help-42

[remote mural]

why is b not true

[upper sparrow]

m x n

e.g. something like $\pmqty{ 1 & 0 & 0 \ 0 & 1 & 0}$ meets the first condition but not the second

[potent lotus]

@upper sparrow

[remote mural]

wow

my mind is eluminated

but if it said if ALL columns are pivots???

then thats c

.solved

[calm coral]

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[remote mural]

i found RREF

the only vector in Nul A is (a), the zero vector is also included since thats that definition of a subspace

so only b and d are left

i know a basis can only consist of linearly independent vectors

[upper sparrow]

,w rref [[3, 1, 4, 0], [1, 2, 3, 5], [2, 0, 2, -2]]

[potent lotus]

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[upper sparrow]

How’d ya get that

[remote mural]

oh no i made an arithmetic mistake

i need to wipe my glasses

[upper sparrow]

These ones can be painful can’t they

[remote mural]

ill redo the whole thing

i did it

(d)

correct!

thank you so much

.solved

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[verbal lintel]

guys would it be the same if i just did the product rule for tan x and 3x2-9 and then did it again for that result and x3

[strange lichen]

@verbal lintel

[strange lichen]

f = x, g = x, h = x

derivative of fg then h method

(x + x)’x + (x+x)x’

2x + 2x = 4x

Now fgh method

x^2 + x^2 + x^2 = 3x^2

Thus they’re unequal, so u can’t do that

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[calm coral]

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[potent lotus]

l'agit

l'agit

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[fresh crag]

School puzzle related somehow to permutations

[fresh crag]

I think it’s not solvable because of lack of information

[calm coral]

@fresh crag Has your question been resolved?

[fresh crag]

How?

[tacit tulip]

if u follow the 5 steps its pretty straight forward

[fresh crag]

Can u elaborate?

[tacit tulip]

I just followed the 5 steps

that are in the assignment

and it fits together from there

[fresh crag]

Yh but explain idk

[tacit tulip]

if u set up a table and go through them one by one it sorts itself out

theres not really anything to explain

[tacit tulip]

but do u have to do permutations or is it just sort it in a table?

[fresh crag]

Teacher just said solve it

[tacit tulip]

ah ok

[fresh crag]

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[surreal patio]

just asking if im wrong or right

[surreal patio]

$$\int^3 _{-1} |x|$$

[potent lotus]

JustToPro

[surreal patio]

so when we split it it is going to be

$$\int^3 _0 x - \int^0 _{-1} x$$?

[potent lotus]

JustToPro

[polar frost]

abs(x) = sqrt(x^2)

[surreal patio]

how would i solve that integral?

$$\int \sqrt{x^2}$$

[potent lotus]

JustToPro

[surreal patio]

cant use u sub , no x present and if i cancel sqrt with 2 then thats wrong

[polar frost]

Hmm lowkey a dumb answer

Hmmm

[surreal patio]

ik its related to piecewise functions and splitting the integral

but im kinda bad at it , so wanted to ask if i split the integral correctly or not

[strange lichen]

This seems right…

[surreal patio]

why the ... tho?

is the question strange or did i do something wrong

[polar frost]

Honestly, I would do something like this

[strange lichen]

No, I was just wondering why u did the sqrt(x^2) stuff

[polar frost]

Its my dumb idea

[polar frost]

....

[strange lichen]

Abs(x) isn’t x from -1 to 0

[surreal patio]

x is negative from 0 to -1

[polar frost]

Oh right....

[strange lichen]

U don’t really need that third case lol

Just define one of the previous conditions to be >=

[surreal patio]

hm

[strange lichen]

But this is good

[surreal patio]

btw another question if u dont mind

whats the difference between
mclaurin series
taylor series

arent they both used to just expand stuff? like ln or cos or sin or smthing

[polar frost]

Isnt mclaurin series is just taylor series but modified so that the expansion center is not at 0 but other point a

[surreal patio]

i dont understand

[polar frost]

Let say u want to convert sin(x) to a series of polynomial using Taylor series

[surreal patio]

k

oh ok i get it

[polar frost]

But the number of polynomial is limited, so u can only expand to 3 terms

[polar frost]

Okay...

[surreal patio]

taylor series goes f(a) + f'(a)(x-a)/2! so on

and maclaurin goes f(0) + f'0)

so on

so are they same thing? just a = 0?

@polar frost ?

[polar frost]

If a = 0 then they're just the same thing yes

[surreal patio]

ah ok

thats good

ty

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[stone leaf]

how do i solve this question and set up the limits for the integrals

[dull geyser]

Its good to draw the bounded region first

It looks like a triangle

[stone leaf]

ive tried that, but still a bit confused,

[dull geyser]

Confused after drawing it or during drawing it

[stone leaf]

after, on how to set up the integral

[dull geyser]

Personally id integrate with respect to y

From 0 to 2

[stone leaf]

and what about x

[dull geyser]

After the y

[hot obsidian]

In a classroom, 1/3 of the students are 10 years old, 1/6 are 11 years old and 15 students are 9 years old. What is the number of students in the room?

[dull geyser]

Actually you'd need to do x first sorry

I'd write the bounds of x in terms of y

And then integrate wrt y

So x would go from y to y√3

And y would go from 0 to 2

[stone leaf]

lemme try that

thx

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[blazing coyote]

The number of values of $c$ for which $y=4x+c$ is a tangent to $\frac{x^2}{4}+y^2=1$ is

[potent lotus]

ƒ(Why am. I here)=misery

[small shuttle]

the number of values where $f'(c) = 4$?

[potent lotus]

b

[blazing coyote]

Yeah, was about to type my thoughts

[small shuttle]

mb carry on

[blazing coyote]

so basically at the points of tangency

$y'=4$

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

I think I'm missing what the C is doing though

Have to figure that out

I know. it's the y intercept

but have to figure out why that matters

[glass heart]

maybe too strong of a hint: ||number of intersections||

[small shuttle]

don't comment on my art but

these two lines have the same slope but different c values

[blazing coyote]

Hmm

so just drew a diagram

so basically I have to find the distance between the origin and the line

and equate it to some parameter of the ellipse?

[glass heart]

you are overthinking

[small shuttle]

keep in mind they don't actually care what the c value is

just the number of solutions

[blazing coyote]

in that case just 2, I think

slope is fixed

the c value can be. changed to translate the line parallel to itself

am I missing something

[small shuttle]

that sounds right

[blazing coyote]

but let's say I wanted to find c

would i still be overthinking it?

[small shuttle]

what's y'?

[winter loom]

dy/dx

[blazing coyote]

$\frac{x}{2}+2yy'=0$

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

$\frac{x}{-4y}=-y'$

[potent lotus]

ƒ(Why am. I here)=misery

[winter loom]

You can use some sort of parametric co-ordinates rather than working with x and y

(doesnt really matter)

[small shuttle]

i would solve for y

you'll get $\pm \sqrt{...}$

[potent lotus]

b

[small shuttle]

ie two cases

[glass heart]

just plug y=4x+c into the other equation

[blazing coyote]

you mean $\frac{x}{-4y}=-4$?

[potent lotus]

ƒ(Why am. I here)=misery

[small shuttle]

oh wow we are overthinking it

[glass heart]

solve x^2/4 + (4x+c)^2=1

[blazing coyote]

oh

[glass heart]

that gives you the intersections of the line and the ellipse

if you only have one intersection that means you have a tangent

[blazing coyote]

so I set the det to 0

and then what?

[glass heart]

det?

what

[blazing coyote]

determinat

[glass heart]

do you mean discriminant?

[blazing coyote]

yes

my bad

[glass heart]

well solve the resulting equation

its a quadratic in c

[blazing coyote]

it's a quadratic in x

mo?

*no?

[glass heart]

quadratic in x, set discriminant equal to zero, get quadratic in c

[blazing coyote]

wait

so

$x^2/4 +(4x+c)^2=1$

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

so I solve this?

as a quadratic in "c"?

[glass heart]

if you multiply this out, you get a quadratic in x

you are interested when this has one solution

so you look at the discriminant of this equation

[blazing coyote]

so that would be when the discriminant is zero

[glass heart]

yes

and the discriminant is some expression involving c

turns out, that expression is again a quadratic

[blazing coyote]

I get

$\frac{x^2}{4}+16x^2+c^2+8cx-1=0$

so how I take the det of this as a function of c and set that to 0?

[glass heart]

still not det

well what is the discriminant

bring it into the form ax^2+bc+c

oh btw, -1 at the end

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

$c^2+8cx+\frac{x^2}{4}+16x^2-1=0$

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

so a=1

b=8x

c= the rest

this looks like it will be rather messy

[glass heart]

for now its still an equation in x

[blazing coyote]

in that case

$\frac{33x^2}{4}+8cx+c^2-1=0$

[potent lotus]

ƒ(Why am. I here)=misery

[glass heart]

so now a=33/4, b=8c and c=c^2-1

with awful naming on my part

now the discriminant is what?

[blazing coyote]

$64c^2-33(c^2-1)=0$

[glass heart]

- not =

[potent lotus]

ƒ(Why am. I here)=misery

[blazing coyote]

this?

[glass heart]

yes

ok we made a mistake somwhere

cause that has no solutions

[blazing coyote]

hmm

[glass heart]

ok 16+1/4 is not 33/4

[blazing coyote]

,w (x^2/4)+(4x+c)^2=1

[potent lotus]

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[blazing coyote]

ah

this is very nasty IMO

I def don't want to find the tangent if the equations are like this

but the idea is the same, right?

[glass heart]

this isnt that nasty

I mean as soon as you have c you have the tangent

the tangent will be exactly 4x+c

[blazing coyote]

yeah, but computing that by hand will be really nasty

finding c that is

[glass heart]

no

well its sqrt(65)

[blazing coyote]

wait, what

[glass heart]

of course computing that would be bad but I mean why should you want to do that

[glass heart]

this should actually be 64c^2-65(c^2-1)=0

[blazing coyote]

oh

thanks

Thanks a lot!

.close

[calm coral]

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[lyric ravine]

Im not sure why any of this was done. The curve is an ellipse and it is quite obvious by a diagram that two tangents of any slope will occur.

[glass heart]

thats why

[calm coral]

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[snow hare]

being $a_n = \frac{\left(n+1\right)^{n}}{k^n \cdot n!}$ for which $k \in \mathbb{N}$ does $\lim_{n \to \infty} a_n = 0$?

[potent lotus]

milanesa de pollo

[hasty fiber]

!status

[calm coral]

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

[snow hare]

how do I get it started¡

for example, I think its known that n! grows faster than n^n

or is it not?

[fleet verge]

How is that?

[formal tulip]

I said n! can be treated as a polynomial of degree n, it doesn't grow faster :P

[snow hare]

my bad

can you guys help me out though?

how comparable is n! and n^n is in here?

[glacial inlet]

i have a way but probably not the intended way: ||sterling’s approximation for the factorial||

no clue how to do it the conventional way tho

[fleet verge]

I dont think he wants to do this

[glacial inlet]

yeah

[snow hare]

<@&286206848099549185>

[calm coral]

@snow hare Has your question been resolved?

[calm coral]

@snow hare Has your question been resolved?

[calm coral]

@snow hare Has your question been resolved?

[calm coral]

@snow hare Has your question been resolved?

[potent smelt]

Because n! = n * (n-1) * ... * 2 * 1 and n^n = n * n * ... * n * n and these two expressions have the same number of terms, we can easily see, because each term in the product of n^n is greater than or equal to the corresponding term in n!, that n^n > n!

As for a method to approach this problem, I can only see using the Sterling approximation. I have no idea how to approach this otherwise.

[ancient thistle]

,align \log a_n & = -\sum_{j = 1}^n \map[\bigg]\log {\f {kj} {n + 1}} \
& = -(n + 1)\sum_{j = 1}^n \f 1 {n + 1} \map[\bigg]\log {\f {kj} {n + 1}} \
& \overset{\mathclap{n\to\infty}}\sim -(n + 1) \int_0^1 \log(kx) \dd x
= -(n + 1)(\log(k) - 1)

[calm coral]

@snow hare Has your question been resolved?

[night dragon]

try using the squeeze theorem maybe?

find two simpler upper and lower functions which bound it

and then try and find values of k

[remote mural]

how doyou propose to find them in this case

[night dragon]

is K supposed to be a fixed value?

[remote mural]

yes

i think

[night dragon]

hmm

[calm coral]

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[vernal blade]

i need help

[vernal blade]

i am unable to answer any part of it

i think we use sin cos and tan?

<@&286206848099549185>

[raven gust]

using the cosine formula,
$cos x = (|AB|^2 + |AC|^2 - |BC|^2) / 2|AB||AC|$

[potent lotus]

ar1tr0

[vernal blade]

yea but do you know what the steps after that are?

im not really good with these types of formulas

[raven gust]

me neither, it comes from the addition rule of vectors which is pretty intuitive.

[vernal blade]

hm alright thanks tho

[raven gust]

plug in the values of the side lengths then it's easy

[vernal blade]

ah okay

[raven gust]

the sine rule is also helpful look into it

.close

[vernal blade]

would you happen to know how to solve this ?

.close

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[solid talon]

is this correct ? im pretty confident about it all until the last 2 steps

[calm coral]

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[analog jungle]

The central angle of the circular sector is 120° and the radius is 15 cm. The circular sector is bent into the shell of a straight circular cone. Determine the angle between the side line and the base of the cone.

[analog jungle]

Idk how to do this............... i need help

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[analog jungle]

<@&286206848099549185>

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[hexed ginkgo]

I don't know how to proceed

[raven gust]

what is G, what is z real complex please detail a little

[hexed ginkgo]

its written
G is a region?

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[plain umbra]

If f0 = 1 f1 = -1, find f9 for a fibonocci series

[plain umbra]

I solved it -13 but the answer was -8

am I wrong or the answer?

I also ran it with chatgpt and it got -13

[calm coral]

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[plain umbra]

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[hoary pebble]

how do I derive 5.4 from 5.1, 5.2, 5.3? i got 5.5 and 5.6 on my own

[hoary pebble]

I got 5.5 and 5.6 on my own

[somber badge]

https://discord.com/invite/physics

[vagrant oak]

some outline is here

either sin(theta) = 0 or sin(90-phi) = sin(theta) (making the second factor 0)

from this just few cases follow

you can verify for which of these cases the 2nd and 3rd condition holds

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[empty marlin]

how did it go from line 2 to line 3?

[calm coral]

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[empty marlin]

<@&286206848099549185>

<@&286206848099549185>

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gilded horizon]

where is the 1/21 coming from?

[winter loom]

what's the integral of y^6/3?

[gilded horizon]

i don't know where the 1/3 y^6 came from either

[winter loom]

Should have asked that first.

what's the integral of x^2

[gilded horizon]

x^3/3

[winter loom]

And your bounds are 0 and y^2

So that's where it came from

[gilded horizon]

but do you not add indices so y^5

[winter loom]

Are you saying (y^2)^3 = y^5?

That is incorrect. Here they multiply.

[gilded horizon]

oh yes i get it now

thank you

.close

[calm coral]

Channel closed

Closed by @gilded horizon

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[calm coral]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[dense thicket]

how do i do this?

[dense thicket]

i know how you're supposed to do it

but i keep getting something wrong with the algebra

[hasty fiber]

show us

[dense thicket]

so first

i took the derivative of sqrtx

wait let me write it out

[hasty fiber]

that would be better

[dense thicket]

wait, do i have to differentiate both sides

when try to find dy/dx?

[hasty fiber]

of course

[dense thicket]

ah ok then that was the issue

i didnt take the derivative of the 2

ok so the answer matches up, thanks

[hasty fiber]

[dense thicket]

do you know how to do this one instead?

[hasty fiber]

i do

[dense thicket]

i only got to the integral of this

and i dont know where to go from there

how do you evaluate the infinity

the integral i got was

-12(x+3)^-1/2

[hasty fiber]

$\int_a^\infty f(x) \dd x = \lim_{t\to \infty} \int_a^t f(x) \dd x$

[potent lotus]

Steakanator

[dense thicket]

uhhh

what is t

[hasty fiber]

some dummy variable

[dense thicket]

so i think i did this and

a is 1

and then

if the equation is -12(x+3)^-1/2

doesnt it go to 0?

[hasty fiber]

seems like it

[dense thicket]

but i checked the answer

and its 6

oh wait

it's 0 for the first only only

when you plug in 1 it gives 6

[hasty fiber]

indeed

[dense thicket]

are all improper integrals solved that way? with a limit

cause i remember our lesson with improper integrals had some nasty formulas that i didnt bother remembering

but i probably have to for the ap test smh

[hasty fiber]

it's slightly more formal than "plugging in infinity"

[dense thicket]

so if infinity was on the bottom what do i do there

do i flip it and make it negative?

[hasty fiber]

infinity on the bottom would be weird

[dense thicket]

and if the bottom was negative infinity and the top is positive

[hasty fiber]

if negative infinity on the bottom then you would want another limit yes

[dense thicket]

ok

ok last one

how do you do seperation of variables with this one

is the integral of dy/dx = y/x?

[hasty fiber]

it isn't

[dense thicket]

hmmm... so how can i split off the dx from the dy/dx

[hasty fiber]

same way as usual, just "multiply" by dx

[dense thicket]

wdym

[hasty fiber]

i mean precisely what I said

dy/dx * dx = dy

[dense thicket]

and then is the integral of dy = y?

[hasty fiber]

sure

but take a look at the right side

[dense thicket]

so then

2x-1/2y^2 dy/dx dx?

[hasty fiber]

where's that coming from?

[dense thicket]

im trying to integrate both sides but idk

[hasty fiber]

i think you need to look at the steps for separation of variables again

[dense thicket]

uhhh

ok

but at this step dy = (2-y)dx

what's next?

to solve for y

[hasty fiber]

well you need to get everything involving y on the same side

[dense thicket]

hmm ok

so

dy + ydx = 2dx

but then how do you get rid of the dx on the left side

[hasty fiber]

you can't

think of another way

[dense thicket]

okay so i integrate it now?

[hasty fiber]

no

what does $\int \dd y + y \dd x$ even mean?

[potent lotus]

Steakanator

[dense thicket]

uhh

[dense thicket]

i dont know what you're getting at here

do i take the derivative of both sides?

[hasty fiber]

you asked if you could integrate. I integrated and am asking you what that even is

[dense thicket]

well

[hasty fiber]

this is why I asked you to re-consult whatever notes you have on separation, because hopefully it contains the proper way to separate them

[dense thicket]

wait

so you dont know how to seperate it?

[hasty fiber]

i do know

[dense thicket]

uhhh

so

what's the derivative of dy

is it y?

[hasty fiber]

that's not a super well defined question

[dense thicket]

im trying to take the derivative of both sides

[hasty fiber]

why?

[dense thicket]

of the equation dy + ydx = 2dx

[hasty fiber]

you're solving a differential equation

[dense thicket]

uhhh

[hasty fiber]

ok no, cease

I've told you that adding ydx to both sides isn't the way forward

you have dy = (2-y)dx

what's another way, aside from adding, that you can move everything with a y to the same side?

[dense thicket]

uh

dividing?

[hasty fiber]

by?

[dense thicket]

2-y?

[hasty fiber]

so you're left with dy/(2-y) = dx

do you think you can integrate these?

[dense thicket]

idk what the integral of dx is

is it x?

[hasty fiber]

$\int \dd x = \int 1 \dd x$

[potent lotus]

Steakanator

[dense thicket]

uhhh

so

x?

[hasty fiber]

surely you can integrate 1 by yourself

[dense thicket]

umm

so i ended up getting y = 2-e^x

[hasty fiber]

it seems you've skipped a step or two

you're also missing the integration constant

[dense thicket]

oh wait

ok

umm

so i got 1-e^(x-1)

but thats not one of the answer options

.close

[calm coral]

Channel closed

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[calm coral]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[strange crown]

assuming we're trying to find m and n for the same function and we're given two different x x-cords, f(3) = 0 and f(2) = 85

Can someone check if my substitution is correct, and confirm if this is the correct answer. I got varying results on my other attempts(not shown here), but I feel confident this is correct

[dusty crescent]

as far as I can tell it looks ok

the way to check this

is to now plug in 3 in the function

and see if you get 0

[strange crown]

ahh yeah and see if I get the correct remainder

[dusty crescent]

and plug in 2 and see if you get 85

[strange crown]

man, all that for me to not think of doing that.. holy I'm so cooked

🤣

ty tho

.close

[calm coral]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[iron cliff]

hello

[cedar sinew]

Help pls

[iron cliff]

SAS they are both right angle with 2 sides of the same value

[cedar sinew]

But the included angle isn’t in between them

The SAS theorem states that the angle needs to be inside the two corresponding sides

no?

[iron cliff]

if u use arccos for both triangles

u will get the same angle between them

arccos(8/17) for both trialges

for find the angle between the two

[iron cliff]

@cedar sinew

[cedar sinew]

someone told me it’s HL

[calm coral]

@iron cliff Has your question been resolved?

[iron cliff]

idk what HL is so theycould be right

[calm coral]

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[calm coral]

@cedar sinew Has your question been resolved?

[calm coral]

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[jolly steppe]

MCQ on absolute inequalities

[brisk pelican]

!status

[calm coral]

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

[jolly steppe]

I've done it and my ans is B, but obiviously the correct one is going to be D)

Don't get how

[brisk pelican]

Can you send your work

[jolly steppe]

Yes

Oh shit I forgot to check critical points

Yeah makes sense, sorry

Also messed up the second case sign, doing stuff mentally when you're a beginner is not good

.close

[calm coral]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[astral breach]

i dont get 2,3,4

[dull wagon]

angle sum to deteremine the third angle
then do whatever you didn in Q 1

[astral breach]

i got 38 degrees

for number 2

and number 3

135 degrees

and for 4

i got 84

could you check if its correct

[dull wagon]

how are you getting 135°

[astral breach]

i added the angles

then subtracted it by 180

[dull wagon]

won't get you 135

[astral breach]

jk 38

wait no

hold

i looked at number

2

61

[calm coral]

@astral breach Has your question been resolved?

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[trail valley]

These 2

[trail valley]

Now what?

Is the cylander 2181370.319 grams?

Anyone?

[fallen linden]

Hi

Again

[trail valley]

Hi i fell asleep by mistake

[fallen linden]

Lolll

[trail valley]

I still need help never got it

[fallen linden]

Oh

Their all like dead

U want me to try?

@trail valley

[shut sierra]

i am not used to seeing inches measurement

wait

[shut sierra]

what are u even stuck at

[fallen linden]

Wait no it’s

[fallen linden]

“Is the cylander 2181370.319 grams?”

[trail valley]

Those 2 questions

[trail valley]

Please help? @shut sierra

[shut sierra]

the first question is not hard, it's just tedious

[trail valley]

Please i never learned this deadass i didnt

[shut sierra]

the second is more straight forward than the first

[trail valley]

My tescher never tought it please help

My teacher never taught the lesson @shut sierra

[shut sierra]

cylinder is 2182485 grams

[trail valley]

R u sure?

Im paranoid imma get it wrong

[shut sierra]

are you taking pi = 3.14 instead or some approximate value?

[trail valley]

Idk tbh 😭

Like i said teacher never taught this

[shut sierra]

so what did you use to calculate the volume of the cylinder

[trail valley]

Idk

Sorry if i sound dumb

[shut sierra]

so how did u get this value?

[trail valley]

Ai app

Im just being completely honest

But idk if its correct

@shut sierra

[shut sierra]

how about you look at the question again and think and tell me how you would attempt to solve it

[trail valley]

Idk

How am i supposed to somve something i didnt lesrn

[shut sierra]

stop saying idk and think

tell me, what does the question ask

[trail valley]

Sorry ok ill try

[trail valley]

Mass in grams of cylandee

Cylander

[shut sierra]

yes

how do we find mass?

[potent lotus]

misha

[trail valley]

Volume of cylinder then multiply by density

[shut sierra]

correct

[trail valley]

??

Oh i was correct

[shut sierra]

do we know any of those values?

do we know either the volume or density of the cylinder?

[trail valley]

Using a calculator i got 133183.422319224

[shut sierra]

for?

what exactly did you multiply

[trail valley]

Mass

[shut sierra]

show your working

[shut sierra]

how is this your working

[trail valley]

Tbh idk sorry

Its 1:18 for me rn im tired its due tomorrow

[shut sierra]

first of all that answer is horribly wrong because you used the wrong units of measurements

[trail valley]

Oh ok

I understand y

[shut sierra]

and i'm not going to help you if you keep searching for the easy way out

[trail valley]

Sorry

[shut sierra]

ask someone else to help you then

[trail valley]

No ill acually try

I didnt mean to sorry

@shut sierra

[trail valley]

I apologize for that and ill acually try now im sorry

Ok whats the step mam @shut sierra

Ill listen to you please 😭

Im sorry bro @shut sierra

Please bro im acually crying irl

I dont wanna fail!!!

[trail valley]

@shut sierra im so sorry i wont do it again

Bro 😭

[fallen linden]

Did u figure out one

[trail valley]

No!!!

She left on me

[fallen linden]

None???

@shut sierra why did u leave

[trail valley]

Cause i tried doing things easy way and i apologize for that

[fallen linden]

Ah

[trail valley]

I acually will listen and do the work!!!

[fallen linden]

Well then why don’t u just google it or is this study material for a test

[trail valley]

I csnt google it it gives me wrong answers i triedd

Bro please im sorry for what i did

@fallen linden idk what imma do

No one can help on any server

[fallen linden]

They left already

[trail valley]

@shut sierra

Im sorry for what i did bro 😭

[fallen linden]

Theirs no point they won’t come back

[trail valley]

Anyone else?

[fallen linden]

Hmm

[trail valley]

Please help

[remote mural]

Hi @trail valley

[fallen linden]

That guy there

[trail valley]

Omg hi!!!

CAN U PLEASE HELP ME I BEG OF U!!

[remote mural]

I put in ur question straight into google and the answer came up

[trail valley]

Bro wrong answer bro

[remote mural]

So idk how you didn’t find it?

[trail valley]

I tried

Wrong answer bro

[fallen linden]

Wait how do u know it’s wrong

[remote mural]

It doesn’t sound wrong

There’s even an explanation

[trail valley]

Thats how

[remote mural]

You checked?

[trail valley]

Yep

Btw my stupid teacher only lets me check once

[fallen linden]

But doesn’t the check have the answer

[trail valley]

I press it again it goes your teacher limit check button

[trail valley]

No

Just tells me if answer correct or wrong

[fallen linden]

That’s kinda dumb

So can u go back after u submit

[remote mural]

So it isn’t 12745987.2 grams?

[trail valley]

No

[trail valley]

Nope

[remote mural]

I think it is

[fallen linden]

Wait so it’s just ur final grade????

[trail valley]

Bro can someone helonme please

I still havnt done my chemistry homework

[fallen linden]

Cant u ask ur teacher

For an extension and help

[trail valley]

Im deadass crying rn irl

[fallen linden]

Does she not do that

[trail valley]

No!!!

[fallen linden]

What grade

[trail valley]

Look anyone help?

[remote mural]

<@&286206848099549185>

[fallen linden]

@harsh siren sorry maybe u can help kinda emergency

@deft grove

@vast trellis

Can y’all help maybe

[fallen linden]

Is it high school

[shut sierra]

@trail valley grab a piece of paper

[remote mural]

10th grade I think

[vast trellis]

What's up

[shut sierra]

and a pencil or pen

[trail valley]

Ok mam

[fallen linden]

Sonic ur teachers are here

[remote mural]

W pfp

[trail valley]

Im sorry for earlier btw

[fallen linden]

Don’t screw it up this time

[shut sierra]

we will attempt to solve the easier question first

[trail valley]

Which is ohh mvm

[shut sierra]

the question asks us to find the mass of the cylinder

[trail valley]

Yep

[shut sierra]

can you tell me again, how do we find the mass?

[trail valley]

One seccond

[trail valley]

calculate the cylinder volume, then multiply the volume by the density

[shut sierra]

that's right

do we know either the volume or density?

[trail valley]

^^

[trail valley]

Yep

[shut sierra]

which do we know

[shut sierra]

yes

the question told us the density

so what do we not know?

[trail valley]

Volume

[shut sierra]

yes

how can we find the volume of a cylinder?

[trail valley]

Da volume lol

[trail valley]

V=πr2h

Is that correct?

[potent lotus]

misha

[trail valley]

Yep thats what i said

[shut sierra]

r is the radius and h is the height

calculate that

the question told us the radius and the height

[trail valley]

With what numbers?

Ohh nvm

Got it

[shut sierra]

the question gave us

[trail valley]

V≈57905.84

[shut sierra]

yes

[trail valley]

Am i correct?

Nvm i am

[shut sierra]

look at the question again

the values for the radius and height were given in inches

[trail valley]

Kk

[trail valley]

Yep

[shut sierra]

but the density is given in grams per cubic centimeter

[trail valley]

Mhmm

[shut sierra]

so we must convert from inches to cm

[trail valley]

So convert this to cm?

[shut sierra]

not quite

[shut sierra]

this value is 57905.84 cubic inches

[trail valley]

Oh ok

[shut sierra]

we convert to cubic centimeters

multiply by 16.387

[trail valley]

This tomes 16.387?

[shut sierra]

yes

[trail valley]

948903.00008

[shut sierra]

did you miss something

948906.70605

[trail valley]

Idk i multiplied like u said?

[shut sierra]

did you miss type or something

my value is different from yours

[trail valley]

Idk

[trail valley]

Using the calculator

[shut sierra]

now that we have the volume, we can find this

[trail valley]

Is this correct?

[shut sierra]

do you not have a calculator

[trail valley]

I do but dont wanna get it

[shut sierra]

calculate mass now

u get 2182485 grams

now enter that as your answer

[trail valley]

Yea look see

It is correct

[shut sierra]

you can see if it's correct or wrong?

[trail valley]

Nope cannot

The check button can only be used once

[shut sierra]

oh

[trail valley]

I already used it once

[shut sierra]

did you use the check button on the first question yet

[fallen linden]

What if u just try submitting it

What’s the worst that can happen

[trail valley]

Its omly 5 questions

[trail valley]

Yep

[fallen linden]

Wait so u only are on question 1?

[trail valley]

No i skipped it

[fallen linden]

So how many u have to do

[trail valley]

5 questions 1 is my last 1

[fallen linden]

So onky 1?

[trail valley]

I think #5 is wrong but im tryna get 3 and 1 correct so i can ger 1 wrong to get 80% B

[fallen linden]

K keep going then.

[trail valley]

Anyways @shut sierra number 1

[trail valley]

Ok number 1 please

[fallen linden]

I think she’s afk

Idk

[trail valley]

Kk

[fallen linden]

What grade r u in

[trail valley]

10

[fallen linden]

So geometry 1?

[trail valley]

Mhmm

[fallen linden]

Ah, for our school that’s 9th grade

We also use big ideas

[trail valley]

Yo she left

@shut sierra

[fallen linden]

I think she left when I came in…

If she did my

mb

[trail valley]

Bruh

[fallen linden]

Can’t u google

The first one

[remote mural]

How’d you go?

[fallen linden]

He still needs to do one more

I think

Why don’t u ask ur teacher for an extension

[remote mural]

You usually can’t for quizzes

[fallen linden]

This doesn’t seem like a quiz

[remote mural]

No?

Is it a test then?

[fallen linden]

It just seems like homework

Or he was studying for a quiz idk

[remote mural]

I’ve never had homework on a website so I wouldn’t know lol

[fallen linden]

Never???

[remote mural]

No

[fallen linden]

I have it so much

[remote mural]

We’d get pdfs with questions

[fallen linden]

Do they print it out for the class

[remote mural]

Or textbook chapters

[fallen linden]

And give it to u

[remote mural]

No

Was just pdfs you download

[fallen linden]

We don’t use our textbook the teacher just gives us worksheets or we can choose the online ver

[remote mural]

Ah I see

[fallen linden]

U use ur textbook to get the questions?

[remote mural]

No no it’s either the questions are found in a long pdf that’s posted

Or questions from a textbook

[fallen linden]

What type of math

And grade

[remote mural]

I’m not in hs anymore lol

[fallen linden]

Ahhh

College?

What are u studying

In math

[remote mural]

University

[fallen linden]

Ah

[remote mural]

I don’t do math

[fallen linden]

But-

[fallen linden]

What does this mean then

[fallen linden]

Or this???

[remote mural]

That was my experience when I was in school

[fallen linden]

Ahhhh hs?

[remote mural]

All of highschool math was like that

[remote mural]

Yepp

[fallen linden]

I’m in middle school math lol

7th grade

[trail valley]

Ok forget #1 please help me with rhis it seems easier

[remote mural]

You have such a long way to go lmao

Good luck

[fallen linden]

Lol ty

[fallen linden]

U might have to find people urself this time sorry

I have to go to bed

[trail valley]

I am

[fallen linden]

K

[remote mural]

Bye noob god

[fallen linden]

Bye 😦

[trail valley]

Yeee

Gn

[fallen linden]

Sorry y’all I have a. Test tmmrw

And it’s 11:06 lol

[fallen linden]

Wait do u have. A test or is it just hw

[trail valley]

Its 2:05 am for me rn

[remote mural]

Ah the good old topic tests

[trail valley]

Hw

[fallen linden]

So no test coming up?

Dang that’s a stressful homework then

[fallen linden]

English write during class period

Okay everyone good night bye.

[remote mural]

Bye

[regal geyser]

hii

@trail valley

[trail valley]

Hello

This question

It seems easy but i dont get it

[regal geyser]

let me see hm

[trail valley]

Also i wanna check if this is correct

[trail valley]

I did this with someone else but checking my answer

[regal geyser]

hm ok

so we are given the radius and the height of the tree

[trail valley]

Yep ^^

[regal geyser]

and its density

[trail valley]

Yep

[regal geyser]

remember what density is?

[trail valley]

What the formula?

[regal geyser]

what if we multiply density with volume?

[trail valley]

Kk

[regal geyser]

we will get

and volume cancels out and we get mass

[trail valley]

We dont have rhe volume tho

In the question

[regal geyser]

but we can find it!

[trail valley]

Kk

[regal geyser]

with the radius and the hight we can find the volume of the cylinder

before operating, lets check the units

[regal geyser]

for this one

[trail valley]

Soo V=πr2h

[regal geyser]

yep

[trail valley]

Ok

[regal geyser]

but be careful with the units

[trail valley]

So hao i plug the # in?

[regal geyser]

we have height in meters radius in meters

and density in meters

i mean kg/meter

[trail valley]

Kk

[regal geyser]

so you can proceed

with plugging the numbers in

[trail valley]

Plug what #?

[regal geyser]

so the volume is

plug h = 40 and r = 0.75

[trail valley]

Ok

[regal geyser]

you should get

[trail valley]

V≈3769.91

[regal geyser]

lol ok

check again, that doesn't seem right

[trail valley]

My bad

[regal geyser]

dsjkfh np

[trail valley]

I acidentaly swapped them

I acidentaly put 40 for radius

[regal geyser]

yeah, 40^2 is a huge number...

[regal geyser]

https://tenor.com/view/poop-cat-meme-dark-fnaf-gif-2491098315710587638