help-42

[trail hawk]

Yess

Got it thanks

👍👍

[spiral dagger]

cool

[calm coral]

@trail hawk Has your question been resolved?

[calm coral]

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[orchid palm]

I am unsure how to proceed once I solve for the limit

Let me post my work 1 second

From here I assume x is not needed since it's no longer relevant to the limit insofar as we are evaluating it in terms of n

So with that removed, the limit would result in 1 according to L'Hopitals besides x

leaving just x

but I don't know how I relate that to an interval of convergence

[rustic osprey]

Make sure to keep your limit notation throughout as well

[orchid palm]

Ok I see what you mean

so I have |x| remaining (not just x of course) < 1 and now I need to plug in -1 and 1 to test for convergence / divergence at each of the endpoints respectively?

Would that be plugged into the original power series?

[rustic osprey]

[orchid palm]

Ok so I get that, but it's hard for me to determine what test I need to use

1 sec I will post what I mean

for -1 this is the resulting expression

I don't know what I should apply here, ratio test again?

[rustic osprey]

Way simpler

||nth term test||

(pretend I did not just fail at spoilering)

[orchid palm]

lol

I see what you mean now

Yeah it's divergent

let me determine the right endpoint now with that in mind

also div. for right endpoint

since its just n+5

not -(n+5)

thank you for the help

.close

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[remote mural]

trying to understand what I cana do here

am I able to substitute 3 and 5 into the equations?

I would assume not because 5+3 isn't 7

[dusty crescent]

no, vector addition doesn't work like that

[remote mural]

yeah

[blazing coyote]

Hint:- Use the law of cosines to add the vectors

[warm warren]

Try draw it out

[remote mural]

$(a+b)=(a)^2+(b)^2-2(a)(b)\cos (\theta)$

[potent lotus]

Remlis

[remote mural]

$|a+b|=|a|^2+|b|^2-2|a||b|\cos (\theta)$

[potent lotus]

Remlis

[remote mural]

@blazing coyote ^ is this a real formula

[blazing coyote]

the square root of the right hand side, yes

[remote mural]

$7=\sqrt{(3)^2+(5)^2-2(3)(5)\cos (\theta)}$

oh right

$49=(3)^2+(5)^2-2(3)(5)\cos (\theta)$

[potent lotus]

Remlis

[blazing coyote]

yup

[remote mural]

$\cos(\theta)=-\tfrac12$

[potent lotus]

Remlis

[remote mural]

$\theta = 60^{\circ}$

[blazing coyote]

it should be +2ABcos(\theta)

not -

[remote mural]

oh apologies

[potent lotus]

Remlis

[remote mural]

there we go

$|a-b|^{2}=|a|^{2}+|-b|^{2}+2|a||-b|\cos (60^{\circ}$

[potent lotus]

Remlis

$|a-b|^{2}=|a|^{2}+|-b|^{2}+2|a||-b|\cos (60^{\circ}$
```Compilation error:```! File ended while scanning use of \trigbraces .
<inserted text> 
                \par 
<*> 1001176911734653009.tex
                           
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[remote mural]

$|a-b|^2 = |a|^2 + |-b|^2 + 2|a||-b| \cos(60^{\circ})$

[potent lotus]

Remlis

[remote mural]

$|a-b|^2 = 9 + 25 + 2(3)(-5) \cos(60^{\circ})$

[potent lotus]

Remlis

[remote mural]

$|a-b|= \sqrt{9 + 25 + 2(3)(-5) \cos(60^{\circ})}$

[potent lotus]

Remlis

[remote mural]

$|a-b|=\sqrt{19}$

[potent lotus]

Remlis

[remote mural]

perfect, thank you

.close

[calm coral]

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[proud osprey]

how do I approach a question like this

[proud osprey]

ohh

i think i its like 345

because every team must play but there wiill be one winner

if that makes sense

i think i can think of it like, every team must be eliminated until there is only one winner

and there are 346 teams

and there is only 1 winner

hence 345 games

[potent igloo]

not quite

Think of 8 players

in the first round, there will be four matches, four eliminations

Then four players left give two more matches, then two players left give one more match. So that's 7, yeah

and let's think of 9 now players now

Oh it's asking how many games, not how many rounds

yeah I think you're right'

[proud osprey]

can I ask one more quick q

I think this is supposed to be uh

inclusion exclusion

im sort of confused on the way they did it though

why are they mutliplying by 4 and 6

[potent igloo]

no clue

[proud osprey]

any idea of what they are doing at all

[potent igloo]

nah I don't do stats

<@&286206848099549185>

[calm coral]

@proud osprey Has your question been resolved?

[proud osprey]

ahh this isnt supposed to be statsss

ill close and ask again

ty.

.close

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[tidal bloom]

counting

[tidal bloom]

https://math.stackexchange.com/questions/4376802/no-of-ways-of-arranging-5-as-7-bs-and-4-cs-such-that-atleast-3-ca-pairs-occur

i dont understand why is it overcounting nd why we must subtract 3 times that

[calm coral]

@tidal bloom Has your question been resolved?

[tidal bloom]

.close

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[remote mural]

hi

[remote mural]

i need help about this question

i will translate it to the person who doesn't know my language

[dusty crescent]

what do you need to calculate here?

[remote mural]

Do you know Pythagorean theorem?

In right triangles, the side opposite the 90-degree angle is called the hypotenuse. In a right triangle, the sum of the squares of the lengths of the perpendicular parts varies according to the square of the length of the hypotenuse.

it says that

5 equilateral triangle-shaped cardboards are placed on the front of the rectangular panel, as above, with one edge and one corner overlapping, so that they do not extend over the face of the panel. The similarity ratio of equilateral triangles with one side on the same line segment and one common corner is 1/2.

If the length of one of these triangles is 96 cm, what is the minimum area of ​​the front side of the panel in square centimeters?

[honest zephyr]

What have u tried till now

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

i didn't tried, because i dont know how to solve it

[gilded quiver]

turkish!!!!

[remote mural]

YES!!

[gilded quiver]

LGS sorusu dimi bu

[remote mural]

evet

2021 lgs

[gilded quiver]

ben çözmüştüm bi dakika

[remote mural]

çözemiyorum

yardım ett

[gilded quiver]

dur anlatcam

hatırlamam lazım

şimdi

en az olan değeri istiyor ya

bize verilen alan 96 bu yüzden en büyük üçgenin alanını 96 varsayalım

[remote mural]

tamam

[gilded quiver]

eşkenar olduğuna göre en büyük üçgenin bir kenarı 32 oldu

[remote mural]

anladım buraya kadar

[gilded quiver]

benzerlik oranı 1/2 ya

alanları oranları da benzerlik oranının karesine eşit olmak zorunda

alanları oranı da 1/4 oluyor

[remote mural]

tamam

[gilded quiver]

hem 1/2 oranından

kenar uzunlukları sırasıyla 32-16-8...

sorunun asıl olayı ise yükseklik

eşkenar olduğu için

yükseklik aynı zamanda kenarortay

dik indiği için pisagor yapacaksın

zaten yükseklikler çıkıyor öyle

[remote mural]

cevap A mı?

[gilded quiver]

evet

[remote mural]

AAAA

ÇÖZDÜM

[gilded quiver]

AAAAAAAAAA

ÇÖZDÜN

AJSHDJLASGDLJASH

[remote mural]

EVETTT

KAÇINCI SINIFSIN BU ARADAA

[gilded quiver]

10

sen

8 galiba

[remote mural]

evett

hangi lisedesin

[gilded quiver]

discord için riskli bir soru

lashdkajshdlka

[remote mural]

HAYIIRRRR

pedofili değilimm

[gilded quiver]

şöyle diyelim

geçen sene 500'le alan bir lisedeyim

[remote mural]

hmm

geçen sene kolaydı

o yüzden puanlar artmış olması lazım

[gilded quiver]

kanka dm geçsek

[remote mural]

olur

[gilded quiver]

insanlar soru çözuyo

aksjhdalksjhal

.close

[remote mural]

.close

[calm coral]

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[calm coral]

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[novel sonnet]

do I have the right idea with this setup

[strange lichen]

idk about the bounds, but shouldnt it be r^2

[novel sonnet]

trye

true

[swift laurel]

for the bounds, i would want to go from smaller to larger theta. so maybe replace the lower angle bound (3π/2) with an equivalent negative angle

[novel sonnet]

woukd that be -pi/2?

[swift laurel]

yes

[novel sonnet]

better?

[swift laurel]

looks good to me

[novel sonnet]

yippeee thanks

.close

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[quaint sapphire]

how to do this

[brisk pelican]

Simplify

Reduce fractions

[quaint sapphire]

so ive put the numerator into

(2x+8)(x-7)

[brisk pelican]

Solid

[quaint sapphire]

then flipped the fraction

[brisk pelican]

Cancel with x+4

[quaint sapphire]

o

how

do i multiply it by 2 to make x+4 to 2x+8?

[brisk pelican]

You cancel x+4 from both and you’re left with 2

[quaint sapphire]

so

what about the bottom fraction

the denominator*

would it be 2(x-7)/2x+3?

taht dont seem right

.close

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[remote crown]

$$ \sum_{n=0}^{\infty} \cos(x)^2 \sin(x)^{2n} $$

[potent lotus]

// 1800

[remote crown]

how do i solve this

[pure kayak]

its geometric

[remote crown]

i see

[pure kayak]

that enough for you to know what to do?

[remote crown]

n o

thi doenst seems to be on ar^n form

[ancient helm]

Sin^2 + cos ^2 = 1

[remote crown]

but its sin^2n

[ancient helm]

Which is equivalent to( sin^2)^n

[pure kayak]

r=sin^2 a=cos^2

[remote crown]

can i not just use ar^n form

and just insert the n afterwards on the sin?

[pure kayak]

? its already in that form

[pure kayak]

as they said

cos^2 ( sin^2 )^n

[remote crown]

so cos(x)^2/ 1-sin(x)^2

?

[pure kayak]

si

[remote crown]

so 1

?

[pure kayak]

si

[remote crown]

nice

just a bit confused why the 2n doenst do much

[ancient helm]

[pure kayak]

?

[remote crown]

there is no n on the cos

[pure kayak]

why does there have to be?

[remote crown]

well i just set a = cos(x)^2 and r = sin(x)^2

but i dont use the sin(x)^2n

[pure kayak]

i mean you do

[remote crown]

oh nvm

im blind

i thought it was meant

as (a*r)^n

[pure kayak]

ah

no

[remote crown]

it is literally only r^n

[pure kayak]

a*r^n

[remote crown]

ok, yeah my bad

[wanton wave]

this only works for cosx =/= 0, no?

doesn't the series equal 0 at those points?

[remote crown]

i think this question is supposed to be between (-pi/2,pi/2)

[wanton wave]

ah okay nvm

[remote crown]

but the second question is if it holds for all x

all real x that is

[wanton wave]

gotcha

[remote crown]

and it says "show that is doesnt not matter x between (-pi/2,pi/2)

so i suppose that means x = 0 also holds?

[wanton wave]

Yeah I just meant if cosx = 0, then the series is 0+0+0+0+...

but at x = 0, cosx = 1

[remote crown]

cos(pi/2) would equal 0 tho

in any case, i gtg

ty for help

[calm coral]

@remote crown Has your question been resolved?

[calm coral]

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[earnest nebula]

if we take the 5/6 out of the equation, why is the 6 still in ln(6x+1)?

[earnest nebula]

I'm not sure that question made sense lol

[dusty crescent]

only 5 is taken out technically

there's a step skipped here

which is causing your confusion

[supple rock]

Integral kx =k integration x

[earnest nebula]

so how did they get the 5/6 out and not just the 5?

[supple rock]

Umm idk how to explain but like differentiate sin(5x) you get 5cos(5x) right

[earnest nebula]

yeah

[supple rock]

So if you integrate cos(5x) you get sin(5x)/5

[earnest nebula]

right

[supple rock]

Same with log

It's log(6x+1) so 6 went to denominator

So it became 5/6

[earnest nebula]

oh okay I think that sort of makes sense

can you show the intermediate step that they didn't write?

[supple rock]

Let 6x+1=t
Dt/dx=1
Dt=dx

Integration 1/t =logt

Resubstituting so we get log(6x+1)/6

Idk how to use the bots to show proper format I would write in notebook and show but I don't have that rn

[earnest nebula]

that's fine

what about this one?

[supple rock]

Use substitution

[earnest nebula]

so u = 3+ ln(x)

and then would I do (1/4 (3+ ln(1))^4 - (1/4(3+ln(e))^4)

that does not seem to have worked

.close

[calm coral]

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[supple rock]

U took limits on reverse it would be - (1/4 (3+ ln(1))^4 + (1/4(3+ln(e))^4)

[calm coral]

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[remote mural]

(37)

What does standard position mean?

[hollow lion]

Based on context, standard position would be such that positive x axis is 0°

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

So basically when it’s 0, 360

Or 360

[hollow lion]

Yes

[remote mural]

Also this question is so confusing for me

I don’t get what it’s asking me

[hollow lion]

Are you able to tell me what the coordinates of a point on the unit circle are?

In terms of theta

[limpid perch]

isnt it (cos(theta),sin(theta)) ?

[remote mural]

Yes

[hollow lion]

Yeah, was looking for dark to tell me that

[hollow lion]

Then, in this q, beta corresponds to theta

Which means the point P can be expressed how?

[remote mural]

Thinking

[hollow lion]

In terms of beta

[remote mural]

The angle of it

Beta

[hollow lion]

Yeah, so what would be the coordinates of P in terms of beta?

[remote mural]

Tan(0.32/0.87)

?

[hollow lion]

Not quite

[hollow lion]

.

[remote mural]

Still thinking

Sorry

[hollow lion]

P is intersection of beta and a circle of radius 1.
This means P lies on the unit circle, so like before you can write P as
P = (cos beta, sin beta)

You also know from the question that
P = (0.87, 0.32)

That means
cos(beta) = 0.87 and sin(beta) = 0.32

Right?

[remote mural]

I had a feeling but I just didn’t wanna annoy you in case I got it wrong

[hollow lion]

Its better to just answer. Being wrong doesn’t matter

[remote mural]

Ok

[hollow lion]

Anyway, you are told f(theta) = tan(theta)

Can you recall an identity of tan that uses sin and cos

[remote mural]

Cot

[hollow lion]

Technically correct, but flipped

[remote mural]

Flipped?

Isn’t that just fan

Tan

[hollow lion]

Yeah, but what is the identity of tan in terms of cos and sin

[remote mural]

sin/cos

[hollow lion]

Yes!

[remote mural]

I was overthinking it or something

So

[hollow lion]

So do you think you can answer the q now?

[remote mural]

Am I suppose to do

x= rcos and y= rsin then translate it into tan style like y/x to get Beta

And that’s my answer

?

[hollow lion]

I think you are over thinking this

You are being asked to evaluate f(beta) = tan(beta)

[remote mural]

Yes

But I thought i need to translate the coordinates to rectangular form first

Since I think they are in polar form

Then use tan

[hollow lion]

Yeah, though we have already done that

[hollow lion]

Here

[remote mural]

Yes

And I just need to put those in tan now to get my answer right

[hollow lion]

Yeah exactly

It seems like you’re fine with the concepts, you just got tripped up by the wording of the question

[remote mural]

Sorry I’ve been trying this for a bit but

Sin0.32/cos0.87 is what I’m supposed to do right?

[hollow lion]

No, sin(beta) = 0.32 and cos(beta) = 0.87

So tan(beta) = sin(beta) / cos(beta) = ?

[remote mural]

1

?

I’m not sure I am so sorry

[remote mural]

I understand this though

[hollow lion]

0.32/0.87

$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} = \frac{0.32}{0.87}$

[potent lotus]

shsgd

[remote mural]

Thank you I understand it now I am sorry for troubling you

May I bless you today for helping me and hope you achieve your future aspirations

May god bless you 🙏

.close

[calm coral]

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[hollow lion]

There’s no need to apologise dw

[calm coral]

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[spice mango]

How is D correct?

I don't get how to prove there are 3 points where the instantaneous rate of change is equal to the avg rate of change

[pure kayak]

you can just imagine the line joining the end points

if you slide it along you can see itll be tangent to the curve at around -3, -1 and 2 roughly

[spice mango]

YOOOOOOOOOOO

Just used my pencil

I see it now

Aint no way

.close

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[spice mango]

bro knows

[calm coral]

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[remote mural]

how did they change (cos(2theta)^2 to that

is that a double angle formula or some formula i dont know?

nvm i got it

.close

[calm coral]

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[spark linden]

can someone guide me on how to answer this

[calm coral]

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[trim nest]

(2x•e^(x^2))•sin(2x+1)+e^(x^2)•2•cos(2x+1)

[trim nest]

How do i simplyfie this?

<@&286206848099549185>

[upper sparrow]

Result of differentiating e^{x^2} * sin(2x + 1)? Is there a particular form you/"they" want it in?

Other than taking out a common factor of 2e^{x^2} not sure what else you can really do with that

[trim nest]

Oh ok, i got It tysm

. close

.close

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[remote mural]

find the missing term :
0,6,24,60,?,210

[remote mural]

nvm

.close

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[shy iron]

Because I don’t get it at all

[shy iron]

Can someone help me with 10 please

[calm coral]

@shy iron Has your question been resolved?

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[supple rock]

Just keep dividing 22/7

[calm coral]

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[tender coral]

<@&268886789983436800>

Name + Troll

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[remote mural]

I am trying to solve the vector differential equation [
\curl \curl \vv \Phi = \vv A +\6[\bigg]\grad{\4{r^2}4\chi}, \q \laplacian \vv A = 0, \q \laplacian \chi = 0
]
where $\vv r$ is the position vector and $\vv \Phi$ is an unknown vector.

[remote mural]

i want a general solution of Phi in terms of A, chi, r

,,, I looked around and by Helmholtz's theorem the solution would be
\e{equation}{
\6{\vv \Phi}{\vv r} = -\grad \6E{\vv r} + \41{4\pi}\int_{\VV'} \4{\vv A{\vv r'} + \412\6\grad{\6\chi{\vv r'}\vv r'^2}}{\norm{\vv r - \vv r'}} \dd \vv r'
}
where we used
\e{equation}{
\vv \Phi = -\grad E + \curl \vv B
}
and the primed coordinates denote the source coordinates. But apparently (1) requires that
\e{equation}{
2\div \vv A + \6\laplacian{\chi r^2} = 0
}
but why is that?

[remote mural]

Oh I meant A(r') for the first term in the integral btw, didnt notice

nvm

.close

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[mighty dagger]

ok how do i know if the set is linear independant or not

[hollow lion]

A set is linearly independent if no element is a linear combination of the others

$\sum_{i} \alpha_i v_i = 0 \iff \alpha_i = 0 \forall i$

[potent lotus]

shsgd

[mighty dagger]

ik that its like

if i have this set {[0,1],[1,0],[2,3]}

ik that 2,3 is a combining of the first 2 vectors

[stark abyss]

do you know reduced row echelon form

[mighty dagger]

multiplied by a scalar

[mighty dagger]

u mean like

L1->L2-5L1 and stuff

[stark abyss]

not sure but you want to get leading 1s

(pivots)

[hollow lion]

Those are just elementary row operations

[mighty dagger]

oh ye and the rest is 0s

[stark abyss]

wait lemme just double check this

if you end up with a column which does not have a leading 1 (pivot), then the set of vectors are linearly dependent

[mighty dagger]

ok wait

i have also this

[stark abyss]

like your example here, we can construct a 2x3 matrix and there will be one column without a leading 1

[mighty dagger]

oh ok gotcha

S1={V1,V2,V3} (they r all vectors) and the set is linear independant

is S2={V1+V2,V2+V3,V1+V3} linear independant ?

now my teacher solved this but i didnt understand

he did this

α1(V1+V2)+α2(V2+V3)α3(V1+V3)=0(vector)

α1=α2=α3=0

[stark abyss]

they should be LI, a general rule to follow is that if you have v vectors in a dimension d, if v = d then most of the time the vectors are LI

[hollow lion]

Distribute a_i and the collect like terms

[mighty dagger]

ye ik that but i shld check if this is LI

[stark abyss]

yeah so your teacher expanded and collected all the vectors together

[hollow lion]

It uses this fact of linear independence

[mighty dagger]

ok wait

if he knows that the alpha is 0

[hollow lion]

Once you have a sum of (a_f + a_g)v_i then you know a_f + a_g has to equal 0

[mighty dagger]

that all the alphas r 0

why did he move all the lphas inside ()

[hollow lion]

Because v_i are linearly independent

[mighty dagger]

oh

so thats like a rule

[hollow lion]

You begin with the fact that $V_i$ are linearly independent. So you want to express
[ a_1(V_1 + V_2) + a_2(V_2 + V_3) + a_3(V_3 + V_1) ]
in terms of just $V_i$

[potent lotus]

shsgd

[hollow lion]

That way you can take advantage of the fact that
[ \sum_i a_i V_i = 0 \iff a_i = 0 ]

[potent lotus]

shsgd

[hollow lion]

And if there is a contradiction, then the set is not linearly independent

[mighty dagger]

hm

[mighty dagger]

u got example for a contradiction?

[hollow lion]

Not off the top of my head, but you will know when you come across it

[calm coral]

@mighty dagger Has your question been resolved?

[mighty dagger]

ok tysm

[calm coral]

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[sinful shuttle]

I thought I could do this, but Idk how to continue T~T. I should be getting -39.2 (unless I messed up my math in the earlier problem, but I'm p confident @~@), and I thought I could just cancel some stuff out (specifically a t_1 and a t_0), but it isn't working. 😭

[slender heron]

a^2-b^2 = (a+b)(a-b)

[sinful shuttle]

?

oh NO is this one of those trig rules

o wait no it isn't, lmao

I am an idiot

stillconfusedtho

[quiet island]

happens

[slender heron]

why trig rules

[dull geyser]

4.9t0²-4.9t1² = 4.9(t0-t1)(t0+t1)

[sinful shuttle]

because I'm in calc and trig is the enemy who is constantly around the corner, sob

uh D:

[slender heron]

ok, but it's factoring rules

[sinful shuttle]

oh it was flipped

[sinful shuttle]

I'msorrythisisfactoring?

this is why we don't do algebra at 6 am without any sleep 😭 I'm so dead

[sinful shuttle]

oh wait

oh frick did not mean to ping, my bad

this

makes sense

Ithink

agh wait that still comes back with -78.4 which is double what I got in question 2 D:

meh, I'm far enough that I can wing it and then go to a tutor after class :') ty guyss

.close

[calm coral]

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[velvet charm]

help

[prime skiff]

what have u tried

[velvet charm]

noting this is smth new for me pls help

[prime skiff]

well you can start by simplifying the root on the left

a cuberoot and cube are inverse functions so they can cancel each other out

[velvet charm]

the root on the left is -6

the awnser is -6

???

[prime skiff]

is the question just asking you to simplify?

[velvet charm]

no

it say i must judt do it

[prime skiff]

so it doesnt say at all whether it wants it to be simplified or evaluated or anything

[velvet charm]

ye says noting it say answer there questions

[prime skiff]

well can't really answer it if it doesnt tell u what it wants you to do with the expression

[velvet charm]

it says to complete the thing

Simplify the following without using a calculator. Show all you
calculations. this is what it says

k

[potent lotus]

faiyrose

[velvet charm]

k

the one on the left is -6

right

k

[potent lotus]

faiyrose

[velvet charm]

and the right one is 3sprt6

kk

how d i put it in steps

bc this question is 3 points

??

ok

got it tysm

i understand now

/close

ok

.close

[calm coral]

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[velvet charm]

.close

ok

.close

ok

ok

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[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[regal bolt]

From your sketch it seems like there's a pretty good guess what the maximal element is, right?

Maybe it's good to think about d) first, because once you have that, it should become clear whether there's minimal elements or not

[calm coral]

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[potent lotus]

hyzae

[pure locust]

k i'ma show you an example of one that's similar

ready?

k we'll prove together the following:

[dusty crescent]

Doesn't this involve induction?

[pure locust]

$1+2+3+...+n=\frac{n(n+1)}{2}$

[potent lotus]

HurricanePanda L204 MSC 32A35

[pure locust]

yeh I'm walkin her through it, we moved over from discussion

[dusty crescent]

Is the question prove for all n?

[pure locust]

this was just a general questino posted

we gud tho

ok so hold that

from what you told me earlier

what's the base case?

yes. start on the right hand side and plug in n=1

yes so does this case hold true?

perfect. just for the sake of the problem, let's do another case. Do n=2 now

why?

correct!

so the thing about induction is that, you can do as many of those singular cases as you need to

to get a hang of what's being proven if it's not completely obvious

so if you wanted to, you could do n=3, n=4, etc. tho most times you don't have to.

so now what's the inductive step?

yeh, so we assume that for n=k, our statement is true, and what do we want to show?

yeh let's write this out carefully

so, we're assuming that if n=k

then

1+2+3+...+k = k(k+1)/2

set this in your back pocket because we'll need that in a second

so now, if we have n=k+1

then that statement above should look something like

1+2+3+...+k + k+1 = (k+1)(k+2)/2, you agree?

so now this is what we have to prove.

Let's start with the left hand side. what do we know about 1+2+3+...+k?

yes!

so we can substitute that in directly

k(k+1)/2 + k+1

what might our next step be? we want to fix up that expression to help us get our desired result

[unique jackal]

I watch this convo with interest.

[pure locust]

well careful

you said that 1+2+3+...+k= k(k+1)/2

so if we have

1+2+3+...+k+k+1

we took the first k terms and replaced it with k(k+1)/2

stare at it for a sec

color helps, one sec

so we have this at our inductive step:

we want to prove this:

so

notice the part I highlighted in green

we said from our inductive step that the green part is equal to what?

yup

so the right hand side is our goal. how do we show that on the left

in other words, what can we do to the left hand side to get the right hand side?

might be somethin you wanna do before that

look at the denominators there

yup, try getting common denominators first

go for it!

well top and bottom by 2 on the k+1

kk

oh nah, you don wanna multiply each term by 2

nah

instead

treat k+1 as a fraction, (k+1)/1, then multiply it top and bottom by 2

no worries haha i wrote it out jus in case

from

well you're going "backwards" so you don wanna recancel

yeh

now what?

same denominators, what can we do with the tops?

ding ding ding you got it!

careful on that last step

it would suffice to show that k^2 + 3k + 2 factors to (k+1)(k+2) but we're assuming your audience knows this

yeh you can simply say

by factoring it, it equals

so that's fine

nicely done!

that's it, you showed the inductive step

therefore

1+2+..+n= n(n+1)/2

ok. i wanted to just show you this, i'll let you try the other one on your own

you can hang out here and have a helper help ya but i'ma go die

indeed! [HPP] Congratulations, you've earned 12 hurricane points!

ok i'ma go stalk duck to see where he at on this

https://tenor.com/view/ドロン-忍法-パッと消える-poof-disappear-gif-15682669

I would say close unlress you want someone to come help ya

up to you

haha ty and likewise!

[calm coral]

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[near crag]

e^x is continuous so that's why the limit of the sequence is indeed e^(i*a)

but that's just a finite value for all values of a

yeah

a can even be 1, why not

yup

[dense fog]

Still doesn’t seem like you understand it. For example $$\lim_{x\to 1} x\cdot a=a$$

[potent lotus]

ScapeProf

[dense fog]

Here the answer is a function of a

Same principle for your questions

The answer is a function of a? f(a)=a

And f(a)=a isn’t?

No just a

Answer is a function of a

Here just the identity

If it was lim(x->1) of x * a^27 * pi * sqrt(ln(a^2+1))

The answer would still be a function of a

[dense fog]

Yes just like this is?

You are taking the limit of n

So answer can’t depend on n

$$\lim_{n \to\infty}\dots=\begin{cases} ? & a<1 \ ?? & a=1 \ ??? & a>1\end{cases}$$

[potent lotus]

ScapeProf

[calm coral]

@remote mural Has your question been resolved?

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[molten lichen]

I dont know how im supposed to do this

got a test tmrw 😦

[verbal torrent]

you need to get the limiting value first i think

[molten lichen]

right so that would be lim as x approaches infinity?

[verbal torrent]

yh but with t instead of x

that's how i understand it

[molten lichen]

yeah, my bad on saying x

so lim as t--> inf = 6000

[verbal torrent]

seems like it

[molten lichen]

and then half of that would be 3000

so do I just do 3000=6000-5500e^(-0.159t) ?

Im getting an approx t of 4

is that right?

[rigid kettle]

plug back in and see

[molten lichen]

P(4) = 3088.268

Thats why Im confused

[rigid kettle]

why

[molten lichen]

even if its an approx the margin is ~88

3000 to 3088

[rigid kettle]

yes but like

say I start a school in january 1st 2020

in april, id say 'im in the 1st year in school'

it wants to know which year you're in when it hits 3k

[molten lichen]

ahhh ok

one more question

since its an even function, would the area be the same

and would it have a sign change

[rigid kettle]

drawing a diagram might be best

write down the definition for even

[verbal torrent]

what does even mean

[molten lichen]

if its graph is unchanged under reflection in the y-axis

[rigid kettle]

yeah

f(x) = f(-x)

so now draw a picture

and that should help you solve

[molten lichen]

So from the given integral I get an answer of -4

[verbal torrent]

I'd get the same

[molten lichen]

thank you

.close

[calm coral]

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[terse swan]

Can someone help me with the rest

[strange lichen]

dont write the integral symbol again after you've integrated

[terse swan]

oh right

[strange lichen]

so we get

$\lim_{c \to \infty} -\frac 1{2\ln 6} 6^u \eval_0^c$

[potent lotus]

Stephen

[terse swan]

yea

[strange lichen]

so when we evaluate that 6^u from 0 to c what do we get

[terse swan]

this the part i struggle w like idk what causes it to be 0 or what cuases it to go to infinity

[strange lichen]

dont worry about that part yet

just evaluate from 0 to c

then we will do the limit after

[terse swan]

alright

so what do we do

[strange lichen]

$6^u \eval_0^c$

[potent lotus]

Stephen

[terse swan]

6^c-6

[strange lichen]

no

[terse swan]

why

[strange lichen]

whats 6^0

[terse swan]

1

[strange lichen]

yes

not 6

so we get

$\lim_{c \to \infty} -\frac {6^c - 1}{2\ln 6}$

[potent lotus]

Stephen

[terse swan]

yea

[strange lichen]

we can rewrite this as

[terse swan]

i kind of have a hard time seeing it that way wolud u be able to write it the way i sent it if u dont mind

[strange lichen]

$\lim_{c \to \infty} -\frac 12 \left(\frac{6^c}{\ln 6} - \frac 1{\ln 6} \right)$

[terse swan]

thanks

[strange lichen]

yea?

o

ok

[terse swan]

now what

[potent lotus]

Stephen

[strange lichen]

lets pull out the -1/(2ln6)

itll make this easier to see

[terse swan]

alr

[strange lichen]

$\frac {-1}{2\ln 6} \lim_{c \to \infty} 6^c - 1$

[potent lotus]

Stephen

[strange lichen]

yes?

[terse swan]

yea

[strange lichen]

ok so now evaluate the limit

as c goes to infinity

what does 6^c - 1 tend to

[terse swan]

infinity?

[strange lichen]

hold on, my spidey senses are tingling

something is off

[terse swan]

yea im confused too

[strange lichen]

oh i think i see the issue

we did the u-sub, but we didnt change the bounds

[terse swan]

is there a easier way to do this where u dont have to change the bounds bc my teacher didnt really teach us that

[strange lichen]

u gotta change the u back to a -2p then

so lets go back

[terse swan]

alr

[strange lichen]

$\lim_{c \to \infty} -\frac 1{2\ln 6} 6^{-2p} \eval_0^c$

[potent lotus]

Stephen

[terse swan]

i also need help with this one

[strange lichen]

yea i hate that stuff

lets focus on this one for now

[terse swan]

alright

[strange lichen]

so this becomes

$-\frac 1{2\ln 6} \lim_{c \to \infty} 6^{-2p} \eval_0^c$

[potent lotus]

Stephen

[strange lichen]

yes?

[terse swan]

yea

[strange lichen]

ok now evaluate the 6^(-2p) from 0 to c

[terse swan]

same thing no?

-2 * 0 is 0 and thn 6^0 is 1

and th othr on is infinifty

[strange lichen]

dont evaluate it yet

just plug in the c and 0

and tell me what u get

[terse swan]

6^-2(c) - 1

[strange lichen]

simplify the 0 one

[terse swan]

alr

[strange lichen]

$-\frac 1{2\ln 6} \lim_{c \to \infty} 6^{-2c} -1$

[potent lotus]

Stephen

[strange lichen]

ok

this is just

$-\frac 1{2\ln 6} \lim_{c \to \infty} \frac 1{6^{2c}} -1$

[potent lotus]

Stephen

[strange lichen]

yes?

[terse swan]

ya

[strange lichen]

now evaluate at c goes to infinity

[terse swan]

0

[strange lichen]

the entire limit

what does it become

[terse swan]

-1

[strange lichen]

good

now what does the entire thing become

[terse swan]

1/2ln6

[strange lichen]

yes

nice

u can simplify this and it becomes

[terse swan]

why is that intgral so long

[strange lichen]

1/ln(36)

[terse swan]

alr can u help me with the area one

[strange lichen]

nah i hate those sorry

just open a new channel

[terse swan]

alr

[strange lichen]

$\frac {\ln e}{\ln 36}$

[potent lotus]

Stephen

[strange lichen]

which is just $\log_{36}e$

[potent lotus]

Stephen

[strange lichen]

thats a nice answer

not too often u see it in that form

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[keen grove]

how do i prove this?

[untold compass]

you kinda just have to expand b.(u-b)=0 all out

everything will cancel and you'll have a1.a2's which are 0 because they're orthogonal

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[terse fulcrum]

how do u tell if hyperbola transverse is vertical or horizontal?

[minor knot]

look at the standard form

if its $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then the transverse axis will be horizontal, and the curves open left and right

[potent lotus]

Renz

[minor knot]

while if its $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ then the transverse axis will be vertical, and the curves open up and down

[potent lotus]

Renz

[terse fulcrum]

so if the y term is on the left its vertical?

[minor knot]

yea

[terse fulcrum]

and x term on the left is horizontal?

[minor knot]

if y² is being subtracted from x² its horizontal
if x² is being subtracted from y² its vertical

[terse fulcrum]

thanks

👍

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[muted shoal]

hey can someone explain this move:

[muted shoal]

based on this:

[marsh valley]

You can pull out a 1/|b_n| , and since that is less than 2/|k| you can replace it in the inequality

[muted shoal]

ughhh, what do you mean by pull out sorry 😅

[muted shoal]

i love u ty

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[glacial raptor]

Tests for independence and the homogenity of proportion

[glacial raptor]

need help caculating this

I caculated everything and got an answer of 8.5 by doing (79)(27)/252

since I'm caculating the expected value

[potent lotus]

leo~

[polar solar]

i think this is either

$\frac{9}{9+15+23}\or\ \frac{9+18}{18+37+24+9+15+23}$

[potent lotus]

leo~

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[vivid rock]

Can someone help or verify if my answer is correct
this is a linear programming problem, formulation

[vivid rock]

I really have a hard time on this subject and would like to somehow master it. I still get confused when theres a lot of variables discussed in the problem. I will be taking this subject again this coming term since I failed my first take.

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@vivid rock Has your question been resolved?

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@vivid rock Has your question been resolved?

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@vivid rock Has your question been resolved?

[snow spire]

One idea is to see how simpler forms work for this problem. We can try one variable with small integer values, and add another variable, or aspect of the problem and see how the values change.

Maybe it would also be helpful to try levels of abstraction, like changing actual number values to symbols.

What if there is one machine only, then two machines.

One given, two given.

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[remote mural]

I don't know how to search these kinds of problem in you tube: ex: A boat is moving with a course of 45grades twoards point B from point A. And from poin t B moving twoards the Points c unitl he meats the northeats of....

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<@&286206848099549185> @sly crane

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[remote mural]

<@&286206848099549185>

Lesson topic please.

The name.

[untold drum]

oh, sorry. i misunderstood. your question isnt a math one.

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[stuck axle]

yo i need help one this pls

[stuck axle]

it talks about high order derivative

[blazing hinge]

which question?

nth derivative is the same as the derivate of the (n-1)th derivative

[stuck axle]

all of it 😭

[blazing hinge]

i dont get the first one

there's no equation there

[pure flame]

1 is just a first order derivative

2 I'm assuming is a second order derivative to indicate rate of change of rate of change

[blazing hinge]

how do you do 1. tho i'm curious

seems impossible

[pure flame]

Am I reading this wrong or is it just taking the derivative and calculating the value at x=-2

Ah nvm I see

Didn't notice y values my fault

[blazing hinge]

i think there's a mistake

there's no =

[pure flame]

True

[blazing hinge]

for the rest just know y'' = (y')'

and when they ask you to find the slope etc plug the coordinates in the equations you get for y' or y''

[stuck axle]

my professor said when we talk about the slope of the tangent line we're speaking of the first derivative

[blazing hinge]

yes

[stuck axle]

but her explanation is too complex i dont understand it

[blazing hinge]

the equation of a tangent line is always going to be the familiar y = mx+b

and when you take the first derivate of a function , you get another function in return

in which you can plug your coordinates to find m

(the slope)

[stuck axle]

ooh ok i see now thanks

ok i see what i have to do now

thanks for the help

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[remote mural]

i am awake

and my brain is working