help-39

[pearl pond]

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[prime bane]

Hi guys, I'm a guy who goes to high school and does computer science but since I like mathematics, I would like to deepen and carry on as a self-taught and I would like to ask you for a list of perfect books from algebra up to calculus and if you can also add something related to computer science such as algorithms, thank you

[brisk pike]

search #book-recommendations

For calc stewarts calculus is a good book imo

u also need to ask, what am I learning this for?

Do you want a maths degree? Do you need to know the maths for comp sci? Is this for a competitive exam? etc. etc.

[prime bane]

@brisk pike Next year I would like to participate in the computer olympics and I would like to know how to solve logical mathematical problems and then I would like to enrol in computer engineering

[brisk pike]

Computer Engineering..Are you by any chance in India? I.e. ever heard of JEE?

[prime bane]

I’m Italian

[brisk pike]

I see

havent heard of Italian computer olympics, so no ideas abt that until u give further info

as for enrolling, again depends on what exam u will be giving

and logical mathematical problems is a very broad term

u gotta narrow it down a bit more

[prime bane]

Do you know codeforces?

[brisk pike]

no not really

[prime bane]

There are many problems that I would like to solve

[brisk pike]

Hmm

Googling suggests that its similar to codewars

correct?

well if its competitive programming then

err 1 sec wait

https://cses.fi/book/book.pdf

https://www.hackerearth.com/getstarted-competitive-programming/

and youtube

that should be enough to get u started

as for the maths required, should be basic algebra, combinatrics, and number theory + some basic linear algebra as well

@prime bane do u understand?

[brisk pike]

any examples?

[prime bane]

Yes

[brisk pike]

alr 👍 if u think ur done do .close

[brisk pike]

alr lemme see

where is shel tho? in each test case?

[brisk pike]

oh got it

we get to choose the position correct?

[prime bane]

I don’t know

[brisk pike]

bruh

[prime bane]

I took the first exercise I saw and sent it to you

[brisk pike]

Well think hard about it.

Analyse the problem

what is it asking you?

Dont think of the solution yet

this is just basic comprehension

everything is explained, even stuff like manhattan distance

if u can come up with even a half-broken explanation of what the questions asking you, in your own words then ur probably good
if no then u need to work on much simpler problems

@prime bane bro u there? atleast give me a yes or no

[prime bane]

Yes

[brisk pike]

alr

[prime bane]

.close

[pearl pond]

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[wanton steeple]

Hi this is my proof can someone give me tips on how I can make it more formal or whatnot I am a proof noob

[pearl pond]

@wanton steeple Has your question been resolved?

[wanton steeple]

<@&286206848099549185>

[pearl pond]

@wanton steeple Has your question been resolved?

[wanton steeple]

<@&286206848099549185> hom les

[wanton steeple]

it may help if i gave the question

[west sapphire]

for (a) you didn't say what you're using for Q

[wanton steeple]

wdym

i said it at the top

[west sapphire]

where?

you can't just use some random invertible Q and expect that A = QAQ^-1

[wanton steeple]

thats just what it means to be similar

a is similar to a

so A=Q^-1AQ

[west sapphire]

that's the claim you're trying to show

that there exists an invertible Q such that A = Q^-1 A Q

[wanton steeple]

am i?

[west sapphire]

so to prove that, you have to show a Q that works

"every matrix A is similar to itself"

is the same as
"there exists an invertible Q such that A = Q^-1 A Q"

to prove that, you need to exhibit such a Q

[wanton steeple]

so how do i go about that

[west sapphire]

there's a very simple Q that works

can you think of one?

[wanton steeple]

identity matrix

[west sapphire]

yep

Q = I does the job

[wanton steeple]

what about the other ones?

symmetry and transitiviy

are they okay

[west sapphire]

yeah those both look fine

[wanton steeple]

okay

.close

[pearl pond]

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[sly python]

guys how do i find the denominator

[grim fractal]

how many total audience members are there

[quiet goblet]

It's written on last right corner of table

[pearl pond]

@sly python Has your question been resolved?

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[frosty lintel]

does anybody have a trick or method for getting the largest square factor

[runic dawn]

If b is an even integer in the equation ax²+bx+c=0, you can use the reduced form. Do you know it?

[frosty lintel]

no

[runic dawn]

If you have an equation written like: $ax^2+2 \beta x+c=0$, define $\Delta/4 = \beta^2-a c$ and the solutions are:
\begin{equation}
x_{1,2}=\frac{-\beta \pm \sqrt{\Delta/4}}{a}
\end{equation}.

This way, you don't spend too much time because, usually, the term in the square root is a smaller number.

Can you give me the equation and I'll show you what to do?

[jolly parrot]

cristorenzo99

[neat harbor]

um

[runic dawn]

Ok, do you set $4-3x \neq 0$ first?

[jolly parrot]

cristorenzo99

[frosty lintel]

yes

[neat harbor]

\frac{1}{2}

how does this work

[runic dawn]

Ok, nice! My students always forget that!

[neat harbor]

$\frac{1}{2}$

got it

[runic dawn]

Did you get 4x^2+4x-25=0, right?

[frosty lintel]

yes

[runic dawn]

Ok, here $a=4$, $\beta = 4/2=2$ and $c=-25$. So:
$\Delta/4 = 2^2-(4)(-25)=4+100=104$ which is a quarter of 416 (hence the name $\Delta/4$!)
Now:
\begin{equation}
\begin{split}
x &=\frac{-\beta \pm \sqrt{\Delta/4}}{a}=\frac{-2\pm\sqrt{104}}{4}=\
&=\frac{-2\pm \sqrt{2^3\cdot 13}}{4}=\frac{-2 \pm 2 \sqrt{2 \cdot 13}}{4}=\frac{-1 \pm \sqrt{26}}{2}
\end{split}
\end{equation}

[jolly parrot]

cristorenzo99

[runic dawn]

Is it clear?

[frosty lintel]

is there a formula for getting delta4

[neat harbor]

i think this is correct

[runic dawn]

See here

,w (x^2-7x-5)/(4-3x)=x+5

[frosty lintel]

(beta)^2 - (a)(c) ?

[runic dawn]

Yes

[neat harbor]

I am not so sure about line four though

so you might have to recheck that

[jolly parrot]

Results provided by WolframAlpha

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

Requested by cristorenzo99

[frosty lintel]

what method is there to further reducing sqrt/104

[runic dawn]

I factorized 104.

Do you want a proof of the reduced formula?

[frosty lintel]

yes please

[neat harbor]

what did I do wrong then

[runic dawn]

Ok. It will make some time to write it in LaTeX

[runic dawn]

There is a wrong sign in 3x^2, 20 and 15 x

[neat harbor]

which one is wrong

[runic dawn]

All three

[neat harbor]

because I removed the brackets?

how about 4x though

how is that correct

[runic dawn]

You know the quadratic formula:
\begin{equation}
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\end{equation}

Here we have $b=2*\beta$, so we have:

\begin{equation}
x=\frac{-2\beta\pm \sqrt{(2\beta)^2-4ac}}{2a}=\frac{-2\beta\pm \sqrt{4\beta^2-4ac}}{2a}
\end{equation}

Now we collect $4$ into the square root:

\begin{equation}
x=\frac{-2\beta\pm \sqrt{4(\beta^2-ac)}}{2a}=\frac{-2\beta \pm 2\cdot  \sqrt{\beta^2-ac}}{2a}

\end{equation}
Then simplify the common $2$ at the numerator and at the denominator:
\begin{equation}
x=\frac{-2\beta \pm 2\cdot \sqrt{\beta^2-ac}}{2a}=\frac{-\beta \pm \sqrt{\beta^2-ac}}{a}
\end{equation}
And you get the formula.

[jolly parrot]

cristorenzo99

[runic dawn]

Because -(x+5)(4-3x)=-(4x-3x^2+20-15x)=-4x+3x^2-20+15x = 3x^2+11x-20

[neat harbor]

Because $-(x+5)(4-3x)=-(4x-3x^2+20-15x)=-4x+3x^2-20+15x = 3x^2+11x-20$

[jolly parrot]

DerTheo

[neat harbor]

just so I can read it more easily

[runic dawn]

You can also sum 4x and -15x into the parenthesis

[runic dawn]

Is that clear? @frosty lintel

[neat harbor]

I think I see my mistake

thank you

[runic dawn]

You are welcome!

[pearl pond]

@frosty lintel Has your question been resolved?

[pearl pond]

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[karmic shuttle]

hey! i need help on this problem — i need it to be explained in simpler terms and hopefully steps to help me solve this!

[runic dawn]

Ok. What is f(1)?

[karmic shuttle]

i think its 0? like cos1 is 0 then 0-0^3

is well. 0

[runic dawn]

The cosine function works with x as radiants

So cos(1)=cos(1 rad), which is something between 0 and 1, but definitely positive because 1 is between 0 and pi/2 and the cosine there is positive

[karmic shuttle]

OH

its like around 0.54

-1 which would be negativd

0.54-1^3 is what i mwan

i think

[runic dawn]

Yes, so f(1) is negative.

[karmic shuttle]

okok i understand that now thank you! but i dont really understand part b

[runic dawn]

Uhm, what you can use? Like, bisector method, Newton's method or something like that is allowed?

[karmic shuttle]

i havent learned newtons method yet, and bisectors theorem wasnt introduced in this class :( but i dont think it like wouldnt be allowed

i more so don’t understand why i have to input it as interval notation

[runic dawn]

Ok, maybe you can use a graphic calculator like geogebra to zoom in and see where is approximately the solution

[karmic shuttle]

im just not too sure how to place it in interval notation

[runic dawn]

Uhm, I don't know how these calculators work. But I can teach you how you could do using the Intermediate Value Theorem (like spamming it lol). This is known as the bisector method I believe

[karmic shuttle]

i just plugged 0.01 as x in the equation to get the exact solution HAHAHA

and thats like 0.9999949 smth

[runic dawn]

cos(0.01) is that number

[runic dawn]

Suppose that x=0.5. If you put that value into your calculator, it would give you something like 0.378, which is positive, so you can say that, using the IVT, there is at least a root between 0.5 and 1. So you restricted the interval from [0,1] to [0.5,1].
So you pick a number in [0.5,1], like 0.8, and you get f(0.8)=-0.103, which is negative, so, for the same reason as before, there will be a solution in [0.5,0.8]

And you repeat the reasoning until you get the desired accuracy

[karmic shuttle]

OH I SEE is this intermediate value theorem right

[runic dawn]

So you continue, you pick x=0.6 and you get f(0.6)=0.225336>0, so the solution is between 0.6 and 0.8

[karmic shuttle]

so the two numbers in the interval notation r gonna equal 0?

[runic dawn]

No, they are some numbers like 0.04 and 0.05 for example. The exercise is asking you to find the solution with 2 correct digits I believe

The reasoning is the following: suppose you know that a solution is in the interval [a,b] and that f(a)>0 and f(b)<0. Compute f(c), with c a number in the interval [a,b]. If f(c)<0, then the solution is in [a,c] (because, for the IVT, f(a)>0, f(c)<0 and f is continous, so there must exist a root between a and c). If f(c)>0, then the solution is in [c,b] (f(c)>0, f(b)<0 and f is continous, so there must be a root beween c and b)

Would you like to do this "algorithm" together? We know that the solution is in [0.5,0.8]. Pick a number here and compute f of this number!

[karmic shuttle]

i think doing it together would def help more than an explanation!! but how did u get [0.5,0.8]?

[runic dawn]

Ok, I'll explain it again. We know that a solution is in [0,1] because f(0)>0 and f(1)<0 for the IVT. If you pick x=0.5, f(0.5)=0.3778>0. So what we can say using the IVT?

[karmic shuttle]

wait haha im so sorry but wouldnt we use 0.1 in this instance? Since it’s what is in the exercise

[runic dawn]

We can try with 0.1 too. What is f(0.1)? (even if with 0.5 it would be faster, but I want you to understand the procedure)

[karmic shuttle]

thank you!1 and with f(0.1) is would be 0.994 but i did try with f(0.5) and i got 0.7526

[runic dawn]

Ok. f(0.1)=0.994, which is positive. What can we say using the IVP as before?

[karmic shuttle]

0<f(0.1)?

[runic dawn]

The statement of the IVP theorem is the following:
Let $f:[a,b] \to \mathbb{R}$ a continous function, with $f(a)\cdot f(b)<0$ (which means that $f(a)<0$ and $f(b)>0$ or $f(a)>0$ and $f(b)<0$).
Then exists a point $c \in [a,b]$ such that $f(c)=0$.

[jolly parrot]

cristorenzo99

cristorenzo99

[runic dawn]

Because if you understand this, you will not have problem with the procedure

[karmic shuttle]

well i do understand that

that sounded like it was mad OOPS!1 sorry haha

[runic dawn]

Ok. f(0.1)=0.99...>0 (and f(1)<0), so what we can say now?
HINT: a=0.1, b=1

[karmic shuttle]

that 0 crosses between these two? im a bit confused
i understand generally that between these two will be where 0 is?

or is there smth im missijg

[runic dawn]

That the point c that will give you f(c)=0 is between 0.1 and 1

[karmic shuttle]

OH I SEE

[runic dawn]

Ok. Now we try, for example, x=0.5. What is f(0.5)?

[karmic shuttle]

0.7526 i think

do i keep doijg it until like i reacj f(c)=0.1?

0.01*

i meant HAHA

[runic dawn]

I have cos(0.5)-0.5 = 0.3778

[karmic shuttle]

ah its meant to be 0.5^3

cos(0.5)-0.5^3

[runic dawn]

You are right, sorry

[karmic shuttle]

nono, thank yoi!! you really helped me break down this concept

[runic dawn]

You keep doing this until you have an interval whose length is 0.01 or less (like [0.67, 0.68])

[karmic shuttle]

thank you so much for ur help!

[karmic shuttle]

AH OKAYOKAY

i can do this

[runic dawn]

So f(0.5)>0, f(1)<0, what can we say about the root?

[karmic shuttle]

that irs between these two?

[runic dawn]

That the root is between 0.5 and 1!

Pick a number between 0.5 and 1 and let's try again!

[karmic shuttle]

thank god for calculators
but would it be these two? since the interval is 0,01 and between em is 0

[runic dawn]

Great work! The solution is 0.865474...

So you know your solution is something like 0.86 and something

[karmic shuttle]

AWESOME!! and i got it write on the exercise too [0.86,0.87]

THANK YOU SO MUCH FOR YOUR HELP!!

[runic dawn]

Because all the numbers between 0.86 and 0.87 are like 0.86....

[runic dawn]

You are welcome!

[karmic shuttle]

.close

[pearl pond]

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[glass meadow]

At most 3 beers and 0 ciders, that means the number of beers and sodas can be (3, 2), (2, 3), (1, 4), or (0, 5)

In the (3, 2) case, you can choose which 3 out of the 5 drinks are beers, and then choose which specific drink is in each spot

Not sure you can do better than just considering each or the four cases, apart from considering the general case of just 0 ciders and subtracting the cases of 5 beers and 4 beers

[waxen sequoia]

(3,2) where 3 represents the beers and 2 represents the sodas right?

[glass meadow]

Yes

[waxen sequoia]

Im not sure how to start tho

[glass meadow]

I laid out the (3, 2) case for you, just do that one and then do a similar thing for the other three cases, and sum it all up

[waxen sequoia]

i feel like this might be better

[glass meadow]

Whichever works for you

[waxen sequoia]

ok, say i start with the (3,2) then the first spot would be 5 x 3 right

you have 5 beer drinks options and 3 spots

[glass meadow]

I don't know what you mean

[waxen sequoia]

second spot would be 5 x 2

[glass meadow]

5 options in each spot and 3 spots doesn't result in 5 * 3 possibilities

[waxen sequoia]

do we add?

[glass meadow]

...no

Have you done (a) through (e) ??

[waxen sequoia]

yes

[glass meadow]

What's your answer for (a)?

Right, so why would you switch to multiplication or addition?

In (a) you have 12 options in each spot, and 5 spots

[waxen sequoia]

5^3??

[glass meadow]

Yeah

[waxen sequoia]

and for the other 2 spots

3^2

[glass meadow]

Yes

[waxen sequoia]

ok so 5 x 5 x 5 x 3 x 3

[glass meadow]

You still need to order them

[waxen sequoia]

wdym by that

[glass meadow]

"the order of the drinks matters"

[waxen sequoia]

right but how do i order them

[glass meadow]

How did you do (d)?

Fair enough, how about (c)?

Oh okay so 5 spots for 5 options for beer, 4 spots for 4 options for cider, and the rest is soda?

[waxen sequoia]

thats right

[glass meadow]

Well you can do something similar for 3 beers and 2 sodas

But in general I think you should learn how to use binomial coefficients, aka "n choose k"

[waxen sequoia]

5^3 x 3^2

[waxen sequoia]

I just got introduced to that yesterday

[glass meadow]

To what? Combinatorics or binomial coefficients?

[waxen sequoia]

binomial coefficients

[glass meadow]

Right so use them

[waxen sequoia]

but what i have rn is also good no?

[glass meadow]

No, because the drinks aren't ordered

[waxen sequoia]

for part c?

[glass meadow]

No for f

[waxen sequoia]

ok this is what i have right now: 5^3 x 3^2

how do i make this ordered

[glass meadow]

You choose 3 spots from the 5 total to place the beers in, and the remaining 2 spots are soda

[waxen sequoia]

right

[glass meadow]

That's 5 choose 3

[waxen sequoia]

ok, im a bit new to this so how do we know when to use binomial coefficients?

[glass meadow]

When you have n spots to place things, and k of the things are special, then you can "choose" k of the n spots to place the special things

Here you consider beers as special, and the rest are sodas

You have 5 spots, and 3 beers

So 5 choose 3

[waxen sequoia]

ok, im getting 10

[glass meadow]

You can do it the other way around too, consider sodas as special, the rest being beers

That gives you 5 choose 2 which is the same thing

[waxen sequoia]

right thats also 10

[glass meadow]

Yeah, so multiply that with what you had unordered, and it gives you the ordered arrangements

[waxen sequoia]

hold up

i had 5^3 x 3^2 where 5 choose 3 is 10

dont i now do 3 choose 2?

[glass meadow]

Why?

[waxen sequoia]

well why are we doing 5 choose 3 and not 3 choose 2?

[glass meadow]

Because you want to place 3 beers into 5 spots

[waxen sequoia]

what about soda? do we not place them for the 2 spots?

[glass meadow]

Once you've chosen where to put the beers, you don't have a choice as to where to put the sodas

It's the 2 remaining spots

[waxen sequoia]

oh

so what do i multiple 5 choose 3 with

[glass meadow]

5^3 * 3^2

Imagine you have three B's and two S's

You first need to order them to form a sequence of 5 letters

BBBSS is one, SBSBB is another

There are 10 different such sequences

Then, for each letter B, you have to choose between 5 options

And for each letter S, you have to choose between 3 options

So that gives you 10 * 5 * 5 * 5 * 3 * 3

That's the (3, 2) case, you still need to do the cases (2, 3), (1, 4), and (0, 5)

[waxen sequoia]

ok that makes sense

do i also multiply each case with the other?

[glass meadow]

No the four cases are like separate problems, you just add them up at the end

[waxen sequoia]

ahh i see

is it ok if i come back in a bit to show my final result?

[glass meadow]

Sure

[waxen sequoia]

id have to also close this chat so is it fine if i ping here once im finished?

[glass meadow]

I might not be there later, so just open a new channel and wait for someone

[waxen sequoia]

sounds good, thanks.

[glass meadow]

[waxen sequoia]

.close

[pearl pond]

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[nocturne plover]

you can remove the (d-e)f one too

but then you've shown it haven't you?

you've shown that when you grab two elements of V and add them together

it won't be an element of V

not always anyway

is there a condition on those numbers more than just (a - b)c = 0 and (d - e)f = 0?

wait

holdup

when is (a-b)c 0?

whenever you are adding two elements of V where one is $\begin{pmatrix} a \ b \ 0 \end{pmatrix}$ and the other $\begin{pmatrix} d \ d \ f \end{pmatrix}$

[jolly parrot]

Katharine

[nocturne plover]

does that result in the sum being in V?

both of them are in V am i correct?

exactly

which means

not closed

both of those are in V right?

and is the sum in V?

there you go :)

[pearl pond]

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[brave jolt]

64.99 is 22% of what number

[light helm]

start with introducing a variable to represent the number you wish to find,
then set up an equation the representsthat statement

[brave jolt]

so like "n"?

[light helm]

yes, you could use n as a variable

[brave jolt]

so would it be 64.99n

[light helm]

no

[brave jolt]

oh

22% = 64.99 x n?

[light helm]

no

[brave jolt]

oh

[light helm]

is is analogous to =
of is analogous to * (multiplication)

[runic shale]

i think there's a simpler way of doing it: 64.99/22 × 100

that'd work, right?

[brave jolt]

i can try

[light helm]

well one should first understand why you'd want to multiply/divide like that

[runic shale]

true

[light helm]

which comes from the first step i'm trying to get them to do

[brave jolt]

nvm i give up sorry for wasting your guys time

[runic shale]

your question is valid. i can try to explain it to you

[brave jolt]

its alr

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[frozen wadi]

do yall know what the 4 means

[frozen wadi]

is it like the cycle

but like it doesnt have pi

so idk

[subtle ginkgo]

remind the general solution of the trig equation of the form: cot( x ) = 1

[frozen wadi]

yes

[subtle ginkgo]

that is cool, then your solution is immediate

[frozen wadi]

huh

i htought the 4 jus meant how many cycles

[subtle ginkgo]

[0, 4) is just a given interval, do not worry about it, first write the general form of the solution here, next you think ab the given interval

[frozen wadi]

ohhh

cuz i got 3pi/4 and 5pi/12

[subtle ginkgo]

so write on a paper, and send readable photo here, to see it

[frozen wadi]

^ and it says i need 4 solutions

idk what to do after

[subtle ginkgo]

that is main solution for cotangent:

$\cot\text{}x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi,k\in \mathbb{Z}$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

and now, if you like, to cope with cot ( 3x) = 1, then replace x with 3x , in a form i gave you now

means:

$\cot\text{}3x=1\Leftrightarrow 3x=\frac{\pi}{4}+k\pi,k\in \mathbb{Z}$

[jolly parrot]

Joanna Angel

[frozen wadi]

wait so 3pi/4 and 5pi/12

[subtle ginkgo]

look at what i wrote

then divide both sides of the last equation by 3

you should receive:

$\cot\text{}3x=1\Leftrightarrow 3x=\frac{\pi}{4}+k\pi,k\in \mathbb{Z}\\x=\frac{\pi}{12}+\frac{k\pi}{3},k\in \mathbb{Z}$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

and now , start to replace integer k, in a such way, to make your particular solutions be in interval [ 0, 4 )

[frozen wadi]

ohhhh

so k=4?

[subtle ginkgo]

k=0, 1, 2, 3

[glacial coral]

You can multiply the interval by 3 so it becomes 0 =< 3x =< 12

[subtle ginkgo]

then for every integer, k=0,1,2,3 you receive, 4 particular solutions in your given interval

[pearl pond]

@frozen wadi Has your question been resolved?

[pearl pond]

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[jolly parrot]

Mycobacterium

[toxic lichen]

well first off eigenspaces correspond to an eigenvalue of a matrix. not to a matrix itself

and second, no, [4,1;0,1] is not the only matrix with that space as one of its eigenspaces

yes of course

[pearl pond]

@midnight haven Has your question been resolved?

[toxic lichen]

i have a feeling you might misunderstand what the word "unique" means and even then the question sounds strange...

[jolly parrot]

Mycobacterium

[pearl pond]

@midnight haven Has your question been resolved?

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[sharp smelt]

what's this theorm called? I'm trying to find a proof for it

[limber oasis]

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

[sharp smelt]

thanks

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[teal flare]

Hiya I don’t understand why the answer is worded like this

[teal flare]

I got the right answer (as in it can be rearranged to the answer in the book) but I’m not sure why they have y-5 at the front

[nimble lily]

it's just slightly nicer than (-2/5)(x-7)+5 I guess

[pearl pond]

@teal flare Has your question been resolved?

[teal flare]

It just doesn’t seem right since they want the answer as an equation

[nimble lily]

it is an equation

sometimes you'll see the equation of a line written as ax + by = 0 rather than y = f(x)

[teal flare]

Ooh that’s true

I’m just confused bc none of the other answers are in that format

[pearl pond]

@teal flare Has your question been resolved?

[pearl pond]

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[vocal dove]

hello, could someone help me with this

[vocal dove]

2^x - 2^y = 960 where x and y are integers

i have to compare whether x-y or 5 is bigger

i've done it one way but it didn't feel like i was applying actual math

[vital estuary]

factor out 2^y and then factor out the largest power of two of 960

[pearl pond]

@vocal dove Has your question been resolved?

[vocal dove]

okay thanks idk why i thought of that but dismissed it entirely

[pearl pond]

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[rough garden]

For the integers a and b, b ≡ a (mod 91) and gcd(a, 91) = 1.

(a) Show that gcd(b, 91) = 1.
(b) Determine a positive number k > 1 such that b^k ≡ a (mod 91).
(c) Determine a^k mod 91 if b = 153.

a) b ≡ a (mod 91) ⇒ 91|b-a expressed b-a = 91k
If we have a number d such that d|b, d|91 ⇒ d|b-a
And now since d|b, d|b-a ⇒ d|a
From this we know d|91, d|a which means that d = gcd(91,a) = 1
Finally, this means that gcd(91,b) = 1

b) Euler's theorem reads mϕ(n) ≡ 1 (mod n) which means that bϕ(91) ≡ 1 (mod 91)
ϕ(91) = ϕ(7 x 13) = (7-1)(13-1) = 7
2 ← theorem
We now know that b^72 ≡ 1 (mod 91)
We can now multiply b^72 ≡ 1 (mod 91) by a and get
a x b^72 ≡ a (mod 91) ⇔ b^73 ≡ a (mod 91) ← b ≡ a mod (mod 91)
Answer: k = 73

c) b ≡ a (mod 91) ⇔ a ≡ 153 (mod 91) ⇔ a^73 ≡ 153 mod (mod 91) ⇔
a^73^72 ≡ 15372 (mod 91) ⇔ a^72^73 ≡ 1 (mod 91) ⇔ 1^73 ≡ 1 (mod 91)

I am at c), does this mean that a^73 ≡ 1 (mod 91) ?

[rough garden]

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[vast berry]

consider the system E of linear equations
x+y+2z=1
-3x-2y+kz=b
x+(k+1)y+9z=0
(a) find the range of values of k such that E has a unique solution

[vast berry]

for (a) i got k =/= -1/3 and k=/= -27 but idk if thats correct

I dont know how to get the matrix into RREF

[pearl pond]

@vast berry Has your question been resolved?

[pearl pond]

@vast berry Has your question been resolved?

[last moth]

Your steps are all correct so far

[last moth]

But why did you decide to set 1+3k and k+27 both to zero?

[vast berry]

I dont know

Idk what i was supposed to do

[last moth]

Okay, so E has a unique solution only if when you row reduce, you get three pivots, right?

[vast berry]

Sorry what is three pivots 😭

[last moth]

Yeah, so when you row reduce a matrix, you end up getting something that looks like this right

These highlighted things in yellow are called "pivots"

[vast berry]

oh ye

[last moth]

So we already found the first pivot

The 1 on the upper left of this matrix here

But we need to keep going with row reduction

How can we clear away the "1+3k" from the third row of the matrix?

[vast berry]

L3 - 3L1 -> L3?

[last moth]

Not quite, that would just make a -3 in the lower left of the matrix

[vast berry]

aaa

hmm

multiplication doesnt work yeah?

[last moth]

Yeah you couldn't just multiply the third row by something to make the 1+3k disappear.

[vast berry]

im not sure how to do that..

[last moth]

Okay maybe it'd be easier if you first multiplied the second row by something to make the leading term 1

[vast berry]

what is "leading term" here?

sorry i haven't learnt these terminologies yet

[last moth]

The leading entry is the first non-zero entry in the row

So that's the -k

[vast berry]

is L2+L1 -> L2 even allowed?

wait

is it that

but then

that'll make first term of L2 non zero

[last moth]

Yeah that won't help.

I mean it's allowed, but it won't help because of what you said

[vast berry]

What should i do then

[last moth]

Try multiplying row 2 by something

[vast berry]

-1/k?

[last moth]

yea

[vast berry]

What do i do next?

[last moth]

Try to get rid of the 1+3k on the third row

[vast berry]

do i just brute force multiply third row by 1/(1+3k)

[last moth]

Try adding a multiple of the second row to the third row

[vast berry]

to 0?

[last moth]

Yes

Remember we want to get it into row echelon form

To figure out if there's a unique solution

[vast berry]

I dont know how to do that sorry

[last moth]

How to do what

[vast berry]

this

[last moth]

Going from the third to fourth line here, how did you get rid of the -3 in the third row?

[vast berry]

3L3+L1

[last moth]

Not quite

[vast berry]

oh i mean

L3+3L1

[last moth]

yeah, can you try doing a similar thing to get rid of the 1+3k in the third row now?

[vast berry]

L3-3L1 -> L3?

[last moth]

No, that would create a -3 in the first entry of the row

Try using L2 instead of L1

[vast berry]

i have no idea how do i do this

[last moth]

Well, you want to get rid of the 1+3k

So probably you should multiply L2 by something to turn it into -1-3k

And then add that to L3

[vast berry]

multiply L2 by -1-3k directly..?

[last moth]

Yes

[vast berry]

If row 3 last term = right side row 3 term then no sol?

[last moth]

Yes, if row 3 is zero except for the last term then there's no solution

[vast berry]

oh wow

thanks

i think i got it now

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[last moth]

Be careful @vast berry, there's something that's still an issue

Remember we multiplied by -1/k

[vast berry]

.reopen

[pearl pond]

✅

[last moth]

That's not possible if k is what?

[vast berry]

0

[last moth]

Yeah

So if k=0, you will have to make a separate case to check if there is a unique solution

But if k≠0, what we just did works

[vast berry]

ah i see

[last moth]

Good luck!

[vast berry]

i think my steps are too long lol, ill try to redo everything

tysm

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[vale turtle]

How to prove that (n^n+n)/(n^2+1) is a bounded sequence?

[magic wind]

bounded above or below?

[vale turtle]

Above

It is bounded below.

Both@magic wind

[magic wind]

i dont think its bounded above

are you trying to prove smthg past its bounded property or just thats its bounded

[vale turtle]

Opps sorry

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[rancid crow]

is f(x) -> y here injective?

[limpid glen]

nope

[rancid crow]

I know that y can't be mapped twice

[limpid glen]

yeah

[rancid crow]

but can a single x value go to two unique ys?

[limpid glen]

if it is mapped twice then it many to one function

[limpid glen]

yeah

many to one funciton

You preparing for JEE?

[rancid crow]

If this is not injective, then what is the difference between bijective and injective?

[rancid crow]

no, are u?

[limpid glen]

Bijective is both injective and onto

[limpid glen]

yeah

[rancid crow]

what would be an example of a mapping that is injective but not bijective

[limpid glen]

If a functions is both injective and onto and the same time

then it is bijective

[limpid glen]

when both codomain and range are equal

[magic wind]

(onto = surjective)

[limpid glen]

you do know what they are right

[limpid glen]

yeah surjective function

[rancid crow]

yh

[terse cedar]

this is not even a function so it's not injective

[rancid crow]

ah ok

[rancid crow]

I think I found an example for this

[limpid glen]

yeah

[rancid crow]

f(x) = 2x

is this valid?

injective but not bijective

so not surjective

[limpid glen]

yeah

that's it?

[rancid crow]

ah, I understand this a lot better now

yh pretty much

thanks a lot

[limpid glen]

k

[magic wind]

god am i stupid y is 2x not surjective

[terse cedar]

it is surjective !

[rancid crow]

uhh

I think I meant if X is all integers maybe, then f(x) = 2x is not surjective?

[magic wind]

yeah that would now be correct i think

[rancid crow]

yeah, sry for the miscommunication

thanks!

[magic wind]

no worries i just got worried id spent a semester not learning anything

[rancid crow]

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[wanton steeple]

kinda confused on how this works and why the only units are 1 and -1

what about like n= 4 and then the inverse is 1/4

does that not make 4 a unit

[magic wind]

because 1/4 isnt an interger i think

[wanton steeple]

oh yeah

right

thanks

[magic wind]

np

[wanton steeple]

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[rancid crow]

is f(Z^2) valid here?

[rancid crow]

wrong screenshot sorry

[hollow cobalt]

You mean f(z) = z^2?

[rancid crow]

I'm not sure of the notation, but I want to square every Z in the domain

[hollow cobalt]

Right, then you are defining f: Z -> Z by f(z) = z^2

But that map is neither injective nor surjective

[rancid crow]

ah

why is it not injective?

[hollow cobalt]

f(-1) = (-1)^2 = 1 = 1^2 = f(1)

Despite -1 =/= 1

[rancid crow]

ah you are correct

hmm

I was thinking f(Z) = f(-Z)? but I think that is injective and surjective

[hollow cobalt]

f(z) = -z you mean?

[rancid crow]

yh sry

[hollow cobalt]

But it's asking for a map that's not surjective

[rancid crow]

ooh

what about f(Z) = 2Z?

[hollow cobalt]

That works

[rancid crow]

cool, thx

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[midnight haven]

i asked this question before but got way too confused -

A wooden block with a mass of 2.0kg is placed on a horizontal table. The coefficient of static friction between the block and the table is μs = 0.4. What horizontal force (F applied) is required to start the block moving?

apparently the answer is 34.. im getting approx 8

[upbeat glade]

Fg = Fn
Sum of forces in x direction = 0
Fapp - Ff = 0
Fapp - uFn = 0
Fapp - ((0.4) * (9.8 x 2)) = 0

Fapp - 7.84 = 0
Fapp = 7.84N —> 7.8N

[midnight haven]

so thats exactly what i did too!

yet my teacher insists its 34

[upbeat glade]

There is something wrong with the answer then. If they posted the solution as to how they got that answer take a look at that and see if they made any errors.

[upbeat glade]

Your teacher or the answer key?

[midnight haven]

answer key

i don't have it its with my teacher tho

question though if static friction was ignored would finding Fapp be possible? and would it be different

[upbeat glade]

How can we ignore static friction? Our goal is to find the minimal force to move it horizontally.

[midnight haven]

thats exactly what i said

how can u ignore static friction if its neede to solve the question

yet my teacher told me no no no

ignore it there's a way to find it

[upbeat glade]

Hmmm.

This is a bit strange yk what best to get a second pair of eyes. Dm.

[plain cedar]

i need help

@scarlet raptor

how do i solve it

[pearl pond]

@midnight haven Has your question been resolved?

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[vocal dove]

Hi, for this question i have to compare quantities

[vocal dove]

Apparently my way is wrong, could someone point out why?

looks right to me tbh

[buoyant ridge]

z = 6, x = 2?

also z = 6, y = 5

[pearl pond]

@vocal dove Has your question been resolved?

[vocal dove]

oh wait

[vocal dove]

z and x can't equal 6 and 2?

wait my brain

this is better right?

[buoyant ridge]

yep

[vocal dove]

thanks

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[small abyss]

hi one question. if α/β+β/α=(a+β)²/αβ–2αβ. What about α/β*β/α?

[glass meadow]

a/b * b/a = 1

Basic simplification...

[small abyss]

Oh cool sorry im just getting confused

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[gilded hollow]

The vectors e_1 and e_2 create a base in the plane. Two other vectors f_1 and f_2 are decided as a new base where:
$f_1 = 4e_1 + 21e_2 \
f_2 = e_1 + 5e_2$ \
a) Make an expression for the vectors e_12 with linear combination to the new bases.
b) Determine the components to the vector u = 4e_1 - 9e_2 in the new base.

[jolly parrot]

Merineth

The vectors e_1 and e_2 create a base in the plane. Two other vectors f_1 and f_2 are decided as a new base where:
$f_1 = 4e_1 + 21e_2 \\
f_2 = e_1 + 5e_2$ \\
a)  Make an expression for the vectors e_12 with linear combination to the new bases. 
b) Determine the components to the vector u = 4e_1 - 9e_2 in the new base.
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.52 The vectors e_
                   1 and e_2 create a base in the plane. Two other vectors f...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 52.
LaTeX Font Info:    Trying to load font information for OT1+lmr on input line 5
2.```

[cosmic torrent]

Hello I can help

[gilded hollow]

I already solved a i think

I need help with b)

$e_{1,2} = \lambda_1 * (4e_1 + 21e_2) + \lambda_2 * (e_1 + 5e_2)$

[jolly parrot]

Merineth

[gilded hollow]

And now i find lambda 1 and 2 by taking the equation: \
$f_1 = 4e_1 + 21e_2 \
f_2 = e_1 + 5e_2$

[jolly parrot]

Merineth

[gilded hollow]

or idk what i'm doing

no clue

Any help would be appreciated

please

[pearl pond]

@gilded hollow Has your question been resolved?

[pearl pond]

@gilded hollow Has your question been resolved?

[gilded hollow]

thanks for the help :(

[pearl pond]

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[fossil lodge]

I have a problem regarding algebraic cryptography

[fossil lodge]

In order to show that $c_k$ is injective, it is the same as saying that we need to show if $c_k(x) = c_k(y)$, then $x = y$

[jolly parrot]

WhoTao

[fossil lodge]

Would that be the correct idea?

In this case, just applying the decryption key as a left inverse would be fine.

I think that is fine..

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[vivid pilot]

whatever i try is always wrong

[vivid pilot]

so far ive done L(A)=36 and U(A)=80

[vital estuary]

check the grid size

[leaden wadi]

The grid sizes are 0.5, not 1.

[vivid pilot]

ohh

thanks

.close

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[vivid pilot]

.reopen

[pearl pond]

✅

[vivid pilot]

how come 18 is still wrong for L(A) then

and 40 for U

[vital estuary]

you have to divide by 2 twice

once in the x direction, once in the y direction

another way too look at it is that 4 squares make a unit square

[vivid pilot]

Oh right

Thanks

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[dusty stone]

What am i doing wrong?

[inland ivy]

how did you get 1\cosx in the denominator??

[glacial sequoia]

you didn't differentiate properly yeah

also there's a way to do it without LH but LH is probably the best method here

[dusty stone]

oh im dumb

[inland ivy]

$\log_{\sin x} \sin 3x$

[jolly parrot]

kheerii

[inland ivy]

[glacial sequoia]

you can use $\lim_{t \to 0} \frac{\ln (1 \pm t)}{t}$

[jolly parrot]

NEON

[glacial sequoia]

multiply and divide by 2, take the 2 inside the log and use the pythagorean identities

[dusty stone]

got it i think

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[viscid pivot]

In Z_17 I have to find a number for 7^2 such so it's in the groups

[pearl pond]

Please don't occupy multiple help channels.

[viscid pivot]

if that makes sense

[fleet sky]

@viscid pivot

.close

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[viscid pivot]

ty

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[opal lagoon]

Is this right 🥲

[opal lagoon]

Nvm is this right

[earnest stratus]

yeah

[opal lagoon]

Alr thanks

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[rancid crow]

pls can I have help answering this question, idk where to start

[warm current]

I would simply make a multiplication table and see whether or not every element has an inverse

[rancid crow]

interesting

so should I start with 1?

I don't understand what Z/6Z means

[tropic saddle]

integers mod 6

[warm current]

Oh you know what? Start with 0

[cinder flower]

making a multiplication table seems a bit overkill

[warm current]

I forgot about 0 🪦

[rancid crow]

so if I let x = 0, I calculate 0 mod 6 which equals 0?

then x = 1, 1 mod 6 = 1

x = 2, 2 mod 6 = 2

I'm not too sure what I'm doing here

[tropic saddle]

good practice

[rancid crow]

what all do I put in this multiplication table?

<@&286206848099549185>

[tropic saddle]

well all elements of Z/6Z

and the products

its a 6x6 table

or rather 7x7 I suppose

[rancid crow]

products modulo 6?

[tropic saddle]

yes

[rancid crow]

ah ok that makes sense

I'll do that now

where do I go from here?

<@&286206848099549185>

[pearl pond]

@rancid crow Has your question been resolved?

[plush bramble]

One of those rows/columns doesn't map to the identity, so it can't have an inverse

[rancid crow]

wdym by doesnt map to the identity

[plush bramble]

a * b != 1

for one a and all b

Find the a

[rancid crow]

0?

[plush bramble]

Sure

That's an example of an element that doesn't have a multiplicative inverse

[plush bramble]

Because there's no b such that 0 * b = 1

[rancid crow]

is this also true for 2 3 and 4?

but not 1 and 5

[plush bramble]

yes

[rancid crow]

ah

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[pearl pond]

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[heady oak]

how would i factor this

[pearl pond]

@heady oak Has your question been resolved?

[heady oak]

<@&286206848099549185>

[acoustic path]

long story short, you wouldn't

[heady oak]

how am i meant to find the x-intercepts then

[acoustic path]

can you show the context surrounding this problem?

[real palm]

are you allowed a computer?

[heady oak]

no

[real palm]

how did you sketch it without a computer?

[heady oak]

find degree

leading coefficient

y intercept

end behaviors

only thing im missing was x-intercepts

but my friend told me its kinda impossible

[acoustic path]

ye i think it just wants you to approximate

like as your diagram is somewhat reasonable that's probably good enough

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[odd halo]

Can someone help me figure out where I went wrong

[dapper kraken]

(x+h)^2≠x^2+4xh+h^2

[odd halo]

Wait so is it x^2+2xh+h^2?

[dapper kraken]

yes

[odd halo]

Ah okay thanks

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[brittle pebble]

https://gyazo.com/a3daca29f06cd85a7e443339b5c8c625
i heard about using the LCM on this but not to sure

[pearl pond]

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

[earnest stratus]

Lcm of what?

Do you have their individual periods?

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[vale turtle]

What can be a short of finding f^(-1)x

[vale turtle]

F(x)=(x+2)/(x+3)

here by general method
I got (-6x+5)/(x-1)^2

[sharp smelt]

in general, there is no shortcut

[vale turtle]

thanks

so did i find correctely?

[sharp smelt]

,w inverse of (x+2)/(x+3)

[jolly parrot]

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[sharp smelt]

doesn't look like it

show your working

[frosty frigate]

first check

if it is invertible or not

[vale turtle]

it is not invertible at x=-3

[frosty frigate]

and then

equate it to y

and find y in terms of x

thats it

[frosty frigate]

sorry

x in terms of y

[vale turtle]

,w d/dx of (2-3x)/(x-1)

[jolly parrot]

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[vale turtle]

,w 1/(x-1)^2=1/(1-x)^2

[jolly parrot]

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[vale turtle]

thanks

.clos

,close

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[agile hamlet]

.open

i have a problem
i need a proof for (b)
i have tried and solved (a)

[pearl pond]

Please don't occupy multiple help channels.

[agile hamlet]

<@&286206848099549185>

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mortal tendon]

what is the area of ABCD?

[agile hamlet]

you just have to prove

it can be anything

[mortal tendon]

yes but what is it in terms of the things given in the question

[agile hamlet]

ar(abcd) = ar(aklm)

[agile hamlet]

just ab*bc

[mortal tendon]

yeah exactly

so now look at the part of the other rectangle that is within ABCD

[agile hamlet]

yeah

okay what about it

[mortal tendon]

oh wait yeah sorry i was looking at the wrong picture

[agile hamlet]

yeah i cud do the first one

this one's a little weirder

[agile hamlet]

can you help?

[mortal tendon]

yeah im giving it a look but it is a little weird

my shape skills arent as great as they should be

[agile hamlet]

i mostly used pythagoras and similarity

and then linked with algebra

[pearl pond]

@agile hamlet Has your question been resolved?

[agile hamlet]

<@&286206848099549185>

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@agile hamlet Has your question been resolved?

[pearl pond]

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[junior imp]

hi there i am confused on this one as in where to get started

[pearl pond]

@junior imp Has your question been resolved?

[midnight haven]

so

u want to find the mean first

and to do that you need exact values whereas the table it giving you a range

so find the midpoints of each income section

30-35, 35-40, etc

[pearl pond]

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[quaint crown]

Subgamma function is exist?
Like for this:
!(-½), !3.5, !8.81, etc.
If this exist, what graph for this?

[pearl pond]

@quaint crown Has your question been resolved?

[plush bramble]

gamma function already exists for those numbers. what are you asking

[quaint crown]

yes

[plush bramble]

?

[quaint crown]

But

[plush bramble]

that doesn't answer my question

[quaint crown]

There are subfactorial

is exist subgamma function?

[plush bramble]

what is subfactorial

[quaint crown]

bruh, just search in google

[plush bramble]

[quaint crown]

Bruh

I search in wikipedia

And

There is no english translate

Only this

https://en.m.wikipedia.org/wiki/Derangement

Idk why

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[rich lark]

Write an equivalent expression to (x-2)(x+3)

[west sapphire]

(x+3)(x-2)
do i win?

[tired violet]

(-2+x)(3+x)

same

[west sapphire]

oh that's even better because it doesn't assume that multiplication is commutative

[rich lark]

i tried writing an expression but i got x^2-2x+3x-6

[west sapphire]

that's valid

[tired violet]

$(x+3)^1(x-2)^1*\cos(x+y)^0$

[jolly parrot]

llspacebarll

[west sapphire]

noice

[tired violet]

ooh i got another $\lim_{b\to 3}\lim_{a\to 2}(x+b)(x-a)$

[jolly parrot]

llspacebarll

[rich lark]

.close

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[rustic gate]

[pearl pond]

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[shy rapids]

how should i approach this integral? i was thinking of trig sub cause it looks kinda familiar but im kinda lost, should i be using partial fractions instead?

[shy rapids]

symbolab says to use two different usubs but our teacher is encouraging us to use the new integration techniques

[west sapphire]

did you try a trig substitution?

[pearl pond]

@shy rapids Has your question been resolved?

[unborn abyss]

partial fractions isn't valid here

trig substitution is probably the way to go

[shy rapids]

yeah thats what i thought, but i got lost because b=0 in the quadratic

so im not sure what i should substitute for x?

[unborn abyss]

you would just do something like x = 1/3 tan theta

or 1/9

or 1 / root3

probably that one

and i'm not sure on the trig function either :D

but something like that

[shy rapids]

okkk ill try that thank you

i think x=√3tan t