help-17

[remote arch]

The final expression makes me a bit suspicious, though I haven't checked it myself. Can you show all of the work?

[bright thistle]

oke let me write it tidy so it can be understandable

i think it is 1/ a+b xd

the base question is here

[remote arch]

What's going on here and here? Are you trying to cancel (a-b)^2?

[bright thistle]

oh nothing forget there

3rd line forget it i canceled (a-b)^2 in last

after =

[remote arch]

k, there's something you need to know then

The thing is that to cancel a factor like that, it needs to be part of every term in both the numerator and denominator.

So this $$\frac{3xy + 7y}{5y} = \frac{3x+7}{5}$$ is ok, because I cancel the $y$ from every term.

[twin meteor]

OneTrackPony

[bright thistle]

oh i get it

[remote arch]

Cool

[bright thistle]

a^2+ab-a^2-ab+(a-b)^2/(a+b)(a-b)^2 if we elimanate a^2 and ab we get (a-b)^2/ (a+b)(a-b)^2 we can cancel (a-b)^2 we get 1/ a+b right?

like this?

[remote arch]

Yeah, looks better

So you would end up with 1/(a+b), as you said.

[bright thistle]

great thank you a lot

.close

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[spring relic]

I am a bit unsure how to approach this question

[mental falcon]

isn't this just limited to x = y = z = 0?

[spring relic]

that's what I was thinking

(0, 0, 0) is the only point that satisfies the equation

still unsure exactly what they are after though

γ: R --> R^3, γ(t) = t*(0, 0, 0) ?

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@spring relic Has your question been resolved?

[spring relic]

would it be reasonable to consider spherical coordinates?

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[zinc salmon]

!15

.15

.timer

<@&286206848099549185>

[zinc salmon]

Someone?

<@&286206848099549185>

ayooo, im here have [5 sqrt(minutes)]^2

<@&286206848099549185>

[winged agate]

Does it say if n is bigger/smaller than a number?

[zinc salmon]

no, only n

an AFA (Brazilian Air Force Academy) question

[winged agate]

Try (1+i)^1+…+(1+i)^n= (31+i) then solve for n

[zinc salmon]

ayo, thats the question

it's not practical to test... 10 possibilities of n in a sum

i already saw it as a geometric progression

[fathom spruce]

y'all help 😭

[zinc salmon]

but we get [a_1 (q^n-1)]/q-1 = (1+i)(1+i)^n -1 / i = 31 +i

[zinc salmon]

.occupied

[fathom spruce]

which channel do I go then

[zinc salmon]

im not an specialist, but qn is the height of the triangle, and mr too

then you get the angles and the sides and use some propriety to afirm congruency

like, the streight that connects the vertex and the midpont of the oposite side is the height, so...

or

you can get a symmetry

p -> midpont qr divides into 2 equal parts...

[zinc salmon]

someone?

[fathom spruce]

so what's the answer

[zinc salmon]

i give up

crowded server

[winged agate]

Lemme get home in a bit

[zinc salmon]

bro, im not a helper, sorry, but you have to do it yourself

[zinc salmon]

k

[zinc salmon]

they share the same side (pq / pr), the same angle (qpr) and another common side (the height of the isoceles triangle)

prove

proved

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[shrewd mason]

Im not exactly sure how to do this

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[fervent wasp]

What is this angle?

[winged agate]

You mean right angle?

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[vocal sleet]

@stuck maple Has your question been resolved?

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[vocal sleet]

Please don't occupy multiple help channels.

[proven garden]

what is the formula for interest?

[plucky carbon]

idkk 😭

what is it

[proven garden]

I = P(1+r/n)^nt-P

P is principle, r is interest rate, n is compound frequency and t is time

so these four affects interest

[plucky carbon]

i got the answerr can u help me w another question

is the answer 65??

[proven garden]

no

[plucky carbon]

what is it 😭

[proven garden]

which angles did you find

[plucky carbon]

i didnt find any

i dont get it

[proven garden]

bruh

you can't just guess lol

[plucky carbon]

i didnt guess

i tried doing

127-62

[proven garden]

for this question you need to know vertically opposite angles and angle sum of straight line

[plucky carbon]

but i think that only works for finding the largest angle

[plucky carbon]

oh

how do i know that

[proven garden]

vertically opposite angle is when you have two straight lines

like a X shape

the opposite angles are equal

and for a straight line the angles sum up to 180 degrees

[plucky carbon]

yeah

[proven garden]

you can find CDE using vertically opposite angles

and DEC using angle sum on straight lines

[plucky carbon]

um

the thing

refreshed

and now it shows different numbers

🌚

can i send them

[proven garden]

sure

[plucky carbon]

okay

wait what

[proven garden]

umm it's the same

[plucky carbon]

yeah i thought it changed

💀

okay then u can continue

[proven garden]

can you find CDE for me

[plucky carbon]

is there a formula

is the answer 127

@proven garden

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[exotic hamlet]

Need help on empirical rule

[exotic hamlet]

Struggling on trying to find the new to solve this

[flat whale]

,rotate

[proven garden]

a mean of 12 hours it's given

[exotic hamlet]

so that's what I would put in place for mew?

[proven garden]

yes

[exotic hamlet]

aight nice

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[ionic marlin]

can we apply limit comparision here

[ionic marlin]

a_n = a_n^2
b_n = a_n

a_n/b_n = a_n^2/a_n = a_n and if we take its limit then it goes to S which is non zero and not infinite

[somber flame]

doesn't a_n converge to 0

[crude arrow]

if a_n converges to 0

[somber flame]

if sum a_n converges

[crude arrow]

we automatically know the limit as n -> inf of a_n is zero

[crude arrow]

so this follows immediately

good job

https://tenor.com/view/ok-all-nice-gif-22927671

[somber flame]

what?

[cerulean lichen]

I'm just saying btw

I think you can also use direct comparison test

[somber flame]

the proof posted above doesn't work

[somber flame]

moreover you should

[cerulean lichen]

what do you mean

do you mean S could be 0?

[ionic marlin]

yeaaa i get it now

[somber flame]

not just could, it is

[ionic marlin]

it should be 0

since its converging

[crude arrow]

mhm ur good

[somber flame]

also

[ionic marlin]

a_n^2<s^2

wher s is limit of sum of a_n

[somber flame]

there is a bit of an issue with a_n allowed to be 0

[ionic marlin]

does this work

[crude arrow]

well ofc

since a_n >= 0

[cerulean lichen]

yep

[crude arrow]

(a_1 + ... + a_n)^2 = a_1^2 + ... + a_n^2 + (extra terms)

sum of these extra terms > 0

[cerulean lichen]

I mean the sum

S

not a_n

[ionic marlin]

aaah

but in limit comparision test we arent working with sum right

[cerulean lichen]

I think you can use limit comparison test if you divide the cases

when S = 0 and S isn't 0

[somber flame]

limit comparison is just a bad idea here

[crude arrow]

I mean why not just state that 0 \ge (a_1 + ... + a_n)^2 \ge S^2

[somber flame]

like there is no way around that

[crude arrow]

and since S converges

[cerulean lichen]

I mean

[crude arrow]

S^2 is finite

so your sum is finite

[cerulean lichen]

of course this is easier way

but using limit comparison test can be also solution

[somber flame]

S is necessarily 0. a_n converges to 0 if sum a_n converges

[cerulean lichen]

I mean S = sum a_n

[somber flame]

mmm ok confusing given some of the earlier messages sorry lol

[crude arrow]

yeah ofc lol

that's p standard notation

[cerulean lichen]

🫠

[crude arrow]

hmm

yeah maF

we can try what you are saying

[crude arrow]

I just don't understand why you couldn't use what my man Heisenberg did here too and use this

part (ii)

cuz lim n-> inf a_n = 0

[somber flame]

notices the hypotheses

[crude arrow]

^ that's why I congraulated my man

[somber flame]

a_n > 0 and b_ n > 0

[ionic marlin]

because c cant be 0 or infinity

[somber flame]

those are different than in the problem

[ionic marlin]

and since lim an = 0 then an/bn =0 if limit of b_n is not 0

[somber flame]

i think you guys are overcomplicating this so i'm just gonna say the simplest proof... if $\sum a_n$ converges then $a_n\to 0$, so for some $N$, $a_n \leq 1$ for all $n\geq N$. So $a_n^2 \leq a_n$ eventually

[twin meteor]

soulgazer

[somber flame]

and direct comparison theorem completes the proof

[crude arrow]

honestly the obvious proof was 0 \le sum \le S^2

it's alr

I think we're good here

[somber flame]

that's not necessarily true

[crude arrow]

ty soulgazer for the help

there you go

[somber flame]

no that's not the issue

[crude arrow]

what's the issue

[somber flame]

wait what does that say now

[crude arrow]

I'm so confused 💀

[somber flame]

0 >= sum a_n >= (sum a_n)^2?

[crude arrow]

I'll just explain the logic

[somber flame]

that's def not true lol

[crude arrow]

ye should be le

all good

[somber flame]

ok but it's still not true, what if S is 1/2

[crude arrow]

so what's ur counter example

I'm sorry you're going to have to spell this out for me

but I will just explain myself when you expand out S^2 you get (a_1 + ... + a_n)^2 = \sum_{k=1}^n a_k^2 + D

where D >= 0

[somber flame]

say a_n = 1/(2^(n+1)) for example. then S = 1/4 + 1/8 + 1/16 + ... = 1/2 right?

0 <= S <= S^2 is false

[crude arrow]

oh ur right cuz \sum{k=1}^n a_k^2 not necessarily greater than \sum{k=1}^n a_k

lol oops

for some reason

I thought I was comparing squared terms

ty for the correction

[somber flame]

🙂

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[vast shale]

So I found 3 cases

The king is at the corners

[signal pendant]

Go on...

[vast shale]

3 cases

Corners, right next to corner (circles in blue) and then in the middle

At the corners the king can only attack 3 ways

The blue circles it can attack 5 ways

Then in the middle, it can attack 8 ways

So now, our universal set is the amount of ways we can place the 2 king s

P(64,2)

So now with the 3 cases

If we place the king at one of the corners (4 of them) then we have 12 other squares where the kings can attack

This is : P(4,1) x P(12,1)

Let me know if my reasoning is any good

[glossy maple]

Not sure what your doing

Your cases are good tho

Think how many ways can you put the first king in the corner

[vast shale]

Ye but for case 1, we have P(4,1) ways to place a king at the corners

[glossy maple]

And how many ways for the second king in that case

Yes

[vast shale]

Then there are P(12,1)

12 squares around the corner

[glossy maple]

4 ways... No need to use permutatioms

Permutations

Ok but the 12

Is like

In total

Let's say you chose to put the king on the top left

[vast shale]

Ah but the kings are diff colours

[glossy maple]

Then how many places are left for the second king?

[vast shale]

Ah then it’s P(3,1)

[glossy maple]

Thats fine because we can place whichever one we want first

[vast shale]

Ok

So 4 ways and 3 ways

[glossy maple]

We want them to NOT attack

And again, no need for P. Just count kt

It

[vast shale]

Ah but I am trying with the universal set

The universal set is P(64,2)

[glossy maple]

Uh ok

Sure

[vast shale]

But we can do it the other way

It’s probably easier

If we place the king at one of the corners

That’s 4 ways

Then to place the king at ways where they don’t attack is 60 ways

Because 4 squares are taken

[glossy maple]

Yup

[vast shale]

So now case 2

We can place a king in 4 areas, then in 6 squares which then leaves 58 squares left

4x6x58

Case 3: there is 1 area, total 36 squares. Leaving 55 squares left so 36 x 55

Total is

240 + 1392 + 1980

= 3612

[glossy maple]

Yes

[vast shale]

Which is the correct answer

[glossy maple]

7^2 tho

Not 36

In third case

[vast shale]

The answer is 3612

In the solution manual

[glossy maple]

,w 460 + 4658 + 3655

Hmm

This should be the answer i think...

Oh no

It is 6

Im dumb

Lmao

36 instead of 49

[twin meteor]

Results provided by WolframAlpha

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

Requested by redstoneplayz09

[glossy maple]

Yeah ok

[vast shale]

Yeah

Thank you very, very much!

.close

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[glossy maple]

Np

[plucky sedge]

@finite hatch

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[exotic pond]

can someone help me understand what this statement means? what i’m understanding is that fsubx gives the rate of change of z as you move along the unit vector î. But won’t the value of z just remain 0 throughout? there literally is no change occurring in the z value of the coordinates as you move along the vector î

[crude arrow]

yees ofc

so z = f(x_1, ...., x_n)

a partial derivative is when you hold all the other variables constant

except x

so ur varying x

[exotic pond]

yes

[crude arrow]

and seeing how that affects z

[exotic pond]

yeah

but on the î vector, isn’t z just 0 for all values of x?… î vector is basically the x axis isn’t it? z is 0 on x axis

[crude arrow]

it's in the direction of i

is <1, 0>

[misty belfry]

z is a function of x and y I assume

so z will change as you change x and y

[crude arrow]

no

[exotic pond]

oh wait yes

i was thinking in terms of z axis

[misty belfry]

yeah z is not independent here

[exotic pond]

so what they’re trying to say is the partial derivative fsubx gives us the rate of change of the function f as you change x.

[crude arrow]

and hold y constant

ur not changing y

[exotic pond]

yes

[crude arrow]

so in the direction of i which is like <1,0>

[exotic pond]

yep got it

thanks!

.closed

[crude arrow]

.close

@exotic pond

[exotic pond]

.close

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[paper yew]

how do i do this and im sorry i didnt mean to send that

and i’m stuck on the entire thing

i wasn’t here for the day of school and teacher didn’t explain

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<@&286206848099549185>

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[hushed pewter]

@paper yew

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[rough barn]

For part b), do we use binomial distribution?

[rough barn]

I think we do bcs the probability is constant and the probability of each outcomes is independent of each other

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<@&286206848099549185>

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[lavish river]

given the power series of e^x, how do I find it's inverse?

[lavish river]

in terms of a power series

[surreal basin]

what center are you choosing for ln(x) (since x=0 isnt possible)

[lavish river]

what centre?

[surreal basin]

the power series of e^x is centered at x = 0

[lavish river]

maybe (e^x) - 1 instead

[surreal basin]

that inadvertently makes things harder

since e^0 = 1,

you can have the center for ln(x) be at x = 1

[lavish river]

let's make the centre be x=1 then

[surreal basin]

the general process for inverting a taylor series is: https://en.wikipedia.org/wiki/Lagrange_inversion_theorem

[lavish river]

thank you

[surreal basin]

but since you really need the "tayloer series for ln(x) at x = 1" theres an easier way to find it

[lavish river]

oh, no I was generally looking for a way to find inverses of power series

[surreal basin]

oh alr

[lavish river]

I don't think there is a formula for a generic power series so I just stated e^x

[surreal basin]

you couldve said sum an x^n

or a power series

not a big deal thouggh

[lavish river]

hm because I don't think there is a general formula? is there? sorry I'm not sure

[surreal basin]

I mean this seems general enough to me

wdym by that

[lavish river]

idkk sorry I'll read up on that thanks 👍🙏

.close

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[lavish river]

why hasn't this closed yet

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[worthy forge]

is there an approximation I should be aware of for ln(1 + x) when x > > 1?

[worthy forge]

Should I just use ln(x) or might there be something better?

I know that ther's a taylor expansion for x<<1

[regal bane]

The Taylor expansion for ln is VERY slow, so I wouldn't even consider that a useful approximation either

Pade approximants are likely to be faster, for small values of x

If x isn't small, you can do something like this:
ln(30) = ln(1.405^(1/10)) = ln(1.405)/10

That shifts the problem a little bit: How might we take an nth root?

[vocal sleet]

@worthy forge Has your question been resolved?

[worthy forge]

it might be relevant to mention that im talking about things on the scale of 10^23

[regal bane]

Like, you're trying to find ln(10^23)?

Don't forget to hit the ❌

[worthy forge]

something like ln (1 + x/ (10^23)

[regal bane]

Ah you're looking for some pretty extreme accuracy

[worthy forge]

yeah. i think i've arrived at a satisfactory answer for this one

but thank you for your help

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[umbral sleet]

can somebody please help me construct a sine/cosine equation for this problem

[umbral sleet]

and walk me through it

it's a homework question, i have my 2 ordered pairs already

(1.5, 2) and (3,8)

[vocal sleet]

@umbral sleet Has your question been resolved?

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[topaz geyser]

What is the equation for this graph?

[topaz geyser]

Im not sure what i got wrong

[proven garden]

v+1 not v+2

[topaz geyser]

Is the rest correct?

I tried submitting it but it says it is still incorrect

<@&286206848099549185>

[proven garden]

Oh the y translation is different too

Since you messed up the x translation

[topaz geyser]

Whats the y translation?

Is it plus 2

@proven garden

[proven garden]

yes

[topaz geyser]

So the final answer is h(v)=tan(pi/6(v+1)+2

@proven garden

I put it in desmos it not giving the same graph 😭

[proven garden]

Rrally

[topaz geyser]

Yeh it still says its incorrect

@proven garden

I also need helo with this one and it should be it

The x values i believe are pi/2

And pi

i figured it out man its v-1 and there is a stretch of 7 for the first question

[vocal sleet]

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[wispy tinsel]

i have this problem

my working so far is:

the 1/3 and the ^3 in the first cancel out

for the last it also cancels out

so

$in(x+2) + in(root(x)) - in(x^2 + 3x + 2)$

[twin meteor]

Micah

[wispy tinsel]

the first two combine to make

$in(x^(1.5) + 2x^(0.5))$

[proven garden]

$\ln{(x^{1.5}+ 2x^{0.5})}$

[twin meteor]

WhereWolf(ping if needed)

[wispy tinsel]

thanks

yah

then we put that over the second to make

honestly i cant put it in you know what i mean

but the answers say

anyone know how to get there/what i did wrong?

<@&286206848099549185>

[proven garden]

You can factor x^2+3x+2

[wispy tinsel]

i can do that but how does it help me?

i cant do some crazy factory shit with the top as well can i?

[proven garden]

It can be factored to (x+2)(x+1)

So you can cancel the x+2

[wispy tinsel]

but you add them together, not multiply

also its 2x^0.5

am i very confused here

$in((x^{1.5} + 2x^{0.5})/((x+2)(x+1)))$

[twin meteor]

Micah

[wispy tinsel]

wait can i take out the x^0.5

@proven garden Thanks very much for your help :) Ive got it now!
:)

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[fallow yoke]

can anyone help me im not sure how to do this

[fallow yoke]

and this

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[livid shuttle]

can someone help me solve this? idk where to begin

[livid shuttle]

.close

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[finite hatch]

Hi

[finite hatch]

To get convert eV to J why do we multiply when an eV is smaller than a J

Wait how come one electronvolt is bigger than a joule

when a joule is bigger

Wait

I am so confused

[fair oasis]

its 10^-19

not 10^19

[finite hatch]

Oh ok thank u

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[wary mantle]

#1195800467880558672 message

[wary mantle]

.close

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[pearl axle]

i know that matrix dimension should by 2 x 2 but not sure where to go from there

[hard atlas]

the columns of the matrix are where the standard unit vectors end up

[pearl axle]

right

i get that but how do the transformations tell me about which entries are in the vectors

like does a reflection around the x1 axis mean G(e1) = [-1,0]

[pearl axle]

and then a reflection of that around x2=x1 mean G(e2) = [1,0]

[hard atlas]

no

why are you only doing one of the steps to each of e1 and e2

[pearl axle]

oh

wait so what am i supposed to do

[hard atlas]

do both things

[pearl axle]

didn't i do both?

there's two reflections right

and one affects e1 and the other affects e2

[hard atlas]

lets do it one reflection at a time

what happens when you reflect e1 and e2 around the horizontal x_1 axis

[pearl axle]

both become [-1,0]

[hard atlas]

no

[pearl axle]

a reflection across the x axis makes the vector negative

right?

so what else would it be

[hard atlas]

a reflection across an axis is not the same as multiplying the vector by -1

[pearl axle]

oh

ok then im not exactly sure

hint pls?

[hard atlas]

have you tried drawing the situation?

[pearl axle]

kind of

im not too sure on how to draw mutliple dimensions

oh wait nvm its 2x2

yeah but im still not sure how to draw them

[hard atlas]

its just R^2

aka the plane

[pearl axle]

yeah but like the reflection

so if we begin at (0,0)

doesen't the reflection across x1 make x negative

[hard atlas]

no

it makes y negative

[pearl axle]

oh

[pearl axle]

so then e1 and e2 become [0,-1]

[hard atlas]

no

e1 is [1,0]

now we make y negative

so we still get [1,0]

[pearl axle]

right

makes sense

so does the same apply to e2

[hard atlas]

depends on what you mean by "the same"

[pearl axle]

wait hold on

so im a little confused

the first transformation occurs

and makes e1 [1,0] and e2[1,0]?

[hard atlas]

no

do you know what reflecting over an axis means

[pearl axle]

yeah

the image gets transformed acroos the axis right

?

[hard atlas]

then why do you think that e2 gets mapped to [1,0]

[pearl axle]

oh

oh

OH

i just re read the question

my bad

i was thinking of e1 and e2 as two points

not vectors

[hard atlas]

that doesnt really change stuff

[pearl axle]

ok

then

im lost

sorry im kinda confused

[hard atlas]

can you open paint and draw the situation

[pearl axle]

hold on lemme try

[pearl axle]

this is the vector post reflection right?

[hard atlas]

well you shouldnt call it x again

but yes

and where are e1 and e2?

[pearl axle]

isnt the vector x below the x axis e1

[hard atlas]

no

[pearl axle]

oh

ok so where would i draw it

[hard atlas]

e1 is on the x axis

e2 is on the y axis

[pearl axle]

right

so then

what would happen

[hard atlas]

so, after the first reflection e1 landed at [1,0] and e2 landed at [0,-1]

now the next reflection

can you draw where the axis of the reflection is?

[pearl axle]

sure lemme try

hold on the problem says x1=x2 line

is that the y axis?

[hard atlas]

no

can you draw a few points (x1,x2) with x1=x2?

[pearl axle]

lemme try

is that the x2=x1 line?

but it goes through the origin

[hard atlas]

no

[pearl axle]

oh

oh is it the other way around

like this

but passing thru origin

[hard atlas]

yes

[pearl axle]

oh

ok so if this reflection occured

then the vectors would swap

e1 would become [0,1] and e2 would become [-1,0}?

[hard atlas]

yes

[pearl axle]

oh

understood

thank you!

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@pearl axle Has your question been resolved?

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[lofty spindle]

I need help with LU factorization

[lofty spindle]

The question is this ^

and i solved it to this

am i on the right track

?

because i dont know if the question implies that there is a way to just use elementary matrices to get to a conclusive answer or if i have to rearrange it somehow to get to the answer

[vocal sleet]

@lofty spindle Has your question been resolved?

[obsidian stream]

Write L and U as the product of elementary matrices

Then use those to write U^-1 and L^-1 as the product of elementary matrices

(Yes, this is very tedious)

[lofty spindle]

Bruh

😭😭

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@lofty spindle Has your question been resolved?

[lofty spindle]

.close

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[lofty spindle]

/reopen

.reopen

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✅

[lofty spindle]

im trying to decompose it to LU but i cannot seem to get it to work

where am i going wrong?

im not getting a lower triangular matrix

[vast shale]

.close

[lofty spindle]

nvm there's no decomposition for LU since row interchanges has to be used

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[vast shale]

Hi

[vast shale]

Tryna find the words to use on this math word bank

I’m on the 4th one

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[karmic osprey]

I did the first one fine but the wording of the second is throwing me off/im not sure how to approach it? Would I just take z= x+1 as my surface and go from there?

[regal bane]

Ye

So for example, the integral in part a) could have represented the volume under the surface z = xy

[karmic osprey]

Thank you so much!!

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[chilly rampart]

Can someone explain part d for me

I don’t understand what he does after writing down the integral with u

How does he get 2(11+18)

[flat whale]

Need to show the entire question to tell

[flat whale]

Read the first sentence

[chilly rampart]

Don’t get it

Still

[flat whale]

Sorry, meant third sentence

"the arc length..."

[chilly rampart]

Ok now I see

But isn’t this h(x)

Is it because we used u substitution

[flat whale]

well the formula for arc length of h is just a multiple of the arc length of f

[chilly rampart]

Oh I get it

That’s why we multiply by 2

Thanks

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[raven ocean]

Is this correct

[raven ocean]

Also is there a shorter way of doing it, if it is correct

[inner osprey]

yes

your method is fine ig but it's slightly long

[pallid forge]

looks correct! not sure if there's an easier route. maybe you couldve added the equations together after multiplying the first by -2 on both sides?

[edgy gulch]

just subtract the two equations

you'll immediately get f + g

[raven ocean]

So instead of subbing it in I can just subtract the entire eq?

[edgy gulch]

yes

[raven ocean]

Ok thank you

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[winged burrow]

Hi

[winged burrow]

I need help with number one question

[jagged cargo]

which one

[winged burrow]

The very first question

[jagged cargo]

so i assume 1 i

[winged burrow]

Yea

[jagged cargo]

do you know the formula for calculating sinx = a

[winged burrow]

No

This is a new chapter

For me

[jagged cargo]

lets say sinx=1/2

[winged burrow]

Okay

[jagged cargo]

what value of x do you think will satisfy such equation

[winged burrow]

Umm

0.7245?

No I mean it's the given value

[jagged cargo]

answer this first

draw out the unit circle

[winged burrow]

Unit circle?

What is that?

[winged burrow]

Omg what is this

[jagged cargo]

in case you dont know, the x axis represents value of cos

[winged burrow]

Ohh

How can I understand that

Are u here?

[heavy yoke]

for angles in the first quadrant, you can draw a right triangle and the side lengths will be 1 (hypotenuse), sin(theta) (opposite, vertical), and cos(theta), (adjacent, horizontal)

[winged burrow]

Right angle?

[winged burrow]

Ohh

Okay

[heavy yoke]

we then define cos(theta) and sin(theta) to be the horizontal and vertical positions on the circle even for angles larger than 90°

[winged burrow]

I just knew abt this

[winged burrow]

Ohh

I got it

Then?

Are u here??

[flat whale]

Try doing a problem with the new information you just learned

[winged burrow]

With that information I can solve them?

[flat whale]

Maybe! We don't know what you know or don't know

Other than what's been discussed in this channel

[winged burrow]

I am not able to solve those

But the information

I understood

[flat whale]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:inverse-trig/v/inverse-trig-functions-arcsin

[winged burrow]

This video will help me?

.close

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[royal kestrel]

factorise completely:

[royal kestrel]

all i did was factor (a/b - b/a)

but i think theres more

[vagrant solar]

correct there’s the rest of the equation

[royal kestrel]

so what i have rn is $(\frac{a}{b} -\frac{b}{a})[(\frac{a}{b}+\frac{b}{a})-(\frac{a}{b}-\frac{b}{a})]$

[twin meteor]

taro

[royal kestrel]

im using this channel

#❓how-to-get-help

[frail crater]

sorry

[gaunt sparrow]

This is good, now what can you do with that right factor?

[royal kestrel]

expand the a/b - b/a?

with the negative in front

[royal kestrel]

like this? $(\frac{a}{b} -\frac{b}{a})[(\frac{a}{b}+\frac{b}{a})+(\frac{b}{a}-\frac{a}{b})]$

[twin meteor]

taro

[royal kestrel]

or am i supposed to

[gaunt sparrow]

Hmm I mean it's not wrong, but there are things you can combine in there

[royal kestrel]

combine the fractions

?

[gaunt sparrow]

Well you have a/b and -a/b

What does that make?

[royal kestrel]

OH

0

and then 2b/a

[gaunt sparrow]

Yes

[royal kestrel]

so

$\frac{2b}{a}(\frac{a}{b} -\frac{b}{a})$

like that

[gaunt sparrow]

Yes

[twin meteor]

taro

[royal kestrel]

and i cant factor that any further yes

no common terms

[gaunt sparrow]

I don't think so either

[royal kestrel]

okok tysm

i have one more that im js hoping to get my solution checked

factorise completely: 5b^4 - 405

5(b^4 - 81)
5(b^2 + 9)(b^2 - 9)
5(b + 3)(b - 3)

then to factor b^2 + 9 i made it into b^2 - (3i)^2 which was (b + 3i)(b - 3i)

so i got 5(b+3)(b-3)(b + 3i)(b - 3i)

is that right?

[vagrant solar]

3 x -3 is -9

[royal kestrel]

what part is this for

[vagrant solar]

nvm it’s correct

[royal kestrel]

ok tysm

.close

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[shut valve]

can I do this or no?

[shut valve]

nvm that makes literally no sense

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[visual marten]

.

[visual marten]

n^3 grows bigger than 2n which is 0 and 1/n is also 0

[somber flame]

why would it be

[visual marten]

sorry

i meant how is it 2 and not 0

[somber flame]

yea it should be 0

[blissful saffron]

,w limit n to infinity ((2n+1)/(n^3 + 2)-1/n)

yes

[twin meteor]

Results provided by WolframAlpha

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Requested by derp_z

[blissful saffron]

it really should be 0

yea

[visual marten]

ok i thought i was really dumb

tyty

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[vast shale]

the logic I'm trying to use here is that f'(x)=2x , so m=2..

and that some point, assigned the variable ("a")

has values (a,a^2)

so what I wanted to do was use my slope and set it equal to the secant of my values

but I don't know if that even makes sense

I feel like I understand the procedure up until the equaling slope to something part

[somber flame]

yea that sounds about right

[vast shale]

I tried it using the $y_2-y_1=m(x_2-x_1)$ formula

[twin meteor]

wyldinwilliam

[vast shale]

to some success

ended up with $a^2-2a+5$

[twin meteor]

wyldinwilliam

[somber flame]

well actually here why m=2?

[vast shale]

well if you sub in one of the given points m=2

or "x"

but that doesn't make sense to me either lol

I feel like I'm using patterns instead of understanding

[somber flame]

that will just give you the slope of the tangent line at the point (1,f(1)) though

the derivative evaluated at x gives the slope of the tangent line at the point (x,f(x))

and among the tangent lines, you want to find the one that passes through (1,-3)

[vast shale]

I see

so those two points should be points you include

in a calculation with x1/y1

[somber flame]

eh not specific enough for me to say yes but maybe haha

[vast shale]

okay

so what's next?

[somber flame]

so you can think of it like, you want to find the x with the property that the tangent line at (x,f(x)) passes through (1,-3)

if you are given an x, you know the tangent line at (x,f(x)) has slope 2x

or we can say a instead of x

to keep your old notation

[vast shale]

okay, so you'd have to equal 2x to whatever slope formula you use

$2x=\frac{x^2+3}{x-1}$

[twin meteor]

wyldinwilliam

[vast shale]

is what I initially used

the y2-y1/x2-x1

[somber flame]

ummm not sure what that does lol

[vast shale]

well hopefully you'd get something useful out of that

but I'm pretty sure you'd have to use the

derivative formula

$2x=lim_h\rightarrow0\frac{f(x+h)-f(x)}{h}$

[somber flame]

i don't think we need that?

we have the derivative already

[twin meteor]

wyldinwilliam

[vast shale]

your right

so what would we do then

we have slopes, we have initial points

[outer warren]

i would've used a dummy variable, but yeh solve
$$2a = \frac{a^2 + 3}{a-1}$$

[twin meteor]

ℝαμΩℕωⅤ

[vast shale]

hm okay

$2a^2-2=a^2+3$

[twin meteor]

wyldinwilliam

[outer warren]

incorrect

[vast shale]

$2a^2-2a=a^2+3$

[twin meteor]

wyldinwilliam

[outer warren]

that's better

[vast shale]

$a^2-2a-3=0$

[twin meteor]

wyldinwilliam

[vast shale]

$(a+1)(a-3)=0$

[twin meteor]

wyldinwilliam

[vast shale]

$$a=-1$$

[twin meteor]

wyldinwilliam

[vast shale]

$a=3$

[twin meteor]

wyldinwilliam

[outer warren]

yes

[vast shale]

that's weird isn't it?

[outer warren]

weird how

[vast shale]

same coords opposite signs lol

[outer warren]

coincidence

[vast shale]

okay so this gives us essentially the new x values

which we use to find the slope formulas

ok not sure I get this

obviously we have two "a" values

but how do these eqns make sense

I'm getting something along the lines of y=2x+3

take a=-1 for example

(-1,1) is the coordinate

$1=-2+b$

[twin meteor]

wyldinwilliam

[vast shale]

$b=3$

[twin meteor]

wyldinwilliam

[vast shale]

$y=2x+3$

[twin meteor]

wyldinwilliam

[vast shale]

wait we have to find the slope first

.close

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[brave ruin]

If i have the floor(-x) and - floor(x) what does the latter mean?

[brave ruin]

I take the former to mean that x would be negative. So
Floor(-x) <= -x.

Buy - floor(x)? Does that mean x is positive?
So - floor(x) <= x ?

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@brave ruin Has your question been resolved?

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[jagged cargo]

translation:

$f(x)$ has a continuous derivative on $[0, 1]$

[twin meteor]

FungusDesu

[jagged cargo]

$\int_0^1[f'(x)]^2dx = 7$

[twin meteor]

FungusDesu

[jagged cargo]

$\int_0^1x^2f(x)dx=\frac13$

[twin meteor]

FungusDesu

[jagged cargo]

$f(1)=0$, evaluate $\int_0^1f(x)dx$

[twin meteor]

FungusDesu

[flat whale]

the cheating way to do this is construct some simple function like linear or quadratic function that satisfies the 3 conditions

probably quadratic since you need 3 variables

[jagged cargo]

well, i do know how to solve this question, and the way is exactly like you said

my goal is to solve this using holder's inequality

how would this be applicable here?

[flat whale]

why would an inequality give you the answer

i mean maybe

[jagged cargo]

yup, thats why im asking

i dont know either

but in my teacher's notes, i saw him using this

and im not quite sure how he did it

ill try my best to translate it

Second way: Recall the Holder's inequality as such:
...
LHS equates to RHS when f(x)...
We have 1/9 ... LHS equates to RHS when f'(x)...
On the other hand...therefore f(x)...
From this:...7/5

[flat whale]

oh then do the first line then

[jagged cargo]

hm wait, i think it just clicked to me

[jagged cargo]

oh yeah i got it now, he was essentially proving if we were to transform this a bit, we should get something equivalent to the first integral

and what he was proving is that it is equivalent

ill ask him more about this, as i have the same goal but used a different theorem in mind

alright, thanks for your help

.close

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[vast shale]

i need some help with this integral

[vast shale]

if it was x^2 on the denominator it would be easy

how should i work on this one?

[tidal dock]

split into two integrals?

then trig sub

[scenic ravine]

I don't think cos(x)/x^2 is integrable