help-39

[versed mica]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[snow sonnet]

It’s alr, it’s log 7

[haughty kraken]

yep

[snow sonnet]

Not just log

[haughty kraken]

i mean yeah, just didnt write it

[versed mica]

sure but

didn’t want to write that

[snow sonnet]

Ok

[versed mica]

how do you isolate x

if u have a log

[snow sonnet]

Do I make the log go to the other side and make it ^-1/4?

[versed mica]

no

[snow sonnet]

7^

[versed mica]

u use yea

7^-1/4=…

[snow sonnet]

x-2

mb

[versed mica]

mhm

so just isolate x

[snow sonnet]

So x= 7^-1/4 -2 ? Is there a way to find the exact value of 7^-1/4?

Without a calc?

[versed mica]

no

u can express it as a fraction

but

yea you’ll need a calculator

for a numerical value

[snow sonnet]

What if I made the -1/4 -> -0.25 and make the base become 0.25/7-2 and calculate the fraction ?

.close

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[feral citrus]

im confused on the whole thing

remember to check for extraneous solutions

[cursive wraith]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[feral citrus]

0=k^2-7k+10

im confused what to do after

stuck

hi

i forgot how to factor

sorry

.close

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[pale grail]

how do we get the last term in equation 2.25

[bronze harbor]

I? I would think it'd be defined elsewhere in the paper

As it refers to a class of individual, presumably based on demographic data

[pale grail]

no not I 😂

this thing

[pearl pond]

@pale grail Has your question been resolved?

[pale grail]

<@&286206848099549185>

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@pale grail Has your question been resolved?

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[naive dirge]

Let’s zoom in with our algebraic brain.

[naive dirge]

Can the x be expressed using only the “l” and some constant.

And gives your reason.

Should I be more specific about my question

[naive dirge]

I wonder why it is possible because I didn’t notice it at the first sight

Better with a geometric view

[wet osprey]

!original

[pearl pond]

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.

[naive dirge]

That’s a physics related question

[wet osprey]

Still show it

[naive dirge]

Unfortunately I do not have the book in my hands rn

[naive dirge]

However, I wonder if that can be explained geometrically.

[wet osprey]

I don’t even know what I’m looking at

I don’t know what 1) and 2) are and what even is the implication

I don’t know what your letters represent

[naive dirge]

I see

[wet osprey]

I don’t know a lot of things and there’s a lot of missing context

[naive dirge]

I would define it then

Is x solvable?

[pearl pond]

@naive dirge Has your question been resolved?

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[molten kelp]

Hii, I'm trying to do this question and I'm getting stuck.

[molten kelp]

So far, I have tried squaring everything

I ended up with x^4+16x^2-c^2-8x^2-32x-8c=0

I turns into x^4+8x^2-32x-c^2-8c=0

But I'm not sure how that helps

Even something like x(x^3+8x-32) - c^2-8c = 0 doesn't quite go towards anything

[pine moat]

so the first thing i notice is that after pulling out an 8 from the sqrt

we see a lot of redundancy

let $w = x^2 - 4x + c$

[jolly parrot]

AlphaNull

[pine moat]

$x^2 - 4x - c - \sqrt{8x^2 - 32x - 8c} = x^2 - 4x - c - 2\sqrt{2} \sqrt{x^2 - 4x - c} = w - 2\sqrt{2} \sqrt{w}$

[jolly parrot]

AlphaNull

[pine moat]

now we want this to be 0

this is a quadratic which is much easier to solve

[sterile coyote]

this is a start

[molten kelp]

I think I'm following along up until w - 2\sqrt{2} \sqrt{w}..

[pine moat]

what part is confusing about that step

[molten kelp]

Oh, I think I see it now

You substituted w for x^2-4x+c

[pine moat]

yep exactly

solving that for w will give you two values

this question is quite difficult because you need a geometric picture as well

of what the function actually looks like

[molten kelp]

I'm not sure what a geometric picture is..

but I assume that with the w - 2\sqrt{2} \sqrt{w} part

You would square everything

and end up with w^2 - 8w?

[pine moat]

yes that looks right

https://www.desmos.com/calculator/aokrh0o5ga

i suggest playing around with this a bit

[molten kelp]

So.. I see that when I turn c more than -6, then the graph is weird

So does that mean that when c = -12 to -6, then there are 2 answers for x?

[pine moat]

hmm not quite

the idea is you solve for w, which tells you what x^2 - 4x - c has to be

then you want that to only have 1 solution

[molten kelp]

Uhh, I think I get the process, but I don't think I have the skills to do it..

I used a quadratic solver and it seems that w = 0 or 8

x^2 - 4x + c = 8?

[pine moat]

yep

then you want to find the c that makes this only have 1 solution

[molten kelp]

so then x^2 - 4x - 8 = -c?

[pine moat]

a quadratic has one real root only if the discriminant is 0

[molten kelp]

so then b^2 -4ac = 0,

16-4(1)(4) = 0

So c = 4?

[sterile coyote]

well do keep the notation separate

the c in question is a part of the constant term

the whole "c" of the quadratic is then (-8+c)

[molten kelp]

So for the third term to be 4, then (-8+c) = 4, and c = 12?

[sterile coyote]

you want to find all real cs, such that the quadratic has exactly 2 real roots

that happens if the discriminant is strictly greater than 0

You're solving an inequality

oh, also, my bad

i dunno where we changed the -c to +c

but it should be $x^2-4x-c-8=0$

[jolly parrot]

🇵🇸Mína🔆

[molten kelp]

Mmm So the discriminate needs to be more than 0, So then b^2 -4ac > 0, then 4^2 - 4(1)(c) > 0, so 16 - 4(c) > 0.. thus c < 4?

So.. I think I'll probably give up on the question right now. I feel that I have too many gaps in my knowledge. Thank you for helping @sterile coyote and @pine moat

[vivid grove]

Type .close

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[placid solar]

What does it mean for a parameter to follow a distribution

[placid solar]

I understand when we say a random variable X follows a distribution but when we want to estimate a parameter what does it mean for that parameter to follow a distribution

[raven grail]

Can i get help on this questions?
i dont understand them...
1There are two positive integers such that the difference of their squares is a cube (of an integer). One of the two numbers is 10.

What is the other one, which is smaller than 10?

2In a class of 34 students there are exactly 8 girls who are insufficient in English. In the same class, 18 students, both boys and girls, are insufficient in English.

How many, among the 14 children in the class, obtained a sufficient evaluation in English?

3If 10 kids from a voluntary ecology association need five days to clean up a forest, freeing it from waste, how many would it take to clean up the same forest in two days?

4Jacopo tries to arrange his 1 euro coins in piles of nine coins and realizes that the number of remaining coins is equal to the number of piles he has formed. He repeats the same operation, forming piles that each contain seven coins and also in this case he realizes that the number of remaining coins is equal to the number of piles formed.

How many 1 euro coins does Jacopo have, at least?

[lime anvil]

!occupied

[pearl pond]

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).

[placid solar]

What does it mean for a parameter to follow a distribution?

I understand when we say a random variable X follows a distribution but when we want to estimate a parameter what does it mean for that parameter to follow a distribution

[lime anvil]

the estimator of the parameter follows a distribution

[placid solar]

how come I don't undrstand what that means

Imagine we estimate a poisson parameter, the estimator follows a distribution in the sense that we can calculate the probability that it takes a certian value?

[lime anvil]

we can find probabilities that our estimate of that parameter lies within certain ranges as we can with other random varianles

[placid solar]

okay I see

thanks!

.close

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[daring wedge]

Prove that r1r2 + r2r3 + r3r1 = S^2
r1, r2, r3 are radiuses of e-scribed circles
S is the constant used in hero's formula which is equal to the half of perimeter of a triangle.
r1= (area of triang.) / (S-a)
r2= (area of triang.) / (S-b)
r3= (area) / (S-c)

a,b and c are three sides of triangle/

[midnight haven]

start by drawing a diagram of the information ur given to help u visualise it

[daring wedge]

done

nvm i solved it

you people so sloww

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[compact token]

Can someone help me with this problem?

Ik a bit about what to do actually, we should define some events, and then taking the expected value of the indicator random variable should give us the estimator.

I am unable to connect the steps altogether, if someone could point out the mistake in my answers it would have been helpful

[pearl pond]

@compact token Has your question been resolved?

[compact token]

<@&286206848099549185>

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[tall garnet]

is this correct

a = f'(x sub n)

[plush bramble]

yea

[tall garnet]

ok ty

it's a little confusing for me still

is this saying that f'(x sub n) is equal to the length A from (x sub n+1, 0) to (x sub n, f(x sub n))??

@plush bramble

[steep lion]

Kind off

[tall garnet]

wdym

[steep lion]

After all, a slope is the ratio of y values over x values

[tall garnet]

okay

[steep lion]

Ohh wait wait. f'(x sub n) is not the length. It is the ratio.

[tall garnet]

it is

[steep lion]

A is the red line right ?

[tall garnet]

yes

[tall garnet]

this is true

doesn't feel right but grab any function

[steep lion]

And ?

[leaden wadi]

Are you learning about the Newton-Rhapson Method?

[tall garnet]

purple line is f'(2)

red line is x^2

[tall garnet]

i am

[leaden wadi]

And you are trying to understand why it works?

Or how it was derived?

[tall garnet]

both

[leaden wadi]

It was derived from the Slope formula with one point set equal to (x_n+1, 0). This where the magic mostly occurs if you know the slope.

This allows you to solve for x_(n+1) which allows you to calculate a new approximation close to a root of the equation you are approximating.

[pearl pond]

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[sand axle]

we haven't done integrals yet but i want to get an early start to it

i understand it with a graph and no unknowns but im clueless at this

and i can't find anything online similar to this

either explanation or a link to something similar would be amazing

(b is real and positive - find b)

[plush bramble]

you would need to learn integrals to understand integrals

[sand axle]

i understand what integrals are but i haven't seen a question similar to this and i can't find anything else online

[regal herald]

do you know derivatives

[sand axle]

yes

[regal herald]

whats the derivative of e^{bx}

[sand axle]

would it just be the same thing multiplied by b?

[regal herald]

yes which is your integrand

your integral will give you e^bx evaluated over that interval, this follows from the FTC essentially

[pearl pond]

@sand axle Has your question been resolved?

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[midnight haven]

tips for 11

we have that second grade equation

m belongs to R

x1 x2 are real roots of the equation

at 11 it is asking me for the sum of the roots x1 +x2 belongs to which interval

now i know this

but doesnt help me

cuz i get

and then if that's the case then it would be C) R

because m belongs to R

is that correct?

[calm wing]

huh?

nvm yes

[midnight haven]

??

[midnight haven]

?

[calm wing]

i think "real" is the hint

the discriminant must be greater than 0

$b^2 - 4ac \geq 0$

[midnight haven]

?

why complicate

that's complicate

we know from viete's

x1 +x2 = s

[calm wing]

yes

[midnight haven]

-b/a

and then boing

we got m

[calm wing]

but those roots can be complex

your roots are said to be real

[midnight haven]

yes i know

[calm wing]

so this needs to be true

oh sorry

like that

[jolly parrot]

artemetra

[calm wing]

in any case it won't hurt to check this

[calm wing]

yeah so what if m is set to such a value that the roots are complex?

it breaks the conditions

[midnight haven]

what is this

wait

no this makes sense

so its

m >= 0

and m >= -4

so that means what

[calm wing]

mhm

[midnight haven]

-4 infinity

[calm wing]

wait whut

[calm wing]

-m >= 0

not m

[midnight haven]

-0 doesnt matter

[calm wing]

m <= 0 then

[midnight haven]

OHHHHHHHHHHHH

eyah

fuck

good one

[calm wing]

it is B

[calm wing]

actually this is a bit wrong

[calm wing]

like the answer is correct but the solution isn't

[midnight haven]

why not

ho

oh

dont tell me

i shouldhave

got

like m

factor

so its like

wait

[calm wing]

$xy \geq 0 \iff (x \geq 0 \text{ AND } y\geq 0) \text{ OR } (x \leq 0 \text{ AND } y\leq 0)$

[midnight haven]

like this

yeah i took the 2 cases

[jolly parrot]

artemetra

[midnight haven]

yes i am aware of that

[calm wing]

ah

[midnight haven]

that's what i did

[calm wing]

i see

[midnight haven]

wait

is my -m factor wrong

should'

no its good

[calm wing]

so the other case must be true

cuz m can't be smaller than 0 and greater than 4 simultaneously

which is what you got

[midnight haven]

yeah wait

[calm wing]

i.e. $-m \leq 0, (m+4)\leq 0$

[midnight haven]

wait why is my factoring messed up

[jolly parrot]

artemetra

[calm wing]

which gives you solution B

[calm wing]

it is not

[midnight haven]

wait

this doesnt

make sense

cuz b its 4

not -4

[calm wing]

^

[calm wing]

it's what you got here

[midnight haven]

oh

so the sign just changes

because of that

[calm wing]

pretty much

[calm wing]

in any case you are supposed to consider this

[midnight haven]

okay now

if you may and i am not asking much

i need help on the other ones

not really help

just spotting the mistake

now i got 4 * m does that mean its 4 times the range of m

actually i messed up at the end

i put range of x1 +x2

but like if m is real

then 4 * m is also real

so

D?

[calm wing]

what is the first symbol

something squared

[midnight haven]

s

from x1+x2

s^2 - 2p

viete's formula

for x1^2 +x2^2

[plush bramble]

<@&268886789983436800>

[timid spindle]

automodded

[pearl pond]

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[visual jasper]

hi, can i get some help on how to get started on this pls

[fathom forum]

Hello guys, can anyone help me with a question pleae

[visual jasper]

bruh

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[digital hinge]

did my physics professor do this integral correctly

I do not think he computed the integral correctly

this is what I got

[pearl pond]

@digital hinge Has your question been resolved?

[digital hinge]

<@&286206848099549185>

[pearl pond]

@digital hinge Has your question been resolved?

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[sharp cobalt]

hello, i need some help trying to prove this

I know i must start out by declaring arbitrary integers x and y

i just am missing the logic to get to x-y=7, is there something else i have to define? am i allowed to let x-y =7 as well?

if anybody could give me a hint or nudge me along the right path i would appreciate it a lot

maybe i can say x - y is also an integer? not sure if this will lead anywhere im still struggling with the logic

[near echo]

this is not a true statement

were you told it is true?

[sharp cobalt]

...oh my god i misread the question

oops

i thought the second quantifier was E

i apologize, let me try to figure it out the other way

[sharp cobalt]

would this be a valid proof?

[near echo]

youd have to flip the order of the quantifiers, but yes this is fine

[sharp cobalt]

wdym flip the order

does de morgan's law make it so that its E y. E x. ?

[near echo]

it should be EyEx yes

[sharp cobalt]

oh interesting i didnt know that

[near echo]

or

wait maybe im tripping

[sharp cobalt]

i was taught the order never mattered

idk ill search it up

[near echo]

the order does matter

[vast berry]

why is this so

[sharp cobalt]

no i meant with de morgan's law and the flipping

[near echo]

ExAy =/= EyAx in general

[sharp cobalt]

in this specific instance the order doesnt matter bc the quantifiers are the same

[near echo]

(\forall x \exists y : x - y = 0) is true, but (\exists y\forall x: x-y = 0) is not

ill be back in a sec, i don't think you need to flip the order, i may have been tripping

[vast berry]

oh

[jolly parrot]

maximo

[sharp cobalt]

rq, im gonna keep this channel open in case i have another question lol sorry

[near echo]

you dont need to flip them

[sharp cobalt]

gotcha, thanks

okay, this was what i thought my previous question was

[near echo]

that looks good

[sharp cobalt]

now actually proving this is kinda slipping out of my grasp, im not exactly sure how to do it

[near echo]

wdym

you're done

[sharp cobalt]

do i just keep doing the algebra and solve for y?

huh

[near echo]

you showed there exists a y so that x - y = 7

[sharp cobalt]

oh, so all i needed to do was plug it in, and show it was an integer?

[near echo]

all you need to do is prove the statement

you have shown that given an arbitrary x

y = 7 - x, which is in Z, satisfies
x - y = 7

you have proved the statement

[sharp cobalt]

one sec sorry, im not sure how to finish the proof so im gonna write what i think down and post it here

[near echo]

your proof is done

what do you think you are missing

[sharp cobalt]

would that be okay?

i was just trying to format it properly lol

[near echo]

sure yeah

looks good

[sharp cobalt]

would this be a valid proof?

[near echo]

= -> =/= 7
you shouldnt be writing this

oh that's -7

[sharp cobalt]

yeah, sorry

[near echo]

np

[sharp cobalt]

i should make my handwriting a little neater

ty for your help btw

[near echo]

should be fine, though i wouldnt write the conclusions like that

i would either put the theorem in words, or just put the box

[near echo]

same goes here

this is, unless your professor/teacher wants it that way

[sharp cobalt]

its how my professor wrote it so i was just following her

but, im more interested in how its actually done

how would a proof work normally?

[near echo]

well it works like this

but you would typically use natural language, rather than symbols

as in, no symbolic quantifiers

[sharp cobalt]

oh interesting

okay

[near echo]

what you have written is also a proof, but formally you dont use the quantifiers is my point

[sharp cobalt]

ah okay

so it needs to be human readable ig?

[near echo]

to be published, typically yes

[sharp cobalt]

wait then what are these symbols and quantifiers even for

[near echo]

since the point of publishing is to communicate it

[near echo]

to formalize foundations

but eventually you forget about them for clarity

[sharp cobalt]

oh okay like in the context of research papers

that makes sense

it also makes sense why theyd teach us these symbols bc its pretty easily to flip them and follow analytical rules

oh, btw are you allowed to just straight up do this

just pick any values to prove it right

[near echo]

if the quantifiers are "exists" then it is up to you to show existence

picking a value that works is sufficient, yeah

[sharp cobalt]

gotcha thanks

i have a couple more proofs to do sorry, thank you so much so far for your help

this is the logic i followed for this, i always feel wrong just choosing a random value but i mean ig it works? would it?

[near echo]

remember

[sharp cobalt]

alr, thanks lmao

[near echo]

if the quantifier is existence

then you get to pick the value

[sharp cobalt]

yeah, what trips me up is that you dont have to set it to a constant, you can include the arbitrary other value in that value

which is helpful but also a little weird to think about

[near echo]

not in this case

since the quantifier for y is after the one for x

[sharp cobalt]

oh i didnt think about that, yeah

so its kinda like local scope lol

[near echo]

sort of, yes

[sharp cobalt]

going back to the scope thing, would this work

[near echo]

this works, but the statement is false

y = 7/x makes a hasty assumption

[near echo]

by "this works" i mean your choice is fine to make, but the argument isn't true for all x

[sharp cobalt]

wait what

all x in rational numbers?

[near echo]

yes, think about it for a bit

[sharp cobalt]

oh, 0

okay im just gonna negate this rq, thanks lol

here are my final proofs, lmao

[near echo]

why is y + 1 > y?

[sharp cobalt]

yeah, i realized i could just set it equal to y

[near echo]

you should also state that if y in Q, then y + 1 in Q

[sharp cobalt]

oh

[near echo]

you cannot since y /> y

[sharp cobalt]

prolly less steps to just set it to y then

[near echo]

same as before, why is y - 1 <= y?
in either case consider what happens when you subtract y from both sides

[sharp cobalt]

i thought it just has to satisfy the condition? y <= y

[near echo]

for that one yes

for the original you need strict inequality

[sharp cobalt]

umm how do i prove this

[near echo]

.

[sharp cobalt]

ohhh i see you just end up getting constants on both sides which represent a true statement, -1 <= 0

so i need to show that?

[near echo]

ye

i would, unless you've explicitly shown in class that x + 1 > x in Q/R/Z/whatever

[sharp cobalt]

is this a valid way

to define this

[near echo]

what is the context

[sharp cobalt]

the second problem

[near echo]

is it the second q

yes that is fine

[sharp cobalt]

gotcha, thanks

thank you so much for your help, do you have any resources you used to get better at proofs

my class just barely scraped the surface of it but im very interested in learning more

past the basic methods, we just covered contradiction and soon induction but i heard if you can get good at proofs you'll be much better off for a career in math

[near echo]

theres tons of good books. if you want good suggestions either ask in #book-recommendations, check #books-old, or look up "logic and proof books math stack exchange"

[near echo]

assuming you are talking about pure math, yes, it is essentially all proofs

[sharp cobalt]

alright, thanks again

.close

[pearl pond]

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[sharp smelt]

a,b,c are in AP, then find $\begin{vmatrix}
x^2+3 & x^2+4 & x^2+5 \
x^2+4 & x^2+5 & x^2+6 \
x^2+a & x^2+b & x^2+c
\end{vmatrix} $

[jolly parrot]

Why am. I here

[sharp smelt]

I'm guessing I start with a few ERTs

so $R_2-R_1$

[jolly parrot]

Why am. I here

[sharp smelt]

so that simplifies to

$\begin{vmatrix}
x^2+3 & x^2+4 & x^2+5 \
1& 1& 1\
x^2+a & x^2+b & x^2+c
\end{vmatrix} $

[jolly parrot]

Why am. I here

[sharp smelt]

now I think $C_1-C_2$?

[jolly parrot]

Why am. I here

[sharp smelt]

which would be$\begin{vmatrix}
-1 & x^2+4 & x^2+5 \
0& 1& 1\
a-b & x^2+b & x^2+c
\end{vmatrix} $

[jolly parrot]

Why am. I here

[sharp smelt]

now do I do $C_2-C_3$?

[jolly parrot]

Why am. I here

[sharp smelt]

which would be $\begin{vmatrix}
-1 & -1 & x^2+5 \
0& 0& 1\
a-b & b-c & x^2+c
\end{vmatrix} $

[jolly parrot]

Why am. I here

[sharp smelt]

again $C_1-C_2$

[jolly parrot]

Why am. I here

[sharp smelt]

$\begin{vmatrix}
0 & -1 & x^2+5 \
0& 0& 1\
a-2b+c& b-c & x^2+c
\end{vmatrix} $

now what do I do?

[prime bramble]

wait, what are you asked to do?

[acoustic path]

compute the determinant

[prime bramble]

oh i see

[acoustic path]

bottom left becomes a - 2b + c

[sharp smelt]

huh, how

ah

ok

yeah, got it

[jolly parrot]

Why am. I here

[sharp smelt]

but a+c=2b

so that becomes 0 too

$\begin{vmatrix}
0 & -1 & x^2+5 \
0& 0& 1\
0& b-c & x^2+c
\end{vmatrix} $

[jolly parrot]

Why am. I here

[sharp smelt]

are there any more transformations I can apply, or is that it?

[acoustic path]

what is the determinant of this matrix

[sharp smelt]

0, I think

[acoustic path]

yes

if a matrix has all elements in a row or column as 0, the determinant is 0

[sharp smelt]

oh yeah, tysm!

[acoustic path]

but there is smth i want to note

[daring wedge]

The resulting answer will be zero as the first column has all zero entries

[acoustic path]

thinking along the lines of arithmetic progression, what's special about these rows?

[sharp smelt]

$x^2+a, x^2+b, x^2+c$ are also in AP

[jolly parrot]

Why am. I here

[acoustic path]

that is correct

anything else?

[sharp smelt]

nothing else comes to mind

[acoustic path]

i c

what about the other rows?

[sharp smelt]

they are in AP too

with a common difference of 1

[acoustic path]

right, then all 3 rows are in arithmetic progression

by now, you might believe that, in general, when the rows or columns of a 3x3 matrix are in arithmetic progression, the determinant is 0

^ (this is a true statement)

then i now ask you if you have any idea why this is true or how it can be proven

[sharp smelt]

That can be proven using these row transformations for any general det , I think

[acoustic path]

more or less

[sharp smelt]

Thanks!

I'll close this now, Have some work to do

.close

[pearl pond]

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[pure belfry]

are my diagram and calculation sufficient?

[rocky bay]

Pretty much

[pure belfry]

what do i have to do to make it fully accurate

i added the angle theta between u and v

[rocky bay]

Like write that component of vector v along vector u is vcostheta

[pure belfry]

ah okay

[pearl pond]

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[mystic linden]

hello

[mystic linden]

so i did the proof of this question, I did prove it. But i'm not sure whether I did it the right way or not. Can someone give me the proof?

[autumn trellis]

Send your proof and someone will check it

[sacred crest]

Just post your question and proof

[mystic linden]

soo it was a question from my exam

and the proof is in the ans sheet

uh

wait i can give it

[midnight haven]

its fine

we can work on it

lets start with the stuff inside the parentheses

[midnight haven]

We cant give you direct answers no matter what. We can reconstruct the solution if you want to

[autumn trellis]

Why not check the proof if they have it?

Just need to wait for a few moments

[midnight haven]

because im assuming they dont have their solution because it was an exam, and they are going to share the reference solution or something

[mystic linden]

EXACTLY!!

[mystic linden]

so this is what I did

[autumn trellis]

The power of x is wrong in the first step

[mystic linden]

what

[autumn trellis]

In all the parenthesis

[mystic linden]

👀

?uh

[autumn trellis]

$\left(\frac{x^{q+1}}{x^{p+1}}\right)=x^{q+1-(p+1)}=x^{q-p}$

[mystic linden]

it also has another one raised to the whole power

uh

[autumn trellis]

Yes, but this part is wrong is what I'm trying to say

[autumn trellis]

It won't be q+1-p+1

It will be q+1-p-1

[mystic linden]

how

[autumn trellis]

Are you aware of the fact that $\frac{a^{m}}{a^{n}}=a^{m-n}$?

[jolly parrot]

smidgin

[mystic linden]

yeah

i did the same thing

but why is there a minus after p??

shouldnt it be p +1??

[autumn trellis]

Use this property, with a=x, m=q+1 and n=p+1

[autumn trellis]

You will get this

[mystic linden]

what

[autumn trellis]

If you do this, you will get $\frac{x^{q+1}}{x^{p+1}}=x^{(q+1)-(p+1)}=x^{q+1-p-1}$

[jolly parrot]

smidgin

[mystic linden]

oh....

fuck

that was a 3 marker

😭

soo

can you give me the proof?

[autumn trellis]

How did the 2 come out of the brackets??

(q-p+2)(q+p)≠2(q-p)(q+p)

[autumn trellis]

No, first start by simplifying the stuff in the parenthesis properly and show us what you get

[pearl pond]

@mystic linden Has your question been resolved?

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
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[vestal venture]

i dont get these coordinate type qns

[indigo pewter]

In how many ways can mr A get to work

[vestal venture]

uh 20?

im guessing 4*5

[indigo pewter]

how many moves does Mr A have to make

To get to work

[vestal venture]

Oh

9

no wait

im trippin

[indigo pewter]

Ur right

He has to make 9 moves

4 right, and 5 up

[vestal venture]

but wouldn't it be the least

he can go different paths and then to 4,5

[indigo pewter]

he can't get there in less than 9 moves

right?

[vestal venture]

yes

[indigo pewter]

So for any arrangement of those 9 moves, we will get one of the shortest paths

[vestal venture]

yes

[indigo pewter]

so how may ways can you arrange RRRRUUUUU

[vestal venture]

Oh damnnnn

[indigo pewter]

whoops

[vestal venture]

9!/5!4!

[indigo pewter]

yes

[vestal venture]

i never thought it this was

[indigo pewter]

which is also 9choose5

or 4

[vestal venture]

much easie

yeee

[indigo pewter]

So there are a total of 9c5 ways for mr a to travel to work

[vestal venture]

yes

[indigo pewter]

How many ways can he go to his friends house

[vestal venture]

5c3

[indigo pewter]

ye

[indigo pewter]

And then he also has to go to work from there

[vestal venture]

4c2

So

5c3*4c2

[indigo pewter]

yes

so its 5c3*4c2/9c5

[vestal venture]

60/

30/63

10/21

Ay

Thanks

.close

[pearl pond]

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[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[finite lily]

hello, im having trouble with this antiderivative problem

[finite lily]

,rotate

[finite lily]

this is the problem

i get an answer of 109.71 ft but the anser is 225 feet

idk how they got that thop

[sharp smelt]

hmm, mind if I solve this in metric to confirm?

[finite lily]

yes no problem

[sharp smelt]

,w 120ft to m

[jolly parrot]

Results provided by WolframAlpha

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

Requested by physicsrocks

[sharp smelt]

ok, so did you use conservation of energy here

and what value of g did you take?

[finite lily]

no this is a math problem

its an antiderivative

i took g = -32ft/s

[sharp smelt]

not too sure, sorry

[finite lily]

its ok 👍

[pearl pond]

@finite lily Has your question been resolved?

[finite lily]

<@&286206848099549185>

.close

[pearl pond]

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[jolly parrot]

Couldn't find an attached image in the last 10 messages.

Feb 24, 2024, 7:01 PM

[pearl pond]

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[steep lion]

@finite lily

[tribal geode]

should've used v^2=u^2+2as

[steep lion]

Yeah 🥲

.close

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[dusky ferry]

help with this

how shld i get started?

sin=y/r

sin^-1(r/y)

Not sure

[subtle ginkgo]

use it, that is all

[midnight haven]

In need of dire help
In dire need of help

[dusky ferry]

alright

[midnight haven]

Yo

[dusky ferry]

yes i am indeed 😭

[midnight haven]

Easy way

Wait

[dusky ferry]

yes

[midnight haven]

Do like this
Will save time

Don't mind anything and everything else

Did you understand kid?

@dusky ferry

[dusky ferry]

looking

no not really

[midnight haven]

K

Okay so

sin inverse something

Gives you an angle right

!?!?!?!!?

[dusky ferry]

yes

[midnight haven]

Good

So just make a triangle

Using whatever you said r/y shit

Idk

Then use that

Angle to get tan

The que is like
Get the feel

[dusky ferry]

no calculator

[midnight haven]

tanx
Where
x = sin(-1) {1/5}

[midnight haven]

Did i use a calculator

[dusky ferry]

just asking

[midnight haven]

No no calculators

Dw

[midnight haven]

???

[dusky ferry]

ai lemme do it

[midnight haven]

K

[dusky ferry]

howd u find root(24)

[midnight haven]

Pythagoras

a²+b²=c²????

When you dont reply i imagine you sitting on a chair and going

[dusky ferry]

lol

ok got that

yeah

[midnight haven]

Yeah

[dusky ferry]

tan=y/x

[midnight haven]

Steps to do this que

Idk what y r x youre doing

So

Explain

[dusky ferry]

sohcahtoa

[midnight haven]

Hypotenuse is r?

The base is x

[dusky ferry]

yes

[midnight haven]

Vertical is y

?

Ok

[dusky ferry]

yres

[midnight haven]

So

Steps

1. given que get feeling

2. tanx where x = sin inverse 1/5

3. make triangle

4. use Pythagoras if necessary

5. now use triangle to find answer

Okay?

[midnight haven]

Yes

[dusky ferry]

que ?

[midnight haven]

Bravo for writing a = ±√24

[dusky ferry]

whats this que

[midnight haven]

Remeber the ±

They can trick you in multiple correct questiosn

[dusky ferry]

bett

[midnight haven]

Question

Is que

Should i give you

One more for practice?

[dusky ferry]

so uh should it be negsative or positive or both?

like the final asnwer

[midnight haven]

You have learnt sin cos tan cot cosec sec?

[dusky ferry]

yes

[dusky ferry]

yeah everyhting

basically

[midnight haven]

Gud

[dusky ferry]

but forgotten most 💀

[midnight haven]

Ok

Imma sleep anyway

[dusky ferry]

so revising/getting back up to sped

[midnight haven]

After giving you the que and ans

[dusky ferry]

question

tyyy

[midnight haven]

Canbe both

±1/√24 is correct

Go on practice imma sleep

DM me for answers prolly tomorrow

[dusky ferry]

bettt

will do

ty

[midnight haven]

K

Welcome

Gn

[dusky ferry]

gn

[midnight haven]

.close if youre done

First ones cot

Btw

[dusky ferry]

alright

[midnight haven]

Cot(cos^(-1) shit)

Yeah

Bye

[pearl pond]

@dusky ferry Has your question been resolved?

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[hazy arrow]

where does the 2 come from and the root 3/2 come from

[plush bramble]

Draw a picture

[hazy arrow]

i did

but the thing is

when i did it

i always had a missing side

[plush bramble]

Show

[hazy arrow]

Ok let me re do cause my diagram loools horrible

[hazy arrow]

im missing a length

[pearl pond]

@hazy arrow Has your question been resolved?

[pearl pond]

@hazy arrow Has your question been resolved?

[hazy arrow]

.close

[pearl pond]

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[tardy delta]

How do I do this type of question

[cursive wraith]

work inwards to outwards

can I just say horrific way to "write" a function

[cursive wraith]

anyways back to the point

inwards to outwards

so we start with w(-2)

[tardy delta]

Yeah idk this was on my test that I missed, but they just gave it to me for practice

[cursive wraith]

what's w(-2)?

[tardy delta]

Would it just be the values in w(x) multiplied by 2

-2

[plush bramble]

w is a function

w(-2) is a number you get by plugging in -2 to the function w

[tardy delta]

Ohh

[tardy delta]

Ok I get that but there’s usually a variable like an x to be plugged into the w,there’s no x except in g

I see that I have to work from w to f

[plush bramble]

You have to answer this first

g(w(-2)) is the function g evaluated at w(-2)

And so on for f(g(w((-2))))

[tardy delta]

Yeah so I need to find what w is to find. G

[plush bramble]

You're given w

[tardy delta]

-2?

-2 is w?

[plush bramble]

No

[plush bramble]

Read this again

[tardy delta]

Ok so when I plug in -2 into the function whatever that is

Is going to be W right

?

The thing I’m confused about is how am I suppose to plug in -2 when there’s no x??

[plush bramble]

Find where there's a -2 in the definition of w(x) and guess

[tardy delta]

Is it 5

W=5 since -2 hits that point

??

-2,5

[cursive wraith]

w(-2) = 5 yes

because the point (-2,5) is in the definition of w

[tardy delta]

I Think f =-1 since 2 hits -1

[cursive wraith]

f(2) = -1 yes

[tardy delta]

Ok that makes sense

Idk why I wrote f=2

[cursive wraith]

maybe intending for f(2)

[tardy delta]

Ohh

Ok so for these types of questions I have to look at where the specific value that’s given hits a specific number in the function

And whatever number it hits, I choose the y

[pearl pond]

@tardy delta Has your question been resolved?

[pearl pond]

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[pale stag]

Hi

[pale stag]

I need help

@limpid lily

With number 10

Anyone here?

[pearl pond]

@pale stag Has your question been resolved?

[pale stag]

close.

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

Now obviously n=1, results in a perfect square

[sharp smelt]

but how do I prove it's the only such integer

[vital estuary]

wicked

i have a simpler challenge for you: can you prove n must be even if n>=2? i think that will help

[sharp smelt]

n must be even if n>2?

my answer key states there's only one such integer though

[vital estuary]

it might help to narrow down your search though

[sharp smelt]

I still don't get it. how can $n \geq 2$ when n can only be 1

[jolly parrot]

Why am. I here

[vital estuary]

you assume n can be greater than 1 for now

[sterile flame]

have u heard of moudlo 4

[vital estuary]

you will disprove it

[sharp smelt]

amodb yes, modulo(n),no

[vital estuary]

its the same thing

[sharp smelt]

hmm, if n is even, then it can be expressed as

[sterile flame]

so yh perfect squares are either 0 or mod(4)

[sharp smelt]

$\left(2^{\frac{n}{2}}+7^{\frac{n}{2}}\right)^{^2}-2\left(14\right)^{\frac{n}{2}}$

[vital estuary]

ehh no

thats a really weird way of doing that

for number theory, try avoiding using fractions as much as you can

[jolly parrot]

Why am. I here

[sharp smelt]

ok, so how should I express it

[sharp smelt]

how?

[vital estuary]

try showing that n cant be odd unless its 1

using quadratic residues mod 4

[sharp smelt]

what;s that?

quadratic residue ?

[vital estuary]

oh

remainders of squares mod 4

[sterile flame]

not sure what u mean by how but the key rule here is a square root of mod(4) would be odd

[sterile flame]

this guys smart gn

[brittle tinsel]

I would just do the mod 4 stuff tbh, just show using induction that 7^n is ... mod 4 and 2^n is ... mod 4

[sharp smelt]

so we take the mod, square it and find the reminder when divided by the same number