help-23

[stuck compass]

sure

x^3(x+8)+x(x-8)?

is that what u were saying?

[stuck compass]

.

[violet cosmos]

It should be something like this I believe if I remember my factoring correctly

[stuck compass]

ive got the answer on the powerpoint

ill have a look

11 is the answer

[twilit hare]

also the second one is wrong

[stuck compass]

-x(x+8)

[twilit hare]

yes.

now you can again take x+8 common

[stuck compass]

does that answer line up with yours?

[twilit hare]

yes it will.

[twilit hare]

what answer do you get after doing this?

[stuck compass]

1 sec sorry

x^3(x+8)-x(x+8)

so thats what we got rn

[twilit hare]

do the next step

[stuck compass]

is that right tho?

before i write it down sorry

[twilit hare]

yes

[stuck compass]

ty

im not sure what the next step is sorry

x+8 is commmon do ii take it out?

[twilit hare]

yes

suppose

[stuck compass]

idk how to write it sorry

[twilit hare]

$y(a+b) -x (a+b)

(a+b)(y-x)$

if this helps

Oops

[flat frigate]

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

[stuck compass]

(x+8)(x^3-x)?

[twilit hare]

last term

[stuck compass]

-x?

[twilit hare]

yes

[stuck compass]

is that the answer?

[coarse jolt]

Yup!

But you'd be thinking "it seems different from the answer key?"

[stuck compass]

lol i was just about to send that haha

[coarse jolt]

That's because it factored about x from $x^3 - x$

[flat frigate]

Kai

[coarse jolt]

so lets break that down

$x(x^2 - 1)$ is what we end up with

[flat frigate]

Kai

[coarse jolt]

does $x^2 - 1$ look familiar?

[flat frigate]

Kai

[stuck compass]

yep

[coarse jolt]

what does it expand to?

[stuck compass]

(x+8)(x^2-1)

[coarse jolt]

Expand $x^2 - 1$

[flat frigate]

Kai

[stuck compass]

x(x-1)

[coarse jolt]

you sure?

[stuck compass]

nope

[coarse jolt]

do you remember how special pair of polynomials have this property

for example, consider $x^2 - 4$

[flat frigate]

Kai

[coarse jolt]

we can make a pair out of it. what does this expand to?

$(x-2)(x+2)$ right?

[flat frigate]

Kai

[stuck compass]

yep

[coarse jolt]

now consider $x^2 - 1$

[flat frigate]

Kai

[coarse jolt]

it's similar. what would that expand to

[stuck compass]

(x-1)(x+1)?

[coarse jolt]

correcto

[stuck compass]

yay

[coarse jolt]

so we have $x(x-1)(x+1)(x+8)$

[flat frigate]

Kai

[coarse jolt]

seems like that's it

[stuck compass]

oh yeah ty

tysm u helped alot

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[noble flicker]

how to prove that the limit of this function does not exist?

i tried taking an aribtrary line y=mx

so i subtitutet into f(x,y) points f(x,mx)

and i get xe^(x/mx)

which is xe^(1/m)

this seems to go to 0 as x goes to 0

but i think there is one problem

if m is very small e^1/m goes to inf

so it doesn't seem right to multiply 0*inf

i don't know if there is any way i can change this function

[left gyro]

try doing y = x^2 instead

[noble flicker]

why?

it works but should't i take a general line?

[left gyro]

the line didnt work

so you shouldnt

I guessed y = x^2 because I graphed f(x, y) and I saw something parabola-like when I looked at it from the top

as for how you make that guess on your own, you could take the more general guess of mx^n instead of just mx

[noble flicker]

thats what i did but its hard to prove it that way

[noble flicker]

but can we have a general conclusion from this?

[left gyro]

the general conclusion of DNE isnt enough?

do you mean a more specific conclusion?

[tame raft]

wrong reply

[tame raft]

the idea is that you need e^(x/y) to increase faster than x decreases

so that xe^(x/y) will not be 0

so you need x/y to be very large

so you need y to be very small compared to x

so y = x^2 works

[noble flicker]

no i mean can we give a general conclusiong by using a particular line(curve in this case)

y=mx is a geeral case

but i think i got it

thank you both

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[fresh spade]

Provide a magma G1 = (M1, ◦) and a sub-magma G2 = (M2, ◦) of G1, such that G1 has a neutral element e1 and G2 has a neutral element e2, and e1 ≠ e2. Justify your answer.

Is this even possible?

[steep lily]

(I think so realised my example didn't work, thinking of another)

[fresh spade]

Well my mental image is that: imagine you have a friend group and every friend has a neutral opinion and only one neutral opinion on a certain topic, now find at least one friend from this group that has a different neutral topic. But there only exist one neutral topic in the entire group and this leads to a contradiction.

Am I not seeing something? It is an exercise that should be possible, but maybe the tutors just made a mistake

[split ether]

It is possible

[fresh spade]

I tried to do this with modulo, but I can‘t think of any solution that makes sense

[split ether]

Can't think of a hint though

[steep lily]

e1 is a neutral element that works for all elements in G1, so all you need is a submagma that such e2 works for everything in G2, but not necessarily G1

[split ether]

One thing you can do is find an nonneutral element x such that xx = x and then think about the subset {x} of your magma

If you can't think of a magma with such x, here is one: ||Consider the set of integers under multiplication||

[fresh spade]

Well I could then just say my set ist {1} and I have (S,) in which case G1 would be 11=1 while G2 could be Ø which can‘t have a neutral element or {0,1} and I could just say. e1=1 and e2=0. but according to my professor a gruppoid can have at max one neutral element

[split ether]

I don't understand what you mean

So 1 is the netural element of {0, 1} with respect to multiplication, right?

[fresh spade]

Yes

[split ether]

And you also know that 0 * 0 = 0

[fresh spade]

And if my subset is 0 it does not matter with which element I multiply, it is still zero

[split ether]

So is there a neutral element in the submagma {0} and what is it?

[fresh spade]

0

Ah i get it now

Clever

[steep lily]

another large class of examples is basically any set with an order, the minimum operation, and a least element (which is then the neutral element), and taking G2 any subset with a different least element

[split ether]

Yeah that's a good family of examples

[fresh spade]

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[feral thicket]

Which q

[little mesa]

part e

what is the given answer for f btw

[feral thicket]

Might be a typo

[little mesa]

it does seem like a mistake

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@sick urchin Has your question been resolved?

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[dusk plaza]

Is this true? I seem to make a mistake when trying to bring them to the same form, I just need to know if this is correct and if you can always change the form from the left side to the right side by taking the square root from "a" and "b"

[dusk plaza]

"a" being s^2 and "b" being 4 in this example

[twilit hare]

are you assuming i to be -1?

even if so, it's still incorrect

[split ether]

s^2 + 4 can be factored as (s - 2i)(s + 2i), yeah

[twilit hare]

even if i is assumed to be -1, you would end up with$s^2-2^2$

[flat frigate]

斯韋裡

[split ether]

Pretty sure they mean i to be the imaginary unit

@dusk plaza Is that so?

[twilit hare]

Waittt

It is correct

lmao I assumed i as -1 when $i^2 is -1$

[flat frigate]

斯韋裡

[dusk plaza]

yes i is imaginary unit

like that?

[twilit hare]

what's this part

the rest is prefect

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[neat prawn]

according to the formula log should be in division right?

[pulsar pecan]

What do you mean

[jolly bridge]

log m^n = n log m
(if i understood your question correctly you are confused why log is not in the denominator , hope the above answers the question )

[neat prawn]

Can you elaborate?

I can't understand why log is the numerator

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[neat prawn]

<@&286206848099549185>

[wooden forum]

As I know log going to denominator is correct

[neat prawn]

yep, but I looked at my solution from last time and it was in numerator

i can't remember why tho

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[tough canopy]

Let the function ( f: \mathbb{R} \setminus {-1} \rightarrow \mathbb{R} ), ( f(x) = \frac{ax^2 + bx + 1}{x + 1} ). Determine the real parameters ( a ) and ( b ) such that the graph of the function passes through the point ( A(2, -3) ) and ( f'(1) = 1 ).

[flat frigate]

Hanee

[pliant shuttle]

help

[tough canopy]

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[vague zinc]

Can anyone help me with this matrix proof? It goes as follows: 'Suppose A,B ∈ dim 2x2. If A and B are symmetrical, then A^2 - B^2 is also symmetrical

[vague zinc]

I don't know where to start.

and then just see if the elements match up?

wait nmv

alright lemme try

thanks for all the info 🙂

I might have some more proofs later so if I struggle with them can I ping you or are you not gonna be online anymore

@proud laurel

alright thanks for the help

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[hardy lark]

I have no idea where I went wrong

[hardy lark]

I’ll send the question in a sec

I rechecked all my work it seemed all right to me, but it says it’s wrong

This is a calculus related rates problem btw

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@hardy lark Has your question been resolved?

[hardy lark]

<@&286206848099549185>

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[hardy lark]

Should I keep this channel open? I’ve never had no response for this long :p

Ok I figured it out

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[crude kite]

how would you get the inverse of this function?

[obtuse plover]

what would you be inverting lol

[crude kite]

cotx

the question just says solve problems related to trigonometric functions and their inverses

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[distant drum]

exponents law! how does (1/2)^2 + (1/3)^2 equal to 13/36

[distant drum]

i got 22/36 and then i simplified it to be 11/18

[pure agate]

1/4 + 1/9 = 9/36 + 4/36

[distant drum]

Oh bruh 😭

LMFAOOOO

okay thank u

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[dusk stratus]

I tried doing it but it didnt work out

[light shoal]

what qualifies as "simplified"?
there are no common factors btw the num and denom so there's nothing to cancel

[obsidian oracle]

ig factor

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[viral bison]

I have forgotten how to do this

please help meh

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[true marsh]

help. how do I approach #2. I'm actually lost

[severe chasm]

Solve for h using cosine, then take the sine and plug in h

[median vigil]

you'll need to find the right trig ratios, then solve for the side length you're interested in. try to find trig ratios that involve only 1 unknown quantity

[true marsh]

7cos(65)=h?

[true marsh]

my brain can't seem to connect the dots idk

[severe chasm]

what is cos(65) equal to

[true marsh]

cos(65)=7/h

[severe chasm]

yes, exactly. now rearrange the equation to the form "h = xxx"

[severe chasm]

not quite

[true marsh]

h=cos65 * 7

[severe chasm]

not quite again

remember you must do the same operation to each side of the equation

cos(65) = 7/h

h*cos(65) = (7/h)h = hcos(65) = 7

h = 7/cos(65)

do you follow that

[true marsh]

up to this

[severe chasm]

yep

cos(65) is just a number right?

divide both sides by cos(65)

now you have a solution for h

which you can use to solve for x

[true marsh]

holy

my number vision has expanded

thank you

[severe chasm]

haha

[true marsh]

I see now

[severe chasm]

Very important to thing to understand in algebra

Now try to solve for x

[true marsh]

alright

top right

I think I did something wrong

[severe chasm]

looks good

You are correct

[true marsh]

seems disproportionate cause it's so close to the hypotenuse

I see, thank you you

[severe chasm]

sometimes that will be the case, no length will ever be equal or greater than the hypotenuse

plug in your own answer is 7^2 + 15.011^2 = 16.563^2?

[true marsh]

yes it is

math is crazy

Thank you harry that's all I have for tonight, you really expanded my view

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[fair warren]

which one is right and why?

this one is trippy

.clode

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[lean otter]

can i please get help and a explanation for question 5bii and 5d

[unique salmon]

show the whole image

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[bold mason]

I'd like to know whether i was right with assuming this problem requires 2 solutions and if i approached it correctly (i do not have answers in my study book)

[hard crest]

yep that seems right to me

you can plug your x value back into the original problem and check it

[bold mason]

Because photomath only gives me 1 solution so i was wondering whether it's correct to also do the second one

Or is photomath not correct here?

[fallen thunder]

i mean isn’t determinant supposed to be a single value

[bold mason]

Determinant can have two values, a positive and a negative, but i am not sure with this particular type of problem because of photomath's answer

[fallen thunder]

no?

[wheat locust]

i think he's trying to say there could be more than one x that makes that determinant equal to 5

[fallen thunder]

but there isn’t?

it’s a linear equation with a single answer

[bold mason]

I-5I and I5I both equal 5?

[fallen thunder]

not sure why op put | | for determinant though

[bold mason]

that's what i meant for example

[fallen thunder]

i know that

why do you have | | is the bigger question

u don’t have those for determinants

if you have absolute values then yes your answer is right

[bold mason]

oh that's the original problem

[fallen thunder]

?

i’m talking about the steps that follow

[wheat locust]

yea there isn't, i was trying to cut him some slack but idk what he meant now tbh lol

[fallen thunder]

the determinant when x = -7/2 is -5

ofc when you abs value that you have 5

but u shouldn’t?

[bold mason]

oh yeah sorry i have absolute values in the problem, sorry english is not the language i'm taking this in

the I I in the problem is absolute value

[fallen thunder]

that thing means determinant btw

not abs value

cuz that doesn’t make sense in context

[bold mason]

oh well then i'm not sure

i was assuming it means absolute value

so then it is supposed to only be one value?

[fallen thunder]

well depends what your question is tbh

but if it’s determinant then yes only one value

[bold mason]

hold on i can show you a similar problem

[fallen thunder]

the one photomath gave you

[bold mason]

This is where i got the problem from, it's like a manual, but the solution is wrong halfway through anyway, so i decided to ask here. Is the whole thing wrong?

[fallen thunder]

well if anything you should follow your manual right

as far as your work goes, it’s correct

(the way it’s intended to be done by your manual)

but the notation is ambiguous tbh because that | | around the matrix is used for determinant

not |det(A)|

but yeah i think your guide wants you to find |det(A)| in which case your work is right

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[bold mason]

Okay, thanks for the help!

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[lean otter]

Hey

[lean otter]

I need help on just 1! Quick question!

2TT

How do turn it into degrees

fyi

I already for questions like 1/1

It's 180/TT

But this is just 1 number instead of 2

if u want to refer to the constant 3.14159.... etc

just say pi

not

TT

What am I doing bro

It's obviously 360 💀

yeah

That was easy

Thank you

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[royal sorrel]

ummm

[royal sorrel]

hello?

just wanted to ask

if there are a 1 to 9, then arranged em in different line

would it be 9! ??

ping me please!

[lean otter]

can you reword this?

[royal sorrel]

im sorry

so there's a number, 1 2 3 4 5 6 7 8 9

how many ways you acan re arrange them?

is it 9! ??

[lean otter]

yes it'd be 9!

[royal sorrel]

okie

thx

see you sooner!

.close

[safe radish]

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[pine igloo]

.rotate

[mental dove]

,rotate

[pine igloo]

Can anyone pls explain where they got k from?

This is the page prior

,rotate

[past birch]

+2kpi?

k is any integer

since cos(x) = cos(x+2pi), it follows that cos(x)=cos(x+2pi)=cos(x+4pi)... = cos(x+2kpi)

[pine igloo]

Trig is my weakest spot, why is it significant to say that cos and sin have period of 2pi?

[past birch]

in the unit circle

one revolution is 360 degrees = 2pi

dont worry, trig is also my weak spot

[safe radish]

@pine igloo Has your question been resolved?

[pine igloo]

i'll look into it more then come back here if i have any questions

thanks!

.close

[safe radish]

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[vagrant edge]

Is there a shortcut to compute y = cos(x+delta) if you have cos(x) but not x

[vagrant edge]

I'm trying to avoid doing the inverse cosine for performance reasons in my code

[thin bridge]

how efficient would that be compared to getting sin(x)

[winter pivot]

Cosine addition formula

You can find sin(x) by using pythag

Then you will need cos(delta) and sin(delta)

But not difficult at all

@vagrant edge

[vagrant edge]

thanks perfect

[winter pivot]

Actually

You will need some info about x for this to work

Because it involves the square root and you don't know if it's + or -

[vagrant edge]

il give some more details. What I'm actully doing is computing the dot products between vectors to get the cosine similarity, and I wanted to perturb the cosines directly without doing any inverse trig

[winter pivot]

Do you have bounds on x? If they're small angles then you should be fine

[vagrant edge]

no bounds, x is the angle between two arbitrary 2d unit vectors

i dont see any square roots in the formulas posted though?

[winter pivot]

You will need sinx

[vagrant edge]

thats ok, I can do the cross product of the vectors

[winter pivot]

Seems kind of inefficient but ok!

[safe radish]

@vagrant edge Has your question been resolved?

[safe radish]

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[ripe tinsel]

for the taylor's inequality why does this hold:

[safe radish]

@ripe tinsel Has your question been resolved?

[ripe tinsel]

yes but I think I start to understand the taylor inequality

but I have 1 question

so I understand for every x and for every n the taylor's inequality hold. but the R_n is a summation of all the n's after the taylor expansion right. is the right side a summation too?

[vagrant ice]

No, the right hand side is just some expression of x and n

Not a summation

Also the centre of convergence is assumed to be x = 0 here

Replace x with x - a for a general Taylor series

Except for the remainder Rn (x)

[vagrant ice]

The left hand side isn't even a summation

It's just an expression for the max error

[ripe tinsel]

but this was my definition of R_n btw

where t_n(x) is this

u can replace the 0 with a

[vagrant ice]

Yeah true then, okay that is a summation

[ripe tinsel]

but

I still don't see how that whole summation is smaller then 1 number

<= btw

[vagrant ice]

The concept is that the error in the Taylor expansion of n terms

Is always smaller or equal to the (n + 1)th term

[ripe tinsel]

uhh why n+1 term only actually?

[vagrant ice]

Cause the Taylor series is the best polynomial approximation to a function

It encodes all the derivatives

[ripe tinsel]

ye

[vagrant ice]

So the 1st to the nth derivatives all match the original function

I'm not sure of the formal proof

[ripe tinsel]

no worries I just wanted to udnerstand the intuive proof

[ripe tinsel]

what does this mean actually?

[vagrant ice]

https://en.m.wikipedia.org/wiki/Taylor's_theorem

[vagrant ice]

They're the same as the original function

[ripe tinsel]

ye if the remainder is 0 it's the same as original I understand that part

is that what u meant?

[vagrant ice]

No, the derivatives are the same

The first derivative, the second derivative etc

[ripe tinsel]

I learned that the radius of convergence of the derivatives are the same

but the derivaties are all the same as the original function value?

[vagrant ice]

Yes, at the centre of convergence

That's what the best polynomial approximation of a function at the point means

[ripe tinsel]

but say I got this as the original function

at the centre of convergence so at x = a u get f(a)

[vagrant ice]

Yeah

[ripe tinsel]

but for the derivative

u get uhm

[vagrant ice]

So actually all the other terms except f'(a)(x - a) vanish

[ripe tinsel]

oh wait

[ripe tinsel]

this one seems a bit weird

so like

let's take this as example

[vagrant ice]

Like one of the terms has derivative 2(x - a), and when you sub a it goes to 0

[ripe tinsel]

so f(a) = c_0

[ripe tinsel]

ye

[vagrant ice]

Yeah

[ripe tinsel]

but if u take the derivative u get c_1

[vagrant ice]

Yep so our Taylor also has c1 x

[ripe tinsel]

so in this case the original function has all the values of the c_0 ... c_infinity

for every x

[vagrant ice]

Yeah so when you take the nth derivative you get n! as the coefficient

But we divide by n!

[ripe tinsel]

ye

[vagrant ice]

That's actually why we divide by n!

[ripe tinsel]

ye I actually understand that part

[vagrant ice]

Just from repeated differentiation

[ripe tinsel]

adn the part where the rest vanishes

[vagrant ice]

Cooooool

[ripe tinsel]

but

the inequality is hard to understand lol

[vagrant ice]

Have a look at some examples

Like sin x and cos x, their maximum absolute value is 1

So you can use M = 1

[ripe tinsel]

ye so for example I understand this

but this is if lim n goes to infinity

[vagrant ice]

No like worked examples with different kinds of functions

[ripe tinsel]

this?

I could follow the steps here too

[vagrant ice]

Yeah

[ripe tinsel]

but where is the theorem used her actualy to me it seems like integrating it a couple times and then u get smth like the squeeze theorem

[vagrant ice]

You'd just directly use the fact that the error for n = 2 is the n = 2 term which is x^3 ÷ 6

They didn't want to substitute pi/180 into x so they made another approximation

[ripe tinsel]

but I mean without the taylor's inequality couldn't I derive that too?

[vagrant ice]

Yeah

They're showing you how to derive it without that

[ripe tinsel]

oh but how will it hlep me understand taylor inequality

[vagrant ice]

Uhhh idk

[ripe tinsel]

do u know how I prove this btw?

because taylor's inequality should be easily derviable after I know this

[ripe tinsel]

https://www.youtube.com/watch?v=lY0LzJXTgeo

I watched this video

I see that it works for some examples

[ripe tinsel]

but he didn't showed this

[safe radish]

@ripe tinsel Has your question been resolved?

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[crimson oar]

If I have this matrix and it has this characteristic polynomial, how can I solve it without just subbing values in?

[crimson oar]

$t^4-12t^3+56^2-128t+128$

[flat frigate]

CharlesWorthingtonShire

[crimson oar]

noting that is does have some imaginary roots

[peak estuary]

well polynomials are hard to solve

unless you wanna use the quartic formula you basically only have the option of the rational root form

aka "plug good candidates in"

[crimson oar]

but you don't know that it has any real roots, do you?

[peak estuary]

no

[crimson oar]

like, it does actually have some real roots, but how would I know that?

[peak estuary]

well you would try the options from the rational root theorem

that would give you the rational roots

for real roots you can try finding the minimum of the function and check if its below zero. that at least gives you the existence of real roots but doesnt say what they are

and of course for that you would have to solve a cubic still

[crimson oar]

hmm, I guess that would be a valid way of doing it. 128 has 16 factors (8 positive 8 negative) and that doesn't seem like an unreasonable thing to test

[peak estuary]

its painful but doable

[crimson oar]

that wouldn't be ideal in a case like this because spoiler alert there is a repeated root at 4

[peak estuary]

well after you obtain one root you use polynomial division

and then go again

[crimson oar]

yup, can't wait

[peak estuary]

the other roots you dont have to try twice

you only have to try 4 twice

[crimson oar]

ohh, you right

and even if 4 wasnt repeated, I would then only have to try up to 32 because it could no longer be 64 or 128

[safe radish]

@crimson oar Has your question been resolved?

[safe radish]

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[proven halo]

Is this enough to prove " gcd(m,n) = 1 iff there exists integers x, y such that mx + ny = 1 "?

[crude bear]

No

Why are you taking m= 17 and n = 12

You have to show for any m,n

[proven halo]

Because it is using Bezout's identity

*The corollary mentioned

Ok, think I see what I did wrong.

.close

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[fluid spoke]

this is what i have now

,rotate

[fluid spoke]

not sure what to do now

[west haven]

you have answer, and the limits, substitute them (upper limit - lower limit) and that's it I guess

[fluid spoke]

idk how to simplify those

[west haven]

take 1/8 outside

what do you have? ln(x)^4

so its ln(4)^4 - ln(1)^4 which is ln(4)^4

so (1/8)(ln(4)^4)

[fluid spoke]

how is this

[west haven]

look

if you have x^2 with limits 2 to 5

then its 5^2 - 2^2

makes sense?

[fluid spoke]

yep

[west haven]

so I just put 4 instead of x (once) and then 1 instead of (x) and subtracted them

[fluid spoke]

ik where the numbers come from im just more confsed about the difference of them

[west haven]

if it is 1 to 4 then its ln(4)^4 - ln(1)^4, if its 4 to 1 then ln(1)^4 - ln(4)^4

difference of them as in subtracting them or what?

[fluid spoke]

yes

[west haven]

well you can keep it as is over here

keep ln(4)^4 like that, ln1 is 0

[fluid spoke]

ohh

[west haven]

because ln 4 would give you an irrational value so no use of computing it

[fluid spoke]

in the key it says that the answer is

$2\left(\ln2\right)^{4}$

[flat frigate]

jacks

[fluid spoke]

how does $\frac{1}{8}\left(\ln4\right)^{4}$ transform into that

[flat frigate]

jacks

[west haven]

ln 4 is 2ln 2

so you get (2*ln2)^4/8

[fluid spoke]

okay i see

thank youu

.close

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[junior wagon]

Is there a way to decrypt affine ciphers without key without using brute force?

[junior wagon]

All I have is an encrypted message

[peak estuary]

a single one?

do you count frequency analysis as brute force?

[safe radish]

@junior wagon Has your question been resolved?

[junior wagon]

No, I was given 10 letters that is already encrypted.

Iirc frequency analysis requires a lot more than 10 letters..

[peak estuary]

hmm

!original

[safe radish]

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.

@junior wagon Has your question been resolved?

[junior wagon]

One sec. Let me get back to it once I get access to computer

.close

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[mighty loom]

How am I supposed to move right pi/8 but it’s in pi/2 intervals?

[mighty loom]

someone please help 😭🙏🙏

[safe radish]

@mighty loom Has your question been resolved?

[safe radish]

@mighty loom Has your question been resolved?

[safe radish]

@mighty loom Has your question been resolved?

[safe radish]

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[safe radish]

@unique basalt Has your question been resolved?

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[coral blaze]

so i have this problem where there is an upside down exclamation mark and i dont know what it means

[coral blaze]

<@&286206848099549185>

[icy lance]

never seen it before, but ill try look until someone who does know appears

[coral blaze]

got it! i did try searching it up online and they said it meant factorials but i wasnt sure

[icy lance]

the only time ive seen excalamations is to denote factorials

[coral blaze]

how accurate that was

[icy lance]

aha

[coral blaze]

yeah me too but it was usually a regular one like this !

[icy lance]

ive just never seen them upside down or subscripted

[coral blaze]

so do you think its just a weird way to write factorial?

[icy lance]

subfactorial is what im getting

sometimes written !n = nearest integer to n!/e

[coral blaze]

i dont know what that means 😰

[coral blaze]

OH I SEE

[icy lance]

ive never seen them either

[coral blaze]

we never covered them before

[icy lance]

im in the second year of my maths degree, and ive never encountered them lol

they dont seem complicated though

[coral blaze]

😭

[icy lance]

just do n!/e and round it to the nearest integer

[coral blaze]

OH I CAN DO THAT

alright thank you!

[icy lance]

np

[coral blaze]

the rest of it is just normal trig so i think i can do it now

how do i uh

close the thing

[icy lance]

'.close'

[coral blaze]

'.close'

[icy lance]

without the ' ' haha

[coral blaze]

OHHH

my bad

.close

[safe radish]

Channel closed

Closed by @coral blaze

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[coral blaze]

thank you!!

[safe radish]

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[random rivet]

,rotate 270

[random rivet]

did i do this right

answer says b is 10?

[delicate sphinx]

What is the question?

[random rivet]

#7

[delicate sphinx]

I got what you got

Not sure where b = 10 came from

[random rivet]

yea me either

okay

tysm

wait

can u help me with another question

[delicate sphinx]

Which one

[random rivet]

this one

[random rivet]

b

i applied the conjugate

but what do i do after

[delicate sphinx]

Expand and simplify

[random rivet]

ik but i got it wrong

-5+5+h

is that correct for the top

or no

[delicate sphinx]

That can be simplified

[random rivet]

is this correct or no

bec the answer is diff

😭

[delicate sphinx]

This wasn't quite it either

[random rivet]

😭

help

what did i do wrong

[delicate sphinx]

Do you know recall what (a + b)(a - b) is?

[random rivet]

yra

[delicate sphinx]

And it is?

[random rivet]

a^2

b^2

[delicate sphinx]

So you are doing this $(-\sqrt{5}+\sqrt{5+h})(-\sqrt{5}-\sqrt{5+h}))$, correct?

[random rivet]

yes

[flat frigate]

CaptainNova22

[delicate sphinx]

So a is?

[random rivet]

wdym a

oug

ouh

[delicate sphinx]

For this

[random rivet]

the - sqrt 5

[delicate sphinx]

So a^2 equals?

[random rivet]

-5

[delicate sphinx]

Not quite

[random rivet]

oh god

25

5^2

25

[delicate sphinx]

It's (-sqrt(5)) * (-sqrt(5))

[delicate sphinx]

Not quite either

[random rivet]

oh my gosh

help

whats my problem

i dont get ot

[delicate sphinx]

You know that a = -sqrt(5) and you are doing a^2

So you are doing $\left(-\sqrt{5}\right)^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

Do you understand that part?

[random rivet]

yea i think

[delicate sphinx]

And -sqrt(5) is the same as -1 * sqrt(5), do you agree?

[random rivet]

yea

[delicate sphinx]

Therefore $\left(-\sqrt{5}\right)^2 = $\left(-1 \cdot \sqrt{5}\right)^2$

[flat frigate]

CaptainNova22
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[random rivet]

yes

[delicate sphinx]

So that equals what?

[random rivet]

5

[delicate sphinx]

Correct

And not -5

Next, what is b?

[random rivet]

sqrt 5+h

[delicate sphinx]

So b^2 equals?

[random rivet]

5+h

[delicate sphinx]

Good

[random rivet]

so

5-(5+h)

?

[delicate sphinx]

Exactly

[random rivet]

okk

[delicate sphinx]

So this equals?

[random rivet]

so this?

no

smthn is wronf

[delicate sphinx]

How did you change the sign from + to - in the denominator?

[random rivet]

ya i noticed

so i sub 0 for h

1/-2 sqrt 5

answer is sqrt 5/10

?

[delicate sphinx]

Are you supposed to sub 0 for h?

[random rivet]

yea

this is lim

h->0

?

waiy

ono

nvm

no

omg

im mixing things up

my answer is diff

than the solution

in solution they made num 1

[rich flame]

Whats the question again

[delicate sphinx]

Not sure what you mean

[random rivet]

thats how my teacher did it

?

[delicate sphinx]

Factor out a negative from the denominator

[delicate sphinx]

Because you have -a - b therefore -(a + b)

[random rivet]

like this?

[delicate sphinx]

Not quite

[random rivet]

oh em gee

this is my end

[delicate sphinx]

$(-\sqrt{5}-\sqrt{5+h}))$

[flat frigate]

CaptainNova22

[delicate sphinx]

You have that

[random rivet]

yes

[delicate sphinx]

Both terms have a -1 in common, correct?

[random rivet]

tea

yea

[delicate sphinx]

So you can factor it out

Factor by gcf

[random rivet]

oky

[delicate sphinx]

If you factor out -1, what do you result?

[random rivet]

so sqrt 5 is part of gcf??

[delicate sphinx]

No

[random rivet]

how did she remove

sqrt over 5

[delicate sphinx]

Looks like a error in writing

[delicate sphinx]

Notice how the step when expanding the sqrt disappears

[random rivet]

this?

[delicate sphinx]

Why is there two negatives in the denominator?

[random rivet]

u said take out

-1

-1 x -

x 1

[delicate sphinx]

You start with this

[random rivet]

yea

i did

i took negtaive 1 out

[delicate sphinx]

Where is the second negative from

You only factored out one -1

For example, (-x - 6) = -(x + 6)

[random rivet]

oh shoot

img

whats wrong with me

mb

😭😭😭

bare with me

okay this

[delicate sphinx]

Yes like that

[random rivet]

okay

omg

why was this so hard what

😭😭😭

[safe radish]

@random rivet Has your question been resolved?

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[fervent sparrow]

How would I factor this?

(Question also asks to state restrictions where necessary, is it?)

[safe radish]

@fervent sparrow Has your question been resolved?

[left gyro]

,,y^2-49+14x-x^2\
y^2-x^2+14x-49\
y^2-(x^2-14x+49)

[flat frigate]

mtt07734

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[tawny knoll]

Hello. I remember once watching a video solving an integral which in order to solve it u had to replace 1 by cos^2(x)+sin^2(x) twice in the process, but i cant find the video/remember the integral. Could you pls share any ideas?

[plucky elk]

Don't use help channels for non math problems

You're just looking for a video with a very vague description

[tawny knoll]

i dont think it is vague.

[plucky elk]

Yes it absolutely is

[tawny knoll]

oc

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[simple oak]

Say I am playing a memory match game. I have 6 attempts where each attempt I can flip two cards(if not a pair, they will be unflipped). In total there are 16 cards (8 pairs). What are the chances I will be able to match three pairs

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[blazing igloo]

when it says A*B = ...

[blazing igloo]

i dont get how to find the resulting supremeum when A and B are interval sets

[devout shale]

it'd be the least upper bound of a*b where a is in A and b is in B

would it not

[blazing igloo]

isnt the product of 2 interval sets like a rectangle

[devout shale]

That's the cartesian product

It describes above A*B={xy, x in A, b in B}

so that's all you need to know about it, it gives the definition right theree

[blazing igloo]

oh alright

[soft lava]

Idk how this is even possible I’m so glad frustrated

[devout shale]

!occupied

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

[blazing igloo]

@devout shale

is this right

also this is the first one

<@&286206848099549185>

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@blazing igloo Has your question been resolved?

[blazing igloo]

helpers

<@&286206848099549185>

<@&286206848099549185>

.close

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[split totem]

how do i do this question

[hasty wagon]

What have you tried?

[split totem]

Im not sure

i think im supposed to use pythag but im idk

[hasty wagon]

yes, you may use pythag

lemme draw something to help

[split totem]

i thought c/a was like the diagonal line on the right

[hasty wagon]

nah, it's the marking for the point on x-axis

[split totem]

ohh i see

what should I do with this information

[hasty wagon]

try using pythag to find the length of the hypotenuse
and then using
SOHCAHTOA to solve for cosine and sine of theta

[split totem]

would the y axis be the hypotenuse in this case

[hasty wagon]

nah

[hasty wagon]

right angle in red

[split totem]

would it be (c/b)^2 + (c/a)^2 = hypotenuse ^2?

[hasty wagon]

yes 🙂

[split totem]

so would the hypotenuse be

[hasty wagon]

correct

since we tooke all a,b,c are positive 😄

SOHCAHTOA to solve for cosine and sine of theta
now this

[split totem]

costheta is c/b / the hypotenuse right

ok i got that part

[hasty wagon]

correct!

so it's part b now?

[split totem]

yeah

[split totem]

how should i attempt this?

[hasty wagon]

Hint: we have found sin(theta) and cos(theta)

[split totem]

im not too sure what to do with the sin(theta) and cos(theta)

do i need pythagorean identities?

[hasty wagon]

nah, we need to use

SOHCAHTOA to solve for cosine and sine of theta

again, but this time we act on the green triangle

[split totem]

How do i find a^2 and b^2 in this case if c/b is the hypotenuse

[hasty wagon]

Second hint: we use sin(theta) in this case, and try SOH
sin(theta) = p.distance/ (c/b)

[split totem]

would sin(theta) still be ?

[hasty wagon]

correct, since the theta is the same theta

[split totem]

Ohhh i got the answer now

thank you

[hasty wagon]

(c) should be easy 🙂

[split totem]

what would be the points im supposed to use?

[hasty wagon]

they are of the same form, for ii, you'll have to draw it out the figure how's negative affecting the calculations

[split totem]

im not exactly sure how to do that

[hasty wagon]

as you can see in (i)
a=3, b=4, c=5

[hasty wagon]

so you can just write the whole thing down as the
perpendicular form is indicated in part (b)

[split totem]

oh so i jsut sub in the values into it

[hasty wagon]

yep

[split totem]

when i sub in for ii am i supposed to be getting root 13/13

[hasty wagon]

it should be, but be careful with the negative signs

[split totem]

ok thank you very much for all the help youve given

[hasty wagon]

Cheers!

[split totem]

.close

[safe radish]

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[safe radish]

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[somber cape]

Am I allowed to solve this in this way

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[flat frigate]

Leno

[hearty egret]

what is S alpha?

[flat frigate]

Leno

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@real swallow Has your question been resolved?

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@real swallow Has your question been resolved?

[flat frigate]

Leno

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@real swallow Has your question been resolved?

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[misty blade]

Help

[misty blade]

Say f:R->R be a continuous function with property such that f(x) is rational iff f(x+1) is rational. How many such functions are there?

The answer says 0, but doesnt any linear function satisfy this?

Say f(x)=x, so x is rational iff x+1 is rational. That's clearly true

[hearty egret]

even f(x)=1 satisfies this property