approach

@hallow plover
There ?

[hallow plover]

yea

so the x^3n = u is really goddamn ugly

[severe owl]

So x⁶n +1 =

(x²n+1)(x⁴n-x²n+1)

So we do this

And then

In numerator

x^(n-1)[x⁴n+1-x²n+x²n]

This gives

x^(n-1)/(1+x²n) +

[hallow plover]

interesting

i dont really get the numerator thing

[severe owl]

[x²n × x^(n-1)]/(...)

[hallow plover]

because how would you write x^(5n-1) as x^(n-1) times something?

[severe owl]

It's x^(5n-1) no?

[hallow plover]

yeye

[severe owl]

So we can

[hallow plover]

how

ill write it on paper

so that i see it better

[severe owl]

We took it common

[hallow plover]

ok

so far so good

i got this:

@severe owl you here?

[severe owl]

Yes

It would be + between them x⁴m + 1

[hallow plover]

wait wait

why

oh

i see

[severe owl]

The harder integral part is separated now.

[hallow plover]

so the last two fractions are wrong

lets not forget about that

from my pic

[severe owl]

Yes

[hallow plover]

ok tell me

[severe owl]

Wait I'm looking for the best substitution to solve this

Ok so now let x^m = tan z

[hallow plover]

aha

but then we would have tan^4 +1

in the numerator

[severe owl]

Wait

[hallow plover]

i think trig subs are tricky here

[severe owl]

I haven't multiples the 1/m you can do em

[hallow plover]

its so confusing what u wrote

let me try it too

[severe owl]

Do this and the last integral is just beautiful

[hallow plover]

so x^n = tan u, is that what u said?

[severe owl]

Yes

[hallow plover]

okk

let me see

[severe owl]

First separate the integrals like I did ok

[hallow plover]

is this correct?

[severe owl]

We don't need to do like this !

x^n = tg z
So, nx^(n-1) dx = sec²z dz

[hallow plover]

oh you took the derivative

smart

so like this?

[severe owl]

: )

I said you to separate the integrals

[hallow plover]

howwwww

and whyyyy

[severe owl]

😭

[hallow plover]

dont give up on me

[severe owl]

Ok let me

[hallow plover]

ok listen

[severe owl]

Explain again

[hallow plover]

do i separate the integral now?

do i make it one with the numerator tg^4 and the other with the numerator 1?

[severe owl]

We had x^n-1 (x⁴n+1)/(x²n+1)(x⁴n-x²n+1)

Right ?

[hallow plover]

no

everything is n

everything

theres no m

[severe owl]

Huh same

Just consider it, typo

[hallow plover]

ok sure

[severe owl]

Now

In numerator

[hallow plover]

aha

[severe owl]

x⁴n+1 = x⁴n+ 1 -x²n+ x²n

[hallow plover]

ok so, can i not just do that now?

[severe owl]

We can

[hallow plover]

with tg^2u?

[severe owl]

Yes

[severe owl]

Surely !

[hallow plover]

ok good

imma do that then

[severe owl]

Did it ?

[hallow plover]

alrighty

we are close

[severe owl]

Nah not that whole square 😭

[hallow plover]

aint no way i did something wrong

[severe owl]

Yes

[hallow plover]

yes?

[hallow plover]

WHAT YES?

[severe owl]

Wait