help-0

[eternal mesa]

Thats so far

[carmine reef]

why did you move the -1 and +1 inside the fraction?

$\sec(2a)$ is $\f 1 {\cos(2a)}$

but $\sec(2a) + 1$ is not $\f 1 {\cos(2a) + 1}$

[ocean seal]

monikanicity

[eternal mesa]

because idk what im doing

[ocean seal]

monikanicity

[eternal mesa]

so the flip part is only for the trig function

[carmine reef]

yeah

sec(anything) is 1/cos(anything)

that's how its defined

but it doesn't work for stuff outside the secant function

you can replace sec(2a) with 1/cos(2a) because when two things are equal, you can replace one thing with the other in an expression without changing it

[eternal mesa]

Is this right?

[carmine reef]

yeah

[eternal mesa]

so if a problem was like 51, where its cos^2 x/2, are you plugging in the function as x and then applying everything to it that was in the original function?

[lone heart]

@eternal mesa Has your question been resolved?

[lone heart]

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[spare fern]

hey guys

[spare fern]

can anyone help me out with this?

I'm trying to find the area of the shaded region between the curve

[alpine sable]

i dont know how to do basic math

[spare fern]

Why are you sending messages on this channel in that case?

[alpine sable]

idk i read it wrong sorry

[mellow zephyr]

what are you looking to do here?

[spare fern]

.

[mellow zephyr]

oh mb

I see

[spare fern]

I found the intersection point to be 3/2

from solving the quadratic formed after solving y= 2x^2 and y = 6-x simultaneously

[mellow zephyr]

good start!

[spare fern]

two roots I got were 3/2 and -2, It can't be -2 so I figured it was 3/2

thanks

[mellow zephyr]

okay, so now area under a curve

[spare fern]

so now I can find the integral with limits 0 and 3/2

or can I subtract?

[mellow zephyr]

integral is good

[spare fern]

I know area = top - bottom

[mellow zephyr]

so on that interval (0 to 3/2) we are integrating which of the functions? (or, which top function minus bottom function)

i.e. which function is the top of the shaded region on that interval?

[spare fern]

y = 2x^2

[mellow zephyr]

so that's the stuff inside the integral for that part

(and to work with your top-bottom thought process -- we're technically integrating (2x^2 - 0) because the top function is 2x^2 and bottom function is the x-axis or y=0)

so now we have $\int_0^{3/2} 2x^2 dx$ for the first chunk

[ocean seal]

becoming_deinos

[mellow zephyr]

can we figure out something similar for the second part?

[spare fern]

hold on I have a question

[mellow zephyr]

yep

[spare fern]

do I always need to find the intersection point in order to get the area under the curve?

why can't I just solve the integral with limits 0 to 6

while doing top - bottom

[mellow zephyr]

which function is on top between 3/2 and 6?

because the "top" function changes, we have to change the function inside the integral

[spare fern]

oh I see

[mellow zephyr]

technically I could define a "top" function that is defined differently on the two intervals, but then I would have to split up the integrals to evaluate anyways

ready to give the second part a try, and then put them together?

[spare fern]

yeah, I'll try it out and let you know how it goes.

ok so I got 9/4 units^2 for A1

A1 = area under parabola

[mellow zephyr]

yep

[spare fern]

A2 = area under straight line

A2 = 81/8

so I got 99/8 units^2 for the total area

[mellow zephyr]

looks good to me

[spare fern]

Alright, thanks for helping me out 👍

[mellow zephyr]

👍

[spare fern]

.close

[lone heart]

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[mossy burrow]

@flint pecan

[mossy burrow]

idk what happened

[flint pecan]

yeah what just happened

[mossy burrow]

idk

[flint pecan]

anyways youre right

[mossy burrow]

ok cool

r u able to stay ??

[flint pecan]

hm not for long

[mossy burrow]

ok

well

#5

she wants me to convert to seconds

[flint pecan]

do you know what ms is?

[mossy burrow]

milisecon?

milliseconds

[flint pecan]

yep

[mossy burrow]

so idk how to do this one

[flint pecan]

1 milisecond is 0.001 second

[mossy burrow]

ok so

358.235?

s

?

[flint pecan]

yeah

i'm in class to gotta go sorry

[mossy burrow]

it’s ok

thank u

.close

[lone heart]

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[twilit relic]

Hi, can someone explain to me how this line works?

[twilit relic]

The first two does not really equal each other!

[lone heart]

@twilit relic Has your question been resolved?

[twilit relic]

.close

[lone heart]

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[dire nova]

;-;

[kindred anchor]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[dire nova]

1

[kindred anchor]

Start with identifying the points.

[lone heart]

@dire nova Has your question been resolved?

[lone heart]

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[grave niche]

Why does pi become negative?

[grave niche]

isnt pi a constant not a variable?

[raven haven]

yes

[grave niche]

then why does pi change signs?
wouldt it just stay as positive pi?

[manic wind]

what's the derivative of x^-1

[grave niche]

x^-2

!

!!!!

[raven haven]

yeah don't give answers

[manic wind]

use the power rule

[grave niche]

thats worrisome, i didnt know that

[raven haven]

if you recall the power rule by

nx^(n-1)

then notice here n=-1

[alpine sable]

you just forgot to multiply the constant by the power

[grave niche]

I see it
I just have my things missed labeled

k i get it now, thank you

[alpine sable]

i know this is like very simple but

im just wondering why

the -0, is included

my friend sent me this obviously they speak a different language and im confused

heres the equations of the lines btw

[grave niche]

.close

[lone heart]

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[rich quiver]

,rotate

[rich quiver]

Is this correct?

[grave niche]

your paper is on life support lol

[rich quiver]

?

[grave niche]

Your piece of paper looks worn out

[rich quiver]

Yeah its very old paper

Is my answer correct ?

[grave niche]

Oof no idea but the helpers should be able to answer your question or at least assist you
<@&286206848099549185>

[rich quiver]

<@&286206848099549185>

[urban estuary]

@sly juniper is it possible to get a clearer picture?

I am not sure where point F is

[exotic belfry]

looks ok for me

[rich quiver]

Right of point E

[urban estuary]

Yes

It is correct

[lone heart]

@rich quiver Has your question been resolved?

[stray light]

Dont have the slightest clue of how to slove trigonometric identities anyone know how?

[upper marsh]

Do you know what sec and csc is

[stray light]

sec is 1/cos

and csc is 1/sin

[upper marsh]

Okay you do know

[stray light]

ye

Im just confused on the process of the question

[upper marsh]

csc times cos is just tan

2tanx= 1

[stray light]

and tan equals sin/cos^2 which equals 1 too huh

[upper marsh]

You got it

[stray light]

what abt the 2nd one

[lone heart]

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[alpine sable]

why is the first one wrong?

yeah why is that false

?

isnt alice wearing glasses and eve has a hat?

wat

[thorny root]

it's like a seperate x for both

[alpine sable]

this is correct i mean like isnt it true though

its asking is there a person with glasses, and is there a person with a hat

so yeah there is

that would be

$\exists x (P(x) \wedge Q(x))$

[ocean seal]

hibyehibye

[alpine sable]

this is 2 diff

[thorny root]

yeh

[ocean seal]

HeyHey

HeyHey

[alpine sable]

Everybody has glasses or a hat

Everybody has has glasses OR everybody has hats

so yeah

universal quantifier only distributes over conjunction so those are diff statements

[lone heart]

@spare sierra Has your question been resolved?

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[wheat crystal]

angular speed problem

[mint thistle]

Hey guys how can i calculate the perimeter of a rectangle given two points. The bottom right and the top left

[wheat crystal]

trig

Oh

This help channel is occupied @mint thistle

[mint thistle]

yeah i see that, it was in available just at first though

[lone heart]

@wheat crystal Has your question been resolved?

[wheat crystal]

no

[stray estuary]

do you remember the formula for switching between linear and angular velocity?

[lone heart]

@wheat crystal Has your question been resolved?

[lone heart]

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[thin pivot]

im not sure if the answers are wrong i am wrong, can someone check for me please:)

[wheat sapphire]

No!

[thin pivot]

i thought so

[wheat sapphire]

Both the real and complex part are of the 3rd quadrant, you can clearly see their signs

[thin pivot]

it seems all of the answers for these questions are wrong

[thin pivot]

thats what i thought

thanks

.close

[lone heart]

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[lone heart]

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[spare fern]

could anyone help me out

[spare fern]

I got the area of the triangle to be 9/2

do I use area = (pi * r^2)/(4) to find the area of the quadrant?

[vale wigeon]

are we to assume that the left half of the graph is a circular arc?

if so, then yes. the region in question is a quarter-circle.

[spare fern]

Yes, I am indeed assuming that is the case.

[spare fern]

so I use this formula?

[vale wigeon]

well... yes? of course?

you're just second-guessing yourself right now are you not

[alpine sable]

oh wait oops, i totally didnt see that line there

[spare fern]

Please post this in another channel; the current one is occupied.

[alpine sable]

delete your message too

it clogs the channel

okay thanks

anyways yeah, @spare fern find the area of the quarter of the circle and subtract it from the area of the triangle formed by that line

[spare fern]

Just curious, why do I have to subtract it wouldn't I need to add it?

[vale wigeon]

i don't know what you mean by "it".

or why this "it" "has to" be subtracted.

[spare fern]

@alpine sable

[alpine sable]

the area formed under the line would be in the negative y axis, it will result in some negative number

[spare fern]

oh true

[vale wigeon]

oh, but you got that the wrong way around

the integral is (quarter circle area) - (triangle area)

but lixera's statement implies it is (triangle area) - (quarter circle area)

so it is the TRIANGLE that you want to subtract.

since it is under the x axis, and therefore counts as negative for the integral.

[spare fern]

oh I see

thanks for clarifying

[alpine sable]

Ah sorry, i meant that. Does sound confusing in hindsight

[alpine sable]

no i meant that

pain

[spare fern]

I got the area to be 2.356

[vale wigeon]

there is nothing in the problem that instructed you to round to three decimal places.

unless you chose to omit such an instruction and leave us in the dark about it.

[spare fern]

actually 2.2

quarter circle area: (pi * 3^2)/ (4)

[vale wigeon]

there is nothing in the problem that instructed you to round to one decimal place.

[spare fern]

alright 2

[vale wigeon]

there is nothing in the problem that instructed you to round to the nearest integer.

what are you doing?

[alpine sable]

,w integral from 0 to 3 sqrt(9-x^2) + integral from 3 to 6 (-x+3)

oops

[ocean seal]

Results provided by WolframAlpha

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

Requested by Lixera#8529

[alpine sable]

[vale wigeon]

@spare fern tell me right now in words what this was all about

[alpine sable]

Ann: Don't round
Lex: Okay let me round more

[spare fern]

ok well 2.5 doesn't appear on my selectable answers

@alpine sable

[alpine sable]

well thats just an approximation

the true answer should be

[spare fern]

oh I see

[alpine sable]

Also, answer Ann

[spare fern]

.close

[lone heart]

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[alpine sable]

Well then okay

[spare fern]

thanks for the help 👍

.close

[vale wigeon]

.reopen

[lone heart]

✅

[vale wigeon]

just gonna make a little remark here

[alpine sable]

umm

[lone heart]

@spare fern Has your question been resolved?

[vale wigeon]

yeah, sorry, i got distracted.

[lone heart]

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[vale wigeon]

i just feel kind of disrespected by the lack of a reply to my questions.

[lone heart]

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[forest scroll]

How is this true?

[vale wigeon]

difference of squares

[alpine sable]

FOIL it?

which is just difference of squares ultimately

[vague coral]

1-x = 1^2 -(sqrtx)^2

[forest scroll]

Ok. Thanks.

Guys

.close

[lone heart]

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[pulsar laurel]

11=1
-√x+√x=0
-√x√x=-x

[lone heart]

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[dense nova]

Does anyone know how to do part a I’m beginning to think it may be impossible

[gray isle]

,rcw

[raven haven]

is that

a picture of a phone taken of a computer of a piece of paper

[dense nova]

Umm yeah sorry for the bad quality

It is still visible for me though

[raven haven]

yeah it's not very visible

[dense nova]

I can write it out

[raven haven]

if you do please use LaTeX as much as you can

[dense nova]

Idk how to use LaTeX lol

[copper epoch]

I dont understand how i find arg(z) Arg(z)

And how to read the circle

[dense nova]

There’s the problem written out by hand hopefully it’s visible enough now

<@&286206848099549185>

[full frost]

should label your figure might help you A LOT

[dense nova]

I have done on a separate more messy copy

However I am beginning to think this may be impossible

For part a

[full frost]

so what do you know about triangle ABC

[dense nova]

It is horizontal

[full frost]

what does it mean for a triangle to be horizontal…

[dense nova]

That it’s sides are horizontal

[full frost]

hmm i’m not sure what that means

[dense nova]

Like if a triangle is horizontal it’s flat along a plane

[full frost]

oh horizontal base

that’s not what i was asking tho

[dense nova]

Oh

[full frost]

from the given info can you tell me anything about the ABC triangle besides it being horizontal

[dense nova]

It’s isosceles

As two side lengths are the same

[full frost]

yeah

what’s the angle of BAC

[dense nova]

Hmm idk that one

How would you find that out??

@full frost

[lone heart]

@dense nova Has your question been resolved?

[full frost]

it tells you that M is the midpoint of BC

first what is angle BMA

[lone heart]

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[copper epoch]

Someone explain

How i find

Arg(z)

arg(z)

[last ether]

it's similar to arctan

[copper epoch]

And how i read this circle

I don t understant how to find this value in the circle

[last ether]

You just take the arctan of Im(z)/Re(z) but recall that arg(z) kicks out values between 0 and Pi

[copper epoch]

2pi - teta

[violet mulch]

!help hi, could someone explain the relationship of standard deviation to the mean. If someone scored 75, the mean was 70 and standard deviation was 3 - is that better than someone who scored 80, with the mean being 70 and standard deviation being 8?

[keen plinth]

,rccw

[lone heart]

Please read #❓how-to-get-help

[last ether]

Are you joking me

Bro

[keen plinth]

[last ether]

Can't you see this one's occupied

[lone heart]

Please read #❓how-to-get-help

[last ether]

Delete this. Read #❓how-to-get-help

[naive nimbus]

None of them were available

And it took me here

[last ether]

There's legit 4 available

[naive nimbus]

When i clicked button

[last ether]

That's the hidden

[naive nimbus]

Ok sorry

[last ether]

My guy there's a section LITERALLY called available

[last ether]

@copper epoch excuse us for the literally 2 interruptions that just happened. Does this make sense to you?

[copper epoch]

I think I understand a little

Thank you

[pseudo ice]

.close

[lone heart]

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[daring elm]

Can you denote a polynomial with a non-capital letter. Like p(x)

[pseudo ice]

You could denote polynomials with whatever letter you want really, capital or lowercase

[however, your course/teacher may require you to do differently so check with them first!]

[daring elm]

Ok thanks!

.close

[lone heart]

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[dense nova]

Is part a possible ??

<@&286206848099549185>

[lone heart]

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[minor steppe]

is this correct?

[vale wigeon]

yes

[minor steppe]

ty

.close

[lone heart]

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[lone heart]

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[unkempt sundial]

no idea how

[lone heart]

@unkempt sundial Has your question been resolved?

[unkempt sundial]

clearer

[lone heart]

@unkempt sundial Has your question been resolved?

[unkempt sundial]

<@&286206848099549185>

[pseudo ice]

Do you happen to know whether it converges or not? I can get to the expansion part but that's about it

[unkempt sundial]

it diverges according to wolframalpha

where did you get with the expansion

[pseudo ice]

So I wrote $\frac{\sin(t)}{\sqrt{t} + \sin(t)} = 1 - \frac{\sqrt{t}}{\sqrt{t} + \sin(t)} = 1 - \frac{1}{1 + \left( \frac{\sin(t)}{\sqrt{t}} \right)}$

[ocean seal]

chartbit

[unkempt sundial]

aha i see

[pseudo ice]

Then the latter, if you have $t>1$ then you have that $\left| \frac{\sin(t)}{\sqrt{t}} \right| \leq \frac{1}{\sqrt{t}} < 1$

[ocean seal]

chartbit

[pseudo ice]

At which point you know the rest

[unkempt sundial]

ok i got to this, just gotta prove the Squared term is divergent

[pseudo ice]

Ohhhhh that's the aim! I forgor about the way you can compare them like that lol

I'm gonna end up learning integral convergence tests from your posts I think 🥹 they're so interesting though!

[unkempt sundial]

🫂

[pseudo ice]

Mind you I got that the squared one diverges and the non-squared one converges when I wolfie'd it

Just didn't know how to argue along those lines haha

[unkempt sundial]

yah that's what i did

hmm

i did end up getting the solutions for the previous series convergence questions

[pseudo ice]

Thinking double angle formula? [for divergence of sin^2(x)/x]

[I'm still kinda cheating lol]

[unkempt sundial]

can we say that sin(x)^(2)/x is greater than or equal 1/x?

[pseudo ice]

I don't think so no? For one you can have sin(x)=0 of course

[unkempt sundial]

aha

[pseudo ice]

Semi random question but how would you show that the integral of cos(2t)/t converges? I see it does but me too small brained to figure out how to show it

[unkempt sundial]

we have similar brains then cause i have no idea

[pseudo ice]

I really wanted to go and say "by the double angle formula, cos(2t)/t converges, and we know that 1/t diverges, hence we must have that the (sin(x))^2/x diverges otherwise a contradiction"

[unkempt sundial]

off topic, but is this correct
[ -\int_a^b{f(x)dx} =\int_a^b{-f(x)dx} ]

[ocean seal]

alihsaas

[unkempt sundial]

me and my friend are having brain farts

[pseudo ice]

Linearity of integrals

[pseudo ice]

It happens, to me much more than it should of course, so you're fine

[unkempt sundial]

alright thanks 👍

[unkempt sundial]

what about -1/x

[pseudo ice]

Well that is true, but then the whole thing is greater than [or equal to] zero of course so

[unkempt sundial]

what is the name of this method

[iron marsh]

can someone help me

<@&286206848099549185>

[unkempt sundial]

[rustic coral]

<@&268886789983436800> @iron marsh posting a test and hijacking someone else's help channel (and also pinging early)

#help-0 message
(Screenshots in case they delete msg)

[iron marsh]

what it said to

go in the

help 0

[pseudo ice]

Not help 0 specifically, but a channel that's under "Math Help (Available)"

Hint: if it doesn't have someone's name on it, you're probably fine

[next brook]

Anyway, this looks like a test you're in the middle of taking -- is that true?

[iron marsh]

I’m home sick so I have to do it at home

[next brook]

That doesn't mean you get to cheat by asking people on the internet to answer the questions for you.

[pseudo ice]

Is it homework or a test?

[iron marsh]

it’s something u have to do for the actual test

idk what it’s called

so I can do a retake

if I complete this

then the actual test is

in a week where u can’t do it at home

[next brook]

Okay, I'll grant you the benefit of doubt on that. Now kindly shoo out of Alihsaas's channel and pick an available channel for yourself instead.

[iron marsh]

how do I go to a avaliable channel

[pseudo ice]

Also what's this the corollary to?

[pseudo ice]

Look here for one - the numbers may be different but they're in that section

[unkempt sundial]

quotient test

so it is that?

[pseudo ice]

I would think so, because you're basically applying the quotient test in that case right?

[unkempt sundial]

explain pls

[pseudo ice]

Could you translate for me please I can see they kinda had my idea I think!

[unkempt sundial]

the translation is the one under the blue text

or green

[pseudo ice]

Oh I'm blind asf sorry!

I was reading the pinkish the whole time and ignored the above for some reason thought it was something else!

[unkempt sundial]

it's alright 👍

[pseudo ice]

Hmmm looks like they're comparing it with the series

Time to look up Abel transformations

Not entirely sure how they got that |sin(n)| >= (sin(n))^2 though, hmm

[unkempt sundial]

the entire page is about series but i assume the idea is similar

https://www.bibmath.net/ressources/index.php?action=affiche&quoi=bde/analyse/suitesseries/serienum_prat&type=fexo

[pseudo ice]

Alright dead the idea of the built in translator

[unkempt sundial]

yah it's really bad 🥹

[pseudo ice]

That's just terrible, "not sin"

[pseudo ice]

Oh at least this is somewhat obvious now, slow moment

When $0 \leq a \leq 1$ you have $a^{2} \leq a$, and $|a|^{2} = a^2$

[unkempt sundial]

mhm

[ocean seal]

chartbit

[pseudo ice]

Also seems like "Abel transformations" [if I've found the correct one] is basically the equivalent of integration by parts for sums

So maybe it's that which could be used

[unkempt sundial]

cos(2n)/2n by parts?

[pseudo ice]

Well if I understand right, that's the aim

[unkempt sundial]

[pseudo ice]

Ohhhhhh fuckkkkk maybe it might be that

Going to the original one, doing by parts on $\int_{1}^{\infty} \frac{\cos(2t)}{2t} dt$

Looks kinda promising still

[ocean seal]

chartbit

[unkempt sundial]

[pseudo ice]

As then it looks something like $\int_{1}^{\infty} \frac{\cos(2t)}{2t} dt = \left[ \frac{\sin(2t)}{4t^{2}} \right]{1}^{\infty} + \frac{1}{4}\int{1}^{\infty} \frac{\sin(2t)}{t^{2}} dt$

[ocean seal]

chartbit

[unkempt sundial]

and sin(2t)/t^2 is div?

[pseudo ice]

Should be convergent right? Bound above and below by, say, 1/t^2 and -1/t^2 and those both converge

The square brackets can be evaluated and you get a finite value

Therefore you have that being convergent, unless I've made a booboo?

[unkempt sundial]

arent we proving that sin(x)^2/x is div?

[pseudo ice]

Yea, we are, but the idea I had was that basically $\int_{1}^{\infty} \frac{\sin^{2}(t)}{t} dt = \int_{1}^{\infty} \left( \frac{1 - \cos(2t)}{2} \right) \frac{1}{t} dt = \int_{1}^{\infty} \frac{1}{2t} dt - \int_{1}^{\infty} \frac{\cos(2t)}{2t} dt$

[ocean seal]

chartbit

[unkempt sundial]

ahh i see

1/t is the div term

[pseudo ice]

You know that the 1/(2t) diverges but if that cos(2t)/(2t) converges then the original must diverge too

(else contradiction)

Damn I feel like I've learned, and this is supposed to be your question

[unkempt sundial]

what did you learn

cause i have no idea if I learned anything

[pseudo ice]

Showing how to show that diverges above

But does all of that make sense and seem logical though?

[unkempt sundial]

yah i think i understood everything

now how do i prove sin(x)/sqrt(x) converges

[pseudo ice]

Think by parts should do that too, right? differentiate the 1/sqrt(x) and integrate the sin, then make a similar bounding argument I would guess?

[unkempt sundial]

i see, lemma try

[unkempt sundial]

i integrate sin x and differentiate 1/sqrt (x) right?

[pseudo ice]

Yep, that's how I did it

[unkempt sundial]

correct?

[pseudo ice]

My thoughts is that you have $-\frac{1}{x^{3/2}} \leq \frac{\cos(x)}{x^{3/2}} \leq \frac{1}{x^{3/2}}$ - that could be negative, but there is a lower bound there

[ocean seal]

chartbit

[pseudo ice]

Then integrating both the left and the right converge, so the middle would need to be somewhere in between them

[unkempt sundial]

1/x^(3/2) isnt enough?

[pseudo ice]

Hmm, well I think you would need to argue carefully, you know like how the standard comparison test requires your functions be nonnegative right

Like my fussiness is because you have stuff like -1/x < 1/x^{3/2}

[pseudo ice]

So showing it's less than something convergent when you could be negative, you gotta be a bit careful

[unkempt sundial]

mmm i see

i think i got it

ill try solving it on my own later

thanks

[pseudo ice]

Been a pleasure shall catch ya next time!

[unkempt sundial]

yall catch me alot these couple of days, gotta grind it hard 🥹🥹

[pseudo ice]

Haha I shall look out for you of course... and hope I can do some of the questions of course

[unkempt sundial]

ill close this now 🫡

.close

[lone heart]

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[crisp elk]

I don't understand how to get the derivative form $$f(x) = sqrt(3x-1)$$

[ocean seal]

vAp

[merry spire]

now try!!

(3x-1)^1/2

[crisp elk]

yeah

[merry spire]

have you done chain rule

[crisp elk]

yes tried it but got nowhere

[merry spire]

show me your work

[crisp elk]

not sure what to plug into the rule

[merry spire]

derive the outside keep the inside the SAME then multiply everything by the inside derivative I will try to get a good image

[crisp elk]

how do i know waht to use as inside and out ?

[merry spire]

of the parenthasis

[crisp elk]

sure

[merry spire]

(3x-1)^1/2

step 1: derive OUTSIDE (ignore the inside):

[crisp elk]

yes

[merry spire]

so that equals

1/2 (3x-1) ^-1/2

[crisp elk]

all the way

[merry spire]

now we multiply this by the derivative of whatever is INSIDE

[crisp elk]

so 3?

[merry spire]

so we have what we just got : 1/2 (3x-1) ^ -1/2 times 3 YES!

that is it!!

[crisp elk]

x^2 * sqrt(x)

[merry spire]

okay this one is more complicated because you start with the PRODUCT rule

[lone heart]

@crisp elk Has your question been resolved?

[crisp elk]

Cant do, missing some understanding somewhere

[hollow sparrow]

[crisp elk]

haha i was desperate

[ocean seal]

Mehdi_Moulati

[hollow sparrow]

use this

[crisp elk]

oh so x^5/2

[hollow sparrow]

don't forget it's -1/2

so 2 - 1/2

[ocean seal]

Mehdi_Moulati

[crisp elk]

x^1.5

^3/2

[hollow sparrow]

yes

now repeat your calculation

[crisp elk]

i have 2x^5/2 * 1/2 x^-1/2

[crisp elk]

<@&286206848099549185>

[hollow sparrow]

how did you get that ?

[hushed wolf]

this

[crisp elk]

I addet befor i multipied

A am unsure how to 1/2 * x^-1/2

[hushed wolf]

can you tell me (3x-1)^-1/2

[hushed wolf]

okay

start from basic then

can you tell me a^-1

hey

@crisp elk

[crisp elk]

1/a

[hushed wolf]

okay, what is a^-2

come here bro you just asked a question

[crisp elk]

sorry

[crisp elk]

not for sure

[hushed wolf]

what

[crisp elk]

i guess 1/2a

[hushed wolf]

1/a²

you must know that a^-n=1/a^n

,rotate

[crisp elk]

.close

[lone heart]

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[alpine sable]

ur notation is funky

are those meant to be the x values?

because if so, no

oh

thats a bigger no then

[alpine sable]

x2

we need to first figure out the structure of what you are trying to write first

no interval notation is not some x and y coordinate

no!! your notation is not appropriate and it doesn't say anything

it is not proper interval notation

yes! that makes sense

honestly, it might be better to use set builder notation and then convert that to interval notation

like

${x \mid x < -1 \text{ and } x \geq 3 } \iff (-\infty, -1) \cup [3, \infty)$

[ocean seal]

Lixera

[alpine sable]

interval notation isn't "(x,y)" which is the first misconception we need to address

with interval notation you are looking at a single variable and seeing the behaviour of that variable

[alpine sable]

as with x here

(-∞,0),(∞,5) does not make sense in interval notation because you first have to ask yourself, what variable are you trying to address with it?

it would be x here obviously, since y would be held constant

you are trying to find the values of x for which y is held constant, and write them in interval notation

which is whats written here

okay but

you are basically saying the purple line is where the function has constant value

with that

for the x values

reorder it to (-infinity, -4), (4, infinity) then yaeh

that would be ur intervals

dont forget the minus doee

yep

u got it

did u get the idea tho

oh okay

thats great

[alpine sable]

really recommend writing it in set builder notation first and see if it makes sense to you

pff yeah thats understandable

[lone heart]

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[kind sundial]

can anyone help me with how to approach this problem?

[lone heart]

@kind sundial Has your question been resolved?

[kind sundial]

i know how to calculate the probability of a certain number of games being played, but im not sure how to find the pmf

[lone heart]

@kind sundial Has your question been resolved?

[tacit arch]

pmf is just P(X=k) for k = games played

and find it out for all possible values of k

[lone heart]

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[vapid shuttle]

I have an ODE and initial conditions, but unsure how to solve for the constant. I think the notation is confusing me.

[vapid shuttle]

,, \frac{dy}{dx}= y+y^3+c_3

[ocean seal]

AustinU

[vapid shuttle]

conditions are y(0)=1, y'(0)=2

How can I solve for c3?

[worn fox]

if y(0) = 1, then y(0)^3 = ??

[vapid shuttle]

1

[worn fox]

right

so now plug in all the information you know

[vapid shuttle]

so when it says y(0)

that means that y=1

[worn fox]

uh well kinda

[vapid shuttle]

I am confused where the 0 is getting inputted into

[worn fox]

it means that when x=0, y = 1

y is a function of x

[vapid shuttle]

yes

Okay thanks, I think I get it., I was just confused because there was no x inputs

.close

[lone heart]

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[knotty mountain]

how am I meant to get the double derivative of this without 2 pages of working out

[alpine sable]

seems like you have to do 2 pages of working out

[last ether]

Well ... suffer

[tacit bobcat]

Just do the working.
If you are unsure of your answer, use a computer after working.

[knotty mountain]

well it was a little sub question in this chapter

so i thought surely it isn't meant to take more than half a page of working out

if that's the case i'm just going to skip it lmao

.close

[lone heart]

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[spare fern]

could anyone help me out with this?

[full frost]

yeah man

[spare fern]

I know the curve on the left will be negative

[full frost]

yeah

and the right

[spare fern]

but can I just find the integral with limits -2 and 0

then after that do the integral 0-2

then add them together?

[full frost]

yeah you can add them

[spare fern]

oh ok thanks

although how would I integrate f(x)?

[full frost]

it’s the area of the shaded region you wouldn’t really evaluate it

[wary stream]

A hint is given, it's an odd function

[spare fern]

so how would I find the area of the two shaded regions?

[full frost]

the fact that it’s odd tells you something

[spare fern]

is there a formula? cause I haven't really learnt about odd functions yet

[wary stream]

Odd functions are symmetrical, correct?

[spare fern]

even functions are symmetrical over the y-axis

odd functions are symmetrical over the origin

[wary stream]

Let me be more specific, odd functions are symmetrical where half of it is below the x axis, the other half is above, correct?

[spare fern]

Yes, but how does this help me find the area?

[wary stream]

Notice how the area below the x axis is equal to the area above, one is just negative

[spare fern]

I can only see the length, not the height, so how would I know the area of both is the same?

[wary stream]

Because it's an odd function about the origin

It's symmetrical

[spare fern]

oh ok

[wary stream]

That's an odd function, can you see how the areas are the same but one us just negative?

[spare fern]

Yes, I can see that.

so in that case they would cancel out

[wary stream]

Yes

Exactly

[spare fern]

Alright thanks, I understand now

How would I do this?

<@&286206848099549185>

[exotic belfry]

do you see the rectangle?

[spare fern]

yes

[exotic belfry]

whats the size of this?

[spare fern]

-k to k

with a height of 4

[exotic belfry]

and this is?

[spare fern]

?

[exotic belfry]

the area of the rectangle is?

no, 5k ist the area of the shaped region, the rectangle is bigger.

[spare fern]

so base * height

but what values would I use to find the area in this case?

for the rectangle

I know 4 is the height

[exotic belfry]

and the base is?

[spare fern]

-k +k

[exotic belfry]

the base is 2k.

[spare fern]

but what about the negative k

[exotic belfry]

if you stand on -k, you have to go k steps to the right to the center, and then again k steps to be on +k., so you went 2k steps.

[spare fern]

oh ok

so 2k * 4

[exotic belfry]

well, so what is the area of the rectangle without the shaped area?

[spare fern]

8k-5k

[exotic belfry]

well. and now describe by your own words, what the area of the rectangel without the shaped area is.

[spare fern]

shaped area?

so then the area of the non-shaded region is 3k

[exotic belfry]

thats right. but i didnt asked for the value, i asked for a description.

[spare fern]

ok so the area of the non-shaded region is equal to the area of the rectangle minus the area of the shaded area

[exotic belfry]

yes, thats right. can you describe it without using the words "rectangle" and "shaped area"?

[spare fern]

ok the definite integral f(x) with limits -k to k is equal to 3k

[exotic belfry]

exactly.

[spare fern]

thanks

[exotic belfry]

you are welcome.

[last ether]

Start with (i)

[spare fern]

for (i) can I write the definite integral 1/x^2 with limits 1 to N

[last ether]

Yeah

(II) should be evident then

[spare fern]

so do I integrate 1/x^2 first

then substitute the limits?

for (i)

[exotic belfry]

just do it.

[spare fern]

ok I got (N^-1)/(-1) - (1^-1)/(-1)

[median oar]

With respect to x

Don’t forget that bit

[spare fern]

which simplifies to (N^-1)/(-1) - 1

[exotic belfry]

no.

[median oar]

Show your working

[exotic belfry]

this becomes 1-1/N

[spare fern]

how?

could you explain it

[exotic belfry]

N^-1 = 1/N -> N^-1/-1 = -1/N

(1^-1) -> 1/1 = 1 -> (1^-1)/(-1) = -1

-(-1) -> 1

[spare fern]

but where does the last step occur?

[exotic belfry]

$\frac{N^{-1}}{-1}-\frac{1^{-1}}{-1}=-N^{-1}-(-1^{-1})=-\frac{1}{N}-(-1)=1-\frac{1}{N}$

[ocean seal]

ThM

[spare fern]

alright thanks

[lone heart]

@spare fern Has your question been resolved?

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[lapis nacelle]

help

[lapis nacelle]

V=< 3cos(theta) , 3sin(theta) >

right?

then the unit vector is V / | V |
this is, 1/3 * V = < cos theta , sin theta >

[exotic belfry]

yes

[last ether]

Save yourself time. You don't need to calculate ||V||. If a vector is hoisted upon an angle, a, then the unit vector is just <cos(a), sin(a)>

[lapis nacelle]

yeah, we can see that in unit circle trig : D

but,

also, i tried with

[exotic belfry]

tried brackets or cos theta insteas costheta?

[last ether]

try cos(theta)

costheta is not a work

It will read it as costheta, not cos theta

[lapis nacelle]

nice!

thanks guys!!

[last ether]

Close the channel if you're done @lapis nacelle

[lapis nacelle]

.close

[lone heart]

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[manic wind]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[vapid steppe]

2

[manic wind]

can you show what you've done so far?

[vapid steppe]

Yea

[tired echo]

complete the square jash 🤔

[vapid steppe]

What do i do with x≥-b/2a

[tired echo]

what does that tell you about a quadratic

[vapid steppe]

well -b/2a is sometimes the vertex formula

[tired echo]

sometimes?

[vapid steppe]

wait does f^-1(x) mean 1/f(x)

[tired echo]

no

[vapid steppe]

oh it’s the vertex

[tired echo]

yes

and the vertex defines the line of... ? for the quadratic

[vapid steppe]

Mirror

[tired echo]

yes, symmetry

[vapid steppe]

The symmetry line

AOS

[tired echo]

so it's asking you for the inverse of the function to the right of that line of symmetry

[vapid steppe]

Does it change the answer

[tired echo]

no

but remember that quadratic have to values that spit out the same answer

and you're considering one half of the quadratic in this case

[vapid steppe]

so it’s one to one

[tired echo]

yes, when you cut it in half like this

[vapid steppe]

i got that 2ax≥-b so a≥-b/(2x)

[tired echo]

no

you don't have to do any algebraic manipulation with that

just choose one +- when you find the inverse

[vapid steppe]

oh

[tired echo]

i would suggest doing it for a concrete example

and then extrapolating

the simplest example being y = x^2

[vapid steppe]

if i have like x²+5x+6 wouldn’t the inverse just be √x+x/5-6

oh wait nvm

[vapid steppe]

ye x=√y

[tired echo]

yes

that's the positive part of it

or in other words

the right of it

[vapid steppe]

ya

[tired echo]

so you've concluded that you're choosing the positive root

[vapid steppe]

wait if you have like √(-x) can you split it to √(-1•x)=√(-1)•√(x)=i√(x)

[vapid steppe]

im pretty sure i got it

[tired echo]

sure?, this isn't really needed here though

[vapid steppe]

is this right

wait

[tired echo]

just complete the square

[vapid steppe]

i forgot how to do that

[tired echo]

relearn

[vapid steppe]

is this right tho

wait don’t i add b²/4 on both side

,w solve ay+b=(x-c)/y for y

[ocean seal]

Results provided by WolframAlpha

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Requested by Jash#2503

[vapid steppe]

.close

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[karmic dragon]

can i get help

[wary stream]

Don't ask to ask

Just post your question

[karmic dragon]

ok

nevermind

how do end it

[ebon hornet]

.close

[tawny condor]

.close

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[dense glacier]

I need help with this identity

I get to (cosx * cos pi/4) + (cosx * cos pi/4)

[fallen verge]

are you familiar with the cofunction identities?

[tawny condor]

are you allowed to use cos(x)cos(y) identity?

[dense glacier]

no just sum and difference

[tawny condor]

alright garlic u can take this

[dense glacier]

but I thought cofunction was just pi/2 - x

[fallen verge]

oh right product to sum would be perfect

[tawny condor]

sum is also fine, then just multiply it together

[dense glacier]

Yeah but after I multiply I get (cosx * root2/2) + (cosx * root2/2)

Where do I go from there

[gray isle]

that doesn't sound right, can you show your work

[dense glacier]

I dont have camera on my computer but Ill try to explain I did cosine of pi/4 and converted it to root 2 / 2 and substituited that in to the (cosx * cos pi/4) + (cosx * cos pi/4)

I'm not allowed to use sum product yet unfortunately

Nvm found my phone

[ashen briar]

hello guys

i want any mathematical suggestions that may help me in this case

i am trying to find the best way possible to find 2 number that results in a number that is given after doing an operation with them

for example i have 471428571

any way to produce this sequence of numbers will be great , eg : 33÷7 would result in 4.71428571

which is the sequence that i want

so the answer is 33÷7

any suggestions or ways to find the 2 numbers?

[dense glacier]

My channel is already being used

I uploaded the sc @gray isle

sorry for ping

.close

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[plucky blade]

How do I figure out how many multiplications are needed?

I know 24 base 10 is 11000 in base 2

<@&286206848099549185>

[lone heart]

@plucky blade Has your question been resolved?

[alpine sable]

2
4
8
16
24.... so answer is 5

[lone heart]

@plucky blade Has your question been resolved?

[plucky blade]

how did u do that

because its x^2^0 then x^2^2 then x^2^3 then x^2^4

Where are u getting 24

<@&286206848099549185>

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[polar fossil]

is 3√8 considered a prime number

[polar fossil]

and is it also considered an integer

[woeful pulsar]

in what ring

it's an algebraic integer, sure

[polar fossil]

ohh

ok thanks

.close

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[wary tinsel]

Yo

[long axle]

Yo

[lone heart]

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[cyan sapphire]

i dont know where to start with this question

[gusty gorge]

can you perhaps write G'(x) = [something involving f' and g']?

[lone heart]

@cyan sapphire Has your question been resolved?

[cyan sapphire]

what do you mean by that

[gusty gorge]

the hint is that you can find an equation relating G'(x) to f'(x) and g'(x)

I suggest you try to see what derivative rules might get you there

if you need a further hint, I'll tell you to try differentiating both sides of G = 6f - g

[cyan sapphire]

if i differentiate both sides

i get

6 and -6

but idk how to use the graph for it

normally i would think to find the graph equation but im not sure how im suppose to find the equation for f(x)

[gusty gorge]

really? how does G' = 6?

where did you get that from

[cyan sapphire]

oh sorry

not -6

-1

[cyan sapphire]

<@&286206848099549185>

[alpine sable]

what is G'(x) in terms of f'(x) and g'(x)?

[cyan sapphire]

thank you for all ur help i figured it our

out

[carmine fog]

should be 5

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[crude marten]

Hello, could someone guide me

[median oar]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[crude marten]

I have tried various of these equations

I've also tried doing it using logic

and

gotten multiple ansdwers

but they're all wrong

[terse spruce]

hi

try to reformulate the question

[crude marten]

wdym

[terse spruce]

The question asks you what's the initial speed when you Deccelerate at -10 m/s2 and you can go 60 meters before coming to a stop

v0 is unknown,
v is 0 m/s
a is -10 m/s2
delta X is 60 m

use the 4th formula

If I am right, the answer is 34.64 m/s

[crude marten]

You are halfway right, that is correct if we were to ignore the reaction time

[terse spruce]

ah true

[crude marten]

so we just change the distance to

50instead of 60