help-27

[marble quail]

I'm not sure how you mean

@drifting sierra Hey Nel ;-; are you here ? :c

.close

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[daring solar]

The context behind this problem is that I have to find out when these two derivates are equal on an interval of -1 to 2

[daring solar]

Idk if there’s another way I can set this up, but idk how to solve that

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[daring solar]

Yea so basically you use a calculator

💀

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[brittle phoenix]

need help with this question

how do i begin?

[restive river]

I think it is easy to show that f is injective

Because f is obviously strictly increasing on R+

For surjectivity I have no ide

Maybe you can try to show that the function g(x)=x^3/3+x^2/2+x-c where c is a postitive real constant always has a positive real root x

[brittle phoenix]

What would it look like?

[restive river]

Okay I think I have an idea

Let g(x) be that function I defined in the message above

Then g(0) is negative

But g(x) is a monic polynomial, so when x becomes large enough g(x) will become positive

But then it goes from negative to positive, so it must pass throught the x-axis because g(x) is continuous

Therefore it has a positive real root

And the surjectivity is proven

[brittle phoenix]

How can we verify if the proof is correct?

[restive river]

I dont know but I see no mistake in my proof

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[craggy sequoia]

Need some help.

[craggy sequoia]

Trigonometry.

[gleaming socket]

I would sub sin(x-1) with another variable

[craggy sequoia]

u?

[gleaming socket]

Sure

[craggy sequoia]

When you say "sin(x-1)", do you mean "sin^2(x-1)"?

so sin^2(u) - 2sin(u)=3?

[gleaming socket]

No

[craggy sequoia]

confused.

[gleaming socket]

u = sin (x-1)

So
u^2 - 2u = 3

[craggy sequoia]

so sin(u) - 2 =3?

okay

u^2 - 2u = 3

u^2 = 3 - 2u

root?

u = sqrt3-2u

?

[gleaming socket]

No

Do you know how to solve quadratic equations?

[craggy sequoia]

box method?

factoring?

[gleaming socket]

Yeah

Rewrite as u^2 - 2u -3 = 0

[craggy sequoia]

okay

[gleaming socket]

Then go

[craggy sequoia]

(u+1)(u-3)?

[gleaming socket]

Yeah

So u = -1 or 3

Now sub back

[craggy sequoia]

okay

so sin(x-1) = -1?

add 1 on both sides

sin(x) = 0?

[gleaming socket]

No that's not how sine works

[craggy sequoia]

I know I need to find values that satisfy the equation

so x = 0,pi

right?

@gleaming socket

sin(x-1) = 3

add 1 on both sides

sin(x) = 4

Sin is undefined.

[gleaming socket]

Again, that's not how it works

How do you solve sin (x) = 1 for example?

[craggy sequoia]

I would find two values on the unit circle where sin(x) = 1

well, one value*

so x = pi/2

[gleaming socket]

Yes

Now what about sin(x)=-1?

[craggy sequoia]

Same thing.

[gleaming socket]

What is the value?

[craggy sequoia]

so x = 3pi/2

[gleaming socket]

Yes

[craggy sequoia]

So where did I go wrong with sin(x) = 0?

[gleaming socket]

This is basically arcsin (-1)

sin (x) = -1
x = arcsin (-1)

Now what is
sin (x-1) = -1?

[craggy sequoia]

0?

[gleaming socket]

No, you can't just separate x-1 inside the sine function

[craggy sequoia]

What do I do?

[gleaming socket]

x-1 = arcsin (-1)

We know arcsin(-1) right?

[craggy sequoia]

3pi/2?

[gleaming socket]

Yes

So x = 3pi/2 + 1

[craggy sequoia]

... I'm still confused.

[gleaming socket]

Confused with which part?

[craggy sequoia]

Can we start from the top?

so (u-3)(u+1)

u = 3

u=-1

sin(x-1) = -1

Do we not add 1 on both sides?

[gleaming socket]

What happens if you add?

[craggy sequoia]

both -1's cancel out

so sin(x)= 0

[gleaming socket]

But -1 is inside the sine function

[craggy sequoia]

So we can't just subtract?

[gleaming socket]

It would have worked if
sin (x) - 1 = - 1

No you can't

[craggy sequoia]

O.K.

So what do we do now?

[gleaming socket]

as you explained, arcsin of -1

Just like you did sin (x) = -1

[craggy sequoia]

So substituse the -1?

[gleaming socket]

sin (x) = -1
x = 3pi/2

sin (x-1) = -1
x-1 = 3pi/2

[craggy sequoia]

we can add one on both sides now?

so x = 5pi/2?

[gleaming socket]

Yes

[craggy sequoia]

But our restrict is [0,2pi)?

so that wouldn't work.

[gleaming socket]

, calc 5pi/2

[woven radish]

Result:

7.8539816339745

[gleaming socket]

,calc 2pi

[woven radish]

Result:

6.2831853071796

[gleaming socket]

Yep

Wouldn't work it seems

[craggy sequoia]

I'm really finnicky with that one.

Can you recommend a video?

[gleaming socket]

Wait I think it can still exist inside

[craggy sequoia]

?

[gleaming socket]

You know sine is periodic right

[craggy sequoia]

right

repeats every 2pi

[gleaming socket]

Yeah

What happens if you subtract 2pi

[craggy sequoia]

2pi from 2pi?

0?

from 5pi/2?

pi/2.

You could just say pi/2 worka.

huh.

[gleaming socket]

https://youtu.be/VMAMARmmDac?feature=shared

[craggy sequoia]

Thanks for your help.

What about U = 3?

sin(x-1) = 3?

[gleaming socket]

No solution coz 3 is beyond the unit circle

[craggy sequoia]

undefined.

Okay.

Definitely need a video.

[gleaming socket]

Only works between -1 to 1

[craggy sequoia]

But thanks for your help

👍

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[wooden dawn]

.Ask

can someone explain how x goes negative here

[wooden dawn]

I dont understand the explaination they have

[strange arch]

so you agree that sqrt(x²) = |x| ?

since sqrt(x²) = x OR -x

@wooden dawn

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[steel mirage]

<@&286206848099549185>

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[calm patio]

Is this correct

[devout snow]

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[calm patio]

<@&286206848099549185>

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[narrow aspen]

You're missing a factor of 2 on the first term since the derivative of 2x is 2, but other than that, yes.

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[spring sundial]

does anyone have a way of remembering that taylor's theorem is squared at the bottom

[spring sundial]

taylor's remainder theorem for multivariable calculus

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[normal vale]

wait nvm my teacher is wrong lol

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[wooden dawn]

in that case can I just pick x instead of -x ?

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[faint hinge]

Why is the answer D?

[faint hinge]

If they're both a "rate at which water __ a tank" then they're basically derivatives. Meaning, we'd choose C to see the total rate of change if it's decreasing?

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[faint hinge]

.reopen

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✅

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[heavy maple]

f(x) = 2x - 4 and g(x) = 3x + 1

f(g(2)) = ?

[heavy maple]

I dont really understand how to solve f(g(2))

[smoky gyro]

Do you know how to do f(g(x))

[heavy maple]

I have nationals tmrw and I'm doing old nationals to study

[heavy maple]

Not really

I got an answer

But idk if its right or wrong

I got it to 14

[unreal eagle]

what’s the value of x in f(x) when g(x) is the input?

[heavy maple]

?

[unreal eagle]

f(g(x))^

what is x equal to in f(x) when you’re using g(x) as the input i.e. f(g(x))

[heavy maple]

Idk

[frozen aurora]

calculate g(2) first

then take that value, and use it in f(x)

[heavy maple]

That is

Like

You calculate that by checking what the y value is

When x is 2

Right?

Or am I tripping

[frozen aurora]

there is no y anywhere in your equation

but sure go ahead

[heavy maple]

Do you not draw it in lines?

😭

[frozen aurora]

no need to draw anything

at all

substitute x=2 in g(x)

g(x) is 3x+1

so sub in x=2 into 3x+1

[heavy maple]

Yes

So

[frozen aurora]

what do you get

[heavy maple]

6+1

7

[frozen aurora]

yes

[heavy maple]

Ye

[frozen aurora]

now put 7 into f(x)

[heavy maple]

Ohhh

And then

[frozen aurora]

and you are done

[heavy maple]

14 - 2

So 12?

[frozen aurora]

close

it's f(x)=2x**-4**

[heavy maple]

?

Oh

I read that wrong

Its 10 then

Right?

[frozen aurora]

yep

that's your answer

[heavy maple]

Thanks man

Thats all I needed

Appreciate it

.close

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[heavy maple]

determine the value of a so that the inequality 2x - a < 5 has the solution x < 7

[heavy maple]

I got it to 4.5

But I'm not sure if thats right

Anyone able to help me out?

[valid iron]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[heavy maple]

I mean

Sure

But its not in english

And i dint really

Hace any work

I just tried calculating it in my head

[valid iron]

how did you get 4.5

[heavy maple]

Uhhhhhhh

I dont remember

Tbh

I dont know how to solve it

So

Could you help me solve it or?

[valid iron]

umm ok so lets start fresh

[heavy maple]

Yea

[valid iron]

,rccw

[woven radish]

Couldn't find an attached image in the last 10 messages.

Dec 13, 2023, 1:14 PM

[valid iron]

nvm

[heavy maple]

determine the value of a so that the inequality 2x - a < 5 has the solution x < 7

[valid iron]

we know that
x<7

so try converting the equation which has a in it , into this format

[heavy maple]

?

[valid iron]

consider
2x -a < 5
transpose the stuff which is in lhs to rhs except x

make sure to change the inequality if you multiple with a negative number

[heavy maple]

What

Whats rhs

And lhs

?

@valid iron

[valid iron]

right hand side and left hand side respectively

[heavy maple]

So

2x / 2?

Wouldn't that also be 5 / 2

?

[valid iron]

?

[heavy maple]

How do I transpose it

[valid iron]

add a to both sides

then multiple both sides by 1/2

[heavy maple]

Yea

[valid iron]

what do you get

[heavy maple]

Or just divide it by q

2

Right?

[valid iron]

yeah same thing

i like the multiplication cause more

[heavy maple]

X - a < 2.5a

?

[valid iron]

huh?

[heavy maple]

I took a on both sides

And divided it by 2

??

[valid iron]

no?

[heavy maple]

Wdym

[valid iron]

consider
2x -a < 5
ADD a on both sides

[heavy maple]

Oh.

[valid iron]

it kinda looks like you multiplied by a but you didnt

[heavy maple]

So x - 2a < 2.5a?

Or?

@valid iron

[valid iron]

not again

consider
2x -a < 5

adding a both sides
2x-a + a < 5+a

[heavy maple]

Oh

• a

So x - a + a < 5 + a

[valid iron]

i mentioned add ' a 'on both sides

[heavy maple]

So pretty much x < 5 + a

Oh.

So a is 2?

[valid iron]

where did the two go

NO?

[heavy maple]

BRO?

[valid iron]

2x <5+a

[heavy maple]

OHHHH

what.

No?

[valid iron]

look at this

[heavy maple]

Is it not x <2.5+a

[valid iron]

look at this

[heavy maple]

Is it not x <2.5+a

[valid iron]

no

[heavy maple]

You said multiply it by 1.2

1/2

[valid iron]

its x< (5+a)/2

[heavy maple]

2x < 5+a

Oh.

[valid iron]

x< 2.5 +a/2

now just compare with the x<7

[heavy maple]

SO I HAD THE RIGHT SNDEER

ALL THIS TIME

IT WAS 4.5

[valid iron]

?

no?

[heavy maple]

.

Brom

BRO

[valid iron]

(5+a )/2 = 7

a =?

[heavy maple]

Compare 2.5 and 7

4.5 difference

[valid iron]

bro 4.5 is a/2

[heavy maple]

Oh

A is 9

?

Thanks man

.close

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[zinc surge]

Hello!
I am unsure on how to start or complete this assignment. Any help is appreciated!
Below you can see the photos of the information needed to complete the assignment ⬇️

[zinc surge]

Below this message is the assignment I need help on ⬇️

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[zinc surge]

<@&286206848099549185>

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@zinc surge Has your question been resolved?

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@zinc surge Has your question been resolved?

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[grizzled yew]

It telescopes

Idk how to give a hint, an easy to prove identity kills this

Try to rewrite arctan(1/(r^2+r+1)) as the sum of two arctangents

[devout snow]

@restive river Has your question been resolved?

[restive river]

idk how to

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[lime walrus]

@restive river

[restive river]

i see

but ig the answer is integral

[lime walrus]

Answer is pi/4

[restive river]

yeah

but

i see lemme check

the answer is 1

@lime walrus

ig you missed one tan

[lime walrus]

Oh ok

So it's 1

[restive river]

:/

[lime walrus]

Tan of pi/4

[restive river]

yeah

alr

[lime walrus]

Yup I missed one tan

[restive river]

noice

.close

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[sullen nova]

hi, just wanted to confirm something for stats -- so ik that mean and standard deviation are denoted differently based on whether its a sample or a population ( x̄ vs μ). is population mean always theoretical?

[sullen nova]

or no

if you have a huge population, for ex every woman on earth, you could never get everyone so youd have to take a sample

but what about for small populations, like the number of students in one class

[restive river]

The key idea is that statistical methods provide a framework for drawing conclusions about a population based on a subset of data (sample), even if the entire population is accessible. This helps generalize whatever you find beyond the specific group you are observinf

[acoustic leaf]

if it's purely about that one class then the population mean can be experimentally determined. even for large populations, that's what a census does (it just gets impractical for larger and larger populations)

[sullen nova]

right

[sullen nova]

yeah

idk my teacher put in the notes that μ is the theoretical mean

[acoustic leaf]

but statistics isn't really needed if you have a population mean. the point of statistics is mainly to approximate information about a population without actually going through the effort of making a census every time

[sullen nova]

you mean sample?

or like actual stats

[acoustic leaf]

a sample is taking a small subset of the population whereas a census is surveying everyone

[sullen nova]

yeah

nvm lol

[sullen nova]

so this is incorrect/?

[acoustic leaf]

depends on context. if your're doing a z-test comparing some statistic to a known parameter than it can make sense to call it theoretical

[sullen nova]

could you give an example?

[acoustic leaf]

say you have a new drug, which treats a disease that normally takes 8 days to recover from. we want to advertise that it takes less than that, so we give it to a bunch of patients and see what their average recovery time is. so far, the only people who've taken our drug are the people in our trial. but our measured mean is still a statistic compared to the hypothetical population mean, which would be obtained by giving the drug to literally everyone in the world who's sick, so the population mean only theoretically exists, but it's still useful to do statistics to estimate it (after all, we don't want this trial to be a fluke)

[sullen nova]

right

i see

alright i get it, thank you so much for your help!

have a great day/night

.close

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[rose night]

someone help me solve 25

[hasty saffron]

What about it

[rose night]

how do we start

[hasty saffron]

What do you have to find

Roots, derivative, integral?

[rose night]

the deririvitve

my bad

😭😭

[hasty saffron]

Ah

Product rule

Take the first function as 3x and second as (2x+1)³

${(\frac{d}{dx} 3x {{(2x+1)}^3} )}= {{(\frac{d}{dx}} 3x)}}{(2x+1)^3} + {(\frac{d}{dx} {(2x+1)^3)}} 3x$

[rose night]

bro w lowkey

[hasty saffron]

I forgot the brackets one sec

[woven radish]

Lorentz
Compile Error! Click the reaction for more information.
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[hasty saffron]

It's fine now

[devout snow]

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[glad scaffold]

How do you solve this

[glad scaffold]

i tried taking the limit inside and solve the denominator

Which gave me e²

forming again 1^(infinity)

oh wait i should take log

[limber atlas]

hm

you can try with a log

[glad scaffold]

yeah i got it

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[onyx lantern]

is the graph drawn correct

[onyx lantern]

And i am not sure how to do c

[limpid nymph]

yes it is

but

[onyx lantern]

ohhh okkay rhanks

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[drifting yarrow]

Can someone please tell me what the J here refers to? i cant understand it the way the book tells me, and when we did this in lectures, we didnt talk about this j at all.

[devout snow]

@drifting yarrow Has your question been resolved?

[stone stump]

well the j doesnt seem to appear anywhere else

so it doesnt seem like it matters at all

ah wait maybe just for each dimension?

I mean we can only guess here, there are quite a few notations for which we dont know what they mean in your course. most notably of course P_j. also V(E; G_m)

[drifting yarrow]

i talked with one of my classmates about this too, and he suggested the same, i just feel like its such a weird choise of notation, But it could make perfect sense. P_J stands for partition, and the V(E,G_m) is the volume of my set calculated given a certain grid.

[stone stump]

well what does the index in the P_j mean

do you maybe have an image of the definition?

[drifting yarrow]

Yes this was what i read and couldent understand.

[stone stump]

ok

you have the square [0,1] x [0,1]

in both dimensions you split the interval [0,1] into pieces by cutting at the points k/2^m

so for m=1 you have the pieces [0,1/2] and [1/2,0]

for m=2 you have [0,1/4], [1/4,2/4], [2/4, 3/4] and [3/4, 1]

and so on

[drifting yarrow]

yes.

[stone stump]

and then you just make that into a grid in the cartesian product [0,1] x [0,1]

[drifting yarrow]

yes, i get that. But what does j mean? if i were to delete j from the problem i could solve it with just k and m, but j is there, and idk what to say if somebody asked me why i ignore it.

[stone stump]

https://i.imgur.com/yYo7T0h.png

well j stands for the dimensions

[drifting yarrow]

well alright then thanks.

[stone stump]

https://i.imgur.com/vuMsNuP.png

you could for example not split up both axes the same way

then P_1 would be {0,1/3,2/3,1} and P_2 would be {0,1/4, 2/4, 3/4, 1}

[olive bobcat]

..

[drifting yarrow]

yeah.

.close

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[sturdy linden]

I need help understanding how to prove the pythagorean identity

[supple knot]

You just plug in 5pi/3 into both trig functions

[sturdy linden]

alright ill try that, Thanks

[devout snow]

@sturdy linden Has your question been resolved?

[sturdy linden]

Im having trouble getting past the equation part now

It would become sin^2(5pi) + cos^2(5pi) =1 correct?

[supple knot]

No

5pi does not equal 5pi/3

[sturdy linden]

then sin^2(5pi/3) + cos^2(5pi/3)=1 ?

[supple knot]

Yes

[sturdy linden]

Is that it?

[supple knot]

Well you have to evaluate the left side

To verify it equals the right side

[sturdy linden]

Alright, could you help with that too?

I just have trouble with trig 💀

[supple knot]

Help how

[supple knot]

You have it correct here

[sturdy linden]

How do I evaluate it tho

[supple knot]

Depends what trig you know

Do you know the unit circle

[sturdy linden]

Ive learn not much about it

[supple knot]

Well how you evaluate them depends what you've been taught

Go look in your notes for special angles that you know sine and cosine of

[sturdy linden]

Oh ok I got it

Thanks for your help

When bringing down sin^2(5pi/3) it becomes 3/4 and with cos^2(5pi/3) it becomes 1/4 which =1

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[slow cipher]

hi

[slow cipher]

how to find the limit of : 1/x power 3

i am supposed to say that the denominator cant be = 0 but its power 3

how to cancel it or smth

[hybrid snow]

Limit as x approaches what

[stone stump]

x^3=0 is when x=0. or what do you mean

[hybrid snow]

as x -> 0?

[slow cipher]

this

[remote tiger]

To find: $$\lim_{{x \to 0}} \frac{1}{{x^3}}$$ ( which I assume to be what you meant) you would have to find $$\lim_{{x \to 0^+}} \frac{1}{{x^3}}$$ and $$\lim_{{x \to 0^-}} \frac{1}{{x^3}}$$ and see if they approach the same value. In this case, $$\lim_{{x \to 0^+}} \frac{1}{{x^3}} = \infty$$ and $$\lim_{{x \to 0^}} \frac{1}{{x^3}} = -\infty$$. Thus, the limit does not exist. The domain is as follows: $$(-\infty, 0) \cup (0, \infty)$$

why latex no work

[slow cipher]

lol

[drifting yarrow]

you have to use $$ around equations

[remote tiger]

ohhhh. thanks!

[woven radish]

Quacker
Compile Error! Click the reaction for more information.
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[drifting yarrow]

is that what your question was or?

you needed the domain too or?

[slow cipher]

but uh how did we know that 0 is a domain

[slow cipher]

yea

[drifting yarrow]

0 is not a domain, its not in the domain

[remote tiger]

The interval notation states that 0 is not part of the domain because it is open brackets

[slow cipher]

ohhh

then whats the domain please xd

0 plus and 0 minus

?

[prime egret]

Everything except 0

[slow cipher]

aha

and other thing

.lock

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[fervent sundial]

https://cdn.discordapp.com/attachments/1081837589780234241/1184896408332009562/image.png?ex=658da3c6&is=657b2ec6&hm=008ab9460b54730851f009aee696a4824cf00619830fac3ce7fd2e5960917082&

[fervent sundial]

Hi, more dumb questions 😭 but how do i realize that the sum of all the curved perimeter is the circumference of one circle

These things

Like how would i know purple = 78 degrees

[uncut crow]

i avoid geometry like the plague sorry

[wooden veldt]

If you follow the line around the circles you make a full rotation and since those curved parts are the only times you change direction and they're all the same they must each be changing direction by 360/5 degrees

[fervent sundial]

geometry sucks! np i think i remember the result from somewhere but not sure how i would’ve done it without that

[fervent sundial]

hmmm nice i think i see

is there a more formal way of showing this?

[wooden veldt]

I would argue that's formal enough

[fervent sundial]

ahaha okay thanks

[wooden veldt]

But its likely :p

I just don't know

I'm on team layla

[fervent sundial]

we’re all on team layla!

okay anyway thanks again~

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[verbal coral]

Can someone tell me why my answer is wrong?

[dense jay]

forgot the ^2

18(6x+2)^2 * x

fix that and youll be sorted

[verbal coral]

lemme try

Got it thanks a lot!

[devout snow]

@verbal coral Has your question been resolved?

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[lavish nimbus]

An infinite arithmetic sequence $a + b + 4 ...$, and a infite geometric sequence $a + 4 + b ...$, where $a \neq b$. Find a and b.

[woven radish]

Good

[lavish nimbus]

I can't seem to find a way to combine the sequences.

[red viper]

are you sure that's full statement?

[lavish nimbus]

That's all.

[red viper]

so we just have that a_1 = a+b+4 and b_1 = a+4+b?..

[lavish nimbus]

a1 = a, a2 = b, a3=4. And a1=a, a2=4, a3 = b.

[red viper]

a bit different from what you wrote at first, but it's fine

[lavish nimbus]

My bad, it should be "series".

[red viper]

so, you know that $a_{n}=\frac{a_{n-1}+a_{n+1}}{2}$ and $b_{n}^{2} = b_{n-1}b_{n+1}$?

[woven radish]

Alisia

[lavish nimbus]

Neither of them have I seen before.

[red viper]

if we assume that a_n if arithmetic progression and b_n is geometric, that's true

so, using that, you can get system of 2 equations with 2 variables

[lavish nimbus]

Is there a way beside using those formulas?

[red viper]

yes, since these formulas were derived from the definition of arithmetic and geometric progressions, you can use definition

[lavish nimbus]

Is there a provement to it?

[red viper]

$a_{n-1} = a_{n} - d, a_{n+1}=a_{n} + d, b_{n-1}=\frac{b_{n}}{q}, b_{n+1}=qb_{n}$ that must be enough

[woven radish]

Alisia

[lavish nimbus]

Got it, thank you:)

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[crisp niche]

you did wrong, let me help you

$\lim_{x \to -\infty } e^{-x}=e^{+\infty }=+\infty$

[woven radish]

Joanna Angel

[crisp niche]

and

$\lim_{x \to -\infty }\frac{-1-x^{2}}{x}=\lim_{x \to -\infty }\left( -\frac{1}{x}-x \right)=+\infty$

[woven radish]

Joanna Angel

[crisp niche]

hence:

$\lim_{x \to -\infty }e^{-x}\left( \frac{-1-x^{2}}{x} \right)=+\infty$

[woven radish]

Joanna Angel

[crisp niche]

and that is all

write it in yourexercise book

to not lose it

[limber atlas]

joanna

something which i always wondered was

can you like invert the terms given in the question

and then change the limit to +oo ?

like just reverse the variablea affected by the limits

[crisp niche]

yes i may write it here too, mmoent

[limber atlas]

thanks, i will experiment myself

dw dw

[crisp niche]

at the end of your limit, you should write: +∞(-∞)(0-1) = + ∞

[crisp niche]

this limit :

[crisp niche]

$\lim_{x \to \infty }e^{x}\left( x-\frac{1}{x} \right)$

[woven radish]

Joanna Angel

[limber atlas]

hm

[crisp niche]

such one you meant?

[limber atlas]

yes yes

like

there are some questions

which i would rather convert in positive oo format

wait let me grab an exmaple

[crisp niche]

ok

yes, and you should remember such two formula, i write now:

[crisp niche]

$\lim_{x \to \infty }e^{x}=+\infty\\\lim_{x \to -\infty }e^{x}=0$

[woven radish]

Joanna Angel

[crisp niche]

and

$\lim_{x \to \infty }e^{-x}=0\\\lim_{x \to -\infty }e^{-x}=+\infty$

[woven radish]

Joanna Angel

[crisp niche]

yvw 🙂

[crisp niche]

let me write generlaly method for limits with cosx when x ->0

[limber atlas]

wouldnt i just create x^2 in denominator and then = 1/2

[crisp niche]

$\lim_{x \to 0} \frac{1-\cos\text{}x}{x^{2}}=\left[ \frac{0}{0} \right]=\lim_{x \to 0} \frac{\left( 1-\cos\text{}x \right)\left( 1+\cos\text{}x \right)}{x^{2}\left( 1+\cos\text{}x \right)}=\\=\lim_{x \to 0} \left[ \frac{\sin^{2}x}{x^{2}}\cdot \frac{1}{1+cosx} \right]=\lim_{x \to 0} \left[ \left( \frac{sinx}{x} \right)^{2}\cdot \frac{1}{1+cosx} \right]=1^{2}\cdot \frac{1}{2}=\frac{1}{2}$

[woven radish]

Joanna Angel

[crisp niche]

this limit above, shows typical apporach, how to cope with such a limit

you're right

[limber atlas]

ohmy

you proved that thing too

yep

this is how i remember it

[crisp niche]

well ) it is just a computatinal proof, just a calculation

[limber atlas]

hey i am still a kid so 😭

[crisp niche]

smiles 🙂

[devout snow]

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[gilded nova]

there’s a for finding midpoints which is (x1+x2) / 2 , (y1+y2) / 2

Can I can use that formula to find the value of p?

or is there another formula

[lavish nimbus]

You don't need that, think of a name for the "distance" from a point to another point.

[stable saffron]

distance formula

\sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}

[gilded nova]

ohh

[stable saffron]

[ \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}} ]

[woven radish]

GigaChad

[gilded nova]

= 5?

and then i just plug everything in but p

and solve for that?

[stable saffron]

yep

[gilded nova]

oh ok

[stable saffron]

u gotta solve the quadratic

p should be -1 or 7 if i didnt do some calc mistake

[gilded nova]

okk

[gilded nova]

did I plug in the numbers right

cuz I usually get myself confused when I put it x2 and x1

[stable saffron]

its correct

[gilded nova]

sorry (7-4)^2

not 3

[stable saffron]

oh yea

[gilded nova]

ok ty

[gilded nova]

wait so is it -1 or 7

cuz why is there two answers

[stable saffron]

25=(p-3)^2 +9
16=(p-3)^2
± 4=p-3
so p= 7 or -1

cuz square of pos or neg no is always pos, so you gotta consider both

[gilded nova]

ohh ok

[stable saffron]

3^2 and (-3)^2 is 9, like that

[gilded nova]

what does ± mean

[stable saffron]

it generally indicates a choice of exactly two possible values, negative or positive

[gilded nova]

ohhh

[gilded nova]

could u help me with another one

[stable saffron]

yep

[gilded nova]

this one

idk where to start

[stable saffron]

ig theres an error in the question, it is not possible without specifying the coordinste, whether its x or y axis

[gilded nova]

huh

wdym

[stable saffron]

i mean theres a way, do you kknow the midpoint formula?

[warped laurel]

doesn't matter

it's just asking for the point

on the axis in general

like any point written as (x,y)

[stable saffron]

pretty sure its 0,2 then

Given points: $(-2, 1)$ and $(2, 3)$

Midpoint formula:
[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) ]

[
\text{Midpoint} = \left(\frac{-2 + 2}{2}, \frac{1 + 3}{2}\right) = (0, 2)
]

[woven radish]

GigaChad

[gilded nova]

yes

[stable saffron]

cool, then plug in the values in the formula

[gilded nova]

ohh ok

thats it?

[stable saffron]

pretty much

[gilded nova]

oh ok

[devout snow]

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[woven radish]

texas

[devout snow]

@real fulcrum Has your question been resolved?

[stone stump]

so if we were in R^2 and not R^3, then our setup would basically be this

https://i.imgur.com/PeNtaIQ.png

so the probability that a ray from the origin hits the sphere is the same as the ratio of the red part and the entire length of the circle its on

[devout snow]

@real fulcrum Has your question been resolved?

[stone stump]

not sure yet, I'm reading through a cople of stackexchange posts

my best guess from how I am understanding the stackexchange posts so far is that the answer should be roughly 1/4 R^2/D^2. which is very far from your experiments. so either your experiments are flawed or my understanding of the posts. probably the latter

well, still quite big from time to time

yeah, not sure. but I have to go. good luck

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[spring saffron]

im stuck on a probabilistic modelling question

[spring saffron]

I was doing this past paper and cant seem to find where ive gone wrong. this is the question

this is my solution attempt however i end up reaching u_1 = u_2 which according to the question shouldnt be the case

[devout snow]

@spring saffron Has your question been resolved?

[spring saffron]

<@&286206848099549185>

[devout snow]

@spring saffron Has your question been resolved?

[devout snow]

@spring saffron Has your question been resolved?

[spring saffron]

<@&286206848099549185>

[devout snow]

@spring saffron Has your question been resolved?

[devout snow]

@spring saffron Has your question been resolved?

[tacit kraken]

Help me

[lime walrus]

Not your help channel

[naive bolt]

oh shit my bad

I didn't even notice

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[leaden plinth]

how do you do this i have no idea, i forgot how to do implicit

[supple pasture]

Try differentiating both sides

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[umbral roost]

Here I am again with the same question ive been asking for what feels like the past week

[umbral roost]

how did they simplify the cosines like that.

[thin inlet]

plugging in pi!

[umbral roost]

how do you get that stupid exponent

[thin inlet]

$\cos(\pi x)=(-1)^{x}$ for $x$ an integer

[umbral roost]

I know that

[woven radish]

🫎 A Certain User(Moosey) 🫎

[umbral roost]

so how do you get 2 in the exponent?

[thin inlet]

-1 out front

[umbral roost]

and?!

[thin inlet]

$(-1)(-1)^{n+1}=...$

[woven radish]

🫎 A Certain User(Moosey) 🫎

[umbral roost]

that to me does not explain the 2 in the exponent. maybe im just too tired

[thin inlet]

,tex .exp rules

[woven radish]

🫎 A Certain User(Moosey) 🫎

[umbral roost]

do you agree with cos(n+pi)=(-1)^(n+1)

[thin inlet]

cos((n+1)pi)

cos(npi+pi)=(-1)^(n+1)

yes

and i moved the -1 out front, into the (-1) exponent.

[thin inlet]

product rule...

[umbral roost]

yeah got it

what about sin(npi/2)

w/ same bounds

[thin inlet]

depends on n

[umbral roost]

an expression for all n,

[thin inlet]

0 for even n

[umbral roost]

makes sense

[thin inlet]

1 for n=1 mod 4, -1 for n=3 mod 4

[umbral roost]

im outta here

this is chinese

thanks anyway

.close

[devout snow]

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[craggy sequoia]

Need some help.

[craggy sequoia]

I converted 1 - sin^2(A) to cos^2(A) via algebraic. manip.

so cos^2A + 1 = 2cos(A)

I then subtracted 2cos(A) on both sides to make a trinomial

so cos^2A -2cos + 1 = 0

Factored.

[zealous harness]

ok that's 90% of the way

[craggy sequoia]

I got (cos - 1) (cos - 1)

am I wrong..

Because they'd be the same answer.

[zealous harness]

Yes, that's not a problem

(x-1)(x-1) = 0 yields x=1 twice, there's no rule against that

[craggy sequoia]

Okay!

so cos = 1

[zealous harness]

Which happens when x =

[craggy sequoia]

and x = 0?

[zealous harness]

Yes !

[craggy sequoia]

Thanks!

.close

[devout snow]

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[craggy sequoia]

.reopen

[devout snow]

✅

[craggy sequoia]

holdon @zealous harness I have another question.

we would turn this into 1/sin^2x = 2 right?

do we multi. each side by its recip?

[zealous harness]

Yes so then you would get sin^2(x) = 1/2

[craggy sequoia]

but sin^2x / 1 * 1/sin^2x equals 1, no?

[zealous harness]

That would do nothing since you would have to multiply the other side by the reciprocal as well

when you have 1/sin^2(x) = 2, take both sides to the -1 power (that is, reciprocate them)

[craggy sequoia]

So you don't multiply?

How do you take something to the -1 power?

[zealous harness]

-1 power means you flip the numerator and denominator

(1/2)^-1 = 2/1

2^(-1) = 1/2

[craggy sequoia]

ahh

so sin^2(x) = 1/2

sqrt.

sin(x) = sqrt1/2

sin(x) = +/- 1/sqrt2

x = pi/4, 3pi/4, 5pi/4, and 7pi/4

pi/4 + 2pin, 3pi/4 +2pin, 5pi/4 + 2pin, 7pi/4 + 2pin

Right?

[zealous harness]

Ya

pi*n

or just pi/4 + pi/2 * n

[craggy sequoia]

Thanks.

+close

.close

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[near rapids]

How do I find the nature and type of a conic? In my case, the conic is 3x^2 + 4xy + 3y^2 + 4x - 4y + 1 = 0. I also have to find out how to find the center of the conic and its axes (i hope I translated everything good)

[rotund bolt]

Can anyone help me with some precall hmw?

[devout snow]

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).

@near rapids Has your question been resolved?

[near rapids]

I don't need the solution, I just need a bit of guidance with what formulas I should use. I tried searching online but I find different answers everywhere so I'm not sure what applies and what not in my case.

[crisp adder]

@near rapids in general, if both the x^2 and y^2 are positive, then you probably have an elipse or circle. In you case since there is an xy term, it is an ellipse. If you have two squared terms and one is negative, then you get a hyperbola. If you only have one square term, then you get a parabola

[near rapids]

and if B was not an xy term would it have been a circle?

[crisp adder]

Not necessarily, no xy term means you can complete the square in both, but it becomes a circle in your case ince the coefficients of both x^2 and y^2 are 3, leading to a circle. If different, then you get an ellipse

[near rapids]

thanks!

[crisp adder]

I'm still not exactly sure how the 4xy "stretches" the circle out though. But I believe everything I said above is correct

The cross term is a bit funky

[thin inlet]

it's always a good bet to try to complete the square

[near rapids]

i'm not sure how to go about doing that

maybe it's the language barrier but I don't understand what completing the square means

found this in an youtube video, going by this it would be a circle, right?

[near rapids]

also can someone please verify if this is correct?

[thin inlet]

that looks good

[near rapids]

alright, thanks:)

.close

[devout snow]

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[subtle vessel]

im honestly just confused how to start my equation at

[subtle vessel]

<@&286206848099549185>

[jade oak]

What does 8 more and 3 times means in mathematical numbers and symbols

[dense jay]

what kind of triangle is it

[jade oak]

Isosceles

[devout snow]

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[devout snow]

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[restive river]

\textbf{Question:} If $A$ is a normal subgroup of $G$ and $B$ is any subgroup of $G$, prove that $AB$ is a subgroup of $G$.

[restive river]

v

.close woops

[devout snow]

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[half wedge]

Hi, can anyone help me with this. I don't know where to start

[wooden veldt]

Try and times the top and bottom of the fraction by the same thing, such that it makes the bottom real

[half wedge]

ok

i tried (1-z) and (1+z) but it doesnt helo

@wooden veldt

[wooden veldt]

Right because really on the bottom you have 1 - x - iy

[half wedge]

oh yh ok

none of the conjugates work

[surreal night]

1+z doesnt work?

(1-z)(1+z)=?

[half wedge]

no

[surreal night]

are you sure?

what is (1-z)(1+z)?

[half wedge]

its $$(1+z)^2/(1-(x^2+2iyx-y^2))$$

[surreal night]

not quite

$(1-z)(1+z)=1-z+z-z^2=1-z^2=1-(x+iy)^2$

[woven radish]

Flappie

[surreal night]

ah wait

[half wedge]

oh i forgot to add the 1

[woven radish]

Hector

[surreal night]

yes

[half wedge]

it doesnt cancel out the i though

[surreal night]

no

this is horribly confusing

sorry

also i realised ive made a mistake

we can take 1-z

and fill in z=x+iy

to get 1-x-iy

then what is the conjugate of this

[half wedge]

1-x+iy

[surreal night]

so we multiply both sides by this

and then we should have gotten rid of the fraction

[half wedge]

ok

yh that works

r/close

[surreal night]

use .close

[half wedge]

.close

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[glossy escarp]

hi please help me understand why 3c is 0.5

[devout snow]

@glossy escarp Has your question been resolved?

[devout snow]

@glossy escarp Has your question been resolved?

[supple knot]

Both rolling 5 and 6 have equal probability

[glossy escarp]

yes i understand that but how does it connect 'at least one 5 rolled before 6'

[supple knot]

Calculate the probability of the complement event

[glossy escarp]

ohh i get it now, thanks

[devout snow]

@glossy escarp Has your question been resolved?

[devout snow]

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[hybrid snow]

the domain of $\log_a(x)$ is $x > 0$

[woven radish]

Umbraleviathan

[hybrid snow]

However what you have, $log_9(0.5) = x$ is an expression and has nothing to do with domain per se