help-23

[hazy elbow]

you did not state so

[languid surge]

2sinθ=4cosθ
sinθ=2cosθ
sinθ/cosθ=2
sinθ/cosθ=tanθ
so
tanθ=2

is it correct till here?

[meager merlin]

YES

bruh

[languid surge]

ok chill

[meager merlin]

The thing is theta is not a single value

Rather a collection of value

[languid surge]

oh so i cant assume theta is same for like rest of them?

[meager merlin]

?

Wdym

I think you're lacking the foundation knowledge of trigonometric

[hazy elbow]

WAIT

@languid surge

[languid surge]

like tanθ , and sinθ both could have different values ?

[hazy elbow]

state what is your question

[meager merlin]

No

[languid surge]

ye?

[meager merlin]

Theta stay the same

[languid surge]

ye so whats the prob

[hazy elbow]

you opened the channel without any question

[languid surge]

ok lemme write it

[hazy elbow]

like what is your goal

[meager merlin]

He did

Find theta that satified 2 sinθ-4cosθ=0

@languid surge Just find theta now

And u're done

[languid surge]

2sinθ=4cos
2sinθ=4(1/secθ)
2sinθsecθ=4
tan/secsec=4

oh wait

mb

[hazy elbow]

there is no mention of whether to solve for theta or a simplification

[languid surge]

i missed *2

ye

really really sorry

dude

[meager merlin]

why....

tan(theta) = 0 is the simpliest form already

[languid surge]

i needed value of secθ

[hazy elbow]

well

WELL

you could've mentionewd it earlier

[languid surge]

but my brain stopped working

i did tan=2 and tan=4

in the same equation

[hazy elbow]

sec^2(x)=tan^2(x)+1

use this

[languid surge]

look its my first time doing trigno

so x means θ?

[hazy elbow]

well, in this case, YES

[languid surge]

ok thx

[meager merlin]

theta, x, y, z, they're just variable

[languid surge]

how did u found this btw?

[hazy elbow]

this is a standard property

[meager merlin]

But usually, variable in trigonometric, we use theta

[hazy elbow]

and can be proved

[languid surge]

ohk

is there anyway that it could be done without it?

cuz i was told i should be supposed to be able to do it without needing anythign other than these

sin2 a + cos2 a = 1
1+tan2 a = sec2 a
cosec2 a = 1 + cot2 a

[hazy elbow]

MY GUY

the second formula is the one

I just wrote

[languid surge]

wait

oh right

im just gonna sleep then

since my brain is dead

i cant do anything either way in this condition

thx a lot

bye have a great day

[hazy elbow]

sure have some sleep

[safe radish]

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[wooden cosmos]

hello

[wooden cosmos]

just want to check my answers

,r

?

fuck how do you rotate again

[safe radish]

@wooden cosmos Has your question been resolved?

[safe radish]

@wooden cosmos Has your question been resolved?

[wooden cosmos]

can you shut the fuck up?

[devout shale]

?

[wooden cosmos]

are you a mod?

can you mute the bot?

[devout shale]

No to both questions, but hit the red "x" if you want to keep the channel open

it will continue asking you, less often every time, to make sure you are still using the channel

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[brisk token]

I'm practicing for a competition and idk if this is right. i will send what i have done so far.

The interior angle at vertex C of right-angled triangle ABC is right, and also angle BAC = 2• angle CBA. The length of side AC is 8 cm. Point M is the bisector of AB, point N is the bisector of BC, point P is the bisector of the distance BM. Calculate the length of the fraction ACMNPB. Intermediate steps should be justified without measurement.

[brisk token]

i saw that CM = AM

and PM = 1/4 AB

and i think that MN = NP = PB?

i this case the ACMNPB = AB + 5/4•AB

[brisk token]

also in the image on the right it says that i said that its 20, but then irealised it was 28cm

NM is 4cm

i now realised

[ionic blaze]

uhhhh
8 + 8 + 4 + 4 + (16-8-4) = 28

[ionic blaze]

could you show ur solution

[brisk token]

i used e as the AB

[ionic blaze]

what's e

and how'd you get that

[brisk token]

like if you have a rectangle

[ionic blaze]

I feel like you went through something complicated when there's an easier concept

[brisk token]

e is the diagonal of A and C

[ionic blaze]

what if
30 60 90 triangle

[brisk token]

angle in C = 90°
angle in A =60°
angle in B =30°

[brisk token]

yes!

[ionic blaze]

yes, and there's a thing about that

[brisk token]

but i turned it sideways

[brisk token]

i asked chatgpt and it used tans/cons?

but we didnt learn them yet

[ionic blaze]

I think you should learn it for the contest

[ionic blaze]

also, never ask chat GPT to solve math problems

[brisk token]

i realised that

thats why i asked here

[ionic blaze]

um
not to be a party pooper but
similar triangle properties ._.

[brisk token]

wym?

[ionic blaze]

a line parallel to the base of the triangle and intersects the midpoint of the hypotenuse has exactly half the length of the base

and then from there just use 30 60 90 triangle

[brisk token]

this?

[brisk token]

i forgot one line at the bottom

so 8cm + 5/4•16cm=28cm

how should i get x√3 in this case?

@ionic blaze ?

.close

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[versed wave]

$3^{x^2-9}+(x^2-9)5^{x+1}<1$

[flat frigate]

ngDe

[versed wave]

does this require lambert W? would appreciate for a cleaner approach

[spice grove]

A very clean approach is just notice how inequality is when x^2 > 9, < 9 and = 9

Like the terms individually.

[versed wave]

hm

when x^2 > 9 then lhs always >0?

wait

yeah it would always be > 0

as when < 9, it would be < 0?

[spice grove]

No.

When x^2 > 9, (x^2-9)5^{x+1} > 0 but 3^{x^2-9} > 1

Because 3^0 = 1

[versed wave]

so when x^2>9 lhs would always > 1

now as for x^2<9

the first term would be > 0 and < 1

the second term would be < 0

so this would always be < 1

same with x^2-9 = 0

so essentially this problem can be reduced to

$x^2-9<0$

is my logic sound so far?

[spice grove]

How'd you get =

[versed wave]

oh wait

my bad, i did 3^0=0

so it doesnt have eq

[flat frigate]

ngDe

[spice grove]

Now you're right.

[versed wave]

yeah holy hell, this is a really nice way of solving it

thanks!

.close

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[torpid mason]

How do you solve this?
9. Ignoring air resistance, if an object is dropped from an initial height h, then its height above ground level at instant t > 0 is given by (photo)
where g is the acceleration of gravity.

a) At what instant does the object hit the ground?
b) If h = 500 ft, compare the impact instants for the Earth (g = 9.8 m/s2 ), Mars (g = 3.7 m/s2), the Moon (g = 1.6 m/s2), and Jupiter (g 24.8 m/s2).
c) Find the formula for the instantaneous velocity y at the general instant t.
d) Use the instants found in b) and the formula found in c) to calculate the corresponding impact velocities for the Earth, Mars, Moon and Jupiter.

[torpid mason]

using limits

this is the formula

[wise anchor]

The instant it hits the ground is the t such that s(t)=0

[torpid mason]

but t can't be 0 so you have to use limits, how do you write it?

just lim t -> 0 (s(t))?

[wise anchor]

At t=0, s = h

No limits required here...

At t=???? , s=0. Rearrange algebraically to solve for t

[torpid mason]

I guess you could do that but that equation only works when t > 0

so you can't replace it

[wise anchor]

t=sqrt(2h/g)

What's the problem here?

[torpid mason]

so you don't have to use limits at all?

because that is work for calculus I

we're studying limits

[wise anchor]

They're calculus coming up, but not in the first part

Here's why they have to specify t>0. When you take the square root, the answer could be positive or negativ time. They're saying forget about negative time

[torpid mason]

alright, i understand

and for b)?

[wise anchor]

Plug and chug with alternative values of g

[torpid mason]

okay, thank you so much

[wise anchor]

Calculus is in part c

That requires taking the derivativ.e. THe derivative is a limit.

[torpid mason]

i see

thank you

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[broken thistle]

Is anyone willing to just double check my answers for this? I'm on vacation so can't contact teacher and there aren't any solutions

[urban coral]

sure ill check

what are your answrs

[broken thistle]

thx, I'm taking a pic of my working out rn

Here it is:

if anything is unclear feel free to ask me

[urban coral]

why did you multiply in question 6

[broken thistle]

I'm assuming the product principle applies, so it's 3 AND 2 AND 4 meaning 3*2*4

what wouldve been the correct way of doing it?

[urban coral]

i believe its just adding

i dont see any key word that says to multiply

[broken thistle]

yeah but wouldn't there be more than 10 combinations?

[urban coral]

why do you think so

<@&268886789983436800>

[broken thistle]

because you can have:
hard drive 1, processor 1, graphics 1
hard drive 1, processor 1, graphics 2
hard drive 1, processor 2, graphics 2
hard drive 1, processor 1, graphics 3
hard drive 1, processor 2, graphics 3
hard drive 1, processor 3, graphics 2
hard drive 1, processor 3, graphics 1

I mean that's already 7

and only for the first hard drive

[urban coral]

it sayd 3 types of hard drives

[broken thistle]

yeah

[urban coral]

says*

so theres only 3

[broken thistle]

so multiply 7*3 which gives 24

or am I doing this wrong

[urban coral]

but where did you get processor from

why are there seven

[broken thistle]

there aren't

[urban coral]

ohh nvm

[broken thistle]

but here there are 7 combinations with just hard drive 1

[urban coral]

youre talking about combinations

[broken thistle]

did u misread the question?

[urban coral]

misread what you were doing mostly

[broken thistle]

oh

yeah I'm learning combinations/permutations, stuff like that

[urban coral]

alright alright

[broken thistle]

question says there are 3 HDs, 2 processors & 4 graphics cards

[urban coral]

so yeah its 24

[broken thistle]

Alright cool, do you want me to repost the images so you don't have to keep scrolling?

[urban coral]

sure if you want

so youre learning combinations etc

makes sense

what grade

[broken thistle]

11th

[urban coral]

cool same

[broken thistle]

oh nice

ight feel free to ping me whenever

[urban coral]

ok

for number 7, the question asks how many choices for you to choose your meal, it means overall right, not like the main meal

@broken thistle

[broken thistle]

uuh

lemme see

I think when it says "how many different choices are there available for you to choose your meal from" it's talking about the whole thing

so how many choices of starter/mainmenu/drink/dessert

[urban coral]

alr

so yeah im pretty sure thats right then

[broken thistle]

[urban coral]

8 too

howd you get your answer for 11

9 and 10 are good btw

@broken thistle

[broken thistle]

for eleven

my logic was:

26 letters in the alphabet, there can be repeats

so each 'letter' slot has 26 possibilities

[urban coral]

hmm alright

[broken thistle]

since there's 3 then that's 26^3

for the digits, there's 10 digits (0 through 9)

so add 26^3 with 10^5

although it might be multiply instead of add

I'm not sure

[urban coral]

but whyd you put 10 if theres 9 possibilities

[broken thistle]

because there's 10 digits

9 and 0

0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is 10 digits

[urban coral]

okok

5 digits

[broken thistle]

10^5 mb

[urban coral]

then looks good

congratumalations

[broken thistle]

alright thank you!!!

[urban coral]

npp

hopefully im not stupid and didnt sign you up for a hella bad grade 💀

[broken thistle]

ahaha nah dw, if we both got the same answers it should be good 👍

It's not graded anyways just work I'm doing to catch up on stuff

[urban coral]

ohh okay

so you cant blame me if you get them wrong

[broken thistle]

I mean I might 👀

[urban coral]

youll be like "DamN hEr"

[broken thistle]

exactly 😂

[urban coral]

😭

[broken thistle]

Its AlL Hr FaUlt

[urban coral]

youll be reporting me to the moderators

tsk tsk

[broken thistle]

With valid reason!!!! you made me fail my high school diploma

those questions were my passing requirements 😭

[urban coral]

i-

[broken thistle]

Alright well this was fun, but I better get back to work, I still have a lot to catch up on 😔

thanks for the help tho! appreciate it

[urban coral]

npp

[broken thistle]

I'ma be closing the channel now

any last words?

[urban coral]

have a good vac

[broken thistle]

Thx!

U 2 (if ur on vac)

.close

[safe radish]

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[urban coral]

i know
(from star wars)

[urban coral]

im not sadly-

but thx anyway

[broken thistle]

I don't get it.... 😅

when in the star wars movies does this happen

[urban coral]

when leia says i love you, and darth vader says i know

[broken thistle]

tf

[urban coral]

🤡

[broken thistle]

oh ok you're trolling me, wow :(

[urban coral]

mb

[broken thistle]

Leia would never say that to vader

[urban coral]

but he does acc say that

i-

[broken thistle]

She would only say that to her lover Luke ofc

[urban coral]

same person

[broken thistle]

what no

vader is anakin

luke is his son 😭

[urban coral]

ehh

apple doesnt fall far from the tree

[broken thistle]

and Leia is luke's twin

[broken thistle]

eh, idk, maybe

and Leia says that to Han Solo wtffff

just rememberd

[urban coral]

ya him

that darth vader

[broken thistle]

wha- whaat njvz

han solo is NOT darth vader

you would be dead rn if I was star wars geek

I am mildly offended that you refer to these characters so lightly 😔

[urban coral]

hehehe

[broken thistle]

what movies do you watch anyways? Twilight 🙄?

[urban coral]

ew tf

i read the books tho

[broken thistle]

nice

[urban coral]

and i cant find the movies anywhere anyway

[broken thistle]

books are fine, haven't read them but if its a book

[urban coral]

i watch mostly dc marvel

[broken thistle]

r/piracy is your friend

[broken thistle]

SO MANY resources to watch movies online

[urban coral]

flash is my fav

[broken thistle]

wha

no

[urban coral]

ya but a bunch of websites are blocked

[broken thistle]

I hope you mean the character

and not the movie

[urban coral]

ya hes my favorite

no the character

[broken thistle]

ok good

[urban coral]

the movie sucked ass

unfortunately

[broken thistle]

Yeah I didn't even watch it but the images I saw, looked horrible

[urban coral]

ya

[broken thistle]

some 1990 cgi right there

[urban coral]

lmaoo

black and white

[broken thistle]

ah goddamit, I really need to get to work, I'm sorry :(, cya, have a nice morning/evening/afternoon wherever you are!

[urban coral]

where are you on vacation if i might ask

[broken thistle]

thx for the fun 20 or so minutes!

Mauritius

[urban coral]

haha no worries byee

feel free to add me too if you want

[broken thistle]

Yeah sure

Last question, do you happen to do the IB by any chance?

[urban coral]

wdym

[broken thistle]

ah ok nvm

IB is the international baccalaureate

[urban coral]

i dont do well with abbreviations

[broken thistle]

it's a different diploma

[urban coral]

i do flvs so

[broken thistle]

and its so much work

idk what that is

[urban coral]

florida virtual school

[broken thistle]

is it any good?

[urban coral]

its pretty difficult lmao

and im taking honors sniff

[broken thistle]

yeah same here unfortunately 😔

[broken thistle]

what is that, does that mean harder?

[urban coral]

yeah

like AP classes

kinda

[broken thistle]

idk what those are either 😭

[urban coral]

uhhh

basically IB

[broken thistle]

alright

[urban coral]

its like advanced

[broken thistle]

ok wtf, I actually need to do work now I'm so sorry, feel free to leave a dm whenever you want!

cya

[urban coral]

byye

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[crude bear]

https://math.stackexchange.com/questions/349738/prove-operatornamerankata-operatornameranka-for-any-a-in-m-m-times-n

My issue is with the last step of the proof. (first answer).

We have N(A) = N(A^T * A). The rank nullity theorem says that
rank A + dim(N(A)) = m and
rank A^T A + dim(N(A^T A)) = n

[crude bear]

I don't see how we get rank A = rank A^T A

[safe radish]

@crude bear Has your question been resolved?

[crude bear]

<@&286206848099549185>

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@crude bear Has your question been resolved?

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[hollow junco]

Hi! I'm confused regarding how to place boundaries for these guys. Pretty sure i did this wrong

[hollow junco]

I'm thinking the boundaries for blue sound be 0 -> pi/2 and for red 0 -> pi

and i think i should then subtract the integrals

[safe radish]

@hollow junco Has your question been resolved?

[hollow junco]

figured it out

.clsoe

.close

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[patent vault]

what p series would i compare this to?

[patent vault]

not sure

[median vigil]

take the highest power of n on the numerator and denominator and discard every other term. then calculate the overall power with exponent rules

[patent vault]

thank yoiu!

.close

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[icy lance]

your derivative is fine

how did you find v

how did you type the answer?

write the exact value

1/3 * ln(1/2)

or equivalent

,w (y=2e^{3x}-3x)

[flat frigate]

Results provided by WolframAlpha

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[icy lance]

1/3 log(1/2)

is fine

sometimes log is used for the natural logarithm

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[nocturne zephyr]

How many 3 digit odd numbers greater than 600 can be formed using the digits 2 3 4 5 6 and 7

[nocturne zephyr]

the answer is 20 but I keep getting 36 idk what I’m doing wrong

[steel stag]

how are you counting

it's kind of weird in that the best way to count is to do 600 and 700 separately, and fill the ones digit before the tens digit

[nocturne zephyr]

Wdym

I’m doing 2 x 6 x 3

Since there’s only two possibilities for the first digit then 6 for the middle and 3 for the last

However it’s not 36 it’s 20 so idk what’s wrong with that

[karmic hedge]

the cases are different since 6 and 7 have different parities

[thin bridge]

there are a different number of possibilities for the last digit depending on whether you start with 6 or 7

starting with 7, you have one less option for the units digit

[karmic hedge]

in other words, if you start with a 6 you can end in 3 choices, but if you start wiht a 7 you have 2

[nocturne zephyr]

So then how would I approach this

[thin bridge]

consider the two cases serparately

[nocturne zephyr]

So would I

Do this

1 possibility for first digit (only can be 6) 4 for second (1 number used) and 3 for third (7 3 and 5 available) + 1 possibility for first (only can be 7) 4 for second (1 number used) 2 for last ( 3 and 5 only)

To get 20

12 + 8

[safe radish]

@nocturne zephyr Has your question been resolved?

[nocturne zephyr]

It was because I didn’t divide out repeats

Thanks

[safe radish]

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[vestal flume]

Looking at this image it demonstrates how translational a_A and the tangential acceleration component perfectly cancel (at point C), leaving only the normal component of acceleration.

However this seems odd to me, because the translational a_A doesn't change with the radius of the circle, however the tangential component does, as its magnitude should increase with the radius.

Anyone able to explain?

[steel stag]

you're asking why the linear speed and the rotational speed cancel to 0 even though rotational depends on the radius?

it comes from the translational velocity coming from the rolling of the bottom point, which is a radius away

assuming no slip and whatever

[vestal flume]

yeah, and apologies im taking dynamics.

oops just unfked my self there

Plurmorant you mind looking at my current HW question that im studying?

[steel stag]

go for it

[vestal flume]

The Explanation is whats confusing me.

mainly this part

[steel stag]

no slip means that the point of contact copies the floor (in the opposite direction I guess)

so like since the conveyor is rolling at a constant speed the disc has to rotate at a constant speed

[vestal flume]

really? even though Point A is decelerating?

honestly any chance your down to voice chat? I Can't type out my confusion clearly enough.

[steel stag]

yea that's kind of confusing me too, but you can always think of rolling and translational dynamics separately

sure I won't talk though

[vestal flume]

alroit thats better than nothing.

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[civic topaz]

if you use 1/2bh and distance between two points formula

you end up with two unknowns

still x and y where (x, y) is the missing coord

[obtuse plover]

ok i did it, but i think i did it in a non-simplistic way

i can show u, but idk if id call it efficient

[civic topaz]

would like to see it

[obtuse plover]

ok, so we know:

0.5bh = 3

thus bh = 6

and using pythagoras, we know b^2 + h^2 = 5

(i got the sqrt(5) to be the length of the hypotenuse using distance formula)

are u still following?

[civic topaz]

yes

[obtuse plover]

so now solve for b and h

[flat frigate]

Stephen

[obtuse plover]

lmk what u get

[civic topaz]

sure

[civic topaz]

i dont get real solutions

[obtuse plover]

show me how u did it

[civic topaz]

and then i used quartic solver on calculator

[obtuse plover]

dont use quartic solver

$h^4 -5h^2 + 36 = 0$

[flat frigate]

Stephen

[obtuse plover]

make a substitution

x = h^2

and then solve

[civic topaz]

i get imaginary roots

with the quadratic

[obtuse plover]

show me

o wait...

WHY

ok

[civic topaz]

wwait my starting question might have been a big vague

or there may be an easier way to approach this

[obtuse plover]

yea i have a feeling

[civic topaz]

one second]

so area of ABO = 1/2(2)(6) = 6

so ACD is half area which is 3

I have the coordinates of D, A, and not C.
HOWEVER, I know that it is a right angled triangle and i know two of the coordinates

do ya follow? @obtuse plover

[obtuse plover]

how do u know DCA is a right triangle?

[civic topaz]

good point

im a fool

ugh. trying to think of another way

[obtuse plover]

it looks like a right angle, but idk if its safe to assume that

[civic topaz]

so now we are left with|:

• area of triangle
• two coords
• need to find the last coord

[civic topaz]

nah its not safe to assume because we dont know the value for m

if the products of the gradients were -1 then we could be safe

but we dont know m

[obtuse plover]

yea

so we still know bh = 6

and we know

[civic topaz]

but we cant use the right angle formula

[obtuse plover]

$\sqrt {(y - 1)^2 + x^2} = b \newline \sqrt {(x - 2)^2 + y^2} = h \newline (x,y)$ is the missing point of the triangle

[flat frigate]

Stephen

[obtuse plover]

but we still need more

hmm

[civic topaz]

could we do something with this

[mellow lodge]

that nasty thing

[obtuse plover]

yea i think we can add it to our equations lol

but i think, as bananahead said, itll be nasty

ig we can try

[civic topaz]

it really shouldnt be this bad

let me step back and rethink the problem

[obtuse plover]

yea we're defo overlooking smth super simple

[mellow lodge]

Maybe doing something with it in relation to the large triangle?

[civic topaz]

unsure what other details could be used

could we do something with it

[obtuse plover]

i dont think it is

[mellow lodge]

yeah but looking at it like that complicates things

[civic topaz]

this is driving me nuts

[obtuse plover]

we know AB is sqrt(40)

[civic topaz]

think about how we can find a coordinate in the first place

usually a solution to an equation of two lines which have been intersected

in this case y=mx+1 and y=-3x+6

so we should find m

[mellow lodge]

maybe find the angles and work backwards from that?

[obtuse plover]

arctan(3) - [arctan(0.5)] is angle DAC

[civic topaz]

wait have you noticed something

OAB area is 6

we are told ACD is half of that which is 3

but then OAD area is 1

that doesnt add up

1 + 3 is not 6

[obtuse plover]

OAD is 1?

[civic topaz]

1/2(2)(1)

[obtuse plover]

oh yea

hmmm

VERY SUS

good observation

[civic topaz]

this is painful

im starting to think only some weird rule or theorem can get you to the answer

[mellow lodge]

you could try to find the boundaries and integrate to get the area. Might be pretty awful though

[civic topaz]

nah dont think its worth going down

thinking about distance formula now

[mellow lodge]

me neither

just spitting out potential options

[obtuse plover]

well if the areas dont add up, how can we even consider proceeding? isnt it a flawed question?

[civic topaz]

nah its not flawed

i might have made an incorrect assumption in my working

man

i think i have to leave this

https://www.chegg.com/homework-help/questions-and-answers/b-diagram-line-equation-y-3x-6-crosses-x-axis-y-axis-b--suppose-m-0-line-equation-y-mx-1-c-q163588076

this is unsolved, if anyone can help, please do!

[civic topaz]

nvm it was a wrong observation

[obtuse plover]

wow ok im really taking L's rn

lol

[safe radish]

@civic topaz Has your question been resolved?

[safe radish]

@civic topaz Has your question been resolved?

[civic topaz]

i think this might work

get x = ... from -3x + 6 = mx + 1
then we know DBC + AOD = 3
AOD = 1
DBC = 1/2 * 5 * x
youll get m from that

[obtuse plover]

@civic topaz howd u get DBC?

[civic topaz]

5 is vertical length

x is horz

length

(from y axis to c)

[obtuse plover]

how do u plan to get the distance from C to y-axis

not seeing it sorry

[civic topaz]

this is x

where x is the missing x coordinate of point C

[obtuse plover]

yes

oh ok so 2.5x = DBC therefore 2.5x = 2, x = 4/5, thus C is (0.8, y)?

is that what ur saying?

[civic topaz]

i worked this out

it doesnt seem right does it

0.8

is closer to 2

i mean when you look at the diagram

[obtuse plover]

angle DAC is 45 btw

[safe radish]

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[safe radish]

@civic topaz Has your question been resolved?

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[tardy violet]

how to do this?

[safe radish]

@tardy violet Has your question been resolved?

[tardy violet]

yes ofc

wait answer is

D

oh

wait wat if i don't know wat the graph looks like

[tardy violet]

like during an exam when i dont hv access to desmos

um no-

oh ok thx

mhm

right

oh ok

[visual pelican]

wats us p:2=2:3 and q:r=4:7 find p:q:r

[tardy violet]

mhm

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[broken thistle]

Can anyone help me solve this?

[broken thistle]

I've tried 5c2 * 21c5 but that's wrong

any other thing I tried wasn't correct either

[round shore]

It says "at least 2" are vowels

not exactly 2

[broken thistle]

ohhhhh alright, hold on let me try to solve it now, thx!

Ok yeah I got the right answer

I substracted the possible arrangements with 1 or 2 vowels from the total amount of arrangments (26c7)

thx!

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[craggy marlin]

Hi! A quick question: to define the domain of a function I only want to know if these two ways are the same --> lR \ {x > -3, x < -2 } and (∞, -3] U [-2, ∞)

[craggy marlin]

Are the same? Thanks!

[dull sequoia]

Man that first one

That is some abuse of notation for real

{x> -3, x<-2} is a set containing elements which are subsets of ℝ

There is no intersection between this and ℝ

[craggy marlin]

oh

In my school we use that to express the domain

Maybe it has so much notation, but express the same than the second one?

[dull sequoia]

Yes

But I don’t like the first one

[light shoal]

assuming the notation in the first one just means $\mathbb R \setminus (-3, -2)$ then yes, they are the same

[flat frigate]

Bungo

[light shoal]

tbf this kind of shorthand shows up a lot in measure theory

[craggy marlin]

I also don't like xdd

the second is more clear

(for me)

[dull sequoia]

It looks like they were just lazy

[light shoal]

would be better to write ${x \in \mathbb R : x > -3, x < -2}$

[flat frigate]

Bungo

[craggy marlin]

thanks to both of you ^^ @dull sequoia @light shoal

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[heady fjord]

why does this series need to have p>0 for it to converge

if an = 3/2(1/n)^p then wouldn't it go to zero if we just increased n anyway

[stable pine]

ok think from a logical standpoint

if we take p < 0

its gonna be 3/2 (-1)^n-1 n^p

thats obviously going to diverge

[stable pine]

exactly it would converge if this was the case

but if p<0

then it would be (n^p)

[heady fjord]

ok thanks

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[real bramble]

right angled paralelogram's dimensions are x 2x and 3x and volume = 16 what is the area of that paralelogram?

[real bramble]

x * 2x * 3x = 16 => x = 2/3powersqrt(3)

but when I checked it on math calculator

waiit I think I get it sorry 😄

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[twilit agate]

In Halmos ("Finite Vector Spaces"), he provides a (weird) proof of the fact that any subspace $M$ of a space $E$ of dimension $n$ has a dimension $\leq n$. He starts with :
"it is possible to give a deceptively short proof of this theorem that runs as follows. Every set of $n+1$ vectors in $E$ is linearly dependent, hence the same is true of $M$ ; hence, in particular, the number of elements in each basis of $M$ is $\leq n$."
He then proceeds with a second part of the proof :
"the trouble with this argument is that we defined dimension $n$ by requiring in the first place that there exist a finite basis, and then demanding that this basis contain exactly $n$ elements. The proof above shows only that no basis can contain more than $n$ elements; it does not show that any basis exists."

What exactly is he talking about ?
From my understanding he takes a basis of the subspace $M$ and then notice it is linearly independent in $E$ as well. So he concludes that the cardinal of this basis is $\leq n$. This, I have understand.

Then, my further understanding (of the second part) is that, while we know the existence of a basis for $E$, we are not ensured that $M$ has a basis.
Is my interpretation correct ?

I haven't included the "solution" he further gives to the "problem", I'll probably post it in a separate question.

Thanks in advance!

[flat frigate]

niobium

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[twilit agate]

<@&286206848099549185>

maybe I should ask it at Stackexchange ...?

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[opal flint]

Hello

[opal flint]

Need help with questions 3 and 4 I dont even know where to start

It is on Factoring Quadratics

[frozen marlin]

GCF = greatest common factor

all you need to do is find the term that they all have in common in this case

[opal flint]

but it has variables and exponents

[frozen marlin]

try splitting them into their components

[frozen marlin]

^

for example

[flat frigate]

ren (what's lemonsaurus?)

[frozen marlin]

try doing the same with the other ones

[opal flint]

whats the 2 2 3 7 for

nvm

i just figured it out

thanks man

[frozen marlin]

np

do .close if you're done

[opal flint]

.clsoe

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[sage cipher]

how many unique combinations i can get for each step here, before it all converge to 1 combination of 6 elements?

[sage cipher]

i would like to find a formula for this, but i can't find it nowhere

order doesnt matter

[granite breach]

This formula?

[sage cipher]

that would be it if i didn't allow for elements to combine with themselfs

as you can see, the second step is 21, not 15

but it would be 15 if elements couldnt combine with themselfs

[dusk basin]

this I believe

[sage cipher]

it works for the first, but it doesnt eventually converts to 1 option at step 6

[dusk basin]

what do you mean?

[sage cipher]

basically, each step asks how many combinations of X elements that don't repeat their members besides equal elements can be formed.

if i have 6 elements, at groups of 1, it's 6

if i have 6 elements, at groups of 2, it's 21

if i have 6 elements, at groups of 6, it's 1

i want to find the steps in the middle

[dusk basin]

it can’t be 21. I’m looking at your diagram and it’s clear that there are repeats

[sage cipher]

i hope this makes it more clear

[dusk basin]

if you are looking for how many where there are NO repeats, then you don’t want replacement

[sage cipher]

same elements can repeat, its just order that doesnt matter

[dusk basin]

draw out what you think 6 looks like. There are more than 1

[sage cipher]

oh yeah i see what you mean now. sorry for the flaw on the question

[dusk basin]

no worries

[sage cipher]

let me check if the formula works for this right now that i realized i was thinking of the wrong result

[sage cipher]

yep the formula is spot on, thank you!!

helped me a lot

[safe radish]

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[safe radish]

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[indigo kraken]

.close

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[rare apex]

solve in /R f(x) =< 0

f(x) = x^2 - 8x + 15

f(x) = (x-3)(x-5)

[thin bridge]

have you done anything after that?

[rare apex]

like what

[thin bridge]

any work / thoughts at all

literally anything

[rare apex]

i know that [3 ; 5]

is the solution

but

i need the process

[thin bridge]

ok

from the factored form there are two main approaches

[rare apex]

(x-3) and (x-5)

[thin bridge]

method 1: the graph/wavy method
do a rough sketch of y = (x-3)(x-5)
all you need the care about is the x-axis / intercepts and concavity
from there it should be clear where the function is greater than, less than or equal to 0

[rare apex]

i use the
x-3 = 0
x-5 = 0
method

[thin bridge]

method 2: do sign analysis
identify the zeros
consider that happens (what is the sign of f(x)) between/before/after those values

if you have a method, try applying it

[rare apex]

i use the

x-3 = 0 or x-5 = 0
x= 3 | x=5

and idk what should I do next

[thin bridge]

I have outlined two methods above

[rare apex]

oh

ok

thanks

close

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[undone python]

help idk where to start

[undone python]

does it have something to do with total derivative being the sum of partial derivatives

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[undone python]

<@&286206848099549185>

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[undone python]

https://media.discordapp.net/attachments/943127908728332289/1222774484335853578/IMG_0153.gif

[granite eagle]

I would solve each equation for u & v respectively then take the partials

[undone python]

but both u and v exist for both equations

do i just take the other as constant when doing one of them?

wait maybe i kinda get what you mean

so i express the first and second in terms of u

take partial

do the same for v

and this will cause both du/dx and dv/dx to popup for both equations

at which point its a linear question?

or maybe i can just take partial directly

[granite eagle]

Yeah expressing then as a set of linear equations might be the play
I’d need to think about it

[undone python]

ill try

yeb

got it to work the linear way

since values for xyuv are given as 1

i forgot abt it lol

aight i got it

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[jade oriole]

Help on how to find formula to be used in this problem

[jade oriole]

I need step on how to solve this

<@&286206848099549185>

<@&286206848099549185>

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[glacial sonnet]

what have you tried

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[quartz kernel]

I've been studying logarithmic functions but I can't understand why z should not be 1

[steel stag]

what's the original problem? like what's w

[drowsy moss]

,rotate ccw

[static ledge]

it would make more sense if it were z should not be 0, becuase ln(0) is undefined (and the equality would only make sense as a limit)

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@quartz kernel Has your question been resolved?

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[lean otter]

hi

[lean otter]

Use the axis of symmetry to find the turning point of the graph and hence express in turning point form:
a y=x2−4x+3

can someone help solve this

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@lean otter Has your question been resolved?

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[quartz kernel]

It was any arbitrary complex number

[quartz kernel]

^ this is the image

[quartz kernel]

That would be obvious
But my prof wrote 1 there for some reason I guess
I can't figure that out

It's something related to this proof also

Cuz here I can only write x+iy≠ x+i(y+2npi)
Which is a very vague thing to say imo
They are both arguments of z

[safe radish]

@quartz kernel Has your question been resolved?

[safe radish]

@quartz kernel Has your question been resolved?

[safe radish]

@quartz kernel Has your question been resolved?

[safe radish]

@quartz kernel Has your question been resolved?

[quartz kernel]

@raven vessel

[azure bane]

bro pinged a user 😭

[quartz kernel]

oof my bad

<@&286206848099549185>

[covert yoke]

@quartz kernel It's almost certainly an error in your notes, there is nothing particularly special about z = 1 in this case.

[quartz kernel]

alright, I am sure he meant z!= 0 instead
well, I'll ask him later, thanks for now
meanwhile, can I ask another question here?

[covert yoke]

!1q

[safe radish]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[covert yoke]

you may ping me when you reopen the question, and I will assist if I am able.

[quartz kernel]

alright

.close

[safe radish]

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[inner abyss]

For part b can someone explain why we don't consider epe

And just deal with ke and gpe

<@&286206848099549185>

[placid oak]

the string is slack since it has natural length 1

[icy sonnet]

i told u to chill right after ur done i am sorry

[obsidian oracle]

Please stop arguing in math help channels

[eternal carbon]

[inner abyss]

It's slack
But since it has v=2 at its natural length
Doesn't it go upwards as its speed decreases
So doesn't the string then compress making it have epe again?

[placid oak]

only springs compress

strings only hold epe in tension, not compression

[inner abyss]

Ah thanks

.close

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[storm ice]

I have still not understood apparently

[storm ice]

What is the idea to solve this?

my approach was the following

[obtuse plover]

instead of dividing both sides by -cos(x), bring the terms to one side, and factor out a cos(x)

this way, you dont lose solutions

[storm ice]

ah so the division is what's screwing me over then

so get cos(x)* ( 2sinx + 1) = 0

and then one solution is when cos(x) is 0 and the other is when 2sin(x)+1 is 0?

[obtuse plover]

ye

[storm ice]

awesome noodles

[iron flare]

yes

[storm ice]

cheers mate

.close

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[obtuse plover]

@iron flare bouta killsteal, not today pal

[iron flare]

i was only going to add a note about dividing out... ill just go to bed

[obtuse plover]

lol im just playing

u good bro

[iron flare]

(yes)

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[lean otter]

(\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2})

[flat frigate]

so flashy

[lean otter]

im new with limits i would like to get my answer checked since i dont have answer sheet

[arctic glen]

Yes your work is good ! And correct

[lean otter]

oh, thank you ! :)

.close

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[solar hazel]

i am trying to prove 1 is a limit point of ${\cos\left(k+\frac{1}{2}\right) : k \in\bZ,k\neq 0}$. (and quick disclaimer, i do not know that this is true but cmon it's gotta be). so my thot was we want to show that elements of ${k+\frac{1}{2}:k\in\bZ,k\neq 0}$ can be arbitrarily close to integer multiples of $2\pi$ and then since $\lim_{x\to 0}\cos(x) = 1$ and $\cos$ is $2\pi$ periodic we are done. i.e. what we want is that for any $\varepsilon>0$, there are $k,n\in\bZ$ such that $|k+1/2 - 2\pi n| <\varepsilon$. or instead, something like $|2k+1 - 4\pi n| <\varepsilon$. the point of that is to make it more amenable to my next idea, which is to use dirichlet's theorem: for any $\varepsilon \in (0,1]$, there exist $p,q\in\bZ$ such that $|p(4\pi)-q| < \varepsilon$. but see the trouble here is i do not know if there are frequent odd $q$'s. i also thought to interpret distance between $k+1/2$ and $2\pi n$ as distance between $k$ and $2\pi n-1/2$ instead but the $1/2$ was still giving me trouble

[solar hazel]

hmm i should change some letters

[light shoal]

i havent read your argument but the claim is definitely true

[flat frigate]

slayla

[solar hazel]

my argument is not complete

i need a kind sir or ma'am to help me complete it

[light shoal]

in fact the sequence $\left{e^{ik}\right}$ is dense on the unit circle in $\mathbb C$, and even stronger than that, it's equidistributed on the unit circle

[flat frigate]

Bungo

[light shoal]

i'll read your shenanigans now

[eternal carbon]

shenanigans™️

[solar hazel]

the trademark slayla shenanigans

[kind tinsel]

a nice result is that integer multiples of irrational numbers mod 1 are dense in [0,1], using the pigeonhole principle on the intervals $(0, 1/n), [1/n, 2/n),... [n-1/n)$ for $n$ sufficiently large, and using the fact that if two values $k_1\alpha$ and $k_2\alpha$ mod $1$ (for $\alpha \in \mathbb{R}\setminus\mathbb{Q}$ and $k_1 \neq k_2$) are in the same interval, you get that $|k_1-k_2|\alpha \in (0,1/n)$, and after that it is not hard to obtain one in every interval. Apply this to $2\pi$, you know that for any $\varepsilon > 0$ then you get for some $k$ that $2k\pi (\text{ mod } 1) - 1/2 < \varepsilon$

oh my god tex

i think this works

[flat frigate]

smay

[kind tinsel]

edit for readability

also, all of the intervals here are open intervals

(0,1/n), (1/n, 2/n), etc.

since if k alpha = i/n then alpha was already rational

idk if this is actually what you're looking for actually, i didn't read the second half of your solution carefully

[solar hazel]

hmm yes ok that works

[kind tinsel]

okay great

[solar hazel]

well i mean it's taking a step back from what i had and giving a different solution but yea it works lol

the 2kpi (mod 1) points cluster around 1/2 because dense in [0,1]

fair enough

thx

[kind tinsel]

melodycreepysmile

[solar hazel]

i'll leave it open for a bit in case anyone has a way to finish what i started

fuck i love the word cluster so much

[mortal ingot]

Welp typical analysis problem

[mortal ingot]

Halving this and taking care of 1/2 seems the best imo

Can always approximate any number with "pa - q" where a is irrational

[safe radish]

@solar hazel Has your question been resolved?

[mortal ingot]

@kind tinsel sorry if I annoyed you and resulting in no response

[safe radish]

@solar hazel Has your question been resolved?

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[karmic zealot]

Could someone help me find the derivative of this function? I’m stuck on how to start

[prisma wren]

quotient rule, chain rule for both the numerator and denominator.

[karmic zealot]

Yeah i tried doing that but i just got confused because it looks so long

[dark cargo]

it will

[prisma wren]

I mean, its a long question. You just have to power through it

[dark cargo]

until you get used to it.

[karmic zealot]

I got this

[prisma wren]

It might help to list out u, u', v, and v' individually before trying to combine everything

[karmic zealot]

I need to get this done quickly 😭

[prisma wren]

close

the derivative of sqrt(t+3) is wrong. Otherwise you're good

[light shoal]

the denom is (t+3)^(1/2), not (t+3)^(-1/2)

[karmic zealot]

Wait which one?

[prisma wren]

the last term of the numerator

[karmic zealot]

Could you remix the picture and show me because I’m lost

Oh

Oh is it -1/2?

Yeah

[prisma wren]

derivative of (t+3)^(1/2), basic power rule

[karmic zealot]

Yeah I noticed

and then I thought of separating the fraction