help-41

[eager shale]

oh wait

[hollow hound]

:/

[eager shale]

mb

[hollow hound]

is fine bro

pls help its my hw 😄

<@&286206848099549185>

no one?

[violet blaze]

ok

so

DBC = DAC

also

ABD = 90

since

AD is diameter

so you can figure out

ABC

so from trinagle

you can figure out DAB

so you can find out

DCB

SO you can gigure out BDC

=BAC

now jst show BDF - ACF

[amber wasp]

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[hollow hound]

👍

.close

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[noble lake]

I need to solve this differential equation, to start I have to check if it is exact. I did this and I am pretty sure it is not exact.

But I have no clue what method to use or where to start to try and solve it

[tired girder]

They say intuition's the best way

ecosydy + esinydx

[winged panther]

Then you need to check again.

[tired girder]

So it's d(esiny)

[noble lake]

It is exact?

[tired girder]

The other side is easier !

[noble lake]

Thanks, made a stupid mistake I understand it now

.close

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[gloomy orbit]

I've got a bit stuck on this one, I'm not very good with standard form and inequalities

[gloomy orbit]

i get that i cube root -1728 and square root 160 and i get -12 and 4 root 10 respectively but

im not sure how to format it

[winged panther]

So the first set gives x ≥ -12 and second set gives -4√10 ≤ x ≤ 4√10. Take the intersection.

[gloomy orbit]

oh ok i get you

[amber wasp]

@gloomy orbit Has your question been resolved?

[gloomy orbit]

i still am not completely sure

im terrible at these

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[frozen temple]

Why is it that when we partially differentiate a 2 deg 2 var equation like a parabola, pair of straight lines etc , with respect to x and y separately and find the intersection of the two resulting lines we get the "center" of the original curve?

[frozen temple]

I was told that this works in class as a shortcut but don't know why this works

[pallid canopy]

Do you have concrete examples

[frozen temple]

Let me send exactly what we did

This is for a pair of straight lines but is supposed work for other curves also

And x1 , y1 is supposed to be intersection of the POSL

<@&286206848099549185>

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@frozen temple Has your question been resolved?

[frozen temple]

<@&286206848099549185>

[viscid linden]

can you type out the writing in the picture?

[frozen temple]

Ok

[frozen temple]

Define S as ax^2 +2hxy +2by^2 + 2gx +2fy +c =0
Differentiate S partially keeping y constant:
ax +hy +g = 0
Now same but keeping x as constant:
hx + by +f =0
Now find point of intersection of these lines and this point is supposed to be the center of S

<@&286206848099549185>

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[noble plume]

Please help me on this question

[noble plume]

So far, I know that in 15 seconds, Andy can run 3/8 of the track

which also means that Andy claps hands with Bob 2 times every 40 seconds

but idk what to do with this info

please help

@hazy bluff

[unborn plaza]

How far does Bob run in 15 seconds?

[noble plume]

3/4 of the track?

[unborn plaza]

Not quite

How far does he run in 15 seconds starting from a clap?

[noble plume]

5/8?

[unborn plaza]

Yeah

👍

[noble plume]

So it takes Bob 25 seconds to run around the trcak

track*

[unborn plaza]

That's not what I got

It's close though

[noble plume]

i meant 24

[unborn plaza]

Yes 👍

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[noble plume]

.reopen

[amber wasp]

✅

[noble plume]

Please help me on this question:

<@&286206848099549185>

[viscid linden]

l/w = 2w/l

you need to solve this

do you see why?

[noble plume]

Because it says that the width of the original sheet becomes the length of the folded one?

[viscid linden]

yes

the new long side is the old short side

and the new short side is half the old long side

but the ratio of the long side to the short side stays the same

[noble plume]

Oh so the ratio will have roots in it

[noble plume]

The ratio of the long side to the short side is 2:1, right?

[viscid linden]

no

that would mean that after the fold, it will become 1:1

[viscid linden]

yes

√2:1 (which becomes 1:(√2/2) = 1:1/√2)

do you see why?

[noble plume]

Unfortunately, no

[viscid linden]

what do you get when you solve this?

what is l/w equal to?

[noble plume]

it's equal to 2w/l

[viscid linden]

l/w = 2(w/l) => (l/w)^2 = 2

right?

[noble plume]

yes

[viscid linden]

so l/w = √2

(and -√2, but who cares)

[noble plume]

oh that makes a lot more sense

thanks for clarifying

[viscid linden]

you're welcome

[noble plume]

So after I rationalize the denominator here, I should get the answer

[viscid linden]

yes

by multiplying both sides by √2, you get √2:1 back

[noble plume]

i put $\sqrt 2 : 1$ into the thing but it said i got it wrong?

[grizzled pagoda]

Eatdonuts

[viscid linden]

wait

[noble plume]

oh wait it's asking to put it as a common fraction

so $\sqrt 2$ im guessing should work

[grizzled pagoda]

Eatdonuts

[noble plume]

bruh it said root 2 was wrong too

[viscid linden]

wikipedia says common fractions are rational fractions

[noble plume]

oh wait a/b has to be both integer

nvm i got the answer

[viscid linden]

what was it

[noble plume]

root 2 over 2

[viscid linden]

bruh

1/√2 would have also worked I guess

it was w/l and not l/w...

[amber wasp]

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[abstract fossil]

stumped by this question

[woeful sable]

1/2A:2/5B:2/7C

2, 5 and 7 have 70 in common, right

nothing before that

you just need to find A:B:C

1/2 of a just means

0.5A

like how you have

2A:3B:4C

except here

the fractions are instead of 2, 3, 4

[amber wasp]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[woeful sable]

mb

[abstract fossil]

honestly i dont understand😭

[woeful sable]

listen

imagine instead of the fractions

[abstract fossil]

i havent done maths since june 2024

[woeful sable]

you have whole numbers

integers

it's a proportion

but instead

the coefficients in the proportion are given as fractions, and you need to get the proportion simplified to an integer form such as 2A:3B:4C

[abstract fossil]

oh

i think i get it

you essentially what to get rid of the fractions?

[woeful sable]

yes

[abstract fossil]

this is how i do it

i nthink

first i times all the fractions by 2

to make it 1A

then by 5 to make it

2b

[woeful sable]

does anything happen with a?

[abstract fossil]

so far

it went from

the original

times all by 2

to become

A: 4/5B: 8/7C
^ incorrect
correct below
A:4/5B : 4/7C

then i think i should times by 5 and then 7

[woeful sable]

yes

[abstract fossil]

so it should have no fractions remaking

[woeful sable]

that's correct

do that and let me know your answer

[abstract fossil]

now i have

35 A: 28B: 40C

there is an error

[woeful sable]

you made a mistake here

if you multiply 2 by 2

it's not 8

[abstract fossil]

silly mistake😡

so it should be

35A / 28B/20C

[woeful sable]

yes

correct

glad to be of service

let me know if u need help w explaining again

[abstract fossil]

thank you so much for the help it was really usefull🙏

[woeful sable]

np

[abstract fossil]

^practice questions, not a real test

.close

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[noble plume]

Can someone please help me on this question:

[noble plume]

So far, I know that x will take m minutes and x+8 will take m-30 minutes

btw x=usual speed and m=minutes

<@&286206848099549185>

[verbal jewel]

usual speed x = 120 / h (h = time for the trip in usual speed) -> h = 120/x
now adjust this formula for an increasing the speed by 8 mph and reducing the time by half an hour.

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

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@noble plume Has your question been resolved?

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[high abyss]

Can you solve for the variable of a quadratic equation without making it into perfect square form

[high abyss]

Instead using and area model and going from there

[sterile swan]

You can use the abc formula

[high abyss]

What is that😭

[sterile swan]

For a quadratic $ax^2+bx+x=0$ has roots at $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[grizzled pagoda]

Bonk

[high abyss]

Thats the quadratic formula

[hollow plaza]

what you mean without making it into perfect square?

send the question

[high abyss]

Im just wondering like in general

[crude kite]

Yes there are many different methods to solving a quadratic, like graphing, quadratic formula, factoring

[high abyss]

Bro what hurts my brain is this

So many different ways which confuses me

[crude kite]

Yeah there's many different ways to solving a quadratic

It just depends on the complexity

Like if you're unable to factor then you can complete the square, else use the quadratic formula

[hollow plaza]

for example for a) u can do this:
(x+3)(x+9) = 0

[high abyss]

Whats it mean to factor?

[hollow plaza]

and u will get -9 and -3

[crude kite]

Literally what you did in part b or d of that image

[high abyss]

So making the x thing then making an area model

[hollow plaza]

on part b u can just divide by 4 and then solve for the sqaure root

[crude kite]

The "x thing" is called factoring

[crude kite]

Nah the easier way is recognizing that it's a difference of squares so it's factorable that way

[high abyss]

Wait so if unable to solve it like what i did in b then you add the c term to both sides then divided and square the b term and if thats bot possible do quadratic equation?

[hollow plaza]

if u can use calculator its ez

I mean if you mean no solution for the equation then the graph is above x

[crude kite]

You don't even have to do that, you can just solve for x directly

[hollow plaza]

like X axis

[twilit night]

learning to factor by expanding a polynomial to (x+1) (x+ 2), is very important later on. its very useful saves alot of time.

just a tip

[crude kite]

https://www.youtube.com/watch?v=puWdJ-s9y_Y&ab_channel=Mario'sMathTutoring

[high abyss]

Alright ima watch this vid its the fact that theres so many different things to this it hurts my brain to have to know all of these

[crude kite]

As I mentioned, it depends on the complexity of the problem

[amber wasp]

@high abyss Has your question been resolved?

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[amber wasp]

@tough mica Has your question been resolved?

[winged panther]

Show your diagram with coordinates of all points P, O, Q, R, S in terms of b

[tough mica]

did you check my solution to the other exercise

[tough mica]

ok

[winged panther]

Did. 7/10 • 12 was correct

∆POQ ~ ∆SRQ => PO/OQ = SR/RQ

[tough mica]

(Area Triangle QRS) / (Area Triangle QOP) = 9/25 = ((PO)/(SR))^2 = ((OQ)/(RQ))^2

(PO)/(SR) = 3/5 = (OQ)/(RQ)

[winged panther]

No it's the other way (@_@;) [QRS]/[QOP] = [RS/OP]² = 9/25

So RS/OP = 3/5

[tough mica]

yeah mb

[winged panther]

Where are the coordinates? I asked you to solve the line equations and give me coordinates of points O,P,R,S

[tough mica]

I'm struggling with that part

[winged panther]

For instance, O being Origin is (0, 0)
P is (0, b)

As P is intersection of line equations: x = 0 and y = b - x

So, (x, y) = (0, b - 0) = (0, b)

Similarly, what is given equation for line RS? What are coordinates of points R and S?

[tough mica]

S = (4, b-4)

R = (4,0)

O = (0,0)

P = (0,b)

0 = b - x
x = b
Q = (b, 0)

[tough mica]

is my Q correct?

[winged panther]

Yes. So, the measurements OP = b and SR = ?

[tough mica]

SR=b-4

(b-4)/b = 3/5

b-4 = (3/5).b
b - (3/5)b = 4
(5/5)b - (3/5)b = 4
(2/5)b = 4
b = 4 . (5/2)
b = 10

what am I even doing bruh

b is between (0,4)

[winged panther]

(b - 4) is a negative number. Measurement of sides of a triangle cannot be negative

[tough mica]

then my S is incorrect

[winged panther]

SR = 4 - b

No, your S is correct. SR = R_y - S_y = -(b - 4) = 4 - b

[tough mica]

ok my bad

[winged panther]

So now, (4 - b)/b = 3/5 is easy to solve

[tough mica]

(4-b)/b = 3/5

(4-b) = (3/5)b
4 = (5/5)b + (3/5)b
4 = (8/5)b
4.(5/8) = b

,w 4*(5/8)

[grizzled pagoda]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

I appreciate the help. you dunno how much i struggle with geo

[noble plume]

number theory is so much worse in my opinion

well atleast the proofs are

[tough mica]

that depends but yeah NT it's maybe a little harder. depending on the problem

but specifically for me even basic geometry is hard for me

but idk, I am just ducking around and finding out ig

[noble plume]

are you studying for amc 10/ amc 12?

[tough mica]

preparing for cemc cayley

[tough mica]

I wish you good luck on your amc preparation

[noble plume]

thx you too

[tough mica]

.solved

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[eager viper]

I’m stuck on the second question, can’t seem to start it right and everything I’ve tried to do ends up being wrong/ not working please help <33

[amber wasp]

@eager viper Has your question been resolved?

[eager viper]

<@&286206848099549185>

[amber wasp]

@eager viper Has your question been resolved?

[winged panther]

Use these:

tan 2x = (2tan x)/(1-tan² x)
log_b² a = 1/2 log_b a

[winged panther]

This will help simplify your equation to f(x)/g(x) = 1/2(x² - 7x - 8)

Which is simply a polynomial equation.. which you can simplify n solve

[eager viper]

Ok let me try that right now ty <3

[eager viper]

Wait are these for the first or the second question?

[winged panther]

Oh, you got through the first part? What x did you get?

[eager viper]

i got 8 or -1

which tbh im not sure its right or not since both of those are also in my restrictions

[winged panther]

Both x = -1 and 8 are incorrect

[eager viper]

wait i just realized i forgot a whole part at the start

give me a minute to redo it lol

[winged panther]

As, for x = -1, f is not defined. And for x = 8, 1/g is not defined

[eager viper]

oh so my restrictions are right then?

[winged panther]

Yes. But you forgot x ≠ 0 is necessary for j to be defined

[eager viper]

i included htat i just didnt say it lol

its because j is a log so it cant be equal to 0/ cant be negative but since it says x is positive (absolute value) we only say it cant be 0 right?

[winged panther]

Solve the equation n let me know what x you get

[eager viper]

ok thank you so much <3

[winged panther]

There's two more restrictions on x, did u get them too?

[eager viper]

• and - 4 right?

[winged panther]

So x can't be {-1, 8, 0, -5, 2/3}

No.. x can totally be 4, in fact x = 4 satisfies the equation

[eager viper]

woahh where did the 2/3 and 5 come from

is it from the g(x) after using factor theorem

[winged panther]

Put g(x) = 0. As g is in denominator, it can't be 0

[eager viper]

thats the same as this right?

[winged panther]

Yeah

[eager viper]

you used factor theorem and then found the roots which are at (x- 8)(x-2/3) and (x+5)

[winged panther]

Hm

[eager viper]

thank you so much by the way you have no idea how confused i was b4 you showed me this

uh i think i did something wrong

i ended up getting a 7 degree polynomial?

right now i have (112.5x/x+1)/(9x^4-39x^3-320x^2+468x-160) = 1/2(x^2 -7x -8)

[winged panther]

Ahahah

[eager viper]

stop i can solve super hard trig identites and yet i cant do 7th grade math 😭

[winged panther]

g(x) is basically (x - 8)(x + 5)(3x - 2)² and (x² - 7x - 8) = (x + 1)(x - 8)

[eager viper]

omfg.

[winged panther]

Can you cross out the polynomial factors?

[eager viper]

i only did that for the restrictions

💀

the x- 8

oh wait

then you move the x+1 over

and cross those out too

ohhhhhhh

[winged panther]

You should be dealing with a cubic now

225x = (x + 5)(3x - 2)²

[eager viper]

right now i have 112.5x= ((x+5)(3x-2)^2)/2

ok we have the same thing

i just didnt move over the 2 yet

do yo get rid of the sqaure root next

[winged panther]

Huh square root?

[eager viper]

the ^2

meant exponent lol

[winged panther]

Yeah, expand and rewrite polynomial

Should be getting ||9x³ + 33x² - 281x + 20 = 0||

[eager viper]

oh

i square rooted it to make it simplier

so i got 3x^2 -2x -10

and that also simpifies my final answers

[winged panther]

?? Square rooted it? Can you show your work?

[eager viper]

yes

one secon

Because 225 is a perfect root

[winged panther]

Nuuuu, what about √x and √(x + 5) ??

[eager viper]

omg

[winged panther]

Taking square root on both sides gives √(225x) = √(x+5) (3x-2)

[winged panther]

.

[eager viper]

im sorry my brain is so fried from staring at this question since like 12

[winged panther]

[eager viper]

not gonna look at that just yet wanna make sure i get it on my own

Ok I think I got it

Oh you were right before 4 does satisfy the equation

Thank you so much for your help I actually understand this now

Now onto #2 lol

[eager viper]

I think I did it right

[winged panther]

Where's the 0 on the number line?

Did you forget the x in the expression?

[eager viper]

Yeah I did lol

Super close tho

So then is this right?

[winged panther]

No, as the expression has (3x - 2)², you need not mark 2/3 on your no. line, as it doesn't affect the inequality

[eager viper]

Oh

Why is that?

Sorry my teacher never explained that to me

Oh wait is it because it’s order 2 so it doesn’t count as a turning point (it bounces off the point)

[winged panther]

Yes

Because (x - 2/3)² ≥ 0 for any real value of x. So you might as well treat it as not affecting the inequality

[eager viper]

So then is this my final answer?

[eager viper]

Ohhhhh

Thank you so much by the way my teacher like barely teaches so this for super helpful

[winged panther]

Yes. (-inf, -5] U (-1, 0] U [8, inf)

[winged panther]

You're doing great if that's the case. You got on the right track every time I corrected you ✓ and you were initially not that clueless either

[eager viper]

Ok thank you again you’ve been a literal life saver <3

[eager viper]

Ahhh thank you so much I’m trying so hard to get above a 90 in this course (I have an 85 right now)

[winged panther]

Btw which stylus do you use for your tablet? If it's one (@_@;) or maybe a writing pad?

I got a Samsung recently but I can't find a Samsung pen for it. So I'm looking for options

[eager viper]

Oh I use an Apple Pencil lol but I’ve had a Adonit note+ before and it was good

Same with the jam Jake palm stylus

[vagrant plover]

hi

[eager viper]

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[idle storm]

Help

[idle storm]

how do I learn desmos ?

To get slopes

[grizzled plume]

Desmos is a graphing calculator

Wdym learn

[mint nacelle]

this page you showed us doesn't have anything on slopes

the slope would be the derivative of a graph (at a particular point)

[idle storm]

One guy said you need desmos for it

So i was concerned

how do I get the solution

For that

??

[woeful sable]

You dont need desmos for graphing it you just need to know the properties of a quadratic function

[idle storm]

Alright

thx

.close

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[quaint jungle]

could u explain me demoivres theorem

[mint nacelle]

https://www.youtube.com/watch?v=bGFMxfO0eK4

this version?

[quaint jungle]

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[quaint jungle]

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[gloomy cedar]

I have to give a presentation on The Ehrenfest Model Of Diffusion in my probability course. So, I don't know that how to prepare for this presentation. Can any of you help me out with it, since I'm confused with the topic that what it says about the probabilities, i mean, i want to know about the main theme of the topic in the realm of probability.

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[night snow]

min value of 1+x, (x>0)
using am gm we get
x+1 >= 2 *(x^1/2)
but x is from 0 to 1, so min value of RHS = 0
whereas min value of LHS =1 (considering x tends to 0+)

[night snow]

did math just break?

[vocal isle]

?

x+1 >= 2 x^/2
What is '2 x^/2"?

[night snow]

2 root x mb

[vocal isle]

OKKKKKKKKKKKKKKKKKKKKKKKKK

[pallid canopy]

The minimum doesn't exist

[night snow]

y?

[frank radish]

wouldnt minimum of 1+x where x>0 be 1.000000000001

[night snow]

ye

even i thouhgt that

[coarse linden]

you can't find a min val in open interval (1, +\inf)

[vocal isle]

x + 1 >= 2 * x^(1/2) only implies that for all x > 0, the sum of x and 1 will always be greater than this specific product: 2root(x)

[frank radish]

1 >= 0 is true, nothing broke

[night snow]

mb

[frank radish]

= means "> or ="

[vocal isle]

You must have a concrete proof step by step for whatever you say. If you don't, even if looks correct, you must not accept it.

[frank radish]

do you wanna prove that theres no minimum or do you wanna find the minimum(you cant)

[vocal isle]

In this case we establish that this is definitely true, and nothing more than that we can say.

What do you mean 'x is from 0 to 1'?

[frank radish]

he probably thought too hard about the = in >=

[night snow]

lol yeah

i got it thanks

looks like i was high

.clsoe

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[loud nova]

How can I be able to identify what rules to use for each specified steps as there are many rules at once and wanted to know for which specific problems to indentify from

[loud nova]

Discrete math

[amber wasp]

@loud nova Has your question been resolved?

[loud nova]

<@&286206848099549185>

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[loud nova]

.reopen

[amber wasp]

✅

[small pendant]

um

[loud nova]

What

[small pendant]

can you tell me whcih one is $\wedge$ and which one is $\vee$ again

[grizzled pagoda]

Arnavutköy

[small pendant]

i forget which one is 'and' and which one is 'or'

[loud nova]

$\wedge$ is and

[grizzled pagoda]

lambda_1

[small pendant]

ok

[loud nova]

$\vee$ is or

[grizzled pagoda]

lambda_1

[small pendant]

so we start with $q\implies u\wedge t$

[grizzled pagoda]

Arnavutköy

[small pendant]

so let us assume $q$ is true

[grizzled pagoda]

Arnavutköy

[small pendant]

this implies that now, $q$, $t$, and $u$ are true, right?

[grizzled pagoda]

Arnavutköy

[loud nova]

Ys

[small pendant]

we know that $\neg s$ is true right?

[grizzled pagoda]

Arnavutköy

[loud nova]

Yes and is given as. apremise

[small pendant]

so now, let us suppose $r$ is not true

[grizzled pagoda]

Arnavutköy

[small pendant]

we want to get a contradiction, right?

[loud nova]

Yes

[small pendant]

or actually, we don't need to do contradiction yet

so we know that $u$ is true

[grizzled pagoda]

Arnavutköy

[small pendant]

we now apply $u\to p$

[grizzled pagoda]

Arnavutköy

[small pendant]

so, we now know that $p,q,t,u$ are all correct

[grizzled pagoda]

Arnavutköy

[small pendant]

right?

[loud nova]

It came from simplification rule and so I am wondering how can identify you need to ue that rule

[loud nova]

Yes as from the premise rule

[small pendant]

basically we try to find all premises which we know are true

we see that there is a rule which requires a signle u

and we know that t\wedge u is correct

[loud nova]

Agreed

[small pendant]

therefore, we decide to use the simplification rule

to just gett u

[loud nova]

Okay then

[small pendant]

so if we know that $p,q,t,u$ are correct

[grizzled pagoda]

Arnavutköy

[small pendant]

then, we know that $p\wedge t$ is satisfied, right?

[grizzled pagoda]

Arnavutköy

[loud nova]

Yew

This is the rule called conjuction to be used for that

[small pendant]

yeah

so now, we apply the premise

and we get that $r\vee s$ is correct

[grizzled pagoda]

Arnavutköy

[loud nova]

Yes

[small pendant]

but, we know that $s$ is false right?

[grizzled pagoda]

Arnavutköy

[loud nova]

How would you know s itself is false or not

[small pendant]

because we know that $\neg s$ is a premise

[grizzled pagoda]

Arnavutköy

[loud nova]

Okay as does make sense for that then

[small pendant]

so if $s$ is false, but $r\vee s$ is true, then $r$ must be true

[grizzled pagoda]

Arnavutköy

[loud nova]

With the use of modus ponens rule

[small pendant]

yes

[loud nova]

Yes

[small pendant]

so this satisfies $q\to r$

[grizzled pagoda]

Arnavutköy

[small pendant]

i gtg now

[loud nova]

Okay

Thank you

[amber wasp]

@loud nova Has your question been resolved?

[loud nova]

.closed

.close

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[night snow]

i dont think my approach is correct

[sterile swan]

!status

[amber wasp]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[night snow]

1

if im not wrong, this is the diagram right

[sterile swan]

yes

very roughly

[night snow]

so what i had done was

[sterile swan]

i can you determine the equation of the tangent line, then you find the intersection points

[night snow]

i considered a pt x1,y1

[night snow]

should i take the pts like A(x1,0) and B(y1,0)?

[sterile swan]

no, the intersection points are determined by the line

if you take A=(10,0) and B=(0,1000) its not tangent to the ellipse

[night snow]

so i need to take a variable pt (h,k)?

[sterile swan]

you need to take a point on the ellipse

and then find the tangent line at that point

[night snow]

ye done

[sterile swan]

whats your equation

[night snow]

xx1/9 + yy1/4 = 1

[sterile swan]

i meant of hte tangent line

[night snow]

isnt it this itself?

[sterile swan]

thats for the ellipse

[night snow]

oh

[sterile swan]

its not a tangent line

[night snow]

y-y1 = m(x-x1) ?

i dont have m

[sterile swan]

and then what are x1 and y1

[night snow]

the assumed pts on the ellipse?

[sterile swan]

yes

[tired girder]

My urge would be to, consider the points as (acosø,bsinø)

[night snow]

after that?

its coming ig

i got some super wierd shit

i used this

[tight bramble]

answer is the option (A)

2

[night snow]

nd this

[night snow]

wrong newbie lol

[tight bramble]

uh you can calculate

how you calculated?

[night snow]

i got m=0 🗿

[tight bramble]

wrong

...

[night snow]

i equated those two equations

[tight bramble]

its wrong

[night snow]

ye ik

[tight bramble]

go check from math solvers internet idk

answer was the option (A)

[night snow]

its wrong

answer is C

[sterile swan]

!nosols

[amber wasp]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[night snow]

i hv the key lol

idk the solution

[tight bramble]

key is wrong

[night snow]

thats why im here

tell how u got

[tight bramble]

like a human being man

you gotta know these shi

i am 9th grade guy

mb

answer is 6

look for options. search which is 6

[night snow]

C

brotha how u got it

[tight bramble]

uh ez

look

lemme break it down

breaking sounds

oops it broke

[night snow]

bruv

[tight bramble]

just know that answer is 6

and i cant send pictures

good night brother

going to solve some ez pz contour triple integrals

[night snow]

helpo

[tight bramble]

another equation?

[night snow]

bruv u sleep

9th grade they teach integration ah

which country

[tight bramble]

iran

country of ease and peace

*for deads only

anyway

send equations brah

[tight bramble]

i am 9th grade but studying college shi

[night snow]

broooo how do i solve this questionn

<@&286206848099549185>

[tight bramble]

which?

[tight bramble]

aight look man

you made us go mad

ill convert my hand writings to latex (my internet is crap and cant send a photo)

[nova briar]

What is O?

[warm burrow]

I would assume the origin

[nova briar]

So it's a right triangle?

[silk sonnet]

not necessarily

[nova briar]

well, it meets at points A and B, both of which lie on the coordinate axes (which I assume are the x and y coordinate axes) so from the origin to any of these points, in order for it to be a triangle, it's gotta be a right triangle, right?

[silk sonnet]

oh yeah

im stupid

[night snow]

ye

[warm burrow]

well actually not necessarily

cause if the point is 0,3 2,0

[night snow]

coordinate axes are at 90 to each other?

[nova briar]

it's still a right triangle

at the origin point

[silk sonnet]

other than that its a right triangle man

[nova briar]

0,3 and 2,0, that sounds like a right triangle.

[silk sonnet]

so basically

[warm burrow]

no it wouldnt be a triangle at those points cause the tangent line doesnt touch the axes

[silk sonnet]

you just have express the area of the triangle by x where x is an x coordinate of a point on the ellipse right?

[nova briar]

it would be because it has two perpendicular sides (line from (0, 0) to (0,3) and line from (0, 0) to (2, 0))

[silk sonnet]

and then find the min where x is between (0,3)

[night snow]

wont it be better if we find the POI on the x and y axes and us 1/2 bh?

[tight bramble]

\begin{array}{l}
\frac{x^{2}}{9} +\frac{y^{2}}{4} =1\
\
\
\
\arrowbullet equation\ tangent\ going\ to\ ellipse\ in\ point\ of\ ( x_{1} ,y_{1}) =\
\frac{xx_{1}}{9} +\frac{yy_{1}}{4} =1\
\
\
\
\arrowbullet \ since\ cordinates\ of\ ( x_{1} ,y_{1}) satisfy\ the\ ellipse\ equation\ we\ can\ say:\
\frac{x_{1}^{2}}{9} +\frac{y_{1}^{2}}{4} =1\
\
\arrowbullet \ x\ and\ y\ intercepts\ when\ we\ set\ x=0\ and\ y=0\
\frac{xx_{1}}{9} =1\ \rightarrowtail x=\frac{9}{x_{1}}\
\
\frac{yy_{1}}{4} =1\ \rightarrowtail \ y=\frac{4}{y_{1}}
\end{array}

[grizzled pagoda]

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

[tight bramble]

this

[tight bramble]

holy yaper of yapers

[silk sonnet]

tf is a yaper

[tight bramble]

person who writes alotta bullsh thing

[silk sonnet]

you mean a yapper?

[tight bramble]

yapper

[tight bramble]

youll go to then area of triangle (i was too tired to write those shits down too) and calculate

[tight bramble]

as 9th grade i calculate this way. opinions?

[nova briar]

How's this not a right triangle?

[warm burrow]

when this happens it is no longer a triangle at all

[night snow]

brooo why arguing over this crap

[tight bramble]

holy yappers of yappingtons of yapps

[night snow]

😭

[tight bramble]

its ez tho

[nova briar]

How come?

[tight bramble]

just a little of equation

[night snow]

hmm ill try tomorrow

[warm burrow]

And in this cause you just get an infinitely long rectangle

[night snow]

its 3 am here lol

gotta sleep ig

[tight bramble]

math guys dont sleep

call an ambulance

[nova briar]

What do you mean? We are talking about the tirangle AOB, which is a point that is on the x axis, a point that is on the y axis, and a point on the origin. The triangle that forms from these 3 points is always a right triangle.

[tight bramble]

ill explain to you man

[warm burrow]

if the tangent to the ellipse is chosen at specific points you dont get a triangle because the line doesnt touch the one of the coordinate axes

[nova briar]

O is the origin. A is just a point on one of the axes (x or y) and B is a point on one of the axes (x or y) and both lie on the perimeter of the elipses - atleast that's what I thought.

[warm burrow]

In 4 cases there is either no A or no B

but there is no use in arguing this as the area doesnt matter anymore in these 4 cases

[tight bramble]

holy yappington💀

[nova briar]

I thought these were the 4 triangles that can possibly be formed

[warm burrow]

Bro went to sleep anyways

oh you just solved the problem

good job

wait nevermind

you mightve gotton close though

Nevermind

The question meant like this

[nova briar]

the area seems to be 3 but the choices don't show that... so I'm either wrong or the key word in the question minimum indicates the answer is 2

[nova briar]

That makes a lot more sense because it has to be a tangeant line to the elipses.

[amber wasp]

@night snow Has your question been resolved?

[warm burrow]

I solved it

but the guy went to sleep

so no point

[nova briar]

Okay, I get the basic idea of how to solve it but solving it is another issue. You have to find the line that is the smallest from the origin point to the perimeter of the elipses. Once you find the line and the ending point of that line that touches the ellipses (C), you run the tangeant along that because it is minimizing the area... and then you have to find A and B based on what points of the axes it will hit from that point (by using maybe derivation). I don't think it matters which quadrant you are working in, so we can just limit it to the top right quadrant...

[nova briar]

just post it at least so he can find it tomorrow or something

and so I can be at peace

I think the answer is B though

[warm burrow]

is B 6?

[nova briar]

either B or C

[nova briar]

No, B was 4. But I knew it had to be > 2 and probably < 8 because it can't be that much of a difference from 3 (which is the maximum area of the triangle inside the ellipses)

[warm burrow]

We need to find the intercepts of the line

[nova briar]

The tangent line?

[warm burrow]

Yes

we can do this in terms of x and y

[nova briar]

So how did you find the tangent line?

[warm burrow]

because it is an ellipse

so set y=0 to get the intercept

[nova briar]

Those are the intercepts of the ellipse, not the tangent line.

We already know all 4 intercepts of the ellipse

[warm burrow]

there are infinite intercepts

[nova briar]

For the tangent line?

[warm burrow]

yes

let me explain

[nova briar]

Yea, for the tangent because there are infinite points along the ellipse.

[warm burrow]

the right triangle will be enclosed by two intercepts of the line and the origin

so we can just find the function for the intercepts

then minimise them

[nova briar]

we don't know the function of the line

Well, we do know that it's the derivative of the ellipse function

because that's what a derivative is

[warm burrow]

$x^2/9+y^2/4 =1$

[grizzled pagoda]

Banana Steeler

[warm burrow]

ok to find the intercept function

for the y intercept

[nova briar]

for x it's 3, for y it's 2

[warm burrow]

We set x to 0

so it becomes $y^2/4=1$

[nova briar]

y = 2; x = 3

[grizzled pagoda]

Banana Steeler

[warm burrow]

and remember we are solving for y

$y=4/y$

[grizzled pagoda]

Banana Steeler

[warm burrow]

so we know it will be (0, 4/y) for the y intercept

now we solve for the x intercept

so set y=0

$x^2/9=1$

[grizzled pagoda]

Banana Steeler

[warm burrow]

solve for x

$x=9/x$

[grizzled pagoda]

Banana Steeler

[amber wasp]

Channel closed

Closed due to timeout

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[warm burrow]

aw man

THE SOLUTION IS 6 before i finish

cause the channel is gonna be locked

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[warm burrow]

The two points we know it will be is (9/x, 0) and (0,4/y)

[warm burrow]

.close

[amber wasp]

Channel closed

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Use .reopen if this was a mistake.

[warm burrow]

nvm if you want me to explain it just dm or something

[nova briar]

y = sqrt(4(1 - ((x^2)/9)))

[amber wasp]

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[tough mica]

DC/BD = 3/4

area ABD = 24 cm^2

(area ADC / area ABD) = (3/4)^2

Area ADC / Area ABD = 9/16

area ADC = (9/16) x 24cm^2

,w (9/16) * 24

[grizzled pagoda]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

so answer is 13.5 cm^2?

[celest cove]

why (3/4)^2?

only one side is scaled

for 1/2 ab

[tough mica]

its not similar triangles, MB

[celest cove]

if i call AB=y

then BD=4x and DC=3x

0.5(4x)y=24
0.5(3x)y=A

[tough mica]

2xy = 24
(3/2)xy = A

,calc 12*(3/2)

[grizzled pagoda]

Result:

18

[tough mica]

A is 18cm^2

[celest cove]

sure, you could make xy the subject or just 'divide one equation by the other'

so to speak

[tough mica]

how so?

0.5(4x)y=24
0.5(3x)y=A

(0.5(4x)y)/(0.5(3x)y) = 24/A
A . (4/3) = 24
A = 24 . (3/4)

,calc 24 * (3/4)

[grizzled pagoda]

Result:

18

[tough mica]

I appreciate the help

.close

[amber wasp]

Channel closed

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[cloud oar]

help this explaination isnt explaining

[cloud oar]

"multiply the quotient of the coefficients"

like what 💀

dont u have to divide first then

[grizzled plume]

Wait what do you need help understanding

[cloud oar]

the explaination

its so confusing

oh wait i get it by looking at the pictures but the explaination is confusing af

so basically i dont need help with anything

.close

[amber wasp]

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[grizzled plume]

Uhh ok

[amber wasp]

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[keen pawn]

Does there exist a divergent monotone seqeunce with a cauchy subseqeunce? I was thinking $a_{2n}= n^2$, $a_{2n+1} = \frac{1}{n}$

[grizzled pagoda]

math rocks(wai)

[stiff jetty]

i dont think so? (not evne in uni yet) so u might have quite a 'hand wavy' answer

from google cauchy subsequence is bounded

if ur sequence is increasing and bounded isnt that now convergent

[keen pawn]

Ah right, I forgot that lemma

Thanks

.close

[amber wasp]

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[stiff jetty]

its not a very good answer but i think u can get the pt

(never heard of rigour in my life)

[pseudo crescent]

It has a cauchy subsequence that's bounded. Does that necessarily mean that the sequence itself is bounded?

[amber wasp]

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[keen pawn]

No

[keen pawn]

Have some work to do, will revisit this in a bit

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[rigid swallow]

How do i find and angle between two 3d vector3's?

[azure plover]

you can use the dot product along with the fact that $\vec u\cdot\vec v = ||\vec u||,||\vec v||\cos\theta$

[grizzled pagoda]

Ham

[rigid swallow]

What do the symbols mean? 😅 i don't work with maths at this high of a level..

[azure plover]

double bars is just the absolute value

theta in this case is the angle between u and v which is what you wanted

[rigid swallow]

And the • thingy? Is it dot or divide? 😅

Or something else 😵‍💫😅

[azure plover]

its called the dot product

[rigid swallow]

Ahhhh

Lhs and rhs?

[sand cargo]

Can you give a demonstration of your problem?

[rigid swallow]

Its hard to but im trying to make a player forcibly look towards a position so i want the target angle and stuff to move the camera towards said position

[azure plover]

in a game engine?

[rigid swallow]

Mhm

Well idk its a game based in unity

[azure plover]

so what are the 2 vectors you have?

[rigid swallow]

The postion and my position

[azure plover]

the position of the camera?

[rigid swallow]

Also like camera forward and forward vectors

Ye

[azure plover]

so you wanna turn the player camera towards a position

[rigid swallow]

Yes

[azure plover]

wdym by camera forward and forward vectors

[rigid swallow]

Like a little ahead of the actual positon yet relative to the coordinates 0, 0, 0

[azure plover]

well, you could use the method i told you to find the angle between the camera facing vector and the vector from player to the position

it's in first person right?

[rigid swallow]

Both 1st and 3rd person

I think i can turn off 3rd person tho let me check rq

Can't find it 😅

[azure plover]

is this what you wanna find?

[rigid swallow]

Ye pretty sure

[fast badger]

Using what Ham said, you can rearrange for theta to get
theta = arccos(dot(u,v)/abs(u)abs(v))

[azure plover]

i feel like there should be a better way to do this than to find the angle and manually turn it that angle

[azure plover]

but if thats the only way then theoretically you calculate this to get the angle

[rigid swallow]

I mean its the best way i can find im just trying to lock the player camera onto a point

[azure plover]

but since it's in 3d, you'd probably wanna find 2 angles for pitch and yaw

instead of a single angle between the 2 vectors

[rigid swallow]

Idk if i can set 2 at once tho

[fast badger]

You have the angle, now you just need the axis to rotate aabout

Which is simply player_fwd x target_dir, where x is cross prod

Like I said yesterday, Unity provides helper maths functions in their scripting api, so use that

Namely, Quaternion.AngleAxis

Read the docs for usage

[rigid swallow]

Im just gonna say this is impossible in the game im coding in 😅

Its too stressful

[azure plover]

whats the method for your camera rotation?

does it rotate along an axis by an angle?

[rigid swallow]

🤷‍♂️

[fast badger]

I believe Unity stores both quat and matrix representations of orientations

So can just rotate quat by quat

No, looks like it’s only stored as mat

But you can just cast the quat to mat4 in any case

[rigid swallow]

Doesn't even work aswell i look at it with left and right not up and down well i can but the up and down is relative to me

My code sucks 🤣

[fast badger]

Switch to the system I described, it is far more flexible

Especially if you decide to animate it with easing functions

[rigid swallow]

Nah its fine ill just add distance to the up and down rotation and then subtract the y height and see if that works

And if it doesn't work ill just give up the idea

I give up vectors/quaternions are stupid im never working with that crap again

[amber wasp]

@rigid swallow Has your question been resolved?

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[keen pawn]

Let $(a_n)$ be a Cauchy sequence. Then determine if $(c_n)=(-1)^{n} a_n $ forms a Cauchy sequence.
\
\
As $(a_n)$ is cauchy, it follows that $\abs{a_n-a_m} < \varepsilon,$ for $n,m>N$. Thus $\abs{a_n-a} +\abs{a-a_m} < \varepsilon$.
We now examine $\abs{(-1)^{n}a_n - (-1)^{m}a_m}$. If $n,m$ are both even it follows trivially that it's $a_n - a_m$, from which we can conclude that $\abs{a_n-a_m} < \varepsilon$. We now suppose that one of $n,m$ is even.Suppose $n$ is odd. It then follows we have $\abs{-a_n-a_m} \leq \abs{a-a_n} + \abs{a-a_m}$. Which is once again less than $\varepsilon$. The case where $n$ is even and $m$ is odd follows similarly. Now suppose both $n$ and $m$ are odd. We then have $\abs{-a_n+ a_m} = \abs{a_m-a_n}$, which is less than $\varepsilon$. We can thus conclude that $(-1)^{n} a_n$ forms a cauchy seqeunce.

[grizzled pagoda]

math rocks(wai)

[silver mist]

What about a_n = 1 though? Don't think c_n is cauchy in that case

[keen pawn]

But is $a_n$ cauchy then

[grizzled pagoda]

math rocks(wai)

[keen pawn]

why not?

[silver mist]

Take epsilon = 1, if n and m have different parity the absolute difference is always 2 which is greater than 1

c_n would look like 1, -1, 1, -1, 1, -1, ....

[keen pawn]

That is true, which is why I split it using the triangle inequality

[vernal surge]

$|-a_n-a_m| \leq |a-a_n|+|a-a_m|$ is wrong

[grizzled pagoda]

rbit

[vestal hatch]

mfw the case doesn't follow similarly

[keen pawn]

oops, yeah

So I just give this counter example and I'm done

right

[vernal surge]

yes

[keen pawn]

thanks

thanks everyone

.close

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[amber wasp]

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[tough mica]

Square $ABCD$ has side lengths of $13$ units. Point $E$ lies in the interior of the square such that $AE = 5$ units and $BE = 12$ units. What is the distance from $E$ to side $AD$?

[grizzled pagoda]

938c2cc0dcc05f2b68c4287040cfcf71

[vestal hatch]

i would recommend drawing a diagram

[tough mica]

how so

we dont where E should be placed inside ABCD

[static eagle]

instead of units, use mm or cm and draw it out with a pencil and ruler

[tough mica]

but we dont where E should be placed inside ABCD

is it in the diagonal?

[static eagle]

in that cause

case *

use coordinates

so for instance.
A is (0,0)
B is (13,0)
C is (13,13)
D is (0,13)

[vestal hatch]

yeah no

[tough mica]

E = (0,5)

[vestal hatch]

that is way overkill

[tough mica]

how to draw E then

[vestal hatch]

what does that even mean

[static eagle]

u can deffo draw it

[vestal hatch]

just put it wherever your heaart desires

if the proportions look wrong then adjust it

[tough mica]

ok

[static eagle]

like place two rulers, one at 5cm and one at 12 cm

whereevr they intersect

is ur point E

[static eagle]

this way could be used if ur numbers are big or in exams when u dont have time to draw it out

[silver mist]

You can use a compass to draw radii of 5 and 12

[vestal hatch]

or you can just eyeball it

ok good enough

next i would find the altitude

[tough mica]

the height of what

[vestal hatch]

the altitude to AB in the triangle ABE

[tough mica]

how

25 = 2x^2

[vestal hatch]

well do you know any formulas involving the altitude?

[tough mica]

25/2 = x^2

h = 5/sqrt(2)

[vestal hatch]

where are you getting this?

[tough mica]

no

[tough mica]

was trying to use pythagoraa

[vestal hatch]

how about involving the height?

[vestal hatch]

then you're probably assuming 2 segments are equal when they aren't or something

[tough mica]

5^2 = h^2 + x^2
12^2 = h^2 + y^2

[tough mica]

I am trying

[vestal hatch]

hmm

i don't suppose you've learned heron's formula

also there's something special about the values 5, 12, 13

[tough mica]

Pythagoras triple

[vestal hatch]

right, which means?

[tough mica]

idk

[verbal jewel]

doesnt matter, draw a point E inside, and draw the given things. then its straightforward.

[vestal hatch]

if the 3 sides of a triangle are a pythagorean triple, what does that make the triangle

[tough mica]

I am trying

[verbal jewel]

or draw a circle center A and a circle with center B.

[tough mica]

right triangle

[tough mica]

okay

[vestal hatch]

how can we find the area of a right triangle