help-41

[jovial field]

1/n^3?

[grizzled pagoda]

bacc

[jovial field]

Ok ok fair enough

n^1/3

Ok fair enough

[patent raptor]

,w plot e^(1/x^(1/3))

[grizzled pagoda]

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[jovial field]

You sure it n^1/2

[patent raptor]

ln(x) > 1/x^(1/3)
x > e^(1/x^(1/3)) if x>2

yea now i am

[jovial field]

No

Like 1/(n)^1/3*(n)^3/2

[patent raptor]

what

[jovial field]

Is equal you to n^1/2

Bro listen

@patent raptor

Apply normal lim n tends to 0

[patent raptor]

to what

[jovial field]

,w lim n tends to infinity (logn)^2/n^(3/2)

[grizzled pagoda]

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[jovial field]

Oof

[patent raptor]

yea

every polynomial is always dominant over log

[jovial field]

No I mean

[patent raptor]

what

[jovial field]

How

How this became 1/2

Pls guide

I am tired

I wanna rest now

And I want you to get rest too

[patent raptor]

bro i realized

i am doing too many mistakes

[jovial field]

Yes

[patent raptor]

i will stop for now

[jovial field]

Ok

Good night/day/afternoon everything

Bye

Let's sleep

I just have 15 due

I will do it it tomorrow

[patent raptor]

aight bet

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[solar sphinx]

what's the amplitude of the wave

[solar sphinx]

I put 4 it didn't work

idk

[hollow night]

maybe different units because its 4mm?

[solar sphinx]

ohh I put meters

I'm so dumb

thanks it worked now

[hollow night]

👍

[solar sphinx]

can someone help I don't understand this question

<@&286206848099549185>

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[round dome]

no matter wich way i try to solve this i always end up with the same thing in a loop ( ln(sin) . cot )

[neat wind]

this is not ibp problem

[night jacinth]

it works tho

but yeah a tad unconvential

[round dome]

not a english speaker, what would be ibp?

and if not, what way would you suggest to solving this problem

[night jacinth]

ibp (integration by parts) is what you did, the u dv stuff

[night jacinth]

!occupied

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[round dome]

thanks, have any ideia whats the best way to approach this problem then?

[night jacinth]

u-substitution is probably the way most ppl would handle that integral

as you did yourself, if you set u=ln(sin(x)) then du=cot(x)dx

so you can rewrite your integral as $\int u\dd{u}$

[grizzled pagoda]

aPlatypus

[night jacinth]

which is easy to integrate

and then when you integrated, replace u w/ what you had at the beginning

[night jacinth]

but you were pretty much done with the ibp tbh

the loop is a good thing

cause it tells you that $$\int \cot(x) \ln(\sin(x)) \dd{x} = \ln(\sin(x))^2 - \int \cot(x) \ln(\sin(x)) \dd{x}$$

[grizzled pagoda]

aPlatypus

[night jacinth]

you have twice the same integral on both sides, just rearrange your equation a bit

$$2\int \cot(x) \ln(\sin(x)) \dd{x} = \ln(\sin(x))^2$$

[grizzled pagoda]

aPlatypus

[night jacinth]

and boom you automagically solved your integral

[round dome]

thanks, that makes a lot of sense

and also, i did for u-sub and it worked as well, thank you

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[split sail]

guys, very simple question but how do u geet the nth term of a sequence

[split sail]

and the common difference

the fastestway

arithmetic sequence btw

[pallid canopy]

You just plug in n if you have the sequence

[spiral zealot]

depends on what you're given

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[rain coral]

.close

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[timid wraith]

Q.6

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<@&286206848099549185>

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<@&286206848099549185>

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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[timid wraith]

1

<@&286206848099549185>

[errant rampart]

A. represent r with p, q, then complete the square in favor of p or q

B. q^2 <= 4p, complete the square

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[storm moth]

a bit lost with conjugacy classes and orbits. not sure how they relate too.

given a group G acting on a set S, for every element s, s' S if we can find a g in G such that gs=s' we say s, s' are in the same orbit.

conjugacy class.. not sure.. what i got is The nontrivial orbits of the action are called the conjugacy classes of unsure how to use it.

[indigo cloud]

there is the specific group action of conjugation

which is so important that its orbits got their own name

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[gaunt vale]

hello

[gaunt vale]

need help with this question in geometry

the tangents in points B and C meet at point F. AF intersects with BC at point D and with the circle at point E.

prove that BD/DC = Area of triangle ABF / Area of triangle ACF

The thing is im getting another known factor that DC is not qual to BD in the next section and i think that if we have to prove what i wrote, it means that they are equal

im kinda confused with the question

<@&286206848099549185>

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[long shard]

Is A some arbitrary point on the circle

I think I got it

U still there?

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[celest badger]

if 2 triangles have 2 lengths the same does it mean theyre congruent or similiar on none

[celest badger]

.close

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[solar grove]

2 sides which are the same?

you cant determine if they are similar or congruent

because the angle between the two sides can vary

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[keen pawn]

Problem 16

a) let $x_2+x_3+ \dots + x_n =t$. Then $f(x_1+t)=f(x_1)+f(t).$

now repeat the same process to $f(t)$ until you're left with $f(x_1)+f(x_2) +\dots+ f(x_n)$ .

At which point you're done

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

whats the mathematical version of "repeat this process"

[keen pawn]

induction>

[indigo cloud]

yes. do it

[keen pawn]

$f(x_1+x_2)= f(x_1)+f(x_2)$
\
let $f(x_1+x_2 \dots + x_n) = f(x_1)+f(x_2) \dots f(x_n)$
\
$f(x_1+x_2+x_3 \dots +x_n + x_{n+1}) = f(x_1+x_2 + \dots + x_n)+ f(x_{n+1})$
\
= $f(x_1)+f(x_2) + \dots + f(x_n)+f(x_{n+1})$
\
Hence completing our proof

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

could use a few more words but ok

[keen pawn]

Cool

thanks

Now part b

$f(x+0)=f(x)+f(0)=f(x) \implies f(0)=0$

[grizzled pagoda]

Veni, vidi, perii

[keen pawn]

or rather consider a linear function cx

f(x)=cx

$f(x_1+x_2 + \dots + x_n)= c(x_1+x_2 \dots x_n) = c(x_1)+c(x_2) + \dots + c(x_n) = f(x_1)+f(x_2) + \dots + f(x_n)$ Hence completing our proof

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

you cant just assume what you want to show

[quick spoke]

well you've shown that if f(x) = cx then f satisfies the functional equation

we're trying to show that if f satisfies the functional equation then f(x) = cx (for x rational)

[keen pawn]

hmm

Could I have a hint

[indigo cloud]

try showing it first for x in N, then x in Z, then x in Q

[keen pawn]

How isn't this equivalent to showing the equation cx satisfies the functional equation

[indigo cloud]

A=>B and B=>A are very different

[keen pawn]

I know

[indigo cloud]

A being "f satisfies the equation" and B being "f(x)=cx for rational x"

[keen pawn]

but surely this is a $A \iff B$ statement

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

even if it were, you only have shown one direction

[keen pawn]

When x is a natural number say n

$f(n)=f(1) \cdot n$

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

why

[keen pawn]

$f(n)= f(1)+f(1) \dots$ n times

[grizzled pagoda]

Veni, vidi, perii

[keen pawn]

so $f(n)= nf(1)$

[grizzled pagoda]

Veni, vidi, perii

[indigo cloud]

you always only write half the steps

[indigo cloud]

why

[keen pawn]

from the defn of the function?

f(1+1+1 \dots ( n times)))= f(1)+f(1) \dots + f(1) n times

[indigo cloud]

not from def, but from (a)

and I wanted to see the n=1+1+...+1 step

[keen pawn]

from f(x+y), it's easy to coclude that f(x-y)= f(x)-f(y)

[indigo cloud]

oh is it?

[keen pawn]

nvm

f(x-x)= f(x)+f(-x)=0

so f(x)=-f(-x)

so f is an odd function

[indigo cloud]

bad order of writing that chain of equalities

0=f(0)=f(x+(-x))=f(x)+f(-x)

[keen pawn]

f(p/q)= pf(1/q)

anf f(1/q)= f(1)/q

so f(p/q)= p/qf(1)

hence completing our proof

[indigo cloud]

why do those equalities hold

[quick spoke]

that's basically the proof yeah, i think like as denascite pointed out, probably more words/reasoning would be good for ur final write-up

i think the easiest way to write it up is to observe that ||f(nx) = nf(x) for n integer, so then u get all the eq. u need||

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[keen pawn]

Thanks

[quick spoke]

as an interesting aside, it isn't true that if f(x+y) = f(x) + f(y) for all real x,y

then f has to be linear
(there are examples of functions where f(x+y) = f(x) + f(y) but f(x) != cx)

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[hexed dock]

Hi, what do I need to fill in the blank?, I know everything about this scheme , how to create this equation, but I don't know what to put in so that the signs are not opposite.... I know that then delta and that's it but... I don't know what to fill here 😭

[pallid canopy]

Try a different grouping

[hexed dock]

by x? not x2?

[slow pelican]

you could prolly just do x^2(x + 3) - 1(x + 3) = 0

and group it like that

[hexed dock]

Thanks!

[slow pelican]

np

[hexed dock]

I'm sometimes so dumb... Thank you so much ❤️

.close

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[celest badger]

if the sum of the lengths of a triangle is 60 and the square of the sum of them is 1352. Find them (a+b+c=60 , a^2+b^2+c^2=1352)

[iron heath]

Do you have the original question, or is it something informally written?

"the square of the sum of them" is a little ambiguous.

[celest badger]

i said that because i translated to english and the other language has different grammer

so kinda hard to translate

so what i said inside the () is the equation formed if not understood

[iron heath]

Are you able to post the original just to get some context?

[celest badger]

wdym original

in that language?

[iron heath]

Yes

[celest badger]

its albanian but sure

Perimetri i nje trekendeshi kendedrejte eshte 60cm kurse shuma e katroreve te brinjeve te tij eshte 1352cm^2.Gjeni brinjet e tij.

[iron heath]

Is this for a homework assignment?

What do you mean by "find them"? What is them?

[celest badger]

find the lengths of the triangle

[celest badger]

no, just working for olympiads

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[feral ore]

forth ticket for the day

[feral ore]

i just want to know if the answer is 1080x or not

[dull pike]

Yes

,w expand (3x-2/x)^5

[vast spade]

yes

[grizzled pagoda]

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[split sail]

yeah its 1080

as samuel said

easy binomial theorem application

[feral ore]

THANK YOU

wait

we dont need to expand @split sail

[feral ore]

it wants the coefficient

[dull pike]

Check the image, you can see the coefficient is indeed 1080

I was just calculating so you can be sure your answer was correct

[feral ore]

thank youus!! i am seriously having mental problem from maths AA thats why i am very unsure of any question i get

i am almost done and its sadly impossible for me to ask all questions

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[calm pewter]

Yo someone help

[trim dome]

do u still need some help?

[calm pewter]

Yes

@trim dome

[trim dome]

sorry, wanna confirm

x . BE = 6 right?

can u take better pic? i missed some parts

[calm pewter]

Yh it is

@trim dome sry my msg are delyaed

[trim dome]

can u re-take the pics?

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[crystal berry]

Decimal number -3 using 3 bits binary:

[crystal berry]

is it 101

[sinful sapphire]

yeah

well it depends if it's 2-complement or 1 complement or whatever

usually it's 101

[crystal berry]

thanks

.close

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[novel quail]

If I have the expression a(b^a) and b isn't e what is the value of W(a(b^a)) where W is the Lambert W function?

[torn furnace]

usually we'd think of using lambert to find the solution to an equation, i think

well wait maybe im dumb gimme a sec

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@novel quail Has your question been resolved?

[novel quail]

It's the inverse function of y=x(e^x) which why I think it would expect there to be a simple expression for the value of W(a(b^a))

[torn furnace]

yea im curious too

i always interpret it like ye^y = x then W(ye^y) = W(x) is just defined to be W(x) = y

then W(x) e ^W(x) = x

but im having a hard time

[novel quail]

Yah that's how I interpret it, my question is what happens when it's not e for the base basically

[torn furnace]

i mean

is it really as easy as like

$ab^a = a(ec)^a = c^a W(a)$?

[grizzled pagoda]

jan Niku

[torn furnace]

ive been playing around because i feel like there must be some problem with this

obviously we have some restrictions and then deal with whatever happens with the branch cut

[novel quail]

I only need to deal with positive real numbers so that shouldn't be an issue

[torn furnace]

,w W(3)

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[torn furnace]

,w Lambert W( 24 )

[grizzled pagoda]

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[torn furnace]

,w (2/e)^3 * lambert W(3)

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[torn furnace]

hrmf

i think im using w function wrong

[novel quail]

I shouldve checked Wikipedia

I can just use formula at the top

[torn furnace]

welp

[novel quail]

Thank you though

.close

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[crystal berry]

The decimal number 5 is equal to 10 in base 5.

[crystal berry]

is this not true?

[neat wind]

this is true

[crystal berry]

thanks

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[frigid jungle]

This is probably a silly question, but my lecturer mentioned that we can't extend the size of a matrix

I wonder why because so far, no matter how I look at it, it shouldn't hurt if we expand the matrix with 0 values :/

[pallid canopy]

extend it for what purpose

not enough context to really understand the statement "can't extend size of a matrix"

[frigid jungle]

well let's suppose a matrix
[ 1 , 1 ]

I would expect that the matrix
[ 1 , 1 , 0 ]
[ 0, 0 , 0 ]

could be used all the same

but as I understood it the lecturer said that the size of a matrix is somewhat constant

as for the purpose, I don't really know, I just thought of what it could represent and didn't see an issue with adding a few zeros

[pallid canopy]

"used all the same" is too vague

[frigid jungle]

are equivalent for all conceivable cases and applications* :|

the reason it could be interesting is that we could add matrixes with different sizes then

and if what my lextrurer says is correct, we can't do that?

[pallid canopy]

then no

"all conceivable cases" you can't even multiply the new 2x3 matrix with the same matrices that [1, 1] used to multiply by

1 x 2 matrices can only multiply on the right by 2xp matrices

for some natural number p

[frigid jungle]

you just add more zeros to the second matrix just as you did with the first one

the result of the multiplication will stay the same just with additional zeros for added "spots" in the matrix

[pallid canopy]

you're just moving goal posts now

find the context your teacher said and ask them.

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@frigid jungle Has your question been resolved?

[frigid jungle]

whatever

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[patent knoll]

Is the limit of 4, not just - infinity?

[patent knoll]

(From the left side)

[pallid canopy]

nothing around x=4 of f(x) is tending to infinity

find those y-values on the axes

[patent knoll]

3, 4, and 5

4 being the solid one

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[thick seal]

plz

[thick seal]

i understand

this i understand

but what is 100^3cm^3??

why couldnt we just use the 1cm^3

like do we always want the numerator 1?

like i notice that pattern

like here for example

but i idont get it

i think i have to watch this

https://www.youtube.com/watch?v=Ymke1rpb3O8

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[regal swift]

need help with part B

[amber wasp]

@regal swift Has your question been resolved?

[regal swift]

<@&286206848099549185>

<@&286206848099549185>

[jaunty shale]

what did you do to find that θ ?!

[regal swift]

cos-1 ( -1.218 / 1.8722)

[jaunty shale]

-1.218 / 1.8722 is the cotangent of the resultant phase angle

not the cosine of it

why not divide 1.8722 / -1.218 instead, and you get to use arctan

[regal swift]

that gives -0.994 rad, do I have to add pi to that?

[jaunty shale]

the equation tan(phi) = -1.536674 has 2 solutions, -0.994 and 2.1477

you gotta determine which one is the correct

not just try 😛

[regal swift]

and how do you determine that?

[jaunty shale]

i know you can just put and it will tell you correct/false

but you must decide yourself

[jaunty shale]

using the system of equations you found

only one of the angles satisfies both of the equations

$\begin{cases} A_R \cdot \cos(\phi_R)=A_1 + A_2 \cdot \cos(\phi_2) \ A_R \cdot \sin(\phi_R) = A_2 \cdot \sin(\phi_2) \end{cases}$

[grizzled pagoda]

Emily

[jaunty shale]

thats what you found, right

[jaunty shale]

only one of these satisfies BOTH

[regal swift]

what's phi 2 here?

[jaunty shale]

phase of 2nd wave...

[regal swift]

only 2.1477 satisfies both eqns

[jaunty shale]

thats right, thats how you decide

dont just plug and see what it gives you

red/green 😛

[regal swift]

[jaunty shale]

yeah

[regal swift]

thanks

[jaunty shale]

spatial and temporal frequencies in the same problem, noice

you dont see that often i guess

[regal swift]

sry one more question

why would doing something like this not work in this case?

[jaunty shale]

i didnt understand what youre trynna do

[regal swift]

solving for theta using the amplitute of the first wave (since it lies on the x axis) and the resultant amplitude by using cosine

[regal swift]

(the numbers here are different than the numbers in my problem)

but if you tried to solve for theta using this method you'd end up with a math error, why is that?

[jaunty shale]

im sorry, you have theta, why are you solving for itttt

you want to make sure?

[regal swift]

no im just asking about a different method

[jaunty shale]

using cosine you mean

[regal swift]

just trying out the method I saw above since it seemed easier but in my case it gives a math error and im confused as to why.

[jaunty shale]

youre getting a different theta because cos(θ) is not equal to A1 / A

the projection of A onto x-axis is not A1

its less than A1

make a right angle

cos(θ) = L / A

not A1 / A

[regal swift]

ah I see

[regal swift]

is there a way to find that yellow angle I highlighted?

[jaunty shale]

90 - θ

[regal swift]

Here why is the amplitude of Yr not equal to the amplitude of the resultant wave?

(in this case its 5 mm)

[jaunty shale]

amplitude of resultant wave is false

[regal swift]

you mean the 5mm is wrong?

[jaunty shale]

not 5, its 3.29

[jaunty shale]

yes, calculate it the same way you did for the previous problem

[jaunty shale]

use these

[regal swift]

so In our case here the length of that Yr vector

[jaunty shale]

you will get 3.29

[regal swift]

is 2.23?

[jaunty shale]

yes

[jaunty shale]

this drawing is not accurate at all

y2 vector should be pointing close to the negative x-axis

angle is 0.8 pi, close to pi

the way they drew it, it looks like 3 pi/4 lol

[jaunty shale]

and even Yr is not drawn well

its so close to being parallel to y-axis

i calculated resultant phase, i found it 1.55, which is close to pi/2 = 1.57

[regal swift]

Why’s this giving me the wrong answer?

oops nvm

my calc was in radians xd

ok so this gives me 2.1488, then how do I check if thats the right angle or if I need to + - pi?

[jaunty shale]

maybe calculate the sine

if its fine, then 2.1488

if its not fine, then - pi

[regal swift]

what should I be expecting when I find the sine of the angle?

[jaunty shale]

to have the correct sign

2nd quadrant, sine is +

4th quadrant, sine is -

[regal swift]

isnt my vector in the first quadrant?

[jaunty shale]

are you kidding, you found Yrx negative

😛

[regal swift]

o wait nvm

im braindead

[jaunty shale]

obviously the angle is correct

no need to - pi

2.1488 is in 2nd quadrant, so gg

[regal swift]

got it thanks alot

[jaunty shale]

youre welcomee

[regal swift]

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[sterile nymph]

Triangles!

Let A be the top of the tree, B be the part of the tree at eye level, C be the bottom and D be your eyes.

DBC is similar to ABD is similar to ADC

Using this and the laws of similar triangles you should be able to first find AD and CD, finally find AC

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@tranquil pasture Has your question been resolved?

[tranquil pasture]

ok thank you!

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@tranquil pasture Has your question been resolved?

[fresh sable]

hi

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[fossil shell]

ik this is chem but i really have no idea how to do this and i hope someone cld help me out

[fossil shell]

ans is a which i have no idea well im not sure how to start tbh

[solemn escarp]

can you figure out how much gas turns into H20?

[fossil shell]

60cm^3?

[solemn escarp]

no

if you have 30cm^3 of hydrogen, then how much oxygen do you need to add to make H20, from the equation

[fossil shell]

ermm im not sure do i take it as 30cm^3 of hydrogen can make 4 hydrogen atoms

and i would need 2 atoms of oxygen

[solemn escarp]

you can do it that way yeah

[sonic ferry]

Just calculate the number of mols and do the bam and you get it

[fossil shell]

or do i say like every 1md^3 of hydrogen used it uses 2dm^3 of oxygen so by the time 15 dm^3 of hydrogen is used theres 15 dm^3 of extra hydrogen left?

[sonic ferry]

It kinda works like that, but not quite.

Volume doesn't work like that in the chemical equations, only the number of mols do, which basically means the number of molecules

[fossil shell]

oh

[rich needle]

but for this case if we assume ideal gas it should behave like "moles"

[fossil shell]

so i convert to mol and solve from there?

[sonic ferry]

I would assume that your teacher or whoever that gave you the question wants you to use PV=nRT

because it says all volumes are measured at room temperatures and standard pressure

[fossil shell]

ermm i have never heard of this eqn before?

[sonic ferry]

Oh um then

[fossil shell]

but how wld i solve it using this eqn tho?

[sonic ferry]

"PV = nRT" is called ideal gas law, go search it up

[fossil shell]

hmm okay

[sonic ferry]

Very easy to use. P = Pressure V = Volume n = number of moles R = ideal gas constant (0.0821 in this case) and T = temperature

[rich needle]

This is not the point of the question, we must assume an equality and solve with that and the limiting reactant to know how much gas is left over from the equation

[sonic ferry]

Just chuck the numbers in

yeah, but to know the limiting reactant we need to find the number of mols

Oh wait

for this question, you don't even need to know anything

[rich needle]

In this case the number of moles is not necessary, this question is focused on mathematics, we use the gas as two equal ideal gases in which in 30 cm^3 there is the same amount of moles

[sonic ferry]

Oh yes. That's correct.

I was trying to approach with the numbers, but that doesn't seem necessary. Sorry for confusion.

So that just means 15 cm^3 of O2 will be left

So A is the answer

[fossil shell]

so we do take it as oxygen and hydrogen both have 30dm^3 of volume at the start

[sonic ferry]

Yes, they both have 30cm^3 at start. Not 30dm^3

[fossil shell]

ohh

[sonic ferry]

The question says cm^3

[fossil shell]

wait so when shld we use mol ratio and find the gas left in volume like heere or like when do we need to convert to mol from volume and compare the mol ratio

[sonic ferry]

Ok, the thing here is, as the other guy said, you don't necessarily need to find the specific number of mols.

Since the the volume of two substances (H2 and O2) are equal at start, that means the number of mol is also same, using the ideal gas law (PV=nRT).

[fossil shell]

ohhhh

so when volume is different thats when we convert to mol and solve from there

[sonic ferry]

Sure, that could be the case.

[fossil shell]

ahh i see

[sonic ferry]

But again, you just need to know the ratio for this question.

It's all about ratios.

[fossil shell]

hmm wil take note!

tks a lot for helping tho!

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[sonic ferry]

Your welcome

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[muted vector]

Suppose k > 0. Let c and d be the roots of x² - kax - (k + 1)b = 0 and a, b be the roots of x² - kcx – (k + 1)d = 0, and a + b + c + d = 180, then k is equal to
A. 11
B. 9
C. 7
D. 5

My approach:
since, sum of roots = -(b / a) [b --> coefficient of x ; a --> coefficient of x^2]
and, product of roots = c / a [c --> constant term]

So, from the 1st equation;
c + d = ka ..1
c * d = -(k + 1)b ..2

From the 2nd equation;
a + b = kc ..3
a * b = -(k+1)d ..4

now ..1 + ..3
a+b+c+d = k(a+c)
so, k(a+c) = 180

and ..2 * ..4
abcd = bd * (k+1)^2
so, (k+1)^2 = ac

i ended up with those equations, i.e,
k(a+c) = 180
and
(k+1)^2 = ac

and further i just didn't know what to do. Please help!

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[fallen vigil]

Can anyone help me solve this idk where to begin

[jovial field]

Yo

Do u know radius

[solemn escarp]

do you mean radians?

[fallen vigil]

Yea

[fallen vigil]

Radius

[jovial field]

Direct help

Arc length (perimeter)=radius X angle subtended at the centre

[fallen vigil]

Is there even a permiter for a circle

[fallen vigil]

Not angle/360 x 2 x pi x r?

[jovial field]

Try finding it with the formula I gave above

No

[fallen vigil]

Ooo

[jovial field]

Angle is in radians right?

Or is it in degree

[fallen vigil]

Im asking my teacher

My friends r all confused too

💀💀

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@fallen vigil Has your question been resolved?

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[pine bolt]

how does one solve this

[pine bolt]

I've tried finding identity of cot but to no avail it's still undefined

so how to solve this without derivative

[true jackal]

Multiply top and bottom by 1 + cos(2t)

Then write tan as sin/cos

[pine bolt]

OHHHH I've never thought of that

thank you good samaritan

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[solar gust]

So the maximum being at 2,11 tell that a is negative

[limpid gorge]

Ok

[solar gust]

y = a(x-2)^2 + 11

It tell this too

So y = a(x^2 -4x +4) + 11

Knowing that it goes through (-1,2) we can replace in x and y

[limpid gorge]

Where did -2 come from?

[solar gust]

For maximum at (b,c) we have a(x-b)² + c

[limpid gorge]

Ok

[solar gust]

So we replace as here

[limpid gorge]

Ok

[solar gust]

2 = a(1 +4 +4) +11

Solve for a

[limpid gorge]

Would it not be
2=a(1+4+4)+11

Ok

[solar gust]

Yeah sure it was a typo i correct it

[limpid gorge]

Np

[solar gust]

So a = ?

[limpid gorge]

-1

[solar gust]

Perfect

[solar gust]

So replace in this

And there you go

Develop at the further

State a,b,c and thats good

[limpid gorge]

What would b be?

[solar gust]

4

[limpid gorge]

Is c =4

[solar gust]

Well you have y = -(x^2 -4x +4) + 11

So y = -x^2 +4x -4 + 11

-4 + 11 = c

[limpid gorge]

Ok

Thank you

[solar gust]

You're welcome

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[limpid gorge]

What is this topic called?

[solar gust]

Quadratic equation and vertex form practice

[limpid gorge]

Ok thank you

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[humble ore]

is the domain of y=sin(x) = {x : 0 <= sinx <= 2pi}?

or [0,2pi]

[strange field]

both mean the same

[humble ore]

but are they correct?

[strange field]

domain of sinx is all real numbers

[humble ore]

so incorrect?

[strange field]

you're just restricted to [0, 2pi]

[strange field]

if you're asking if domain of sinx is [0, 2pi] then yes thats incorrect

[humble ore]

hmm, can u explain?

how is the domain all real numbers

[strange field]

do you know unit circle

sin(x) is opposite/1 in unit circle

and opposite can be defined for all points on circle

so angle x can take any value

hence domain is real numbers

[amber wasp]

@humble ore Has your question been resolved?

[humble ore]

thank you, what about the range?

[strange field]

-1 to 1

[humble ore]

so it will be always -1 to 1?

[strange field]

yes

[-1,1]

[humble ore]

but the domain?

depends on the function

?

[strange field]

what

every function has its own domain and range

sinx's domain is all real numbers and range is [-1,1]

[humble ore]

oh ok

thank you

now how do i know if this is an odd or even func?

[strange field]

a function is even if f(-x) = f(x)

and odd if f(-x) = -f(x)

[humble ore]

how do i use that in my function

[strange field]

substitute x with -x

and see what u get, the original funtion or negative of original function

[humble ore]

negative?

[amber wasp]

@humble ore Has your question been resolved?

[keen topaz]

What is sin(-x)

[humble ore]

-sinx

[keen topaz]

So is it even or odd

#help-41 message

[amber wasp]

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@humble ore Has your question been resolved?

[humble ore]

odd as hell

[amber wasp]

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[split sail]

how can i determine the nature of this 😔

[split sail]

its not an alternating series so i dont really know what to do

and idk if i can prove that its absolutely convergent

[pallid canopy]

Do comparison first, then alternating

Or do limit comparison test

[split sail]

compare it to whatt

[pallid canopy]

That's the majority of the problem so try something first

Use the fact that sine and cosine are between -1 and 1

[split sail]

i did and all i got was that the absolute value of the general term was O(1/sqrt(n))

[pallid canopy]

You're right. That's basically it just don't take the absolute value because you're supposed to use alternating series test

[split sail]

we can't conclude anything from this though

the series (-1)^n/sqrt(n) is convergent but we can't say the same for (-1)^n * |un|

the comparison thing needs a constant sign

[pallid canopy]

Then don't compare the absolute value

[foggy fossil]

<@&286206848099549185>

I forgot the a ^ { 2 } -b ^ { 2 } formula can anyone help

[amber wasp]

@split sail Has your question been resolved?

[split sail]

what should i do then

[split sail]

(a-b)(a+b)

[pallid canopy]

Compare with absolute values in tact

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@split sail Has your question been resolved?

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[stone wigeon]

the question is to find the mod of z

[stone wigeon]

which is pretty easy

but I made a mistake in the workings and I don't know where

[tulip tapir]

,rccw

[stone wigeon]

because I would have expected the cos isin to go to 1 but the symbol is the wrong way round

[spiral zealot]

if going that route, you want to square the Im part which doesn't include the i

[stone wigeon]

ah thanks didn't know that command

[tulip tapir]

you aren't calculating a^2 + b^2 here but rather a^2 - b^2

which is not what you want

[stone wigeon]

then what do I do with the i?

[spiral zealot]

leave it out

[stone wigeon]

huh... I can do that?

[spiral zealot]

|a + bi| = sqrt(a^2 + b^2)

[tulip tapir]

yeah

imagine the complex plane

and a number on it

you can make a right-angled triangle

with a and b as the sides

then the modulus is simply the hypotenuse of said triangle

[stone wigeon]

ah yeah good point

[tulip tapir]

which is, via the pythagorean theorem, a^2 + b^2

and NOT a^2 - b^2

because the sides are a and b, not a and bi

[stone wigeon]

yeah gotcha

thanks for the correction!

[tulip tapir]

nw

[stone wigeon]

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[ancient raft]

Find the continuity of $$\frac{x+1}{x+3}$$

[grizzled pagoda]

Good

[ancient raft]

I got to $$\frac23 \cdot \frac{|x|}{|x+3|}<\epsilon$$

[grizzled pagoda]

Good

[ancient raft]

I'm a bit confused of the part, where assuming delta < 1, which is |x| < 1, such that |x+3|>2.

How did the solution came to |x+3|>2?

[weak zinc]

Continuity where?

Anyways, $\abs{x} < 1$ implies $-1 < x < 1$, hence $2 < x + 3 < 4$, so $x + 3$ is greater than 2, and in particular is positive (so $x + 3 = \abs{x + 3}$)

[grizzled pagoda]

@weak zinc

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@ancient raft Has your question been resolved?

[ancient raft]

x=4, but this makes sense.

So it's not worthy to mention < 4?

Since 2 is delta's minimum?

[weak zinc]

Hmmm, strange if it's continuity at x = 4, how comes it's |x| < 1 then

[ancient raft]

x = 0, my bad.

[weak zinc]

You won't need that, what you want to use is that $2 < \abs{x + 3}$ so that $\frac1{\abs{x + 3}} < \frac12$, hence you get to bound what you have by saying $\frac23 \frac{\abs{x}}{\abs{x + 3}} < \frac23 \cdot \frac12 \cdot \abs{x}$ and just need to force that latter to be less than $\varepsilon$

[grizzled pagoda]

@weak zinc

[ancient raft]

So the expression containing x, is going to be less than something? Or I mean the variable in denominator. If it we're in numerator, it's going to be greater than something, if it were in denominator, it's going to be less than something?

[weak zinc]

Well, for the denominator, you want the denominator to be as small as possible to make the fraction as big as possible, and the smallest the denominator can be is 2

[ancient raft]

I think I got it.

Thanks for the help

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[grizzled iron]

What did I do wrong?

[pallid canopy]

Why is iw in modulus here

[grizzled iron]

They're parenthesis, but I just draw them weird

I don't know if the way I'm computing w2 is correct or not

[sinful carbon]

when you divided by (z-1)^3 you put (z+1)^3

not sure if this helps too much since it looks like you put the negative back in

[grizzled iron]

Imma start from 0 and see if I get my textbook answer

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[balmy quiver]

small question; the coefficient is the term with the highest exponent right?

[balmy quiver]

and the degree is the highest exponent right?

but the degree isnt in the coefficient? im very confused

[elfin vale]

The degree is the highest exponent

A coefficient is the number that multiplies the x

And the leading coefficient is the number that multiplies the x with the largest exponent

[dull pike]

The degree is explained already by d

[balmy quiver]

so 5x^2 is the leadign coefficient and 2 is the degree

[dull pike]

5 is the coefficient

2 is another coefficient

The 2 that multiplies y

In 2y

The degree is 2 also because the highest exponent is 2

[elfin vale]

Coefficients can't have x

[balmy quiver]

5 is the leadign coefficient

[dull pike]

Yes

[balmy quiver]

thx

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[crimson stratus]

how do you know which vector is longer between w(-2,6,-8) and r(3,0,6)

[crimson stratus]

do i have to calculate the norm simply

[neat wind]

yes

[crimson stratus]

k ty

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[sudden jasper]

im solving a rational inequality word problem. Do i still have to shade the interval (-500, 1000)? It is true but I put no solution set because the problem is based on real life and about production of something, and since you can't produce a negative amount, I put no solution set.

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[hollow shoal]

How do I use tool kit functions to complete transformations?

[hollow shoal]

I'll paste the problem

the graph of g(x)=-sqrt-x+1 end sqrt _4 can be obtained from the graph of y= by reflecting the graph through the A Origin B X axis C Y axis
then shifting the graph A left B right
Units and A up B down

and it says to only use tool kit functions which I know what those are but what is it asking me to do?

<@&286206848099549185>

Hello?

[split sail]

this is unreadable, do you have a picture?

[hollow shoal]

One moment

Uh I do but I don't think it's that legible give me one second...

Yeah that's definetly not going to be legible

[split sail]

if you pasted this from an online source a picture of that would work too

[hollow shoal]

Okay.....let me see

[split sail]

like a screenshot

[hollow shoal]

Give me another moment

[split sail]

dw im still here

[hollow shoal]

Does this work?

[split sail]

yeah

it says "from the graph of y = " then ends

the equation for y needs to be known for this

[hollow shoal]

So it connects to the first problem?

Because I'm trying to solve problem no 2

[split sail]

i think so

[hollow shoal]

Wait wait wait I think I got it

It's asking me what transformations I need to get to that state

[split sail]

because it's asking for the series of transformations needed to convert the graph of y into the graph of g

[split sail]

yes esxactly

which can only be destermined by examining y

[hollow shoal]

Oh okay hold on give me one second.....

[split sail]

yee

remember to

!done

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[hollow shoal]

So I just need to know the different transformations then.....

[split sail]

ye they're given though you just have to pick the right one

[hollow shoal]

Ah okay

[split sail]

!done

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[long peak]

not really queastion but at the same it is

[long peak]

but oh my gosh

i literally

cannot read this

what is wrong with my math teacher

i just want someone to tell me what hes trying to say 😭

[dull pike]

[celest cove]

all i can focus on is the missing bracket

[long peak]

thank you

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[thick seal]

is it 9861g

[thick seal]

like

1255cm^3 times 0.7857g ?

[dull pike]

Why 1255?

[thick seal]

like

i converted ml to l then l to cm^3

[dull pike]

1 mL = 1 cm^3

[thick seal]

o

ah i see

question

cm^3 can convert any si prefix?

[shadow stump]

if by si prefix you mean that cm^3 can be converted to m^3, km^3, mm^3, etc, then yes

[thick seal]

oh ok

perfect then

so this would just be

multiplication

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[pliant parcel]

I get a different result from WolframAlpha. Did I do something wrong?

[patent raptor]

You cannot pull out cos(theta)

[grizzled pagoda]

bacc

[patent raptor]

This is what you have to integrate, and you can use an identity for cosine square

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[upper parcel]

<@&286206848099549185> plz sav e me i ben doing this fofr long time

part c

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[final matrix]

Well for the first one you could either substitute values and prove both sides of the equation are not equal

Or make a statement based on the generally ellipse form

*circle

[upper parcel]

yea mainly part b and c i dont understand

[final matrix]

Ok well part C

You have a point T at an arbitrary location outside of the circle

And two lines drawn from T tangential to the circle

And then a third line from the centre of the circle C to the point T

I believe there's a rule that states the angle between the tangents must be 90 deg iirc

Or something along those lines

[upper parcel]

yea

[final matrix]

So if there's that 90 degrees

You just need to show CT bisects the angle

With 45 deg on both sides

Given that the angle is 90 deg

[upper parcel]

these are the only methods tho

[final matrix]

It's hard to tell without seeing more of the question

Sorry

[upper parcel]

alright could u do part c?

[final matrix]

I don't know what A and B refers to

[upper parcel]

oh my bad

@final matrix

[final matrix]

oh nvm you do this using your information about the point T

sorry I thought T was an arbitrary point

I would suggest drawing it out to determine lengths

[final matrix]

here you would calculate the gradient via the rise/run from point T by subtracting your tangent locations from T

and then via the gradient and knowing points on the line you can use slope intercept form to solve for the equation of the gradient

[upper parcel]

could u show

[final matrix]

I can't show rn sorry

[upper parcel]

where do i get the tangent locations from?

[final matrix]

correction you already have the gradient given it's a 45 degree angle

which you proved

so then you can just find the equations using that and your point T

you don't need to find the tangent locations I think

my bad

[upper parcel]

gradient of AB right?

[final matrix]

that's for the tangents

that gives you TA and TB

and then you have a triangle for AB to solve where hopefully you know enough information

[upper parcel]

where the flip do I get TA and TB from?

all i know is t has two tangents that touch the circle at A and B

and that the angle between one of the tangents and CT is 45

[final matrix]

TA and TB are the tangents

correct?

[upper parcel]

yes

[final matrix]

I might have rushed through a little but

you know the form of the tangent will be y = mx+b

and given that we have proven that each tangent is at a 45 degree angle

we know m will just be equal to 1

so you have y=x+b

and y=-x+b

[upper parcel]

ok so i find the two equations

and then i plug them into the circle formula?

[final matrix]

you then know both run through the point T (-6,6) so you can substitute those values into both equations to find B (with x and y)

that will give you the equations of both tangents

and then you can draw this out

you have a triangle with known TA, TB and yoou can find AB

[upper parcel]

so after i get TA and TB I solve for A and B right?

[final matrix]

yeah you can set the equations equal

as in the circle equation and the tangent equation

you can then find the locations of A and B

then you can subtract one from the other to find the gradient

and then resubstitute points to find the b value in y=mx+b

this might be more complicated then you're meant to do tho

this is just how I'm thinking about it

[upper parcel]

alright thanks

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[night stratus]

I can't think of the equation I have to make with this problem

I know I have to use same variable letters to solve it

[cobalt stump]

30(6x)+20x=2600

thats the equation

but then simplify

[night stratus]

Ah thanks I see it now. I kept writing the equation the wrong way. I was trying every way. I know the 2600 had to be on the equal

[cobalt stump]

what did you get

[night stratus]

13?

x = 13