help-33

[elfin berry]

USS-Enterprise

[waxen dust]

Just solve for x from here

[ionic mountain]

k

so it be 3x+25-15x=10
then 3x-15x=10-25
then -12x=-15
then x=-15/-12
then it be x=15/12
then it be 5/4

so x=5/4

right

[waxen dust]

Correct.

Now what?

[ionic mountain]

then subite 5/4 into the y

right

[waxen dust]

Well the phrasing is incorrect

You are trying to find y

So you can't substitute y for 5/4

[ionic mountain]

i meant x

be 5/4

[waxen dust]

Yes.

[ionic mountain]

then you slove for y

right

[waxen dust]

Into one of the earlier equations

[ionic mountain]

so it be 15/4+y=5

right

[waxen dust]

Correct.

[ionic mountain]

then it be y=5-15/4

[waxen dust]

Yes

[ionic mountain]

how to slove from here

[waxen dust]

Well

You are substracting 15/4 from 5

Convert to fractions with the same denominator

[ionic mountain]

so then it be `15/4 -20/4

right

since 5/1

time 4

is 20/4

[waxen dust]

Yes but you switched the numbers around

You've got 5 - 15/4 not 15/4 - 5

[ionic mountain]

then it be 15/4-20/4

right

so that be -5/4

[waxen dust]

No, again, you've got $y = 5 - \frac{15}{4}$

[elfin berry]

USS-Enterprise

[waxen dust]

You convert 5 to a fraction

[ionic mountain]

that be 5/1

sicne 5 has a demotir of 1

[waxen dust]

$y = \frac{5}{1} - \frac{15}{4}$

[elfin berry]

USS-Enterprise

[waxen dust]

Then bring them to a common denominator

[ionic mountain]

then 1 times 4 is 4

,w so it be 1*4

[elfin berry]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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Requested by pratyush1234567

[ionic mountain]

so it be 1*4=4

then i have to do 5*4

also

[waxen dust]

Yes

$y = \frac{5 \cdot 4}{1 \cdot 4} - \frac{15}{4}$

[elfin berry]

USS-Enterprise

[waxen dust]

$y = \frac{20}{4} - \frac{15}{4}$

[elfin berry]

USS-Enterprise

[waxen dust]

Notice it's 20/4 - 15/4 not 15/4 - 20/4

You switched the numbers for some reason

[ionic mountain]

okay

idk why

then

[waxen dust]

Me neither 😅

[ionic mountain]

then it be 5/4

wight

[waxen dust]

Correct

$(x, y) = \left(\frac{5}{4}, \frac{5}{4}\right)$

[elfin berry]

USS-Enterprise

[waxen dust]

Is the solution to your system

[ionic mountain]

so it be $y=/frac{5/4}

so it be ,w y=5/4

,w y=5/4

[elfin berry]

Results provided by WolframAlpha

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Requested by pratyush1234567

[ionic mountain]

right]

[waxen dust]

Correct

$x = \frac{5}{4} \\ y = \frac{5}{4}$

[elfin berry]

USS-Enterprise

[ionic mountain]

okay

thx

for your help

[waxen dust]

No problem

[ionic mountain]

https://tenor.com/view/thank-you-thank-thanks-thank-you-images-thank-you-so-much-gif-10869781250610919366

[waxen dust]

Have a good day

[ionic mountain]

bye

.close

[marsh citrus]

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[marsh citrus]

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[copper ice]

does (y-x) always have to be possite? or negativ is also okay?

[spark otter]

positive, 0 and negative are okay

[copper ice]

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[thorn marsh]

for this question, i got 1/(1+2e^2)

i took the derivative of x+2e^x which i assume is 1+2e^x right?

what am i doing wrong

[still temple]

@thorn marsh You're confusing 2 different meanings of ^-1, and you're also messing up the order

First of all, the ^-1 comes before the '

Second, here it means "the inverse of the function", not "the function raised to -1"

[thorn marsh]

yeah i know

im following this

[still temple]

o shit

[thorn marsh]

is my derivative wrong?

[still temple]

@thorn marsh Okay so notice that in the formula you were given there's a y on the left side and an x on the right side

They are not the same variable

[thorn marsh]

ohh

[still temple]

Since y=f(x), x=f^-1(y)

[thorn marsh]

okok

[still temple]

So the full formula is (f^-1)'(y)=1/f'(f^-1(y))

I had somehow never come across this formula before

[thorn marsh]

alright ill try this question again

ty

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[topaz dock]

is x = -2 the only zero? because i was working on this question with someone and we only found that zero

[jagged relic]

The only real one yeah

[topaz dock]

oh

the question was to find all zeros of the polynomial

[night mica]

$x^5+x^3+8x^2+8=x^2(x^3+8)+(x^3+8)=(x^2+1)(x^3+2^3)=(x^2+1)(x+2)(x^2-x+1)$

[jagged relic]

Do you know complex numbers?

[elfin berry]

everg

[night mica]

the two quadratics have no real roots (study $\Delta$)

[elfin berry]

everg

[topaz dock]

wait im so confused

[night mica]

by what ?

[topaz dock]

so -2 is the only zero? yes or no

[jagged relic]

If you don't know complex numbers, then yes

[night mica]

yes i gave a proof

[topaz dock]

ok. but you just said that they have no real roots

[night mica]

the quadratics*

but (x+2) has x=-2 root, ahd its a linear polynomial

[topaz dock]

im just confused because its a 5th degree polynomial, so wouldnt it have 5 roots?

or zeros?

[night mica]

only in the complex plane

[jagged relic]

It does, the other roots are complex

[topaz dock]

so for my answer i could just say -2

[night mica]

but in the real number a polynomial might not have roots

[jagged relic]

It depends on whether you've learned complex numbers are not, like I've mentioned twice

[topaz dock]

ugh the topic is finding complex zeros

ughhh but i have no more space

[jagged relic]

Well, then you need to find the four other roots

[night mica]

good luck!

[topaz dock]

alright

[night mica]

hint: $\pm i$ might be useful here 😉

[jagged relic]

That's not a hint lmao

[night mica]

😭

[elfin berry]

everg

[night mica]

is it better now ?

[jagged relic]

That's like saying "this has to do with complex numbers"

[night mica]

😦

[jagged relic]

Just let them ask for help if they need it

[marsh citrus]

@topaz dock Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[wheat needle]

what the difference between subscript as infinite and without

[marsh citrus]

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[azure bone]

Idk

That's not in standard form right

Idk

70+6(n-1)

Idk

<@&286206848099549185>

[tepid salmon]

hi

[azure bone]

Henlo

[tepid salmon]

I believe the answer should be f(1)

[azure bone]

So true

Wat is the second thing

6

[tepid salmon]

wait a minute

f(1) = 70

f(n) = f(n-1) + 6

[azure bone]

Ty

.close

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[tepid salmon]

sorry, was paying attention else where.

ok

was it the right answer?

[tepid salmon]

was it correct?

[azure bone]

Yes

[marsh citrus]

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[severe spire]

I need help determining when the function is increasing after getting the first derivative and after finding the critical numbers

[severe spire]

well increasing and decreasing

here's the original function

<@&286206848099549185>

<@&286206848099549185>

[still temple]

@arctic hare

[severe spire]

hello

.close

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[sullen ice]

what is happening here?

[copper raven]

what specifically don't you get ?

[spark otter]

A = 2A^T

so substituting A by what it's equal to

[sullen ice]

why did they have to do that?

to get rid of the transpose?

[spark otter]

why did they want to do that

and yes to find some equation on A that doesn't involve the transpose

[sullen ice]

okay ty

[marsh citrus]

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[neon bridge]

how do i do this

[marsh citrus]

@neon bridge Has your question been resolved?

[neon bridge]

<@&286206848099549185>

[tepid salmon]

don't cheat.

[neon bridge]

i’m not cheating, this is a practice exam paper

[terse turtle]

you could construct a line segment OB

[neon bridge]

like this?

<@&286206848099549185>

[marsh citrus]

@neon bridge Has your question been resolved?

[fresh oyster]

yeah

now use the inscribed angle theorem
on this diagram, 2*<AOB = <ACB

[fresh oyster]

Because DO and OB are both the radius, so its isosceles

So we can find all the angles of the triangle you drew and apply this theorem

[marsh citrus]

@neon bridge Has your question been resolved?

[marsh citrus]

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[fathom mountain]

$z^3=\frac{1}{3}i|z| \overline{z}$

[elfin berry]

DeFra

[fathom mountain]

How to develop this

[copper raven]

we pretty much solved it for you already

are there some steps you didn't understand ?

@fathom mountain

[fathom mountain]

I need to try developing it mormally

[fervent yarrow]

you mean (x+iy)^3??

[fathom mountain]

?

[copper raven]

wdym by "develop this normally" then ?

[fathom mountain]

Usually this exercise end up with z=something

With only 1 z

[copper raven]

if z=re^(ix), then
r^3 = 1/3 r^2 [radius]
and 3x = (pi/2 - x) + 2kpi [angle]

taking back what we did before

r^3 = 1/3 r^2, i.e. r^2(r-1/3) = 0, so either r=0 or r=1/3

3x = (pi/2 - x) + 2kpi, i.e. 4x = pi/2 + 2k*pi, i.e. x = pi/8 + k*pi/2

so your solutions are either z=0 (for r = 0)

or z=1/3 e^(i * pi/8 + k * pi/2) (for r = 1/3)

@fathom mountain

[fathom mountain]

Is this the only way to solve it? @copper raven

[copper raven]

it's the most straightforward way to do it

[fathom mountain]

What other way are you thinking of?

[copper raven]

as adam chebil mentioned, you could go the z=a+ib way, but it's a huge pain

[fathom mountain]

Im not sure what he meant

[copper raven]

well every complex number, you can write as z=a+ib right

[fathom mountain]

Yes

[copper raven]

then you plug it in the equation

(a+ib)^3 = ...

and you find conditions on a and b such that the equation is satisfied

[copper raven]

it's the same thing as the other method, except with cartesian form instead of polar form

[fathom mountain]

Like this?

[fervent yarrow]

i don't recommend doing that i just said that cause he said that he wants to develop something...

[copper raven]

yeah ik

[fervent yarrow]

but if u're ok with developing (x+iy)^3 then go ahead

but exponential form makes everything simpler (most of the time)

[fathom mountain]

The other we did were developing right?

[fervent yarrow]

u mean older problems i helped u solve?

[fathom mountain]

Ye

[fervent yarrow]

i don't recall

if z is not raised to a high power then z=x+iy is the way to go imo

[fathom mountain]

Why is ø conjz = pi/2 - x

[copper raven]

no

arg(conj(z)) = -arg(z)

i just incorporated the i = e^(i pi/2) afterwards

so arg(i conj(z)) = pi/2 - arg(z)

@fathom mountain

[fathom mountain]

Oh its i

Ok

So its x= 2pi?

[copper raven]

why are you saying that ?

[fathom mountain]

3x=pi/2 - x

Sry

3/2pi

[copper raven]

right so 4x = pi/2 + 2kpi

[fathom mountain]

Oh ok

Wait why

Why 2kpi

[copper raven]

it's to get all the possible solutions to the equation

since e^(ix) = e^(i(x + 2kpi))

[fathom mountain]

Whats k here

[copper raven]

an integer

[fathom mountain]

What integers

[copper raven]

it's exactly like when you're solving trigonometry equations

[copper raven]

any integer

[fathom mountain]

Ih wait

Its just 2kpi

[copper raven]

if you don't account for that, you might miss a lot of solutions

[fathom mountain]

So it stays the same

[copper raven]

yeah

[fathom mountain]

But if u solve z=1+i

U dont do 2kpi

Why

[copper raven]

why would you do it

there's nothing to solve for here, you already have the solution

you already have z=1+i

there's nothing to do

[fathom mountain]

What about r

[copper raven]

maybe you typo'd

[fathom mountain]

R^3 = 1/3 r^2

[copper raven]

did you mean z^2 = 1+i or something

[fathom mountain]

Nono nvm

[copper raven]

there's no 2kpi action going on here

if two numbers have different radiuses, they'll never be equal

but for the angle you have to be careful

[fathom mountain]

So uhh

No radius?

[copper raven]

wdym no radius?

[fathom mountain]

We have to find r

[copper raven]

yeah so just r^3 = 1/3 r^2

you don't have to add 2kpi or something else

[fathom mountain]

We have 2 radius

[copper raven]

yes there's the radius of the left side of the equation (r^3)

and the radius on the right side (1/3 r^2)

those two have to be equal if you want the left and right side complex numbers to be equal

[fathom mountain]

Yes but r solutions are 2

[copper raven]

wdym ? @fathom mountain

[fathom mountain]

Solutions of r

Are 2

0 amd 1/3

[copper raven]

yeah

what's the problem with that

@fathom mountain

[fathom mountain]

Nothing just check

.close

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[cobalt lodge]

Weight = Lenght^3 x 9630
Lenght 0.09 meter

Why does the fish weigh 268gm when I get 800gm

Length in meters and weight in grams of the fish in the water so that the weight is proportional to the third power of the length. The proportionality constant is 9630.

[marsh citrus]

@cobalt lodge Has your question been resolved?

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.

[cobalt lodge]

@patent pelican

can you help?

tyy

no

i did 0.09^3

wtf now i got 7

wait

its 0,3^3

wait

0.03^3

now i get 0.26001

wttfffff

i dont udnerstand

error

lengt is 30cm

and 30cm in meter is 0.03

nvm 0,3

ye i got right answer now

i got 1 more question

'

V/A

How do I do that?

It is

But how?

r/3

How do I do this?

whatttttttttttttttt???????????

i will send u picture

1 sec

'

now i didint understand anything

like?

[cobalt lodge]

would i get right answer here?

Thats what I did bro

3(4 x pi x r^2)

what is that?

12 x 12pi x 12 r^2?

[cobalt lodge]

💀

12 x 3pi x 3r^2

ok

okkkkkk

super easy

i understood it now

thanks alot

i did 2 pages of my math book since last time we talked

🙂

hope i can get quicker

.close

[marsh citrus]

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[last tartan]

bro is a good person

W

[marsh citrus]

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[spring sleet]

"we consider the in the set E ....1)show that.....2)show that......3) is ]0,2] included in E? justify"

any idea how to begin?

[stoic saddle]

do there exist natural numbers x and y such that (x+y)/(xy) = sqrt(2)?

[bleak dome]

x = y = sqrt 2

[stoic saddle]

don't give out answers esp. wrong ones

[bleak dome]

mb

[marsh citrus]

@spring sleet Has your question been resolved?

[spring sleet]

no

[stoic saddle]

can you prove that?

[spring sleet]

whatever i replace x and y with it gives me in this format "a/b"

[marsh citrus]

@spring sleet Has your question been resolved?

[spring sleet]

@stoic saddle

[spring sleet]

<@&286206848099549185>

[spring sleet]

.

[spring sleet]

.close

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[ionic mountain]

this rught

since 40*100=4000

4000/40=100

than 100*300=30000 salior she can make for

right

[marsh citrus]

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[marsh citrus]

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[cunning jackal]

it wont occur only at 270

this function is periodic

but taking it as 270 will give you a valid solution

since sin(270 degress) is -1

as i said your answer will also be valid

unless he restricts you to a range of theta

no

theres only 1 minimum value

which occurs at more than 1 points

thats how a periodic funciton works

it repeats itself

after a certain interval

which in this case is 2pi

so by plugging every angle which is of the form

-pi/2+2npi

where n is integer

youll get an x at which you get the minimum value

yes you will ofc

and that answer will also be correct

because minima need not necessarily be obtained at one point

for example

minimum value of sinx is -1

you get that at infinite values of x

-90, 270, 270+360, 270+720 and so on

yes

[marsh citrus]

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[white lion]

If Card(E) = n, how would you go about calculating this :
$\sum_{Y\subseteq E}^{}2^{Card(Y)}$

[elfin berry]

Azenx

[wet holly]

that's basically same as calculating $\sum_{i = 0}^n \binom{n}{i}2^i$ cus there are nCi subsets of size i

[elfin berry]

992qqoloy

[wet holly]

which u can do with binomial theorem

[white lion]

ok thx

.close

[marsh citrus]

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[azure storm]

Is this correct? Thank You

[white lion]

3²sqrt(3) = 9sqrt(3) and there is no further simplifying

[azure storm]

why?

[white lion]

sqrt(ab) = sqrt(a)sqrt(b)

So you'd have

sqrt((3²)²)sqrt(3)

= sqrt(3^4*3)

= sqrt(3^5)

Which is not simplifyable

[azure storm]

i see, thank you very much, though I still have a question

[white lion]

yes

[azure storm]

It appeard an exercise wich the top one was correct

if the top works, why doesn't the lower one also work?

wait

what

never mind. . .

the f?

wow, it was just a coincidence

jeez. . .

Thanks gor your help

[white lion]

np

[marsh citrus]

@azure storm Has your question been resolved?

[marsh citrus]

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[ionic mountain]

this right

[desert dirge]

seems legit

[ionic mountain]

becuase t is a

5 is h and -9 is k

rgiht

<@&286206848099549185>

[still temple]

yes it is

[marsh citrus]

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[stark field]

I asked someone when do we meet, 9AM or 14PM. They responded: Ok

[stark field]

does anyone understand if they meant 9AM ?

[summer trench]

#discussion perhaps

[frozen gate]

Most mathematical question ever

[stark field]

yes

[still temple]

sadly we are mathematicians not mind readers

[fervent yarrow]

it means he's not gonna meet you

from my experience

[stark field]

hmmm

[naive sage]

prob heard you wrong

[crystal wraith]

That's why use 24 hour system

[stark field]

it was messages

[still temple]

this guy gets it

[summer trench]

24 hour system is trash

[still temple]

💪💪

[frozen gate]

Why don't you ask again for clarification :3

[stark field]

i dont want to come off as anoying

[crystal wraith]

You'll change your opinion in ||445 days||

[stoic saddle]

i mean enjoy the rest of your life being wrong lmfao

[frozen gate]

This question ironically has the most amount of helpers helping

[crystal wraith]

Close?

[crystal wraith]

EXACTLY!!

[stark field]

hmm ok

[naive sage]

lmao

[stark field]

there is no scientific concensus

[stoic saddle]

12h time is fine in like, casual speech

[still temple]

it's fine in spoken language

[summer trench]

wait no, it really is, its just confusing

[still temple]

and in short intervals

[naive sage]

so like you look are your phone and go Oh its 13'o clock

[crystal wraith]

It's*

[stoic saddle]

how the fuck is a straight count from 0 to 23 more confusing than the am/pm stuff

[naive sage]

idk maybe im just used to 12 hours

[still temple]

you are not going to be saying "let's meet at 21:30" to someone in your voice

that's cursed

[stoic saddle]

like if you want to call your friend and say "meet me at the mall at six"

that's fine

[naive sage]

so like clocks are all confusing then to you guys?

[stoic saddle]

in written correspondence you should always use 24h imo lest there be any miscommunication

[still temple]

agreed

[naive sage]

thats why am and pm exist

but ok

[stoic saddle]

24h users don't have issues with analog clocks

at least i don't

[summer trench]

idk but i never used 24 hours system so whenever someone says like 17 i gotta add 12+.... all this

[stark field]

I agree

[stoic saddle]

if anything you would need to SUBTRACT 12.

17:00 is 5:00 pm.

[stark field]

Ann is right

[naive sage]

yeah whenever I look at something like that I have to do the mental math first

[summer trench]

ik

[crystal wraith]

Annihilator Ann. Where's your honorable at?

[summer trench]

lol i just prefer normal 12 hour one

[naive sage]

it takes me like an extra half second just to read the time

[stoic saddle]

@crystal wraith nyb

[stark field]

High intelectual discussion about hours

[crystal wraith]

Fair enough.

[stark trail]

He will meet you at both times

clearly

[stark field]

hmm

prior to this he said: whenever (yesterday)

when I suggested meeting in (monday)

today (sunday) i say 9 or 14?

[wet holly]

u should try asking ur teacher to clarify

[naive sage]

just choose one

[stark field]

hmmm

yes this causes stress

i think should I respond like

I say: Ok so 9? 👍 🥂

[wet holly]

nah just ghost

[stoic saddle]

09:00 or 14:00

ask him that

[gilded smelt]

as is clear language relies on defaults and implications

I believe the first option is meant to be the default

[stoic saddle]

also

"this or that" "ok"

is just bad communication

communication 101: when somebody asks you a question, you answer

[stark field]

ye

[naive sage]

but that is an answer

[gilded smelt]

they did give an answer tbf

it's just the wrong type

[stark field]

hmmm

[gilded smelt]

I suggest we extend types to language

let C and its children claim its first human language

[wet holly]

the only obvious answer is to ghost after replying with an "ok"

[stoic saddle]

communication 102: when somebody asks you a question, you give them an INFORMATIVE answer.

[gilded smelt]

hmmm ann is right try to extract metadata from the answer (like their IP and real name)

[stoic saddle]

<@&268886789983436800> troll.

[stark field]

the guy uses drugs

[naive sage]

okkk then dont meet them

[stoic saddle]

how critical is it that you meet with him?

[stark field]

so he doesnt care like

maybe tomorow in 9 we just meet

and 1 hour prior i just check

[gilded smelt]

but yeah in all seriousness if they said ok you pick the time (and tell them) and let them deal with it

[stark field]

should i just react to his message with 👍

[naive sage]

thats even more confusing

[stark field]

yes 😄

i give it back to him

[gilded smelt]

do you want to meet him?

[stark field]

taste his own medicine

[gilded smelt]

if so, then you need to be the helpful one, sadly

[stark field]

yes we meetup in coffeshop

[frozen gate]

Is this a date

[stark field]

no

old friend

[frozen gate]

oh then he probably just misread your text

[naive sage]

why are you so anxious about clarifying if it's an old friend?

[stark field]

wym

[naive sage]

just ask again to clarify

[stark field]

oh

[naive sage]

thats not rude

[frozen gate]

if my old friend (or even any friend) asked me for clarification after I didn't answer their question I wouldn't be upset

[stark field]

ye fr

[frozen gate]

In fact I'd be more upset at myself for not answering properly

[stark field]

most likely he dont care what time we meet

but my college is 10-14

[stark trail]

I know a solution

[stark field]

so i say 9 or 14?

[stark trail]

just tell him you have to reschedule for a diferent time

and when he asks what time

[naive sage]

he prob read your text as can you meet either 9 or 14?

[stark trail]

say yes

[stark field]

😄

[naive sage]

damn

[stark field]

ok i go send msg

[idle ridge]

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[idle ridge]

This isn't really a mathematical question, so please move the discussion to one of the discussion channels like #discussion

.close

[marsh citrus]

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[manic pawn]

Can someone explain to me what is going on with this chain rule proof

[manic pawn]

The parts i dont understand, is the first line have f(g(x)) - f(g(a)) over x - a

I see that this area highlighted means its derirative of g(x) i guess

but like how, it looks similar to principle where f(x + h) - f(x) over h

[marsh citrus]

@manic pawn Has your question been resolved?

[marsh citrus]

@manic pawn Has your question been resolved?

[manic pawn]

https://tenor.com/view/sad-gif-7523306793289960933

[glass silo]

It is basically the same thing, just written slightly differently (if you chose $h = x - a$, then it retrieves you the definition of the derivative you’ve seen before, with the limit becoming $\lim_{h\to 0} \frac{ g(a + h) - g(a) }{h}$)

[elfin berry]

@glass silo

[glass silo]

Either way the derivative is basically “taking some other point away from the one you’re considering, working out the gradient, and seeing what happens as that point gets closer and closer to the original one”

The equivalent definition $\lim_{x \to a} \frac{ g(x) - g(a) }{ x - a}$ (the gradient between the points $(a, g(a))$ and $(x, g(x))$ is just easier to work with in this case really

[elfin berry]

@glass silo

[manic pawn]

Oh ok thanks

https://tenor.com/view/ryan-reynolds-kiss-handsome-smile-gif-17092194

.close

[marsh citrus]

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[pulsar flume]

How can I even begin this? for the forward implication, I assume G acts faithfully on X, so the identity is the only element of G that leaves every element of X fixed. What direction can i take that assumption to lead toward getting that no two distinct elements have the same action of each element of X?

[marsh citrus]

@pulsar flume Has your question been resolved?

[pulsar flume]

.close

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[marsh citrus]

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[loud hamlet]

Why are the other 2 not inflection points?

[hollow sparrow]

can you show your work on how you ended up with those answers?

oh wait

actually yeah those are correct

but they're in the wrong order

1 > -2 + sqrt(3), so 1 should be the largest value

[marsh citrus]

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[hearty ferry]

someone help how to factor

[marsh citrus]

Please don't occupy multiple help channels.

[hearty ferry]

<@&286206848099549185>

[mighty tide]

what is the problem?

[hearty ferry]

i need to factor the equation but i’m not sure how to

on top left

[mighty tide]

do you know the quadratic formula

?

[hearty ferry]

i dont think so

i’m using a different way called the magic x

not sure if it’s working tho

[mighty tide]

hmm

I tried to factor it

but it is not factorable

there are no integer solutions for this equation

[hearty ferry]

can you give me a example of one that is factorable

and i’ll try and see if i’m doing it the right way

[mighty tide]

okay hmm

4x^2 + 8x + 6

[brittle vessel]

are you sure its not +4 at the end?

[modest bluff]

If +4 itd work

[hearty ferry]

wym?

[mighty tide]

solutions in C ig..

[brittle vessel]

what?

[mighty tide]

just the quadratic formula, it would be negative if it was +4 at the end... just wondering

[brittle vessel]

yes, it would have two neg solution, so?

[mighty tide]

so complex numbers !!

[brittle vessel]

nope

[mighty tide]

like

wouldn't be i there?

[brittle vessel]

no

[brittle vessel]

anyway @hearty ferry are you sure its not 6x^2+11x+4?

[hearty ferry]

idk i just made it up practice i thought it would be factorable

[hearty ferry]

did i do this right tho?

[mighty tide]

nice

[mighty tide]

yup

[mighty tide]

can you explain to me... pls ;-;

[hearty ferry]

how would i do it with a trinomial tho

[mighty tide]

well the magic x is working only for x^2 apparently

[brittle vessel]

https://www.youtube.com/watch?app=desktop&v=CH4jYzajkp0

[mighty tide]

for trinominals, you should separate terms, factor some parts and find the common factor in all terms

[brittle vessel]

,w 6x^2+11x+4

[elfin berry]

Results provided by WolframAlpha

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

Requested by spreeter

[hearty ferry]

i’m too dumb for this

[mighty tide]

ohhh

okay I see it now

thanks

[brittle vessel]

try it with the 6x^2+11x+4

you need 2 numbers left and right which add up to 11 and multiply to 24

and then group and take out the common parts

[marsh citrus]

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[marsh citrus]

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[still temple]

How is this wrong?

here is my work

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[cedar flame]

hey guys

Someone

I need help with some mathwork

[still temple]

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[still temple]

Anyone

[marsh citrus]

@still temple Has your question been resolved?

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[tardy idol]

im not sure why this is wrong, i used the geometric series formula of a/1-r

[still temple]

This series has $u_1 = -\frac{5}{3}$

[elfin berry]

Renegade

[still temple]

and $r = -\frac{5}{3}$

[elfin berry]

Renegade

[tardy idol]

yes

[still temple]

since $|r| < 1$ this converges and you probably made a small error when computing the sum

[elfin berry]

Renegade

[tardy idol]

it would be (-5/(3(1+5/3))) right

which would be -5/8 i thought

because -5/(3+5)

[still temple]

oh yes you are correct

[tardy idol]

oh wait but the abs(-5/3) is >1

[still temple]

it does converge, and it does converge to -5/8

oh

right

[tardy idol]

hahaha

[still temple]

hahahaha

oopsies

[tardy idol]

so the other equation

[still temple]

no no

the other equation is for a sum up to $n$ terms

[elfin berry]

Renegade

[still temple]

i know what you are talking about

but this series just diverges

since $|r| > 1$

[elfin berry]

Renegade

[tardy idol]

oh yes true

i have one more question

this would be geometric too right? with a =1 and r= (1/6) ?

[still temple]

n = 1 you have 1/6

wait

hold up

i think it's supposed to be $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{6^n}$

[elfin berry]

Renegade

[tardy idol]

why so?

[still temple]

because otherwise the power is redundant

since 1^n = 1

[tardy idol]

i agree

[still temple]

for any real number n

[tardy idol]

thats what i thought but maybe theyre just testing me

[still temple]

i doubt it

but we can do it both ways

[tardy idol]

well either way i should solve the way its written

yeah lol

we can do both

[still temple]

if it truly is 1^(n-1)/6^n

then we can simplify that to 1/6^n

so $u_1 = 1/6$ and $r = 1/6$

[elfin berry]

Renegade

[tardy idol]

oh a= 1/6 ohhh this makes sense

[still temple]

in which case it does converge to $\frac{1/6}{5/6} = 1/5$

[elfin berry]

Renegade

[tardy idol]

that was the answer

i guess they just wanted to test that id figure out it was redundant to simplify for a geometric equation?

[still temple]

perhaps

[tardy idol]

could i ask for help on 1 more, i struggle with the ones where they ask of series and sequence

[still temple]

yes of course

[tardy idol]

so for the sequence, it would be 2/5 right?

[still temple]

okay i'd start with part (b)

indeed

[tardy idol]

when the seq converges.. i believe that means the series diverges right?

[still temple]

not quite

it just means it may diverge or converge

[tardy idol]

ok.....

so what does knowing the seq give me? nothing? (genuinely asking)

[still temple]

do you want an example?

the sequence $(a_n) = \frac{1}{n}$ clearly converges as we approach infinity

[elfin berry]

Renegade

[still temple]

in fact, it converges to $0$

[elfin berry]

Renegade

[still temple]

but $\sum_{n=1}^\infty \frac{1}{n}$ is a well-known harmonic series which diverges

[elfin berry]

Renegade

[still temple]

yes nothing

[tardy idol]

OH YEAH i remember hearing harmonic series somewhere

i forget about it all the time tho

[still temple]

for the series

i'd look at the partial sums

and see where that converges

[tardy idol]

partial sum

[cobalt rover]

Why is only this channel running

[still temple]

partial sum means $P_1 = A_1, P_2 = A_1 + A_2, P_3 = A_1 + A_2 + A_3, \hdots$

[elfin berry]

Renegade

[still temple]

because i can't answer everyone's questions at the same time

[cobalt rover]

Since when is there only 1 helper

[still temple]

There isn't, but the others could be busy

[tardy idol]

oh yeah i see i see

[still temple]

after all helping here is volunteer work

[still temple]

yes

so

[cobalt rover]

Ahh ok

[still temple]

$P_1 = 4/11$

[elfin berry]

Renegade

[still temple]

$P_2 = 4/11 + 8/22 = 0.72727272723$ something like that

[elfin berry]

Renegade

[tardy idol]

yes

okok i see isee

what does knowing the partial sum do? just show where its going?

[still temple]

yes

[tardy idol]

would the ratio test work here too?

and is the partial sum thing work well for most series?

[still temple]

the ratio test is a test for convergence

not for evaluation of sums

[still temple]

ehh

i wouldn't say most

[tardy idol]

but it would tell if the series converges i though?

[still temple]

yes

that is correct

[tardy idol]

I think i understand

well thank you

i wish you luck with your helping of people

im sorry for that guy that was shitting on you 😭

youre doing very well

[still temple]

thank you very much, best of luck and hope to see you here again

[tardy idol]

.close

[marsh citrus]

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[lost stratus]

"Given is a certain inhomogenous linear system of equations over the real numbers with four equations and three unknowns. It is revealed that the vector v_p is a particular solution. Is the vector 2 * v_p a solution to the system?"

My answer:

For it to be possible for 2 * vp to be an answer to the system, it should be so that:

c1 * x1 + c2 * x2 + c3 * x3 = k
while also:
2 * (c1 * x1 + c2 * x2 + c3 * x3) = k

If this is true, then 2k should be equal to k. But we know from the description that this is an inhomogenous system of equations, which means k =/= 0 and that k = 2k is not possible. Because of this, it means that 2 * vp is not a solution to the system of equations"

I would like some feedback on this answer since I want it to be 100% correct and up to standards with little to no room for criticism about not explaning in depth

[quaint elm]

didn't you already ask this

[marsh citrus]

@lost stratus Has your question been resolved?

[lost stratus]

Yes but I keep getting different answers

So I’m asking one last time

[marsh citrus]

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[untold wyvern]

For part(i) regarding whether VunionW is a subspace, is there any smarter way to solve this question? What I did was to manually find counter-example. (4/3,2/3,0,0,0) is an element of VunionW since its an element of V while 2(4/3,2/3,0,0,0) is not an element of VunionW since its neither an element of V nor W

[marsh citrus]

@untold wyvern Has your question been resolved?

[marsh citrus]

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[cloud field]

find the quadrant for when sin is positive and cos is negative

thats your answer

[marsh citrus]

@wind plover Has your question been resolved?

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[low mica]

how to solve

[low mica]

i have t.p

t.p is

(17.5, 12)

y = a(x-17.5) ^2 + 12

how do i find 'a'

[cloud field]

you could try the point (0, 4.65)

[marsh citrus]

@low mica Has your question been resolved?

[marsh citrus]

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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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[waxen dust]

[floral linden]

?

what's your question

what

[stoic saddle]

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[stoic saddle]

so just a few things regarding how the help channels work:

1. your first msg should contain your question
2. that's not how you use .reopen -- it is for reopening a channel you already had because you want to ask something else in it. in fact, putting a . at the start of a message in an available channel SUPPRESSES the usual behavior of opening the channel.
3. no, we don't give out answers here. we can, however, guide you through the solution. and we will, if you cooperate.

2 years ago, Katie was 3 times her sister's age. Four years from now, Katie will be older than twice her sister.

what are we looking for, exactly?

are we looking for katie's current age, the sister's current age, or both?

ok, katie's current age.

alright.

have you made any progress on this?

(it's ok if not, but we need to know)

ok, show what you've got so far.

that's it?

weh. not good.

plus the ratio would be 3:1 anyway

but this won't be of much use to us

do you know your way around algebra?

like basic stuff, solving linear equations

4th grade...?

how old is that?

like 10, no?

so you are 10 years old?

sorry kid, you can't be on discord at all

not just on this server but you are too young to use the app

<@&268886789983436800> we have an underage

[floral linden]

lmfao

[stoic saddle]

no, your first statement about your own age is taken at face value.

[unborn condor]

then this should be obvious

[floral linden]

which curriculum?

[stoic saddle]

you have claimed to be 8 years old

[floral linden]

which curriculum are you in dUbakur?

you say you're in 10th grade

in india

what

[unborn condor]

india doesnt have grades, as far as i'm aware

[stark cipher]

🤨

[floral linden]

bio science is not a curriculum 🤣

???

[stoic saddle]

hi dghost

[floral linden]

bro

[stark cipher]

Hey, just looking now

[floral linden]

we do

im from india

are you in ICSE, CBSE, or what

[stoic saddle]

op claimed to be from the uk, initially said he was in 4th grade, when asked about his age claimed to be 8
then changed his story to be a 10th grader from india

[unborn condor]

they are not called grades are they?

[floral linden]

what textbook does your school use?

[floral linden]

no, they're called standards

[stoic saddle]

okay so

[unborn condor]

precisely. So my point stands, and as Ann is saying, he's just lying from the start :/

[floral linden]

yeah

[stoic saddle]

@still temple pending moderatory judgment, let me ask you this to see whether you know it or not:
if i gave you the following problem:

7x + 11 = 92, solve for x.

could you do it? Y/N/R

that is incorrect, how did you get that?

[floral linden]

can you take a picture of this textbook with your hand next to it

[stark cipher]

I feel like you weren't being serious about being 8 but it's not something to joke about on Discord either

[stark cipher]

if you can do ^ then I'll be content

[stoic saddle]

i on the other hand would like to know what you did to get from 7x + 11 = 92 to x = 72 in my example problem

because i would like to actually guide you through that age problem you came here with

i'm also going to be busy for the next... hopefully no more than 20 minutes

[stark cipher]

Yep, that's all good

[stoic saddle]

ok so

right yeah of course. i need to do some paperwork irl, but i'll be back with you in 20.

while i'm gone, please show your work for solving the equation 7x+11=92 -- i want to know what mistakes you've made so i can address them.

[marsh citrus]

@still temple Has your question been resolved?

[young sage]

no

im dUbakur from my brothers account

hes in college

ann are u back

and the answer to your question is

73

[stoic saddle]

@still temple @young sage whichever account you respond from, i asked you to show your work and you still aren't answering... if you just guessed x=73, then say you guessed it.

[young sage]

no

[stoic saddle]

also no i'm still a bit busy, sorry.

[young sage]

k tell me when ur back

[stoic saddle]

if the channel closes before i'm back, open a new one and ping me there.

[young sage]

11 + 7-92 + x

friend me

[marsh citrus]

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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>.
• Type the command .close to free the channel when you're done.
• 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.

[woeful shadow]

need help

[woeful shadow]

could someone help?

i dont understand the working out for Q1 a and b

[echo bough]

its pretty straightoforward for 1)a)

what have you tried?

[woeful shadow]

absolutely nothing

i dont understand it

i look at someone else’s working out and it didnt make sense

[still temple]

for a), try subbing in r = n-r into nCr

[echo bough]

try writing both nCr

and nC(n-r)

and see what you get for both