help-28

[hot knoll]

And y is the height

[empty trail]

wait should it be 2x

cause its symmetrical

[hot knoll]

Yes 2x

Good

[empty trail]

oh wait i dont needa find the dimensions

just the max area

[hot knoll]

Well the area is found using those dimensions

You do height * width

Which correspond to y and x

[empty trail]

cant i just take the derivative

[hot knoll]

Yes

That's exactly right

Once you have the area formula you take derivative and look for critical.points

A = 2x*e^(-x^2)

Simplify and take derivative

[empty trail]

oh yeahhh

i forgot abt critical points lmao

havent done that in a while

[hot knoll]

xD need those for maxes and mins

Dont forget them

[empty trail]

ok i got 1/sqrt2

ima plug that back in to

2x*e^(-x^2)

[hot knoll]

That should be the max

Technically you should do either sign test or 2nd derivative test to confirm it's a max

But in this case given the setup you can be confident it is

[empty trail]

oh yea i forgot i needa do that for max and min sometimes

[hot knoll]

Every time in fact

[empty trail]

okk i got sqrt(2/e)

after simplifying a bunch

im pretty sure its right

thank you for help

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[twin wolf]

Don’t know how to start

[onyx glen]

well you HAVE started by finding the gradient of line AD.

[lofty bane]

Ad perpendicular to ab

[onyx glen]

so you're not "unable to start" but "unable to progress from here".

[onyx glen]

@twin wolf

[twin wolf]

i dont know what to do with that information

[twin wolf]

only because i was told to like i dont know why i had to do it i just did it

[onyx glen]

do you know the relationship between the gradients of two perpendicular lines?

[twin wolf]

i just know its -1/m to find the perpendicular gradient

[warm kite]

m1 * m2 = -1

where m1 m2 are orthogonal

[twin wolf]

i dont get it?

what am i supposed to do with that?

[warm kite]

You have the gradient of equation of AD

use the relation to find the gradient for equation of line AB

[twin wolf]

oh

okay

[onyx glen]

don't use the letter x as a multiplication symbol.

[warm kite]

Okay

[onyx glen]

so you're a calculator in a meat suit, then.

[twin wolf]

i usually dont bother with understanding the actual meaning of a formula

only that i understand what it finds and how to find it

but actually now that i think about it

i reckon i can find it

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[onyx glen]

bad!

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[smoky plover]

wdym by solve?

not sure

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[mellow horizon]

I need help pls, with part c)

[mellow horizon]

<@&286206848099549185>

<@&286206848099549185>

The marscheme said it's wrong

And I don't understand how

Why does it multiply 10 with two etc.

Please helppppp I have an exam on Further Pure maths tmrrrr

[little mica]

Hey

Where you stuck

[mellow horizon]

On Part C

Thank you so much

[little mica]

You get equation in n

[mellow horizon]

When I write the eq Sn+1 why do I have to multiply 10 with 2

[little mica]

Is this solution

[mellow horizon]

yes

[visual furnace]

remember the equation for arithmetic series, [2a+(n-1)d]

since a=10 then 2x10

[mellow horizon]

Ohhhh but wait, isn't it the sum

[visual furnace]

yes, the full eq for sum of arithmetic series is this:

[mellow horizon]

OHH

OHHHHHH

[visual furnace]

good luck on your exam

[mellow horizon]

I got it!! THank you SO SO SO SO SO MUCH

[visual furnace]

welcome

[mellow horizon]

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[lime sandal]

What is the limit of the function f(x) = (-4x^2 + 6x + 18)/(-5x^2 + 11x + 12) as x approaches 3?

The first thing I did was I found the values of x for the top and the bottom equation and got this:

lim x -> 3 = ((x - 3)(x + 3/2)) / ((x - 3)(x + 4/5)).

The (x - 3) cancels out, and I'm left with:

lim x -> 3 = (x + 3/2) / (x + 4/5)

[lime sandal]

Now I can simply put 3 into the equation and I should get my limit

I get 45/38 but apparently the correct answer is 18/19

who's in the wrong here?

[hot herald]

you didn't factorise correctly
specifically you're missing the factors of 4 and 5

[lime sandal]

I'm sorry, I don't understand what you mean

[hot herald]

-4x^2 + 6x + 18 = -4((x - 3)(x + 3/2)
-5x^2 + 11x + 12 = -5(x - 3)(x + 4/5)

[lime sandal]

oooooooooooooh

sheesh, I really missed that one

thank you

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[tired blade]

can you simplify $\vec{a} = k\vec{F_{net}} + k \cdot \frac{1}{m}$ to $\vec{a} = k\frac{\vec{F_{net}}}{m}$

[glossy valve]

rynite

[tired blade]

$\vec{a} = k(\vec{F_{net}} + \frac{1}{m})$

[glossy valve]

rynite

[tired blade]

I don't know if we can add $\vec{F_{net}}$ and $\frac{1}{m}$

somehow

my algebra skills are a bit rusty

[glossy valve]

rynite

[tired blade]

I believe we can do it with a common denominator.

im not sure

[eager obsidian]

what's m here

mass?

[tired blade]

yes

[eager obsidian]

not sure adding a vector and a scalar makes sense here

[tired blade]

ah forgot, (still learning linear algebra too lol)

[eager obsidian]

your initial equation itself seems a bit off to me

[tired blade]

im just trying to prove this

or I mean, im not sure how both proportionalities are exactly combined

I know that $\vec{a} \propto \vec{F_{net}}$ is like $\vec{a} = k{\vec{F_{net}}}$

[glossy valve]

rynite

[tired blade]

and $\vec{a} \propto \frac{1}{m}$ is like $\vec{a} = k \cdot \frac{1}{m}$

[glossy valve]

rynite

[tired blade]

so I thought I could add both of these to find an equation that relates mass, acceleration, and net force

[eager obsidian]

well first of all the two constants should be k1 and k2 really

and k1 (assuming k1 is for the first proportionality) isn't even a "constant", but actually dependent on m

[tired blade]

ah

makes sense

[eager obsidian]

here's a more in depth explanation from SE (https://math.stackexchange.com/questions/433754/how-does-one-combine-proportionality?noredirect=1&lq=1)

but I think you should be talking about the magnitude of F, not the vector itself in these equations

[tired blade]

yeah, my bad

[tired blade]

ill look into it

thanks

[eager obsidian]

np

[tired blade]

I don't quite understand this phrase.

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[uneven cargo]

A small plane with 4 rows of seats with 3 seats in each row. Eight passengers were counted on board of the plane and were randomly assigned to the seats. A married couple follows to board the ship. What is the probability that there will be two adjacent seats in the same row for the couple?

[uneven cargo]

A 8/15
B 32/55
C 20/33
D 34/55
E 8/11

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[little mica]

I am getting 28/55

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[novel dirge]

pls help

[novel dirge]

<@&286206848099549185>

[hollow cave]

this is pretty cool

you can calculate the area

[hollow cave]

hello?

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[plush tiger]

its 30*10

NOO

omfg

im so dumb

wtv

[plush tiger]

can u help my question pls

[hollow cave]

ok

[plush tiger]

ty

[hollow cave]

the formula for the area of a parallelogram is $\frac{1}{2}ab$
where $a$ and $b$ are the two side lengths

[glossy valve]

Gamer Sans The Gamer Skeleton

[plush tiger]

can u asnwer my question in help-32

[plush tiger]

this makes no sense btw

its lenght*height or a times b times sin(x)

[hollow cave]

wdym sin(x)

[plush tiger]

of the angle

acute angle

[hollow cave]

you dont know the angle...

so why

[plush tiger]

thas why u cant calc

[hollow cave]

you can

[plush tiger]

how??

[hollow cave]

wait no

you cant

neither of our formulas are correct

[plush tiger]

yes

[hollow cave]

you need the height

so you cant calculate it

[plush tiger]

u can calculate it by doing height*lenght OR ab times sin(x), but we dont knoiw x

so uz cant calc

but my forumla is correct

[hollow cave]

no it isnt

[plush tiger]

yes it is

[hollow cave]

you need the height of the parallelogram

[plush tiger]

formula 1: height*lenght

formula 2: ab * sin(x)

[hollow cave]

ohhh

there's 2

well

you dont know the angle

you dont know the height

[plush tiger]

yea

[hollow cave]

so you cant calculate it

[plush tiger]

thats why u can5t calculate it

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[hollow cave]

answer

[plush tiger]

but the formulas are correct

i never said u can calculate it

i just said the formulas

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[plush tiger]

@hollow cave now can u help me in help-32

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[plush tiger]

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[hollow cave]

one sec

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[river wave]

How do I start 41

[twilit leaf]

break it into sec^2*sec^2 then use pythegorean identity

[hallow walrus]

u=tanx may prove as a useful substitution

[river wave]

Alright

Right now I have this

[strange basalt]

uhh

[river wave]

But

[strange basalt]

dont do both

[river wave]

It doesn’t seem right

Ok

I kept

[hallow walrus]

$\sec^2\theta=\tan^2\theta+1$

[glossy valve]

kheerii

[hallow walrus]

not -

[river wave]

Oh

Ok

Should I distribute

Or

Sub

Not sure

I integrate by parts

Right

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@torn jolt Has your question been resolved?

[heady cape]

is that f(t)=3 or f(1)

!status

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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

[heady cape]

where are you

he pulled the -2 out from the demominator

did you learn the limit definition of a derivative

yea so hes algebraically manipulating it to look like the limit def

so he can sub in the f'

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[heady cape]

i actually rlly like that question

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[austere lotus]

Could someone help me with this

[austere lotus]

I don't know where to start

[void nova]

You're given what 5•10⁸ is, so you can replace it in the bottom expression

[austere lotus]

I tried that though

I got 52 * 10^7

[void nova]

Yep that's it

[austere lotus]

It says thats wrong

[polar radish]

It's probably $5.2 \times 10^8$

[glossy valve]

David K.

[austere lotus]

Oh yes thanks mate!

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[austere lotus]

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✅

[austere lotus]

What do i do for this one

[empty sapphire]

same thing as the last problem.

[polar radish]

Same thing

[empty sapphire]

convert 4x10^7 to (something) x 10^6

then do your arithmetic

then put it in standard form

[austere lotus]

I got 7*10^6 it said it was wrong

[empty sapphire]

that's because that is wrong. Take a look what you did here

[austere lotus]

7*10^7 then?

[dim compass]

also wrong

[polar radish]

3 × 10^6 + 4 × 10^6 × 10

[dim compass]

It uses the fact that when in a sum or substraction, you have a term in product that CAN be taken common. Example ab+by=b(a+y).
It uses the concept of taking commons or factoring in.

[polar radish]

Yep exactly

[dim compass]

10 * 3 - 10 * 91771 = 10(3-91771) another example of taking common

xyz-abz+pqz=z(xy-ab+pq). Here you took "z" as common or rather factored z in.

50-45=5(10-9), you factored 5. Another example

[austere lotus]

Can you just give me the answer I don't understand a thing you're on about

It's ok i've done it now

Thank you for all your help though

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[fluid prawn]

How I do this

[fluid prawn]

7pi/6 and 11pi/6

I know those r when sin is -1/2

[candid fable]

,tex .unit circle

[glossy valve]

onion

[fluid prawn]

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[worldly geyser]

How are these two equal? I can't seem to figure it out

[gritty rose]

,tex .exp rules

[glossy valve]

riemann

[worldly geyser]

I know these rules, but I still wasn’t able to reach the answer. Here’s what I tried.

Wait

Let me try something rq

[grim skiff]

Try simplifying first before adding them

[worldly geyser]

I got it thanks for the help, can't believe i missed that my brain's been getting rusty

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[tired tinsel]

I need help with this wuestio

[tired tinsel]

*question

[static perch]

so

basically

[atomic blade]

Use properties of similar triangles. Are we assuming the roof and the gray triangle are right triangles?

[tired tinsel]

alr

[candid fable]

$\frac{49}{75}=\frac{15}x$

[glossy valve]

onion

[candid fable]

similarity

[tired tinsel]

I got the answer 110. I didnt assume the two triangles were right triangles but assumed they were similar

[atomic blade]

i mean you really cant do much unless you assume the triangle is either a right trianlge or an isoceles triangle

one is easier than the other

because heron's formula sucks egg

[tired tinsel]

I used trig

[tired tinsel]

you can solve it with out assuming its a right triangle

split the bigger triangle in half turning it into right triangle. get the side legths of the triangle. use tig to find the angles. aplly the angles to the smallr triangle then use trig again to get the area

[atomic blade]

that only works if its isoceles

[tired tinsel]

true just realized that. poorly written question

[atomic blade]

you can give the answer in terms of some angle, t, and then have an unknown in the answer

[tired tinsel]

its multiple choice and its for practice so I wont go through the hassle but thats a good idea

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[primal sable]

Hi I need help with this question about fraction decomp for integrations

[primal sable]

I am confused what they mean by a/x +a2/x^2 etc

I did the decomp and i didnt get all of those

[thick hedge]

$a_1,a_2$ etc are constants

[glossy valve]

Why am. I here

[granite torrent]

how did you do the decomposition?

you might have erred there

[primal sable]

ok i can show you

one sec

did i do anything wrong?

@granite torrent

[granite torrent]

oh

to decompose fractions with powers of terms, you'd use the general form is C/x + C/x^2 + ... + up until the highest power

[primal sable]

oh

[granite torrent]

i don't think the result you received actually equals the original expression

[primal sable]

ok so i have to do 1/x + 1/x^2 etc

till i get to x^4

[granite torrent]

pretty much

[primal sable]

I am kinda knew to this so i ahd no idea

ok but i still have to do x-2 ight?

[granite torrent]

yes that's still necessary

[primal sable]

ok

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[torn jolt]

given that a and p are some vectors, I cant figure what w would represent

[torn jolt]

<@&286206848099549185> sorry;;

I think I found w to be the orthogonal projection of p?

but then I cant figure out what the meaning of w dot our projection would be

I know the dot product is the

"measure" of how parallel 2 vectors are

should I simply

I think I got the answer nvm

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[misty fractal]

hi, i would one do this more rigorously? It made sense to me in my head but couldn't ocme up with an actual argument to support that

[silver lintel]

the main insight is that multiples of k>1 cannot be next to each other

so if k divides n then k does not divide n+1

for k>1

in your case 2,4,6,8,...,100 divide h(100)

so 1,2,3,4,...,50 divide h(100)

[misty fractal]

wait i don't follow

well i get the idea you mentioned though

if k divides then k does not divide n + 1

how'd you get this though?

[silver lintel]

the first part is from the problem description

[misty fractal]

yes

[silver lintel]

for second part, if 2n divides h(100) then n must divide h(100)

does that make sense?

[misty fractal]

sure yes

but it's talking about h(100) + 1, no?

[silver lintel]

yes

so first we know 2,3,4,..,50 divide h(100)

[silver lintel]

using this we know that none of 2,3,4,...,50 divide h(100) + 1

[misty fractal]

oh yeah i think you're onto something

thanks thanks let me try

[silver lintel]

so in particular no prime factor less than 50 divides h(100) + 1

sure

[misty fractal]

okay so first of all h(100) = prod(2n) from n = 1 to 50

so obviously h(100) | 2n

and transitively h(100) | n

h(100) + 1 thus can't divide n

and since h(100)|n for n = 1 to 50

then (h(100)+1)|n can only divide n for some n>50?

[silver lintel]

the idea is correct but you made some mistakes in the notation

[silver lintel]

its the other way around

[misty fractal]

oh yeah sorry

[silver lintel]

2n | h(100) (meaning 2n divides h(100))

[misty fractal]

2n|h(100)

indeed

[silver lintel]

yes

[misty fractal]

okay this makes sense

[silver lintel]

same here

[misty fractal]

i'll probably forget after some day

but thanks lol

[silver lintel]

n cant divide...

[misty fractal]

[silver lintel]

no worries

[misty fractal]

yes

[silver lintel]

and also you would have to say what n is in a formal proof

e.g. Take $n\in \mathbb{N}$ such that $1 \leq n \leq 50$, then $2n | h(100)$ ...

[glossy valve]

LF

[misty fractal]

right thanks

this is in an algebra class though

so no proof work yet i guess

[silver lintel]

oh alright

its cool that you want to try and prove things tho

you get a deeper understanding that way

[misty fractal]

indeed indeed but i don't think i'm equipped to write out fleshed out proofs

:C

[silver lintel]

no need

[misty fractal]

might need some prereq course for that ig before i could get to that

oh

but yeah i get the idea

[silver lintel]

as long as you can convince yourself its good

thats it

[misty fractal]

which could serve as a proof (For me)

yeah

[silver lintel]

the idea is the main part

[misty fractal]

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[minor sail]

my answer didn’t match up with the calculator, any mistakes in this?

[open igloo]

2+ (1/2) = 5/2 (second last line)

[minor sail]

yeah that’ll do it

thanks

[open igloo]

Np

[minor sail]

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[minor sail]

any issues in this integral? sorry for the messy handwriting

[minor sail]

<@&286206848099549185>

<@&286206848099549185>

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[digital pike]

Stuck on this! I have used the formula (x^2 +a^2) then use a*sinh(x). i understand that, however the answer states that a in this instance is sqrt(3)? im unsure as to how they got this value for a. any help appreciated.

[digital pike]

<@&286206848099549185>

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@digital pike Has your question been resolved?

[digital pike]

<@&286206848099549185>

[little mica]

Yo tried x=√3tant

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[toxic rain]

how'd it go from step 1 to 2?

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[next basalt]

Let OACB be the parallelogram to the right. The sides OB and BC are divided into three lines of equal length where the dividing point P ∈ OB is closest to O, while the dividing point Q ∈ BC is closest to C. Show that the intersection points P′, M and Q′ of the diagonal OC with the line The angles AP, AB, and AQ, respectively, divide this diagonal into four equally long parts. −→ −→ Hint: Choose a = OA and b = OB as bases. Due to the symmetry, it suffices to show that …

Express OP′ in two terms and take advantage of the fact that a and b are linearly independent.

How do I prove this?

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[silver fern]

test?

[silver fern]

If my plan is to find the antiderivative of this

https://cdn.discordapp.com/attachments/903463354654404658/1198974962246111292/image.png?ex=65c0db72&is=65ae6672&hm=68ddaeb6d7a35e7526fbc7f68013b24e91430e44e06c48b1fdb06fe2f323d2fb&

is this the correct notation:
$F(x)=\int_{}^{}cos(x)*sin(x)dx$

[glossy valve]

Nox

[spiral vigil]

well we normally wouldn't write the star, but yes that seems to match what your problem states

i'm not quite sure what your question is

the way i'd type that is $\int \cos(x) \sin(x) \dd{x}$

[glossy valve]

ヘイリー

[silver fern]

Im just trying to make sure im 100% correct on notation since my prof is very stingy

[strange basalt]

the dot/star works to, not that significant tho

[silver fern]

so if I wanted to solve the integral towards getting F(x) ud just do

$\int \cos(x) \sin(x) \dd{x} = ... = .... = F(x)$?

[glossy valve]

Nox

[strange basalt]

if f(x) = cos x sing x
F(X) = integration of f(x)

like a general notation

for deivative some like f'(x), some preffer a bar over it , many other ways too

like i would suggest check your lecture notes by the proffessor , or ask him in person / by mailing
generally they dont mind if you used another notation , but make sure your proffessor is that guy

[silver fern]

Idk im losing my mind over this, I spend more time worrying about notation than actually solving everything zz

[strange basalt]

lmfao ok

[silver fern]

Ive already found out hes in fact not that guy

[spiral vigil]

if i were computing that antiderivative i wouldn't even mention f or F

[strange basalt]

fair

[silver fern]

so if the integral solves for

$(x^2-2)sin(x)+2x*cos(x)+C$

[glossy valve]

Nox

[silver fern]

youd just leave it like that?

[spiral vigil]

yeah

it doesn't, in this case

but yes that's how i'd write my answer

[silver fern]

oh i took the answer from a different case woops

but yeah thanks, I guess ill try my luck and pray to god my prof doesnt have a bad day

[strange basalt]

praying good for your prof 🔥 , gl !

[silver fern]

tyty

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[hazy helm]

first step of induction for P(1) and letting n=1

LHS i get x

RHS i get 1+x

[opal pewter]

r u sure about that ?

[hazy helm]

idk what im missing

[opal pewter]

and r u trying to prove that my induction ?

[opal pewter]

1+x+..+x^n at n=1 ... So wheres the 1 at the beginning going ?

[hazy helm]

oh i thought that it isnt used

cause in other questions it just dissapoear

s

such as this

solution says this

[opal pewter]

at n=1 the only value in lhs is 1

[opal pewter]

But here at n=1, 1+x

at n=0, it becomes only 1

[hazy helm]

im not getting it

[rapid rain]

x^0 + x^1 + ... + x^n

when n= 1, the sum stops at 1

so x^0 + x^1

so 1 + x

[hazy helm]

ohhhh

[rapid rain]

when n=0, the sum stops at 0

so 1

[hazy helm]

all this proof is hurting my brain

that makes a lot more sense

wait so in other words u can either sub 1 into n or look at the first term?

or just look at the first term

[rapid rain]

when you do "n=1", yes you literally sub in 1 into n

but you have to be careful

on the left the sum goes from 0 to n

so if n=1, the sum goes from 0 to 1

and so there are 2 terms

0 and 1

[hazy helm]

ah ok

[rapid rain]

(in this sum)

[hazy helm]

so in general just be careful for the starting term

[rapid rain]

yes, UNLESS :

[rapid rain]

in this case, the sum goes from 1 to n

so if n = 1

the starting term and the ending term are one and only one term

so there's just 1 term

[hazy helm]

ahhh this is making a whole lot more sense

thanks

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[cosmic crag]

please help me with this. im going crazy after being stuck on this for over an hour. Been trying random stuff with the normal distribution formula at this point

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@cosmic crag Has your question been resolved?

[upper anvil]

consider p(X>b)+p(Y<b)

[cosmic crag]

uh a hint on how to carry forward with that?

[upper anvil]

wait i retract that

instead

notice that they both have the same variance

and then consider a graph of the normal distribution of both of them

if you combine them into one graph (without adding, just put them side by side with overlaps)

theres going to be a point where the two graphs meet

in which case that is the point where the probability is the same

[cosmic crag]

oh

so

we just use the equation of normal distribution

equate 2 graphs, the x coord of intersection is just b?

[upper anvil]

yeah

you can just graph it out and you'll see something

[cosmic crag]

thanks that makes sense

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[misty fractal]

hi, how would i solve for integer solutions for x here

[misty fractal]

x^x -10x = -3

(aside from testing numbers)

is there a way to do it via number theory

or only something related to Wlog works?

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@misty fractal Has your question been resolved?

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@misty fractal Has your question been resolved?

[wise gate]

x must be positive, so -3 = x(x^(x-1) -10) and therefore x is a divisor of 3

[misty fractal]

wait what

how did you infer x must be positive

[wise gate]

x can't be negative, because x^x = 10x-3 is integer

And it can't be zero because 0^0-10 0 equals 1

[misty fractal]

also (x^(x-1) -10) could be a divisor of 3 instead right

[wise gate]

Both are divisors of 3

[misty fractal]

i’m confused, why should they be?

only one has to be right?

[wise gate]

No, if a times b is 3 then both a and b are divisors of 3

[misty fractal]

hmm you’re right, this makes sense with an example

what is this called btw?

it’s probably elementary but i might get some context just by looking at it from other pov as well

[wise gate]

It's the definition of divisibility

a divides c if ab=c for some b

Since ab = ba, b divides c too

[misty fractal]

oh okay i see now

[misty fractal]

why is this one? 😭

isn’t 0^0 undefined

[wise gate]

Well

[misty fractal]

well you could also define it as 1 ig

[wise gate]

Yes, that depends on who you ask

It's similar to whether 0 is a natural number, it is but some people say it is not

[misty fractal]

i see, u defined it as 1 right?

[wise gate]

Yes

[misty fractal]

okay and for the next thing i still don’t get how you see this by observation

[wise gate]

10x-3 is integer

And x^x

is integer if and only if x >= 0

Oh

Wait

[misty fractal]

oh yeah you’re right i just got it when you were typing

wait

okay yeah makes sense ig

[wise gate]

Except for x = -1

[misty fractal]

oh yeah that too

that’s integral

[wise gate]

But

Oh nevermind

Yes, x^x is integer if and only if x >= -1

[misty fractal]

wait

and integral too right?

i mean x>=-1 and x in Z?

[wise gate]

Yes

[misty fractal]

okay so finally i get this i guess

so i’d just try x = 1 and 3 right?

[wise gate]

Yes

[misty fractal]

nice, thanks-

i guess the key takeaway is the divisibility thing

didn’t think of that tbh

[wise gate]

You could also have done it without that, but I think it would have been more difficult

[misty fractal]

i tested numbers and got 3 too

i was just curious what i could’ve used for other questions

also i just wanted to learn

[wise gate]

You could have used the derivative too

Look for real solutions

And see which are integers

[misty fractal]

how does that relate

[wise gate]

The derivative is (log(x)+1)x^x -10

Somehow show that it only has one root

Therefore the equation can have at most two roots

3 is one of them

[misty fractal]

idk where that results comes from too

could you elaborate

not on the showing one root part

[misty fractal]

but this

[wise gate]

And since 0^0 -10 0 = 1, and 1^1 - 10 1 = -9, the other one is between 0 and 1 and is not an integer

[wise gate]

It is a consequence of Rolle's theorem

https://en.wikipedia.org/wiki/Rolle's_theorem

[misty fractal]

😮😮i remember learning it okay okay i need to review all this stuff lol

wait so

i think somebody else told me

for the previous question right

x^x = 10x -3

they said they mod x both sides

[wise gate]

Oh

[misty fractal]

and then they have -3 (mod x)= 0

that seems illegal lol

is that a thing?

[wise gate]

At least it works for x > 0

[misty fractal]

i’m confused

x^x (mod x) is 0 right?

[wise gate]

Yes

[misty fractal]

10x -3 mod x is -3

okay i see

[wise gate]

It actually is the same as this

[misty fractal]

why not for x>=-1?

[misty fractal]

right

[wise gate]

Do you mean for x < 1?

[misty fractal]

no unedited

no i edited *

[wise gate]

Maybe it actually does

[misty fractal]

[wise gate]

But seeing an equation mod 0 is weird

[misty fractal]

you can have mod negative numbers?

[wise gate]

Of course

[misty fractal]

oh yeah

okay i think for all i can gather from this

[wise gate]

Do you know what complex numbers are?

[misty fractal]

i need to learn number theory asap lol

[misty fractal]

yes

[wise gate]

Well, you can consider complex numbers a+bi with a and b integer, and have (mod c+di) with c and d integer

[misty fractal]

yeah that’s above my pay grade

[wise gate]

And you can have (mod v) where v is a vector

And many other things

[misty fractal]

is that all part of elementary number theory?

[misty fractal]

this is how you can have mod negative numbers right?

u can’t over the reals?

[wise gate]

Actually a lot of advanced mathematics is about seeing things modulo other things

[wise gate]

No, you don't need to

[wise gate]

No

[wise gate]

It's not exactly the same

For example you can have (mod 1) in the reals, so you just look at the decimals

But then you can no longer multiply numbers

And if you want to multiply numbers, then all numbers equal 0

It's weird

[misty fractal]

interesting but sounds hard

lol

,w 10.1 mod 1

[glossy valve]

Results provided by WolframAlpha

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

Requested by kannaaaaaaaaaaaaaaaaaaaaaaa

[misty fractal]

oh

[wise gate]

Yes, it is hard the first time you study it

[misty fractal]

honestly didn’t even know you could get decimals as results to mod lol

[wise gate]

But again, that mod only works with addition

You can't multiply or divide

[misty fractal]

i see

thanks thanks

anyway so basically the mod x thing works too right

same conclusion about divisors of x

weird tbh but that’s a new “trick” for me then

[wise gate]

For example, 1.5 times 0.5 = 0.75, but also 0.5 times 0.5 = 0.25 (mod 1)

[wise gate]

It is an extremely useful trick

[misty fractal]

you mean mod x both sides?

okay okay, nice

[wise gate]

Or other numbers

[misty fractal]

right i use mostly

mod 3 and stuffs

never mod variable

but yeah u mentioned u can mod anything lol

so that’s a new insight

[wise gate]

As long as that thing is integer

[misty fractal]

how is a vector an integer

u said mod v is a thing

[wise gate]

No, but you can do mod anyway

For example if v = (1,-1,1)

Then (2,0,0) = (1,1,-1) (mod v)

And you lose 1 dimension

[misty fractal]

[misty fractal]

so you can mod anything?

isn’t mod 0 undefined

[wise gate]

You have to define what mod means

[wise gate]

No, it is defined, and it does nothing

[misty fractal]

[wise gate]

a = b (mod 0) if and only if a = b

Because a=b (mod 0) means 0 | a-b, or 0c = a-b for some c, so a-b=0

[misty fractal]

hmm yeah makes sense

all this time division by zero was undefined for me

0^0 was also undefined

and yeah i’m aware the divides thing uses multiplicative inverses and not actual division

[wise gate]

No, it is not division by zero, but rather divisibility by zero

[wise gate]

^

[misty fractal]

right fair enough

thanks again

[misty fractal]

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[burnt monolith]

Im not sure how to get started on this question

Parametric Differentiation

[torn jolt]

okay first objective is to solve for t in both equations

can you do it

[burnt monolith]

Solve for t

Oh wait

so make

t the subject?

[torn jolt]

yes

[burnt monolith]

alr i can do that

t = 1+x/2

and

t = y-1

[torn jolt]

its t = (1+x)/2

but alright

[burnt monolith]

alr

[torn jolt]

ok so like

equate both equations

[burnt monolith]

(1+x)/2 = y-1?

[torn jolt]

yes

[burnt monolith]

alr

[torn jolt]

now solve for y

[burnt monolith]

y = (x/2)+2

[torn jolt]

yuh

[burnt monolith]

alr

[torn jolt]

wait

uh

you did something sus

[torn jolt]

how did u get this

[burnt monolith]

sus?

[torn jolt]

wrong is what i meant

[burnt monolith]

so 1+(x/2)=y-1

bring 1 over

[torn jolt]

ok but

u r wrong

its not 1+

you have $\ds \f{1+x}2 = \f12 +\f x2$

[burnt monolith]

uhh

i am confused

[torn jolt]

you have this yes?

[burnt monolith]

yeeesss

[torn jolt]

and like

[burnt monolith]

oh

[torn jolt]

you also have this

ao

so

\

,, \f{1+x}2 = y -1 \Implies \f12 + \f x2 = y -1

[burnt monolith]

ohh yes i see what i done

idk why i changed (1+x)/2 to (x/2)+1 during my calculations

me dumb

(x+1)/2+1 = y

[torn jolt]

its okay lmao

but yuh

[burnt monolith]

al

r

[torn jolt]

thats your equation

[burnt monolith]

yes

[torn jolt]

so its just an equation of the line

[burnt monolith]

y=mx+c

[torn jolt]

i trust you can graph it on ur own?

[burnt monolith]

i think i can do it

[torn jolt]

okie dokie

[burnt monolith]

thxs

[torn jolt]

yea seems legit

just like uh

remember that t is between 0 and 2

so like ur graph needs to be consistent with that

plug in t = 0 and t = 2 in both the x and y equations you had in the beginning

anr then accordingly restrict your graph

[burnt monolith]

alr

(-1,1) when t = 0

(3,3) when t = 2

[torn jolt]

yuh

[burnt monolith]

alr

[torn jolt]

so your graph is restricted between those two points

n then like

notice rhe direction of motion would be like

diagonally up

[burnt monolith]

yes

up

[torn jolt]

because that's in the direction of how t grows

ok sweet

[burnt monolith]

mmmmmmm

[torn jolt]

guess that's it

[burnt monolith]

I see

[torn jolt]

good job

[burnt monolith]

alr

thank you

[torn jolt]

no worries

[burnt monolith]

thank you so much

you are a great tutor

[torn jolt]

lmao much appreciated

you are a great student too 👍

anyways you can close with .close

[burnt monolith]

.close

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[proud veldt]

what is this function exponential? x^3+5+2x

[torn jolt]

No. It is a polynomial

[candid fable]

you mean cubic?

[torn jolt]

yeah

[proud veldt]

when is it exponential

[proud veldt]

of which degree?

[torn jolt]

when it is of the form $b^x$

[torn jolt]

third degree

[proud veldt]

so not with a number instead of the x?

how do you call that btw i am not english native

wait so exponential is 3^x but x^3 is polynomial

?

[odd flicker]

hate to intterupt, but can i borrow u when yr free?

[proud veldt]

lmao

hahaha

[torn jolt]

yes

[torn jolt]

no

[proud veldt]

thanks

[proud veldt]

:/

would this be exponential: 3^x+2x+5

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@proud veldt Has your question been resolved?

[proud veldt]

.

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@proud veldt Has your question been resolved?

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[granite yew]

how do I derive the equation of a polynomial function, P(x) when all the information I'm given is:

• it is a degree of 5
• has exactly two x intercepts at x = 2, x = 3
• P(0) = 6

[wise gate]

I think that's not enough information

[wise gate]

There may be more than one polynomial with those properties

[granite yew]

yeah I thought so as well

I guess it was asking for one possibility of an eqn, but it was so vague

[wise gate]

Well, since it has exactly two roots, one of them must be multiple

So for example we can do P(x) = (x-2)^2 (x-3) Q(x), and Q(x) without roots

Like Q(x) = 1+x^2

But P(0) = 6, so (-2)^2 (-3) Q(0) = 6, that is, Q(0) = -1/2

So Q(x) = -1/2-x^2 could work

[granite yew]

thats true, but wouldnt we still need to find the coefficient?

[wise gate]

Which coefficient?

[granite yew]

if there is one thats 1<

when I learned about deriving polynomial functions, we had to find the coefficient as well

so for example a(x-b)(x-c), so a would be the coefficient and we'd have to input a coordinate to solve for it

not sure if that applies here

[wise gate]

Well, that is implicit in the Q

[granite yew]

oh I see

[wise gate]

If you say which roots a polynomial has, it will be determined up to a coefficient that multiplies it, like a(x-b)(x-c), and if 0 is not a root, then the value at zero is a(-b)(-c) = abc, and therefore the value at zero determines the value of a

Since I chose Q such that P(0) = 6, I somehow chose something that was equivalent to that coefficient

[wise gate]

But are you sure the degree is 5 and not something smaller like 2?

[granite yew]

yeah it was 5

[granite yew]

ohh that makes sense

I see how urs could work since it fulfills alll of the listed characteristics of the graph, thanks for explaining

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[brisk coral]

help

[brisk coral]

it says

Look closely at the figure showing a map of a park.

The distance shown is 7 cm on the map.

How many meters is the stretch in reality?

i still dont get it

[wide wind]

That's the ratio for figure:reality

1 cm in the figure equals to 600 cm in reality.

[brisk coral]

ah

so i have to use an ruler for this one?

how do i do it then

i still really dont get it..

[brisk coral]

so what is the asnwer to this question?

so its 4200

7x600

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[cinder wind]

how would i find the modulo inverse of a matrix?

[cinder wind]

like i have no idea how to do it

[gritty rose]

do you mean determinant

[cinder wind]

yes

[gritty rose]

depends on the matrix

do you have one

[gritty rose]

the determinant of the inverse is the reciprocal of the determinant of the matrix

[cinder wind]

wdym?

i don't understand

[gritty rose]

which part

reciprocal of a number x means 1/x

[cinder wind]

so is the determinant of the inverse 1/54?

[gritty rose]

,calc 13 * 6 - 24

[glossy valve]

Result:

54

[gritty rose]

yes

[cinder wind]

wait but from there how to i find the inverse then

if my modulo is 26

hello?

[gritty rose]

what is your actual problem

[gritty rose]

you just asked for the determinant

[cinder wind]

so for the decryption of a hill cypher

i have a key

and the code

but i dont get how to fully decrypt it

[gritty rose]

!original

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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.

[cinder wind]

now i need to decrypt it

[gritty rose]

this was wrong

you don't need the determinant at all