help-27

[blissful herald]

which question

and what did you do

[strange igloo]

i did 270x6

to find the outside area

i did 270x12 to find the outside and inside

i addded the base area to both and sstill got it wrong

i calculated the whole rectangke

i did everything

[blissful herald]

you should add the area of the base too

[strange igloo]

i did

[blissful herald]

what did you do to find thr area

the*

[strange igloo]

this one

[blissful herald]

wait what

oh wait nvm

[strange igloo]

there are multiple slides

[blissful herald]

one sec

[strange igloo]

ok

[blissful herald]

so [3sqrt(3)/2]a²h is equal to 1894

[strange igloo]

what.

im in 7th grade bro 😭

[blissful herald]

oh

how did you find the area of the base in the previous question then

[strange igloo]

i divided the total volume by the height

[blissful herald]

how did you find the volume

[strange igloo]

it was on there

[blissful herald]

bruh

I asked for the previous question

not this one

[blissful herald]

also this must work

[strange igloo]

ik i did that

[blissful herald]

wait no nvm

[strange igloo]

and itr is correct on the other slide

[blissful herald]

you forgot to subtract the remaining volume

[strange igloo]

but this is the one i need help on

everything else is correct

execpt this one

[blissful herald]

so you found the area of the base right

[strange igloo]

yes

[blissful herald]

then just add the area of the rectangles

to the base

[strange igloo]

ok

i did it

the answer was 1809.4

and it was wrong

im telling you this thing is very hard

i also did 270 times 12

so she can paint the inside and outside

but it wwas wrong too

[blissful herald]

the you must multiply the area of base times 2 as well

cause outside and inside

[strange igloo]

oih

oh

its worng

wrong

[blissful herald]

ok so the area of a hexagon

is 3sqrt(3)a²/2

[strange igloo]

wait

my friend got it somehow

[blissful herald]

what did he do

[strange igloo]

hes texting me it rn

[blissful herald]

alr

[strange igloo]

he added the area of the base to the 6(27*13)

since the whole height not partial is 13

.close

cclosoe

vclosde

.close

[devout snow]

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[plain schooner]

I keep getting 1/10pi

but the answer is 5

I'm not sure what im doing wrong

[winter patrol]

1 sec

multiplication mistakes,
cancellation mistakes

[plain schooner]

sorry

[winter patrol]

4 * 2* 5 isn't 20

[plain schooner]

ohh

lol

[winter patrol]

you're also failing to cancel the factor of pi at the very end

and you messed up the multiplication of 4 * 5^2 at the end too

[plain schooner]

alright wait let me attempt it again

[winter patrol]

and i don't know where 4 * pi * 5 is coming from

[plain schooner]

5 is radius

because diameter is 10

[winter patrol]

but surface area of a sphere isn't 4pi r
nor is 4 * 5 equal to 10

[plain schooner]

fuck lol

my brain is fried haha my bad

oh okay i got it

This is the right way?

.close

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[safe knoll]

which question?

[compact narwhal]

wait nvm i figured it out

mb

[kindred agate]

These are some painful computations ur gonna do

Gl man

[autumn fossil]

letter e looks fun

[compact narwhal]

yea im cooked, i have an exam on this tomorrow

[kindred agate]

Oh e is the easiest one

[autumn fossil]

the best one

[kindred agate]

Those questions look like painful computations, but atleast they’re conceptually easy

You just need paper and a calculator

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[still zephyr]

Already got the first implication

Not sure about second

[devout snow]

@still zephyr Has your question been resolved?

[devout snow]

@still zephyr Has your question been resolved?

[faint zinc]

So you showed that if l = AB then AB < l, but not if AB < l then l = AB?

This seems like one of those problems which is so basic that we kinda need ground rules to know what it is that we actually are supposed to know.

[still zephyr]

How would that be 🤔

Since we are working from the axioms, definitions and primitive terms

[faint zinc]

Well, the first standard axiom of the euclidean plane is that a straight line may be drawn between two points.

If we construct the line AB from the point A and B then we get a line.

Two distinct straight lines are either parallel and intersect at 0 points or are not parallel and intersect at exactly 1 point.

(might need to prove the above)

However we have that l contains both the point A and B, which are also in the line AB, therefore, AB and l cannot be distinct and therefore must be the same line.

[still zephyr]

Is this sufficent justification?

[faint zinc]

As long as you prove this or you are allowed to use this fact

@still zephyr

[still zephyr]

Will do

thanks

[still zephyr]

What if they are not in the same plane?

[faint zinc]

Skew lines you mean?

Then they intersect at 0 points

[still zephyr]

We can consider two straight lines that are neither parallel or intersect

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[normal forum]

Can we find the coordinate A using (2 dimension) vectors if we know the information vgl1: 3x-y = 9, vgl2: x-3y = 3, vgl4: x-y = 3. Where vgl4 is the angle bisector. Also we know that point A lies on vgl4 and has a perpendicular distance of sqrt(10) from both lines vgl1 and vgl2.

[devout snow]

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[deep vortex]

#1195800467880558672 message

[deep vortex]

.close

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[small hornet]

I think I got a wrong saturation ?

[devout snow]

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[remote vigil]

So the min-max theorem doesnt apply to this function becuz its discontinuous at x=3 but is the minimum still -4?

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[karmic hound]

Due in like an hour

[karmic hound]

Anyone wanna help me through it real fast please??

[covert root]

You have made no attempt

[turbid current]

Do you know what i^2 is

[karmic hound]

I have no idea where exactly to start

[turbid current]

Start with the first question

[karmic hound]

Oh crap yes I totally forgot about that

[turbid current]

Do you know what i^2 is

Ok

[karmic hound]

Isn’t i^2 just 1

Or -1

[turbid current]

i is sqrt-1

So squaring it will be -1

[karmic hound]

Ohhh

[turbid current]

That should be enough to get you through q1

[karmic hound]

Option 3

For problem 1

[glossy dew]

eeeeeeeeyup

[karmic hound]

Yes!

Now number 2. I’m not really sure 😭

[glossy dew]

hint: convert 8 and 4 to powers of 2

and use power rule

[karmic hound]

I got option 4 as my answer for number 2

[glossy dew]

ye 👍

[karmic hound]

That’s correct?

[glossy dew]

yes

[karmic hound]

Lmao I got my answer as -15/13 and -14/5 was the best option so I guessed

[glossy dew]

hmm

thats not going to help elsewhere though

especially when it isn't mcq

[karmic hound]

True lmao

That’s a problem for another time , finding right answer the right way

Maybe we can revisit that

I have to hurry to finish these tho lmao

Number 3, where would I start to go about this one

[glossy dew]

again hint: (2³)^(3k+4) = (2²)^(2k-1)

bruh

in quadrant III, both sin and cos are negative

and sin^2 + cos^2 = 1

hope that would help in solving

[karmic hound]

I did this yes

[glossy dew]

and you'll get 2^(9k+12) = 2^(4k-2)

[karmic hound]

Yes I have this too #

[glossy dew]

and you'll get 9k+12 = 4k-2

[karmic hound]

I got it

I figured it out lol

I had all the way up until that step and then did the easy part WRONG

But I got it now

Thank you!

But now # 3

I’m a bit lost ngl

[glossy dew]

ok

do you know that sin² + cos² = 1

[karmic hound]

I have briefly learned that yes

[glossy dew]

then you can use that

you know what cos is

[karmic hound]

Yes

-3/4

Would the answer be #1?

No

#2

The answer is option 2. Correct?

[devout snow]

@karmic hound Has your question been resolved?

[glossy dew]

nope

[karmic hound]

Crap

[glossy dew]

9/16 + sin² theta = 1

sin² theta = 7/16

and then what's sin

[karmic hound]

Sqrt 7/4

[glossy dew]

eeeyuo

[karmic hound]

Isn’t that what I said?

I’m confused

[glossy dew]

but in 3rd quadrant sin is negative

[karmic hound]

Oh

So

-sqrt 7/4

[glossy dew]

yes

[karmic hound]

I got that answer a different way

[glossy dew]

oh

ok we'll see 4th one now

you just do polynomial long division here

[karmic hound]

I will do this thank you

I just added 3/4 to both side so I had sin^2 = 1 + 3/4 which is just 7/4 but then I rooted both side so I had sin = sqt7/4

Which

Doesn’t make sense now that I think about it

But it kind of worked for me so??? Win win maybe. 🤔

[glossy dew]

it does

thats right too

[devout snow]

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[karmic hound]

.reopen

[devout snow]

✅

[karmic hound]

Interesting

[glossy dew]

its the same thing lol

you did the exact same steps

wait no

wait what

nvm it's not the same

yeah it doesn't make sense

[karmic hound]

Lmao

I got option 3 for question 4

[glossy dew]

lemme try

[karmic hound]

Okay

[glossy dew]

nah doesn't work

not option 3

try converting it back to a fraction

[karmic hound]

What

My work is ass but where did I go wrong?

[glossy dew]

you divided by 3x-1

the denominator is 3x+1

bruh

[karmic hound]

Don’t you change it though?

Reciprocal? Or whatever

So is it the opposite? Number 2?

[glossy dew]

what

not that either 😔

stop guessing

[karmic hound]

Omg

[glossy dew]

now you know 9x² - 2 = (3x+1)(3x-1) -1

from long division

now divide both sides by 3x+1

on the LHS you'll get the original expression

on the rhs is your desired answer

[karmic hound]

Oh. I understand now!

So it’s option 1.

[glossy dew]

yes

all complete 🥳

[karmic hound]

There is a 5th question but I got it lmao

Thank you so much for the help!

[glossy dew]

welcome

[devout snow]

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[tepid brook]

why do we change the power of v isnt it a constant

this is a bernolli equation btw

[devout snow]

@tepid brook Has your question been resolved?

[hot steeple]

what do you mean?

they state y = f(v)

and then they try to find dy/dx in terms of v

[tepid brook]

shouldnt vs power stay the same

dv/dx

my problem is here

exactly

[silk night]

they are changing y to v

so power is changing

[tepid brook]

the thing i dont get is

oh i get it know

i watched a video about implicit differentiation

thank you

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[hard knot]

Question

[hard knot]

finite region B, is bounded by curve y = g(x)

the coordinates axes and line x = 4

Find area B

My understanding is I do

am I supposed to differentiate g(x) or integrate?

or just leave it

[bold drum]

you literally used the integration sign here hahaha

also don't forget dx at the end 🤓 ☝️

[hard knot]

yeah lol

true

i just dont remember

100%

[hard knot]

oh so integrate?

I thought differentate to find area of curve no? just bad memmory

[bold drum]

yeah when you write the integration sign write dx at the end

[hard knot]

yeye

[bold drum]

no differentiation is to get the rate of change, so the gradient

[hard knot]

okokok

bad memmory

ty!!!!!

:3

[bold drum]

haha npnp

[hard knot]

)

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[orchid laurel]

What point C on the fence will make the trip between A->C->B the shortest? We are not allowed to use mirroring to do this, so what I have done is made a function, differentiated it to find it's minimum value and that was my answer, 10. I can't see any issues with it but my classmate says it is wrong some help would be greatly appreciated

[heady plinth]

It is 10, your classmate is wrong

[orchid laurel]

thanks bro you're the best

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[formal gazelle]

I'm doing polar decomposition of matrices and I have no idea how my teacher obtained U

i see that V* is both eigenvectors, but normalized
i see that sigma is the root of the elements in the diagonal matrix (the elements being the eigenvectors)

but what is U?

[devout snow]

@formal gazelle Has your question been resolved?

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@formal gazelle Has your question been resolved?

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@formal gazelle Has your question been resolved?

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[fiery hedge]

i know this is a math server but i think some of you help with this one.
I have two objects A and B, where A is moving with velocity $v_A$ and B is at rest. A collides with B and gets embedded into B. Now both of them are moving together with velocity v. Using conservation of momentum i get $v = \frac{m_A}{m_A + m_B} * v_A$ but with conservation of energy the result is $v = \sqrt{\frac{m_A}{m_A + m_B}} * v_A$. Where did i go wrong?

[woven radish]

zelphanine

[restive river]

energy is not conserved in inelastic collisions

so It makes sense you are not getting the same answer

[fiery hedge]

why would energy not be conserved, doesnt that only happen when we have some kind of external force doing work? i see no force here

[vast bone]

Kinetic energy is conserved only in elastic collisions

[restive river]

well to be pedantic, energy is always conserved everywhere and for sll time. kinetic energy itself is what's not conserved

[fiery hedge]

so if i have to write the net energy when both of them are moving together, what would that look like?

and if kinetic energy is lost, is it converted into heat or some other energy?

[restive river]

well think about it. You have an object moving with some velocity, then that velocity turns 0 and all that kinetic energy goes away. Where did it go? it got converted to internal energy, and the kinetic energy itself is lost

[fiery hedge]

i see, and by internal energy does this mean that energy was increased at the atomic level? like the small particles are vibrating but the object as a whole isnt really moving

[restive river]

The objects like become deformed as a result of the collision, and that changes the interactions between some of the atoms or molecules. The atoms and molecules also on average move a bit faster. The result of those internal changes is more internal energy

[fiery hedge]

yeah that makes sense. thanks a lot for the help, i apreciate it

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[lament schooner]

how do we trig sub this

if theres no squre root

[urban harbor]

i think x + 1 = tan theta should work

[lament schooner]

Yeah okay I wasn’t sure before

Tysm!

[urban harbor]

denominator becomes (tan^2 t + 1)^2 = sec^4 and dx = sec^2 dt so you end up with an integral of cos^2 t dt i think?

[lament schooner]

Yes it is

Okay I got same thing

So it’s good now ty!

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[tall sinew]

plss help >_<

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[desert estuary]

is 5) for all x in A or any x in A

[desert estuary]

im leanng towards any since {x} should be one single element

[heady plinth]

is there a difference?

[arctic harness]

I nned to know how to calculate rotation of a circle and area of a circle

[desert estuary]

yes? look at the relation

it would fial symmetry

[desert estuary]

i think?

[heady plinth]

uh idk but i agree it fails symmetry

i guess if x isnt in B then the relation isnt even defined for the reverse

[desert estuary]

like for instance A = {1, 2, 3} and B = {1, 2} right

we can pick 3 in A, such that A = B U {3}

but the converse doesnt hold clearly

[heady plinth]

so its not an equivalence relation

i still dont get the distinction between any and all

either way theres no way its an equivalence relation

[desert estuary]

Yeah, I think it was worded poorly tbh

it just for x in A

[desert estuary]

also equivalence symbols can be arbitrary right

[heady plinth]

wym

like the symbol used to represent it?

[desert estuary]

yeah

[heady plinth]

yeah ofc

[desert estuary]

im used to seeing R

like xRy

so this threw me off a bit

[heady plinth]

= is a relation

<

[desert estuary]

true... that is a good point

also what is the difference between strong and weak induction, and when do you use either

[heady plinth]

bro honestly idk but they asked me to use strong induction on a test last week

gonna look it up rn

[desert estuary]

LMAO

You give off good vibes

what are you studying?

[heady plinth]

generally math but it was a cs course

"discrete structures"

[desert estuary]

Like algebriac structures?

I see

[heady plinth]

i cant think of a time you would ever need to use strong induction tbh

[desert estuary]

also can you explain what a quotient is in relations

[desert estuary]

hmm

[heady plinth]

honestly this explains it better

[desert estuary]

what textbook are u reading

[heady plinth]

fucking stackexchange lool

[desert estuary]

im going to kill yself

[heady plinth]

:)

[desert estuary]

mean af

typical math major

[heady plinth]

im dead

[desert estuary]

from?

[heady plinth]

mirth

[desert estuary]

no clue what that is

[heady plinth]

deadly disease

contagious too

[desert estuary]

mono?

[heady plinth]

i had that shit a few months ago and it was the worst

0/10

[desert estuary]

math majors get hella bitches

[heady plinth]

tru

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[restive river]

my problem with this is

I can't take the derivative

and have positive roots

wait a min

.close

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[quartz plank]

yo anyone help me with calculus

[quartz plank]

Find the area shared etween r=1 and r=-2sin3theta

[wicked turtle]

that sounds exciting

can you show your work?

[quartz plank]

wait

[quartz plank]

2

[old falcon]

t=1/162

[quartz plank]

the answer is 2pi/3 - surd(3)/2

u got another answer?

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[dry iron]

For part C, what do I sketch? Like do I sketch the originally curve of Y or am I meant to sketch y’ or y”

[sweet mantle]

y most probably

original curve

[upper schooner]

using the knowledge gained about the stationary points and points of inflexion you found from the previous parts

[dry iron]

So it it not the original y graph?

[upper schooner]

You sketch the orignal curve - and show where the original curve has its stationary point/s and point/s of inflexion (along with the roots it has)

[dry iron]

So I draw a graph with a max at 00 and min at 2,4 and inflexion at 1 -2?

Oh okay

[upper schooner]

(and @uncut crow )

[dry iron]

Thank you

[upper schooner]

Also don't forget the other x-intercept that the curve has, make sure you note that too as they asked for(!)

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[jagged scaffold]

part d

[jagged scaffold]

No clue how to do part d with a curve that looks that complex

tried using desmos for better understanding but failed that too

[upper schooner]

What did you get as the equation of the line through A and B?

[jagged scaffold]

y=x-4

[upper schooner]

Cool

Do you have the Desmos you did from before?

[jagged scaffold]

yep

like what is going on here

[upper schooner]

Cool, then basically, the area they want you to find, if you give me one second...

...is basically this (but above the x axis, ignore the part that's below )

[jagged scaffold]

ah okay so not the stuff below the x axis

gotchu

[upper schooner]

If I asked you to find the area under the curve $y = x^{2/3} - \frac2{x^{1/3}} + 1$ and above the $x$-axis from $x=0$ to $x=8$, you can do that right?

[woven radish]

@upper schooner

[jagged scaffold]

Can I tell you how I think Id do it?

[upper schooner]

Sure this is the integration chapter after all

[jagged scaffold]

I'd integrate with limits 1 and 8 to find the area under the curve from x=1 to x=8

and then id subtract this from the area of that traingle from x=4 to x=8

[upper schooner]

Yep that's pretty much the idea, you know how to do it

[jagged scaffold]

thanks bro

[upper schooner]

(also yea should have done x=1 there )

[jagged scaffold]

but how on earth am i supposed to envision this curve without desmos

[upper schooner]

It is a bit annoying to do, but you know that when x is between 1 and 8, the curve is above the line (they don't intersect between those x values other than at x=8)

[thin mist]

you can solve algebraically for the intersection of the curve with the line and for the x intercept of the curve

[upper schooner]

You also know that the line goes above the x axis from x=4, that's pretty easy to see

[jagged scaffold]

aight got it

thanks

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[tame otter]

Anyone wanna play Minecraft Education?

[devout snow]

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[spare crypt]

the usual first trick with bounces is you just copy the square a bunch of times and travel through the wall

[devout snow]

@frigid estuary Has your question been resolved?

[pastel pasture]

that's so cool i never thought about that

it's basically just a light ray and mirror

[heady plinth]

This seems like a very tricky problem

You can reduce it to “how many combinations of x,y are there such that x and y are coprime and x+y=2026

But that’s still so hard idk what to do from there

[frigid estuary]

ok but then how do we know that number of bounces are 2024

[spare crypt]

try to figure out how many walls it passes through to get to each vertex, it's a pretty easy pattern

[heady plinth]

Yo I just got out of the shower

I figured this one out lmoa

if x and y aren’t coprime, they share a factor

and 2026 would have to have that factor too

And 2026 is just 2x1013 which is a prime

So we only have to worry about 2

So every odd number up till 2025 works

1,3,5…

So there are 1013 ways

I can elaborate if that doesn’t make sense but I think I’m right

Ah sorry

Of course 1013,1013 doesn’t work

So there are 1012 ways

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[kind shoal]

I need help with a

[soft umbra]

Do you know the coordinate of that?

[kind shoal]

(7.5,0)

i found it

i got an equation

but for some reason the answer in the textbook shows the same equation but the bx isnt there

[soft umbra]

May I see your solution and what the textbook given?

[kind shoal]

my solution was 225y=-32x^2+272x+1800

[soft umbra]

There shouldn’t be 272x

[kind shoal]

ik

[soft umbra]

May I see your progress?

[kind shoal]

idk if my writing is readable lol

[soft umbra]

It’s okay

I’ll try

[kind shoal]

ok

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[kind shoal]

.

[kind shoal]

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[restive river]

pls help with this qn

[restive river]

i tried many ways but didnt work

[floral ridge]

why did straight substituting not work?

[restive river]

0^0

?

its indeterminate i think

[floral ridge]

oh right

try log

[restive river]

i dont have the energy to take so many logs

and besides what will happen inside the cos

theres the same {x}^{x}

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[faint orbit]

Is this correct?

[devout snow]

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[faint orbit]

.reopen

[devout snow]

✅

[faint orbit]

<@&286206848099549185>

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[devout snow]

@faint orbit Has your question been resolved?

[bold drum]

your method seems correct

if your calculation is correct your answer should be right too

[faint orbit]

It's all good then? Aside from it mot being totally simplified

[faint orbit]

Alright, thanks!

Appreciate the help.

.close

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[mighty coyote]

Either $ABC$ a triangle and $\Gamma_B$ (respectively $\Gamma_C$) a circle passing through $A$ and tangent to $BC$ in $B$ (respectively in $C$). If the diameter of $\Gamma_B$ $=56$ and the diameter of $\Gamma_C$ $=126$ . what is the radius of the circle circumscribed by $ABC$? I used the law of sines and got a result of 8.001 but this is wrong, I also tested with equations but impossible.

[woven radish]

habibjr.

[devout snow]

@mighty coyote Has your question been resolved?

[fossil locust]

You might want to ask in #geometry-and-trigonometry as the help channels close every 1 or 2 hours

It's not a question someone could easily do

And if you ask there, you'll get pinged in that channel instead of a random help channel that will get lost in the channel shuffling

No worries

[mighty coyote]

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[indigo stratus]

Hello, could anyone help me solving this? Thank you

[soft umbra]

!status

[devout snow]

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

[indigo stratus]

3

[soft umbra]

What’s your answer?

[indigo stratus]

I would do what you did

but it's not correct

I'm wondering why, though

[soft umbra]

yeah, ofc it’s incorrect

Because sqrt5 is greater than 2

[indigo stratus]

why is that important in that case?

[soft umbra]

You should change their position after removing the sqrt and the square

[indigo stratus]

but why can't I just did what you did on your image (I would assume that would be the first step).
why do we have to consider that sqrt5 is greater than 2?

[soft umbra]

You don’t get negative stuff after removing sqrt

[indigo stratus]

how is that what you did first called in english, if you know? (removing sqrt and that exponent)?

[soft umbra]

Plus, you don’t get negative real numbers after applying a square to a real number

You always get positive

[indigo stratus]

thank you for the above example. yeah, that makes it more clear.
are there any resources about that... I couldn't look up on internet because I wasn't sure what to look for exactly.

[soft umbra]

Just remember that every real numbers become positive after applied power of even numbers

That’s all you need to know

[indigo stratus]

okay, thank you for help!

[soft umbra]

np

[indigo stratus]

but still, is there any specific rule for that? removing the sqrt and exponent would be completely fine in other cases (maybe not really similar like those) so I'm not sure what I have to consider always then before doing it

if you understand what I mean

[cyan oracle]

consider the thing be x

when you solve you will get both positive and negative solutions

but you go back to x and see that it is a positive qty

[indigo stratus]

Okay, that makes more sense.

Thank you!

[soft umbra]

!done

[devout snow]

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[indigo stratus]

.close

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[restive river]

<@&286206848099549185>

[restive river]

<@&286206848099549185> need ur help

[lavish radish]

ur supposed to ping them if u dont find help from someone else after 15 minutes

u dont ping them immediately

[restive river]

im really sorry i didnt know that until now

[devout snow]

@restive river Has your question been resolved?

[fossil locust]

It's easier to think about it as the percentage who passed in each

So 30%, 28%, 20%, and 15%

I honestly want to say 15% but I don't think that's correct

Draw 4 circles in a Venn diagram

The trick is that there are so few conditions on the areas that you can let the middle one be 7

Woops should be 23 not 21

So I didn't realise this at first but you can let all the regions with multiple circles intersecting be 0

And it would work out

Wait so it can't be determined

There's just too many possibilities as I said

Like the middle one could be 9 or anything

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[foggy hamlet]

I'm attempting math tutoring, completely free
https://chalmers.zoom.us/j/3570051594

[wary ruin]

How do I describe the following area in polar coordinates. (One square = one unit of length.)?

[foggy hamlet]

.close

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[wispy berry]

what is 1+1?

im having trouble

[soft umbra]

!help

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[gilded nova]

what is this question asking for

like what do it mean measure of angle from tower 3 to tower 1

which angle is it asking for

[winter patrol]

did you ignore the last three words?

[devout snow]

@gilded nova Has your question been resolved?

[gilded nova]

oh

so it’s asking for tower 1?

sorry

[craggy dagger]

there are 3 points:
tower 3, tower 1, caller
they form an angle at tower 1

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[cold cloak]

why cant i do this? im tryna restrict it so that all numbers are positive

[acoustic leaf]

delete the comma

[cold cloak]

thanks

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[cunning roost]

if you multiply 36 x (5/x^2)

why would it become 180x^-1

are you allowed to cross the x's out bc i thought u cant do that in an addition equation

I thought

it would become like this

[acoustic leaf]

,, \frac{1}{x^2} = x^{-2}

[woven radish]

cloud

[acoustic leaf]

when you multiply together you add the exponents

,,180(x)(x^{-2}) = 180x^{1-2}

[woven radish]

cloud

[cunning roost]

ohhh

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[hybrid snow]

Typo probably

Uh

[vestal dirge]

thats wrong

[hybrid snow]

Lmao yeah that's a typo I mean

Fuck man

Time to complain to your professor

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[restive river]

yeah

its goood

[vestal dirge]

yeah works, personally I would do a proof by contradiction but this is clever

[tender coral]

okay awesome

thanks

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[livid carbon]

If we substitute y = x/x-1, we get f(y) = 2f(1/y), and if we substitute y = (x-1)/x, we get f(1/y) = 2(y), now we can eliminate f(1/y) from these two equations and get f(y) = 0, but isn't it incorrect to substitute y = x/x-1 once and y = (x-1)/x one more time in the same eqn?

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[maiden tree]

x

[maiden tree]

i need help with this question

how do i first find the transformed matrix B?

[devout snow]

@maiden tree Has your question been resolved?

[maiden tree]

<@&286206848099549185>

[devout snow]

@maiden tree Has your question been resolved?

[devout snow]

@maiden tree Has your question been resolved?

[maiden tree]

.

just need an answer by monday. i have a lot of time since it's not a mandatory assignment. just trying to understand the steps

[devout snow]

@maiden tree Has your question been resolved?

[maiden tree]

i know that P^-1 reverse the transformation done from the linear transformation

but why do we get B from B=P^-1^TAP^-1

im confused about that

thanks

[maiden tree]

Not resolved

[lavish prawn]

In this kind of questions, it's sometimes a bit unclear if you want to transform the quadratic form forwards or backwards with the transformation

If knowing nothing else, I would think that this question asks me "If you define a new quadratic form $q_B$ by $q_B(x,y) := q_A( Tx, Ty)$, what is the matrix $B$ associated to it?"

[woven radish]

Lartomato

[lavish prawn]

And then it's an easy manipulation, because $q_A(Tx,Ty) = (Px)^T A (Py) = x^T P^T A P y = x^T (P^T A P) y$, so clearly the matrix $B$ is $P^T A P$

[woven radish]

Lartomato

[lavish prawn]

But, from what you wrote, it sounds like this exercise doesn't want to transform the quadratic form "forwards" along $P$, but "backwards", along $P^{-1}$

[woven radish]

Lartomato

[lavish prawn]

And then the form you'd be considering is $q_B(x,y) := q_A(T^{-1}x, T^{-1}y)$

[woven radish]

Lartomato

[lavish prawn]

That's something you'd have to know from the circumstances. In a "real-world" scenario, this becomes clear when you ask the question "Do I want my quadratic form $q_B$ to have transformed vectors as an input, or untransformed ones?" -- in this situation though, it's a matter of understanding the definition given in the textbook

[woven radish]

Lartomato

[lavish prawn]

Lemme know if that helps

[devout snow]

@maiden tree Has your question been resolved?

[maiden tree]

That helped a lot

Based on answer keys

It took the inverse of it so I was confused

But for now, since it’s unclear

I’ll just go with backwards

[maiden tree]

One question

How do I know if it’s forward or backwards

It’s usually explicit but when it’s not mentioned

Should I just go with backwards. I’m just doubting myself

I can send you another instance of the same question

Thanks for the help

[lavish prawn]

In principle, both always makes sense, but within your class it sounds like the transformation is always the "backwards" one

[maiden tree]

perfect

thank you

makes sense

[lavish prawn]

good luck!

[maiden tree]

.close

[devout snow]

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[primal tundra]

idk where to start, i imagine f is some exponential function (maybe e^x), but other than that im not sure on how to make an initial value problem from this

[devout snow]

@primal tundra Has your question been resolved?

[primal tundra]

.close

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[wicked sequoia]

Having trouble with this, never done a system of linear congruences before

[outer marten]

chinese remainder theorem?

[wicked sequoia]

No idea what that is

[outer marten]

or in this case just try numbers ahahah

[wicked sequoia]

I need to find all solutions

[outer marten]

oh well then you need some version of Chinese remainder theorem

[wicked sequoia]

Clearly I don’t given the book hasn’t taught me it yet

[outer marten]

i can tell you how to get the result the book wants

consider mod 12

and see which ones will work

[wicked sequoia]

What do you mean

[outer marten]

the intuition of mod 12 is as it is 3x4

[wicked sequoia]

Yes

But how does that help

[outer marten]

which numbers mod 12 satisfy the equation

if x=10 mod 12

then it is a solution

why is this the only solution? because u can easily prove that none other numbers mod 12 work

[wicked sequoia]

But I don’t think that’s the method the book wants

It did one example linear congruence but it didn’t do anything like that

[outer marten]

what did it do

[wicked sequoia]

I just don’t know how to generalize this

[outer marten]

you do the same thing x=3a+1=4b+2 looking mod 3 we get 4b=-1 mod 3 and so b=-1 mod3. now plug b=3t-1 and finish the proof

[wicked sequoia]

I have x being congruent to 0 mod 2 but that doesn’t sound right

[outer marten]

you have to use mod 3 here or mod 4

coz the question is not mod 2 mod 5 anymore

[wicked sequoia]

I did

[outer marten]

so u gotta switch those numbers

so how you getting mod 2?

[outer marten]

x=12c+10

which means the answer is 10 mod 12

[wicked sequoia]

Oh my god

How did I miss that

[outer marten]

you had it all along 😭

[wicked sequoia]

.close

[devout snow]

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[amber sluice]

can someone help me go over this problem for sketching a polynomial

-9/2x + 6x ^2 - 2x^3

what i did so far is rewrite the whole thing -2x^3 + 6x ^2x -9/2x

[amber sluice]

i got the degree which is 3 and the leading coefficient which is -2 so im assuming the graph would not be symmetrical

so end behavior should look like up down?

left
as x - > negative infinity y -> infinity

right

as x -> infinity y -> negative infinity

[dense jay]

dont think so

actually, ill let you add your signs to your x infinities before i say so

[amber sluice]

wym

[dense jay]

youve said x to infinity approaches two things

alright that seems fine now

[amber sluice]

yeah i just realized

[amber sluice]

now im having trouble getting my x intercepts

but so far is my thought process correct?

[dense jay]

let me just trigger my telepathy

[amber sluice]

whaaaaaaaat

[dense jay]

(I dont know what your thought process is my man)

[amber sluice]

i meant the things i typed above

[dense jay]

what do they have to do with finding x intercepts

[amber sluice]

bcause i need to sketch the polynomial

[dense jay]

finding x intercepts doesnt require sketching by any means, but sure, your current observations are correct

[amber sluice]

okay thats all i needed to know

.close

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[urban matrix]

The lowest frequency of sounds that can be heard by humans is 20hz. The max pressure p produced with a freq of 20hz is 0.02minipascals write a sine model that gives the pressure P as a function of time

[tropic hill]

!onechannel

[devout snow]

Please stick to your channel.

[urban matrix]

LOL u left

someone help me plsssssssssss

[misty crest]

didn’t this already get answered

[urban matrix]

it did not

I closed my previous channel cuz it went dead

@misty crest and this one is dead too

smh

[urban matrix]

alr

[urban matrix]

Chatgpt

[urban matrix]

I tried making an equation but I got stuck on where to find the amplitude the rest I understand

[misty crest]

well you know what the maximum pressure is

that should help you

[urban matrix]

https://tenor.com/view/matthew-mc-conaughey-crying-sad-gif-12671400

[urban matrix]

I get that but how is that the amplitude

I thought the min is 20 hertz

[misty crest]

noooo

we’re making a pressure function

[urban matrix]

can u explain what the ampliotude is my mind aint working tday

[misty crest]

the amplitude is related to the pressure

not the frequency

[urban matrix]

oh and how so

[misty crest]

the amplitude is just how much pressure essentially

from zero

so the midline will be zero

zero pressure

[urban matrix]

no like in terms of math

[misty crest]

if no sound

[urban matrix]

what is the amplitude

[misty crest]

oh

i mean

it’s just the distance from the midline

like

[urban matrix]

like

[misty crest]

to the max

or min

[urban matrix]

so the amp is the max and the min right?

[misty crest]

yea

so it would be 0.02

[urban matrix]

then what is the hz in this context

[urban matrix]

Oh

[misty crest]

frequency ig

[urban matrix]

so its just there for the sake of the problem

ic

so lemme make my equation now

[misty crest]

well you need it for the inner part of the sine function

asinbx +c

[urban matrix]

oh ya

one sec

P(t) = 0.02sin(1/20t)

@misty crest

wait

[misty crest]

sorry

sorry

[urban matrix]

P(t)=0.02\sin(40\pit)

[misty crest]

it does 20 cycles per second

[urban matrix]

yes

[urban matrix]

all g

[urban matrix]

all g

wait so am I right or no

[misty crest]

where’d you get that

[urban matrix]

I gotchu

lemme explain

so the amp is 0.02 since its given the freq is the reciprocola of the periord

hence,

[misty crest]

why

[urban matrix]

wdym why

thats the definition

[misty crest]

yes true i’m just think P=2pi/f

[urban matrix]

henceforth, 20 = |b|/2pi = 40pi

thats right, right?

[misty crest]

you’re goated

[urban matrix]

alr then

yesir

thanks

[misty crest]

mhm

[urban matrix]

have a good one man

.close

[misty crest]

you as well

[devout snow]

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[devout snow]

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[restive river]

hey there, i Just wanted to know how to solve this type of equation for X:

C= ax+b mod(n), here n=26 but i dont know if that matters (a,b,c are all real) Thanks 🙂

You can dm me if thats easie

[tropic hill]

!onechannel

[devout snow]

Please stick to your channel.

@restive river Has your question been resolved?

[spare crypt]

C-b=ax mod n requires you finding the modular inverse, which doesn't have a nice formula so you'd have to use say the euclidean algorithm https://www.expii.com/t/solving-linear-congruence-ax-b-mod-n-3389

[restive river]

okay, thank you ill look into it 🙂

does it still work if i have it as c=ax+b mod(n)?

[spare crypt]

yea you can move over b in mod arithmetic

with n=26 I might just brute force what x is

[restive river]

why can you do that tho?

[spare crypt]

uhh the whole point of modular arithmetic is you can do addition/multiplcation just like with regular equations

visually you're just like shifting along a clock or whatever

[restive river]

okay okay okay, i just played around with it to make it make a little more sense

and when youve got it in the form, c-b=axmod(26), a and 26 must be co prime otherwise axmod(26) will be 0 which isnt neceserily congruant to c-d right?

[spare crypt]

nah that a isn't a multiple of 26 I think

er they have to be coprime to ensure a solution yea

[restive river]

which is the same as saying they are coprime right?>

[spare crypt]

coprime is a little stricter like 13 and 26 are coprime

but you're right they have to be coprime or else sometimes there's a problem

[restive river]

also is c-bmod(26)=ax ? like is it okay to move the mod(26) around?

because the link you sent me says ax=bmod(26)

[spare crypt]

oh in math (mod 26) is always on both sides of the equation

it's like just saying everything written is in mod 26 world

[restive river]

OHHHHHHH

i kept thinking of it as a function

[spare crypt]

yeaaa % is like a whole different thing mechanically

[devout snow]

@restive river Has your question been resolved?

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