help-27

[unkempt quiver]

Ok let me prove it to you

[open pendant]

nah, if i remove 5 from the left, i rmeove it from the right

[unkempt quiver]

2 = 2 right?

[unkempt quiver]

Of course

[open pendant]

yea bro but theres no parenthesis

u dont multiply everything by 5

ur wrong

[unkempt quiver]

1 + 1 = 2 right?

[open pendant]

bruh lets gets someone in here

[unkempt quiver]

So now let's multiply 5 on both sides

[open pendant]

@supple knot

u around?

[unkempt quiver]

5 + 5 = 10

[open pendant]

nah, i want someone to see this

i know u goofed

[unkempt quiver]

But if we just multiplied one term it'd be
1 + 5 = 10

Which would be 6 = 10

[open pendant]

<@&286206848099549185>

[unkempt quiver]

LMAO

[open pendant]

bruh this is completely wrong

entirely

[unkempt quiver]

LITERALLY anyone you ask is going to tell you that you multiply every term in an equation

[open pendant]

nah dude well see

[unkempt quiver]

Alright

I give up sorry

Not dealing with this today

[open pendant]

its all good to be wrong sometimes

[unkempt quiver]

Review ur basic algebra

[open pendant]

buddy its ok

we all get the basics wrong sometimes

no need to be embarrassed

[unkempt quiver]

I was trying to show you that you multiply every term in the equation not just the fraction

[unkempt quiver]

You're a troll lol

LITERALLY anyone you ask is going to tell you that you multiply every term in an equation

[open pendant]

take it in stride

have some modesty

relax

ok, then we wait i see

[unkempt quiver]

Alright alright

[open pendant]

im okay being wrong

are u?

[unkempt quiver]

I apologize for telling you the way to do it

And for you not listening to me

[open pendant]

bruh have some humility, jesus

[unkempt quiver]

I apologize for dealing with such stupidity

Cya

[open pendant]

u need to study the self

too much time spent studying the other

[unkempt quiver]

<@&268886789983436800> someone is trolling when I'm trying to explain basic algebra

[open pendant]

thats not a good use of mods

were discussing this equation he gave me

i do not think its right

he is quite angry that he thinks it is

that it all

[unkempt quiver]

sigh I was showing you you had to multiply EVERY term

[open pendant]

and u dont

its ok, just relax

[celest dock]

uh

if youre so unconvinced

why not just try an example

take 1 + x = 2

obviously the solution to this is x = 1

now multiply both sides by 5

if we do what you said and take 5 + x = 10, well, now x = 1 no longer works

[supple knot]

Don't ping me out of nowhere thanks

[celest dock]

whereas if we do 5 + 5x = 10, x = 1 still works

indeed, the algebraic justification here is that, what you do to one side, you have to do to another

given 5/5 - 6x = 7, we multiply both sides by 5

5 * (5/5 - 6x) = 5 * 7

[open pendant]

both sides by 5, not everything by 5 though right

[celest dock]

and then the 5 distributes over the LHS

how would you multiply the LHS by 5 without multiplying the 6x by 5?

like explain to me what algebraic operation youre doing to 5/5 - 6x

[open pendant]

we were using a made up example

so in this eaxmple though

i multiply both sides by 4

to get

5-6x=35 right?

[celest dock]

but you didnt multiply both sides by 5

[open pendant]

sorry 5*

[celest dock]

you multiplied one side (the RHS) by 5

and only multiplied part of the other side

[open pendant]

yes

ok

[celest dock]

5/5 - 6x = 7
multiply both sides by 5
5 * (5/5 - 6x) = 5 * (7)

now distribute in the 5

5 - 30x = 5 * 7

[open pendant]

ahhhhhhhh

ok that makes sense

@unkempt quiver

we got out answer

thank u Namington

very wise

[celest dock]

and fwiw, you can check that the same solution works

the solution to 5/5 - 6x = 7 is x = -1

and this also solves 5 - 30x = 35

[open pendant]

very cool thank u again!

[unkempt quiver]

I gave him that exact example too

[open pendant]

sorry for being mod summoned, it definitely wasnt necessarry

nah jowo

this was ur example

[open pendant]

completely different

it was a solid try though, i appreciate you help ❤️

[unkempt quiver]

Gj! Hope u understand it now

[open pendant]

yea namington cleared it up

plz remember not to ping mods for invalid reasons

[unkempt quiver]

I'm glad he did! And please remember not to troll or be disrespectful

[open pendant]

awww, u say the sweetest things ❤️

have a good sleep

.close

[devout snow]

Channel closed

Closed by @open pendant

Use .reopen if this was a mistake.

[unkempt quiver]

Troll ^^

[devout snow]

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[scenic surge]

$\lim_{x \to 2} \frac{\sqrt{27-x} - 5}{\sqrt{18-x} -4}$

[woven radish]

Worth

[scenic surge]

for this isnt it just 0/0

so undefined?

[sick fulcrum]

um

how did you get 0/0

[tame palm]

0/0 is one of the indeterminant forms for which you can use L'hopital.

[sick fulcrum]

you cannot just substitute to compute limits like that btw

[scenic surge]

wait

its

4/5

l hopital rule

yes

sorry

[sick fulcrum]

*with exceptions

you can if its in the domain of the function

but here 2 is a limit point that lies outside the domain

[scenic surge]

got it I forgot abt the indeterminate form

.close

[devout snow]

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[thin gulch]

how do i do rachel took 1/2 hour to paint a table and 1'3 hour to paint a chair how much time did she spend painting

[thin gulch]

<@&286206848099549185>

[queen niche]

you'd do common denominators again

[devout snow]

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

[queen niche]

then multiply other side using the other denominator

[thin gulch]

ok i got 3/6 and 2/6

UGH

[queen niche]

yup then add the numerators and leave the denominators alone and that's your answer

[thin gulch]

3/6 AND 2/6

oh sry i thought i accidentally typed the wrong thing

its 5/6?

also was the other question's answer only 12 or 12/9?

[queen niche]

yup

[queen niche]

just 12 you got rid of the denominator 9 when you turned 9/9 into 1

[thin gulch]

ok

how do i do a warehouse has 1 and 3/10 metres of tape. if they separate the tape into pieces that are 5/8 metres long, then how many pieces will they have

its division right but idk what to d

do

[thin gulch]

thx

[queen niche]

just follow your steps

re write it

turn the mixed number into a improper fraction

so 1 and 3/10 becomes 13/10

then 13/10 / 5/8

[thin gulch]

what do i do with 13/10 and 5/8

[queen niche]

cross multiply then flip

[thin gulch]

13x8 and 10x5? its 104/50?

[queen niche]

yup

[thin gulch]

wdym flip it

50/104

[queen niche]

wait nvm i think you did it write

right sosrry

i remembered the method wrong

keep the first fraction as iss

change the ssign to multipcation

then flip the second fraction

then multiply them across

giving you 13/10 * 8/5

104/50

you were right

[thin gulch]

i can simplify and make it to a mixed fraction? idk

[queen niche]

yup

so 50 goes into 104 twice

with a reminder of 4

[thin gulch]

i could've simplified the fractions of the other questions too?

o

idk

ok

[queen niche]

yea

how many more problems do you hgave

[thin gulch]

7 more

i have another question ]

a chef bought 7 6/7 pounds of carrots to make a carrot cake he only used 5 2/7 pounds of the carrors how many carrots are left over

is the answer 1 3/7

its subtraction right

or division?

[thin gulch]

its division right

how do i tell if a question is subtraction or whatever

[queen niche]

they'll use words like, difference, decrease, are left, remain, change, fewer than, subtract in the problem

[thin gulch]

ok what about for division and addition and multiplication

idk what i did wrong

[queen niche]

for division they'll say remain dividend the quotient or like group

addition theyll ssay sssum all, together, add, increase, more than

multiplication double, product, per, each, twice, quadruple, area

[thin gulch]

is my answer right tho

wasnt the soup question 4 4/10 not 4 8/20 or theyre the same thing?

[queen niche]

your right same thing but you dindt fully reduce it

4/10 is also 2/5

which is also 8/20 or 16/40 or 12/30 or 6/15

[thin gulch]

oh ok

[queen niche]

your getting a lot better at fractions tho you almost got it down

[wild sonnet]

it's 16/40

[thin gulch]

how do i do this

[thin gulch]

is it also 4 2/5 idk.

[queen niche]

Yup that would be the mixed number

[thin gulch]

is it all simplified or something

[queen niche]

Yea

For the carrot cake one you just fid subtraction wrong but you did your fractions right

It would be 2 and 3 7ths not 1 and 3 7ths

Because when you subtract the fractions you get 3 7ths

Then you do 7 minus 5 then add 3 7ths

Giving you 2 and 3 7ths

Also make sure you write your units they get more picky about that because it gets more important when you get into like science classes like physics and chemistry

So 2 and 3 7th pounds of carrots

[thin gulch]

oh um i got 454/49- 406/49

434*

[queen niche]

Y[u dindt need to find a common denominator since they are both 7

So they already have a common denominator

Being 7

Imma go to bed if you need more help or need to remember refer to this

[thin gulch]

nooo ok bye

[queen niche]

For divide keep the first fraction as is change the sign to multiplication then flip the second fraction

Then solve like multiplication

Remember Keep Change Flip

[thin gulch]

ok

.close

[devout snow]

Channel closed

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

If $ax^2 +bx^2 + c = a(x-d)^2$ with $a \neq 0$, find d in terms of a, b, c

[woven radish]

Worth

[scenic surge]

what does it mean find d in terms of a, b, c?

Can I just solve for a, b, c, d seperately?

[winter patrol]

d = stuff with a,b,c,x (and numbers) and no d

[scenic surge]

what wdym

I am confused

Do I just solve for d?

[winter patrol]

yeh...

[scenic surge]

oh

shit

tysm lol

can I send my work here once done and get it checked with u aswell?

[scenic surge]

could I say that $\frac {a^2}{a} = a$ ?

[woven radish]

Worth

[winter patrol]

yes

[scenic surge]

ok ty

[devout snow]

@scenic surge Has your question been resolved?

[scenic surge]

This is what I did so far

fixed there

Is that right

@winter patrol

<@&286206848099549185> pls help

[supple knot]

that's mostly all wrong

[supple knot]

ramonov made a mistake here by including x. your expression for d should not have x

[scenic surge]

oh shit

[scenic surge]

so do I solve for x first?

wait no

it was c-d

not x-d

I accidnetly put an x there

so should I just solve for x and then plug that in d(x)

[sage burrow]

you mean $ax^2 +bx^2 + c = a(c-d)^2$?

[woven radish]

ThM

[scenic surge]

yep

u can see in "simplified" I went back to c-d lol idk why I put an x in the beginnin

[scenic surge]

but do I just do this?

[sage burrow]

you have x^2 two times on the left -> you can write this in an other way so that x^2 appears only once. then you can put the scalar c on the right side. then you should see the rest.

[scenic surge]

$x^2(a + b + \frac{c}{x^2})$

[woven radish]

Worth

[scenic surge]

like this?

[sage burrow]

not exactly. let c alone in the first step

you have still x^2 twice.

[scenic surge]

oh

so

$x^2(a + b) = a(c-d)^2 -c$

[sage burrow]

yes, but you missed the right side

[woven radish]

Worth

[sage burrow]

yes, see you now the next steps?

[scenic surge]

mhm Im still solving for x right

[sage burrow]

yes

sorry. stop

[scenic surge]

ok

[sage burrow]

i have to go back to the start. are your sure it is $ax^2 +bx^2 + c = a(c-d)^2$?

[woven radish]

ThM

[scenic surge]

wym if Im sure

wait

a min

OMGGGG

OMGGGGGGGG NOOOOOOOOOOOOOOOOOO

where the hell did I get c from

It's $ax^2 +bx^2 + c = a(x - d)^2$

[woven radish]

Worth

[scenic surge]

shit im so sorry idk wth is goin on with me

[sage burrow]

and it should be bx^2 and not bx?

[scenic surge]

bro omg

wth

It's $ax^2 +bx + c = a(x - d)^2$

[woven radish]

Worth

[scenic surge]

wth am I DOINGGG

do I solve for x now

[sage burrow]

$ax^2 +bx + c = a(x - d)^2 = ax^2 - 2adx + ad^2$

[scenic surge]

damn i gotta solve for d all over again too

[woven radish]

ThM

[scenic surge]

why do tha?

[sage burrow]

now compare the coefficients at x^2 at x and the scalar.

[scenic surge]

wait over here

u solved for?

oh wait

rhs

got it

so

$bx + c = -2adx + ad^2$

[woven radish]

Worth

[scenic surge]

$bx + 2adx = ad^2 - c$

[woven radish]

Worth

[scenic surge]

$x(b + 2ad) = ad^2 - c$

[woven radish]

Worth

[sage burrow]

no,

[scenic surge]

shit

I thought u wanted me to get rid of ax^2 - ax^2

[sage burrow]

i said now compare the coefficients at x^2 at x and the scalar. -> you have on both sides a by x^2

[scenic surge]

yeah so cant I get rid of em

[sage burrow]

then b and -2ad at x and c and ad² as scalar.

[scenic surge]

wait what

[scenic surge]

like this?

[sage burrow]

no, if $ax^2+bx+c = dx^2 +ex +f \forall x$ then a=d, b = e, c = f

[woven radish]

ThM

[scenic surge]

where did u get e and f from....

Im lost rn

[sage burrow]

lets say it in words. if two quadratic terms have the same values for all x, the coeffizients have to be the same.

if $ax^2 +bx + c = a(x - d)^2 = ax^2 - 2adx + ad^2$

[woven radish]

ThM

[sage burrow]

-> c = ad^2

[scenic surge]

ohhhhhhh

I see

and b -> 2ad?

[sage burrow]

yes

[scenic surge]

I see

[sage burrow]

sorry, -2ad

[scenic surge]

oh yea

so how will this help me

solve for x

[sage burrow]

you should express d in terms of a,b,c. if c = ad^2 then ...

[scenic surge]

oh

but

wait

im kind of confused rn

I will still have x in my answer doe

?

[sage burrow]

if b = -2ad, so d = b / (-2a), and you have expressed d in terms of a and b (c is not needed)

[scenic surge]

oh

why isnt c needed doe

wont c be $\pm \frac{\sqrt{ca}}{a}$

[woven radish]

Worth

[scenic surge]

I mean d =

in terms of c

oh u meant for b

i express d in terms of a and b

and for c i express d in terms of and c?

[sage burrow]

you expressed d in terms of a and b. c depends on a and b, so it is not necessary.

[scenic surge]

uh so my answer is literally just, In terms of a and b $d = - \frac {b}{2a}$ ?

[woven radish]

Worth

[sage burrow]

yes, i would say so,

[scenic surge]

but

If $ax^2 +bx + c = a(x-d)^2$ with $a \neq 0$, find d in terms of a, b, c

It says c aswell

[woven radish]

Worth

[sage burrow]

$d = 0c - \frac {b}{2a}$

[woven radish]

ThM

[scenic surge]

oh why 0c thats literally just 0?

So I could just say that since c is dependant on a and b I don't have to conclude c in my answer?

[sage burrow]

my answer would be d = -b / 2a, c doesnt appear since it depends on a and b, but if you insist for c in the result, this would be a way.

[scenic surge]

mhm I see

[sage burrow]

c depends on a and b in the way $c = \frac{b^2}{4a}$

[woven radish]

ThM

[scenic surge]

Got it

So would this be right?

like would this be good enough

[sage burrow]

looks ok for me.

[devout snow]

@scenic surge Has your question been resolved?

[devout snow]

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[coral crescent]

determining the nature of integral of (1+sinx)/(1+sqrt(x^3)) from 0 to infinity

[fossil garden]

so what do u need?

@coral crescent

[coral crescent]

i think my work is correct so far, but idk how to continue with integral of 2/(1+sqrt(x^3)) from 0 to c

[fossil garden]

not sure what im missing, but why doesnt the same bounding work like u did for the integral from c to infty

[coral crescent]

I can't use the comparison test with 2/sqrt(x^3) at 0, no?

[fossil garden]

[coral crescent]

is the integral divergent?

[fossil garden]

alright so split the bounds as 0 to 1, and 1 to infty

we know 1 to infty its convergent(p series), and for 0 to 1 case, try the 1/(1+x^(1.5))< 1/(1+x)

$$\int_0^1\frac{2}{1+\sqrt{x^3}}<\int_0^1\frac{2}{1+x} $$ where $x \in (0,1)$

[woven radish]

clemenson

[coral crescent]

ah i see

[fossil garden]

and u can evaluate that

[coral crescent]

mmm got ot

it

what difference is it between putting c or 1?

[fossil garden]

nothing really

we can use the theorem that establishes convergence status between the series from n=1 to infty directly

thats all

[coral crescent]

got itttt

[coral crescent]

gott it

thankk you ❤️

.close

[devout snow]

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[devout snow]

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[spiral violet]

K-field
V- line space on it
What $S_K(V)$ may mean

[woven radish]

Михаил Колесников

[pseudo basin]

@spiral violet а с чего бы нам знать? может, лучше заглянуть в учебник, откуда Вы это обозначение взяли?

кстати, переход на новую строчку в Техе — два бэкслэша подряд

[spiral violet]

А вдруг

[pseudo basin]

так-то, давайте попробуем обойти трудности перевода. буквой K вы обознаете поле, а V -- линейное пространство над K?

[spiral violet]

да

[pseudo basin]

по-английски это называется "vector space" или "linear space", если хочется указать на поле — "linear/vector space over K" or "K-linear/vector space"

тем не менее, скорее всего где-то в Вашей книжке обозначение S_K(V) должно вводиться

[spiral violet]

в том то и дело что нет(

[pseudo basin]

опаньки

может, покажете, где вы его увидели, чтобы из контекста можно было хотя бы догадки сделать?

[pseudo basin]

хм. судя по всему, A — какое-то кольцо

[pseudo basin]

может быть это кольцо симметричных многочленов над V? хотя тогда эта штука зависит от базиса последнего...

так что это не точно

но вряд ли обозначение нигде не вводится. не может такого быть

[spiral violet]

Я подумаю над этим

.close

[devout snow]

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[fresh horizon]

nope

also it's a scaling transformation from ABC to A'B'C'

<@&268886789983436800>

[mellow panther]

I think they meant it very loosely

[fresh horizon]

what does that mean

[mellow panther]

As in dilation has a very precise meaning in maths but here they just meant shrinking

[main gull]

It's not shrinking

[fresh horizon]

ah I see

also that

[main gull]

ABC is the smaller triangle

[mellow panther]

wait, what. How is that dilation then?

[main gull]

Dilation is a transformation, which is used to resize the object. Dilation is used to make the objects larger or smaller

Dilation is the term that describes a "growing" or "shrinking"

[restive river]

It’s 1.5

[main gull]

Don't give out answers

[mellow panther]

I see. So, any transformation that changes the size without changing the shape is a dilation

[restive river]

There’s nothing hard to explain, they are just similar triangles

[main gull]

Still don't give out answers

That is not the purpose of the server

[mellow panther]

If you just post the answer that kills the point of "explanation"

[fresh horizon]

maybe add a little more explanation than just the answer even if it's so simple

[restive river]

The explanation is similarity, nothing more. This is not a server to study a field. So just say the way

[fresh horizon]

kinda, it is a scaling respect to the origin

[main gull]

Well dilation describes the scaling factor where if 0 < x < 1, then it's a shrink. x > 1 it's a growth

[main gull]

And you still do not give answers

That's not how people learn by just stating answers

[mellow panther]

ANYWAY, where's OP who posted the question?

[main gull]

Not here

Apparently

[fresh horizon]

also this is certainly a server to "study fields" if it were to help with the answer

[mellow panther]

We should close this channel then

[main gull]

Nah let the bot do it

[fresh horizon]

I don't see how giving a brief explanation is studying the whole field of geometry but sure....

[main gull]

If they don't respond to the bot message, it'll auto close

[restive river]

It is not. The best way to study a field is just by having a quiet place and understand it alone.

[fresh horizon]

I think it is not up to you to decide that one for anyone

[restive river]

Nobody can teach whole Euclidean geometry here

[fresh horizon]

you don't need to?

this argument is just pointless

let's cease

[fringe olive]

thx for help

.close

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[untold vector]

Prove that 8|(3n–1) by Mathematical Induction.

[mellow panther]

What have you tried/done so far?

[untold vector]

I solved the question but not sure if it is right

[mellow panther]

huh? You wrote (3n - 1) above but in the paper you have written 3^(2n) - 1 ?

[mellow panther]

I'm assuming this correct. So, I'll check this

[rotund heron]

I am guessing the one on the paper is correct?

[untold vector]

This one

2 is missing there

it is 2n

[mellow panther]

okay

[untold vector]

I got the solution, but not matching with mine

[mellow panther]

This is a wrong way to do this. You assumed P(k) and then you have to prove P(k+1)

So, you can't write 3^(2(k+1)) -1 = 8m, you have to show that!!

Also, in the step marked (2) you can't separate the addition of powers into addition of numbers

$a^x \cdot a^y = a^{x + y} \ne a^x + a^y$

[woven radish]

numbpy

[untold vector]

yeah, that is a multiply

[mellow panther]

Are you sure, that looks very much like a plus to me

[fresh horizon]

and you treated it as a plus further down the paper

[untold vector]

It's plus, but a typo

[mellow panther]

yep

[untold vector]

yeahh

[mellow panther]

It should be a product, make it a multiplication and proceed as usual

[untold vector]

the further steps are fine i think

[mellow panther]

Although, honestly there was no need to write that as 8m + 1. It's slightly more tedious now

[untold vector]

I treated it like a multiply

[mellow panther]

alright

[untold vector]

What should I do instead?

[mellow panther]

Just leave it as 8m. As in for P(k) it would be -

$3^{2k} - 1 = 8m$

[woven radish]

numbpy

[untold vector]

okay

What about this
#help-27 message

I didn't get the steps there

Is it the same thing you are telling that not to move -1?

[mellow panther]

which line you didn't get

[mellow panther]

yesp

[untold vector]

Okay, understood. Many thanks @mellow panther 👍

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[untold vector]

How should I solve the question further?

[mellow panther]

you have to prove this for all n, natural numbers?

[untold vector]

I'm confused as what to add in RHS

[untold vector]

yeah

[mellow panther]

Do you know the sum of GP formula ?

[untold vector]

No, haven't got there yet.

[nimble badger]

a( r^n - 1)/(r-1)

[untold vector]

Should I add this on RHS, with 2 ?

[feral agate]

replace the left hand side with the equation he gave

[mellow panther]

nah, the thing is that this might be tricky to do with just induction

[feral agate]

do you know what a r and n stand for

[nimble badger]

This is the formula for sum of GP where a is term 1, r is common ratio and n is number of terms

[nimble badger]

Sorry I forgot to mention that earlier 😅

[mellow panther]

If he hasn't been taught this, I doubt he'd be allowed to use it

[untold vector]

The given explanation

[mellow panther]

That's clever

[untold vector]

I didn't understand

there is a multiplication with 1/2 both sides

but then how +1 is added in the last step?

[mellow panther]

yes, exactly

for the +1, they added 1 to both sides of the inequality

basically, $a < b \implies a + c < b + c$

[woven radish]

numbpy

[untold vector]

but is it the 1 with (k+1) on LHS ?

[mellow panther]

wdym?

[untold vector]

they added 1 on RHS

[mellow panther]

and on the LHS also

[untold vector]

but where is +1 on LHS

[mellow panther]

It's right there, in the front

[untold vector]

with K?

[feral agate]

the first term

[mellow panther]

^^^^^^^^

[untold vector]

What happened to this one?

[mellow panther]

In the second last step we got -

$\frac12 + \frac{1}{2^2} + ... + \frac{1}{2^{k+1}} < 1$

[woven radish]

numbpy

[mellow panther]

you forgot to multiply the 1/2?

$\frac12 (1 + \frac{1}{2} + ... + \frac{1}{2^{k}}) < 1 \implies \frac12 + \frac{1}{2^2} + ... + \frac{1}{2^{k+1}} < 1$

[woven radish]

numbpy

[untold vector]

yeah, got it.

But is it legit? I mean decrease RHS and then increase again?

It was 2, it was decremented to 1

[mellow panther]

yes, this is why I said it is clever, although cheeky is a better word

[mellow panther]

It was halved to 1 and then increased by 1 back to 2

There's a small difference

[untold vector]

yeah

[mellow panther]

In fact, let me in on a small secret. If this sum is extended to infinity it will exactly be equal to 2

[untold vector]

The next chapter is Series, so maybe I'll see GP and stuff

[mellow panther]

ie $\sum_{i=0}^{\infty} \frac{1}{2^i} = 2$

[mellow panther]

sure, sure

All the best

[untold vector]

I'm new to math, this E sign is scary

[mellow panther]

ah, now worries it's just a notation to represent a sum

[untold vector]

Don't even know the name

But I guess that it is in the syllabus

[mellow panther]

It's called a summation, you'll meet it soon enough

anyway, all the best

close this channel if you're done

[untold vector]

Thank you so much @mellow panther @feral agate @nimble badger

.close

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[paper wren]

State the domain and range for: $y=2|x-4|+3$

[woven radish]

Zyme><SOL

[paper wren]

not so sure what you mean

[tame palm]

2(x-4) and 2(-(x-4))

[paper wren]

what about the 3?

sorry I'm trying to relearn past course material for an exam and have forgotten about this so I don't have that much background on this lol

[tame palm]

Well, first, what is the minimum possible value of |x-4|?

[paper wren]

I'm just guessing here, 0?

[tame palm]

Correct. Now solve for y to find the minimum of the range.

[paper wren]

y=3 then

[tame palm]

[3, infinity)

There is nothing restricting the domain as well.

[paper wren]

oh

could you maybe link a video or something that'll get me on track

[tame palm]

You can never go wrong with the Organic Chemistry Tutor.

https://www.youtube.com/watch?v=djT6-YamHaA

[paper wren]

oh I actually skimmed through that one

I didn't find an absolute value example which is why I decided it wasn't helpful

[tame palm]

https://www.youtube.com/watch?v=ld4UD98yHio

[devout snow]

@paper wren Has your question been resolved?

[devout snow]

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[livid iris]

Just would like to verify my answer

Answer: (1-x) (log3/2) + 1 all over 4

equals x

[wooden veldt]

And what is the question

[livid iris]

.close

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[hasty ice]

Help

[hasty ice]

Provide me details on this one

[late oriole]

Rectangle is dilated from the origin by a scale factor of 13 to create S'

Okay so, what do you know about scaling an object? What does it do to an object?

[hasty ice]

Wdym scaling

[late oriole]

Ah, I mean dilating, or "dilated"

[hasty ice]

Oh

Well

When the scale factor of a shape is 2 it increased by 2

Nevermind

Scale factor > 1 enlarge

scale factor < 1 reduce

is what i know

[late oriole]

Okay

What happens if scale factor is 1, just curious?

[hasty ice]

and a scale factor of x will change the permimeter by a factor of x and the area by a factor of x^2

[late oriole]

Also, just to help you visualize what "scaling" or "dilating" is, here's a video of scaling an object in 3D (but you can apply the same concept in 2D)
https://youtu.be/FPp3ClfDYqI?t=17

Anyways, you can see that the object itself can get larger (scale factor >1) or smaller, but the object always is a cube still

[hasty ice]

i have no clue

Or i forgot

[late oriole]

Now that can be applied in 2D as well - if you have a rectangle that is scaled, it will also remain as a rectangle, just a bigger version of the rectangle, or smaller

[late oriole]

So you said a scale factor of x will change the perimeter by a factor of x
So if a scale factor is 1, then that means a perimeter will change by a factor of 1

So what will that do to an object?

[hasty ice]

Nothing?

If a scale factor is 2 it will multiply by 2

if the scale factor is 0.5 it will multiply by 2

0.5*

[late oriole]

Yes

[hasty ice]

so

[late oriole]

So that means scale factor of 1 does nothing

[hasty ice]

Exactly

ye

[late oriole]

Okay so do you know what congruent vs. similar is?

[hasty ice]

Yea

but but

for me its

congruent vs supplementry

[late oriole]

Wait, wdym, your answer choice says Rectangle A is supplementary to rectangle A'?

[hasty ice]

No no

What is simialr

i only known congruet and supplemenary

[late oriole]

Oh

Congruent = exact same
Similar is not the same lengths/area, but the shape is still the same

[hasty ice]

Oh ok

i see

i see

[late oriole]

https://youtu.be/FPp3ClfDYqI?t=17
At the timestamp I linked, just watch for 3 seconds
you can see even though the cube is getting larger and smaller, its still a cube, meaning it is still similar

[hasty ice]

U can find it out from have two degrees and adding it then subtracting it by 180?

[late oriole]

So:
If you scale an object larger = the sides got larger, so do you think Rectangle S' is congruent to Rectangle S'?

[late oriole]

I don't know what you mean by that,

But if you have a straight line with 1 unknown angles but 2 known angles, doing the above will find out the missing angle

(it's an unrelated concept)

[hasty ice]

That

[hasty ice]

No?

[late oriole]

You're right, if you have a side length of 2, and the other rectangle has a side length of 4, clearly 2 is not equal to 4, so it is not congruent

[hasty ice]

so its similar

[late oriole]

However, if you make the rectangle larger, but it still looks like a rectangle, is it similar?

Yes

Because when you scale a shape, you don't affect the angles, just the sidelength

You see the small rectangle on the left has 90 degree angles, just like the larger scaled rectangle on the right

[hasty ice]

ok

[late oriole]

So now you answered A&B, I covered everything you need to know about C-F

[hasty ice]

ok

B is wrong

and A is right

[late oriole]

yes

[hasty ice]

And there equal

so c

[late oriole]

No

[hasty ice]

?

[late oriole]

So you say the shapes aren't congruent
meaning they do not have the same side lengths
then how can their side lengths be the same?

[hasty ice]

So there not equal

??

[late oriole]

Yeah, the side lengths are not equal

[hasty ice]

So F?

[late oriole]

Do you think the left rectangle's side length = the right rectangle's side length?

It's clear that it is not equal, meaning that it is not congruent

No, the angles are equal

[hasty ice]

Wait so the angles are equal but not at the same time?

[late oriole]

The side lengths are not equal, the angles are equal

[hasty ice]

oh

So the angles measurements are equal

[late oriole]

Yes

[hasty ice]

So this should be my final answers

[late oriole]

Yes

[hasty ice]

ok

[hasty ice]

could u also help me with this one

would they have the same area

if they were translated and reflected

[late oriole]

If you move a piece of paper on your desk from your left hand to your right hand, does the size of the paper change?

Translated, reflected = movement of shape

[hasty ice]

no

it wont

[late oriole]

Correct, so you will say yes, the rectangles HAVE the same area
But you will need to explain why yourself

[hasty ice]

I don't believe i could use you're explation

[late oriole]

You cannot use my explanation, but you should google yourself why translation and reflections do not affect the area of a shape rather than me telling you the answer outright, otherwise next time you see this problem you may not know how to do it again

[hasty ice]

so reflecting and translating will not affect the area of the shape

[late oriole]

yes

[hasty ice]

what about rotating and dilating

[late oriole]

First of all, I think you need to watch a video on reflection vs translation vs rotation vs dilating

I also explained everything about dilation above

For rotation, if you turn a piece of paper sideways, does the size of the paper change?

[hasty ice]

It wont change

[late oriole]

yes

[hasty ice]

But i wont say it would have the same area

Since not only its being rotated its being dialated

[late oriole]

Yes

The dilation (with a scale factor other than 1) means that the paper is being made bigger/smaller

Remember that if scale factor = 1, then nothing will happen to paper

[hasty ice]

also

Area = w × h

w and h are being changed when dilating

so the area should change

if the width and height is 4 meaning the area is 16 but when dilated by 2 the width and height are now 8 and the area is now 64?

same goes with < 1

So the area wont the same

@late oriole

[late oriole]

Uh

yes, all correct

[hasty ice]

Alright

Thanks

I done

[late oriole]

so if dilation is 1/2
then 4x4 becomes 2x2

[hasty ice]

Yes

changing the area

[late oriole]

yes

[hasty ice]

Alright thanks @late oriole

.close

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[timid siren]

I'm trying to understand the difference between geometric measure and outer measure. From the source I've read it feels like this:

[timid siren]

ah crap I'll do some research before opening this, can't even from my question right

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[green walrus]

$((x^3+1) \cdot \frac{1}{2}x^{-\frac{1}{2}}) - (x^{\frac{1}{2}}(3x^2)) = \frac{1}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}} - 3x^{\frac{5}{2}}$

[woven radish]

Willow

[green walrus]

correct, so far?

[small raptor]

Yep

[green walrus]

ok, now what's the rule for adding and subtracting variables with fractional exponents?

I don't see this topic covered here:

https://mathinsight.org/exponentiation_basic_rules

[upper schooner]

If they have the same exponent you can "combine" them, if that's what you're looking for?

So something like $ax^n + bx^n = (a + b)x^n$

[woven radish]

chartbit

[upper schooner]

If [they're] not [the same exponent], then there isn't much you can really do with them

[green walrus]

$\frac{1}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}}$

^ this

[woven radish]

Willow

[upper schooner]

For those, the exponents aren't the same, 5/2 != -1/2 so doesn't apply here

[green walrus]

so, i just leave them as is?

[upper schooner]

For now I would at least - is there like a form they want the "final answer" in?

[green walrus]

I'm just working through this problem:

https://www.calcchat.com/book/Calculus-11e/2/3/13/

so far, my answer doesn't look like theirs

but I assume they've omitted a number of steps

so hopefully, i'm on the right track

[upper schooner]

Keep going for now - you can get towards theirs with a bit more work from here tbh

[green walrus]

i'll write out each step then, since this is difficult

$\frac{1}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}} - 3x^{\frac{5}{2}} = -\frac{5}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}}$

yes?

[woven radish]

Willow

[upper schooner]

Yep that's good, nice!

[upper schooner]

That looks quite a lot like the numerator in the final line tbh

[green walrus]

next:

$\frac{-\frac{5}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}}}{(x^3+1)^2}$

[woven radish]

Willow

[green walrus]

and at this point, I'm stuck

i think I could bring down $x^{-\frac{1}{2}}$

[woven radish]

Willow

[upper schooner]

You could maybe try multiplying the numerator and denominator by the same thing?

[green walrus]

multiply it by what?

[upper schooner]

Something that would make that $x^{-\frac{1}{2}}$ (and while we're at it, those fractions) in the numerator disappear?

[woven radish]

chartbit

[green walrus]

$\frac{1}{2}x^{-\frac{1}{2}} \cdot 2x^{\frac{1}{2}} = 1x^0 = 1$

[woven radish]

Willow

[green walrus]

yes?

[pale patrol]

Math so hard

[upper schooner]

Yep, that's the idea that will make things nice

So the idea is to multiply both the top and bottom by that and things should neaten up nicely!

Reason both the top and bottom of course is that it's like equivalent to multiplying by 1, so it won't change the value of the fraction

[devout snow]

@green walrus Has your question been resolved?

[green walrus]

$\frac{2x\frac{1}{2} \cdot (-\frac{5}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}})}{2x\frac{1}{2} \cdot ((x^3+1)^2)} = \frac{-5x^3 + 1}{2x^{\frac{13}{2}} + 2x^{\frac{1}{2}}}$

[woven radish]

Willow

[green walrus]

this is what I have

looks like the same answer in the final link excepted I simplified everything since there weren't any instructions to not simplify everything

[upper schooner]

Numerator is fine, denominator though, be careful! Looks like you did $(x^{3} + 1)^{2} = (x^{3})^{2} + 1^{2}$, and it doesn't work that way unfortunately

[woven radish]

chartbit

[green walrus]

ah, right

I always make that mistake

so, it is:

$\frac{2x\frac{1}{2} \cdot (-\frac{5}{2}x^\frac{5}{2} + \frac{1}{2}x^{-\frac{1}{2}})}{2x\frac{1}{2} \cdot ((x^3+1)^2)} = \frac{-5x^3 + 1}{2x^{\frac{1}{2}}(x^3+1)^2}$

[woven radish]

Willow

[upper schooner]

Yep, and that's exactly what they have there!

[green walrus]

ty

also, because everything is in simplified form, it should be implied that I don't evaluate $(x^3+1)^2$

[woven radish]

Willow

[green walrus]

and I misspoke earlier, I should have said evaulated instead of simplified.

[upper schooner]

Haha to be honest, I didn't even notice you're fine!

[green walrus]

I can't be too careful with my meticulous professor

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[final lynx]

what does the hat mean

or the line above the z1

is it magnitude?

and how do you solve for it

i know how to do z1

just -z1 (with the hat or dash)

[violet wind]

the hat means conjugate

and we'd say bar

[final lynx]

ok

[violet wind]

$\hat{i}$

[final lynx]

so if it is 3+4i

the conjucate is 3-4i?

[woven radish]

monikanicity

[violet wind]

^ this is hat

yeah

[final lynx]

ok

what about the negatie

say z = 3+4i

would -z = -3=4i?

[violet wind]

-3-4i

[final lynx]

oh ye that

[violet wind]

yes

[final lynx]

and the conjucate is -z bar = -3 + 4i?

[violet wind]

that would be $\bar{-z}$ yes

[woven radish]

monikanicity

[violet wind]

oh that looks horrible

but yeah

[final lynx]

ok thank u

alg ill do it

if i have questions ill ask again

thank you tho

[violet wind]

np

[devout snow]

@final lynx Has your question been resolved?

[devout snow]

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[foggy wave]

i think in meters what do you think? is my answer correct?

[devout snow]

Please don't occupy multiple help channels.

[wooden veldt]

meters does seem like the most sensible choice yes

[foggy wave]

alright thanks, i just wanted to make sure.

.close

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[chrome gulch]

What is the general term of this?

[chrome gulch]

(n!)/?

[viral sluice]

I guess n!/(2n-1)!!

[chrome gulch]

wait what is !! ? how does it work ?

[viral sluice]

It's called double factorial

[chrome gulch]

aight, going to google that one

[viral sluice]

You multiply every number from 1 to n but only the ones that have the same parity as n

so, say you have 5, yiu multiply every odd number from 1 to 5

so 1*3*5

[chrome gulch]

Damn i have never heard about this before

Thank you!

[viral sluice]

Yw!

[chrome gulch]

.close

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[lavish wind]

so i got a

but dont know how to do b at all

i was told to do it with the e functionn but i dont really understand

the task is to conclude the binomnial theorem from the formula in a

(f*g)^n is the devirate

n th devirate

[sleek yacht]

(e^ax * e^bx)

the nth derivative is (a+b)^n e^(a+b)x

and using the leibnitz thing will give you the sum you want

then simplify with e^(a+b)x

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[dull otter]

why is that these points are exactly 180 degrees from each other and are the same distance away from the x axis?

[dull otter]

the 2 functions are a sin function and a cos function

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[dull otter]

the second one

i think

[earnest stirrup]

if it says so, then prob only one of them has no solution

[dull otter]

the angle is greater than 90 in the second one

so the red side has to be greater than the green side

but since the red side is given to not be greater than the green side

the parallelogram cannot exist

second one

ya but the angle is less than 90

[earnest stirrup]

in a triangle, opposite to a greater angle, lies a longer side

[clear jay]

Yea exactly

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[dark dawn]

is there a relationship between the trace and the A_12 and A_21 elements?

[wooden veldt]

i would just explicitly calculate each trace

[dark dawn]

Is it correct to say 0=b+c?

[wooden veldt]

yeah that checks out

[dark dawn]

soo

there is a relationship between the diagonal elements and

the upper right and lower left?

I thought tr(A) was the sum of A_ii

[wooden veldt]

unclear what you're asking

the trace is the sum of the A_ii yes

[dark dawn]

and is there a relationship between those elements and the A_21 and A_12 elements?

OH

nevermind

we're all good

[wooden veldt]

im not sure why youre looking for such a relationship yeah

you answered the question

[dark dawn]

I was convinced that I was looking for it to equal 2 for some reason

got it mixed up

do you reckon you could answer another one of my questions?

[wooden veldt]

can always try

[dark dawn]

is it possible to solve a 3x2 matrix ?

for 2 variables?

[wooden veldt]

depends on the matrix

[dark dawn]

uhh lemme post working

at the bottom

[wooden veldt]

,w rref {{10,-1,0},{-10,-1,0},{22,-3,-4}}

[woven radish]

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Requested by iCaird#6258

[wooden veldt]

eh you might have messed up somewhere, bc this tells us 0=1 and there are no solutions

[dark dawn]

Painn and suffferrrrinnngggg

I dont understand why I cant do it with a matrix

because backwards subbing works

just above it

and it gives a correct intersection point on the line

[wooden veldt]

check second equation

[dark dawn]

I dont see anything wrong with it

[wooden veldt]

2 - 10t = 2 - s

right?

for y coords of each line

ah i just saw the pic above

you wrote 2+s

[dark dawn]

I think i'll just have to accept my fate

[wooden veldt]

just fix your typo

its just a minus error

1 + 10t = 1 + s
2 - 10t = 2 - s
4 + 22t = 3s

[dark dawn]

-10t+s=0

[wooden veldt]

which is the same as the x coord equation 10t - s = 0

so you really have two equations in two unknowns

[dark dawn]

wait

this +1

its possible now

5 minutes and I shall be back

[wooden veldt]

yes and you'll get your 1/2 and 5 you wanted

[dark dawn]

LETS GOOOO

thanks

was blind asf