help-42

[chrome robin]

is J 37

coressponding angles id kwhat

i need to find these angles

[cedar quartz]

Yeah

[chrome robin]

k?

180-37?

[cedar quartz]

And k also 37

[chrome robin]

damn

u sure

[cedar quartz]

Yeah

Im sure

[chrome robin]

huh so is k 37

or 180-37

[cedar quartz]

All this angles right here are 37

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[chrome robin]

thx

[calm coral]

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[chrome robin]

hi

[chrome robin]

what about s? 😢

[cedar quartz]

R = 153

S = 180 - 153

[chrome robin]

y?

[reef nimbus]

s is the same as the complement to 153

[chrome robin]

huh

isnt 180- for angles on staright line

oh shi

they are

my bad 💀

thx guys have a blessed day

or night

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[chrome robin]

pls help

idk how to find hypotenuse

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[mossy wigeon]

how do we do a)

is it just adding a_1 to a_10

of S_N

[chrome robin]

is it 180 - 20 - 90 then angleso n straight line so 110 then 180-110 /2
x = 45?

[swift path]

Aren't S_N the partial sums?

[hasty fiber]

is it just adding a_1 to a_10
of S_N
what does this mean?

[chrome robin]

whats this shit

[upper sparrow]

!occupied

[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).

[mossy wigeon]

im confused on the notation

like im not too sure what

thisis asking

[swift path]

a_n are some numbers

Which we don't know

There is a_1, a_2, a_3, a_4,...

Then there are the S_N

[mossy wigeon]

what is 4-8/n^2

[swift path]

Wait

[swift path]

The S_N can be written the following way:

S_1 = a_1

S_2 = a_1 + a_2

S_3 = a_1 + a_2 + a_3

Et cetera

[swift path]

This is saying, that
S_1 = 4 - 8/1^2
S_2 = 4 - 8/2^2
S_3 = 4 - 8/3^2
...

So for example a_1 + a_2 + a_3 = S_3 = 4-8/9

[mossy wigeon]

so a_3 would be 4-8/3^2 right?

[swift path]

No

S_3 is 4 - 8/3^2

But S_3 is not a_3

S_3 is a_1 + a_2 + a_3

Do you understand the difference?

[mossy wigeon]

yeah but a_3

is it

[swift path]

No

[mossy wigeon]

ohh

[swift path]

You must find a_3 from the S_N

For example, to find a_4

You can do the following:

S_4 = a_1 + a_2 + a_3 + a_4

and

S_3 = a_1 + a_2 + a_3

So, S_4 - S_3 = (a_1 + a_2 + a_3 + a_4) - (a_1 + a_2 + a_3)

And a_1 + a_2 + a_3 cancels

So S_4 - S_3 = a_4

Therefore, a_4 = S_4 - S_3 = 4 - 8/16 - (4 - 8/9) = 40/144

Can you extrapolate that to a_3?

[mossy wigeon]

ahhh okay

yeah i was confused by my book

tysm!!

.close

[swift path]

Also

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[swift path]

Wait

[mossy wigeon]

yeah?

[swift path]

This

[upper sparrow]

.repoen

[mossy wigeon]

.reopm

.reopen

[calm coral]

✅

[swift path]

The symbol $\sum_{n=1}^{10}a_n$ is a different way of writing $a_1+a_2+\cdots+a_{10}$

[potent lotus]

d

[swift path]

Thanks

[mossy wigeon]

oh why is that

[swift path]

Someone invented it many years ago to avoid writing ... every time there was a sum with many terms

It is a widely used notation

[mossy wigeon]

oh i

i thought you meant

it is a different thing

than

yeah okay yes i know that

okay LOL

mb

[swift path]

So it is asking for the result of that sum

[mossy wigeon]

wait i did dthat though

like

[mossy wigeon]

i added a_1 until 1_10

[swift path]

I think you added S_1 until S_10 instead

[mossy wigeon]

ah yeah

okay ill redo it tysm!

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[grand harness]

Hi, I have no clue how to solve this limit

[grand harness]

I know that, if it exists, it must be zero but if it does I just don't have the tools available

so any help would be appreciated

[swift path]

To show it is zero, bound its absolute value

[grand harness]

I tried it already but got nowhere

though it's because I'm not good at that method

I don't "get" it yet

[swift path]

In the numerator you have x y sin(x)

Forget about the sin, as x -> 0

You deal with it later

Now you have x^2 y

It has degree 2 on the x and 1 on the y

Now, a denominator like |x^2 + y^2| can be bounded in three useful ways:

|x^2 + y^2| > x^2

|x^2 + y^2| > y^2

And |x^2 + y^2| > |2xy|

And which one you choose depends on the numerator

You want to choose one that leaves you with not negative powers of x and y after dividing

In this case, both x^2 and |2xy| are useful

[grand harness]

I think I got it

also the sin doesn't matter if I'm bounding because the highest value sinx can have is 1

[swift path]

No

[swift path]

It is a different argument

You have $\dfrac{|xy\sin(x)|}{x^2+y^2}$

[potent lotus]

d

[swift path]

Then multiply and divide by |x|/|sin(x)|

You obtain $\dfrac{|\sin(x)|}{|x|}\dfrac{|xyx|}{x^2+y^2}$

[potent lotus]

d

[swift path]

The first fraction goes to 1 as x -> 0

[grand harness]

I now understand what you mean

you take advantage of the identity lim x->0 sinx/x = 1

that's cool

after that it's easy to prove the rest of the equation is equal to 0

because you get y x^2/x^2+y^2

|y| tends to 0 and x^2/x^2+y^2 is bound between 0 and 1

so the limit is 0

that's a real theorem

[swift path]

If you justify why x^2/x^2+y^2 is bound between -1 and 1, then yes, that works

It is easy to show anyway

[grand harness]

Sorry, I mean 0

but yes

x^2 =< x^2 + y^2

[swift path]

Yes

[grand harness]

epic, thank you

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[late scroll]

Part 2 question 1

I got 1/2

I don’t think I did something wrong

Is the question wrong

[hybrid agate]

I got 1/2 and 3/2 so they might be “wrong”

It’s not really wrong it’s just that using this technique you would get 1/2 and 3/2

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[knotty mulch]

is this exponential decay or exponential growth? why?

[knotty mulch]

.close

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[late scroll]

How do I prove that this series converges or diverges

I can’t use the integral test bc the derivative is positive

The nth term test/limit test return 0 so it is inconclusive

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[late scroll]

Help

[late scroll]

Scroll up a bit

[hybrid agate]

The derivative is in fact negative

$\cfrac{1}{\sqrt{x}}= x^{-1/2}$

[potent lotus]

alex

[steep mirage]

Help pls😭

Is my work wrong?

@late scroll

<@&286206848099549185>

[hybrid agate]

Use your own channel

This one is occupied

[steep mirage]

I was assigned to this one

[lunar trench]

you literally were not

[steep mirage]

Figured it out. Sorry

Got it

[calm coral]

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[calm coral]

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[slim coyote]

anyone can help me out for part (iii) and further

[slim coyote]

(i) should be 12 and (ii) should be 6

[stiff root]

SOMEOME HLEP

:)

[dense falcon]

<@&286206848099549185>

[calm coral]

@slim coyote Has your question been resolved?

[untold compass]

I thiiink DGM is parallel to ABFE, and since ABFE has the plane equation 6x + 0y - 1z = 36, it has the equation r dot (6,0,-1)=c and any plane parallel has the same equation too

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[mossy wigeon]

im not too sure where to start with this

this is a geometric serie right?

[dull geyser]

Sure is

[civic dirge]

But it doesn't have the form of one, as written. You'll need to change it a bit

It'd be nice if both powers were k+1

[mossy wigeon]

uhm how do i do that

[upper sparrow]

Rules of indices

[mossy wigeon]

like 2^k+1+4?

i cant think of anything else

[verbal finch]

make it

2^4*(2/9)^k+1

[upper sparrow]

Yep

[verbal finch]

is it 64/63?

the ans

@mossy wigeon

[mossy wigeon]

if its 64/63

i would be divergent no?

[verbal finch]

no

if it was divergent u wouldnt have an answer

[mossy wigeon]

r>1 is divergent

[verbal finch]

r is 2/9 here

[mossy wigeon]

oh

so 64/63

you mean the sum right

then yeah its that

i got that by getting a_1 a_2 a_3

but idk why they wanted to make it k+1

[upper sparrow]

(it would probably have been better to have it as k - 1, as that's closer to the form ar^{k - 1}, but regardless what matters is that you can find the first term and common ratio)

[mossy wigeon]

why r^k-1 specifically? is there a reason

[mossy wigeon]

also what should i do after i find this

if i use this appraoch

[upper sparrow]

Cause then your sequence/series is then a, ar, ar^2, ..., so your first term, when k = 1, is a

[mossy wigeon]

ah okay

[upper sparrow]

If that's the form you use, of course, some index from 0 and then you can ar^n, with n = 0 getting you a and all

[upper sparrow]

And the idea is that you should hopefully be more able to identify what the common ratio is when you get those to the same (first term is a tiny bit of work, putting k = 1 in directly, if you don't rearrange as per before)

[mossy wigeon]

ahh tysm!

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[dawn sparrow]

Why is it that I need to add the two?

[lucid umbra]

Consecutive even numbers

[swift laurel]

it might be more clear with parentheses:
x + (x + 2) = 74
x is the first number, (x + 2) is the second number

[dawn sparrow]

but 36+38?

[dawn sparrow]

Yes but why cant you do 36+38=74

thats two even numbers

[swift laurel]

you are trying to set up an equation to describe the situation, not find the right number by guesswork

[dawn sparrow]

yes so how would i find it out?

the equation is x+x=74

simple as that

where does 2 play a role

[swift laurel]

the equation is "two consecutive even numbers that add to 74"

let's call x the first number

[dawn sparrow]

k

[swift laurel]

then, the next even number must be x + 2

[dawn sparrow]

why tho

[swift laurel]

the next number after x is x + 1, but that's odd, so we have to find the number after that, which is (x + 1) + 1 = x + 2

[dawn sparrow]

my brain aint braining

just explain it simpley

please

[swift laurel]

if i have the number 12, then the next even number after that is 14, which is 12 + 2

[dawn sparrow]

nah ur actually a legend

thanks bro

this server be saving my life

🙏🏻

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[lament tendon]

don't solve it for me just tell me how can I use <FCG to find arc FG?

[upper dew]

draw radiuses PF and PG

and notice that they r perpendicular to BC and DC respectively

which makes a quadrilateral

[lament tendon]

huh

[upper dew]

it makes the quad CFPG

in which u have 3 of the angles

[lament tendon]

ok im lost, can you solve it and explain

[upper dew]

it would look like this

but they gave u P

so u can figure out O

[lament tendon]

aha then

[upper dew]

did u find angleFPG from the original diagram

[lament tendon]

no?

what if we use this <FCG= 1/2((360- GF)-GF)

lets make gf x

then its 61=1/2((360-x)-x)

122= 360-2(x)

122-360=-2(x)

-238=2(x)
119=GF

is this right?

then we can 360-119 to find arc GHF

[upper dew]

ye

thats correct

[lament tendon]

oh

okaay then, thank you so much

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[tepid oriole]

I am trying to use my TI-84 Plus CE to calculate this, but it keeps giving me a syntax error. Although i can calculate it on integral calculators online

[tepid oriole]

how will i go about calculating it in the TI-84?

.close

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[blazing coyote]

is the answer to this (b)?

[glass heart]

why

[vital raven]

well it's not D

[blazing coyote]

cos^n (x)for nπ is either 1 or -1

[vital raven]

and C doesnt restrict enough values

so it cant be C either

so its B or A

it suffices to show there is at least 1 solution lol

[blazing coyote]

or do I use the fact that there are as many negative ones as positive ones

[lyric ravine]

It is, yes

[blazing coyote]

thanks

[vital raven]

process of elimination is the way i think

[blazing coyote]

I just use the fact that |r|<1

[vital raven]

yes but how do you know there is no solution, i.e. (A)

x = pi/2

[blazing coyote]

cos(pi/2)=0

[flint condor]

Can't you evaluate it as GP?

[blazing coyote]

that's the idea I used, yes

for an infinite GP to converge |r|<1

[vital raven]

cot^2x = cot^2x LOL

so its literally just the domain of cotx

[blazing coyote]

thanks everyone !

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[winter marlin]

im supposed to solve this using definition of integration

[winter marlin]

i keep getting different answers, can someone help me with this?

thing is i dont understand how the summation distributes

is this correct? or an even possible way of distributing the summation?

if so what happens next? i keep getting different answers like infinity but if i integrate regularly its b'2/2 - a'2/2 and it just wont give me that answer

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@winter marlin Has your question been resolved?

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[amber wedge]

is -(a/b) = -a/b = a/-b ?

[pseudo wedge]

yes

[amber wedge]

ok thanks

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[amber veldt]

why does the Area integral have tangent?

Area of polar curve: is just sqrt(R^2) within an integral

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[remote mural]

no progress doned by me since i have zero idea where to start

please ping me when youre responding

any help would be appreciated thanks

[rustic osprey]

Hint: Draw perpendiculars to BC, from D, E, F

[remote mural]

done

what next ?

[rustic osprey]

it's a hint

[remote mural]

yeah yeah im thinking

[remote mural]

dont know what to do next

FR(DP+EQ)-DP(EQ)>=0
is it the next move ?

actually im not sure

[rustic osprey]

lemme look at what you have rn

[remote mural]

sure sir

[rustic osprey]

this idea seems fine, I'm assuming the specifics are correct

I would instead substitute into y^2 > xz directly (then just write the proof backwards)

you can find x and z in terms of y and the lengths using the last two equations

[remote mural]

since i showed DP>FR and EQ > FR
i took random values and getting RHS as >=0
should i put like this?

[rustic osprey]

i took random values and getting RHS as >=0
?

as in you picked specific values for the lengths?

that's showing it's true for a specific case, not in general

[remote mural]

yuss

ohh then would i do this further

is it weong attempted ?

[rustic osprey]

uhhh idk wym tbh

if you want to keep choosing specific values

that's not a proof

[rustic osprey]

.

[remote mural]

yea

[remote mural]

then how can i show this

[rustic osprey]

...

[remote mural]

means you want me to find y^2 from 1st eqn
and put x and z in terms of y from 2nd and 3rd equations ?

[rustic osprey]

yes

[remote mural]

$$\frac14 (BC)^2 (FR)^2 \geq \frac14 (BC)^2 (DP)(EQ)-Y \frac12 (BC)(DP+EQ)+Y^2$$
$$\frac14 (BC)^2 (FR^2 - DP \cdot EQ) \geq y^2 -\frac{y}{2} (BC)(DP+EQ)$$

[potent lotus]

praee2k

[remote mural]

sorry capital and small y are same

[rustic osprey]

I'm lazy so I'm assigning variables

and $a+b+c+d=\ell$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

$x+y=\frac{1}{2} \alpha \ell, z+y=\frac{1}{2} \gamma \ell, y=\frac{1}{2} \beta \ell$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

$y^2>\frac{1}{4} \alpha \gamma \ell^2 -\frac{y}{2}(\alpha \ell+\gamma \ell)+y^2$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

oh that's much better

b/c now you need to show that $$y>\frac{\alpha \gamma \ell}{2(\alpha+\gamma)}$$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

or simply $$\beta>\frac{\alpha \gamma}{\alpha+\gamma}$$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

yeah that's much nicer

[remote mural]

no i need to y^2 > zx

lemme do once from yours method

[rustic osprey]

the point is that we're trying to find something simpler to prove, then we can simply work backwards.

[remote mural]

ohh great idea

[rustic osprey]

I'm pretty sure it's trivial if you consider ||EQC|| and ||FRC||

[remote mural]

lemme do try

by yours method once

[remote mural]

how you get the idea of EQC and FRC ?

[rustic osprey]

the configuration screams similarity

[remote mural]

why its not sounding to me

ik these triangles are similar lemme think

,,\frac{\alpha}{\beta}=\frac{b+c+d}{c+d}

[rustic osprey]

you forgot abt d

[remote mural]

aiyoo

[rustic osprey]

._.

that's right, but I'd stick with the (b+c+d)/(c+d)

[rustic osprey]

b/c remember you want to piggy back off of this

[potent lotus]

praee2k

[remote mural]

man im done

dont know what to do

my exam is about to happen in 12th may

im still a dumbshit trahs

[rustic osprey]

think of how you can rearrange this to get a beta/alpha

[remote mural]

do i need to take other side similarty case too ?

[rustic osprey]

yup

[remote mural]

i got alpha / beta = (b+c +d)/(b+c)
and gamma / beta = (a+b)/(a+b+c)

but dont knwo how can i manipulate furtheer

[rustic osprey]

.

[remote mural]

im not getting

still

my brain aint brainin anything

[rustic osprey]

Hint: $\frac{\alpha+\gamma}{\alpha \gamma}$

[potent lotus]

Civil Service Pigeon

[remote mural]

bro im so dumb i literally find it and didnt doned naything further related with this

[rustic osprey]

?

$\frac{\beta(\alpha+\gamma)}{\alpha \gamma}>1$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

now use this

the rest is just annoying algebra

[remote mural]

just backward proof right ?

[rustic osprey]

yeah just do it backwards again to prove the lemma

anyway ping me if you're still stuck

I gtg

[remote mural]

yussir

1st part doned

thank you so much

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[calm coral]

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[remote mural]

hi, currently working on line/surface integrals of vectors fields but stuck on the parametrization of the curve? Once I am given the parametrized form of the curve I can compute the integrals and cross products perfectly fine, just stuck on the easy step 🙃 was wondering if somebody could help explain how to parametrize a curve? I’ll send three examples with the step I am confused on highlighted

[remote mural]

please @ me when you respond so I don’t miss it! Sorry if I don’t respond quickly, it’s quite late in my timezone

[inner zealot]

You mean how they got s, t, and s^2/8 from y^2 = 8x?

[remote mural]

yup! silly question probably

[remote mural]

so I can just pick whatever variable I want to equal s and then solve the equation for the other variables from that definition?

[inner zealot]

Basically, you could solve for y=sqrt(8x) but that adds a ± into the equation which you generally want to avoid with parametric equations.

z is independent so you can set it equal to its own parametric variable.

[remote mural]

okay perfect thank you so much! my professor never explicitly explained that step or why he chose the variables so even though it was super simple I was lost but could do the rest of the steps haha

.close

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[calm sapphire]

what does this mean

[calm sapphire]

<@&286206848099549185>

[calm coral]

@calm sapphire Has your question been resolved?

[untold compass]

find a matrix that squares to A

[calm sapphire]

thank u

so far

i started by finding the eigenvalues

now im finding the eigen spaces

is that the correct approach

[untold compass]

yea eigenvalues are the usual way

the way I know is you get the eigenvalues/eigenvectors and the square root becomes Vsqrt(D)V^-1 where D is diagonal with eigenvalues and V is columns of eigenvectors

called the eigendecomposition

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[warped sparrow]

can someone help me with this

[calm coral]

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[fervent cargo]

only own non-verified existence fronts the verified.

[calm coral]

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[calm coral]

@idle fractal Has your question been resolved?

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[idle fractal]

.reopem

.reopem

.reopem

[cerulean cradle]

.reopen

[rough ferry]

$4^{2x} = 5^{2x-1}$

[potent lotus]

rynite

[rough ferry]

I've tried solving this

and ended with

$\frac{-\log5}{2\log{\frac{4}{5}}}$

[potent lotus]

rynite

[rough ferry]

is the simplest form?

or can I still go further

answer must be exact

[blazing coyote]

unless you can use certain special functions, you'll have to resort to graphical means me thinks

[rough ferry]

what do you mean certain special functions

I'm just asking if I can simplify further

[wary path]

im not sure what brother is talking about either, but i'd go about this by applying the log first then do the other manipulations after, i think it'd be cleaner

[rough ferry]

im not sure I follow

[quaint sphinx]

yes

this answer is good enough

only thing i might suggest is you can remove the negative sign and have 5/4 instead of 4/5 on the bottom

[wary path]

ah idk why i thought doing it another way would remove the necessity for a fractional part

[rough ferry]

do you mind showing me how?

[wary path]

it's the log property -log(a/b) = log(b/a)

[rough ferry]

ah

$\frac{\log\frac{1}{5}}{2\log{\frac{4}{5}}}$

so that

[potent lotus]

rynite

[wary path]

Mqnic's recommending you to absorb the - to log(4/5)

so it becomes
$\frac{\log{5}}{2\log{\frac{5}{4}}}$

[potent lotus]

feeds

[rough ferry]

how did the fracs flip

[wary path]

move the negative sign to the denominator first, then apply the appropriate rule

so $\frac{-\log{5}}{2\log{\frac{4}{5}}}$ becomes
$\frac{\log{5}}{-2\log{\frac{4}{5}}}$ becomes
$\frac{\log{5}}{2\log{\frac{5}{4}}}$

[potent lotus]

feeds

[wary path]

i hope that's clear enough, and yeah i agree with this suggestion of removing the negative sign

[rough ferry]

is that done by multiplying the whole thing by -1

[wary path]

yeah you can think of it that way

[rough ferry]

alright thank you!

.close

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[cosmic wraith]

help plz

[cosmic wraith]

all ive gotten up to is x(x+2) = 8pi

[wary path]

x(x+2) is the area of the rectangle

[blazing coyote]

they are talking about perimeters here

[cosmic wraith]

oh shoot

thx

another question

when it says the difference between succesive rings was 1 point more than monday

does it mean that each ring will be 3x - d + 1, 3x - 2d + 1, etc

[wary path]

wait

it becomes x+3, x+3 - (d+1), x+3 - 2(d+1),... the middle point increases by 3 and the point difference increases by 1

[cosmic wraith]

bruhh

i accidently did 3 times x

not 3 + x

tx

[wary path]

❤️

[cosmic wraith]

so for tuesday it would be 3(x+3+1)

@wary path

[wary path]

one arrow on x+3-(d+1) and three on x+3-2(d+1)

[cosmic wraith]

ah i c

thanks

[wary path]

👍

[cosmic wraith]

bruh i still cant get the snwer

[wary path]

what's your total for monday and tuesday

[cosmic wraith]

monday is

2(x-d)+2(x-2d)

= 4x-6d

[wary path]

yup

tuesday?

[cosmic wraith]

tuesday is -5 - 7d?

no 4x - 5 - 7d

[wary path]

,w x+3-(d+1) + 3(x+3-2(d+1))

[potent lotus]

Results provided by WolframAlpha

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

Requested by therealfeeds

[wary path]

hm hold on

yeah youre right

okay

and James got the same score on both days

what does this give us

wait no

[wary path]

• 5

4x + 5 -7d

[cosmic wraith]

?

o

[calm coral]

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[scenic merlin]

I stuck at proving an upper bound for $\beta$ in the following context:

$\alpha,\beta,\gamma,\delta\in\mathbb{R}$ satisfy the following:

If $x\in[0,\frac{1}{4}]$, then $\alpha x\ge 0$.

If $x\in(\frac{1}{4},\frac{1}{2}]$, then $4\beta x+(\alpha-\beta)\ge 0$.

If $x\in(\frac{1}{2},\frac{3}{4}]$, then $4\gamma x+(\alpha+\beta-2\gamma)\ge 0$.

If $x\in(\frac{3}{4},1]$, then $4\delta x+(\alpha+\beta+\gamma-3\delta)\ge 0$.

$3\alpha+2\beta+\gamma=16$

$-2\alpha-\beta+\delta=-16$

Now I am able to show that $\alpha\ge 0$ and $\beta\ge -\alpha$.

For the upper bound, I found $\beta\le 16-2\alpha$ but I was told that it is not the least one. I want to show that $\beta\le \frac{16-4\alpha}{3}$.

Can anyone help please.

[potent lotus]

Trenton

[calm coral]

@scenic merlin Has your question been resolved?

[scenic merlin]

<@&286206848099549185>

[calm coral]

@scenic merlin Has your question been resolved?

[calm coral]

@scenic merlin Has your question been resolved?

[calm coral]

@scenic merlin Has your question been resolved?

[calm coral]

@scenic merlin Has your question been resolved?

[calm coral]

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[gaunt pilot]

is the limit of e^-n sin(n) zero? as n tends to inf

[quaint sphinx]

yes

the exponential e^n outgrows the constant-bounded sin(n)

[gaunt pilot]

thx for confirming

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[median cloak]

What happened from Mickey to Minnie

[hasty fiber]

differentiation and isolating b'(t) it looks like

[median cloak]

Chain rule?

[hasty fiber]

that's involved

[median cloak]

I don’t know how to differentiate it

[hasty fiber]

well you're correct to use the chain rule

you have some function a, and you're squaring it. How would you differentiate that?

[median cloak]

Is it correct

[hasty fiber]

i'm not sure I follow where the *t bit comes from

[median cloak]

That’s according to the chain rule

[hasty fiber]

that would imply that a'(t) = t

[median cloak]

I see

I see

Smart ppl

Smart ppl

We bang that t out of the equation

Then it should be correct

Agree

[hasty fiber]

so what's your resulting expression for this derivative?

[median cloak]

2* a(t) after Mickey put that t into the casket and beneath the ground

[hasty fiber]

well now you're not applying the chain rule

[median cloak]

I am!

d/dt * t = 1

[hasty fiber]

but the inner function isn't t

i think you need to look at what the chain rule says once more

[median cloak]

Wut

[hasty fiber]

top line is correct, but again your inner function (g in your above notation) isn't t. The inner function is a

[median cloak]

What a Mickey Mouse

Ohhh

I should consider []^2 as a function

The action of square a function

[hasty fiber]

indeed

[median cloak]

And thereby the inner function to be a(t)

What a Mickey Mouse

[hasty fiber]

quite so

[median cloak]

This should be correct

[hasty fiber]

that is correct yes

[median cloak]

Thank you Steak

You’re the smartest steak out of the world

[hasty fiber]

https://tenor.com/view/breaking-bad-walter-white-youre-goddamn-right-gif-14600753

[median cloak]

What a Mickey Mouse!

.close

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[paper marsh]

does anyone know how to solve this?

[paper marsh]

integration problem

[remote mural]

integration?

[paper marsh]

i mean

derivatiev

lol

[remote mural]

chain rule

[paper marsh]

oh wait I see

double chain rule

[remote mural]

ye

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[spiral forum]

The txt book says my answer is wrong so far

[polar frost]

Your textbook may simplify it

[spiral forum]

8. B)

It's different no?

[polar frost]

Nvm

[polar frost]

You're wrong

I compare ur answer with the solution and it doesnt match up, u probably made a mistake somewhere

[spiral forum]

.close

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[polar frost]

@spiral forum

U miss the term

Then they match up perfectly

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[lilac oar]

HI

[orchid ridge]

Whats your question?

[lilac oar]

what's the perimeter of half an equilateral triangle?

[verbal finch]

what is perimeter of equilateral triangle side of a?

[lilac oar]

2x

oh you mean 2x + x + x root 3 oh okay

ty igtg 👋

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[verbal finch]

what?

[lilac oar]

u mean 2x + x + x root 3

perimeter

2x + x + x root 3

[verbal finch]

what no

3a is perimeter

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[bitter monolith]

Can someone tell me where the cosinus went in the solution?

because we have the form y = ax + f but f = 30sin(x) - 10cos(x)

so yp(x) should be equal to a*cos(x) + b*sin(x)

[winter herald]

Differentiate the result and you will see it works.

[bitter monolith]

hm, i'll try

oh yeah, we get -10cos(x)

but then, where did the 30sin(x) go

[calm coral]

@bitter monolith Has your question been resolved?

[marble escarp]

RHS is 3y + 30sinx so 3(-10sinx + c1e^() ) + 30 sinx = -30sinx+30sinx

[calm coral]

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[crimson vault]

I have this execise where I need to prove that the product ideal is an ideal. I manage to do that part but now they ask me what is wrong with just defining the product ideal as {xy , x in I1, y in I2}

[calm coral]

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[crimson vault]

.close

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[remote mural]

Let a,b,c be three distinct positive real numbers such that (2a)^ln a =(bc)^ln b and b^ln 2 = a^ln c. Then 6a + 5bc is equal to

I tried solving but got nowhere

[chilly lodge]

This seems to be missing some info

I got a=b and c=2

[remote mural]

There's no solution given too

Maybe I should skip it?

[chilly lodge]

wait I got a solution

[remote mural]

Ok

Guide then

[chilly lodge]

in both the equations take ln on both sides

[remote mural]

I tried on that way

After that

What should I do?

[chilly lodge]

substitute lnb=(lna*lnc)/ln2 in the first equation

[remote mural]

Now?

[chilly lodge]

the RHS is wrong

[remote mural]

What? oh I am not good

[chilly lodge]

$\ln\left(a\right)\left[\ln\left(a\right)+\ln\left(2\right)\right]=\ln\left(b\right)\left[\ln\left(b\right)+\ln\left(c\right)\right]$

[potent lotus]

B-eard

[chilly lodge]

$\ln2\ln b=\ln c\ln a$\\$\ln b=\frac{\ln c\ln a}{\ln2}$

[potent lotus]

B-eard

[remote mural]

Yes that equation I sent Is after I put eqn 2 in eqn 1

What should I do after that step?

[chilly lodge]

$\ln\left(a\right)\left[\ln\left(a\right)+\ln\left(2\right)\right]=\frac{\ln\left(a\right)\ln\left(c\right)}{\ln\left(2\right)}\left[\frac{\ln\left(a\right)\ln\left(c\right)}{\ln\left(2\right)}+\ln\left(c\right)\right]$

[potent lotus]

B-eard

[chilly lodge]

Now factorize as much as possible

[remote mural]

Ok

a=c and a≠1 and a≠1/2

[chilly lodge]

show me the factorized equation

[remote mural]

Ok wait let me write it clearly 😭

C=2

Lol

Sorry

@chilly lodge

[chilly lodge]

$\left(\ln2+\ln a\right)\left(\ln a\right)\left(1-\frac{\ln^{2}c}{\ln^{2}2}\right)=0$

[potent lotus]

B-eard

[chilly lodge]

$\ln a=-\ln2 ;\text{or} ;\ln a=0 ;\text{or} ;\left(1-\frac{\ln^{2}c}{\ln^{2}2}\right)=0$

[potent lotus]

B-eard

[chilly lodge]

now

lna=0 is rejected

since a !=1

so,

[remote mural]

Yess

[chilly lodge]

lna=ln(1/2)

and 1-(ln^2(c))/ln^2(2)=0

so you can solve for a and c

[remote mural]

And b I will find by putting values

[chilly lodge]

yes

[remote mural]

Right

Thanks

[calm coral]

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[calm coral]

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[remote mural]

can somebody tell me in which quadrant csc(-1410*) [or CSC(-330 degree) will lie
it should lie in 1st quadrant but a YouTuber teacher told me that it would lie in 4th quadrant and i got confused

[remote mural]

and:
is it okay if i write "cosec(4 * 360° - 1410°)" as "-cosec(4 * 360° - 30°"
because i get the same answer on both

[calm coral]

@remote mural Has your question been resolved?

[dense falcon]

First quadrant

Try to make them positive first

Here add 360 to -330

You'll get 30 which is in first quadrant

[remote mural]

🙏 thank you so much, got it
can you answer my 2nd question

[dense falcon]

Yeah

One min

Hey @remote mural

[remote mural]

yes

[dense falcon]

First 4×360 can be ignored

Can u understand that?

[remote mural]

yeah

[dense falcon]

Yeah so it's just cosec(1410)

If u subtract by 360 over and over

You'll get cosec(330)

[remote mural]

its cosec(-1410)

[dense falcon]

Mb then

Since it's cosec (-1410)

Add 360 to it till u get positive value

[remote mural]

okay..

[dense falcon]

You'll get cosec 30

That's it

[remote mural]

🙏 yes, got it
thank you so much

🌹

!end

.close

[calm coral]

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[sinful shuttle]

????? Wouldn't it be (x+1)(x+5) because we want to
plug in the -1 and -5 to get 0?

[topaz raft]

What’s the full question

[sinful shuttle]

Omg how did I forget that part-

[topaz raft]

You want two numbers that multiply to the last term but add to the middle

[sinful shuttle]

that would be -1 and -5 right? D:

[topaz raft]

yes

[sinful shuttle]

so why are the x values counted as 1 and 5 here ?

or am I just exhausted enough that I'm forgetting how factoring works @~@ (this is very possible)

[topaz raft]

Wdym?

[sinful shuttle]

like (x-1)(x-5) implies that the x vals are 1 and 5 cause it's supposed to equal 0, and that's the only way to make the equation fit, but since -1 and -5 were the factored numbers, I thought it would be (x+1)(x+5) D:?

[reef nimbus]

multiply it out, you won't get the numerator if the factors are (x+1)(x + 5)

[sinful shuttle]

I know that (x-1) and (x-5) are right, but like

does factoring not solve for the x val ??

[topaz raft]

It gives you the roots of the quadratic

[sinful shuttle]

............ a h

oki ty! :D I probably could've just asked it that way from the start, I apologize 😭

[topaz raft]

Nw

If you set (x-1)(x-5)=0 you get the x values for which make the quadratic equal to 0

[sinful shuttle]

woopsie T~T

[calm coral]

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[calm coral]

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[tame spoke]

Hey I’m stucked on exercise 3, question 3a

I showed that on the question before

And I need to deduce the inequality now on 3.a

[mortal orbit]

D'après la question précédente, que vaut le terme du milieu dans l'inégalité?

[tame spoke]

In+1/n+1

[mortal orbit]

yep

donc on pourrait pas appliquer le résultat d'une autre question ?

[tame spoke]

Comment je peux utiliser le in+1 ?

[mortal orbit]

In+1 c'est juste I_k avec k un entier, qui vaut n+1

Et n'y a-t-il pas des inégalités sur I_k pour n'importe quel entier (naturel) k?

Peut-être qu'on a déjà prouvé ?

[tame spoke]

Celle d’en haut ?

[mortal orbit]

càd ?

[tame spoke]

La 3.a

[mortal orbit]

?

3.a c'est ce qu'on veut prouver

[tame spoke]

1a je voulais dire

[mortal orbit]

oui 1a

Du coup

[tame spoke]

Ah ouiii

Pour tout n elle fonctionne

Donc pour n+1

[mortal orbit]

oui

[tame spoke]

Comment je pars de cette inégalité ?

Comment je manipule les 2

[mortal orbit]

eh bien

T'as In+1 compris entre...

[tame spoke]

Je pense pas que ça avance grand chose

[mortal orbit]

non j'ai pas compris ce que tu essayais de faire

[tame spoke]

Jsp

[mortal orbit]

Le terme du milieu

on a dit que c'était quoi

[tame spoke]

In+1/n+1

[mortal orbit]

oui

et maintenant on utilise la question 1a

? <= In+1 <= ?

[mortal orbit]

pourquoi In - 1/((n+1)e) apparaît à gauche ?

bon on prend une autre approche

D'après la question 1a, ? <= I4 <= ?

[tame spoke]

Jsuis trop nul pour montrer des inégalités c’est aberrant

I4 ?

[mortal orbit]

oui, $I_4$

[potent lotus]

rafilou2003

[mortal orbit]

d'après la question 1a, c'est compris entre quoi et quoi ?

[mortal orbit]

c'est ça, comment tu as fait ?

[tame spoke]

Beh j’ai juste remplacé le n

[mortal orbit]

par 4

[tame spoke]

AHHHH

[mortal orbit]

Donc

[tame spoke]

Ok

[mortal orbit]

$? \leq I_{n+1} \leq ?$

[potent lotus]

rafilou2003

[mortal orbit]

ah bah voilà

bon

maintenant comment on termine ?

[tame spoke]

Bonne question

[mortal orbit]

comment on obtient l'inégalité mentionnée dans la question à partir de ça ?

Sachant ce qu'on a déjà dit avant

[mortal orbit]

ouaip c'est ça

[tame spoke]

Je suis vraiment trop bête

Mais je galère trop à montrer des inégalités

Alors que j’ai tout pour le faire

Y’avait rien de dur en vrai

T’as des idées pour le début du 4

C’est une vraie torture

Juste pour la 3b je dois réutiliser la même inégalité ? @mortal orbit

[calm coral]

@tame spoke Has your question been resolved?

[calm coral]

@tame spoke Has your question been resolved?

[calm coral]

@tame spoke Has your question been resolved?

[calm coral]

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[calm coral]

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[remote mural]

can someone help me finish this

[remote mural]

thats what i thought

[tardy mauve]

do you know the answer

yeah

because

[remote mural]

ya its c

[tardy mauve]

see it like

[remote mural]

i just dont know how to get there

[tardy mauve]

$$ 9x^3 / 6x^2 $$

[potent lotus]

Not Scammer

[tardy mauve]

because the terms

like -3x^2 or 3x or -8x

doesn't matter that much

while you're taking n to infinity

they don't really matter.

the thing that matters is

the terms with the biggest power

(exponent)

[remote mural]

mk

.close

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

@idle fractal Has your question been resolved?

[calm coral]

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[opaque plume]

Hello there. How could we make that first manipulation on the given expression? ( from (M^-1)T to (M^-1) )

[opaque plume]

<@&286206848099549185>

I think it's somehow possible represent this simplification with triple product inside determinant, cause right side is triple product, dunno how to represent this

[calm coral]

@opaque plume Has your question been resolved?

[opaque plume]

<@&286206848099549185>

[tame quail]

We're trying to prove something like this. Edit: C should be mxq for some q, not necessarily mxp

Notice how we need to change AT to A so that the dimensions line up

That's not really a proof, but it illustrates why we need to transpose A

[opaque plume]

Don't see here problem with dimensions

[tame quail]

C has dimensions m x p, so we can only multiply it from the left by a matrix with dimensions _ x m

[opaque plume]

c is vector 3x1

[tame quail]

oh I see, I thought it was a matrix

@opaque plume Here you go

x dot y = xTy

and (AB)T = BTAT