help-10

[vital dew]

No

The lowest common multiple is 60

How do you make 10 into 60?

How do you make 20 into 60?

And how do you make 30 into 60?

[crystal plinth]

LCM

[vital dew]

60 is the LCM

[crystal plinth]

yes

[vital dew]

10*6

20*3

30* ?

[crystal plinth]

OHHHHH

[vital dew]

30 * what number?

[crystal plinth]

bruh its so simple how did i not get that

[crystal plinth]

2

[vital dew]

Well Done!

You multiply top and bottom by the same number

[crystal plinth]

and then how do i get 11 from that? 1*60=60 divided by what to get 11? what is Rges exactly? Rges is the total circuit

[vital dew]

6+3+2 is what

[crystal plinth]

oh 6+3+2

yeah

and then 1*69=60/11=5.45 or whatever

60 not 69

i get it now thank you Hasbulla

bye bye

.close

[obtuse pebble]

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[idle tide]

hello, i have this very simple math problem that i think i know how to solve but google (in red) shows that I somehow missed a +1? (I know y is technically 1y but doesn't it get removed here?) thanks for the help!

[idle tide]

i forgot to add a squared at the end of the first answer just a heads up

[vital dew]

can you send just the question

[idle tide]

=

thats the whole thing

i don't get it cause im a bit slow lol

[vital dew]

make the denominators equal

[idle tide]

so i square the first y?

making the x be squared also?

so it comes out to this

this is what google tells me thats why im skeptical

i have an app that i can take pictures of equasions and it sends me the answer

ill retake the pic 1sec

yeah look at the top

that's what it gives me

i'll trust you that my answer is correct tho lol

thanks!

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[versed quiver]

given 10 people around a table who know 2 or 3 others, how many ways can they be seated so no one is sitting next to anyone they know?

[tired shell]

not enough information

[obtuse pebble]

@versed quiver Has your question been resolved?

[versed quiver]

what else can i provide?

[soft sail]

i'd imagine you have to pick between them knowing 2 or 3 ppl

[obtuse pebble]

@versed quiver Has your question been resolved?

[versed quiver]

the question states 2 or 3

well, rather

the question asks whether this is possible to seat people such that no one knows eachother

which intuitively i know is correct

but how do i prove this?

i assume i do so by assuming every person knows 3 people and proceeding from there since that is worst case

but how many ways can 10 people, each of which knowing 3 others, can be seating around a circular table such that no one knows the persons next to them?

[obtuse pebble]

@versed quiver Has your question been resolved?

[obtuse pebble]

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[hybrid gull]

Does anyone mind checking my work for this WebWork Laplace Transform problem?

[hybrid gull]

I really tried to show every step

This is for part a, t > pi/2

[obtuse pebble]

@hybrid gull Has your question been resolved?

[obtuse pebble]

@hybrid gull Has your question been resolved?

[obtuse pebble]

@hybrid gull Has your question been resolved?

[obtuse pebble]

@hybrid gull Has your question been resolved?

[obtuse pebble]

@hybrid gull Has your question been resolved?

[placid kayak]

@harsh moat read #❓how-to-get-help

[obtuse pebble]

@hybrid gull Has your question been resolved?

[timid silo]

i do not know what the Gauss Jordan method is

but that can be solved with basic algebra though

let number of strawberry creppes sold be x
let number of org creppes sold be 2x
let number of gr tea creppes sold be g

i) 3x+g = 405
ii) 2x(4.5) + g(8) + x(2) = 2320

comes down to simultaneous eq

you can solve the rest i hope

[harsh moat]

Thanks for the clarification. I'll try to look at it

[harsh moat]

ok sorry

[hybrid gull]

.close

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[slender plank]

2 2 2 2 2 = 28
You can use +-()*/

[slender plank]

write on other chunnel

[dapper geode]

wow mb

we send at the same time

[slender plank]

np <33

2 2 2 2 2 = 28
You can use +-()*/

[dapper geode]

seems to be lacking a 2 in every possible solution I've tried

[spiral maple]

you done need multiple channels.

[unique solstice]

Doing a brute force check for basic stuff it doesn't look like it'll be any kind of straight forward operation

It'll have to have some kind of parentheses and pre calculation

I might write a better program later that can group together the terms in different ways

This doesn't seem to be much of a math problem though

It's just brute force

[slender plank]

?

[unique solstice]

Where did you find this problem? @slender plank

[slender plank]

my teacher

[unique solstice]

Hmm

And you're sure it's only five 2s?

[obtuse pebble]

@slender plank Has your question been resolved?

[slender plank]

.close

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[polar gulch]

Need help with meaning here, the word through here has the same mean as the line that contains/ has the point A ?

[polar gulch]

or passes by point A ?

.close

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[hasty harbor]

I'm having trouble using the epsilon delta definition to prove that the limit of 1/x-1 as x->1 does not exist

[dawn yew]

can you try writing out, using quantifiers, the statement you're trying to prove

[hasty harbor]

Yeah, there exists epsilon>0 such that for all delta>0 there exists x st 0<|x-1|<delta and |f(x)-L|>=epsilon

[dawn yew]

I would stick a 'for all L' on the front of that, but yeah

[hasty harbor]

I'm just looking at the limit approaching 1, but yeah that would be nice to know.

[junior shard]

#help-8, ty

[dawn yew]

to disprove 'limit of 1/x-1 as x->1 exists', you need to disprove 'there exists some L such that (limit of 1/x-1 as x->1) = L', right?

[hasty harbor]

I don't see why it would be necessary since it could exist in other places.

[dawn yew]

let me try TeKing this, I think It'll make it clearer

[warm shale]

TheZachMan

TheZachMan

[hasty harbor]

Oh, I see

[dawn yew]

which means that you can make epsilon a function of L, basically

[hasty harbor]

Ok

So epsilon equals something in terms of L?

[dawn yew]

yeah

[hasty harbor]

What about finding an x?

[dawn yew]

so you need to show that, no matter what L is, you can pick an epsilon such that, no matter what delta is, you can pick an x such that 0<|x-1|<delta and |f(x)-L|>=epsilon

so x can depend on one or several of L, epsilon, and delta

[warm shale]

TheZachMan

[hasty harbor]

Where do I even start? I tried messing with |f(x)-L|>=epsilon and got x>=1/(L-e) +1 and x<=1/(L+e)+1

[dawn yew]

drawing a picture might help

[dawn yew]

that should work (and those are two conditions of which at least one has to hold, right? is one of them easy to satisfy for numbers that are very close to 1?

you might want to seperate into cases where L is positive, negative, or 0

[hasty harbor]

What's up with this?

[dawn yew]

looks right to me, what are you asking what's up with?

[hasty harbor]

So, use this to intuit an answer?

[dawn yew]

yeah

oh, that region you've parenthesised you mean? that's the region where if x is in that region, f(x) is e-close to L

so you're trying to pick x outside of that region

but still have x be d-close to 1

[hasty harbor]

So look at x-1<=1/(L+e) and x-1>=1/(L-e) to get an idea of d?

[dawn yew]

you don't get to pick d, you have to show that whatever choice is made for d, you can pick an x that makes those equations hold

a helpful way to think of these very quantified statements can be as a "game" that you're playing against an opponent - they pick an L, then you pick an e, then they pick a d, then you pick an x, and you win if 0<|x-1|<delta and |f(x)-L|>=epsilon

the statement is true if you can always win this game

[hasty harbor]

Ok so given any L you have an e, so e is dependent on L. Then given any d pick an x, so x is dependent on d.

[dawn yew]

yes

x might depend on L and e as well, but it should probably depend on d

[hasty harbor]

Ok

If L>=0 then we still have the same inequalities for x, but if L<0, then x>=1-1/(|L|+e) and x<=1/(e-|L|)+1?

[dawn yew]

yeah that looks about right

are you writing 'and' to say that either one of those two conditions needs to be satisfied, or both need to be satisfied at the same time

[hasty harbor]

I figured both since it came from |f(x)-L|>=e

I guess it could be one or the other. I think given the same conditions both will hole

[dawn yew]

$|f(x)-L|\geq\epsilon$

should be equivalent to

$f(x)-L\geq\epsilon$ OR $-1*(f(x)-L)\geq\epsilon$

[warm shale]

TheZachMan

[hasty harbor]

Ok

[dawn yew]

try and see if you can figure out what regions of that picture those inequalites correspond to

[hasty harbor]

So these are equivalent to the L+e, L-e around L, which correspond to some x

But actually the function outside if that

So these are the x-values outside of the parenthesis?

[dawn yew]

yes, exactly

[hasty harbor]

Woo! That's cool

Ok so these are the ones we want given any delta, but in terms of delta

[dawn yew]

yeah

and in fact, the only reason x needs to be in terms of delta is to make sure 0<|x-1|<delta is true, as it turns out

[hasty harbor]

Oh

[dawn yew]

(because if you said "pick x = 10", that wouldn't work for small values of delta")

[hasty harbor]

So do we need to do x=min(what have you)

[dawn yew]

yeah, that's the right idea

[hasty harbor]

Which of the inequalities do I choose for any given L?

Since it's or does it matter?

[dawn yew]

it turns out it does matter - in your picture, what happens if I pick a very small delta?

[hasty harbor]

The gap shrinks towards the limit

[dawn yew]

yeah, but if delta is small, will either inequality work? or only a specific one?

[hasty harbor]

If delta is small then the range x can be is small, which means we can only pick x>=1/(L-e)+1?

[dawn yew]

why that one

the two inequalities correspond to either side of the parenthesised region, right? which one of those is near x=1

[hasty harbor]

Ohh it's the other one

x<=1/(L+e)+1 is closest to 1, which we want because we are trying to approach it. Is that reasonable?

[dawn yew]

yes (for positive values of L)

[hasty harbor]

So, the other side for negative L

So these respective x's will be part of the min, or also max for the negative L case. If they are bigger, then just regular |x-1<d?

[dawn yew]

yeah

[hasty harbor]

Then I guess just something in terms of delta that gives epsilon?

So also in terms of e?

[dawn yew]

yeah, you'll probably need to use epsilon and L as well

to figure out what value to choose for x

[hasty harbor]

Right, so if L>=0 then x=min(1+1/(L+e), something with delta)?

This something with delta if that's x then it's closer to 1 than the Le expression, but it still needs to yield e when doing the proof

So would the something with delta still have e in it? I don't think it would

Could I have it be 1+d/2 for the positive L case and 1-d/2 for the negative L case?

For the negative L case, should I do x=min(1-d/2, 1-1/(|L|+e)) or the max?

[obtuse pebble]

@hasty harbor Has your question been resolved?

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[dusty fjord]

Anyones help would be appreciated trallu

really

[wary vigil]

which one

[dusty fjord]

the whole thing 😿

going from a to e if you have time

or we can just do a if you odnt

dont

[wary vigil]

do you know how to differentiate?

take the derivative?

[dusty fjord]

i do!

i think

i know how to find the derivative and the rules

would f’(2) be = finding the derivative of sqroot 0?

[wary vigil]

no

[dusty fjord]

or would i find the derivative of the og function first

[wary vigil]

f'(2) means filling 2 into the derivative

[dusty fjord]

gotcha so should i find the derivative of sqrt x-2?

[wary vigil]

yes

[dusty fjord]

thank u ill be back! would that be chain rule btw

ill ping u

i just needa solve the problem

[wary vigil]

technically yes, but since the inner function is just x it doesn't matter

[dusty fjord]

oh really..?

so just do it like normal

[wary vigil]

yeah the derivative of x is just 1

so at the end you'd just be multiplying with 1

[dusty fjord]

ahh true

so isnt it just (1-0)^1/2

or am i missing something

[wary vigil]

we have the function

$(x-2)^{\frac{1}{2}}$

[warm shale]

Katharine

[dusty fjord]

ahhh

[wary vigil]

the derivative of this is?

do you know?

[dusty fjord]

yeaj!

(1-0)^1/2

so 1^1/2

[wary vigil]

no

[dusty fjord]

ah how’d i mess up

[wary vigil]

it's $\frac{1}{2}(x-2)^{-\frac{1}{2}}$

[warm shale]

Katharine

[dusty fjord]

ahh i see

so i wouldn’t find the derivative of x or -2?

[wary vigil]

you take the number in the power

put it up front

[dusty fjord]

i understand!

[wary vigil]

and -1

and you then multiply this by the derivative of what's inside

[dusty fjord]

i thought you had to find the derivative of x and -2

[wary vigil]

but because that's x

[dusty fjord]

but i completely understand

[wary vigil]

you'd be multiplying by 1

you do

but after

[dusty fjord]

ohhh

[wary vigil]

not inside

[dusty fjord]

gotchaa

[wary vigil]

multiply afterwards

anyway

[dusty fjord]

so now i just use this

[wary vigil]

now we fill in x = 2

[dusty fjord]

to find a

[wary vigil]

x = 0

and x = 4

let me write it in a better way

$\frac{1}{2}\frac{1}{\sqrt{x-2}}$

[dusty fjord]

f’(2)=0

[warm shale]

Katharine

[wary vigil]

if we fill in x = 2

in this

what do we get

[dusty fjord]

0

[wary vigil]

no

we get 1/0

dividing by 0

[dusty fjord]

is it becuase 0^-1/2 becomes 1/0

[wary vigil]

anything to the power of a minus

[dusty fjord]

flips

[wary vigil]

because 1/ that

becomes

[dusty fjord]

gotcha

so its undefined?

[wary vigil]

yeah

our derivative is undefined at x = 2

and x = 0?

[dusty fjord]

1/2(1/-2^1/2)

thats alot of 1/2’..

[wary vigil]

$\frac{1}{2}\frac{1}{\sqrt{0-2}}$

[warm shale]

Katharine

[wary vigil]

we get a square root of -2

does that exist?

[dusty fjord]

no that’d be an i?

[wary vigil]

that'd be complex

yeah

[dusty fjord]

i see!

[wary vigil]

so that's not possible either

how about x = 4?

[dusty fjord]

1/2(1/2^1/2)

is that 1/2?

[wary vigil]

$\frac{1}{2}\frac{1}{\sqrt{4-2}}$

[warm shale]

Katharine

[wary vigil]

not quite

its 1/2sqrt(2)

so that one does exist

[dusty fjord]

ahhh ok

should i simplify it more

or good enough

[wary vigil]

now we know that the derivative of the function f(x) = sqrt(x-2) doesn't exist at either x = 0 or x = 2

but it does at x = 4

thats a

that's what theyre asking for at a

what the results of filling in those values are

[dusty fjord]

Ohh i see

Thank u so much

do u know how to do the others too?

[wary vigil]

i'm not entirely sure if i do

at b

why can we say that the derivative of $\sqrt{x-2}$ is $\frac{1}{2}\frac{1}{\sqrt{x-2}}$

[warm shale]

Katharine

[wary vigil]

i think more specifically what they want to know is

why can you use powers

[dusty fjord]

would it be easier if its like this

[wary vigil]

instead of square roots

well they want you to explain why

it's $\frac{1}{2}(x-2)^{-\frac{1}{2}}$

[warm shale]

Katharine

[wary vigil]

this

is the derivative of sqrt(x-2)

[dusty fjord]

Oh ok gotchu

so which would be easier or better,

[dusty fjord]

the way u wrote it or this

[wary vigil]

i think the point of the question is why is the derivative of a square root

a negative fraction power

[dusty fjord]

Ohh ok

@wary vigil im back sorry about that

[wary vigil]

ok

[dusty fjord]

hmm

wait

remember when f(0)= a negative sqr root

so cant we say anything thats less than or = to f(2) will not exist?

@wary vigil i think its tryna ask how the function works

and we know that it didnt exist at 2 and 0

so if the number keeps going lower wont it not exist

[wary vigil]

the domain of sqrt(x-2)

is [2, infinity)

the domain for the derivative

is (2, infinity)

[dusty fjord]

yeah!

we can say that

that’d be correct them?

then

[wary vigil]

that's is correct

[dusty fjord]

https://tenor.com/view/tanjiro-kamado-kamado-tanjiro-demon-slayer-kimetsu-no-yaiba-gif-21599496

cool!

do you have time to do the others?

@wary vigil if u dont have time tho i understand

thank u so much🖤😭

[wary vigil]

oh i have a big screen i didn't see it

[dusty fjord]

no worries!!

can help with the rest?

ill resend

@wary vigil

[wary vigil]

you know the general equation for a line?

[dusty fjord]

not really🙁

[wary vigil]

y = ax + b

[dusty fjord]

i probably do but its bee a while

Ahhhh

[wary vigil]

or y = mx + b

[dusty fjord]

yes i do then!

so what i would do after

x would replaced 6?

[wary vigil]

a or m is the slope of the line

[dusty fjord]

gotchu

[wary vigil]

we need to find the line that is tangent to the function at x = 6

tangent means?

[dusty fjord]

the line of the slope of the curve?

[wary vigil]

yeah

a line that touches the curve

meaning it has the same slope as the curve

so we need to find what $\frac{1}{2}\frac{1}{\sqrt{x-2}}$ is at x = 6

[warm shale]

Katharine

[wary vigil]

to find the slope

[dusty fjord]

so

1/2(x-2)^-1/2 x=6. Would 6 replace the x then?

[wary vigil]

yes

[dusty fjord]

lemme solve it

[wary vigil]

^-1/2 though

[dusty fjord]

right!

i got 1/2

1/4*

[wary vigil]

yeah

so the slope of the line has to be 1/4

[dusty fjord]

How would i find the equatioxnwith that

equation with*

ah icsee

[wary vigil]

but to make it be a tangent

the line needs to be moved up or down

so that it touches sqrt(x-2)

at x = 6

[dusty fjord]

how would i do that?

ohh

y-y0

=m(x-x0)?

would we use that formula

or not?

[wary vigil]

not sure

how i would normally do it would be to do mx + b = sqrt(x-2) at x = 6

and then find b

[dusty fjord]

i see

[wary vigil]

m being 1/4

[dusty fjord]

how would i find b?

[wary vigil]

fill in x and m

and then solve for b

[dusty fjord]

m is 1/4

what would x be

y=mx+b

y=1/4x+b

[wary vigil]

x = 6

[dusty fjord]

i see thank you

so do we need to find y and b. Or do we already have y

@wary vigil

[wary vigil]

we have y because y = f(x)

[dusty fjord]

ohhh

so

sqroot x-2?

[wary vigil]

yeah at x = 6 in this case

[dusty fjord]

aqroot 6-2?

sqroot6-2

[wary vigil]

yeah that's y

[dusty fjord]

what about x?

x is 6 too?

i have this so far

[wary vigil]

x = 6 yes

and then find b

[dusty fjord]

-1/2

so 1/4x -1/2?

[wary vigil]

yeah

that's it

that's the line that touches sqrt(x-2) at x = 6

[dusty fjord]

so how should i writr it as

write

f(6)=1/4x-1/2?

or just f’(x)

ill just write it as y=1/4x-1/2 actually lmao

how would d be different?

[wary vigil]

d is the normal line

which is perpendicular to the tangent line

[dusty fjord]

can you remind me how to do that😞

like this,

?*

[wary vigil]

yeah

but there's an easier way to do it

if you just use y = 1/4x - 1/2

we know that the 1/4 has to be -1/(1/4)

because if m is the slope

then the new slope is -1/m

[dusty fjord]

oooo

so

lime this

like

which is -4 right

[wary vigil]

yes

[dusty fjord]

is -1/2 the same?

[wary vigil]

no

it's a different number

[dusty fjord]

oh

how would you get it,

[wary vigil]

you have to fill in x = 6 and y = 2

to find the new value

[dusty fjord]

oooo

makes sense

wait

why y=2?

[wary vigil]

that's f(2)

you calculated that before

[dusty fjord]

is it because of 1/2?

oh wait

ur right

lmaoo my bad

26?

-4x+26

is that correct

e’s an interesting one lmao,

[wary vigil]

that's correct

i have no clue about e

at all

[dusty fjord]

and finally e, do u know that as well

LMAO

gotcha

ill do the rest then

thank u so much Katharine😭

I really appreciate you sorry for all the bother!🖤

thank you again @wary vigil

[wary vigil]

wait

i made a mistake

it's y = 1/4x + 1/2

not -1/2

[dusty fjord]

AH

wait

oh no worries

i know thsg

that

ohhh

nvm

its postive how come?

why wouldnt it be negative? @wary vigil

3/2- 2 is -1/2?

[wary vigil]

wrong way around

it's 2 - 3/2

[dusty fjord]

yep

just realized

[wary vigil]

oof

[dusty fjord]

LMAO

[wary vigil]

i didn't see it

until late

[dusty fjord]

so we just need to change C?

haha its ok! Thank u so much for the heads up

D is good right?

[wary vigil]

yeah

because we didn't use y = 1/4x - 1/2

we used sqrt(x-2)

which means we got the right answer at d

even though we got the b wrong at c

[dusty fjord]

yeah!!

Thank u so much again!

i appreciate you

[wary vigil]

no problem

[dusty fjord]

.close

[obtuse pebble]

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[dusk nexus]

could anyone help me with this?

[dusk nexus]

ive tried

cos^2x=cos2x+sin^2x

but idk where to go from there

[spiral maple]

cos(2t)=2cos^2(t)-1

[dusk nexus]

oh wait

oops

i forgot about that

thanks

.close

[obtuse pebble]

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[frosty stag]

Yoo can someone help me and asnwer these questions

[idle thunder]

???

Very sketchy account

V = lwh

[frosty stag]

???

Ya so would #1 be this

[frosty stag]

Also How am I sketchy lol

[idle thunder]

Look at your username

[frosty stag]

i guess whatever

so

is what i said right

[idle thunder]

Looks like it

[frosty stag]

for the laundry detergent question

ok

how about the 2nd one that 1s kinda confusin

u there dawg?

[spiral maple]

what's the formula for surface area of a rectangular prism?

[frosty stag]

(wl + hl + hw) x 2

i think

[spiral maple]

yes.

So solve the problem.

oh it's triangular

[idle thunder]

@spiral maple It's a triangular prism

[frosty stag]

lmao

i thought i was trippin

[spiral maple]

anyway, same idea

just know the formula and apply it

or derive it if you must.

[obtuse pebble]

@frosty stag Has your question been resolved?

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[sweet hatch]

.closed

.ask

.open

how do I do this question

[sweet hatch]

i'm stuck on hwo to change the integration of ((3x√(x+1) + 2(√x+1)) all over 2(x+1) to get the integration of (x√(x+1)) all over x+1

this is my wokring out so far

<@&286206848099549185>

[obtuse pebble]

@sweet hatch Has your question been resolved?

[obtuse pebble]

@sweet hatch Has your question been resolved?

[sweet hatch]

<@&286206848099549185>

[obtuse pebble]

@sweet hatch Has your question been resolved?

[jolly ginkgo]

Perhaps they still want you to find integration of sqrt(x+1)

[sweet hatch]

no

ur suppose to make it in terms of the other

[jolly ginkgo]

Ik and I still think that needs to be done

Because it doesn't show up in that form

[sweet hatch]

no idea

;-;

[jolly ginkgo]

;-;

[obtuse pebble]

@sweet hatch Has your question been resolved?

[mild ocean]

@sweet hatch rewrite the integral in terms of u = x+1

[sweet hatch]

which integral

im confused can u pls show me the working out

[mild ocean]

$\int \frac{u-1}{\sqrt{u}} \dd{u}$

[warm shale]

quantum

[sweet hatch]

im confsued

[mild ocean]

what’s confusing you

[sweet hatch]

why did u do u-1

at the top

[mild ocean]

i did a u sub with x+1

so i could rewrite it as that

u = x+1, x = u-1

$\int \frac{x}{\sqrt{x+1}} \dd{x} \\ u = x+1 \ x = u-1 \\ \int \frac{u-1}{\sqrt{u}} \dd{u}$

unfortunate

[sweet hatch]

my brain go pop

[warm shale]

quantum

[mild ocean]

with this substitution, i was able to rewrite the integral

into this form

@sweet hatch have you covered u substitution yet?

it’s the middle of the night for me, so sorry if i don’t respond, i might have fallen back to sleep

[sweet hatch]

yes. same

sorry i took a nap

[mild ocean]

ok, i’m still awake @sweet hatch

sorry for the ping again, wanted to make sure i got to you

[sweet hatch]

allg

[mild ocean]

so are you still confused?

[sweet hatch]

nope

I spent some more time

looking at it

[mild ocean]

ok that’s good

do you have anymore questions

[sweet hatch]

yep

[mild ocean]

ok

[sweet hatch]

how do I find the area of the shaded

nvm

im done

tysm ❤️

.closed

[mild ocean]

it’s .close

also no problem

[sleek galleon]

Hi

[sweet hatch]

.close

[obtuse pebble]

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[native narwhal]

I need help

[native narwhal]

can u solve this question

[obsidian isle]

Just use the compound interest formula haha

@native narwhal what is the formula

[obtuse pebble]

@native narwhal Has your question been resolved?

[obtuse pebble]

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[glacial mesa]

did i do sth wrong here

[glacial mesa]

wait

that x

[native narwhal]

I am still looking about the formula but I have too many formula

[glacial mesa]

how tf

[native narwhal]

huh?

[glacial mesa]

nono I'm referring to my work, I accidentally integrated 1 to x^2

[native narwhal]

oh okay

[glacial mesa]

but I still didn't get the right answer even after fixing that

[native narwhal]

<@&286206848099549185>

can u help me

[native narwhal]

I hope you will did

[glacial mesa]

what formulas do u have

[native narwhal]

https://cdn.discordapp.com/attachments/907384558578712586/920975409754046464/C90B6758-ECA6-4BF8-B542-977039DD8F3C.jpg

I m not sure

the answer

guys will u help me

omg

[native narwhal]

.reopen

[visual lotus]

put it in https://www.integral-calculator.com/

it should work it out with steps better than someone can explain

[obtuse pebble]

@glacial mesa Has your question been resolved?

[obtuse pebble]

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[clear sinew]

Let Q be an orthogonal matrix $$Q \in \mathbb{R}^{n \times n} $$. Then $$ QQ=Q^2=I $$. Can someone explain to me why that is?

[warm shale]

mrbrown

[clear sinew]

I know that $$Q^T Q = I$$ and $$Q^{-1} = Q^T $$.

[warm shale]

mrbrown

[high quiver]

It's simply not true in general

(e.g. Take a rotation by 90 degrees)

[clear sinew]

ohhhhhhh

i read the excercise again, it was speaking of symmetrical matrices

Let Q be a symmetric orthogonal matrix. Ok, if Q is symmetric then Q^T = Q, otherwise it is not generally true. Thank you!

.close

[obtuse pebble]

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[austere torrent]

Question

[austere torrent]

Z^100=1

Under the condition that Imz >0

How many roots are there?

I've gotten many answers such as 49, 98,50 and my own personal one 0

Which answer is wrong and if all of tel are wrong which one is the right one

[high lily]

what have you treid so far

[austere torrent]

Nothing

But I think that since Imz has to be above 0

Then the real part can't be the only part left therefore

There cannot be a root

So what's the answer? And explanation?

[high lily]

Then the real part can't be the only part left therefore
wdym

[austere torrent]

Well if if Imz can't be 0

And has to be above 0

Then the real part won't be the only part left right?

[high lily]

Then the real part won't be the only part left right?
wdym

dunno what you mean by that

[austere torrent]

Idk how else to explain it lmao

Do you mean the (real part) part?

Real part = Rez

Maybe the translation was wrong

[high lily]

it feels like you're making a bad assumption that raising a complex number to certain powers won't get you a real value

e.g z = -0.5 + sqrt(3)/2 * i
is a complex number with Im(z) > 0
where z^3 = 1

you should read up on roots of unity and their representation on the argand diagram

[obtuse pebble]

@austere torrent Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[reef girder]

pls help

[obtuse pebble]

@reef girder Has your question been resolved?

[upbeat plinth]

.close

[obtuse pebble]

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[timid silo]

32/52

[high lily]

where's 32 coming from

[timid silo]

The cards

[high lily]

what cards are you looking at thats leading you to 32

you have an image of all 52 cards in a standard deck
can you use an editing tool of your choice and mark/highlight all cards you think are hearts or face cards

• not +

[high lily]

also just use a normal P

[timid silo]

Ohhhhh ok

[high lily]

fancy Ps are usually used for powerset

[obtuse pebble]

@timid silo Has your question been resolved?

[royal basin]

fancy usually means mathcal or mathscr

[obtuse pebble]

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[obtuse pebble]

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[vocal pond]

35 part b

[vocal pond]

Can someone help me with 35 part B

[mild ocean]

@vocal pond it’s just a weird version of the chain rule

[vocal pond]

I know

That’s what I get

[mild ocean]

the outside function is $g(u) = \int^{u}_2 \frac{1}{2t^3} \dd{t}$

then what’s confusing you?

[vocal pond]

I’m doing something wrong. It should be a 1 in the numerator

[warm shale]

quantum

[mild ocean]

oops

well i can’t really help, because you didn’t really show any work

because i don’t know what process you went through

[vocal pond]

Do you agree with the first thing I did

On the top line?

[mild ocean]

no

[vocal pond]

Why

[mild ocean]

because it’s incorrect

[vocal pond]

Oh it see

37 part B. Can someone remind me how to calculate a particles stopping point

[mild ocean]

pretty sure it’s when the derivative of the position function is 0

[vocal pond]

Ok thanks

Do I need to close this channel now that I’m done?

[mild ocean]

yeah

the command is .close

[vocal pond]

.close

[mild ocean]

gotta resend the message

[vocal pond]

.close

[obtuse pebble]

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[wheat dragon]

https://cdn.discordapp.com/attachments/910139039376621568/921038073687572550/unknown.png

[wheat dragon]

can someone kindly help me for (b) (taken from hubbard & hubbard)

[obtuse pebble]

@wheat dragon Has your question been resolved?

[wheat dragon]

<@&286206848099549185>

[obtuse pebble]

@wheat dragon Has your question been resolved?

[obtuse pebble]

@wheat dragon Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[cobalt locust]

I'm trying to find the solution for the following

[cobalt locust]

find the values of x, y and z

such that

x+y+z = 120

and

xy + xz + yz is maximal

[mild ocean]

don’t immediately ping helpers

[cobalt locust]

sorry

[obtuse pebble]

@cobalt locust Has your question been resolved?

[cobalt locust]

<@&286206848099549185>

[obtuse pebble]

@cobalt locust Has your question been resolved?

[obtuse pebble]

@cobalt locust Has your question been resolved?

[obtuse pebble]

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[timid silo]

I have the right answer, just need the steps

[brave bramble]

What's the slope of the line they gave you?

[timid silo]

You have to find the slope

Can u write the steps on paper so it's easier

@brave bramble

[brave bramble]

Sorry no. I can guide you to the correct solution, however!

[timid silo]

Ok

?

[brave bramble]

What's the slope of the line they gave you?

[timid silo]

That's the point, you find the slope

[brave bramble]

Good luck

[timid silo]

😔

[obtuse pebble]

@timid silo Has your question been resolved?

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[waxen fog]

can someone help me

[short spruce]

#❓how-to-get-help

does nobody know how to read

we need to change this to a literature server

[solemn geyser]

whoops sorry

[waxen fog]

😂

ok so

[obtuse pebble]

@waxen fog Has your question been resolved?

[obtuse pebble]

@waxen fog Has your question been resolved?

[mild ocean]

@waxen fog find out which of the choices has the square root of two numbers that are perfect squares

[waxen fog]

i alr solved it by my sef

self

lol

i got another question tho

[mild ocean]

then close the channel

oh

[mild ocean]

it’s not gonna be fun, but solve for x

[waxen fog]

bro i feel so dumb honestly

[mild ocean]

keep it as a fraction as well until you’re done

[waxen fog]

all my classmates said its easy

and solved it

im just sitting there

like a dumbo

:/

[mild ocean]

multiply both sides by 5/3

[waxen fog]

reciprocal?

[mild ocean]

do you understand why i say to do that?

[waxen fog]

to get the x by its self

lol

[mild ocean]

well yes, but do you understand the step

[waxen fog]

yea

i do

[mild ocean]

ok

[waxen fog]

so u flip 3/5 to 5/3?

why though

[mild ocean]

(x/y)*(y/x) = 1

there’s a 3/5 on the left

so i multiplied both sides by 5/3

[waxen fog]

oh ok

one se

sec

so do i do

3/5 x 3/5?

and then

1.04 x 3/5

?

[mild ocean]

no

do you understand why i multiplied both sides by 5/3

[waxen fog]

now i dont

im confused

wait do you mind if you can do it on paper

[mild ocean]

you see how the left is being multiplied by 3/5

[waxen fog]

and send a picture..?

[mild ocean]

no sorry my handwriting sucks

[waxen fog]

oof ok

[mild ocean]

also i don’t feel like it right now

[waxen fog]

okay

yea

x X 3/5

distribute

right?

[mild ocean]

and you under that (x/y)*(y/x) = 1, right?

[waxen fog]

huh?

[mild ocean]

because this just turns into xy/(xy)

this is just saying that any non zero number multiplied by its reciprocal is 1

basically (3/5)*(5/3) = 1

[waxen fog]

ohhhhh

ok

[mild ocean]

so that’s why i multiplied both sides by 5/3

[waxen fog]

so multiply 5/3 on both sides?

ok

[mild ocean]

to get rid of the multiplication on the left side

ok

[waxen fog]

and then what

so now thats gotten rid of

1.04 x 5/3

?

[mild ocean]

so now we have $x = \frac{5.2}{3}-\frac{4}{3}$

[waxen fog]

yep

so we multiply

?

[mild ocean]

it would be helpful to simplify the right side, but that’s not super necessary

actually whatever

[waxen fog]

-4/3 on both sides?

[mild ocean]

it’s 5.2

[waxen fog]

yea ik the answer, my friends told me

but im studying for a test

and wanna know how to get the answer

[mild ocean]

ok

yeah

[waxen fog]

,w 5/3(1.04)

[warm shale]

Results provided by WolframAlpha

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

Requested by nakruf#3318

[waxen fog]

waaa

[mild ocean]

ugh i’m dumb

[waxen fog]

so howd u get 5.2???

[warm shale]

quantum

[waxen fog]

im dumber man...

[mild ocean]

1.04*5 = 5.2

we had 1.04*(5/3)

so i just simplified it

[waxen fog]

ok

can u help me with one more thing?

[mild ocean]

yeah

[waxen fog]

he didnt give us question examples for this

he just said this

so how do i study for it??

[mild ocean]

is that statistics

or something

[waxen fog]

no

algebra 1

lmao

im in 8th grade pre regents

[mild ocean]

i saw data and had flashbacks

anyways

i recognize linear regression but i’ll look it up really quick anyways

[waxen fog]

okay

[mild ocean]

apparently it is statistics, so i can’t help you, because i won’t know what i’m doing

unless you just want help on finding the equation of a line

because i could just be having a small brain moment

[waxen fog]

nah man thanks for all your help

:)

[mild ocean]

no problem

whenever you’re done, do .close

[waxen fog]

.close

[obtuse pebble]

Channel closed

Closed by @waxen fog

Use .reopen if this was a mistake.

[waxen fog]

have a good day

[mild ocean]

thanks, you too

[obtuse pebble]

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[obtuse pebble]

@gloomy carbon Has your question been resolved?

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[cosmic kiln]

Ahem

[cosmic kiln]

Im trying to calculate my final grade to see if i have to turn in this assignment or not

and im confused with the math

If I get a 100 on a assignment worth 25 weight then another 100 on a assignment weighted 10 then a 0 on a assignment weighted 25. While my current final grade is a 64 would that bring my grade up to at least a 70

and If it does reach 70 would 2 60's with a weight of 35 bring it down below a 70 or bring it up still

[royal basin]

what's the weight of the final?

[cosmic kiln]

What do you mean?

[royal basin]

er

so let me try to understand what you have here

Assignment 1: weight 25, grade 100
Assignment 2: weight 10, grade 100
Assignment 3: weight 25, grade 0
Final: weight ???, grade ???

is this correct?

[cosmic kiln]

yes

[royal basin]

so what's the weight of the final?

[cosmic kiln]

the final what?

Their is no final test or anything

[glossy yacht]

Final exam?

[royal basin]

the final test? the other thing not yet graded? whatever you call it

...

[glossy yacht]

Then there's no final

[cosmic kiln]

The other thing not graded is the 3rd assignment

with the 0

[glossy yacht]

tbf went they said "final grade" they meant "final grade" and not "grade of the final exam"

the final wouldn't have a weight, it'd just be sum of weight * grade

[royal basin]

oh wait

there's a difference between 0 and ungraded just saying

[cosmic kiln]

Its not getting done

because im not doing it