help-42

[pseudo spear]

Sorry, I made numerous mistakes in the old channel and I was confusing people.

[dull wagon]

try subbing again
you made a few errors with parentheses,
forgot /2 for one of the terms
and some sign errors

actually, before subbing you didn't distribute that - sign properly

[pseudo spear]

Was my indefinite integral correct?

[dull wagon]

wdym

[pseudo spear]

Before the substitute

[dull wagon]

no

[pseudo spear]

oh

[dull wagon]

you didn't distribute that - sign properly

that - is applied to each term inside, not just the x^2/2

[pseudo spear]

ohhhh

So it should be 3x -x^3/6 + 3x^2/2 + 1/2x over the evaluated interval of -1 to 7?

[dull wagon]

yes

[pseudo spear]

So that becomes 3(7- (-1) - (7^3 - (-1)^3) + 3(7^2 - (-1)^2)/2 + 1/2(7-(-1))

[dull wagon]

missing a )

and a /6

3(7- (-1)) - (7^3 - (-1)^3)/6 + 3(7^2 - (-1)^2)/2 + 1/2(7-(-1))

[pseudo spear]

Great, that’s what I have written down, I just mistyped it. Tysm!!!

[calm coral]

@pseudo spear Has your question been resolved?

[calm coral]

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[pseudo spear]

The answer key has a different answer, can someone help me check for mistakes? I ended up with 432 and the answer key says 204.8.

[olive lion]

What is your area function?

[dull wagon]

you forgot the ^2 on the x for the base

[pseudo spear]

Thanks

[olive lion]

and i would recommend instead of the brackets

make it 2(-x²/9 +4)²

instead of the first line

but that is also preference

[pseudo spear]

Do I get the same answer either way?

[olive lion]

ye it is the same

as what you put

(-x²/9+4)(2)(-x²/9+4) = 2(-x²/9+4)²

but your only mistake is the ^2 i think ye

[pseudo spear]

Hmm

[pseudo spear]

It’s still not correct, and I don’t know why 😭

[calm coral]

@pseudo spear Has your question been resolved?

[calm coral]

@pseudo spear Has your question been resolved?

[calm coral]

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[indigo grove]

!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

[mint vigil]

1

I understand what a function is but I don't really understand what I'm being asked to do

[indigo grove]

you know what is domain and range ?

[mint vigil]

Yes

Is it the y value?

[indigo grove]

yep the set of y values of that the function gives

you know parabolas ?

or quadratic graphs

[mint vigil]

Yep

Do I need to make a graph?

[indigo grove]

not really but it'll make the question alot eaiser to understand

so a parabola with a positive number before the x^2 points upwards always
so it'll have a minimum point at it's vertex

[mint vigil]

So it's a quadratic?

[indigo grove]

yea

[mint vigil]

oh

[indigo grove]

and the vertex's y value is the start point of the range

so to find the vertex of the quadratic
the x value can be found with x=-b/2a
and to find the y value of the vertex you plug the value you found (for x) in the original quadratic equation

[mint vigil]

Ah okay

So the range is essentially just the vertex?

[indigo grove]

the range starts from the vertex
and continue increasing

[mint vigil]

So the range would be everything greater than 2?

[indigo grove]

you did a mistake not sure where tho

[mint vigil]

negative 2?

[indigo grove]

should be negative 4
not sure where you went wrong ?

[mint vigil]

I divided 4/2(1)

?

[indigo grove]

that's not the b value

ax^2+bx+c
the b value is 0 (so the x terms never shows)

[mint vigil]

oh okay

Thank you

[indigo grove]

and if you want you can plot the function to check you'll see the y values never go lower than -4

[mint vigil]

Okay

.close

[calm coral]

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[calm coral]

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[strange crown]

a little confused as to how the 1-1+sin^2x simplifies to 2sin^2x

[subtle compass]

It doesn't

the mistake is further up

[strange crown]

oo okay

[subtle compass]

$\cos{2x}=1-2\sin^2{x}$

[potent lotus]

Jay

[strange crown]

oh.....

thank you I see it now, I spent so much time confused near the bottom

[subtle compass]

So, the second line should be $\frac{2\sin{x}\cos{x}}{1-\left(1-2\sin^2{x}\right)}$

[potent lotus]

Jay

[strange crown]

yeah I kept looking back at the cos2x identities

and not seeing it

yeah and then in the end the 2's cancel out

anyways cheers

.close

[calm coral]

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[calm coral]

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[ancient grotto]

im bored

[ivory pilot]

I agree, math for fun right?

what do you think is the answer

[ancient grotto]

yeap

no clue

(im lowkey too lazy to do this lol)

[old falcon]

lazy

[ancient grotto]

lmao

ok i am being stupid

[ivory pilot]

if we considered that the question didn't even mention the upper part what do you think the value of sin^2 + cos^2

[ancient grotto]

i realized what it is

ok let me post smth else

smth that im actually struggling with

i cannot find

there

[ivory pilot]

hmm what did you try with that

[ancient grotto]

this was like late at night yesterday

im revisiting it

[ivory pilot]

maybe taking a common factor would allow you to see the answer

did you get it?

[ancient grotto]

n!+(n+1)!=n!(n+2)?

ill try it

[ivory pilot]

yeah

[ivory pilot]

i think that you would be able to see the answer just by looking, substitute n = 19 and you will see what would be the remainder

[ancient grotto]

2?

bc of 19!*21

[ivory pilot]

no
ok let's look at something easier
what is the remainder of dividing 6 by 2

[ancient grotto]

0

$6\equiv 0\mod 2$

[potent lotus]

;(

[ivory pilot]

yeah,

can you see why
because 6 can be written as 2 * 3
2 here cancels with the 2 on the denominator
keeping all the integers on the nominator the same

applying the same logic with the original question can you see what would be the answer

remember that 19! = 19 * 18 * 17 * 16 * ... * 2 * 1

[ancient grotto]

i gotta eat ill brb

[ivory pilot]

Ok
by the way the answer would be 0 if you thought about it carefully you would see why

[ancient grotto]

i got it as 1

[calm coral]

@ancient grotto Has your question been resolved?

[ancient grotto]

hmm

[calm coral]

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[calm coral]

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[river frost]

Hello, I'm confused about how they integrated sin(theta) d(sin(theta)). I don't think I fully understand what d(sin(theta)) actually means?

[swift dragon]

$d(\sin(\theta))=\cos(\theta)d\theta$

[potent lotus]

Bonk

[glad parrot]

They integrated wrt sin(theta)

[swift dragon]

the derivative of sin(theta) is cos(theta)dtheta

[glad parrot]

Unfotunately no

Thats not what they meant

[swift dragon]

they turned the cos(theta)sin(theta)dtheta into sin(theta)d(sin(theta)), no?

[river frost]

Ohhh awesome I didn't know you were allowed to do that

[glad parrot]

They integrated sin(theta) wrt sin(theta) so its like sin²/2 (same as x dx)

[river frost]

How come the limits are still in terms of theta though?

[pseudo wedge]

yeah this is what they did

they did a u-substitution in a complicated way

[river frost]

Oh I thought it was like if you replace sin(theta) with x and integrate x wrt x then you'd have (1/2)x^2 si

*so

[river frost]

Yeah that was earlier though not the part I'm confused about

[glad parrot]

Yes

[river frost]

Then how come the limits are still in terms of theta?

[pseudo wedge]

they can be in terms of sin(theta) as well

[glad parrot]

It could have been between 0 and 1 for sintheta

[pseudo wedge]

$\int_{\theta = 0}^{\theta = \pi/2}\sin\theta\dd{(\sin\theta)}=\int_{u=0}^{u=1}u\dd{u}$

[potent lotus]

kheerii

[river frost]

Ohhh I was wondering why they felt the need to specify the tersm for those limits but not any of the other limits

Awesome, thank you so much!

.close

[calm coral]

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[glad parrot]

Yes in that case, but here it was asking for what do they do after, i should have a better wording in my answer

Soz

[swift dragon]

they were confused about the d(sin(theta))

and where it came from

[river frost]

No I wasn't haha

I was confused about the line after :P

[glad parrot]

"How they integrated"

[swift dragon]

ah i see now

[river frost]

No worries

[swift dragon]

you did say that you didnt fully understand what d(sin(theta)) meant

[glad parrot]

Soon green i bet

Under 3days

[swift dragon]

so i interpreted that as how they got to there

[river frost]

That is true but I get the stuff you said

[swift dragon]

nah i still need 2k messages i think

s?me

[river frost]

I mean what is a d by itself with anything under it anyway

[swift dragon]

oh, bots dont work here, nvm

[river frost]

But the example of treating it like x dx helped so I get that now too

[swift dragon]

yup

[river frost]

.close

[swift dragon]

its already closed

but we just decided to keep talking 😄

[glad parrot]

Bot ain't boss here

[swift dragon]

3 days is rly fast, im just barely halfway in over a week

[drifting seal]

yea i’m not so sure how it works but i think 3 days is too quick

probably more like a couple more weeks

[swift dragon]

just one more week is enough

[drifting seal]

i’ve only seen you for a day or two

[swift dragon]

ive been here for over a week

so you havent been looking ig

[calm coral]

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[topaz raft]

how can i go about starting this?

[swift dragon]

!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

[topaz raft]

1

how do i start

[ancient grotto]

it’s a direct application

denote F to be the antiderivative of f

we are given that $\lim_{x\to\infty} F(x)-F(-x)$ exists

[potent lotus]

;(

[topaz raft]

yes

[ancient grotto]

the integral given is just $\lim_{t\to\infty} F(h(t))-F(g(t))$

[potent lotus]

;(

[ancient grotto]

can you finish it now?

[topaz raft]

am i supposed to show this algebraically?

[ancient grotto]

you can use the composition rule here

$\lim_{x\to a}f(g(x))=f(\lim_{x\to a} g(x))$

[potent lotus]

;(

[topaz raft]

o i dont think ive seen that one for limits before

[calm coral]

@topaz raft Has your question been resolved?

[calm coral]

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[exotic cosmos]

is this thales?

[ancient grotto]

i dont know what thales is but ill try to answer this to the best of my ability

note that there are 2 similar triangles

i will draw them out

[exotic cosmos]

what

[exotic cosmos]

we don't know that

[ancient grotto]

if thats the case there is too little information to determine it

because those lines need to be parallel

[leaden thunder]

@exotic cosmos did you leave out that any of the lines are parallel from your problem?

[exotic cosmos]

okay fine yeah i think they might be parallel

[ancient grotto]

what was the original question

[leaden thunder]

this is how you indicate them on a drawing

[exotic cosmos]

that's not my drawing 😭

[exotic cosmos]

basically i was told that ? = 2x

by splitter theorem

i'm trying to reason out why

okay yeah they're parallel lines

[ancient grotto]

hes providing an example

[exotic cosmos]

i know, and i'm saying that's not my drawing

[leaden thunder]

then you should give the full context however this is communicated with you

[ancient grotto]

this is all the information i can get from the drawing

[leaden thunder]

whether in a drawing or with words

[exotic cosmos]

all those lines are parallel

the ones in the drawing

the vertical ones

and the horizontal ones

[ancient grotto]

alright

[ancient grotto]

hmm

okay, can you imagine flipping the diagram onto the 24 base?

you can apply some transversal theorems there

[exotic cosmos]

i solved it

[exotic cosmos]

what 😭 does this mean

[tranquil wasp]

what was your solution

[exotic cosmos]

i deleted

hmm wait

[ancient grotto]

transvervals

uh

[exotic cosmos]

so ? = 2x

[tranquil wasp]

yeah i got that too

[exotic cosmos]

okay anyway thanks

[ancient grotto]

oh, you misread my diagram.

[exotic cosmos]

.close

[calm coral]

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[ancient grotto]

@exotic cosmos its 24 and 4, not 2y and y.

...

[fickle musk]

I think she wrote 4

[ancient grotto]

im confused now

was it 24 and 4 or 2y and y

[calm coral]

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[remote mural]

Now suppose that....

Could someone explain it please

[leaden thunder]

what is "it"

[remote mural]

Here is axiom 3

[remote mural]

The proposition

[leaden thunder]

do you know the meaning of disjoint

[remote mural]

Yeah

A intersection B = phi

[leaden thunder]

which part of the proposition or the proof confuses you first?

[remote mural]

this doesnt make any sense

the last part

[gusty walrus]

h old ru mr?

[leaden thunder]

this one?

[remote mural]

17 why do you ask ?

[gusty walrus]

idk

bc

i didnt study those yet

[remote mural]

yeah

[gusty walrus]

yay excitn

[remote mural]

they both can be equal only if k is infiinity ?

[swift laurel]

well we can split the sum as from 1 to k, then from k+1 to infinity. but each term in the second sum is 0

recall in the paragraph above that all events for i > k are null events

[remote mural]

ohh

lol dumb me

thanks

.close

[calm coral]

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[empty moss]

Help how do I start

[vital blade]

take rootx common out from the denominator

[empty moss]

How does that work

[vital blade]

like this

[empty moss]

Oh

[vital blade]

then think of a substitution

[empty moss]

How did you know to do that

I need the intuition

[vital blade]

instincts

lmao jk practice

i solved many questions like these

do uk the next step?

[empty moss]

Yea

[vital blade]

the next step will help u understand

[empty moss]

U substitution

[vital blade]

what wud u substitute?

[empty moss]

Du/dx

U= root x+1

[vital blade]

yeah

[empty moss]

What happens if u did root x = u

[vital blade]

so then when u do this, the other term also combines without any trouble

[empty moss]

Yea it cancels

[empty moss]

Goat

[vital blade]

u can still do it

u can convert it into a form of partial fraction

but that wud be lenghty

[empty moss]

Yea

Okay thanks man

See u around

.close

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[calm coral]

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[tepid wedge]

What are you stuck on?

[swift dragon]

!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

[delicate quest]

1😭

I think I have an idea of like solving y using the 2 equations then plugging the value of y in y=x²-4x to see if x gives me real roots or not

I just donno where to begin from to find the value of y

[tepid wedge]

Find y^2 and sub that in first, then y

That should hopefully reduce the thing to a quadratic in y

[delicate quest]

Where should I sub the y²

y²=x⁴-8x³+16x²

[swift dragon]

how would oyu be able to fill that into the second equation?

[tepid wedge]

Subtract 2 times the right side from the quartic and add 2y^2

[delicate quest]

I have -45x²+180x+63+2y=0

[tepid wedge]

not quite
$2x^4 - 16x^3 + 77x^2 - 180x + 63 = 2(x^4 - 8x^3 + 16x^2) + 45x^2 - 180x + 63 = 2y^2 + 45x^2 - 180x + 63$

[potent lotus]

LooseEthics

[tepid wedge]

now do the same with y

[delicate quest]

Lemme see

Where did that equation come from

[tepid wedge]

That's just what's left over after subtracting the 2y^2

77x^2 = 45x^2 + 2*16x^2

[calm coral]

@delicate quest Has your question been resolved?

[delicate quest]

Ok

[delicate quest]

I have this now

@tepid wedge

[calm coral]

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[reef gale]

If the 4-digit number a072 leaves a remainder of 1 when divided by 3, and is less than 6000, what are the possible values of a ?

[fiery crater]

if a072 leaves a remainder of 1 when divided by 3 then a072-1 = a071 is divisible by 3

and if a number is divisible by 3, then the sum of its digits is also divisible by 3

so check for all possible values of a between 1 and 5 where the sum of digits of a071 is divisible by 3

[main musk]

hint: if you sum up the digits of a number and take the remainder when divided by 3 of that sum, that is equal to the remainder of the original number when divided by 3

[reef gale]

Ah so simple

How do you guys come up with the solution?

I literally can't think

[worn depot]

same but i fix this sometimes by asking myself why and how when reading a question

[calm coral]

@reef gale Has your question been resolved?

[reef gale]

Note

[calm coral]

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[boreal cloud]

what have you tried

[slim cove]

subsitute 3 + x^1/7

[boreal cloud]

okay then

[slim cove]

u = 1/7x^-6/7

[boreal cloud]

yup

[slim cove]

but then its negative..

[boreal cloud]

what does a negative exponent mean

[slim cove]

you move it to denominator??

[boreal cloud]

yup

[slim cove]

ohh

wait

multiply both sides by 7

ohh i get it

if u dont mind

i also need help with this one

😔

[slim cove]

ik it says to subsitute 5 + sqrt2x

but i watched like

a vid with an example

and i got kind of confused bc they manipulated the numbers around

so u = 5/2sqrt5x??

[boreal cloud]

after substituting 5 + sqrt 2x

what do u get

[slim cove]

5/2sqrt5x

[boreal cloud]

are u sure

[slim cove]

1/2x^1/2

[boreal cloud]

okay now put that back in

whats the integral now

[slim cove]

wait mb isnt it

1/sqrtx

[boreal cloud]

it should be 2/sqrt2x

[boreal cloud]

you cant cancel the 2 because its under the ^1/2

[next wren]

why is there the 1/2 in the first place

inner derivative should be in the numerator

[boreal cloud]

thats from power rule

but yes you need to multiply by derivative of 2x

[next wren]

oh my bad

[boreal cloud]

yup

[slim cove]

hmm

so

we need to bring it over to other side?

[boreal cloud]

so du = 1/sqrt(2x) dx

[slim cove]

uhuh

[boreal cloud]

try to get that in terms of u

[slim cove]

ill try 😭

[boreal cloud]

,ccw

,rotate

[slim cove]

Wait hold on

I made a mistake

Okay

I'm not confident 😭

[boreal cloud]

uh

$u = 5 + \sqrt{2x}$

[potent lotus]

ashy!

[boreal cloud]

so what is sqty 2x

[slim cove]

huh

u - 5 = sqrt2x

(u-5)^2 = 2x

(u-5)^2/2 = x ??

😭

is that waht u mean

[boreal cloud]

you want to remove all instances of x

currently you have $\sqrt{2x} du = dx$

[potent lotus]

ashy!

[boreal cloud]

so you have to get sqrt 2x in terms of u

[slim cove]

huh

so dont we use

u = 5 + sqrt2x

and then find x??

[slim cove]

and then we replace it here

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[raven crest]

Can someone please help me on this question

[raven crest]

I know that if the last three digits of the number is divisible by 8, the entire number itself is also divisible by 8

I also know that 664 is the biggest possible last three digits of the 8 digit number

[sand sky]

What are other combinations for the last three digits ?

Figure that out and apply permutations on the first 5

[ancient grotto]

alcumus?

just wondering

[raven crest]

it's challenge homework

[sand sky]

The last digit is 2,4,6.
100 mod 8 gives 4 and
200 mod 8 gives 0
...
10 mod 8 gives 2
20 mod 8 gives 4
30 mod 8 gives 6
40 mod 8 is 0
...

we need the sum of modulo to be 8, so that the no. Is divisible by 8

Now start by fixing the last digit

Okk?

[raven crest]

okay

[sand sky]

The question surely needs some calculation I don't see a direct approach!

[raven crest]

can't we just list out the possible last three digits and go on from there?

i don't think there should be that much

[sand sky]

Listing out without any methodology, is like brute forcing, although my approach is quite the same but easy to visualise

[raven crest]

ok

[sand sky]

See like

Let me give you an example

Let's fix the 100th digit to an even multiple and the last digit to be 2

So the middle one would be 3,7

so you see it is easier to figure out

0,2,4,6,8 = 5 ways , middle 2 ways so total 5*2 = 10 ways !

[raven crest]

oh i see

[sand sky]

The rest is just calculations

After you figure out the ways it's divisible by 8

You can permute the first 5

6^5

So probability=

6^5 *(combinations divisible by 8)/6^8

[raven crest]

I see what you mean, but 372 is not divisible by 8?

[sand sky]

I said 100th is even multiple

0,2,4,6,8

072, 272,472,672,872

[raven crest]

oh ok

[sand sky]

You need to fix the 100th to odd

[sand sky]

If you do so the modulo gives 4

So the 10th place must yield modulo of 8-4-2

[sand sky]

Try and do that, make a table like this to make it simpler to observe

[flat gorge]

4a + 2b + c = 40, 32, 24, 16, 8
4a + 2b = 38(2), 36(4), 34(6), 30(2), 28(4), 26(6), 22(2), 20(4), 18(6), 14(2), 12(4), 10(6), 6(2)
2a + b = 19(2), 18(4), 17(6), 15(2), 14(4), 13(6), 11(2), 10(4), 9(6), 7(2), 6(4), 5(6), 3(2)

(•) denotes the value of c for that number

Clearly 19 is not achievable. The rest can be counted without much hassle

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[copper notch]

how to determine absolute convergence of this series?
[\sum_{n=1}^{\infty}\left(\sin\left(\frac{1}{n}\right)-\tan\left(\frac{1}{n}\right)\right)]

[potent lotus]

Slowaq

[copper notch]

i tried comparing it with $\sin^2\left(\frac{1}{n}\right)$ but i can not figure out how to show that $\sin\left(\frac{1}{n}\right)-\tan\left(\frac{1}{n}\right)\leq \sin^2\left(\frac{1}{n}\right)$

[potent lotus]

Slowaq

[copper notch]

@lyric ravine

[lofty jolt]

hi

teache me how to use texit

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[copper notch]

<@&286206848099549185>

[worthy adder]

sin(x) ~ x when x tends to 0

You could refactor by sin(1/n)

[copper notch]

wdym refactor?

[worthy adder]

And to prove this it is sufficient to show that X( 1 - 1/sqrt(1-X²)) ≤ X² for X in -1,1
but it only holds for 1 ≥ X ≥ -0.83 so this is the wrong way I believe

[worthy adder]

Your general term is in fact
Sin(1/n)(1-1/cos(1/n))

Now one can use Sin(1/n) ~ 1/n when n gets to +infty

[copper notch]

ah alright yes i aproiximated it with taylor and compared it with 1/n^3 which yields 1/2 thus my limit is absolutely converging thanks!

.closeä

.close

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[finite oasis]

Consider the following transformations
\begin{align*}
x: \R &\to \C \[1ex]
T: (\R \to \C) &\to (\R \to \C) \[1ex]
\Theta : \R &\to \R
\end{align*}
Consider the specific mapping $\Theta: t \mapsto kt$ for $k\in\R$. Is there a special name for the following property: [
T(x\circ \Theta)(t) = T(x)(\Theta(t))
]

[potent lotus]

Aero

[finite oasis]

(reopening for further insight from others)

[placid heath]

does T transform a transformation into another one?

[finite oasis]

yes

it indeed does

[unique jackal]

I don't know of a name for this property; in fact, I've never seen it come up before in any setting

[exotic cosmos]

[unique jackal]

is there somewhere where it comes from?

[finite oasis]

Well, it should be something resembling time-invariance, but it is invariant under scaling rather than time shifts

change it to Theta: t -> t + k and that should be time invariant

[sick lark]

this is related to time-scale equivariance or scale invariance. These properties often arise in systems or operators invariant under specific transformations

[unique jackal]

I see. unfortunately, I don't know of a name for this property

[finite oasis]

Unfortunate

I'd think it would be pretty useful

[unique jackal]

there's a chance that you might find an answer if you ask in #math-discussion, me thinks

you can leave this open and also ask there

[sick lark]

What's the question?

[unique jackal]

for anybody viewing the channel now, here's the question

[sick lark]

A transformation T is equivariant if applying a transformation (e.g., scaling via Θ) before or after T yields the same result. Here, T "commutes" with the transformation Θ

So ye this property is a form of covariance under transformations or equivariance.

@finite oasis

[calm coral]

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[shy estuary]

For what values of a and b is the matrix diagonalizable?

[shy estuary]

For eigenvalues λ=1 (double) and λ=2 (also double)

So the algebraic multiplicities are am(1)=2 and am(2)=2 respectively. So the geometric multiplicities should be 2 and 2 for the matrix to be diagonalizable

Substituting λ=1 and bringing the system to matrix form (to find the bases of eigenspace in order to find the geometric multiplicities for λ=1) we get this

How do I solve this system and come into conclusion from this point onwards?

(For the eigenvalue 1)

[calm coral]

@shy estuary Has your question been resolved?

[shy estuary]

<@&286206848099549185> please 🙏 help

[undone leaf]

Hell nawh

No jk

Oh eigenvalues ?

Cool

Ok so

Just give me your matrice

[shy estuary]

.

[undone leaf]

Uh

What the fuck

Ok

I have a paper with all my shit on it

Wait a sec

Bc my dumbass dont remember the characteristic polynomial of a 4 matrice

Ok so

Imma call my friend

@craggy crater

Dumbass come here please

You remember the characteristic polynomial of a 4*4 matrice ?

[craggy crater]

lol dumbass

whats up

[undone leaf]

Im good

[craggy crater]

A-lambdaI

[undone leaf]

Yeah

[craggy crater]

det(A-lambdaI) = 0

solve for lambda

[undone leaf]

@shy estuary

You listen ?

[undone leaf]

Helping guys ( and insulting them also ) is too much fun tf

[craggy crater]

1 - lambda 0 0 0
a 1 - lambda 0 0
3 5 2-lambda 0
4 6 b 2-lambda

take determinant set to 0

[undone leaf]

Bah it’s not (A-lambda)*vector=0

[craggy crater]

its different forms

[undone leaf]

Oh yeah ok

Wanna help other guys ?

[craggy crater]

sure

[undone leaf]

Cool

Go for the one just behind it

Banana

[shy estuary]

You two can't be serious

[undone leaf]

We do are

[craggy crater]

?

[craggy crater]

.

[undone leaf]

I think he needs a bit more help than that

[shy estuary]

Read my whole description of the problem

[craggy crater]

gauss-jordan elim

[shy estuary]

Can't... too complicated because we have 2 parameters a and b

Isn't there an easier solution?

[craggy crater]

not that i can think of on the top of my head

[undone leaf]

Im cooking your answer dawg

That’s all was in my head

Idk if it works for u tho

@craggy crater you think it’s correct?

[undone leaf]

@craggy crater ?

[calm coral]

@shy estuary Has your question been resolved?

[craggy crater]

thats right

[shy estuary]

Thanks, I'll look into it and I'll reply later!

[calm coral]

@shy estuary Has your question been resolved?

[shy estuary]

Very sexy

Thanks bud

.close

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[vital blade]

idk how to start

my dumbass brain tried doing something with log but nothing came

[leaden thunder]

Did you do any algebraic manipulations

Exponent rules etc

[vital blade]

i cud only proceed 1 step

by taking 16 = 2^4

[potent igloo]

That's a start

[vital blade]

after that/?

[flat gorge]

after that? AM-GM

[flat gorge]

.

[vital blade]

oo

i got it

i got 2 conditions

.close

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[strange sparrow]

hi all! question on quantifiers

[strange sparrow]

i just wrote this, I believe it is true, can someone confirm it?

[swift dragon]

you cant just swap for all and exist quantifiers

in the first case, y can only depend on x

in the second case, y can depend on x and z

[strange sparrow]

but my professor said $\exists x \forall y p(x,y) \implies \forall y \exists x p(x,y)$

[potent lotus]

Minnie

[swift dragon]

ah, i see what you mean now

yes, this direction works, the implication

[strange sparrow]

i know the converse isnt true

[swift dragon]

but the other way around wouldnt work

[strange sparrow]

but for my situation with the three quantifers above, would it hold true?

[swift dragon]

yes

[strange sparrow]

cuz i see it as brackets

I can just swap it

[swift dragon]

yeah exactly

[strange sparrow]

kk thank you

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[calm coral]

Please don't occupy multiple help channels.

[empty moss]

Where did I go wrong

[ancient grotto]

you forgot your +C

also, you can just let $u=\arcsin(e^x)$ next time, rather than extending the steps

[potent lotus]

;(

[empty moss]

Oh okay thanksss

.close

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[long oasis]

Hello

[long oasis]

I need some help as to what I’m doing wrong

Cause if I input -cos(pi/2) I’ll get 0

[inner zealot]

Once you take the integral, it's just y = -cos(x) + C

[long oasis]

So would that just mean I would get -cosx+1

[long oasis]

Oh so not dy?

[inner zealot]

No.

[long oasis]

Kk

[long oasis]

?

[inner zealot]

Yes.

And it's not dy = integral(...).

[long oasis]

Oh

[inner zealot]

An integral is a sum.

[long oasis]

It’s just dy =

[inner zealot]

It's just I = integral(...).

[inner zealot]

,rotate ccw

[inner zealot]

Looks good.

[long oasis]

Alrighty

Thanks

.close

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[long oasis]

Hello

[nocturne heron]

yo

[tough hinge]

No.

[long oasis]

Ok sorry

Anyways my question

So for 1.

To find the distance over [0,3] would I just do 0.5 x 10 and 1 x 25 and add it all together

I’m not really sure though because I tried that and was wayyy off

[pure kayak]

its because the velocity is in km/h

and your time is in min

you also had to do the rest of the time, that was only up to 1.5 minutes

[long oasis]

So convert it to min

So everything x 60?

[pure kayak]

yeah, or convert your time to hours, either works

[pure kayak]

not quite

[long oasis]

Well not everything

I would take 10, 20 etc and multiply by 60?

Or divide by 60

[pure kayak]

if its say, 10 km in 1 hour
then youre saying it would be 600km in a minute?

[long oasis]

It would be 1/6 km a min

[pure kayak]

si

[long oasis]

Ok so divide the km by 60

And then multiply by minutes to find the distance?

[pure kayak]

yup

[long oasis]

Alright

I’ll try it and show

I got like

1.04

But the answer key says .96

[pure kayak]

show me your steps

[long oasis]

Well

I did 10/60 then multiplied by 0.5

Then (25/60) x 1

And so on

And added them together

[pure kayak]

,calc (100.5+25+150.5+20)/60

[potent lotus]

Result:

0.95833333333333

[pure kayak]

you did something wrong

[long oasis]

Oh

How did you solve it

[pure kayak]

the same way you were

so you probably typed something wrong

double check

[long oasis]

Ok

[calm coral]

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[calm coral]

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[sand sky]

just divide by 2 then ! To get a = 1

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[west horizon]

One message removed from a suspended account.

[calm coral]

@west horizon Has your question been resolved?

[west horizon]

One message removed from a suspended account.

[leaden thunder]

Again it's separable

[west horizon]

One message removed from a suspended account.

[calm coral]

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[cloud estuary]

is it c?? very confused from my calc test

[cloud estuary]

helloooo? is there anyone

<@&286206848099549185>

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[cloud estuary]

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[blazing wadi]

Any hints

[clear delta]

write out the first few steps

[blazing wadi]

Well my question is that why aren't the two sequences ap
The flea advances 1cm each time and hare 1km each time

[swift dragon]

well, when the rubber band stretches

it also stretches a small part behind the flea

pushing it forward a tiny bit

[blazing wadi]

Hmm but well how would I find out how much rubber band stretches each time

[clear delta]

well.... it tells you

[swift dragon]

the rubber band stretches uniformly

so if the flea is at 1cm and the rubber band is 1km

is it at 0.00001

on the band

so if it stretches to 2 km, it stays at 0.00001 on the band

which is 2cm

and then it jumps another cm

so at the end it is at 3cm and the band is stretches 2km

[blazing wadi]

Ohhh fck man I'm just dum

[swift dragon]

so then it is at 0.000015 on the band

[blazing wadi]

Thanks alot

[swift dragon]

its not rly an intuitive question

[blazing wadi]

Tru other ques in the sheet are much more difficult and straightforward atleast

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[orchid ridge]

i have no idea what to do with this

[orchid ridge]

have only drawn a diagram so far

[clear delta]

harry potter.....

can you find some way to relate the width of the triangle to the height?

it says R but tbh you can just assume it's 1 if you want

[orchid ridge]

ill keep it as R cause ill forget it later uhhh

i can use trig

but

hmm

lmao i have no idea

[clear delta]

draw a line from the center of the circle to one of the corners

that will make you a couple of right triangles

[orchid ridge]

?

[orchid ridge]

i did it this way, is there a possibly better way to do this?

[clear delta]

tbh i haven't done this but that seems reasonable to me, it has the shape of what i'd do

[orchid ridge]

aight thanks

.close

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[compact dirge]

Where do i start

[calm coral]

Please don't occupy multiple help channels.

[tall moon]

do you want to closr this orr..?

[compact dirge]

Nah its just I didn’t see anyone on here

I’m so lost

.close

[calm coral]

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[calm coral]

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[mellow breach]

PELALSLELALSE HELP

[mellow breach]

TRIGONOMETRIC PARAMETRIC EQUATIONS

x = sinz + tanz
y = sinz - tanz

therefore

(x + y) / 2 = sinz
(x - y) / 2 = tanz

[keen flower]

shouldn't parametric equations have a third variable?

[mellow breach]

i used x by accident just pretend its theta

But

1 + cot^2 x = csc^2 x
therefore
1 + (2 / (x - y))^2 = (2 / (x + y))^2

[oak elm]

Which is theta

[mellow breach]

The trig

stuff

sintheta tantheta

[abstract wave]

can you use more brackets

[mellow breach]

hold on ill shwo u my working on a napkin

[potent smelt]

Looks fine enough to me

[potent smelt]

This part

What's wrong with it?

(I don't think there is anything wrong here)

@mellow breach what's your question?

[mellow breach]

Imm not sure whether im supposed

to simpkify further than that

It looks wrong

Theres no ans given in the textbook

[potent smelt]

It's correct, you can simplify further though, but what are you trying to find exactly?

!original

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[flat gorge]

Proceed in this direction: 1/4 = 1/(x + y)² - 1/(x - y)²

[mellow breach]

it says Eliminate theta from the equations

Thags all

[mellow breach]

wait whatd u do

take reciprocals and multiply by 4?

[potent smelt]

If that is what it wants, then you are done

[flat gorge]

take 4/(x - y)² from the left to the right, and divide whole equation by 4

[mellow breach]

im assuming its supposed to form into a quadratic or something

[flat gorge]

Idts, (x² - y²)² + 16xy = 0, does not look like a quadratic to me

[mellow breach]

wtaf whered u get that one

from

[flat gorge]

,w graph (x² - y²)² + 16xy = 0

[potent lotus]

Results provided by WolframAlpha

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[flat gorge]

simplify this...

[keen flower]

you've already eliminated theta

[mellow breach]

,rotate

[keen flower]

you have an implicit equation in x and y

no need to solve for y

[flat gorge]

Yes. As Axe said, you're good enough at this point

[mellow breach]

Hm alright then

Thank you

!

ill close the thingy

.close

[calm coral]

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[calm coral]

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[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

well, to find a, we just need to

find <(1,a,1,1)> ∩ T^{perp}

I guess?

otherwise since S ⊂ H
and S ⊂ T^{perp}
maybe we can find a by plugging (1,a,1,1) in H

okay so lets find a first

<(1,a,1,1)> ⊂ T^{perp}

so lets find the intersection between (1,a,1,1) and the orthogonal complement of T

[mortal orbit]

well you can do that if you want

but you can immediately say (1,a,1,1) is orthogonal to...

[simple musk]

T

(1,a,1,1) is orthogonal to (1,1,1,1)

,w (1,a,1,1)*(1,1,1,1)=0

[potent lotus]

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[simple musk]

what about S?

[mortal orbit]

well

you know S is a subspace of H

and S is also a subspace of T^perp

so S is a subspace of ...

[simple musk]

H ∩ T^perp

,w Nullspace {{1,1,1,1}}]

[potent lotus]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

x(-1,1,0,0)+y(-1,0,1,0)+z(-1,0,0,1)

T^perp = <(-1,1,0,0),(-1,0,1,0),(-1,0,0,1)>

,w rref {{1,-2,0,-1,0},{1,1,1,1,0}}

[potent lotus]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

x1 + (2/3)x3 + (1/3)x4 = 0
x1 = (-2/3)x3 + (-1/3)x4

x2 + (1/3)x3 + (2/3)x4
x2 = (-1/3)x3 + (-2/3)x4

(x1,x2,x3,x4) = ((-2/3)x3 + (-1/3)x4, (-1/3)x3 + (-2/3)x4, x3, x4)

x3(-2/3, -1/3, 1, 0) + x4(-1/3, -2/3, 0 ,1)

H ∩ T^perp = <(-2/3, -1/3, 1, 0),(-1/3,-2/3 ,0,1)>

H ∩ T^perp = <(-2, -1, 3, 0),(-1,-2 ,0,3)>

S = <(-2,-1,3,0),(-1,-2,0,3)>

[calm coral]

@simple musk Has your question been resolved?

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[molten sphinx]

I need some help with this

[fathom solstice]

how many milligrams evaporated?

[calm coral]

@molten sphinx Has your question been resolved?

[molten sphinx]

8701

[fathom solstice]

what is the volume of the room in m^3

[molten sphinx]

162

[fathom solstice]

so we can divide the numbers (8701mg)/(162m^3)

luckily, we can rearrange this to (8701/162)(mg/m^3) which is in the units that we want

[molten sphinx]

53.7098

[fathom solstice]

does your teacher care about sig figs?

,calc 8701/162

[potent lotus]

Result:

53.70987654321

[calm coral]

@molten sphinx Has your question been resolved?

[molten sphinx]

Yes

[calm coral]

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[molten sphinx]

.reopen

[calm coral]

✅

[molten sphinx]

He only wanted 53.7

But I got it

Thank You

.close

[calm coral]

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[calm coral]

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[torn surge]

,w sqrt(x ^ 3 - 2x ^ 2 + x) + sqrt(x ^ 3 + 6x ^ 2 + 9x)

[torn surge]

number 305

( i tried to usethe bot but it's giving bad solutions) Existance conditions are a must

[oak elm]

Uhh what exactly are we supposed to do

Like what's the goal

[torn surge]

solve the expression number 305 and verify for wich values the expression exists

tried it way 2 many times but it ain't it

[oak elm]

So where the expression ≥ 0

[torn surge]

yeah

[oak elm]

Well you can put it into an inequality

then factorize them

[oak elm]

what have you tried

[torn surge]

can't rlly understand why in the results of those existance conditions it is supposed to be x≥1

[torn surge]

lemme show

Here

[oak elm]

put the expression within the roots into the inequality

[torn surge]

isn't it possible only in the radicals with shots numbers?

[oak elm]

shots numbers?

[torn surge]

i mean

[oak elm]

what does that mean

[torn surge]

3 5

i found it on google translate

3 5 7

yk those who can't be divided

[oak elm]

primes

[torn surge]

oh okay

[oak elm]

I'm not quite sure what you meant but in this case it works

[torn surge]

could you show me the procedure ( if you've done it)?

[torn surge]

our teacher told us we can do that type of thing( putting the expression in the roots into theequity) only when the radical's index are primes numbers

as the number sign doesn'tmatter

[oak elm]

uhhh I think it's just a language barrier because I'm not quite sure what you mean

[torn surge]

hm

[torn surge]

could this be more clear?

[oak elm]

So we have $\sqrt{x(x-1)^2}+\sqrt{x(x+3)^2}$
put both of them into inequalities
$x(x-1)^2 \ge 0$
$x(x+3)^2 \ge 0$