help-33

[hot mantle]

-32 is the minimum

[flat raft]

yeah I see

I guess what you've done is basically another way of plugging in -b/2 for x

[hot mantle]

The boxed equations above were the only given sets up information fyi

[flat raft]

yeah I see

the natural thing to do here is just to plug in (-b/2, -32) into the equation for x and solve

but I suppose you could also formulate it getting the curves to coincide, but that's basically implicitly plugging in x = -b/2, y = -32 since if you're asserting that's the vertex form of the equation, you're basically saying that this is a point on it

[hot mantle]

Well you would first have to establish -b/2 as the x value

Everything following that step is relatively simple

It’s just a matter of substitution

[flat raft]

well, in what you wrote for the vertex form, you're basically implicitly assuming that the vertex occurs at x=-b/2

because the vertex form is just y = x^2 shifted by the vertex coordinates

so it's just potatoe potato

[hot mantle]

Right, but in order to produce that value, I had to complete the square and factor for x^2+2ab+b^2

[flat raft]

no, you wrote this by just saying that it's the vertex form

[hot mantle]

I skipped a few steps

[flat raft]

okay but I'm not talking about that

I'm talking about writing the second equation you did

the point is that you have to somehow show that every quadratic is a translated and rescaled version of y=x^2 to actually say that's the vertex form

I'm sure you're just taught this fact and told to accept that every quadratic can be written as a(x-b/2a)^2 + c = y, but this has to actually be proven lol

[hot mantle]

Proving stuff pfft

Lame

Alright thank you

[flat raft]

and that proof yields also that x = -b/2a is the x-coordinate of the vertex

so it's not really a biggie that you have to actually find it

[hot mantle]

I came up with it tho

I wasn’t familiar with that formula

Did I assume by inserting the y in the equation?

[flat raft]

I just set up the equation in vertex form
wdym? I'm saying that saying that you wrote something in "vertex form" is implicitly assuming what I said

[hot mantle]

Oh my bad my bad

You’re absolutely right

[flat raft]

like there's no reason to believe that (x-b/2)^2 - 32 = y is the same curve as the one you're talking about unless you know that (-b/2, -32) is the vertex

[hot mantle]

Yup

I see what you’re saying now

Took me a minute 😅

Anyways thank you

.close

[marsh citrus]

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[hot mantle]

.reopen

[marsh citrus]

✅

[hot mantle]

But is vertex form not in fact applicable to all quadratics?

[flat raft]

it is

[hot mantle]

Okay thx

.close

[marsh citrus]

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[random gust]

I am supposed to simplify this to a+bi form

which I did

but

I got separate answers from the answer key

Answer key answer

My answer

I just need someone to show me their work.

*oops there is also a i multiplied to the end of both answers

[devout mauve]

I wouldn't call either of those simplified

[random gust]

just "change to a+bi form

[devout mauve]

uhm that seems wrong

[random gust]

??

[devout mauve]

that is a purely complex number with no real part

[random gust]

the first part is the real part

2+5 square root of 2

[devout mauve]

that's not how you write this

[random gust]

a+bi?

[devout mauve]

written like this you mean (2+5sqrt2)*(sqrt5+sqrt10)*i

so what you mean is I guess (2+5sqrt2) + (sqrt5+sqrt10)i ?

[random gust]

answer is textbook

in

[devout mauve]

so the textbook has the plus but you just decided to ignore it?

[random gust]

oh shoot yeah forgot the plus

[devout mauve]

also the imaginary part in that picture is again different

[random gust]

oof im just typing them onto desmos for the photo

the point is my answer is different

than the textbook give one

anyone can do the work??

[devout mauve]

well of course we could. but that's not exactly what we are here for

how about you show your work

and we can check where you went wrong

did you write the original picture correctly?

[devout mauve]

this one?

[marsh citrus]

@random gust Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[pure bloom]

Need help with a

Someone posted in the other channel same time as me oops

[proud ice]

oh joy conics

Okay well first, do you know how the standard form looks?

[pure bloom]

😭

yes its x-h^2/a^2 +y-k^2/b^2=1

right

[proud ice]

Ooh parentheses would be nice

But yes, I can make it out

[pure bloom]

sorry im mid problem solving 😭

[proud ice]

nw

[pure bloom]

ive grouped together like terms and have it equal to -185

[proud ice]

It would be +(y... or -(y... depending on the type of conic

[pure bloom]

it should be +y right

since the y^2 is positive

[proud ice]

But the x^2 is negative

[pure bloom]

wait nvm

its a hyperbola

okay

right

it would be y first then x

[proud ice]

Fun fact but irrelevant, it could also be a straight line.

[pure bloom]

oh really

but we can assume it isnt since it says conic right

[proud ice]

Conic's usually write x^2 term first and positive

[pure bloom]

yes

[proud ice]

lines and points are conic equations too in the extreme case

Edge cases

[pure bloom]

ah gotcha

prob wont run into that if its not too prevalent

so i currently have the y's and x's grouped and it = -185

[proud ice]

okay

And what next?

[pure bloom]

i want to complete the square correct

[proud ice]

yes precisely

[pure bloom]

so i would add 184

so it equals -1

right

[proud ice]

why are you adding 184?

[pure bloom]

isnt that part of completing the square?

or no

getting it =1

[proud ice]

You want to complete the squares for the x and y terms

That is you want some kind of (x-h)^2 and (y-k)^2

[pure bloom]

oops my bad

okay yes

so if its 9y^2-90y

how do i go about figuring that through

[proud ice]

If you have ax^2+bx, to complete the square, you need +(b/2a)^2

Then you have (x+b/2a)^2

If it's ax^2-bx, then you still add +(b/2a)^2, but now the squared term is (x-b/2a)^2

[pure bloom]

okay let me track

talk me through it a little please sorry 😭

[proud ice]

sure

You have 9y²-90y

So you need to add (b/2a)² or (90/2*9)²

[pure bloom]

okay and is this just a typical part of completing the square? you always add b/2a^2?

[proud ice]

If you have the first two terms, yes

[pure bloom]

you mean the squared value and the y value

i think all of mine have that

except the last one which is y^2-4y-12 but thats pretty easy

[proud ice]

indeed

[pure bloom]

okay and so i add (90/2*9)^2

to which part of it

like how do i physically add it to the 9y^2-90y

[proud ice]

Add the value to both sides of the equation

[pure bloom]

so its (9y^2-90y)+(90/2*9)^2

and then i add that on the -185 side aswell

[proud ice]

mhm

gotta keep the equation equal

[pure bloom]

okay and so i do 8/2*4)^2 for the x portion?

And the y looks like this?

[proud ice]

It's 90/(2*9)

And yes for x

I gtg so good luck

[pure bloom]

okay no worries ty for helping

[marsh citrus]

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[woeful viper]

Hello

[woeful viper]

0.2, 0.3, 1.1, 1.4, 4.8, 0.3, 1.9, 3.8, 2.1, 0.1, 1.6, 2.2, 2.5, 1.6, 1.8, 2.4

1. The table at the right shows the time spent waiting in line (in minutes) at a local supermarket

a) Perform a one-variable analysis of this data by finding the mean, median, mode, variance, standard deviation, range, and first and third quartiles.

b) Determine if there are any outliers. Explain how you did this.

c) Create a modified box-plot for this data.

[marsh citrus]

@woeful viper Has your question been resolved?

[marsh citrus]

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[cobalt sentinel]

Yea gimme a sec

First write the 8^(x-1) with a base of 2

@fast crest

Nice now calculate this

We’ll get to that

Just do this first

Now how can we write a square root as an exponent

[elfin berry]

Stephen

[cobalt sentinel]

Yea 2^[(2x-3)/2]

= this

Np

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[torpid anvil]

Can anybody explain to me what an arch length is

[proper zodiac]

in what context

[torpid anvil]

Like the relationship between arch length radius and theta

In a circle

Oh wait no

I get it now

.close

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[young garden]

Any ideas pls

[open shale]

I am pretty sure there is missing info

but if a = 0

we have an arithmetic progession

and sin a = 0

[young garden]

Nah there isn't unfortunately

[young garden]

How?

Sin 0 = 0

So everything will cancel?

[open shale]

0<a<180

[open shale]

yeah, but it is not an option

[young garden]

Ahh true☠️

[open shale]

ok

what i know

is that it must be an decreasing sequence

[open shale]

wait not true

3sin x, 4sinxcos x, 3sinx-4sin³x

[young garden]

Idk how u got the last one

Sin x + 2x?

[open shale]

yes

[young garden]

Ahh ok then

[young garden]

What kind of sequence would this be?

[open shale]

Well

0<a<180

so sin x >0

so 3sin x > 3sinx -4sin³x

The problem is 4sinxcosx

[young garden]

Yeh

Hmm

I don't have a clue 😭

[open shale]

We have 2 possibilities

3sinx-4sin³x< 4sinxcos x < 3sin x

or

3sinx-4sin³x<3sin x < 4sinxcos x

[open shale]

If this is the case

4sincosx/2 = 3sinx-4sin³x +3sinx

sinx!=0

[young garden]

Is that x factorial or not equal to?

[open shale]

2cos x = 6-4sin²x

[open shale]

yes

[young garden]

Which one?

[open shale]

cos x = 3-2sin²x

[open shale]

not equal to

[young garden]

Ahh right

[young garden]

So are they the 3 possibilities?

[open shale]

wait what

[young garden]

,solve 4sincosx/2 = 3sinx-4sin³x +3sinx

[open shale]

nah

delete that

[elfin berry]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 12)

[open shale]

it is easy

[young garden]

Oh lol

[open shale]

look

what is the maximum value of 2sin²x

[young garden]

4cosxsinx=0?

[open shale]

why

[young garden]

2sin2x =0

[young garden]

Differentiate and equate to zero?

For max point?

[open shale]

what?

[young garden]

Wait that's wrong?😭

[open shale]

dude

[young garden]

That's the max value? I'm probably overcomplicated it

[open shale]

$-1\leq \sin x \leq 1 \implies 0\leq \sin^2 x \leq 1$

[elfin berry]

sopinha

[young garden]

Ahhhh

[open shale]

so the maximum is

[young garden]

1

[open shale]

no, of 2sin²x

[young garden]

2

[open shale]

okay

cos x = 3-2sin²x

what is the lowest value of 3-2sin²x

lowest

[young garden]

1

1

[open shale]

what is the domain of cos

[young garden]

-1 to 1

[open shale]

so?

[young garden]

Wait

[young garden]

The lowest value of this should be 1??

[open shale]

Yes

[young garden]

3-2?

[open shale]

And the range of cos is -1 to 1

so sin²x must be 1

[young garden]

Yea

[open shale]

We have x = 90

only solution

(to this case)

now we wish it would be an arithmetic progression

[young garden]

So now we have a solution of x = 90

[open shale]

well

we have to check if it fits

3sinx-4sin³x,4sinxcos x ,3sin x if x = 90

3-4, 0, 3
-1,0,3

nope

either i did a mistake

or

idk

ah

It sould've benn 4sincosx = (3sinx-4sin³x +3sinx)/2

The solutions on wolfram aren't beautiful

[young garden]

What were they?😂

,calc 4sinxcosx = (-4sin^3x +6sinx)/2

[elfin berry]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 11)

[open shale]

My internet is shit so the site isn't loading

the only readable solution was x=0

[marsh citrus]

@young garden Has your question been resolved?

[young garden]

😭😭

[marsh citrus]

@young garden Has your question been resolved?

[young garden]

@ionic lava sry abt the @ but u got any clues?

U helped with something similar b4

So I was wondering if u had any ideas abt this one

[ionic lava]

@young garden you have cos(a°) and you want sin(a°) ,Right ?

[young garden]

Nah I don't have any of them😭

Well we do have cos a° actually but it's in terms of k

[ionic lava]

you looking for a° ?

[young garden]

Nah we want sin a°

[young garden]

Yes but cos a° is in terms of K so idk how that's gonna work😭

[ionic lava]

it's easy to find sin(a°) in term of k

since cos²(a°) + sin²(a°) = 1

[young garden]

Oh Yeah that's pretty easy

But how do we get the actual value?

[ionic lava]

i will try to solve it

if i found something i will send it

[young garden]

Great, thanks!

[marsh citrus]

@young garden Has your question been resolved?

[ionic lava]

@young garden i guess K = 5

[young garden]

ill try it ty

so whats sin a?

[ionic lava]

$sin(a) = \frac{sin(3a) + 2 sin(2a)}{K - 2}$

[elfin berry]

Mehdi_Moulati

[ionic lava]

this is what i found

[young garden]

whats a though?

ohhh nvm nvm

[ionic lava]

since $cos(a) = \frac{\sqrt{k} - 1}{ 2}$

[young garden]

so, 0.786

[elfin berry]

Mehdi_Moulati

[ionic lava]

$2cos(a) = \sqrt{k} - 1$

[elfin berry]

Mehdi_Moulati

[ionic lava]

$2cos(a) + 1= \sqrt{k}$

[elfin berry]

Mehdi_Moulati

[ionic lava]

so $K = (2cos(a) + 1 )² = 4 cos²(a) + 4 cos(a) + 1$

[elfin berry]

Mehdi_Moulati

[ionic lava]

we know that : $sin(3a) = sin(a) (4cos²(a) - 1)$

[elfin berry]

Mehdi_Moulati

[ionic lava]

so $4cos²(a) = \frac{sin(3a)}{sin(a)} + 1$

[elfin berry]

Mehdi_Moulati

[ionic lava]

and we know that
$2 sin(2a) = 2 \cdot 2cos(a)sin(a) = 4 cos(a) sin(a)$

[elfin berry]

Mehdi_Moulati

[young garden]

i got 0.786 for sin a

you think thats right?

[ionic lava]

$sin(a) = \sqrt{1 - cos²(a)} = \sqrt{1 - \frac{\sqrt{K} - 2}{2}}$

[elfin berry]

Mehdi_Moulati

[young garden]

great, ill try it

i got 0.939

[ionic lava]

i make a mistake

$sin(a) = \sqrt{1 - cos²(a)} = \sqrt{1 - (\frac{\sqrt{K} - 1}{2})²}$

[elfin berry]

Mehdi_Moulati

[ionic lava]

,w sqrt(1 - (sqrt(5) - 1)²/2²)

[elfin berry]

Results provided by WolframAlpha

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

Requested by Mehdi_Moulati#1210

[young garden]

😭 thats wrong for some reason

how did you get k = 5

[ionic lava]

i proved that $K = (2cos(a) + 1 )² = 4 cos²(a) + 4 cos(a) + 1$

[elfin berry]

Mehdi_Moulati

[ionic lava]

$K = 4 cos²(a) + 4 cos(a) + 1$

[elfin berry]

Mehdi_Moulati

[ionic lava]

first of all K must be positive (because K is inside the square root)

so K>= 0

[young garden]

yeahhh

did you get an arithmetic sequence?

[ionic lava]

i just try to find sin(a) in terms of sin(3a) and sin(2a)

[young garden]

😭

any luck

[marsh citrus]

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[still temple]

whats the answer?

[still temple]

Do you still have the lessons with you? If not, I can (try to) teach you how to do it from scratch

pls do

Alright, first of all, do you remember what m and c represent visually?

m is gradient

c is y intercept

right?

True! (because if you input x=0, you get m*0+c=c, so f(0) is just c)

ok

So, in this case, what would c be?

-4

Alright, that's the easy part - Now, you need to pick any 2 x values

In theory they can be anything but I suggest you pick 0 and 1 in this case because it makes the math easier

okay

0 and 1

First:

• What is f(0)?
• What is f(1)?

what's f?

The function of the line

Oh, I guess I should've said t

Because it's the line T, my bad

np

So...

• What is t(0)?
• What is t(1)?

idk

Hmmm, alright, let me show you

[still temple]

wait is it t(0)=0

Is this not how you reason through it?

[still temple]

Nope, t(0) is the value of y when x is 0, a.k.a. the y-intercept (sorry, I'm not sure how to explain this well, it might sound like word mash)

wait

r u explaining the American way

The American way?

I had no idea there was a difference-

Pretty sure this is standardized internationally

there's a moderate difference when u teach it, however the answer is the same

Yeah, there's different ways to find it

In fact I was gonna end this by showing you another way to get m

Which doesn't work as often but is still nice to know

Everything on the purple line is on x=2

[still temple]

yes

And everything on this line is on y=3

yes

And, as you can see, the point that is both on x=1 and y=1 is also on the line

Hence, when it comes to this line, when x is 1, y is 1,

Which in turn implies that t(1)=1 (in general it should be t(x)=y)

t(x)=y)

or the t(1)=1

?

[still temple]

ohh ok yea

didn't c that

Alright, do you now see how to read the graph?

yes

So, do you know what t(0) is?

yes

👀... So what is it?

-4

True, which is also the y-intercept, because the x=0 line is the same as the y line (try to wrap your head around this, I know it sounds confusing)

So we now know that our function is of the form mx-4

Okay my brain just did a 360

You don't need to x values for this step, I just realised, you only need one

Let's pick x=1 because that'll make the calculations easier

If we plug in x=1, we get m*1-4=m-4, and we know that this should equal 1

Hence we get the equation m-4=1, which should be easy to solve

@still temple Everything good so far? Note, math class is confusing sometimes, but at least the teacher has experience in teaching. I don't, I'm just a volunteer

I'm taking this silence as a no-

Do ping me when you come back, and feel free to ask as much questions as possible to make this crystal clear

@still temple

I'm very sorry, I couldn't find my laptop charger

[still temple]

mood

[still temple]

Yeah, that's why I really don't like it

[still temple]

I never avoid the oportunity to have long conversations with my math teacher about the education system, but oh well, not here... So let's get back to the subject

Is everything clear?

And, if not, which message first gave you any confusion?

Everything tbh

[still temple]

So, which message was the first to be confusing?

Even if it was just a slight confusion, that still counts

I have another question if u dont mind

Sure

Ah, you see, in theory if I teach you how to solve the first problem correctly you should easily learn how to solve the second

https://cdn.discordapp.com/attachments/907022703230345237/1050155279154679918/image.png

So, let's go back to that

Do you get that t(0) is -4 and t(1) is 1?

Yes,

And do you get why the function must be of the form mx-4?

may u remind me why pls

Well, do you see why all linear functions are of the form y=mx+c?

yea

And do you remember how we found out that c is -4 in this case?

Yeah

the y intercept

Yeah, a.k.a. the value of y when x is 0

[still temple]

Is that part also clear?

[still temple]

ye

Ah, see now why math teachers hate it so much when you say that what you didn't understand was "everything"? ;) Clearly you understand quite a bit in fact

[still temple]

Because of that, we can substitute -4 for c

And we get y=mx-4

The last piece of the puzzle is that m value

[still temple]

yeah lol

The trick here starts by picking any x value

In this case, I'll pick x=1 because that tends to make the calculations easier

[still temple]

But when we have to sub

i find it kinda confusing

[still temple]

When you know the value

Once you know that c is -4, it's easier to replace all instances of c with -4

It means you have less unknowns, less pieces of the puzzle to find

[still temple]

By looking at the graph, we can see that where x is 1, y is also 1

[still temple]

So therefore, you can replace x with 1, in which case you also have to replace y with 1 for it to stay true (similarily, if you replaced x with 0, you would have to replace y with -4), which gives you m*1-4=1 (obviously m*1 is just m)

@still temple Everything clear so far?

Ping me when you come back

Alright, I'll have to go. When you come back, ping the "Helpers" role, and ask them to explain anything I said which was confusing

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[broken wadi]

Hey, I'm currently learning about Taylor and Mauclarins approximations. Is the encircled part the remainder?

[broken wadi]

Reason why I'm asking is that the remainder should essentially be how off the approximation is

The quesiton in my book states to estimate the error when approximating ln(1+x) approx 0.1

However, I'm a bit unsure on how to do this

[main idol]

not enough context to answer your question.

can you show the whole page

or you could try calculating the remainder for that function and see if it matches

[broken wadi]

It's in Swedish.

[main idol]

yea it is

"maclaurins" gives it away

[broken wadi]

Ahh

My question, after thinking for a bit more, is when should you know whether theta should be 0 or 1?

or in your example, c?

I know that a = 0 since it's mauclaurins but how about theta? Textbook never demonstrated on how to decide its value, only that it's between 0 and 1

ahh solvedi t

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[crimson dragon]

Need help on this problem^^

[untold spruce]

find the derivative of E, and set it equal to 0

[crimson dragon]

I don't really know what that means, they just want me to be able to identify it off of a graphing calc

[untold spruce]

which class is this for sorry, I am unsure which mwthod would be bestfor you

[crimson dragon]

Precalculus

[untold spruce]

oh, hmmm when it says minimize, does it mean most negative?

[untold spruce]

you should probably just graph it and find minimum that way

[crimson dragon]

V means speed so it can not be negative

Yes I just do not know which one it would be because the graph is like the one I put above

How can I identify a minimum speed of a graph like that?

[untold spruce]

this is beyond the scope of pre calculus, you would need a derivative. If your calculator has a local minimum function, you should try to use that

[crimson dragon]

Yup, just did it and got ir right

Ty @untold spruce

[marsh citrus]

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[crimson dragon]

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[deft apex]

how do I know i've selected the correct vectors for a and b for the cross product

so I would use OR and OQ

but my head gets clogged in 3D space if I try to imagine too many points

anyone know how tf to get it each time?

[whole hazel]

Doesn’t a parallelogram have 4 vertices?

Or is the origin included

[deft apex]

origin is one

[whole hazel]

oh ok

Just do R and Q

[deft apex]

yeah I got OR and OQ

but I had no idea to select those without graphing it on a calc

[eternal tundra]

doesn't the vector product give us the area formed by the two vectors?

i mean the absolute of the product

[deft apex]

yes

[eternal tundra]

meaning
A=|v x w|

[deft apex]

its |a x b|

[eternal tundra]

yeah

[whole hazel]

Yea

[eternal tundra]

a and b are given

[deft apex]

no no the issue is selecting the vectors by knowing which ones are the edge

[whole hazel]

yea the parallelogram spammed

spanned*

[eternal tundra]

a=(1,1,0)
b=(2,2,2)

[deft apex]

if I was given like 2 other vector points

[whole hazel]

@deft apex you just choose two adjacent sides

It doesn’t matter which

[eternal tundra]

a x b=(2,-2,0)
|a x b|=sqrt(2^2+2^2)=sqrt(8)=2sqrt(2)

[deft apex]

yep

[eternal tundra]

,calc sqrt(8)

[elfin berry]

Result:

2.8284271247462

[deft apex]

got 3

like

how did you look at those points and know which ones are the adjacent sides

or

guess ill just draw 50 thousand things till I get it

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[brisk light]

Can someone explain how to do this to me?

[silk sable]

There is a point L between J and K which create two segments called JL and LK.

The sum of those two segments must equal JK.

JL + LK = JK.

You are given JL = 2/3 LK. Substitute that into the above equation and solve for LK.

[brisk light]

oh i see

thank you

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[still temple]

4 2 3
4 4 4
4 4 4

guys is this a lineraly dependent matrix?

[marsh citrus]

Please don't occupy multiple help channels.

[main idol]

.close

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.close

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[silver star]

heres how i set it up i got the intengral from 10 to 1

(2 pi x (x^3 - 4 pi x ^2 - 4 pi x )) and its wrong ?

<@&286206848099549185>

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[golden sun]

Can someone please explain to me how this works?

[marsh citrus]

@golden sun Has your question been resolved?

[indigo nacelle]

ok basically you need to break this into parts (at least thats how i understand it)

$\lim_{n\to \infty} \frac{1}{n} = 0$
$\lim_{n \to \infty} \frac{n+1}{n} = 1$
so,
$\lim_{n \to \infty} 8(4^{\frac{1}{n}} - 4^{\frac{n+1}{n}} = 8(4^0 - 4^1) = 8(-3) = -24$
and
$\lim_{n \to \infty} n(1-4^{\frac{1}{n}} 😒

[elfin berry]

AJ_l0l
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[indigo nacelle]

how did they get ln

lol

[golden sun]

so I'm looking at the process my teacher provided and he split it into denom and numerator

[indigo nacelle]

nvm -- i tried 😭

[indigo nacelle]

yea like that makes sense -> you get -24

but then howd he get ln(4)

[golden sun]

but Idk how he got 4^1 from 4^(n+1/n) when subbing in infinity

ngl

[indigo nacelle]

cuz the denom turns into n(1-4^0) = n(-3)

[golden sun]

he lost me at the 0/0

:')

[indigo nacelle]

oh yea

now i rmmr

i tried using latex and its confusing

let me write it on paper

[golden sun]

tysmm

[marsh citrus]

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[indigo nacelle]

this took me too long

but like you just apply lhr

[golden sun]

.reopen

[marsh citrus]

✅

[indigo nacelle]

but to do that you turn the inf * 0 to 1/inf * 0 and stuff

for a detailed summary: see https://www.symbolab.com/solver/limit-calculator/\lim_{x\to\infty}\left(n\left(1-4^{\frac{1}{n}}\right)\right)?or=input

cuz the top is easy to get, its -24

and then you apply lhr to the bottom and get the -ln4

[golden sun]

ohh ic tysm

I think I just need to review the lhr bc it's been a while but thanks!

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[vivid tusk]

not an actual question, just need someone else's opinion

[vivid tusk]

the answers in this multiple choice should be written using OR, not AND, no?

like when stating solutions to a quadratic, you use OR. Using AND wouldn't make sense cus you'd need both x at the same time

[stoic saddle]

you are exactly right

all these ANDs should be ORs

[marsh citrus]

@vivid tusk Has your question been resolved?

[vivid tusk]

thanks, now to convince my friend to change their and into ors lol

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[whole jetty]

Find the dimensions of these

[marsh citrus]

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[elfin berry]

Basudev

$P(F \cup A) = \frac{6}{52} + \frac{2}{52}$

[thin bison]

Should be this right??

Question umber 10a

[stoic saddle]

surely there are 12 face cards in the deck, not 6?

[thin bison]

Oh ye

F

[stoic saddle]

it says "face card or red ace", not "red face card or red ace"

[thin bison]

Ye my mind slipped

[elfin berry]

Basudev

[thin bison]

This right??

[stoic saddle]

aye

[thin bison]

Thx

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[celest shadow]

can anyone explain why 6 turns to 1/6 in my professors notes?

[glass silo]

Think of it more as $\int_{0}^{\frac{4\pi}{3}} \sin(u) \cdot \frac{1}{6} du$

[elfin berry]

chartbit

[glass silo]

Replace the dx with 1/6 du, if that makes it clearer why?

[celest shadow]

no im still kinda confused

sorry

so i replace dx with 1/6 but where does the 1/6 come from

[latent jay]

u = 6x
therefore du/dx = 6
therefore du = 6 * dx

[late geode]

the work is coloured in lime

[latent jay]

therefore dx = (1/6)*du

so they substitute that in

and they swap out 6x for u

[marsh citrus]

@celest shadow Has your question been resolved?

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[orchid oracle]

How do I solve this?

[orchid oracle]

This is the answer if it helps:

[late geode]

consider the first few derivatives,
leave stuff in expanded/exponent form
and you should see a pattern

[orchid oracle]

So, I have to calculate, lets say the first 3 derivatives, and I should see a pattern?

Let me try 😼

Okay I got it, thx

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[tardy basin]

Mates English isnt my first language someone give me answer 💀

[tardy basin]

Is it 26?

[deft apex]

please dont tell me this is some type of survey thing

[tardy basin]

No

It isn't

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[still temple]

should i use product rule for finding d/dx(xy)??

[strange hazel]

yes

[still temple]

k

[strange hazel]

just replace u=x and v=y you will get your answer

[still temple]

i am getting x*dy/dx + y

should i further convert it and write dy/dx = -y/x?

[raw saddle]

yes

depends on whether you intend on doing anything further with it and what your school requires

[strange hazel]

i think no

[raw saddle]

if for instance you need to differentiate it again, its easiest to leave it in this form

[strange hazel]

cuz you are not given any function

so leave it like this is better option

[still temple]

i need to apply this in a question so ig i should just leave it this in the first form..

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[still temple]

How do I find the antiderivative of this?

[fleet dagger]

$2\sqrt{t} - \frac{4}{\sqrt{t}} + 3$

right the bot was down

[elfin berry]

Doggo
Compile Error! Click the reaction for more information.
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[still temple]

do I turn sqrt t into t^1/2?

[eternal epoch]

split numerator

and integrate separately

so f(t) = 2root(t) - 4/root(t) + 3

now integ separately @still temple

[still temple]

how do you take the antiderivative of 2t/sqrt t?

[eternal epoch]

notice that t is (root(t))^2

so it becomes 2root(t)

then use intx^n formula

[fleet dagger]

I was thinking of the $\frac{-4}{\sqrt{t}}$ for a sec ignore me

[elfin berry]

Doggo

[still temple]

wait how did we get 2sqrt(t)?

[still temple]

^

[fresh salmon]

my math bot said

To find the antiderivative of a function, we can use the rules of integration. To start, we can write the given function as the product of two simpler functions:

$$\frac{2t-4+3\sqrt{t}}{\sqrt{t}} = \frac{2t-4}{\sqrt{t}} + \frac{3\sqrt{t}}{\sqrt{t}}$$

doesnt work

smh

[elfin berry]

dqvidutzul
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[eternal epoch]

@still temple

tell me can u understand

[still temple]

ok i see what you mean

[eternal epoch]

did u understand it tho

[still temple]

yea, you turned t into sqrt(t)^2 so you could cancel

[eternal epoch]

yeaaa

good job

[still temple]

thanks for the help

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[sullen sable]

Hi, I got the following question

There are 2 halls the sum of the people in them is 900
If 100 people will move from hall A to hall B the amount of people in hall B will be twice as large as the amount of people in hall A so here's what I've already done

[sullen sable]

$a+b=900$
$a-100= $

Why is this not working?

[main idol]

bot's borked

[sullen sable]

Well anyway I don't know what shall I put on the 2nd side

Like I do

2(b+100)

But I don't know why is it true

[bitter warren]

Did you get
A + B = 900
A-100 = 2(B+100)
?

[sullen sable]

Yes

But

About the last equation

Idk why is it working

But that's what are teacher told us to do

I just don't understand the logic behind it

[bitter warren]

So we know that 100 people moved from A to B, so we get (A - 100) and it says that when 100 people move into B (B+100) the amount of people in B is twice as large (2*(B+100)) as the amount in A

[sullen sable]

Right

But why not 2b+100

Wouldn't that make more sense

Oh

Wait

Nope I thought I understood but I didn't

[marsh citrus]

@sullen sable Has your question been resolved?

[marsh citrus]

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[tight crest]

This question is under the topic Rolle's theorem and Lagrange theorem.
Show that $x^5+5x=5$ has exactly one solution. Any help?

[elfin berry]

!Mike

[main idol]

mean value theorem?

set $f(x) = x^5 + 5x - 5$

[elfin berry]

riemann

[main idol]

and find an interval where the zero occurs

then apply MVT to the interval

[tight crest]

ig, my teacher calls it Lagrange

[devout mauve]

that takes care of at least one but not exactly one

[main idol]

oh balls

[tight crest]

all answers says is suggestion is consider the interval [0,1]

[main idol]

yea apply MVT to that

[tight crest]

I have another question. How do I know what interval to use?

I did the following, I don't see how this proves there is exactly one solution. Any help?

[main idol]

one way is to show the function doesn't have any turning points / critical values

or equivalently (i think), the function is either increasing/decreasing

[tight crest]

$5x^4+5=0, x^4=-1$

[elfin berry]

!Mike

[tight crest]

Therefore f`(x) has no real solution therefore no turning points therefore strictly increasing?

[marsh citrus]

@tight crest Has your question been resolved?

[tight crest]

.close

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[soft flare]

/help

[soft flare]

may I ask why the f of X the has a exponent which is negative -1 and I wanna know the reason to that

sutdying the basics of inverse function rn

[swift cypress]

it's just the notation for inverse

similarly a^(-1) is the inverse of a

so mathematicians just copied that notation

[soft flare]

wym by that me no understand 😦

monkey words please

like

[swift cypress]

???

[still temple]

It's the definition of inverse function

You can check the wiki page or some online course

[soft flare]

so u basically just write it when u get the FA when finding the inverse of the function??

okay nvm I get it know

mb HAHAHAHAH

so basically when finding the inverse the negative 1 there is the proper notation for the inverse function?

[still temple]

yes f the function f-1 the inverse function of f

[soft flare]

alr ty

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[split fulcrum]

Hey

[split fulcrum]

Can anyone help with contour integrals?

[main idol]

[split fulcrum]

i have found a = 3 and b = 2

and know i need to do the contour integral

and i have made it this far

And i feel like the first part is 3(2pii)

but i dont knoe

know

[main idol]

use residue theorem

[hushed egret]

[split fulcrum]

hmmm

[hushed egret]

cauchy's integral formula

[split fulcrum]

so i have this example from my book

which kinda look a like

[split fulcrum]

how do i apply this

[hushed egret]

oh gosh

the wikipedia entry on residue theorem

[hushed egret]

use cauchy's integral formula

or like

just the 1/z integral

[split fulcrum]

what is the last integral

[hushed egret]

+2

but by cauchy-goursat its 0

since its not in the interior domain of C

[split fulcrum]

where i get confused is, i know the result is 10 pi i

[hushed egret]

oh wait

no it is in the interior domain lol

you evaluate it in the exact same way

[split fulcrum]

so its 2(2 pi i) ?

[hushed egret]

yeah

[split fulcrum]

Thank you so much. Can i ask one more question before you leave?

it is a nearly identical problem

how do i see if it is in the interior domain?

so z+1 is not in the domain?

[marsh citrus]

@split fulcrum Has your question been resolved?

[marsh citrus]

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[thorn pebble]

Can someone help me with this

[marsh citrus]

@thorn pebble Has your question been resolved?

[marsh citrus]

@thorn pebble Has your question been resolved?

[nova crescent]

well the problem is saying cos t is the x value and sin t is the y value

so i guess they just want you to state the values on the unit circle that are on the bottom half

[thorn pebble]

yea

[thorn pebble]

This is an example of what the answer could look like

[nova crescent]

that would be like all the values that are from

sorry the point should be c(t) not t

i mean the answer is just the angle from pi to 2pi

because that's literally how the unit circle is defined

if you plug in values of t from pi to 2pi, you will end up drawing the lower half of the unit circle

if you plug in values of t from 0 to 2pi, you will end up drawing the entire unit circle

sorry i dont know how to explain it good

[thorn pebble]

but how would i figure out its those values

[nova crescent]

you know how angles work in the unit circle

you start at 0 at the extreme right

and the top is pi/2

the left is pi

the bottom is 3pi/2

[thorn pebble]

ok that makes sense

thank you

[nova crescent]

and the x values are cos and the y values are sin

[thorn pebble]

there was one more question can you help me with it

[nova crescent]

if the angle is defined as pi, which is 180 degrees, then the x value is cos(pi) and the y value is sin(pi)

cos(pi) is -1 and sin(pi) is 0, which you can infer from the chart since you just look at

[nova crescent]

do you still need help with it

[thorn pebble]

yes I do

[nova crescent]

i think you got the first question wrong btw

its not 11pi/4

wait

let me check again just to make sure

[thorn pebble]

yea its not 11pi/4

[nova crescent]

yeah i dont think its 11pi/4

[thorn pebble]

so what would it be

[nova crescent]

then you can just add 2pi to the answer because the period for sin and cos are 2pi, so after you add 2pi they just revert back to the same value

i got -sqrt2 by finding the normal coordinate from the polar form

the r and theta given in the problem are -2 and pi/4 so the x coordinate is rcos(theta) which means -2cos(pi/4)

same procedure for sin

and you find that both x and y coordinates happen to be -sqrt2

[thorn pebble]

ok

[marsh citrus]

@thorn pebble Has your question been resolved?

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[rapid geyser]

hey

[rapid geyser]

I was able to get the partials but am unsure how to show all the solutions

here is what I have so far

[marsh citrus]

@rapid geyser Has your question been resolved?

[rapid geyser]

<@&286206848099549185>

[thorny slate]

I thibk

Im Not sure i just started partials

But get the deritive in each plane

And set them = to 0

[rapid geyser]

didnt i already get all the derivatives

in respect to each axis

[rapid geyser]

?

[thorny slate]

Looks like you’ve already solved it? Lmao

[rapid geyser]

I got the derivatives just not all the CP's

cause Im not sure how to fully solve a 3 varaible equation

[thorny slate]

Yeah , we only did 2 variables is well

A critical point is when the deritive is 0 right?

[rapid geyser]

yes

[thorny slate]

And you’ve solved for x,y and z

[rapid geyser]

thats what i was trying to do

but it seems like there are more solutions

[thorny slate]

Do yk like how you have

X(y+1)=0

That also implies x=0 Is a solution

[rapid geyser]

yes i have that written down

and -1,-1,-5

but that one can become an infinite differnt version of itself

so how do i get all or a funcion of it?

[thorny slate]

WOLFRAM ALPHA TIMEEEEW

[rapid geyser]

i tried

did it work for you?

[thorny slate]

Gimme a sec

[rapid geyser]

ok

[thorny slate]

I’ve asked chatGPT

[rapid geyser]

lol same

but not always right

[thorny slate]

No sols

[rapid geyser]

what you get

and i will say what i got

[thorny slate]

Yeah I dont think it calculated the deritives correctly

[rapid geyser]

i had it solve the three equations but it said not possible

[rapid geyser]

which is wrong

yeah thats wronf

shoudl I just ping for someone else?

[thorny slate]

Yeh get help jaha

[rapid geyser]

lol

thx tho

<@&286206848099549185>

[proven timber]

People need to stop using chatbots for math

[rapid geyser]

real

[proven timber]

Can you post the images again so I don't gotta scroll up

[rapid geyser]

https://media.discordapp.net/attachments/907022703230345237/1050563700202946610/0E01ECCF-A7F8-4B80-A657-466267CAA5D6.jpg?width=1019&height=1359

[proven timber]

How do you get x(y + 1) = 0

[rapid geyser]

that was a mistake i see

but still would that give me a definite set of solutions?

[proven timber]

Idk I haven't worked it out

[marble glacier]

Should be y(x+1)

[rapid geyser]

yeah

but wont that still give inifnite solutions

anyways

[proven timber]

Work it out and see

[rapid geyser]

ok

well think about it

cause if you have an x and y

and you use them in equation

1

then you can just alter the z to fit the numbers

i will show

look

so do y(x+1)=0