help-19

[glad anchor]

your saying we shouldnt assume b < a

[nimble blaze]

no

stop putting words in my mouth

[glad anchor]

sorry? im just trying to understand

[naive agate]

b has to be less than a tho, no?

[nimble blaze]

yes

[naive agate]

So for example, if b was 5 and a was 9, it would be 6+7+8?

[nimble blaze]

yes

[naive agate]

Ok cool. That makes things easier. Cheers.

.close

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[buoyant solstice]

i'm trying to prove rigorously that the Gaussian integers are not a field. I know all I have to do is just identify one nonzero element that is not a unit, but is there a way to rigorously prove something is not a unit other than just saying "clearly no product with this element and another from the ring will give the multiplicative identity" ?

[buoyant solstice]

like I was just like "take the complex number 2 from the gaussian integers. clearly, no other gaussian integer can be multiplied by 2 such that their product is 1"

is this sufficient? or should i say that in the regular set of complex numbers that the "inverse" of 2 is 1/2, and since that inverse is unique to the set of complex numbers and the gaussian integers are a subset that no such unit exists in the gaussian integers?

[odd edge]

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[buoyant solstice]

.close

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[jaunty magnet]

I need to be able to assign numbers a value between 0 and 100 based on their position between two numbers.

In this situation I have a given min and max, for example 32.5 and 78.2. If my input number was 32.5, it should output 0 and if my input number was 78.2 it should output 100. Any value between these two should give a value indicating its position between the two. For example the halfway point should output 50. The term linear interpolation comes to mind but I might be completely off base, can't figure this out either way.

[dim thicket]

if your min is a and your max is b

you might want to assign 0 to a and 100 to b

so you would have two points (a,0) and (b,100)

do you know how to find the equation of a line between two points?

[jaunty magnet]

I don't. Ideally for my use case I need an equation with 3 places I can drop in min, max and my variable number and have the result be that 0-100 number I am after

[dim thicket]

do you need the explanation or just a formula?

@jaunty magnet

[jaunty magnet]

Just the formula, I wouldn't mind an explanation but I don't want to waste your time if unecessary

[dim thicket]

okok well the slope of a line is given by its rise over its run

[jaunty magnet]

The gradient?

I genuinely need like an equation where I can just plug those 3 numbers in and get the output 0-100

[dim thicket]

yes

for a linear interpolation (a line that is) the gradient is constant

and is given by rise over run

so in your case your equation is gonna look something like $y=\frac{100x}{b-a} + c$

[clever fjord]

LF

[dim thicket]

where x is your input and y is your output

and c is a constant that we need to figure out

when x=a we want y=0

so we can solve for c :

$y=\frac{100x}{b-a} + c \implies 0=\frac{100a}{b-a} + c \implies c = -\frac{100a}{b-a}$

[clever fjord]

LF

[dim thicket]

putting all this together you get $y = 100 \cdot \left( \frac{x-a}{b-a}\right)$

[clever fjord]

LF

[dim thicket]

@jaunty magnet

[jaunty magnet]

It's beautiful 😌

Thank you!

[jaunty magnet]

I'm curious though, with the latex noation. Why is it that if I put this into Google and I want the correct result, I need to wrap both subtraction operations separately e.g. 100 * (154.234 - 100) / (200 - 100)

Is that equivalent to what that notation actually means?

[odd edge]

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[dim thicket]

yes when you write a fraction a parenthese is implied on the numerator and the numerator

[dim thicket]

.close

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[keen hearth]

do you have the "given triangle"

yk that perpendicular point

give it a name

ok now call AX = x and XC = y

then see if you can do it

try plugging in tan alpha and tan gamma into your original equation

b tan alpha tan gamma / (tan alpha + tan gamma)

[keen hearth]

you can prove it equals h if you sub in tan alpha and tan gamma into this and get h

Its a "show that question"

so if you start with the RHS and get the LHS you win

Well done

Yep

you just proved LHS = RHS

I mean

you can "derive" it in the sense that

you just start with b (h/f) ...

and then end up with just h

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[mystic saffron]

I was wondering why the answer was 4 and 3. I thought I would be able to use arc length/radius = radiant for this.

[mystic saffron]

I found the radiant to be 5 pi/12

[fiery smelt]

75 degrees to radians in 5pi/12

[mystic saffron]

Yea

[fiery smelt]

the arc length formula that you're using states s = rθ

where s is the arc length and r is the radius and θ is your angle in radians

so basically

multiply your radians by the radius

[mystic saffron]

5pi/12 * 13?

[fiery smelt]

yea

[mystic saffron]

It ends up being 3

On the questions.

[fiery smelt]

5pi * 13 is 65pi

[mystic saffron]

Does the question have an incorrect response as the answer than?

[fiery smelt]

no ?

[mystic saffron]

Oh sorry. When I answer 65pi/12 originally on the quizizz, it marked it wrong.

I was wondering how it was four, but it seems like a error on the quiz part.

[fiery smelt]

well i mean technically A is right too

its an approximation of 65pi/12 as a decimal

B and D are not right though

[mystic saffron]

Yeah. Thank you for confirming it.

[fiery smelt]

no prob

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

Yes

[fiery smelt]

click the checkmark so the bot can close the channel and reopen for someone else

whenever you're done with your channel, type ".close" so the bot can do its work

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[vapid flint]

I need help on number six, im still a bit confused on this lesson so i dont quite understand the question

[vapid flint]

pls ping

[odd edge]

@vapid flint Has your question been resolved?

[vapid flint]

<@&286206848099549185>

[glass trench]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vapid flint]

1

or something idk how to do it

!1

[glass trench]

i mean

can it just be any function?

because then you just draw some random curve with those properties

[vapid flint]

idk man probably i just woke up from a hangover

[glass trench]

💀

if you don't knwo the questoin

i can't really help

[vapid flint]

ok

ill just close it first

.close

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[vast crag]

Can anyone help

[south plume]

woah

1 by 1

which one are you struggling with

[vast crag]

I understand most of the questions except these

Can we start with this

[south plume]

okay

triangles ABE and DBC are similar

do you get that?

[vast crag]

Yes

[south plume]

great

that means that all lengths within those triangles are proportional

[vast crag]

Ok

[south plume]

what's the ratio of AE to DC

you know their lengths

[vast crag]

Yes 20 and 10

[south plume]

also is this a test?

[vast crag]

No

[south plume]

okay

[south plume]

divide them

what do you get?

[vast crag]

2

[south plume]

great

that means that ABE is twice as big as DBC

or to put it differently, DBC is half the size of ABE

the height of ABE is h

[south plume]

what's the height in DBC, using this fact?

[vast crag]

18

[south plume]

no

[south plume]

in terms of h

[vast crag]

I’m confused now

[south plume]

if the height in ABE is h, and DBC is half the size of ABE, then what's the height in DBC?

don't use the 36

[vast crag]

2/h

[south plume]

almost

[glass trench]

so close

[south plume]

1/2 h

half of h

[south plume]

yes lol

[south plume]

so you get why it's half of h, right?

[vast crag]

Yes

[south plume]

great

[south plume]

now, 36 is the sum of those two heights

right?

[vast crag]

Yes

[south plume]

great

[south plume]

so you have h+0.5h=36

now, solve for h

[vast crag]

24

sorry i thought i entered the answer

Can anyone help with this

[south plume]

the longest "length" of the square is its diagonal

if you show that the diagonal is less than the diameter, then it fits

otherwise, it doesn't

[vast crag]

ok

Can anyone help with this

[odd edge]

@vast crag Has your question been resolved?

[glass trench]

alright

so

[tawdry sierra]

Find the length of CD

Ok

Who's answering this

[glass trench]

oh do you wanna take this

[tawdry sierra]

Uh sure

[glass trench]

ok

[tawdry sierra]

Alright

First you have to find the length of CD

To do this, construct BE

let me open paint

Do you understand how the green line is 7

[vast crag]

Yes

[tawdry sierra]

Do you know the pythagoras theorem?

You have to use the to calculate the red line(I'll send a pic)

[vast crag]

Yes

[clever fjord]

diaas_(yt)

[tawdry sierra]

which is also equal to CD as they are paralell

Now you have to use Pythagoras's theorem one last time to calculate $\sqrt{11^2+(\sqrt{207})^2}$

[clever fjord]

diaas_(yt)

[tawdry sierra]

Which will be your AC

[clever fjord]

diaas_(yt)

[vast crag]

Thank you

[tawdry sierra]

That should be your answer.

[vast crag]

Can you help with another question

[tawdry sierra]

Sure why not

[vast crag]

I got one mark for this

[tawdry sierra]

Ah man

Let's see

[vast crag]

Can you help with this instead

[tawdry sierra]

which

[tawdry sierra]

Yeah sure

So they've given AC AB BE and AD

First prove that the triangles are similar

They have 2 equal angles, A and the ABE ACD pair..

So they are similar

next use the property AB/AC = AE/AD=BE/CD

Work it one at a time, let me do it for you

AB/AC=8.2/12.3=BE/CD=3.8/CD

82/123=38/CD

[vast crag]

Shoot

I’m so sorry

[tawdry sierra]

what happened?

[vast crag]

I finished that question

[tawdry sierra]

Oh lol 😂

[vast crag]

It is this one

I sorry

I’m

[tawdry sierra]

Smart people, your teachers

if you assume ABE and ACD to be similar

10/x=15/3=5

x=2

That's one of the answers, give me a sec.

Idrk about the second one

It's a good question

Or maybe I'm dumb today

Ok I think I got it

Assume CDA and BEA to be similar

You'll get (x+10)/10=18/15=6/5

oh shoot, you get x=2 again...

My bad

yeah sorry i don't get it

[vast crag]

Thank you for the help

[late dust]

Could be that AEC and ABD are similar

Would be a pretty stupid question

[vast crag]

Can anyone help

I got this but it is only 2/3 marks

[odd edge]

@vast crag Has your question been resolved?

[vast crag]

Can anyone still help

[odd edge]

@vast crag Has your question been resolved?

[vast crag]

?

[novel finch]

1. THE DIAMETER IS 11CM NOT 9.899

2. YOU DONT REALLY NEED SQUIGGLY EQUAL SIGNS, √ DOESNT TAKE AWAY FROM THE ACCURACY OF THE MEASUREMENT

3. INCLUDE UNITS I GUESS;
   DIAGONAL = √(98CM^2) = (√98)CM

4. A BETTER ENDING EXPLANATION ?

THOSE ARE THE 4 WAYS I CAN SEE U LOSING MARKS

[vast crag]

I tried them it didn’t work

Can anyone think of another assumption other than abe and acid are similar x=2

[odd edge]

@vast crag Has your question been resolved?

[bronze fox]

Not drawn, but you could say triangles ABD and CBD are similar? Then you get a different x.

[bronze fox]

@vast crag

[vast crag]

It didn’t work

@bronze fox

[bronze fox]

What did you get for x?

[vast crag]

Or I don’t know what to put for x

I didn’t put one

[bronze fox]

Let’s find x. Are the triangles similar if CD and AD have the same length?

[vast crag]

Can anyone help

[grim pulsar]

ok im here

think about the triangle

there are two possibilities

the first possibility is: CA,DA is similar to BA,EA

and the second possibility, which is kinda tricky

is that DA is similar to BA, and CA is similar to AE

now here you go

so for the first possibility, which CA is similar to BA, and DA similar to EA, you easily find that x= CA-BA = 12-10 = 2

@vast crag

and for the second possibility, when DA is similar to BA, you can find that the proportion is 1.8

therefore x = CA-BA = 18-10-8

I hope this help

<@&286206848099549185>

[queen talon]

why did you ping helpers lol

isn't that the solution

[grim pulsar]

i dont know

we have to ask him

i mean @vast crag

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[vast crag]

Thank you I didn’t see the notification

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[oblique rapids]

SOMEONE HELP

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[cosmic schooner]

Let $d_1$ and $d_2$ be two topologically equivalent metrics on $X$. As we know, a sequence converges with respect to $d_1$ if and only if it converges with respect to $d_2$. But then it also holds that every Cauchy sequence with respect to $d_1$ is a Cauchy sequence with respect to $d_2$?

[clever fjord]

Levens

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[deft garden]

can someone help me

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@deft garden Has your question been resolved?

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[fringe sail]

why is theta in Q3?

i dont get it

cant it literally be in any quadrant cause it does a full revolution

[high mango]

it says in the problem that theta is between π and 3π

Now if cos theta is negative, it's in quarter 2 or 3 right

but π ≤ theta ≤ 3π means that theta is after quarter 2

so it has to be quarter 3

you have to learn what the signs of each trig function are in each quadrant, a negative cos has to be in quadrant 2 or 3

[odd edge]

@fringe sail Has your question been resolved?

[fringe sail]

but cant it loop around

back to

q2

?

cause its 3pi

[high mango]

You’re right it totally can

[fringe sail]

i think my teacher might have messed up

[high mango]

That’s a poorly posed question then

Yea haha

[fringe sail]

alr

.close

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[hollow plaza]

Let X1..Xn be a random sample from a population with CDF F(x) = x ^ theta , 0 < x < 1. find a sufficient statistic for theta

[hollow plaza]

<@&286206848099549185>

are you a helper

seems like ur tagged as one

can you help me with this one?

[hollow plaza]

this one

Let X1..Xn be a random sample from a population with CDF F(x) = x ^ theta , 0 < x < 1. find a sufficient statistic for theta

i replied to my earlier message

nice

.close

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[light zodiac]

This is what I got

Would the answer be 5???

It says round final answer which kinda scares me fr

But maybe if it's someone picks the wrong one fr

Someone please verify

[mint mirage]

It just says that, incase you get a decimal answer

If you get a whole number, you get a whole number and that's fine

[light zodiac]

I'm correct right

[mint mirage]

If you did the math right, sure

[light zodiac]

🙏

I'm scared😭

Yay

[light zodiac]

I think this one won't be too hard

(-4, 5.6)

[light zodiac]

@mint mirage can u verify this

NVM it was correct

It's A right <@&286206848099549185>

Oki ty

My ultimate enemy

I'm so lost

I don't get the ab=1/3bc part

[lapis willow]

let me draw this out

[light zodiac]

Okay ty

Can I rewrite that as a ratio

Idk how

Is the whole thing 4/3s

I'm not sure

I will afk for like 10 min (prob a little more)

[lapis willow]

So explanations are in, this method is a more visual-efficient way of understanding the question

@light zodiac

[light zodiac]

Back

Uhhh

[lapis willow]

U have the answer?

[light zodiac]

So AB is a third of BC right

[lapis willow]

Yes

Absolute Value wise yes

[light zodiac]

Idk how to do it😭

[lapis willow]

? Which part u do not understand?

[light zodiac]

Wait lemme take a stab at it I'll ping you when I'm ready fr

[lapis willow]

ok

[light zodiac]

Back

Yeah I have no clue what I'm doing

I get to as far as like the middle of your second page

But I get lost from there

[lapis willow]

ah i see

u start from the right then u look to the left

u know the formula to get the hypotenus of a triangle right?

[light zodiac]

Pythagorean theorem

[lapis willow]

u use that to get a figure of square root 80

[light zodiac]

And distance formula

[lapis willow]

which is the same as square root 5 times 16

that can later be simplified into 4times square root of 5

so u just find x as the hypotenus is 4x/3

then u substitue back to find AB

nvrmind i just do the explaination

since AB can form a smaller triangle inside of the triangle made from AC

the would have the same shape but different ratio

[light zodiac]

This is what I get 😭 😭 😭 😭

[lapis willow]

only works if it is on the tip

[light zodiac]

Oh wait

I messed up

Fr

[lapis willow]

this is different tho

[light zodiac]

Fr

[lapis willow]

u try doing it part by part

[light zodiac]

I cant😭

Idk what to do after getting 4 and 8

@lapis willow I'm lost😭

[lapis willow]

u know how to get 4x/3 right?

[light zodiac]

Is the answer (3, -5)

Idk😭

[light zodiac]

No

[lapis willow]

the answer is (1,2)

[light zodiac]

Oh wat

[light zodiac]

But that says bc is 1, 2 fr

Not B

[lapis willow]

sorry bad handwriting

is B(1,2)

[light zodiac]

But it looks way off

[lapis willow]

wait

i notice

i missing a step

yeah

my bad

[lapis willow]

should be 5,-3

[light zodiac]

Oh why

That doesn't look like one third of BC fr

[light zodiac]

It looks like it's off the line

<@&286206848099549185>

[lapis willow]

wait

what?

sorry then my bad

[light zodiac]

1-2 is not -3

(5,-1) looks right

@lapis willow wat ur thoughts

Or uh B I mean

But this looks like fourhts

Fourths

@onyx totem

Yo wrong person

My bad😭

I pinged the wrong guy😭

[lapis willow]

@light zodiac

[light zodiac]

HI

[lapis willow]

my bad wrong method

i got the answer

[light zodiac]

YAY

[lapis willow]

(16/3,-1/3)

[light zodiac]

Wat the

[lapis willow]

To find the position of point B on line AC such that AB is 1/3 BC, you can use the following method:

1. Find the coordinates of point B, denoted as (x, y).

2. Use the midpoint formula to find the coordinates of the midpoint M between points A and C:
   M(x_m, y_m) = ((x_A + x_C) / 2, (y_A + y_C) / 2)

3. Since AB is 1/3 BC, you can find the coordinates of point B by moving 1/3 of the distance from point A to the midpoint M:
   B(x, y) = (x_A + 1/3 * (x_m - x_A), y_A + 1/3 * (y_m - y_A))

Now, you can plug in the coordinates of points A and C to calculate the position of point B.

[light zodiac]

It looks like less than a third of BC

[lapis willow]

Do u have the answer

[light zodiac]

No

[light zodiac]

https://brainly.com/question/16564510

Yss

Yes

[lapis willow]

so asnwer correct?

[light zodiac]

5,-1 is correct

[lapis willow]

ok

guess i need to do more Questions now

oh

how the hell i went from 1-2=-3

LOL

my bad tho

[light zodiac]

That's what I said😭

[lapis willow]

glad that it helps

my bad

[light zodiac]

Yes

.close

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[lapis willow]

sorry been braindead from doing too much calculus

[light zodiac]

.reopen

[odd edge]

✅

[light zodiac]

RAAAA

[lapis willow]

nice

[light zodiac]

RAAAAHHHHHHH

[lapis willow]

good luck on the journey

[light zodiac]

https://tenor.com/view/rocky-gif-24362132

[light zodiac]

Ty

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[last mountain]

Help

[last mountain]

Forgot how to do thu

This

[fresh sphinx]

you substitute the number inside the bracket for x in the function

so for 6) it would be (-2)+7

[odd edge]

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[thin void]

hello

[thin void]

i’m confused how they got 3x^2

here is the previous work

[glass trench]

$\frac{dy}{dx}=(x^3+2)'(x^2-1)^4+((x^2-1)^4)'(x^3+2)=(3x^2)(x^2-1)^4+((x^2-1)^4)'(x^3+2)$

[clever fjord]

大野雄大 👻

[glass trench]

because $(x^3+2)'=3x^2$

[clever fjord]

大野雄大 👻

[maiden badger]

well said

[thin void]

oh right

okay one sec

[maiden badger]

basically subtracted one from the power to get 2

and multipled the 3 to x

[thin void]

okay wait

[maiden badger]

the 2 leaves because the dx of a constant is zero

product rule is f'g+g'f

[thin void]

so why did they decide to use the chain rule for this then? because it takes a lot or steps to solve this:?

they used the chain rule for that then went to product rule

[glass trench]

well for this one

its a composite function

so its much easier to use the chain rule than use the product rule 4 times

[maiden badger]

because higher powers above 2 its easier

[thin void]

ohhh

[maiden badger]

imagine foiling (x^2-1)(x^2-1)(x^2-1)

its like a unwritten calc rule

[thin void]

yeah

okay i got it

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[placid wagon]

help with part a please

[lost helm]

You have dA/dt, and A=pi×r^2. So you have d(pi×r^2)/dt. Can you simplify this in terms of dr/dt?

[placid wagon]

would it be dr/dt = 1/(pi x R)?

[lost helm]

Just check the calculation again

[placid wagon]

wait ignore the =-0.5 at the end

[lost helm]

Shouldn't d(pi×r^2)/dt be 2pi×rdr/dt?

[placid wagon]

u have to derive it when taking it out?

[lost helm]

Yes. Product rule

[placid wagon]

oh yh mb

[lost helm]

d(fg)/dt=fdg/dt+gdf/dt

So now you can work out dr/dt when r is 8 cm

[placid wagon]

is this right?

[lost helm]

The method looks right. You can check the calculations using a calculator.

[placid wagon]

ok thanks

is part B the same

?

like this ?

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[proper ivy]

I need help in terms on how to face this

[odd edge]

@proper ivy Has your question been resolved?

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@proper ivy Has your question been resolved?

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@proper ivy Has your question been resolved?

[queen talon]

Translation pls?

[proper ivy]

Given a sequence (an) that satisfies (an+1 -1/2an) tends to 0, prove that (an) tends to 0

[odd edge]

@proper ivy Has your question been resolved?

[clever fjord]

Tardis

[quasi sparrow]

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[mystic saffron]

The assignment says to convert that to interval notation, how would I do that?

wait actually nvm had a brain fart lol

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[cinder prism]

can you please check my si unit

[odd edge]

Please don't occupy multiple help channels.

[cinder prism]

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[vale zinc]

i got it wrong and idk what

this was the solution

idk where u get the 1/2n+1

[quiet berry]

you forgot to integrate

[vale zinc]

👍

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[hardy panther]

Use implicit differentiation

$\frac{d}{dx}\left(xy-2y = 1-x^3\right)$

[clever fjord]

Disorganized

[hardy panther]

Oh, there's a formula for that

You need this derivative though, the dy/dx

Haven't noticed what

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@gloomy tinsel Has your question been resolved?

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[pine drift]

Idk how to solve this. My attempt equations was

N + D = 44
0.05n + 0.10d = 3.10
And then I multiplied the second equation by -10 so the dime values cancel out and solve but I get -3.25 which is wrong….

[brittle beacon]

Can you show the intermediate steps, if you've done them, please?

[pine drift]

Like what I have done on paper?

[brittle beacon]

Yep, if you did anything on paper (if not don't worry!)

[pine drift]

This is what I did

[brittle beacon]

That should be -0.5n

[pine drift]

Ohhhh

OH YES

TYSM

[brittle beacon]

A pleasure that should hopefully sort it for you!

[pine drift]

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[opal osprey]

need help solving

[odd edge]

@opal osprey Has your question been resolved?

[mint mirage]

Do you know the quadratic formula and the general form for a quadratic

[opal osprey]

yes but how do I work backwards?

[mint mirage]

What's the quadratic formula?

[opal osprey]

x=-b+/- the sqaure root of b^2 - 4(a)(c) over 2(a)

[mint mirage]

Good now you can compare it with the problem

You have (-4 +/- sqrt(-124))/(-14)

Do you see how -14 relates to 2a?

[opal osprey]

yes a = -7

[mint mirage]

Can you do the same with b?

[opal osprey]

the +/ -4 sqrt(-124) is making me stuck

[mint mirage]

What -4sqrt(-124)?

[opal osprey]

yeah

[mint mirage]

I'm asking what because I don't see a -4sqrt(-124)

I see a -4 +/- sqrt(-124)

[opal osprey]

this part

[mint mirage]

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ as mentioned, that's the quadratic formula
You have $x = \frac{-4 \pm \sqrt{-124}}{-14}$ and as mentioned, the 2a is related with -14 so you found that a = -7, what does b relate to?

[clever fjord]

CaptainNova22

[mint mirage]

Do you see how the stuff under the root, the b^2 - 4ac is related to -124?

And how -b relates with -4?

You can compare terms like that

[opal osprey]

ok I see

i got it now

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[opal osprey]

thank you

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[astral kindle]

I have no idea how the Domain was found

[cold urchin]

What values of x are not allowed

[astral kindle]

I don’t know

[cold urchin]

Long day

Look at the denominator

[astral kindle]

I see -1

[cold urchin]

We know the denominator can’t be 0

well x^2 isn’t allowed to be 1

[astral kindle]

How you determine that?

[cold urchin]

the denominator isn’t allowed to be 0

x^2 - 1 = 0

When x^2 = 1

[astral kindle]

Oh I see

[astral kindle]

Exactly

[cold urchin]

So solve for x

And the domain is R \ {those values of x}

[astral kindle]

For the denominator X? It could be 1 and -1

[cold urchin]

Right

So the domain is R \ {-1,1}

[astral kindle]

I see

totally true

why is there a negative infinite to -1

[astral kindle]

the first parenthesis group

[vapid temple]

because that's where the domain starts

R is like (-infinity, +infinity)

[astral kindle]

i see

i understand it

thxs

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[nocturne basin]

I'm confused on where to continue from this point

[nocturne basin]

(calc 2 trig integration)

[stone cipher]

ah, that's not the ideal approach here

what you generally want to do is use like double-angle type formulae to reduce the powers

for example, sin(x)^2 cos(x)^2 = 0.25sin(2x)^2

so that reduces it from power 4 to power 2

and then if you reduce it to power 1 then you can just integrate it easily right

[nocturne basin]

oh I see
I've never done that before
I think I will search a youtube video on it
is there any way to continue from the point where I'm at, though? or am I at a dead end

[stone cipher]

uhhhh tbh it's easier to go from where you started

you could continue

[nocturne basin]

oh so it's very hard?

[stone cipher]

not really

it's just slightly more annoying

like essentially sin^4 is about as annoying to deal with as sin^2 cos^2

essentially there's two things in where you ended up and only one thing where you started

so the thing you started with is easier

[nocturne basin]

could you tell me where to go with the slightly more annoying route? unless it's like 10 more steps

I think it would deepen my understanding

because I get really confused when it comes to converting dx to du

[stone cipher]

wdym

the slightly more annoying route is just like, what i said but applied to sin^4

still need to turn sin^4 into like a power 1 thing

u-substitution isn't the way here

i can't easily justify it but yeah trust me lol

[nocturne basin]

ohhh nvm then lol that sounds out of my curriculum

[stone cipher]

it's not actually that hard

like ok

i basically turned it into sin(2x)^2 already didn't i

so call that sin(y)^2

[nocturne basin]

I'm not sure where you are going nor where you're coming from

[stone cipher]

then the point is that

cos(2y) = cos^2(y) - sin(y)^2 = (1 - sin^2(y)) - sin^2(y) = 1 - 2sin^2(y)

so sin^2(y) = 0.5(1 + cos(2y))

so sin^2(2x) = 0.5(1 + cos(4y))

so i already showed that what you started with was 0.25 sin^2(2x)

so like

it's actually 0.125(1 + cos(4x))

and now you can integrate that right

[nocturne basin]

yeah

[stone cipher]

it's just trig identities

[nocturne basin]

so I should have used the purple ones on the bottom

[stone cipher]

yes

and the first pink one

[nocturne basin]

oh wow
so much stuff to remember

[stone cipher]

yeah integration is a bit of an art

with a bunch of associated tricks

[nocturne basin]

is there a way to know when you're supposed to use certain methods or is it just something you subsciously figure out?

[stone cipher]

it's an art lol

you learn to recognise things

so it's like, ok

so for anything that looks like this with powers of trig functions

you can basically always reduce them like this

so that's one tell

but also if you have like sin^3(x) cos(x)

then that's clearly just u^3 du/dx where u = sin(x)

so you should do that instead, it's easier

the u-sub works in that case

it doesn't work in this case because like

the powers

even

but yeah there are certain indications that you should use some method or trick or whatever

varyingly

[nocturne basin]

thank you for helping me understand it better

salut 🫡

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[civic folio]

so the first one I used distribution and then I used FOIL

[astral kindle]

are you confused with e. or you don’t know how to do b.?

[civic folio]

E

[brittle beacon]

Looks like you mixed up trying to common denominator the and cross multiplying

[civic folio]

wait

like

don’t multiply it by x?

[brittle beacon]

Yea, you don't need to multiply by x and x + 2 as you did

Just common denominator inside the brackets to start with, and then afterwards, you basically have the product of two fractions

[civic folio]

Ok so

I have

,rotate

[brittle beacon]

Yep yep, looking good, then you can also simplify as the numerator and denominator have a common factor

[civic folio]

Math is fun

I may be struggling but I’m not giving up

[brittle beacon]

Alternatively, when you're multiplying them, look what you have in common

[civic folio]

Oh yes

GCF

[brittle beacon]

The most you can do is try, and seems like you're getting there

[civic folio]

ok

I got

Final

x(-x+2)

I have a clarifying question

Problem C

oops I forgot to put little 3

so if it’s 2^3+2^3 then

we take both of that out of that symbol thing

so then it becomes 2 times 2 times 3

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[upper veldt]

Can someone help me proof this by induction?

[upper veldt]

n≥2

[brittle beacon]

Have you shown the base case to be true? Have you given the induction step a go?

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@upper veldt Has your question been resolved?

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[tall cairn]

can anyone make out how we're getting R, Z, and whatever the fuck that other symbol is in the denominator?

[radiant merlin]

because you're dividing by the product of those three

since each term only concerns a particular variable, the ones that are not affected by the particular derivative of that term will get cancelled out, leaving you with terms that only depend on r, phi, and z

for example, the first term only involves rho derivatives, so phi and z functions will "pass through" it and will cancel with the ones in the denominator

the R in the numerator, on the other hand, still gets acted on by the derivative so it can't cancel the R in the denominator

[odd edge]

@tall cairn Has your question been resolved?

[tall cairn]

i actually get everything past this particular point, little confused as to where the dividing comes from

[quasi sparrow]

They just want to get back to the original phi

[tall cairn]

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[dire spoke]

Help on C

[mystic saffron]

what did u find the extrema to be

[dire spoke]

I found the dervs and second derv so far

[mystic saffron]

or have you not found them yet?

[dire spoke]

Idk what an extrema really is tbh

[mystic saffron]

just a maximum or a minimum

[dire spoke]

Oh

X=0

X=400

Those are the local max and mind

Mins

Or global idk the difference actually

[mystic saffron]

have you found any other critical points via the derivative

[dire spoke]

How do you know if its a global or local point

I found the second derv

And plugged in

Stuff

And ik which ones the max and min

But not global or local

[mystic saffron]

we'll get to that. Just calm your horses for a bit because we're jumping between stuff very fast

[dire spoke]

LOL sorry

[dire spoke]

Here ill show my work

[mystic saffron]

yes that's better

[dire spoke]

I dunno if its actually local

Im just guessing

[mystic saffron]

okay so you got pretty lucky in that you found the critical points to be your bounds anyways

[mystic saffron]

Extreme value theorem, have you studied it yet?

[dire spoke]

No

[mystic saffron]

ok

so

on a finite continuous interval you are guaranteed to have a global maximum and minimum

that's the gist of it

you are considering a finite continuous interval 0<= x <= 400

you have found two points f(0) and f(400) to be a minimum and a maximum

so by the extreme value theorem, they are....

[dire spoke]

Im lost

[mystic saffron]

you are guaranteed to have a global minimum and maximum in a finite continuous interval

you have found a maximum and a minimum

the theorem tells you you will have global extrema

so like, in conclusion what can you say?

[dire spoke]

Wait

So

I dunno

Do i have a finite continuous interval?

[mystic saffron]

yeah

you do

[dire spoke]

So

I do have a global

[oblique echo]

Mhmmm, so its a finite interval and if you determined its a maximum usinx extreme value theorem if [a,b] is continuous we ensure there exists an absolute and local maximum, as well as an absolute local minimum.

[dire spoke]

Im confusion

Mhmm

So

How do ik if 400

Is local

[oblique echo]

Well I mean plug it into the function and rank them in order

[dire spoke]

I plugged in 400

And got 1600

[oblique echo]

First calculate the derivative and determine the critical points.

once u have plug those points in and find those values and contrast it to the values of ur endpoints.

[dire spoke]

I plugged in 400

Got 1600

And 0

Got 0

[oblique echo]

Mk find the critical points and do the same thing

[dire spoke]

Aren’t the crit points

400

And 0

?

[oblique echo]

Critical points are where f' = 0 or DNE. But A maximum obviously has to a finite point and a value so DNE depending on the scenario if its a cusp then maybe, but if its a vertical asymptote the values would be infinite as we get close to that vertical asymptote.

Aren't you on Application of Derivatives?

[dire spoke]

Ummm

I think so

Yea so

The crit points

Are 400 and 0

Because thats when f’ is 0

[oblique echo]

Alright.

Then x = 400 is where ur absolute max is and 0 is where absolute min is.

For the EVT by the way you always want to evaluate the endpoints, but a common mistake is people just evaluate the endpoints and then rank them. No, don't do that. You always want to evaluate the critical points as well and then plug in and find all the values.

[dire spoke]

This one has me lost

I dunno the first step

D/dx

[oblique echo]

Ima do the question and show you.

[dire spoke]

Okayy ty

I got to like

(dy/dx)x + dy/dx= dy/dx(90x)

But thats me totally guessing

[oblique echo]

evaluate that

[dire spoke]

You lost me at when

You

Lined in f and g

[oblique echo]

what confuses you?

[dire spoke]

How did it all become f’g

[oblique echo]

Well if we are differentiating a product that is the formula to differentiate a product.

[dire spoke]

So is this completely wrong

[oblique echo]

Yes.

[dire spoke]

💔

[oblique echo]

Do you understand Leibniz's notation I don't think you do based on how you stated that.

[dire spoke]

Nop

I was trying

To go off my teachers example

[oblique echo]

You shouldn't be doing problems if you don't understand the basic things with all due respect. You should fill in the gaps of knowledge and then move onwards.

[dire spoke]

Thats my teachers example

[oblique echo]

Alr ima solve it too.

Yk I can help you, but I gotta go soon I got stuff to do.

[dire spoke]

Okk

[dire spoke]

Wdym solve it too

Its solved

[oblique echo]

To ensure she did it correctly, because I need to understand why you are misinterpreting it.

[dire spoke]

Mhmm

So

Did u see

What I misinterpreted

[oblique echo]

dy/dx)x + dy/dx= dy/dx(90x)

Yeah its evident.

So in implicit differentiation, you cannot actually just solve for dy/dx like that. Although you may believe that initially. Its called "implicit" beacuse its implied we find dy/dx after differentiating both sides. Picturize it as confusing differentiation.

What your first mistake was you should have done

d/dx (xy+y) = d/dx (90x)

d/dx (xy) + d/dx (y) = d/dx (90x)

[dire spoke]

Ohhhh i did do that

But i forgot

To

[oblique echo]

Uhhhhhhh I don't think so.

Because you did dy/dx90x that is impossible

dy/dx is not an operator.

You are confusing the meaning of the notation.

[dire spoke]

Oh yea

Idk why i put the y

[oblique echo]

dy/dx = SOMETHING
(The derivative) = SOMETHING
d/dx (derivative operator)

d/dx is a derivative operator you have to differentiate something

think of it like a machine

sin(x) for example

sin(x) by itself doesn't mean anything unless we plug a value in for x

[dire spoke]

Mhm yea I understand idk why i did that lol

[oblique echo]

it will then return an output

Don't excuse your mistakes, because that is quite a silly mistake. You should understand this notation crystal clear at this point if ur on implicit differentiation.

No needa feel embarassed or upset abt it, but if u don't understand the true meaning which I believe is the case ur digging urself in a deeper hole.

Alright Jessica, I can assist u for 15 more mins but after that I am leaving sorry got things to do.

[dire spoke]

No yea I get it

[oblique echo]

Ur sure?

[dire spoke]

If it was 90yx tho

I could do dy/dx 90x?

Oh

Then maybe i dont

🪦

[oblique echo]

No worries.

So your confused on the notation.

Lets address that.

So note dy/dx = something. In written terms we would say The derivative equals something.

Now d/dx is completely different than dy/dx in the sense that we cannot put an equals sign.

Because d/dx is a derivative operator

it does something, which is it differentiates a function/polynomial.

Thing of d/dx as a machine

we put something in the machine (OUR INPUT)

the machine does something to our input and returns an output