help-10

[neat mist]

oh ok

i think have it

thank you

if you want to input on any of the other problems that would be cool

otherwise i am good

[coral hollow]

cool

i don't know enough about this course to make any meaningful input though

[neat mist]

oh ok

.close

[obtuse pebble]

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[paper sparrow]

For wich λ does that set produce a base of R^3

[timid silo]

wow language barrier

[paper sparrow]

?

[timid silo]

its probably easy

[paper sparrow]

I keep getting it wrong

[timid silo]

but i dont get the language

[paper sparrow]

Yeah, I translated

[median dome]

They translated the question..

[iron edge]

[timid silo]

oh

[iron edge]

All good

[paper sparrow]

det(A) must not be 0

[brave bramble]

You'll collect them into a matrix and row reduce

At some point you might have to divide by an unknown. Remember you can't divide by 0

[paper sparrow]

by row reduce you mean the gauss method

[brave bramble]

Or, if you feel more comfortable taking the determinant, that works too

Just see when the det is 0

[paper sparrow]

I did, but I got other results

I got λ must not be 0, -1 and 1

but the answers in the paper says otherwise

Is it a typo in the paper or am I wrong

[brave bramble]

,w matrix {{x,1,0},{1,x,1},{0,1,x}}

[warm shale]

Results provided by WolframAlpha

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

Requested by kaynex

[paper sparrow]

yup, i am wrong

well let me show you my work if you can find my mistake

<@&286206848099549185>

[obtuse pebble]

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[obtuse pebble]

@paper sparrow Has your question been resolved?

[paper sparrow]

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[brazen cedar]

can anyone help me to solve this prob. The answer is 125, however I tried to used Trapezoidal formula and my answer was 120 which is wrong

[obtuse pebble]

@brazen cedar Has your question been resolved?

[brazen cedar]

<@&286206848099549185>

[silent saddle]

I'm not sure but I think the answer will be the area under the curve, all you need to do is graph the data you have and calculate the area bounded by the curve and the time axis this will most likely be your answer or really really close to it

[brazen cedar]

ahh i know how to solve it

i forgot to sum with v0

v(t) = v(0) + intergral(0 -> t) a(x)dx

btw tysm

.close

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[humble aspen]

please help me I need to write a project about equations. I have already written about square, biquadratic and linear. I'm in 8th grade

[sage dagger]

too broad

[humble aspen]

what else can you write about, methods, etc.

[zenith raft]

yes we need a more specific doubt

@fathom flicker

[sage dagger]

yes maybe sir austin can help specificate your doubt

[obtuse pebble]

@humble aspen Has your question been resolved?

[obtuse pebble]

@humble aspen Has your question been resolved?

[kind hawk]

what are you expecting here

"project about equations" is about as unspecific as you can get

[obtuse pebble]

@humble aspen Has your question been resolved?

[winter cargo]

hes asking what to write for his project about equations

and yeah true

[obtuse pebble]

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[lost widget]

If we have N numbers a1,a2,…,an for every i<j with ai<aj we can erase from the sequence one of this two elements. Describe an efficient method to see if we can remain with only one element or not.

[lost widget]

<@&286206848099549185>

[obtuse pebble]

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[pastel wren]

(-2r)^4=(-2)^4*r^4

(-2)^4=16

you can see as well that (-2r)^4= -2r* -2r* -2r* -2r =16r^4

beacuse the exponent is odd

[obtuse pebble]

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[timid silo]

Hi anyone know how to find the area

Also PT = QT we prove early on

[strong aspen]

hkdse?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Sorry?

if ur referring to the exam then no

[mighty geyser]

try checking if you can find area of quadrilateral in terms of the diagonals

[timid silo]

Hm like breaking it down into two triangles or something?

[mighty geyser]

,w area of quadrilateral given diagonals

[mighty geyser]

breaking it into the 4 triangles

[timid silo]

so ptq
trs
stq
rtp

area of pqt : trs = 1:4

like do i continue like this or?

[obtuse pebble]

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[mighty geyser]

yeah you can continue

[obtuse pebble]

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[mighty geyser]

you'll have to realise more ratios of areas from some ratios of lengths

[mighty geyser]

.close

[obtuse pebble]

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[timid silo]

.reopen

.reopen

[timid silo]

Hey could u see this and tell me if this is like going correct

[mighty geyser]

.reopen

[obtuse pebble]

✅

[mighty geyser]

area of TQR is 0

[timid silo]

Oh mb

Wrote that wrong

[mighty geyser]

probably just a small mistake

you have some ratio of lengths, you can also get the area of PTR

[timid silo]

meant STQ

[timid silo]

i dont understand how

Like half sides sides sin theta?

[mighty geyser]

you can do that too

[timid silo]

but wont that give me value in terms od x

And also i dont know theta

[mighty geyser]

you can try it to see if it cancels out neatly

Another way of thinking about ratio of areas is via the ratio of the bases when the height is the same

[timid silo]

Oo

So for like

PRS

TR is common height for PTR and RTS

yasss i think i did it

[mighty geyser]

not TR, but it's the altitude from R onto PS

[timid silo]

Could u confirm

[mighty geyser]

you might want to use the 1/2 sin a b formula

otherwise it would be a conceptual error

answer is still okay

[timid silo]

So common angle instead?

[mighty geyser]

the angle is supplementary

[timid silo]

Oh ok

Alright thanks

imma close the channel

.close

[mighty geyser]

.close

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[copper owl]

Help

[opaque mesa]

I don't understand this

wdym he finds 98 minus the result he needs to find

[copper owl]

The answer is 5 but i don't know the process

Think about 5

5³= 125

Student takes (5-2)³

3³=27

125-27=98

But i dont know how can i find 5 with process

[opaque mesa]

make an equation

let this number be x

so the number 2 less than it would be x-2

[copper owl]

Yepp

So

(x-2)³=x³-98

Right?

[opaque mesa]

yeah

[copper owl]

How can i find 5 now

[opaque mesa]

simplify this

$(x-2)^3 = x^3 - 6x^2 + 12x - 8$

[warm shale]

General_Jacob

[copper owl]

=x³-98

Then -6x²+12x =-90

Right?

[opaque mesa]

yeah

[copper owl]

And now?

[opaque mesa]

-6x² + 12x + 90 = 0

do you know the quadratic formula?

[copper owl]

Nope

[opaque mesa]

$ax^2 + bx + c = 0$

[warm shale]

General_Jacob

[opaque mesa]

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[warm shale]

General_Jacob

[copper owl]

What is that

[opaque mesa]

quadratic formula

[opaque mesa]

this is a quadratic equation

[copper owl]

So how can we find 5

[opaque mesa]

-6x² + 12x + 90 = 0
your a is -6,
your b is 12,

and your c is 90

and now you substitute

[copper owl]

I can't

[opaque mesa]

why?

[copper owl]

±?

[opaque mesa]

• or -

[copper owl]

İ know

But + or -

[opaque mesa]

$2 \pm 1 = 2 - 1 or 2 + 1$

[copper owl]

Which one

[warm shale]

General_Jacob

[opaque mesa]

• or - means that there are two solutions

both are correct

[copper owl]

My teacher said that question is easy for 13 y.o student bruh what is that process

Nvm thx

[opaque mesa]

@copper owl

[copper owl]

Thx @opaque mesa ❤️

[obtuse pebble]

@copper owl Has your question been resolved?

[obtuse pebble]

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[meager palm]

for this question can someone show me the second part of the hint

[meager palm]

like im not sure what that is used for

I used (f o g) [A] = F[g[A]] to show symmetry holds but im not really understanding: for any f, g in L f o g in L

[tired shell]

it's just asking if a(cx+d)+b is a linear function

[obtuse pebble]

@meager palm Has your question been resolved?

[meager palm]

yes?

isnt that just proving it is invertable

that is g?

[tired shell]

umm I think the invertible part isn't the major part it's mostly just about the definition ax+b with a!=0

[meager palm]

sorry could you elaborate then, im not sure im completely understandong

[tired shell]

L is defined as ax+b with a!=0, so fog in L means that f(g(x)) can be written as ax+b, which you can show by plugging g(x) into f(x) and multiplying it out

[obtuse pebble]

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[shadow lava]

ummm, I found this image on Google

is it nonsense?

afaik "down" is correct but to find magnitude || u x v || you just multiply both magnitudes together?

[teal turret]

yo avid u didnt respond to my dm bro : (

[shadow lava]

it's not me, i don't play Fortnite

[shadow lava]

also i don't know why that game is so popular, or how it got to be so popular, but that's for another discussion

.close

[obtuse pebble]

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[timid silo]

Hi super quick one, what's this called please?

[timid silo]

inverse expontent or what sorry?

[queen briar]

negative exponent

[timid silo]

thank you! 😄 I was googling inverse etc.. thanks 🙏

.close

[obtuse pebble]

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[wary vigil]

negative exponent is a type of inverse, the multiplicative inverse

[obtuse pebble]

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[tough wraith]

Can someone please tell me how this is inconclusive? I am so confused

[obtuse pebble]

@tough wraith Has your question been resolved?

[wary vigil]

the sum of 1/n does not converge

which means that even though the cos(n)/n is always smaller than or equal to that it says nothing about whether it converges or not

[tough wraith]

but how am I supposed to know if it converges or diverges? The integral test works well in conjunction with this from what ive noticed, but without it, how am I supposed to know?

[wary vigil]

you can't

not if the function you're comparing to doesn't converge

if instead it was less than or equal to 1/n^2

then you would be able to say that the sum converges

because the sum of 1/n^2 converges

[tough wraith]

yeah but then how am I supposed to pick a answer if I don't know if the other series converges or diverges, given I know the simpler of the two converges or diverges?

[wary vigil]

i'm not sure i understand your question

[tough wraith]

I know one of them diverges. I need to know if the other converges or diverges. But how?

in this case, the "other" is cos(n)/n

[wary vigil]

some other test

maybe comparing it to some function you know converges

or some other convergence test

[tough wraith]

wait what is the comparison test trying to find? It's trying to find if the simpler series is applicable to compare to the more complex series in how they diverge and converge right?

[wary vigil]

it's a way of showing convergence or divergence by comparison to a known sum

often a simpler one

[tough wraith]

what does it mean then if it's conclusive?

does it mean that the simpler sum isn't a good comparison to determine convergence or divergence?

[wary vigil]

in a way yes

in this case

since it is smaller than or equal to a divergent series

it is inconclusive

since it could be convergent or divergent

either is still possible

if you compared a different series to 1/n

and it turns out the terms of that series

say n

is always bigger than 1/n for n > 1

then you KNOW for a fact that the sum of n from 1 to infinity diverges

[obtuse pebble]

@tough wraith Has your question been resolved?

[obtuse pebble]

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[half carbon]

Hey, I have this problem, but I'm just a bit confused on finding the critical points.

[half carbon]

I'm at the point where I have this:

[wary badger]

so you differentiated yes

[warm shale]

Someone

[half carbon]

Yes, that's correct

[wary badger]

did you check the end points

they want absolute max/min on the interval so you have to include the endpoints

[half carbon]

not yet, just trying to first find the critical points from the differentiation

[wary badger]

ok

so derivative should be 12sin^2(x)cosx +6cosxsinx

factoring out 6cosxsinx

6cosxsinx(2sinx+1)

yea looks good

so what’s the trouble

[half carbon]

$12sin^2xcosx+3sin(2x)$ is what I got for the derivative, yeah

[warm shale]

Someone

[wary badger]

you set them equal to zero

[half carbon]

So I set that to 0 and I'm at the point of

[warm shale]

Someone

[wary badger]

and you’re stuck?

[half carbon]

yes

[wary badger]

well do you know when sinx=0

where on the unit circle

[half carbon]

pi/2?

[wary badger]

no that’s when sinx=1

[spice chasm]

thats when cosine is 0

[wary badger]

so

for unit circle

[half carbon]

let me pull up the unit circle

one second hahaha

[wary badger]

sine corresponds to the height

[half carbon]

oh, so just 1?

[wary badger]

and cosine corresponds to the base or horizontal distance

so what angle makes no height

[half carbon]

oh my lord wait

[wary badger]

on the unit circle

[half carbon]

0 hahaha

[spice chasm]

pi n?

[wary badger]

yes

[half carbon]

sin is related to y, right

[wary badger]

mhm

so 0 is one point

[half carbon]

then pi/2 is another

[wary badger]

there are 2 more in the interval

[half carbon]

oh?

[spice chasm]

no

at pi/2 sine is 1

[wary badger]

no pi/2 is essentially on the y axis

which has a height of 1

so there is height

[half carbon]

oh, sorry i was moving on the cosx=0, there's more than 1 critical point at sinx=0?

[wary badger]

but there is no horizontal length

so cos would be zero there

[wary badger]

yes

[half carbon]

would it be pi?

[wary badger]

because it rotates around

[wary badger]

pi and 2pi

2pi is the same as zero

[half carbon]

right

should i include 2pi in this question or nah

[wary badger]

yes because it’s in the interval

[half carbon]

got it

[wary badger]

it’s just after one rotation

or revolution

whatever you want to call it

[half carbon]

so then cosx=0 is pi/2 and 3pi/2?

[wary badger]

yes

now for the last one

[half carbon]

it would be sinx = -1/2

which is

[wary badger]

yup

which quadrants

[half carbon]

3 and 4?

[wary badger]

yes

[half carbon]

7pi/6, 11pi/6

[wary badger]

and what angle makes sinx=1/2

yea

so there’s 7

[half carbon]

sooo now, i do the thingy with all 7 of those critical points + the interval, right?

plug the critical point into original function blah blah blah

and find max/min

[wary badger]

ugh this is evil lol yea

[half carbon]

ahhhh busy work, gotta love it

[wary badger]

always the trig equations

have to check 9 values of x lol

[half carbon]

luckily 2 are duplicates

😂

just to make sure, I do

[warm shale]

Someone

[wary badger]

yes

[half carbon]

whew

[warm shale]

Someone

[half carbon]

Should I put

@wary badger?

[wary badger]

sorry

yes

[warm shale]

Someone

[half carbon]

got it, thank you!

[wary badger]

and plugging in 2pi will give the same thing as 0

but make sure you write that if it were a test

[half carbon]

will do, thank you again :D

[wary badger]

you’re welcome

[half carbon]

.solved

[obtuse pebble]

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[dusty spoke]

i have a question
Know that
m²=n+4
n²=m+4

Without using calculator
What is the answer for
7mn-m³-n³

[iron edge]

!status

[obtuse pebble]

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

[dusty spoke]

1

[merry peak]

substituting, we get n^4-8n^2-16=n+4

work from there

(got that by substituting m from the second equation into the first equation)

[dusty spoke]

But it is too much need for calculate and there is only one answer I can submit

[merry peak]

is no using calculator part of the question?

thats stupid

[dusty spoke]

Yep

It like quadratic formula that alpha and beta and calculate alpha×beta²

[merry peak]

sorry mistyped

wait

[dusty spoke]

I tried for 4 hours without any progress

[merry peak]

lemme rephrase it

[merry peak]

4 hours

holy fuck

it would be faster to try all values for n

at that point

[dusty spoke]

The answer for n and m is not integer lol

[merry peak]

oh righ

t

well in questions like this, the format of the answer is often trying to tell you something

you want something to simplify to 7mn -m^3-n^3=(some number)

[dusty spoke]

Those are the progress I have made

[viscid gull]

got the sol I think

Start off by subtracting the first equation by the second

u get m^2 - n^2 = -(m - n)

Problem doesn't specify if m = n or not so I guess assume it doesn't

then m + n = 0

Then compute (m +n)^2 to find mn and (m + n)^3 to find m^3 + n^3

[dusty spoke]

Seems true I will try it Hemmmmmm

[iron edge]

Solved

[misty fractal]

m^2-n^2 = n-m
Therefore
(n-m)(n+m) = n-m

If n = m,
7mn - m^3-n^3
= 7n^2-2n^3

Otherwise, we can divide by n-m to get:
n + m = 1
Which means
7mn - (m^3+n^3)
= 7(1-n)n - ( (1-n)^3 +n^3)
= -10 n^2 + 10 n - 1

[viscid gull]

it's easier to not solve for n/m imo

that said idk how else you'd do it for the n = m case but

if this is like a comp math problem

Then u can just do wishful thinking when they don't specify this type of thing

[iron edge]

I don’t think so

I guess it’s his assignment

Tho

[dusty spoke]

True I take it for practice

[iron edge]

nvm XD

[viscid gull]

Yeha seemed pretty comp mathy to me lol

[dusty spoke]

The ans is -11

I can't get it

[iron edge]

The daddy has given you the solution

Just follow and you’ll get through it

[viscid gull]

Hmmm

is n rational in the not equals case?

[dusty spoke]

How can I use those equation to get the ans -11 ,sorry I don't understand xd

[viscid gull]

cus mn and m^3 +n^3 are

So if n isn't rational then that's why I was thinking solving for mn and m^3 + n^3 is easier

[iron edge]

Quadratic function

[dusty spoke]

It's not equal to 0 huh?

[viscid gull]

but yeah I messed up m + n = 1

Still should be rational

[iron edge]

Since m=n
Set m=n=k
m²=n+4, n²=m+4 —> k^2=k+4

[viscid gull]

but then roots are irrational and that's messy

[iron edge]

True, it’s better not solve the root

[dusty spoke]

Wow thanks for that

[viscid gull]

hmm

[dusty spoke]

I still can't calculate without calculator hahah

[viscid gull]

For m = n I get irrational answers and for m not equals n I get -15

ok so for the second case

I do like (m + n)^2 = 1^2 = 1 = m^2 + n^2 + 2mn

Adding the two equations you have m^2 + n^2 = m + n + 8

= 1 +8 = 9

So 2mn = -8, then mn = -4

theen

(m + n)^3 = m^3 + n^3 + 3mn(m + n)

So 1 = m^3 + n^3 + 3(-4)(1)

then 13 = m^3 + n^3

so actually it's 7(-4) - 13 = -41 which is even more far off

Maybe if it were m^3 + n^3 + 6mn the answer would be - 11?

[dusty spoke]

I also get this ans....

[viscid gull]

hmmm

[dusty spoke]

How dare ..

[viscid gull]

Maybe it's the answer sheet that's wrong then

[dusty spoke]

But calculator don't show -41 is a solution

[viscid gull]

It should tho

I think hmm

-10n^2 + 10n - 1 includes - 41 in the range

n^2 - n - 4 = 0

[dusty spoke]

Yep the 12.3152 is one solution but another is not

[viscid gull]

,w n^2 - n - 4 roots

[warm shale]

Results provided by WolframAlpha

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

Requested by 992qqoloy

[viscid gull]

,calc 1/2(1 + sqrt(17))

[warm shale]

Result:

2.5615528128088

[viscid gull]

, calc (2.56155)^2

[warm shale]

Result:

6.5615384025

[viscid gull]

,calc 1/2(1 - sqrt(17))

[warm shale]

Result:

-1.5615528128088

[viscid gull]

ok so then m = n?

,calc 7*(2.56155)^2 - 2*(2.56155)^3

[warm shale]

Result:

12.315351427652

[dusty spoke]

4 solutions well

[viscid gull]

oh wait

m + n = -1

Then mn = -3

[dusty spoke]

Damnnnnn ohh

[viscid gull]

then m^3 + n^3 = -10

So 7mn = - 21, +10 for -m^3 - n^3

-11

[dusty spoke]

True!!

Thank you very much

.close

[obtuse pebble]

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[humble abyss]

I'm trying to grasp the proof for the undecideability of Turing Machines.

[humble abyss]

During the proof for the undecideability of Turing Machines, is it necessary to take as input the turing machine itself? What is the point of that during the proof.

For example, for the step where we say D(<M>) = { H<M, <M>> == accept -> reject, reject otherwise }. Why not have H<M, w> for some arbitrary w.

[merry peak]

It’s so that we can prove that neither of the cases could be true

[humble abyss]

Just for reference, here's the proof I'm looking at, Micheal Sipser page 207

[merry peak]

Bc if it were true, it would be false, but if it were false, it would be true

[humble abyss]

hold on before I say something stupid lemme try to see your point

[merry peak]

Basically, we could phrase it another way

We could say define C as the negation of M

And do C(C, <C>)

But its easier and more compact to define the function like this

[humble abyss]

I'm starting to get it by seeing the problems that come when we don't have <C> in there

actually nvm I still don't fully get it

is it a notation compactness thing or is it fundemental to the proof

like lemme see

[merry peak]

Notation compactness

The proof with C is equivalent im pretty sure

The important thing is plugging in c to itself

[humble abyss]

nah the thing I"m confused about is a bit different I think

[merry peak]

Why we plug D into itself?

[humble abyss]

say I have

H(<M, w>) = {
  accept if M accepts on input w
  reject if M rejects on input w
}

Then D would be defined as follows

DEFINE W = 01234903 (SOME CONSTANT)
D(<M>) = {
  accept if M rejects w
  reject if M accepts w
}

Then

D(<D>) = {
  accept if D rejects w
  reject if D accepts w
}

Ok by the time I reached this point writing my argument I realized the flaw

the recursive nature only appears if the input to D was <D> to begin with

D(<D>) = {
  accept if D rejects <D>
  reject if D accepts <D>
}

this part

[merry peak]

Yes

[humble abyss]

I see it nowww

took me some time but thanks for the hints

[obtuse pebble]

@humble abyss Has your question been resolved?

Channel closed

Closed by @humble abyss

Use .reopen if this was a mistake.

[misty fractal]

Well yes if $m = n$ then we can use the equation $n^2=n+4$ to compute the values of n for $7n^2-2n^3$

Regarding the other case (m not equal to n, therefore $m+n = -1$) yes it was -1, not 1
We could perform symmetric reduction, noticing that:
$7mn - (m^3+n^3)
= 7mn + 3mn(m+n) - (m+n)^3$

So
$F(m,n) = 7mn - 3 mn + 1
= 4 mn + 1
= 4(-1-n)n+1
= -4n-4n^2+1
= -4(n^2+n-3) - 11$

In the last step, I performed a polynomial long division of
$-4n-4n^2+1$ by $(n^2+n-3)$ which I know makes n equal to zero.
In conclusion $$F(m,n) = -11$$

[obtuse pebble]

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[warm shale]

Daddy_314

[misty fractal]

@dusty spoke this is a detailed solution in case you still need it

[merry peak]

His channel is still open

Anyway close this

[merry peak]

.close

[wary badger]

do this

.close

[obtuse pebble]

Channel closed

Closed by @wary badger

Use .reopen if this was a mistake.

[misty fractal]

Dont make fun of the elderly

Who cant use discord properly

[wary badger]

[merry peak]

I was telling him to close mate

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[willow swan]

How is this not (-infinity, infinity)

[merry peak]

Well what’s f’(w)

[wary badger]

undefined at w=0

[willow swan]

I got 49/10

[wary badger]

huh

what

[willow swan]

for the direivtice

[merry peak]

f’(w)

[merry peak]

Uh

[wary badger]

how could it be a number?

[merry peak]

Show tour work

[wary badger]

it’s a function

[willow swan]

It was just a hard guess

I suck at math lmap

[wary badger]

💀

[merry peak]

Do you know what a derivative is

[willow swan]

Shows change

[merry peak]

Ok

so f’(w) is

[willow swan]

How do I find it?

Do I need to plug numbers in?

[timid silo]

first find it then well see

what will the derivative be?

[willow swan]

Thats what im trying to find

[merry peak]

Do you know the power rule

[willow swan]

−
90
w
11
+
4

[timid silo]

so you are having problem finding the derivative?

[willow swan]

Yes

[merry peak]

?

[timid silo]

then i think you should study your textbook

[timid silo]

rather than asking qn here

[willow swan]

That is what I got

[merry peak]

????

Do you know the power rule

[timid silo]

its not possible to correct basics here bro

[willow swan]

Im 3 weeks away from never need to do this bs again so I am just fighting through

I dont know why you feel the need to be a dick

[timid silo]

just spend 2 mins

and open the book

and read

[merry peak]

Do you know the power rule

[willow swan]

It shows you how to differentiate correct?

[merry peak]

In what way

[willow swan]

as is like n=n*x?

[merry peak]

How have you gone almost a semester through a calculus course without knowing the power rule

[willow swan]

I dont know lmao. I have an A and I am playing school my man

I just figure it out on the exams

[merry peak]

You’re playing yourself mate

[willow swan]

I dont know how to explain it lmao

Im not at all

[merry peak]

You dont learn shit

[willow swan]

Final math course of my life

dont need this shit whatsoever

And I have an A in the course

[merry peak]

Good luck with your art major man

Gl

[willow swan]

I dont know why you feel the need to be so hostile

I was just trying to get some help

Have fun being a slave to Honda buddy

.close

[obtuse pebble]

Channel closed

Closed by @willow swan

Use .reopen if this was a mistake.

[merry peak]

Slave to honda?

What?

[willow swan]

Dude its over

We are gonna go our seperate ways here

I hope you find something else to do besides belittling people

[merry peak]

I help people here in my free time

[willow swan]

I know and I appreciate that

[merry peak]

It aint my fault you played yourself and managed to not learn anything in your class in a few months

[willow swan]

Look, man. Some classes matter to me and ones that don't. This is a Gen Ed to me. I am a marketing major. I do not need to find f(x) in the real world ever

I understand it is different for everyone.

And like I said

I have an A in the course

[merry peak]

Ok serious question

How did you manage to get an A without learning anything

Thats like a delicate balancing game

[willow swan]

There is alot of home work and essays. I am very good writer. Its a weird ass math class

[merry peak]

Essays?

Wait hold on

Essays

[willow swan]

Swear to god

[merry peak]

Lmfao

[willow swan]

Its hilarious

[merry peak]

“Heres my 2 page essay about the social implications of derivatives on our society”

[willow swan]

Pretty much.

Shit is dumb

[merry peak]

Lmfao

[willow swan]

College Algebra is all you need for Business

Its a dumb concept in the first place to make anyone outside of STEM take it

[merry peak]

You are forced to take calculus?

That’s dumv

[willow swan]

Correct

That is Ohio State for you

I appreciate your help

Have a good night my man

[merry peak]

Gn man

[willow swan]

.close

[merry peak]

Good luck

[willow swan]

You as well

[merry peak]

Sorry for the mean stuff i said

[willow swan]

Youre good and youre not wrong at all lmao

If i had to take more of this I would be fucked

So keep doing what you are doing

. close

.close

[obtuse pebble]

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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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[cobalt beacon]

Very stuck on this problem. Have no work that’s valid

[cobalt beacon]

Pls help!

[brazen viper]

So this area can be thought of as the area of two circular segments

To get the area of a circular segment you can use the area of a sector minus the area of a triangle

@cobalt beacon let me know if you need more help

The rest should be trigonometry

[cobalt beacon]

I might need more tbh

Gotta jog my memory abt this again

[brazen viper]

Ok, well, give it a try with this approach, and let me know what you get stuck on

[cobalt beacon]

Which part you talking about when you say “area of the sector”?

[brazen viper]

Pizza slice

[cobalt beacon]

Oh so everything in that circle that isn’t green, expect for the triangle thing

Or am I way off

[brazen viper]

Yeah the wedge

Including the green

https://en.wikipedia.org/wiki/Circular_sector

[cobalt beacon]

Wait so to find the circular segment area I have to find all the white space except for the circular sector and subtract that by the circular sector?

Wait no that wrong

Hold up

[cobalt beacon]

Now I’m really messed up

[gusty gyro]

If you don't mind ask im curious how find the area for the green part?( cant we like the area of SR and take of it the area of the Triangle🥲? )

[gusty gyro]

I mean this.

[brazen viper]

So you can find the area of the segment by finding the area of the sector and subtracting the area of the triangle

[brazen viper]

If you look closely, your solution is two segments, one for each circle, joined at their flat side

[cobalt beacon]

I’m sorry I think I sort of understand, but I need to see some calculations or at least what to do with them because I don’t fully understand

[brazen viper]

So the area of a circle sector is θ/2 r^2

With θ in radians

Do you see why/how?

[cobalt beacon]

So I take what I get from this equation and subtract what exactly?

[brazen viper]

The area of the triangle.

A = 1/2 bh

b = 2 * (b/2)

b/2 = r sin(θ/2)

h = r cos(θ/2)

1/2 bh = 1/2 (2 sin(θ/2) cos(θ/2) r^2)
= 1/2 sin(θ) r^2

[cobalt beacon]

But that triangle includes the thing in green right?

[brazen viper]

No

That's just the isoceles triangle

With height given by "d" in the diagram from Wikipedia

[cobalt beacon]

Man I’m so confused I need something visual or I’m not gone get this

[brazen viper]

Sorry, normally I'd draw it out, but I'm on my phone on the couch and

[cobalt beacon]

Lmaooo

[brazen viper]

My cat is sleeping ssh

One sec

[cobalt beacon]

No problem

[brazen viper]

So the triangle is triangle APB, and AOB

We can drop an altitude to C so the height is the length of PC

And the base is the length AB

AB is twice AC

And APC is a right triangle

So we can do trigonometry on it

That's where the sine and cosine with angle θ came from

We want the circle segments these are joined together on by the green line AB

And the one from the red circle is green and red

The one from the purple circle is green and purple

[gusty gyro]

So to find area of circular segment?
(θ/2 r²π) - (dc/2)?

[brazen viper]

Yes, but you can simplify this to just in terms of the angle θ

And the radius

(θ - sin(θ))/2 * r^2

[brazen viper]

Oh, my mistake no pi in that

[cobalt beacon]

What is theta supposed to be?

[brazen viper]

The area of a circle segment that we developed

[cobalt beacon]

What do I do to calculate that tho?

[brazen viper]

From here #help-10 message

And here #help-10 message

You need the radius of the circle and the angle

And that's it

You'll use it twice, once for each circle

Then add the results

[cobalt beacon]

Dude I don’t understand

Idk how ur finding the angle

[brazen viper]

So let's talk about that now then

What information do you have?

I see

[gusty gyro]

(just a curious student :>)

[cobalt beacon]

The smaller circle has radius of 6.18 cm, and larger circle has radius 14.6 cm. The distance between the centers of the circles is 18.22 cm

[brazen viper]

So, you'll need to use the law of cosines to get the angles

You have the side lengths of the triangle PAO and PBO

You can use either triangle to get θ/2

For either circle

[cobalt beacon]

I still don’t get how to find the angle using law of cosines with just radius’s

[brazen viper]

Apply it to triangle PAO

We have a^2 + b^2 - 2ab cos C = c^2 right?

Solve for C

[cobalt beacon]

Yeah

Tf is a and b tho

[brazen viper]

a and b are the sides adjacent to the angle C and c is the side opposite it

So if we're trying to find angle APO then a is PA and b is PO while c is AO

(alternatively you can swap a and b, doesn't matter)

[cobalt beacon]

Look man I’m sure you’re explaining it well, but I have no idea what is happening. I don’t have time to sit here all night on this one problem, and I don’t think I’ll understand it until someone solves it, and I can learn from their work instead of being told what to understand about it. I’m going to close this thread if you are not comfortable with solving it, and I would understand why.

[brazen viper]

I'm not going to solve the problem because it is literally against the rules for me to.

!nosols

[obtuse pebble]

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

[cobalt beacon]

Okay then I’m closing it

.close

[brazen viper]

Ciao

[obtuse pebble]

Channel closed

Closed by @cobalt beacon

Use .reopen if this was a mistake.

[obtuse pebble]

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Send your question here to claim the channel.

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[lilac pebble]

dont understand how he got that unit circle

how is 1.88 that angle

and how is it even possible too

1.88 is almost a full revolution

how is it that small

im so confused

[cinder basalt]

1.88 rad are 107 deg

the picture shouldn't matter either, you can do it purely algebraically, the picture is a visual aid

[red ice]

That's not true: 1.88 is less than pi which is around 3.14

pi/2 is around 1.57

So it's closer to pi/2 than pi, which is what we would expect given the x-coordinate is only -4/13

[lilac pebble]

wait

isnt 1.88 = 1.88π

[red ice]

It could be even smaller like -1/2 or -1

[red ice]

No ofc not

[lilac pebble]

then how do i even convert 1.88 to degreees

huh

[red ice]

Good question

pi radians is 180 degrees

So first do 1.88 rad = 1.88/pi * pi rad

Then that's 1.88/pi * 180 degrees

[lilac pebble]

huh

why

[red ice]

Which part are you confused by?

[lilac pebble]

oh nvm

i got 107.72 deg

[red ice]

Yeah that's correct

[lilac pebble]

but how is the angle on the outside though

shouldnt it be inside the right triangle

ik its obtuse so it cant be inside

but why would it be like that

[red ice]

Unit circle definition: cos x is the x-coordinate, sin x is the y-coordinate

[lilac pebble]

how does that help

[lilac pebble]

isnt it a angle

[red ice]

Yeah, so if cos x is negative, the x-coordinate of the point must also be negative

Which would give you quadrants II and III

[lilac pebble]

wait wtf

isnt -cosx

always in quadrant 1 or 2

[red ice]

-cos x could be anything

Cause cos x can also be negative

So -(negative) = positive

-cos x could be both negative and positive and 0

[lilac pebble]

oh nvm i confused it with arccos

[red ice]

Yeah

[lilac pebble]

so like

if theres ever a triangle in quadrant 2

the angle is always that angle from the triangle to the positive x aaxxis

[red ice]

Yes

It's always the angle starting at the positive x axis

And going counter-clockwise

Btw if you draw cos x and reflect it across y = x, that's arccos x

We choose 0 to pi for the domain of cos x (and range of arccos x) cause it's the most convenient

[lilac pebble]

wait but arccos(-1/2) is at the second quadrant and the angle is 2π/3

[red ice]

Imagine drawing a vertical line x = -1/2 and seeing where it intersects the unit circle

Yeah so one of the intersections is in the 2nd quadrant

The other intersection is in the 3rd quadrant

[lilac pebble]

but why isnt it π/2 then

because thats the angle from the triangle to the positive x axis

[red ice]

That's about right, cause -4/13 is about -1/2
And 120 degrees is around 107 degrees

[red ice]

It's not the angle from the hypotenuse

You need the angle to the hypotenuse from the x-axis

[lilac pebble]

so wouldnt this diagram be incorrect

because that angle isnt 2π/3 but 2π/3 - π/2

[red ice]

Flip your triangle so that the base is on the negative x-axis

The correct triangle should complete the other half of the rectangle

[lilac pebble]

so this is wrong?

[red ice]

Yes

[lilac pebble]

huh but i copied it frmo a video

[red ice]

You found a bad video then

RIP

[lilac pebble]

but then the angle inside is still not 2π/3

2π/3 is the other half of the angle frmo 0-π

so the correct angle in the triangle is 1.047??

π - 2π/3

[red ice]

The angle is drawn like that

[lilac pebble]

wtf

huh

and that applies to al triagles in quadrant 2

?

[red ice]

Yes

And this applies to all angles in quadrant 3

For all 4 quadrants, you start from the positive x-axis, and then draw an arc counter-clockwise until you hit the hypotenuse

[red ice]

What you're doing here is converting it to a principal angle

Which is really nice and how you should proceed

But that's not the original angle

Is what I'm saying

[lilac pebble]

ok

thanks

[red ice]

Npnp

[lilac pebble]

.close

[obtuse pebble]

Channel closed

Closed by @lilac pebble

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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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[reef gorge]

can someone help me out with 3b?

im not exactly sure how to start

[hollow island]

for 3b, try disecting it into a rectangle and triangle

[reef gorge]

ok

so i got a rectangle and two triangles now

what then?

[hollow island]

you should get a rectangle and one triangle (not two triangles)

[turbid scaffold]

try 1 rectangle and 1 triangle

[reef gorge]

how?

[reef gorge]

OOOH

like that

[hollow island]

yes

[reef gorge]

but then

ok

so the top side is

7.5m

and the left side is

1.8

can i just do cosine rule?

in order to find the area

[red ice]

Don't overthink it

[reef gorge]

oh wait

it has a 90 degree angle

so the area of the rectangle is 6.75

and 47.46 of the rectangle

corect?

then its 54.2cm^2

alright, thanks!

.close

[obtuse pebble]

Channel closed

Closed by @reef gorge

Use .reopen if this was a mistake.

[turbid scaffold]

what values

did

you use

[reef gorge]

.reopen

[obtuse pebble]

✅

[reef gorge]

wait

area of the triangle

is 6.75

meters

and the rectangle is 47.46m

[turbid scaffold]

uh

what did you use

[reef gorge]

wdym?

[turbid scaffold]

for length and width

of rectangle

[reef gorge]

11.3 for width

[turbid scaffold]

yes

[turbid scaffold]

ohh\

[reef gorge]

4.2 for length

[turbid scaffold]

ok

u switched them

i got confused

ok yeah no issue

[reef gorge]

thanks for the help!

.close

[obtuse pebble]

Channel closed

Closed by @reef gorge

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[haughty talon]

where do i start

[timid silo]

find dbc

n acb

using ab

[haughty talon]

dbc is 60?

[timid silo]

no

its b/2

[haughty talon]

yea it is

[timid silo]

now find acb

[haughty talon]

using ab?

[timid silo]

use a

[haughty talon]

im confused

[timid silo]

u see aob and abc?

the triangle

[haughty talon]

which a?

A or a

[haughty talon]

how can i find the angle?

[timid silo]

same ting

its a/2

you have to find bca

[haughty talon]

lemme see

which angle is a/2?

[timid silo]

bca

[haughty talon]

ok

im still stuck

[timid silo]

u have bca and dbc

but 90 + bca + dbc = 180

[haughty talon]

ok ty

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[halcyon sparrow]

If I have the equation of a circle and the equation of a line going through it, if I input one into the other do I get the points of the circumference

[halcyon sparrow]

But that line doesn’t go through the centre

[inland matrix]

Yeah

[surreal forge]

you get the intersection of the circle and the line

that is to say, two points that lie on the circumference of the circle

[halcyon sparrow]

thank you

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[heavy bobcat]

stuck on this.

[heavy bobcat]

Need help

[inland matrix]

!status

[obtuse pebble]

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

[heavy bobcat]

I dont know where to begin

With different values of height 3.75 and width X 5.5m the answer is 11.8, but i dontk know where to start nor how to get to that value

<@&286206848099549185>

Anyone?

[timid silo]

is the answer 14.14?

@heavy bobcat

[obtuse pebble]

@heavy bobcat Has your question been resolved?

[heavy bobcat]

sorry for the late response, i do not know what the answer is but if u could tell me how you got it i can try and work it out

[heavy bobcat]

.

[timid silo]

F along y component = F sin theta

[heavy bobcat]

The answer to the other question with different values was 11.8, there values were height 3.75 and width of C 5.5 m

if you want to try use those values and try get 11.8

[timid silo]

same diagram?

[heavy bobcat]

yup

[timid silo]

what was the load

[heavy bobcat]

20kN

[obtuse pebble]

Channel closed

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[heavy bobcat]

same exact diagram but just different values

.reopen

[obtuse pebble]

✅

[timid silo]

ok

[heavy bobcat]

hey akira are you free for a call to help me ?

[heavy bobcat]

.

[timid silo]

i don't do calls

[heavy bobcat]

cool, could you tell me how you got 14.14

and ill double check using the other values to get 11.8

[timid silo]

ok

from this,F along y component = F sin theta , F= 20kN and sin theta is 45 degrees which is 1/sqrt(2)

it may be wrong, so let me check with the answer u provided

[heavy bobcat]

ok

<@&286206848099549185>

[timid silo]

@heavy bobcat hey, i am sorry , i have to go now

also, at the bottom, the two bases are different, what does it mean?

[heavy bobcat]

the bases are the same