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[chrome geyser]

Hello, it's me again, how do I solve this question? In advance thank you very much.

[chrome geyser]

Volume of rhombus and square are the same. What is the sharp angle of rhombus if it's area is two times smaller?

[ornate condor]

um

[gray isle]

volume of a 2d shape?

[ornate condor]

squares and rhombus dont have volume

[keen plinth]

mmm but maybe they mean area

[gray isle]

but then what area is two times smaller

[keen plinth]

could be a language barrier issue

[gray isle]

think one of those is supposed to be perimeter

[chrome geyser]

Yes.

How do I calculate the angle if I don't have any data?

[gray isle]

is this translated?

[chrome geyser]

For example it isn't written how much a contains.

[chrome geyser]

It is because I'm from Serbia.

[lone heart]

@chrome geyser Has your question been resolved?

[chrome geyser]

Not yet.

I will translates better.

I found that (a root 3) /2 = 2a, and I've got idea that sin 60 = root 3 / 2.

[lone heart]

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[gray heart]

Why is i^3 equal to -i but i^45 is positive?

[minor needle]

because i = sqrt(-1), hence i^2 = -1

so (-1) * i = -i

and i^45 is "positive" because:

$$i^{45}=i^{44} \cdot i = (i^{2})^{22} \cdot i = (-1)^{22} \cdot i = i$$

[ocean seal]

Modus

[gray heart]

thanks for the explanation

.close

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[misty sluice]

Can I ask how did the numerator turn to -5? When I solve it, it will result to 1.

[abstract fractal]

Show your work

[misty sluice]

du/dy*

[abstract fractal]

You might want to properly place parenthesis

(3y - 1)(2) instead of 3y - 1(2)

[lilac nest]

You should use the parentheses properly; anyways, the mistake is that there was a minus sign in the middle but you didn't change the sign of (1)3, leaving it as +

[abstract fractal]

The issue is a distribution error, likely caused by the lack of parenthesis

[misty sluice]

ohh thxx ill keep this in mindd

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[mellow tusk]

cuz

chain rule

what did u try

yh use this

[ocean seal]

Curium0x247

Curium0x247

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[fierce prairie]

what is your question?

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[ember pelican]

Yello I am just confused how how -2pi,pi would work in this question

[minor needle]

do you know how to solve tan^2(s) = 1 for s? [-2π, π) is only an interval

[ember pelican]

Yep

Just don't understand the interval

[minor needle]

so what's your solution

[ember pelican]

Tan =1

[minor needle]

but that's not a result

result should be s = ...

[ember pelican]

S = 1

[minor needle]

nope

it's wrong

[ember pelican]

For the final answer? That's why I wanna know how [-2pi,pi) works

I get how [0,2pi) works

It goes around once

Does [-2pi,pi) do that but backwards?

And if so does that change the answers?

[lone heart]

@ember pelican Has your question been resolved?

[ember pelican]

Nothing I do seems to work

[minor needle]

you know tan^2(s) = 1 has infinitely many solutions, yea? you're supposed to choose these ones which belong to the given interval

[ember pelican]

The problem IS the interval

[ember pelican]

This is what I put down as the answer and it's wrong

But according to the unit circle it really shouldn't be?

[minor needle]

but you've said s = 1

[ember pelican]

Yes after I square it

I have to square tan and 1 first

[minor needle]

and now s = pi/2 etc. is also wrong, firstly u need to solve it correctly

[ember pelican]

Then do the rest

[minor needle]

then we can talk about the interval

[ember pelican]

I thought the circled ones where right?

[minor needle]

and btw. [-2π, π) means u take solutions which satisfy -2π <= s < π, so why did you take 2π

[ember pelican]

Oh I just replace it with 0

[minor needle]

tan(π/2) doesn't exist

look here

it means values refer to cos and sin

not tan

[ember pelican]

Yea I divides the 0 and 1,-1s

[minor needle]

so 0/-1 = 0 (not 1)

hence it's not a solution

[ember pelican]

Oh

Wouldn't the answer not exist then?

Nvm

[minor needle]

there are solutions to this, I'd give you a hint

$$\tan ^{2} s = 1$$
$$\tan s = -1 \vee \tan s = 1$$

[ocean seal]

Modus

[ember pelican]

Forgot 2sqr/2

[minor needle]

so now using your figure you should be able to find solutions in two first quadrants

basically in [0, π]

[minor needle]

two of them are correct

π/4 and 3π/4

interval ends at π
you have 7π/4 and 5π/4 (they're bigger than π)

[ember pelican]

Are there more answers?

Or just 2

[minor needle]

but you've found to 2π

just subtract 2π from 5π/4 and 7π/4

and then solutions will be in the interval

[minor needle]

and two more answers, yes

[ember pelican]

Got -3pi/4 and -6pi/4

[ember pelican]

I got it wrong but there are way more numbers than I thought

Nvm it just included negative pi/4

i subtracted wrong

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[dense wave]

Hello, this is a last resort after attempting to understand Z-scores. I am struggling to get a hold of a through f. 😭 I've searched everywhere online, I can't find an answer for anything. And if I try I keep getting frustrated and unsure. If anyone could just explain z score / how to calculate the percentages between long distance numbers that would be great. Thanks.

[tacit arch]

2f?

Did you do a-e already?

[lone heart]

@dense wave Has your question been resolved?

[dense wave]

I could do 2f , sorry, my apologies! It's just doing the percentages.. I'll send a pic of the problems I'm struggling to understand. I haven't done them yet. I'm finding difficulty on how to start

[tacit arch]

Pick one and attempt it

Probably start with a

What are you stuck on? Should just be reading off z table values

[dense wave]

Just trying to find the percentages between the numbers on the empirical rule

[tacit arch]

Write 60 and 80 in terms of mean and std dev

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[short veldt]

ab/a+b=2, ac/a+c=5, bc/b+c=4

[short veldt]

how do I solve this

[last ether]

Use parenthesis if needed

If that's what you're given verbatim, you can cancel out terms

Like ab/a = b, a≠0

[ornate condor]

lol

im guessin its not

[last ether]

I'm guess that either but you never know

[ornate condor]

lazy notation~

[short veldt]

yea sorry I forgot the parenthesis

ab/(a+b)=2, ac/(a+c)=5, bc/(b+c)=4

so how should I do it

[ornate condor]

now

that looks like a proper question

lol

so

solve for a,b,c?

[short veldt]

yes

[ornate condor]

hm

work with the first eq first

so

assume a+b is not zero

we multiply both side by that

ab=2(a+b)

cool?

then from there u can write a in terms of b

try doing that

[short veldt]

alright

then?

[ornate condor]

yea just

write the 2nd equation in terms of a

[short veldt]

I did both

[ornate condor]

then use the 1st equation to convert the a's into b's

then the last equ also just

just write all in terms of b

lol

[short veldt]

I don't get what you mean

@ornate condor

[ornate condor]

so like

each equation can be written in terms of

anyway

what have u gotten so far

[short veldt]

a=(ab-2b)/2

and a=(2a+2b)/b

and ab^2-2b^2=4a+4b

@ornate condor

[ornate condor]

um

the

first equ

has to be

all the variables a on 1 side

[short veldt]

oh

how do I do that

[ornate condor]

ab=2(a+b)

ab = 2a+2b

ab - 2a = 2b

a = 2b/(b-2)

a = 2 + 4/(b-2)

[short veldt]

2+4?

[ornate condor]

$a = 2 + \frac{4}{b-2}$

[short veldt]

I got confused after ab-2a=2b

[ornate condor]

we divide both side by b-2

on the left side we can write

ab-2a = a(b-2)

[short veldt]

oh

[ornate condor]

ok actually

my method can solve it

but

i think

it takes a bit too long

leme

think of another

[short veldt]

a = 2b/(b-2)
how did you get that

[ornate condor]

yea sorry

i

found

a

way faster method

so sorry

hahaha

so

um

ab/(a+b)=2, ac/(a+c)=5, bc/(b+c)=4

for the first equ

we can take its inverse

so

$\frac{ab}{a+b}=2$

[ornate condor]

$\frac{a+b}{ab}=\frac{1}{2}$

[ornate condor]

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$

[ornate condor]

got it ?

@short veldt

[short veldt]

uh not really

[ornate condor]

which bit

[short veldt]

yea I understand the steps

but not sure how that helps

[ornate condor]

so

we can repeat the same for each

and we already know how it goes so we get

$\frac{1}{a}+\frac{1}{c}=\frac{1}{5}$

[ornate condor]

$\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$

[ornate condor]

can u see how to do it now

[short veldt]

I'll try

ok I'm pretty sure I can replace 1/a, 1/b, 1/c, with other variables?

[ornate condor]

yea

its like

https://tenor.com/view/substituir-kakashi-hatake-naruto-clássico-gif-18550808

substituition jutsu

[short veldt]

ok that's what I learned in class ty

[ornate condor]

[short veldt]

whats that

[ornate condor]

just

substituition

just throwing in a meme

[short veldt]

ok lol

[ornate condor]

[short veldt]

thanks I got it!

.close

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[candid compass]

How do I go about answering 5?

[cold hinge]

@candid compass

use this rule

you can then simplify some of the radicals

do you see how

[candid compass]

ah yes i do now ty

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[graceful bobcat]

Hel

[graceful bobcat]

Me

Help

[real gazelle]

what's your question

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@graceful bobcat Has your question been resolved?

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[graceful bobcat]

Uh

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[cunning trout]

How do i write vector AB?

[ornate condor]

ok so

to get that we take

B - A

another way of saying it is

with O being the origin

[cunning trout]

How do you know we minus B from A?

[ornate condor]

AB = AO + OB

[median dirge]

Draw position vectors.

[ornate condor]

yea so like

like taht

AB is that

then its the same as

AO + OB

[median dirge]

Umm

[ornate condor]

and AO is the same as just -OA

or negative A

[median dirge]

The origin is the initial point of both vectors.

[ornate condor]

yea u could

do that too i guess

lol

just tryna

find another way to explain but idk it complicates it a bit :c

[median dirge]

Yeah does work tho

[cunning trout]

Hmm i see

[median dirge]

So basically when you draw a vector between two points. There are two ways. Either subtract the position vectors of those points, or do it this way

[cunning trout]

So with questions like these, we subtract?

[ocean seal]

Muhammad Hussaini

[median dirge]

Which is B - A

[cunning trout]

I’m still confused with how we get B-A

Let me try understand

[median dirge]

See

B vector would be what?

And A vector would be what

[cunning trout]

Ok I understood now, thanks

.close

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[sonic raptor]

If I am on an island where one hour equals one week in normal time, and I am on that island for 71 years normal time, how many days am I on that island in island time?

[abstract fractal]

What have you tried

[sonic raptor]

its not years divided by hours

i think

idk

i really havent tried alot

im stumped

i dont even know where to start

[abstract fractal]

We have a way to convert from weeks in normal time to hours in island time

We have 71 normal years

How many weeks is that?

[sonic raptor]

3702

point 14

[abstract fractal]

How did you get that?

Nvm

[sonic raptor]

LOL

[abstract fractal]

Fuckin leap days

[sonic raptor]

ignoring leap days

[abstract fractal]

Ignoring leap days?

Nvm again

Why isn't there an even 52 weeks

Whatever. It's 3702.14 normal weeks

And 1 normal week = 1 island hour

So how many island hours is that?

[sonic raptor]

3702.14 hours

[abstract fractal]

Yes

How many days is that?

[sonic raptor]

154.256

[abstract fractal]

Yes

Tada

[sonic raptor]

thank you sm

i wish i was good at complex math

😭😭😭

tysm again

.close

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[gray isle]

express the pairs of fractions with the same denominator

then it would be clear which one is greater or if they're equal

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@alpine sable Has your question been resolved?

[gray isle]

you need to answer the bot by commenting

or it'll close on you

you'll also need to explain where you're still stuck and/or which part of the explanation you don't understand

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[ocean seal]

marc

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@zinc cosmos Has your question been resolved?

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@zinc cosmos Has your question been resolved?

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[pastel bane]

Im very stuck on the inductive step for 2 part b and would greatly appreciate any assistance

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@pastel bane Has your question been resolved?

[echo socket]

What's r though?

[pastel bane]

r is the remainder in the Euclidean algorithm. Like a=bq+r but as a sequence of steps until the r becomes zero and the algorithm terminates

[pastel bane]

<@&286206848099549185>

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[pastel bane]

.close

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[quartz cave]

Why can I not take the square root to simplify this further?

[minor needle]

you can, but it won't simplify it

[wise oracle]

you have everything you need to know

[quartz cave]

Hmm alright

[wise oracle]

radius is 5 and it moves 3 on the x and 2 on the y

[quartz cave]

I thought removing as much as possible is always recommended

[wise oracle]

what was the question

[quartz cave]

Using the distance formula to find the equation for the given radius at a certain point

But when I worked it out I took the square root here

[wise oracle]

yeah well your radius is root 25

[quartz cave]

And the tutor stopped at this point

So getting to this point of the equation is good enough?

[wise oracle]

yeah im actually not sure why it doesnt go more forward

maybe it separates it from other functions

im not too sure

[quartz cave]

Hmm gotcha

Modus says it does not simplify anything

So I guess it is a aesthetic difference

Cant imagine I would being wronged for taken the square root on any test

[wise oracle]

yeah i guess it depends on the question

do you know probability

[quartz cave]

probability is left out at the moment

[wise oracle]

how old are you

because in australia we do what do in uni in high school

[quartz cave]

haha too old for this

I am a programmer but am trying to get into AI

So I am spending a lot of my free time fixxing up my math skills

This is also high school stuff over here, pretty early aswell if I remember correctly😂

[hollow shale]

You could, but then you get two functions and thus two equations

Nicer to have both solutions as one equation

[quartz cave]

How would it turn in 2 functions?

[hollow shale]

x^2 = 9

=> x = ± 3

If you model y as a function of x, or vice versa, then there are two functions for y that satisfy that equation

[quartz cave]

You would end with (x - 3) + (y - 2) = 5 right?

[hollow shale]

y = 2 ± sqrt(25 - (x-3)^2)

Nooo.

You end up with 5 = sqrt((x-3)^2 + (y-2)^2)

[quartz cave]

Ah my bad

[hollow shale]

You can't sqrt each term

[quartz cave]

That was where I was going wrong then

[hollow shale]

sqrt(a) + sqrt(b) ≠ sqrt(a + b)
but
sqrt(a)*sqrt(b) = sqrt(ab)

[quartz cave]

Taking the square root of an exponent does not reduce the exponent

[hollow shale]

Euh, if you have one term, then the exponent is divided by two.

Otherwise, you can't split the sqrt like that

[quartz cave]

Aha

So if there are multiple terms

Then you cannot take the square root of the whole side

That makes sense then

[hollow shale]

You can, but not of each term

[quartz cave]

No right right

You can

But it will get messy

[hollow shale]

You apply the whole thing to both sides

[quartz cave]

Right

On the 25 it would be nice

But on the other side you would end up with something more complicated than before

correct?

[hollow shale]

Yes

[quartz cave]

Thats what I needed to know

Thank you

.close

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[alpine sable]

ABC is a triangle.

D is mid point of side BC.

[desert tusk]

is line AD supposed to be straight?

[alpine sable]

BD=DC

[alpine sable]

Yes. It is a line joining A and D.

How do we find the value of angle "x" ?

[lone heart]

@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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[weary wyvern]

Yes

Search sum of squares formula

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[sharp moat]

no hopital

[vale wigeon]

$\lim_{x \to 1^+} \frac{\pi - 2 \arctan(\frac{1+\sqrt{x}}{\sqrt{x-1}})}{x-1}$

[ocean seal]

Ann

[vale wigeon]

this?

[sharp moat]

Yea

[vale wigeon]

first most reasonable step seems to be to substitute t := sqrt(x-1)

[sharp moat]

also the numerator is a function and the image of 0 is 0 according to the exercise

k

[sharp moat]

but using the definition of derivatives won’t help

[vale wigeon]

should get $\lim_{t \to 0^+} \frac{2}{t^2} \paren{\frac{\pi}{2} - \arctan\paren{\frac{1+\sqrt{t^2+1}}{t}}}$

[ocean seal]

Ann

[vale wigeon]

does this help us at all?

not sure but it looks promising

[sharp moat]

Maybe we can use arctanx + arctan 1/x = pi/2

nvm

I think I know what to do

I got -inf

But it’s not true

I got it

thanks

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[modest tulip]

I have to proof that this equation(first ss) is a solution of the following equation(second ss), for context this is an equation for the launch of a rocket while neglecting air resistance and changing g. M= mass in kg, v is speed in m/s, u is de speed of the exhaust gasses in m/s in function of the rocket, g is a constant 9.81 m/s²

[modest tulip]

i think the best way to start is to do the derivative of best side ofthe equation of the first ss

[slender marten]

If it's a solution then differentiating v(t) with respect to t and substituting it into the left side of equation (2) should give the right side.

[modest tulip]

this is what I have right now but I'm a bit clueless right now

I simplified it a bit

[slender marten]

It might have made more sense to write ln(M(0)/M(t)) = ln(M(0)) - ln(M(t)) to simplify calculations. But not necessary to do, of course.

[modest tulip]

yh but I have to get rid of the ln

[slender marten]

Writing it as
v(t) = u_out[ ln(M(0) - ln(M(t)] - gt
and taking it's derivative with respect to t gives
v'(t) = -u_out[M'(t)/M(t)] - g.

[modest tulip]

yh that's what i hae here

[slender marten]

True.

[modest tulip]

but i don't know how to go further

i'm close bc i already have v'(t), -u_out and m'(t) which i do need

the only thing I'm confused about now is how to get from -u_out/M(t) -g to -M*g

[slender marten]

$M(t)\left(-u_{\text{out}}\frac{M'(t)}{M(t)} - g\right) + u_{\text{out}} M'(t) = -Mg$

[ocean seal]

stabulo

[slender marten]

Just plug it into the LHS of this:

[modest tulip]

can you explain how you got to the +u_out*M'(t) = -M g part

the first part i get

[slender marten]

All I did was substitute the solution we got into the left hand side of the image.

[modest tulip]

oh yeah I get it thank you

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[ocean seal]

Chochan

[silver onyx]

should we do by doing each part

~Please Ping me while you answer so I can come :)) ~

[rigid smelt]

what operator is []?

[gray isle]

is that the floor function?

[modest tulip]

i think it's jsut supposed to be normal brackets no?

[rigid smelt]

depends on context

[gray isle]

that's why we're asking

[silver onyx]

greatest integer function

[silver onyx]

no

[modest tulip]

oh mb

[silver onyx]

GIV

[silver onyx]

MB means?

mean binomial exp?

[gray isle]

mb = my bad

[rigid smelt]

i suppose you want the domain of (floor(x))^2 - 5floor(x)+6, starts by finding the domain of floor(x) and then the domain of the whole thing

domain of floor(x) shouldnt be too hard

[silver onyx]

ooh

[silver onyx]

yep

[modest tulip]

just my bad lmao

[silver onyx]

that's [0,1) ryt

oops

[rigid smelt]

no...

remember how the function is defined

[silver onyx]

that was fraction, that's where I made mistake

[gray isle]

it also seems like you're mixing up domain and range

[silver onyx]

oh mb

[vale wigeon]

domain, range and solution set...

what is shown here is in fact an equation

[silver onyx]

yes

so we have to do part by part ryt!?

[vale wigeon]

what we have to do first and foremost is to understand what is actually ASKED of us.

and that is very unclear right now because of your poor restatement of the problem.

or maybe your book is just that shitty and asks for "range" when it means solution set.

[silver onyx]

ok if it is to find "x belong to ____"

[vale wigeon]

ok so you want to SOLVE the equation floor(x)^2 - 5 floor(x) + 6 = 0

yes?

[silver onyx]

yes

[vale wigeon]

right...

so there's some obvious factorization to be done here

(floor(x) - 2)(floor(x) - 3) = 0

[silver onyx]

ooh ye

then!?

[silver onyx]

is it [2,4)

[vale wigeon]

well that's where floor(x) equals either 2 or 3 so yes

[silver onyx]

oh

I put answer as [2,3], I am dumb lol

ok thanks

.close

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[silver onyx]

.reopen

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✅

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[alpine sable]

I need help with all of these

Even if i watched tutorials i keep scratching my head

[modest dragon]

do you know the formula of the elipse?

[alpine sable]

yes

x^2/a^2 +y^2/b for vertical

x^2/b^2 + y^2/a for horizontal

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[glad vessel]

bro anyone wanna peer review my shit

[glad vessel]

low brain hw

[ornate condor]

[minor needle]

$$\boxed{\bold{JUST \ ASK}}$$

[ocean seal]

Modus

[glad vessel]

ok bro

in R^2

its bad

because the one who did it has low number of brain cell

[lone heart]

@glad vessel Has your question been resolved?

[glad vessel]

peer review pls

shit bro

it lack so much brain cell no one willing to take a look

fuck this shit man

[silver badger]

where can i geht help

[sonic raptor]

#❓how-to-get-help

[silver badger]

thank you

[lone heart]

@glad vessel Has your question been resolved?

[glad vessel]

broo send help

[lone heart]

@glad vessel Has your question been resolved?

[glad vessel]

hell no

[lone heart]

@glad vessel Has your question been resolved?

[glad vessel]

stfu

[remote heron]

its unclear what you are asking ppl to do

if you are done with the exercise then what do you want 👁️

[glad vessel]

someone to check if it is correct

[remote heron]

on the off chance you got something wrong?

or what are you concerned about

[glad vessel]

ye bro

I found 2 typos already

@remote heron bro give opinion kurwa

[clever folio]

The stuff about vector magnitudes is wrong but they have the right idea

Idk if this is your work or not

But p=|p|(cos(a),sin(b))=(cos(a),sin(b)) only works when p is a unit vector.

So, whoever wrote this kinda fudged it when they said |p| or whatever factors out then proceeded to ignore it which leads to some kinda wong equations later.

But you can also just kind of avoid that issue by just saying you're only considering rotations of a unit vector and noting rotations of a non-unit vector can be found by scaling rotating then scaling back or whatever.

Also the last equation isn't true unless z is 1.

(Just my first impressions from a quick skim)

[glad vessel]

why u doubt that though

[clever folio]

Because you keep referring to the person who wrote this in the third person.

[glad vessel]

oh ok

ye bro I made it

@clever folio bro i dont many brain cell

[clever folio]

That is irrelevant to the question you're asking about.

[glad vessel]

bro you know flammy

[glad vessel]

.close

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[ruby vault]

i figure dout the answer to my problem but could someone explain to me why exactly people rearrange terms like this?

[ruby vault]

Like could someone somewhat explain the purpose of rearranging it like this

[ancient saddle]

I don't think there's a specific purpose in terms of getting the correct answer. It's probably just the algorithm they use

Maybe they want the x at the left

[rose sigil]

yea there are plenty of ways to do whatever you are trying to do, it doesn't have to be done like that

[ruby vault]

Oo oki tysm

how in the world do i close this now

.close

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[alpine sable]

4d is -2 over -2, i got n(-2 over -2) or n+1 (-2 over -2)

[alpine sable]

draw it out

gradient of AB is (0-4)/(-1-1)

so 2

if it is parralell the gradient of CD will be the same which is 2

[dusky turtle]

how tf is two minuses turn into one minus

[alpine sable]

can we just like

use an unoccupied channel

anyways u got CD?

i mean t

yeah

yes

it's okay

https://tenor.com/view/drunk-venti-gif-23048214

[alpine sable]

like if u draw a number line

[dusky turtle]

nvm i figured it out

nothing to do with numnber line or sum shit

[alpine sable]

kk

<@&286206848099549185>

sorry forgot to mention the last killer question under question4

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[rocky summit]

@prime badge

[rocky summit]

Now that I've had a second to stop moving, and I've gotten some food in me, I finally understand what you're referring to

I'm just not sure how I'm going to put it into words

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[lone heart]

Please don't occupy multiple help channels.

[hallow moth]

???

.close

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[sly comet]

Helpppp

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[vernal lance]

I don't even know where to start

[alpine sable]

do you know what quotient rule is

[runic trench]

id use power rule tho

[vernal lance]

is quotient rule the [ f'(x)g(x)-g'(x)f(x) ] / g(x)^2

I think I have to factor out x^2

[runic trench]

yeah do that firs

first

[vernal lance]

I ended up with this

now do I do sum rule?

[runic trench]

yeah

then differentiate each using power rule

[vernal lance]

https://tenor.com/view/batman-thinking-thinking-about-you-think-thinking-of-you-gif-20503421

[runic trench]

is that the final ans ??

[vernal lance]

I think so

[runic trench]

ik itll start with 9x^2

[vernal lance]

oh wait my bad I messed up with the math

[runic trench]

so fast ??

[runic trench]

i still have to spend some time to write on paper

ok im almost there

yeah its correct

[vernal lance]

woohoo

thank you so much

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[languid ermine]

help

[languid ermine]

Hey i dont understand this question

i dont understand counter examples at all

[deep grove]

i think all u need to do is give an example where the statement isnt true

for example

in a,

its not true if u take 4 and 4

[languid ermine]

can i write something like 5 and 5

for a

[deep grove]

no because then the area of the rectangle wont be 16

basically the statement says, if the area is 16, then the length will always be 8, u have to give an example showing that the length will not always be 8

[languid ermine]

would it work if I just swaped it

8 for the width and 6 for the length

[deep grove]

but 8*6 is not 16

[languid ermine]

oh

man

oooh i mixed

i meant swap the 8 and 2

8 for width and 2 for length

does it disprove that its has to be 8 for the length and 2 for the width

[deep grove]

that would work, but i don't know if the teacher would accept it or not, i guess its better to be on the safe side unless you dont have a choice yk

[languid ermine]

its ALEKS

im not sure if it goes through

ill just stick to it how about .B

do i have to put 3 angles that add up to 180 but all of the angles be obtuse?

[deep grove]

no not all

just not all can be acute

u can have 2 acute one obtuse

like u can have
1, 1, 178

[languid ermine]

that works as a counter example?

because it has one obtuse angle

since it says it all must be acute

i also have to write the reasoning on paper

[deep grove]

yes

[deep grove]

it says all acute, and in this case, only 2 of them are acute

[languid ermine]

i see

what about C

is it trying to say Y is the midpoint of XZ

[deep grove]

it says Y is on the line XZ
but it can be anywhere

in the example Y is the mid point

if its not the midpoint then the statement could be proven false

so just choose any other point for y

[languid ermine]

i still don't understand the number part of it tho

[deep grove]

wdym

[languid ermine]

i mean the lengths

[deep grove]

literally just choose any lengths that add up to 40 and that arent 20

[languid ermine]

aaa

i see

[deep grove]

see here, they just chose 20 and 20, those give a total of 40 so they work

[languid ermine]

wording is so tricky for me

[deep grove]

u can choose 1 and 39, or 5 and 35 or anything really

[languid ermine]

i see

.close

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[deep grove]

wdym, whats the question

[wise thistle]

b+c

ez

b-c

ez

[deep grove]

a+c = b here
so a = b-c

[thorny patio]

The direction of c affects the answer here

b + c would be further "down"

Because both b and c point "downward"

[deep grove]

b-c

[thorny patio]

b - c

Subtracting with c will pull "upward" from the end of b

Up and to the left in opposite direction of c's orientation

[deep grove]

2a + b = c
so b = c - 2a

[wise thistle]

bro stop giving answers

[deep grove]

do you know how to add vectors

[deep grove]

is that not what we are meant to do

[wise thistle]

no

[wise thistle]

it isnt

[deep grove]

oh

[keen plinth]

lol did you read the rules

[wise thistle]

read the rules

[keen plinth]

it says to try not to give answers

[wise thistle]

feels like ur using him for homework answers

[deep grove]

welp

mb

[wise thistle]

💩

@deep grove guide him ig

thx

[vale wigeon]

can we not do the whole hostility thing please

[deep grove]

do you understand why b = a+c

yeah, and a and c are also in those directions

so its all positive

[wise thistle]

alr peace

[deep grove]

so when you have b = a + c
then just move the c to the other side
and you get b - c = a

i mean he wasnt wrong, i shouldnt have straight up given the answer

?

uhh sure? why

just close the channel why delete evrythign

bro 💀

[keen plinth]

don't be toxic if you don't want to leave behind toxicity

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[sage quartz]

so for the second question, I've been estimating numbers, but none of them seem to be correct to the system

[gray isle]

how have you been estimating, what have you been entering

[lone heart]

@sage quartz Has your question been resolved?

[sage quartz]

uh 7 im kind of confused on how to approach the problm

[gray isle]

how did you get 7

[sage quartz]

tbh i guessed

[gray isle]

you're pretty being asked to estimate the slope of the tangent line at 24°C

[sage quartz]

ohh thank :)

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[scenic wing]

Is PEMDAS an axiom ?

Isnt 3+4 times 4 with pemdas
3+16 = 19

But if you did it without PEMDAS 7 times 4 = 28

Aren't both technically correct ? the only explanation I can see for PEMDAS is as an Axiom to maintain a standard order and law among maths

[carmine reef]

This isn't really "without pemdas" as much as it is with a different system of performing operations

That being, just go left to right

[scenic wing]

ok

but am I still right in PEMDAS being an axiom that exists for a standard order in math ?

[carmine reef]

Pemdas isn't really considered an axiom since it's about the way we represent statements and not about the underlying rules of math

I suppose you could call it one

[scenic wing]

but the different system of performing the operation is still technically right ?

[carmine reef]

Yeah it would still be a valid way of doing math

But not what the rules currently are

It's like "if i were to change the meanings of the words red and blue is it still a valid language"

[scenic wing]

ha

I knew it so it is an axiom

[carmine reef]

it's not the standard everyone else uses and in that sense it's "wrong" to say like the sky is red

but there's nothing deeply objectively true about pemdas, it's just useful

[scenic wing]

so its an AXIOM

[carmine reef]

In that it doesn't rest on anything yes

It's not derived from simpler rules

[scenic wing]

Math really is just a language

long live geometry

[carmine reef]

That's one way to see it

[scenic wing]

it's going to save me so many headaches knowing this

when I try to deeply look into the true meaning behind stuff

[carmine reef]

Axioms deal more with fundamental truths then pemdas probably lol

I would just call pemdas a convention

As opposed to like, "there exists a line between any 2 points"

[scenic wing]

convention is literally a synonym to axiom

read both their definitions

[carmine reef]

Hmm

Doesn't quite seem the same

[scenic wing]

both generally saying an agreed upon fact or formula

if you look at the math definition of axiom sure it differs

yeah ig you're right

[carmine reef]

I see the similarity

They're both things we decide on using as rules

[scenic wing]

wait but if the "statement or proposition" in the math definition is reached through the original definition then yeah it is

[lone heart]

@scenic wing Has your question been resolved?

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[native nebula]

Help

[ornate condor]

[native nebula]

How to factorise it further?

[ornate condor]

um

i think uve to expand full first

and group like terms

[carmine reef]

You can take an n out of both terms first

not that it's really matters that much whether you do that before or after grouping like terms

[ornate condor]

no i mean in this form its factored into 2 chunks which makes it hard to group like terms when they are half factorised

[vale wigeon]

say, why do we need to factor this anyway

maybe there's a way to do without for the problem at hand (which is currently unknown)

[native nebula]

The question is
3^3+5^3+7^3+...+n^3

[carmine reef]

if you just do a bunch of expansion you'll have a polynomial

[native nebula]

This one i got

[carmine reef]

Are you sure n is the last term in the sum here

[vale wigeon]

show the ENTIRE problem

[carmine reef]

Did you find 3^3+...+n^3

[native nebula]

Yeah

[carmine reef]

or did you find $\sum_{i=1}^n (2i+1)^3$

Because it does not look like I get 27 when I plug in n=3

It looks like I get something around 500

funny enough 3^3 + 5^3 + 7^3 is right around 500 too

[native nebula]

Yeah

[carmine reef]

wait hang on let me fix the sum

[native nebula]

495

[ocean seal]

monikanicity

[carmine reef]

there we go

[native nebula]

Yeah I found this one

[carmine reef]

That's not the same as 3^3+5^3+...+n^3 though

[native nebula]

Why not?

[carmine reef]

That's 3^3+5^3+(2n+1)^3

Well, what's 3^3+5^3+...+7^3

(it's 495 evidently, now plug n=7 into your polynomial and see if that's what you get)

[carmine reef]

The last term in this is (2n+1)^3

[ornate condor]

woah i is not an imaginary number, rare

[carmine reef]

i is my favorite index

some people use k but it just doesn't hit the same

[native nebula]

8127

[native nebula]

Same for every computer science student

[carmine reef]

yess

so when you write 3^3+5^3+...+n^3

It doesn't mean loop n times

It's more like a for(int i=3; i<=n, i+=2)

which notably will not loop n times

[native nebula]

Then what should be the notation?

Like i mean the expression?

[carmine reef]

is there an original problem you're working on?

[carmine reef]

This seems like good notation to me

[native nebula]

Yeah that's true

[carmine reef]

As long as you dont use the +...+n^3 thing unless you're gonna say (2n+1)^3 instead

[native nebula]

Ok yeah

And i wrote it this way...

[carmine reef]

ah yeah so n terms

Do you have some sort of rule for sums of power series you're using here?

[native nebula]

Yeah

[carmine reef]

I think it might have been a little easier to simplify first

[native nebula]

Like what?

[carmine reef]

(n+1)^3-n^3

Or well

(2n+1)^3-(2n)^3 is the nth term

[native nebula]

Actually I have done this way and i got the result

[carmine reef]

you could expand that out, you would get 12n²+6n+1

ah

[native nebula]

But i am trying to solve it this way

[carmine reef]

Fair enough

iv is fun here because it's just T(T(n)) lol

T(n) being the nth triangular number

[native nebula]

What is triangular number?

[carmine reef]

nth triangular number is 1+2+...+n

general formula n(n+1)/2

[native nebula]

Yeah

Know this one

But it's called triangular number?

[native nebula]

But it's not triangular number

In what sense you saying it's triangular number?

[carmine reef]

O

[native nebula]

They all have the same common difference of 2 in the sequence
3,5,7,...

[carmine reef]

. O
OO
OOO
OOOO
OOOOO

triangle

[native nebula]

Yeah got ya

But in this case it's not like that

[carmine reef]

Nah just for the 4th one lol

Not the problem you're on

[lone heart]

@native nebula Has your question been resolved?

[native nebula]

Then what would be the general expression for the series
3^3+5^3+7^3+...+n^3?

[carmine reef]

something like $\sum_{i=1}^{(n-1)/2}(2i+1)^3$ ig

[ocean seal]

monikanicity

[native nebula]

Why (n-1)/2? It could have been (n+1)/2

[carmine reef]

but then the last term would be (n+1)^3

[native nebula]

Sorry i don't get you

Can you please explain why?

[native nebula]

First of all you are considering this series right?

.close

[lone heart]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silk steeple]

help

[runic trench]

[silk steeple]

question 5

thé wet surface part is confusing

[lone heart]

@silk steeple Has your question been resolved?

[silk steeple]

nope

[runic trench]

find v

[silk steeple]

I got the V and V’

V as (1/3)πr²h

[runic trench]

ok

find base of the cone

[silk steeple]

Is it πr²

[runic trench]

yes

[silk steeple]

What do I do with the base