help-27

[robust bobcat]

To solve for tensions what you should do is to break up all our forces into forces in the x direction and forces in the y direction

The free body diagram is a good tool to see in which direction the forces are acting

Since both tensions are not purely in the x or y direction you would need to break them into x and y components. This is where the angles will be useful

[vapid veldt]

i will be honest, i don't think the 50 60 70 purple triangle is needed to solve the problem

[blazing gyro]

i cant use a free body diagram is there anther way 💔

[robust bobcat]

Why not?

[vapid veldt]

is that something your teacher drew?

[blazing gyro]

yeah thats what im confused on thats what my teacher did

[blazing gyro]

yes

[blazing gyro]

cuz my teacher only likes it HIS way 😭

if we do a different way or like use a different formula he said he will deduct marks

im so lost

[rare kernel]

<@&268886789983436800> hloo

[blazing gyro]

This is what i got down i js rewrote whay my teacher did on the board

[vapid veldt]

ahhh i see now

so, i'll explain a bit first

T1 and T2 have vertical and horizontal components

since the sign is unmoving, the horizontal components cancel each other out, and thus have the same "length"

and the vertical components cancel gravity, so their "lengths" added up is the "length" of the gravity force

so, starting from the point where the sign is, you draw another T1 vector. since it's the same horizontal length as T2, you can draw a line straight down from the origin of T2 to the end of the new T1

and that's the purple triangle

let's label A the origin of T2, B the position of the sign, and C the end of the new T1

[blazing gyro]

wait so are u saying the yellow highlughted and pink highlighter are teh same then?

[vapid veldt]

no

[blazing gyro]

im lost

[vapid veldt]

in the original problem, with only the 20 30 130 triangle, the horizontal components of T1 and T2 cancel out

[vapid veldt]

AB is T2

and BC is T1

[blazing gyro]

wait wait hold on how are they canceld out is it because its sideways instead orrrr what are teh horizontal components of it

[blazing gyro]

OH yeah eyah i see that

AC represents gravity right

and the weight?

[vapid veldt]

yes

[dry oak]

Well the beam isnt moving side to side right?

[blazing gyro]

its not yes

[dry oak]

So its not accelerating yes?

[blazing gyro]

yes

[dry oak]

So theres no net force right?

[blazing gyro]

oh wait does the horizontal componetns only show when its OOHHH waittttt

im so confused sorry i just taught myself force jsut a couple houes ago

[dry oak]

Well look if the net force total is equal to 0

[blazing gyro]

that means that one of the components have to = 0 right

[vapid veldt]

if the object is not moving, then the total force on it is 0

[dry oak]

Not one of the components but the sum of the components

So the sum of horizontal forces must be 0 and sum of the vertical forces must be 0

[vapid veldt]

gravity only pulls down, but the wires are pulling diagonally. together, the wires cancel out gravity vertically, but horizontally they cancel each other

[blazing gyro]

Wait how would this be rearranged so that f and h are on opposite sides i understnad better with the formula if there is one

[dry oak]

Theres no formula to use here its just net forces=0 when in equilibrium

[vapid veldt]

draw a line straight up from point B

[blazing gyro]

Is there a diagram u guys can show its hard tryna visualize thr cancelation 😭

[blazing gyro]

Pint b in this?

[vapid veldt]

nevermind we are working with forces not distances

[dry oak]

Did your teacher write this or is it a textbook?

[blazing gyro]

all the information is form the textbook but my teacher likes to reword it

so i believe the formula is that same form the textbook

the coloured writing is mine

but hte examples i got are form him

i copioed it down from the board

[dry oak]

Ye so your teacher is responsible I was just wondering cause its a very nice method of teaching he sounds like a good teacher

[blazing gyro]

he talks REALLY fast so thats why im lost 90% of the time 😭

but yeah hes nice

[dry oak]

Anyways so as we said before since this system is in equilibrium so the both the horizontal components and the vertical components must be 0

But the tension is at an angle so you have to find out there respective horizontal and vertical components

[blazing gyro]

even when its 0?

[vapid veldt]

unfortunately the professor said not to use a free body diagram

[dry oak]

Yep cause then the sum of the horizontal components must be 0

[dry oak]

How?

[blazing gyro]

oh yeah i have to use his method

was with the diagram

[dry oak]

Which diagram

[blazing gyro]

thbis is exactly what i got from the board

[vapid veldt]

this one

[supple knot]

draw the FBD and then erase it

[blazing gyro]

lol bnut how would i show mh work

[dry oak]

Your teacher is tryna put my mind in a box

[blazing gyro]

THATS WHY IM SOI LOST

[dry oak]

By making me use his method

[blazing gyro]

he also told the class that stuff might be similar to physics but to NOT use the physcis method

like a thrid of my class are physics students

they were tweaking

😭

[vapid veldt]

so, your vectors T1 and T2 have horizontal and vertical components

[dry oak]

Oh alr hes basically just using the sin law its an intresting method and it doesn't look that bad

[blazing gyro]

yes

[dry oak]

I dont think you need the components in this case tbh

[blazing gyro]

yeah i jsut dont get the value he put for sinB with gravity

tahts what i dont understna\

[dry oak]

You mean why he put 12g?

[blazing gyro]

nono

why he put

sin50

underneath

[dry oak]

Oh are you familiar with the sin law?

[blazing gyro]

yes

sinA/a = sinB/b

or a/sinA = b/sinB

[vapid veldt]

because

[dry oak]

Yep and whats the angle opposite to 12g on the diagram

[blazing gyro]

OOOOOOOOOOOOHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[vapid veldt]

on the purple triangle, b is the angle opposite to AC, which is the force of gravity

[blazing gyro]

OHHHHH SOT AHTS WHAT IT WAS omg

i feel so dumb

THANK YOU SO MCUH

[dry oak]

Mp

Np

[vapid veldt]

cheers

[blazing gyro]

ok i can clsoe this now

[dry oak]

Idk how you gonna use this method when the system is not in equilibrium

[blazing gyro]

wdym

[vapid veldt]

if it was accelerating, one of the forces would be greater

[dry oak]

Like when the net force isnt 0 you gonna have to use a free body diagram so why wouldnt he just let u use it on cases when the net force is 0 to get your free body force diagram drawing skills solid before jt becomes diffucult

[vapid veldt]

simply, the vectors wouldnt form a closed triangle

[blazing gyro]

oh 😭

um hopefully he wont give a question like that

this si for calculus and vectors 💔

[dry oak]

Ye I mean I suppose he wont Im just saying if he made you draw free body force diagrams it will develop the skills needed for harder problems that could come up later

[blazing gyro]

yeaj i guess

WAIT

I HAVE ANOTERH

QUESTION

[vapid veldt]

sup

[blazing gyro]

WHY IS IT SIN70 instead of sin30

or

[vapid veldt]

because in the purple triangle, T1 is opposite to the 70° angle

[blazing gyro]

OOOHHH so completly forget the first triangle

[vapid veldt]

it's useful for 20 + 30 = 50

but beyond that yeah

[blazing gyro]

i se i see

okay THANK YOU

[vapid veldt]

cheers

[dry oak]

Blud

[blazing gyro]

.close

[devout snow]

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[blazing gyro]

my other things not closed either i think

unless thats js on my end

[ebon coyote]

It takes a while, dw about it

[blazing gyro]

okok

[devout snow]

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[blazing gyro]

Hi guys im back um can someone explain this to me like what the hell im supposed to do wtih compression I TRIED DOING IT BUT I ACTUALLY SONT KNOW WHAT IM DOING

[hollow jolt]

Im confused by lower and upper angle

[blazing gyro]

i think its the diagram

but i dotn get the diagram at all

😭

[vapid veldt]

in which direction are the 60 kgs being pulled?

[blazing gyro]

in the question it just says lift

so thats why im confused ont eh diagram being sideways

its what i got from my teacher

[vapid veldt]

the way it makes sense to me is if the top force T is from a wire, and the bottom force C is from a solid rod

and the object is being held in place

[blazing gyro]

its supposed to be c is for compression i dont get it tho

[vapid veldt]

that way, the rod and rope add their vertical components to counteract gravity, and their horizontak components cancel out

[blazing gyro]

idk about tension at all isnt compression the opppsoite of tesnion

like tensions being pulled

compressed is squished

[vapid veldt]

yes

the bottom diagonal part is being compressed by the weigh of the mass, as well from the tension pulling left

so the bottom part pushes up and right to balance out the forces

[blazing gyro]

wait so is the pully being pulled to the left or 😭

[vapid veldt]

what i imagine is happening here is a drawbridge situation

[blazing gyro]

OOOOOHHH wait so only the top one is a wire and the weigh tis connected with the compression line?

[vapid veldt]

that's what i think

[blazing gyro]

it says compression in the pole

waht does that mean

[vapid veldt]

i am honestly not too sure

[blazing gyro]

😭 😭

[vapid veldt]

in your notes, C is the lower diagonal force opposite to 65°

that's what it is being considered as in the law of sines

[blazing gyro]

My teacher used g i think gravity in it

What do i need gravity for

[vapid veldt]

if the 6o kgs being lifted means they are motionless midair, then the vertical components of C and T add up to counteract G

so that's why you can draw the triangle on the right. the right-most vertical line is G

[blazing gyro]

so i just double C and T

with G

to get G i mean

[vapid veldt]

yes, remember to be careful with the angles

[devout snow]

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[delicate ledge]

how can I find the cumulative probability of a muon going through two detectors that are parallell behind one another with different detection efficieny. what is the overall/cumulative probability of detecting a muon when it hits the detector?

the detector has 4 sectors 53% 65%, 84%, 71 %

[devout snow]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[delicate ledge]

1

[winter torrent]

is there anything important about it being a muon or is this just two independent events

[devout snow]

@delicate ledge Has your question been resolved?

[delicate ledge]

nothing important about the muons

just two independent events

so basically, either it can be detected by both A and B, A, B or neither right?

that's our probability space?

[winter torrent]

yes... i see 4 of them though?

[delicate ledge]

no its two detectors, but each detector has 4 section

[winter torrent]

oh ok. figure out the overall chance that one detector actually detects the muon

[delicate ledge]

one section has the probabiltiy of 53%, second is 65%, third is 84% and fourth is 71 %

[delicate ledge]

don't we have to consider all possible outcomes?

like case a,b,c and d

[winter torrent]

you just need at least one to detect right

it is easier to determine the probability that all four of them miss

that is, the probability that sector a misses AND sector b misses AND sector c misses AND sector d misses

[delicate ledge]

it says "find the cumulative probability of a muon being detected when it hits the detector" but then it says in the tip section: "Consider all the possibilities for when the muon will be detected, and calculate the probabilityfor each possibility that the muon will be detected when it hits the detector. Then consider the probability of each event. Calculate the total probability from these values"

[delicate ledge]

yeah i think so

[devout snow]

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[delicate ledge]

.reopen

[devout snow]

✅ Original question: #help-27 message

[delicate ledge]

99.24% chance to detect a muon?

idk if i did it right

@winter torrent

[winter torrent]

i get something slightly different but not by much. How did you get that number?

[winter torrent]

oh this is for one detector. yes thats right

[delicate ledge]

so if a muon hits a specific section then it is 1-p of it failing, right? our muon is missed if it fails on both detectors so 1-(miss x miss)?

i got 88% now @winter torrent

i just took the average

of the 4 sections

[winter torrent]

wait you had it with 99.24%

[delicate ledge]

isn't this what happens if i go through all sections?

when i did the 1-(miss x miss) i got 88

[winter torrent]

thats the chance of at least one section catching you

[delicate ledge]

oh im confused now

[winter torrent]

each detector has a 99.24% chance to detect a muon, right?

[delicate ledge]

i think so

im so lost lol

[winter torrent]

okay, so then theres a 0.0076 chance that a muon does not get detected by one detector

so the chance that a muon evades both detectors is that squared

[delicate ledge]

0.005776?

that's gonna be 99% again

or am i stupid

[winter torrent]

decimal point isnt in quite the right spot there

should be like 0.00005776 or i guess about 0.005776%

[delicate ledge]

hmmm i think i get it

thank uu

[devout snow]

@delicate ledge Has your question been resolved?

[devout snow]

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[frosty crescent]

solve for c?

[covert shale]

multiply both sides by something to get rid of the denominators

[frosty crescent]

||in particular, least common denominator is the best||

[covert shale]

mhm! (you're looking for ||lcm(2,4,8)||)

[fluid stag]

do be careful when multiplying the LHS though, cuz there is one common mistake I see people do when multing multiple terms in slightly more complex situations like this

[frosty crescent]

reminder to use the ||distributive property|| and ||multiply on both sides||

[pseudo basin]

!status

[devout snow]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mild nebula]

so I did the GCF and multiplied everything by 8 except 1

oh wait lol

[frosty crescent]

everything? lol

[mild nebula]

i did a similar problem to this too

[frosty crescent]

also it's LCM

[pseudo basin]

can you show your work though

[covert shale]

(should be multiplying the 1 as well lol to balance LHS/RHS)

[pseudo basin]

sounds a bit sus the way you phrased it

and i, for one, want to see the execution of it so that we can tell if you messed up or not

[mild nebula]

i'll do it real quick with the new info lol you will be embarrassed for me what i wrote

[covert shale]

believe in yo self

[mild nebula]

ah whatever this is before

[frosty crescent]

c/8 and c/8???

you copied the problem wrong 😛

[pseudo basin]

yeah well ok

you would have needed to multiply the 1 by 8 as well

having attempted to multiply BOTH sides by 8...

also wait

how did c/2 become c/8

and how did c/4 become c/8

that's also kinda sus

waiting for the redone version

[mild nebula]

ok so c = 8 is one solution if we were to add c with 3 since it has negative sign and c = 8/3 is the other solution we could of gotten of we subtracted c with c

[frosty crescent]

the equation is still wrong, since you have c/8 and c/8 again

also

linear equations can only have zero, one, or infinite solutions

how did you get 2? are you doing a quadratic?

[mild nebula]

i got 2 by adding negative c with and so that leaves us with c - 2 or also 1c - 2c

[covert shale]

there is only one solution here because this is a linear equation

[frosty crescent]

...adding -c with what?

[mild nebula]

the 3

[pseudo basin]

dude you've still got the 2 and 4 magically transmogrifying into 8's

[covert shale]

and again you have written (c/8) - (c/8) + (3c/8) when it ought to have been (c/2) - (c/4) + (3c/8)

[frosty crescent]

...the 3 is a coefficient tho

[covert shale]

redo this question from the start, it will help you

[frosty crescent]

ok uh do you know the distributive property???

[mild nebula]

they are all c's so they all go together

[pseudo basin]

yeah scrap everything youve written there

[mild nebula]

alright i will

[pseudo basin]

and try to be more careful ig because you're also messing up collecting like terms

[frosty crescent]

ok, try your solutions, do they even satisfy the original equation???

you're doing something wrong if they don't

[covert shale]

let him redo it first

[frosty crescent]

ok

i forgot about the c/8 problem 😛

[mild nebula]

my goodness my adhd is all over the place lol

[pseudo basin]

your work shows a bit of all-over-the-placeness too

but now it seems actually correct

,w c/2 - c/4 + 3c/8 = 1

[woven radish]

Results provided by WolframAlpha

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Requested by ann_dec

[pseudo basin]

@mild nebula anything else to ask atm?

[mild nebula]

no thats all i need thanks

[pseudo basin]

ok then you can .close the channel

[mild nebula]

ahh so thats how it works lol

.close

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[devout snow]

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[radiant ginkgo]

can someone tell me how to find the three angles in the triangle thing? I know where to go after finding that

[radiant ginkgo]

this triangle

[violet wind]

But yeah, what does the measure of alpha tell you about the triangle

[radiant ginkgo]

one of the angles is 45 deg?

[violet wind]

Yeah

Once you write out all the angles and sides you know it becomes a trig problem

[radiant ginkgo]

I think I just figured out why the top left angle is 45

is it because of this?

[violet wind]

Yeah

Exactly, nice

[radiant ginkgo]

oh

77 degrees

nice

thanks!

.close

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[sharp crater]

only sin 60 is known

[glossy dew]

wasnt there a weird identity like\
$\sin(x)\sin(60^{\circ}+x)\sin(60^{\circ}-x) = \frac{1}{4}\sin(3x)$

[glossy dew]

use that

[sharp crater]

so

sin 20 * sin 40 * sin 80=(sin 60)/4

?

so i get

3/16?

ok thanks

.close

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[modern bronze]

Can someone verify my answer?

[pseudo basin]

you have a typo in the penultimate line

you wrote ln(10-4) when you should have had ln(10) - 4

aside from this you have the correct answer

[modern bronze]

mk thanks

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[tight peak]

i am doing b now

[tight peak]

I am assuming U also is a subset of M, otherwise there is little info as for what U is to begin with

[pseudo basin]

yes, so one would think

[tight peak]

but yeah okay so I have to do this

<@&286206848099549185> if this is a thing still

[devout snow]

@tight peak Has your question been resolved?

[valid silo]

Try using an inductive process and then union.

[static ember]

can you show there exists an open ball whose complement is infinite?

[devout snow]

@tight peak Has your question been resolved?

[devout snow]

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[night rune]

if we use TOTAL = TV +PHONE+READ- double overlap - 2(all three) + neither , i can basically fill everything aside from phone , double overlap and "all three"

can someone guide me through this.

[devout snow]

@night rune Has your question been resolved?

[native gyro]

Look at 2

What does it say(i suggest venn diagram if you cant just imagine it)

[devout snow]

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[tight peak]

i do not think so, since that does solve the problem

cause like what if the open ball has a finite complement

[tight peak]

(I am doing b btw)

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[devout snow]

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[tropic sparrow]

Can somebody please help me get the answer walk me through it the AIs can’t get it either

[frosty crescent]

ok

have you heard of pfd?

aka partial fraction decomposition

...@tropic sparrow? where are you?

[sonic nova]

haha this ones easy

[scarlet sequoia]

polylongdiv can help as well

[onyx torrent]

correct me if i am wrong but isnt that the wrong method for this question?

[frosty crescent]

it is

[onyx torrent]

then why mention it? wont it just confuse the op?

[devout snow]

@tropic sparrow Has your question been resolved?

[devout snow]

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[quaint hearth]

what is nu_2

[signal ruin]

$mu_2$=0

[woven radish]

Frannah⍫

[frosty crescent]

ok

,texsp ||i think it's $\sqrt[3]5$, but i don't really know this problem is super hard||

[violet wind]

Oh noooo spoilers

Hint: additive identity

😂

[woven radish]

nadat12

[signal ruin]

Look at the frickin picture its $\sqrt[3]{5}+\mu_{2}

[hollow sorrel]

if you have $\cbrt{5} apples$ and you get $0$ more apples, how many apples do you have now?

[woven radish]

Avery
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hollow sorrel]

oops

[frosty crescent]

\sqrt[3]{5} not \cbrt

[signal ruin]

But canth helper help me already i am losing time

[supple knot]

This surely is a troll right

[scarlet sequoia]

Yes, you're losing our time

[frosty crescent]

yeah, it's the cube root of 5???

[signal ruin]

….

[frosty crescent]

what other answer could it give???

[signal ruin]

What the frick

I need the answer brih

[frosty crescent]

yeah

[supple knot]

!noans

[devout snow]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[frosty crescent]

what answer do you waant

[signal ruin]

Wasnt this helping server?

[supple knot]

But also you were given the answer already

[frosty crescent]

you're not being very clear, so we can't help

[signal ruin]

So is the answer “spoiled”?

[frosty crescent]

idk

[supple knot]

<@&268886789983436800> yea troll

[violet wind]

What if I don't accept the legitimacy of the reals

[signal ruin]

Bruh

This supposed to be helpers?

Can o ping to get helped bruh?

[violet wind]

First we need to make the algebraic numbers into an ordered field

[frosty crescent]

ok you need to just read the rules 😭😭😭

[worn moat]

I don't think this question was asked in good faith

[frosty crescent]

just, please, take a second

[supple knot]

Did you forget peano.

[worn moat]

@signal ruin Please don't use the help channels for joke or troll questions.

Similarly, everyone else, this isn't an opportunity to dunk.

[signal ruin]

Not using trolls bruh

I just need help with:

[worn moat]

I think this has already been answere.

[frosty crescent]

what is the context of this problem???

[bitter drift]

peak ragebait🥹

[signal ruin]

…

[worn moat]

Yeah, ok, I can see you've done this before.

[violet wind]

Honestly I think it might just be a crank

[worn moat]

I'm closing this now

,close

[violet wind]

There's some cranky behavior in help-11 that doesn't seem funny enough to be a troll

[worn moat]

I disagree but alr

.close

[devout snow]

Channel closed

Closed by @worn moat

Use .reopen if this was a mistake.

[supple knot]

German close

[worn moat]

So many commands, so little space in memory...

[scarlet sequoia]

close to close the channel

[supple knot]

True. RAM is expensive now

[devout snow]

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[rustic sapphire]

I'm struggling with the reference angle

[vital sedge]

well do you know what a reference angle is?

[rustic sapphire]

yeah, from my understanding, it's the x amount that is remaining

from the reference angle

[vital sedge]

well we have 390 degrees

1 full cirle is 360 degrees

[rustic sapphire]

right

but since it went over the 360

and its stuck at qadarent I, do we add 360 + 90 which 450

[vital sedge]

Imma be honest with u, I've never heard of a "reference angle" until now

But apparently this is what it is

[rustic sapphire]

😂

yes thats it

[vital sedge]

yes, so now imagine. the little arrow going around 390 degrees

where does it end up

[rustic sapphire]

mhm it passes 360

[vital sedge]

yes

[rustic sapphire]

were now in quadrant 1

[vital sedge]

yes!

[rustic sapphire]

so do we like 360 + 90 = 450, 450 - 390 = 60 ?

[vital sedge]

well it's the angle to the x-axis right

[rustic sapphire]

mhm

[vital sedge]

at least that's what im getting form the picture

so if we have an arrow that starts on the right side, moves a complete circle

[finite fern]

[vital sedge]

aka 360 degrees

and then some more

[rustic sapphire]

yes

[vital sedge]

what would be the red angle

[rustic sapphire]

thats why i added 90

Apparently, the correct answer is 30

but im not sure how it got it

[vital sedge]

I think it's 30 degrees, as 390 - 360 is 30

oh hey

look at the picture again

[rustic sapphire]

i understand but why did we do 390 when its already passed it

[vital sedge]

cos(390) = cos(30)

cos(360) = cos(0)

cos(360 times any integer) = cos(0)

[rustic sapphire]

oh okok

gonna be honest, did not understand

thank you for trying

.close

[devout snow]

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[devout snow]

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[restive river]

Anyone wanna play
H mu

[devout snow]

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[hushed oyster]

After doing separation of variables, I get $$\int \frac{1}{x \log(1+x)} dx = \int dt$$

[woven radish]

Ajay

[hushed oyster]

But the LHS integral isn't possible in terms of elementary functions

,w \int \frac{1}{x \log(1+x)} dx

[woven radish]

Results provided by WolframAlpha

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

Requested by ajayym

[lunar harbor]

whoever wrote this messed up

[bold harbor]

Just leave it as that and then add a +C lol

Thats like what most people actually do when they meet nonelementary functions in de

[hushed oyster]

I think there might be a typo

[hollow sorrel]

write your answer as an infinite sum maybe?

acc no

[lunar harbor]

Yeah probably this integral looks jigged https://math.stackexchange.com/questions/713330/the-indefinite-integral-int-fracdxx-lnx1#answer-719895

[hollow sorrel]

probably wouldn't work

[hushed oyster]

This comes from a first year undergrad differential equation class

[hollow sorrel]

such an elegant solution 😊

[hushed oyster]

Unless there is some god like trick idk about

[lunar harbor]

I think someone just fucked up lol

[https://math.stackexchange.com/questions/2659673/confusing-to-find-the-integral.-can-suggest-any-method]

[bold harbor]

My reflection aaaaa

😭😭

Ask your teacher about it ig

[lunar harbor]

$\displaystyle \int^{x}_{0} \frac{\dd{r}}{r \log(1+r)}$

[woven radish]

Civil Service Pigeon

[frosty crescent]

this isn't defined

the integral should at least be from 1 to x

[lunar harbor]

eh you gotta do the best you can with what you have

[hushed oyster]

Firgured as much lol. When my sister first showed me I thought I was going crazy coz I couldn't solve a simple ODE

[lunar harbor]

rip

[bold harbor]

I would go crazy too

[hushed oyster]

Anyways, thanks guys

.close

[devout snow]

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[devout snow]

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[rustic sapphire]

So I know im using a 30-60-90 triangle for this

sec is H/O

H is 2

O is 1 which is 30

oh i solved it myself lmfao

.close

[devout snow]

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[hollow sorrel]

good job!!

[cyan hazel]

hi, I know this is really basic but I really don’t know what to take in account for making the equations in this type of problem

4.3

A bottle contains 750 ml of juice with a concentration of 50% fruit pulp. Clara drinks 100 ml of the juice and then refills the bottle with juice that has a lower pulp concentration. If the final concentration is reduced to 48%, what was the concentration of the juice used to refill the bottle?

Answer: The concentration was 35%.

⸻

4.4

A flight of 555 miles is completed at two different speeds and takes a total of 5 hours. During the first part of the flight, the plane travels at 105 miles per hour, and during the second part it travels at 115 miles per hour. How long does the plane fly at each speed?

Answer: 2 hours at 105 mph and 3 hours at 115 mph.

⸻

4.5

Juan has $40,000 and Ana has $35,000. They go together to a bookstore and buy the same book. After paying, Ana has five-sixths of the money left that Juan has. What was the price of the book?

Answer: The book cost $10,000.

⸻

4.6

Seven people buy a store by contributing equal amounts. If there had been two more partners, each person would have paid $1,400 less. What was the total cost of the store?

Answer: The store cost $44,100.

⸻

4.7

Pedro and Juliana start walking from the same point in opposite directions. Pedro walks 1.5 km/h faster than Juliana. After 4 hours, they are 38 kilometers apart. How fast did each one walk?

Answer: Pedro walked at 5.5 km/h and Juliana at 4 km/h.

[devout snow]

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[zinc siren]

What conditions must vectors a,b and c fulfill in order for them to form a triangle?

[devout snow]

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.

[lunar harbor]

https://math.stackexchange.com/questions/2917974/check-whether-the-three-vectors-form-a-triangle

[zinc siren]

is there a proof for the a+b+c=0 thing?

[lunar harbor]

this should be fairly intuitive if you visualise adding the vectors from tip to tail

adding all three vectors means you go around the perimeter

and you should thus end back up where you started

aka going along all three vectors gives no change in position

[zinc siren]

I mean it kind of makes sense, like a graph with a loop

[lunar harbor]

and hence the sum of those three vectors is the zero vector

[lunar harbor]

yes

[zinc siren]

but don't vector a+b, a and b form a triangle?

sorry, a figure with vectors wasn't defined to me

[lunar harbor]

notice how all three vectors don't go the same way

if you take it in this order, you get a sum of zero though (as consistent with earlier)

a, b, and -a-b

[zinc siren]

yeah, so a figure is defined as a path of vectors such that you end up where you started?

[lunar harbor]

ig you could say that

if figure means closed loop

then yes the vectors for the sides (going either clockwise or anticlockwise) sum to zero

[zinc siren]

okay, ty

.closed

.close

[devout snow]

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[lunar harbor]

[devout snow]

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[bold harbor]

Hi, is (b) possible? My relative in 5th grade asked me for help and Im about to go insane

[lunar harbor]

yeah I'm having doubts too

I doubt the unstated assumption is the two having the same area

since 52^2 / 173 is ugly

[bold harbor]

Like there has to be some missing information

[lunar harbor]

yeah definitely

idk are they supposed to reconstruct some kinda scale with a ruler?

I doubt it since this is giving me sg vibes

and idt sg does that

[bold harbor]

HOW DO YOU KNOW

[lunar harbor]

it's obvious

[bold harbor]

Ok but yea

Was just trynna make sure I wasnt going crazy

Oh I have another actual question

I need some hint for 7.53

[lunar harbor]

[bold harbor]

The only observation Ive made is either Δ isnt a perfect square and n-a cannot be written as the sum of factors of a

[lunar harbor]

vieta's formulas

[bold harbor]

Wait i dont get how vieta will help here

The sum of roots and product of roots is always gonna be rational

Since youll have the square roots cancel

[lunar harbor]

think more

[bold harbor]

🫡

[lunar harbor]

How’s it going

[bold harbor]

Still lost

[lunar harbor]

||how can you revwrite rs+r+s||

[bold harbor]

I tried expressing n in terms of the roots

[bold harbor]

2a-n??

[lunar harbor]

oh I meant rs-r-s oops

[bold harbor]

Oh yea thats just n

[lunar harbor]

mhm

[bold harbor]

Uhhhh

[lunar harbor]

think about how you can rewrite rs-r-s itself

independent of the context of the question

[vast drift]

also i feel like u shouldn't completely abandon this observation either

[bold harbor]

Factoring r or s?

Like r(s-1)-s

[lunar harbor]

the expression rs-r-s is basically factorable if that's what you're saying

[lunar harbor]

not this

[vast drift]

im assuming u made this observation because u are aware that integer coefficient quadratics will have perfect square discriminant if the roots are rational

[bold harbor]

Yea

[vast drift]

this should probably help pt u towards what pigeon is trying to tell u

[lunar harbor]

Wait how would you use this? I was thinking ||roots r, s are integers since sum and product are both integers, so (r+1)(s+1)=n+1 is a product of two integers, and finish is triv||

[lunar harbor]

oh um

are we on diff wavelengths lol

[frosty crescent]

wait this seems like an AoPS question

aren't those copyrighted?

[bold harbor]

They are?

[frosty crescent]

yeah

[bold harbor]

Wtf thats not good

[lunar harbor]

plenty of people ask aops questions on here all the time

[frosty crescent]

oh i didnt know

[lunar harbor]

it's fine

anyway back to the question

do you see how you can rewrite rs-r-s

[bold harbor]

I thought of using the +1-1 trick

[lunar harbor]

yes yes

[bold harbor]

But i still have no idea how this helps

[lunar harbor]

well what do you get after doing that

[bold harbor]

(r-1)(s-1)=n+1

[lunar harbor]

mhm

ok maybe just sit there and let brain do brain things lol

[bold harbor]

Oh

[lunar harbor]

I think I've given you enough to work with

[vast drift]

I was thinking ||this tells u if the roots are rational, then they must be integers. So then using ur factorisation (r-1)(s-1) = n+1, for any given n, rational roots exist if there is an 'a' between 1 to n such that n+1 = (r-1)(s-1) has integer solution for r and s. Since r, s = a >= 1, they must both have same sign.

Then, if r, s > 0, r+s = a - n => r+s <= 0 which isnt true. Then r, s must be less than 0.
so making a dummy substitution for ease of reading, i guess -r => p, -s => q, we have n+1 = (p+1)(q+1) where p, q >=1 (skipped a bit of algebra)

So essentially, n+1 must be composite for rational root to exist for at least some a. Then we can just choose the non-prime n+1 for our solutions
||

[frosty crescent]

||oh yeah, integer/rational root theorem is a big w on this problem||

[lunar harbor]

oh I swept all that under the rug of trivial 😭 (trivial = annoying ig lol)

[vast drift]

yeah it is just bit of a logic exercise ig

[bold harbor]

Ok if this just the entire solution ill try continuing from here and looking at this

[vast drift]

well before u do that, do you have a rough idea of what the approach should be

i think (not necessarily true but just my opinion) that it is easier to find the cases where there is a rational root

then its complement is the answer you are looking for

[bold harbor]

Something with the rational root theorem yea

[vast drift]

Perfect square discriminant gives you the conditions u need and it tells you information about what your roots looks like so then you can work towards what that makes n to be

[vast drift]

if u haven't learnt it u dont necessarily need it

[vast drift]

this gives u the equivalent information

[devout snow]

@bold harbor Has your question been resolved?

[bold harbor]

I got it lol i had to peek at your sol a little though

Thank you!

[devout snow]

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[velvet coral]

If we are given a function $f: X \to Y$ and $S \subseteq X$, would it be correct to say $y \in f(S) \iff \exists x \in S: y = f(x)$?

[woven radish]

LXDL

[void fox]

yes

[fluid stag]

correct.

[velvet coral]

So then here for this proof, (after showing that permutations preserve convexity and inverse image of perspective function preserves convexity, how come they don't have any sort of existnetial statement?

or at least when I do this deinfe $h$ as the correspodning persecptive function and $p$ as the permutation function, want to show $epi ; g = P(h^{-1}(epi ; f))$

$ (x, s, t) = a \in epi ; P(h^{-1}(epi ; f)) \iff \exists b \in h^{-1}(epi ; f) : P(b) = a$ and then I am kinda stuck

[woven radish]

LXDL

[devout snow]

@velvet coral Has your question been resolved?

[uncut crow]

if we take an element (x,t,s) in epi(g), then we have h(x,t,s) = (x/t, s/t) \in epi(f) by the equivalence shown in the proof

that shows epi(g) \subseteq h^{-1}(\epi(f))

[devout snow]

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[uncut crow]

.reopen

[devout snow]

✅ Original question: #help-27 message

[uncut crow]

for the other direction, take an element x in h^{-1}(\epi(f)). then it has the property h(x) \in epi(f). and there exists.... you can try to fill it in from there

[devout snow]

@velvet coral Has your question been resolved?

[devout snow]

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[quartz bloom]

My idea was to find the point of intersection of the line and the trajectory of rock .

When i did the calculations , i found a quadratic in tan θ , but the values given in hint doesn't satisfy my answer.

[quartz bloom]

The quadratic in tan θ was :-

tan² θ +( 2 tan φ) tan θ -(1+tan φ )=0

[devout snow]

@quartz bloom Has your question been resolved?

[scarlet swallow]

Could you pls reopen your question 2 hours later? I'm currently busy but I could help you later :)

[marsh trail]

I am unsure as to how you got this equation..you said you solved the two equations you wrote but the equation of trajectory has unknown variables right?

[devout snow]

@quartz bloom Has your question been resolved?

[tame jungle]

@quartz bloom, what exactly are you allowed to use here?

Like what level of math?

Because this question becomes easy to do with multivariable calculus.

[devout snow]

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[slate crater]

Im struggling with 3a. I was thinking of substituting a=3^x and b=2^y^2, but i went nowhere with this. Any hints?

[hollow jolt]

Maybe convert 3^x/2 into sqrt3^x

And square that equation

So you can substitute 3^x

[slate crater]

That's what I tried to do

[sturdy shell]

What about trying to use the index laws so you divide and essentially get one equation

That’s what I’m thinking

[hollow jolt]

Instead isolate sqrta first

Then square it

It should be easier

[slate crater]

Aren't you dividing difference by difference? I cannot apply law of indecies there

[safe knoll]

could try a=3^x/2 and b=2^(y^2)/2

[hollow jolt]

good shout

[safe knoll]

top equation becomes a^2 - b^2 while the 2nd one is a-b

difference of squares and substituting gets u 2 equations , a-b and a+b

[slate crater]

Oh wait how i didn't think of that

Thanks

.close

[devout snow]

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[versed marten]

Hi im confused between compressing and stretching a graph mean

[versed marten]

because it can be either horisontally and vertical but my question only states compress

[fluid stag]

show your question please.

[hollow jolt]

scaling up x compresses while scaling down stretches, the other way around for y

or was that your question

[obsidian nebula]

strech horizontally -> compress verticaly I think

[versed marten]

but how do ik which way is the question

[sinful fable]

If you multiply something next to x,say like 4, for ez sin(4x), than in sin(x) you have a value of 1 at pi/2, but you would have that same 1 at pi/8 in sin(4x) because ultimately its sin(4*pi/8)=sin(pi/2), hence every value gets squished by a factor of 4 in the graph

So if you muiltply numbers greater than 1, it squishes the graph, if you divide, its expands the graph/streches it

[devout snow]

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[neon wagon]

hello, can someone explain me the locus of an extreme point or a point of inflection?

[neon wagon]

i dont understand it

[versed juniper]

a locus is a set of points so they would just be asking what points are extrema and points of inflection

[neon wagon]

yes but

its some stuff with parameters

[versed juniper]

show the original question

[versed juniper]

what's f then

[neon wagon]

just a random function

ok wait

this is f nvm

[versed juniper]

so what's the derivative and second derivative

[neon wagon]

wait so its something like this, generally

but you couldnt do this with functions that are x^3 or bigger right?

[versed juniper]

you would find the derivative and such still

[neon wagon]

oh shit yeah

ok hold on let me find the derivatives

[neon wagon]

so f'x = 2ax-8 and f''(x) = 2a

[supple knot]

yea find such points using that extrema and inflections occur when f' = f'' = ?

[neon wagon]

wait so now what

[fallen depot]

What did you get?

[supple knot]

it was a question

do you know what values f' and f'' equal at extrema and inflection points

[neon wagon]

oh

i mean

[supple knot]

?

[neon wagon]

id have to do stuff with D again?

[supple knot]

no idea what D is

[neon wagon]

You mean do f‘=0

the discrimination or something

[supple knot]

where is D in your question

[neon wagon]

no like cases where D<0

etc

[supple knot]

this is correct for extrema

[neon wagon]

yea and f‘‘=0

[supple knot]

,w discrimination

[woven radish]

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[neon wagon]

Diskriminante

here

or

[supple knot]

this is correct for inflection points

no discriminant involved at all

[neon wagon]

oh

okay

so do that for my fa(x)?

[devout snow]

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@neon wagon Has your question been resolved?

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[quasi jacinth]

example of dependent clauses in graphs

[quasi jacinth]

graph theory

[small jackal]

<@&268886789983436800> spam

[faint zinc]

was it a 4 image spam?

[small jackal]

Yes

[faint zinc]

chartbit got it

[small jackal]

ok

[quasi jacinth]

anyone knows?

[devout snow]

@quasi jacinth Has your question been resolved?

[rotund umbra]

<@&268886789983436800>

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[lunar harbor]

wrong server

try looking at #old-network

.close

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[drowsy parcel]

I can't understand a property of Geometry...Can anyone help me?

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[drowsy parcel]

where the angles BHC, CHA and AHB will form

in an obtuse angles triangle

i can't visualise

it is give that orthocentre of an obstuse angles triangle is outside and left downward side of the triangle

like this

but i dont get it.. what angle to be subtracted from 180 degree

[drowsy parcel]

do i have to give more context or do you understand ?

[drifting sierra]

The blue angles are equal, and so by similarity, the green angles are equal

Same with the red and orange ones

Now the angle BAC is green + orange

The angle BHC is red + blue, which is 180 - (green + orange)

[drowsy parcel]

how

and i don't mean this...

I want to know this in an obtuse angles triangle.

where the formulas fit

[drifting sierra]

Not sure what you mean

If ABC is obtuse, it's the same picture but with A and H swapped (and also E and F swapped I guess)

[devout snow]

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[velvet coral]

I'm pretty lost on how to even show this. I'm not sure what the monotonicty is going to be used for. How would I even start a proof for this?

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[severe anchor]

I have $\int{\sin{2x}\cos{2x}dx}$ and 5 answer options we won't worry about.
Using $u=2x$, I have turned the equation into $\int{\sin{u}\cos{u}\frac{du}{2}}$ then $\frac{1}{2}*\int{\sin{u}\cos{u}du}$

I'm now stumped because I have no clue where to go from here as this integral isn't in my formula sheet and I don't know how to integrate when two factors have the variable

[woven radish]

Flappy the Turd/Frost

[limpid dove]

what have you tried

[severe anchor]

Nothing further, really. Hence being stumped. I can't figure out what to try

[hushed oyster]

You could use a double angle identity

[scarlet swallow]

Alright, do you notice that the derivative of a sine function is a cosine function and vice versa?

[severe anchor]

Yes, I know sine and cosine are each other's derivatives and integrals (sneaking in a negative)

[limpid dove]

this is what i was thinking

[severe anchor]

What exactly is a double angle identity?

[scarlet swallow]

Sure, there are two ways to solve this identity

[hushed oyster]

Which ones are available that you can use?

[limpid dove]

yeah

[limpid dove]

so that hints towards some substitution which you could do

[hushed oyster]

For this question, I think ,texsp ||$\sin(2u) = 2 \sin u \cos u$ ||

[woven radish]

Ajay

[scarlet swallow]

Yep, so there are two ways to do this:

1. Substitution of sine/cosine
2. Double angle formula

[severe anchor]

I have the feeling the substitution one will be what I'm supposed to know

[hushed oyster]

Then just use a u-sub again

since you know cos and sin go back and forth

[severe anchor]

...OH WAIT A SECOND

[hushed oyster]

You have been told so before too

[severe anchor]

w=sinu
dw=cosudu
So now I have $\int{wdw}$ right?

[woven radish]

Flappy the Turd/Frost

[severe anchor]

Or did I interpret this wrong

[hushed oyster]

You're correct

[severe anchor]

Alright cool! And the only integration easier than when f(x) = x is when f(x) is a constant so that's nice

$\frac{1}{2}\frac{w^{2}}{2}+c=\frac{1}{4}\sin{u}^{2}+c=\frac{1}{4}\sin{2x}^{2}+c$

final answer?

[woven radish]

Flappy the Turd/Frost

[severe anchor]

It's supposed to be sine squared not 2x squared

[hushed oyster]

Yes, correct

[severe anchor]

Cool, thank you'

.close

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[dapper fable]

For a natural $n$, let $S(n)$ be the sum of all the positive integers less than $n$ that are coprime to $n$. Determine naturals $m$ such that $S(m)$ is prime.

[woven radish]

Copter

[dapper fable]

i dunno how to begin

oh im stupid

.close

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[grizzled iris]

I am literally stuck on a worked example, after spending 2 days on this topic, I can't believe it

[fluid stag]

what would be the issue with it?

[grizzled iris]

I don't know how to find the points of the function with horizontal stretch by factor of 3

[fluid stag]

this is a worked example, isn't it? did you peek at the working?

[grizzled iris]

Yes it is a worked example, but the issue is that I have done other worksheets and I am sitting in this topic for 4 days and still cannot solve such basic question

And I have no idea what is missing in my conceptual knowledge

[fluid stag]

you didn't answer my question.

[grizzled iris]

It's literally this

I don't understand this

[fluid stag]

yeah, this is what I expected but kinda terse

[grizzled iris]

What does that mean

[fluid stag]

,w terse

[woven radish]

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Requested by leafwingtowa

[grizzled iris]

Ok what does this imply?

[fluid stag]

that whoever wrote this text gave a quick and short explanation

if you have to ask this my comment doesn't matter

anyway, point is

[grizzled iris]

The point is I shouldn't be learning from this book

Because its terse

And there won't be a lot of explanations for actions

[fluid stag]

if that's what you want to take away from my comments, I suppose I won't stop you, but I will also step back to prevent any other misinterpretations

[grizzled iris]

I am literally asking you for help and advice because Im too bad in math to state anything correct

[fluid stag]

the point is that I'm not criticizing the book

[grizzled iris]

Please criticize it if needed

[fluid stag]

it's not my position to

and it's not why I'm here!!

what do you stand to gain from my criticism of your book anyway

[grizzled iris]

That I won't use it because it's poorly written

[fluid stag]

then what are you gonna use?

[grizzled iris]

Other resources?

[fluid stag]

you have other resources ready?

[grizzled iris]

Yes

[fluid stag]

also, Swifter, I see you typing a paragraph there

do you intend to take over?

[grizzled iris]

[brazen halo]

Basically, how I would solve this just reading this from face value:

1. Find the quadratic equation of y, (in the form of ax^2 + bx + c = 0), we have info on highest point, x and y intercept.

2. After finding the quadratic equation, in the form of ax^2 + bx + c, replace all x in that equation with 1/3x. Then you will have a new quadtratic equation in the form of y=(f(x)). From there, you can sketch the graph.

😄

[brazen halo]

Yea lmao XD

[fluid stag]

I'll let you take over then

[grizzled iris]

The point is that they are not asking for finding the quadratic equation

[brazen halo]

Woklay

[grizzled iris]

They are just asking for the new points

[fluid stag]

would be nice if you inform me beforehand but aight

[brazen halo]

The new points are from the quadratic equation amigo

[grizzled iris]

So I have to find the quadratic equation first, then get the new points?

[brazen halo]

I just suddenly dropped by and saw this. Looks simple enough to help

[brazen halo]

Yeps

First, find the quadratic equation of f(x)

[fluid stag]

I mean that's fine but usually if a helper is engaging... well nvm

[grizzled iris]

What if I just multiply the x cordinates of each point by scale factor 3?

[fluid stag]

that works

[grizzled iris]

So why would I find the whole equation

Instead just multiplying the x cordinates

[fluid stag]

because I honestly don't know what the other helper is trying

[grizzled iris]

Can you take over? You were the best help I got, also by Ann and Nicole, and Hanoku

[fluid stag]

ok in a way I do, but it's super long

[brazen halo]

Just giving my two cents on how I would solve it. From what I read, it just says sketch graph of y=f(x/3)

[fluid stag]

did you misspell my teacher's name

[grizzled iris]

Ooops

Flowery child

[fluid stag]

that's miss Hanako, but nvm

either way you have identified what needs to be done

[grizzled iris]

Yeah thanks

.close

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[gritty saddle]

If Hausdorff dimEk ≤ a for every k how can i show that dim of union ≤ a

[devout snow]

@gritty saddle Has your question been resolved?

[devout snow]

@gritty saddle Has your question been resolved?

[void fox]

I guess you have a better chance of getting an answer at #advanced-analysis

[gritty saddle]

.close

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[karmic basalt]

no clue from where to start

[zenith jacinth]

bah, tu commences par u_n+1 - u_n et u_n+2 - u_n

[devout snow]

@karmic basalt Has your question been resolved?

[karmic basalt]

Cest bien?

,rotate 90

[zenith jacinth]

oep

[karmic basalt]

oep?

[zenith jacinth]

oui je veux dire

[karmic basalt]

aaa ok

[zenith jacinth]

$$u_{n+1} - u_n = \int_0^\pi \frac{\sin(u+n\pi)}{u+n\pi} du$$

ah

[woven radish]

Herels

[zenith jacinth]

tu evalues ensuite u_n+2 - u_n

[karmic basalt]

@zenith jacinth cest logique?

[zenith jacinth]

jarriv pas à voir

ah tu as introduit u_n+1 dans le truc
oui ca marche

[karmic basalt]

,rrcw

[karmic basalt]

Mais je ne Vois Pas ou celui cela va nous aider

la dexieme partie specifiquement

[zenith jacinth]

avec les suites extraites ?

[karmic basalt]

oui

pourquoi meme utiliser les suites extraites?

[zenith jacinth]

Je suis un peu rouillé des maths mais de ce dont je me souviens, si les suites extraites d'une suite u_n convergent vers l, la suite u_n converge aussi vers l
Ca peut etre utile de passer par les suites extraites pour simplifier certains trucs

surtout que si on passe par du 2n ou 2n+1, on peut utiliser la périodicité de sin

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[karmic moon]

Hi, the instruction was to solve for x and i got (3±3sqrt5)/2
Is it right??

[fluid stag]

,rccw

[vital edge]

Would help to show what exactly you've done

[fluid stag]

um...

oh sorry

[vital edge]

So that we can actually check

No worries, all yours

I'm heading to sleep anyway

[fluid stag]

but yea please show the full image

cuz I'm getting a negative number under the square root