help-38

[burnt mulch]

So you want to get from f'(x/2) to f'(x) somehow

And this is the most natural way to do it

[vocal cobalt]

i got the new expression 2 * int 0 to 10 of sqrt (1+(f'(u))^2) du

what do i do now?

[burnt mulch]

b/c atm, you only have info about the arc length of f, and hence about the integral of sqrt(1+(f'(x))^2)

What do you have about arc length in the question rn?

[vocal cobalt]

the final answer should be of numbers

[vocal cobalt]

how do i get to it?

from the expression i got

[burnt mulch]

????

I'm asking you to read the question

and tell me what it tells you about the arc length of the graph of f

[vocal cobalt]

0 to 5 is 11 and 5 to 10 is 18

[burnt mulch]

mhm

[vocal cobalt]

so i just do that?

[burnt mulch]

what exactly are you doing with those numbers

[vocal cobalt]

since its like in the arc length formula style i just get the arc length thats given? but i dont get what length im getting with 0

[burnt mulch]

You want to find $2 \int^{10}_{0} \sqrt{1+(f'(x))^2} \dd{x}$.

[solid kiln]

Civil Service Pigeon

[burnt mulch]

Currently, the arc length information they gave you is in words.

[burnt mulch]

Since you want to find an integral in the end, you may find it useful to convert this into some integrals

this way you're saying "ok I have these integrals, and I need to find this integral"

[vocal cobalt]

int 0 to 5 sqrt (1+(f'(x))^2) dx? like this?

[burnt mulch]

sure

[vocal cobalt]

int 5 to 10 sqrt (1+(f'(x))^2) dx

how is this gonna give me answer?

i combine it?

[burnt mulch]

try it out

[vocal cobalt]

int 0 to 5 sqrt (1+(f'(x))^2) dx + int 5 to 10 sqrt (1+(f'(x))^2) dx = int 0 to 10 of sqrt (1+(f'(x))^2) dx

and i multiply 2?

[burnt mulch]

yes

[vocal cobalt]

ok got it

thanks for helping 😀

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[tender flame]

should i rearrange for x because its revolved about y-axis?

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[simple panther]

!status

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[odd tapir]

is this wrong?

because don't they both approach 0

[split chasm]

yeh, its wrong

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[drifting reef]

me

[drifting reef]

23x3

how do you do 23x3

[solid sundial]

use a calc

or so 20x3 + 3x3

[drifting reef]

ok

[mortal kestrel]

ts some uni lvl shii😭💔

[worldly wing]

too hard

[mortal kestrel]

Frfr

[stuck plover]

Use column addition

[drifting reef]

ok

[stuck plover]

2 3
2 3
---
? ?
---```

[drifting reef]

😮

i see

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[silent kiln]

Would it be incorrect to say $n^3$ because there could be $f(n) \geq n^3$?

[solid kiln]

allarkvarkk

[queen beacon]

What is the question

[silent kiln]

It's asking $\Theta(???)$

[solid kiln]

allarkvarkk

[queen beacon]

Well, what does the master theorem give you

What’s the statement and does it apply, first off

[silent kiln]

wait i got it nvm

ty for the help

:)

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[queen beacon]

So real

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[verbal linden]

i'm trying to calculate the probability of a portfolio value within a certain Z-score range at a certain time t, or alternatively calculate a probability density function at time t. here is an example scenario:

say i have $100 principal to invest in a portfolio. the portfolio's expected annual return is 8% with a standard deviation of 15%. i will be withdrawing $7 from the portfolio every year.

Now, for every year, i would like to know the PDF of the total portfolio value at that time. or for simplicity, i would like to know the values at Z=1 and Z=-1.

[verbal linden]

finding the mean (Z=0) at any time, is simple even with the fixed withdrawal rate. but finding any other value for Z becomes non-trivial, because of the fixed withdrawal rate (not being a percentage of the total)..

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[verbal linden]

.reopen

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✅

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[verbal linden]

thought i should mention, i am trying to get a closed-form mathematic solution, not monte-carlo simulation.

[zinc raptor]

finding the values at Z=1 and Z=-1 is equivalent to finding the variance

you should be able to set up a recurrence for the variance in the same way you solved for the mean

i can't guarantee it will result in anything nice

your random variable after n years is a sum of products of up to n independent gaussians, which i don't think has a nice pdf

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[verbal linden]

thanks, trying to understand this bit..

[verbal linden]

hmm, the way i solved for the mean is like this:

P = principal (100)
W = fixed annual withdrawals (7)
R = exopected rate of return (0.08)

adding volatility into the first term (the principal amount) is easy:

Z = Z-score
V = volatility (0.15)

but i still need to account for volaility in the second term (the withdrawals).

because withdrawals don't simply apply in a deterministic way like that. withdrawals continuously reduce the size of the effective principal amount that would otherwise be subject to volatility..

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@verbal linden Has your question been resolved?

[verbal linden]

when i started trying to figure this out, i thought it would be a simple problem, but it might in fact be impossible to do..?

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@verbal linden Has your question been resolved?

[simple haven]

It's actually pretty difficult because you're withdrawing a fixed value, rather than a percentage.

[verbal linden]

if i was withdrawing a percentage. then it could just simply be factored in at the same place as the expected return (R) i'm assuming?

[simple haven]

Hmm... Perhaps this counts as a semimartingale

Yeah, this counts pretty sure

[verbal linden]

if withdrawals are fixed nominal values, then it's a random walk with a positive drift (in percentage, so compounded) combined with a linear negative amount over time.

if withdrawals are percentages, then it's a random walk with a positive or negative drift (depending on the value after R - W), so yes much simpler.

[verbal linden]

i will look this up

[simple haven]

https://en.wikipedia.org/wiki/Semimartingale

[verbal linden]

ngl, most of that went over my head. i'm kind of a math pleb.

[simple haven]

It's ok, this is not math I'm super familiar with either.

[verbal linden]

but seems like they say brownian motion is a semimartingale. and i think this process is brownian motion; volatility in finance can be treated like that.

still unsure how exactly to combine a fixed linear rate of depletion to an otherwise typical brownian motion.

[simple haven]

That's part of the càdlàg process

[verbal linden]

hmm, does cadlag work for something that is continuous? or only in the case of withdrawals being discrete events happening once per year. maybe it wasn't clear from my explanation attempt, but the withdrawal rate is $7/year but it happens continuously, if that makes sense.

$0.01917808219 /day for example

the timeframe doesn't matter, just that the total withdrawn is a linear thing over time.

the rate of withdrawal is independent from the portfolio value, but it is still continuous.

[simple haven]

It does work

Sorry, I stepped away for a moment

I thought you were modeling a discrete thing, but continuous makes it substantially easier actually

But yeah, changing it to continuous still is càdlàg

[verbal linden]

ok, thanks. well i have some reading to do on cadlag!

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[onyx forge]

close the channel

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[round minnow]

how to simplify this?

I've learned basic derivative rules and the hardest part is simplifying by far

[vapid grove]

distribute things first?

rewrite roots as exponents if you need to

[round minnow]

alr bet lemme try

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[crimson halo]

Can someone help me idk know what its asking me to do

[crimson halo]

I quite litteraly dont understand what it expects me to do

[dapper swift]

the parent function is y = x^2

[empty orchid]

Try to expand $-\qty(\frac14x)^2$ first.

[solid kiln]

;(

[dapper swift]

yeah you save a step by doing that

[crimson halo]

Whats “parent fubction” even mean?

[empty orchid]

The function that was “originally there”.

i.e., the most basic form of the transformed function.

[crimson halo]

Its this vocabulary its stupid to me

Whats it mean by transforming the given the graph of the parent function

[empty orchid]

So if I were to, say, consider $g(x)=\frac12(x-5)^2+4$, and $f(x)=x^2$, then $\frac12f

[solid kiln]

;(
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crimson halo]

I dont get ot whats it mean by transforming

[crimson halo]

You there?

Wait so all its asking is what transforms the given line to be the line that the equation goves?

[dapper swift]

you should say function or curve

but yes

so it's like a little game: each step, you replace or add something to y = x^2 to get it closer

the trick is you only need two steps

the first step is $y = x^2 \mapsto y = -\frac{1}{16} x^2$

[solid kiln]

south

[crimson halo]

Im fed up dude my math teacher assigns us work over spring break

[dapper swift]

I'm not counting the minus as an extra step btw, should be pretty obvious what that does

[dapper swift]

oh no, it's spring break for you!!

I thought you already had yours lmao

fuck I have spring break only after next week, oh well

[crimson halo]

Nope ive had it with math every day i do it toll 10 pm and now i have to do it over the one fucking week i have to myself

[dapper swift]

noooooooooooooo

[dapper swift]

why is Easter so late this year

[crimson halo]

Idk

[dapper swift]

it's to do with the full moon actually

so that means it's lunacy (ba-dum-tss)

[crimson halo]

Ive had it bro two years ahead of my whole school and i have to do those

[dapper swift]

wait what grade are you at rn?

[crimson halo]

I cant say bruh

Wayy to many people on this server

Old enough tho dont worry

I need help on this i needa graph these

The second one doesnt make sense

Is the negative 1 applied to x or is it just negative 1

[crimson halo]

You there?

[dapper swift]

it's just y = -1

so that would be just the point (x, y) = (-2, -1)

[crimson halo]

How so

[dapper swift]

cause the piecewise domain there is only one x value

x = -2

usually it's a horizontal line, correct

[crimson halo]

I thought the left side is the equation and the right side tells you the limits

[dapper swift]

yes, so y = -1 for x = -2

that's a single point

[crimson halo]

Ohhhh and the right side can tell you snything?

[dapper swift]

the right side tells you the domain yes

so if the domain is an inequality, which it usually is

you'd sketch that function between those x values

[dapper swift]

so this is a special case

[crimson halo]

But this osnt an inequality so what is ot?

[dapper swift]

a point!

you have 1 value of x and 1 value of y

so a filled-in circle on your graph

[crimson halo]

But if its x=-2 how can it be -2,-1 if the -1 is the y intercept

[dapper swift]

a point doesn't have a y-intercept...

you're thinking about it as a line which you shouldn't be

[crimson halo]

Im confused i thought the left side was the equation

[dapper swift]

yes

[crimson halo]

Ye so if its equation the only thing there is the y intercept

@dapper swift

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@crimson halo Has your question been resolved?

[dapper swift]

.。。。。。。

idk what to say anymore

[crimson halo]

bruh pls i dont understand if the -1 is the y intercept and x is -2 how can it be -2,-1 if the negative one is the y intercept?

[dapper swift]

x = -2 and y = -1

[crimson halo]

but i thought the left side is an equation and the reg number is the y intercept?

[dapper swift]

it was for y = -1 and x = -2

not anymore

[crimson halo]

so in that its not an eqution?

[dapper swift]

yeah it's a point

[crimson halo]

ohhhh i understand it now

so f(x) just think of it as y

so y=-1

[dapper swift]

yes!

[crimson halo]

I only got that cus i asked gauth😭

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[crimson halo]

thx tho

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[narrow jewel]

hello im trying to prove that this is true
i took a function h(x) = f(x)-g(x) (continuous) and showed that its integral is 0.
Im planning on showing h(x_1)>0 for some x_1 and h(x_2)<0 for some x_2 and using the IVT to get x_0
but im kind of confused on how to get there

[narrow jewel]

the idea is to assume for contradiction that for all x \in (a,b) h(x) \neq 0

BWOC let $\forall x\in (a,b), h(x)\neq 0$

[solid kiln]

kirsch8

[narrow jewel]

then Case 1: $\forall x\in (a,b), h(x)>0$

[solid kiln]

kirsch8

[narrow jewel]

$\int_a^b h > 0$ which is a contradiction. Same for case 2 where $h(x)<0$

[solid kiln]

kirsch8

[narrow jewel]

maybe closed interval instead of open

so there exists $x_1,x_2 \in [a,b]$ s.t $h(x_1)>0, h(x_2)<0$

[solid kiln]

kirsch8

[narrow jewel]

then use IVT

[simple haven]

We know that there is some average value for both these integrals

[narrow jewel]

we havent learnt that yet

[simple haven]

This value is ∫f(x) dx/(b-a)

[narrow jewel]

i mean yeah we can say that

i guess?

[simple haven]

It is possible to prove (using continuity) that f(x) must equal this value (call it x0) somewhere on the interval [a,b]

Similar argument for g

These are the same value

[simple haven]

This might be a direct consequence of the IVT, or it might need some massaging. I don't really recall it off the top of my head

Hth

[narrow jewel]

hmm. is the way i wrote above wrong?

[simple haven]

No, it's not wrong, now that I read it more closely

Sorry, actually I rather like your approach better than mine now that I understand it.

[narrow jewel]

i was wondering if id need to add more detail to it, or if it isnt super clear

[simple haven]

Seems ok to me?

You demonstrate that h(x) > 0 for some x, and h(y) < 0 for some y, then use the IVT to show that h must equal 0 at some point between x and y. This point is where f and g are equal

[narrow jewel]

:)

thank you so much

.solved

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[chilly crypt]

ifff theres two points that i need to plot in a scatter plot how do i label it? Liek for exambl i have two ponits that are [1, 96]

[chilly crypt]

can i label it in some way that indicates theres two of it?

[trim lichen]

!xy

[trim jolt]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[trim lichen]

(why would you be making a scatterplot with only 2 data pts)

[simple haven]

I think the idea is that the scatterplot had many different points, but one of the points was repeated once.

@chilly crypt what plotting software are you using?

If you have many such points then you might want a different plot style than simple scatterplot

Especially if the points have an associated uncertainty

For instance, if you have a lot of logically overlapping data, points with a size and transparency might be useful: https://matplotlib.org/stable/gallery/lines_bars_and_markers/scatter_with_legend.html#sphx-glr-gallery-lines-bars-and-markers-scatter-with-legend-py

https://matplotlib.org/stable/_images/sphx_glr_scatter_with_legend_001_2_00x.png

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[chilly crypt]

Hello helooo sorry i had to do something hold onn

context: We already had a scatterplot made of jamovi but for our final performance task we had to draw it digitally or on paperrrr

I have two repeating points bcs my bivariate data is study hours and exam scoresss, two students had 1 hour study time and got 96 for their final grade

so im drawing mine digitally on ibispainttt

[chilly crypt]

hereheree

[trim lichen]

oh like

oh you mean like two duplicate data points in a bigger set

hm...

[chilly crypt]

Yesyesss i also have 5 more thats repeated thats why i only have 25 points in my scatterplot hehe

[trim lichen]

https://stackoverflow.com/questions/21081453/drawing-scatter-graph-using-matlibplot-pyplot-when-points-are-duplicate

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[chilly crypt]

.reopen

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✅

[trim lichen]

you should have clicked ❌ on this one

[chilly crypt]

thank you i will open thiss

[trim lichen]

so it wouldn't slam the door in our faces

[chilly crypt]

OH I SEE

[chilly crypt]

ahahahah 😭

[trim lichen]

anyway it looks like there's no universal solution, but one method is to draw the points with some transparency so that duplicates appear darker

[chilly crypt]

OHH THANK YOUU!!!

thank you so muchh, i will close the channel now if its alright hehe

[trim lichen]

sure

[chilly crypt]

.close

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[lone apex]

Guys is my answer & working out correct? I'm trying ai but I'm getting confusing Answers.

The working out and answer is mine original.

But after I did it

I confirmed with ai and its telling me all sorts of things.

Is my answer & working out correct?

Is my answer & working out correct

Or mostly correct?

[split chasm]

no

[lone apex]

Am I mostly correct?

[split chasm]

no

after perimeter of circle being pi * D

[lone apex]

How many marks I get out of 4?

[split chasm]

nothing else after that made sense

[lone apex]

How many marks I will get out of 4?

[split chasm]

1

[lone apex]

3

[split chasm]

maybe 0.5

[lone apex]

!

3 yes?

[split chasm]

no

[lone apex]

2?

OK

So how do I get the correct answer?

[split chasm]

apply what's actually stated on your formula

[lone apex]

It's a semicircle

The formula is for a circle

[split chasm]

you have 2 semi circle arcs which would together form a full circle
so for simplicity, you can use the circumference of a full circle = pi * D

you pi looks like an R, but regardless not what you want

you seem to be applying the circumference for a full semi-circle,
however that straight part doesn't contribute to the cirfumference of this shape

[trim lichen]

this kind of paper is designed to award only integer numbers of marks

i think

[lone apex]

Yes

Will I get 2 marks?

[split chasm]

1

and you also seem to be ignoring the straight parts of the final figure

[lone apex]

@trim lichen

Is it 2 marks?

[split chasm]

and not converting units correctly

and having cm pop up somewhere

[trim lichen]

i'd have to see the mark scheme to tell you with certainty how many marks you would earn.

[split chasm]

having pi * D present would at most get 1 mark

[lone apex]

In ur opinion is it 2?

[split chasm]

but to me nothing else made sense and imo wouldn't get any more marks than that

[trim lichen]

i have no opinion and will not comment on this any more.

i noticed this issue looking through your work, by the way.

regardless of how everything else is (which i did not look through in full), this conversion is incorrect.

it's 1 m = 3.3 ft but you treated it as if it were 3.3 m = 1 ft, which is wrong.

[split chasm]

the way i'd allocate 4 marks would be:
lengths of circular arcs (1 mark)
total perimeter of figure (1 mark)
conversion from m to ft (1 mark)
conclusion (1 mark), if properly justified

[lone apex]

The conclusion is correct

She does have enough

Conclusion is correct, she does have enough

[split chasm]

based on your flawed calculations it would appear that way, but no

[lone apex]

Based on the work you showed, your final answer is correct, which is key. However, the working is a bit unclear and contains some unnecessary steps and unit confusion. In an exam setting—where clear, logical working is expected—you might lose a mark for clarity. So overall, you’d likely score around 3 out of 4 marks.

[trim lichen]

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[lone apex]

You correctly:

Recognized that the shape's perimeter includes the two long sides of the rectangle plus the circumference of a full circle (from the two semicircles).

Calculated the straight sections as 2.8 m + 2.8 m = 5.6 m.

Used the circle's circumference formula (π × diameter) correctly, obtaining approximately 6.28 m for the circle.

Added these to find the total perimeter as about 11.88 m.

Converted 40 feet to meters using the conversion factor (40 ÷ 3.3 ≈ 12.12 m).

Compared the two values accurately to conclude that she has enough garland.

These are the main parts you got right in your solution.

See ?

@trim lichen

Its 3 marks

[trim lichen]

ok look

if you are going to run to your beloved GPT no matter what we say

then it will not matter what we have to say

because you will just slap us with more GPT output

plus GPT actually gets a lot of shit wrong ESPECIALLY from images full of text.

[split chasm]

if you don't trust me that it'll net you 1 marks, perhaps even 0 depending on how strict the teacher is,
and trust gpt,
then by all means continue what you're doing.

first four statements are wrong

[lone apex]

@Ramo

@split chasm its telling me

My work is worth

3/4 marks

" Based on a close look at your working and how marks are typically awarded in Functional Skills Level 2 Maths, here's a detailed breakdown of how many marks you'd likely get:

Total marks available: 4

The marks are usually awarded as follows:

MarkCriteria1 markIdentifies the need to calculate the perimeter using 2 straight sides + 2 semi-circles1 markCorrect use of formula for circumference (π × diameter or π × radius × 2)1 markConverts between feet and metres correctly using 1 m = 3.3 ft1 markCompares the required and available lengths correctly and gives the correct conclusion

Your likely score: 3 out of 4

Mark 1 — Perimeter structure: ✓

You recognised that the shape consists of a rectangle and two semi-circles, and you're trying to work with the perimeter — this earns 1 mark.

Mark 2 — Circumference formula: ✓

You referred to π × D for the circle — even if your working was messy, it’s clear you knew the formula — 1 mark.

Mark 3 — Conversion between feet and metres: ✓/✗ (partial)

You attempted to use the 1 m = 3.3 ft conversion, but your use of it was confused (you divided in the wrong direction: “8.8 ÷ 3.3 = 9.6 ft” is backward).
This could be seen as a partial method, but most strict marking schemes would not award this mark — 0 marks here.

Mark 4 — Final conclusion: ✓

You said: “Yes, she has enough.” That’s the correct final conclusion, so 1 mark.

Final score: 3 / 4 marks

With a cleaner explanation of the conversion and fewer unclear steps, you'd easily get full marks.

Would you like me to show you a model 4-mark version of the answer?

"

[willow urchin]

@lone apex continuing to waste helpers time with gpt will earn u a swift ban

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@lone apex Has your question been resolved?

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[vocal cobalt]

how do I know if theres gonna be a sign change at x=0 or not? without calculator (desmos)

[dull pilot]

of the last thing in blue?

[vocal cobalt]

yea basically when does that thing gonna have a sign change

[dull pilot]

do yk wavy curvy

[vocal cobalt]

whats that

[trim lichen]

do you actually do your work by scribbling directly on your question paper like that?

and nowhere else?

[vocal cobalt]

yea

[trim lichen]

😭

you're torturing yourself mate

get yourself a file in your notes app or whatever. copy the question. and then do your work properly and properly notated

cause like

if it's gonna get even a smidge more complicated than "compute this one integral"

you're gonna have a bitch of a time w/ it

and when all you write is this

you make it impossible for yourself to keep track of what is what, you know?

[lunar stirrup]

wavy curve he meant

[trim lichen]

so like the thing is

[lunar stirrup]

so plot the critical points for where the terms are = 0

[trim lichen]

this work should have been written out as something like this:

$f'(x) = 15x^4 - 20x^3 \ f''(x) = 60x^3 - 60x^2 \ f''(x) = 60x^2(x-1)$

[lunar stirrup]

ykw watch a video on it its too hard to explain over text

[solid kiln]

Ann

[trim lichen]

at the bare minimum

so you do not accidentally use the first derivative where you were supposed to use the second, and so on...

[vocal cobalt]

ok but how to solve it

[trim lichen]

you know that a point of inflection for y=f(x) is a point where f''(x) changes sign, yes?

[vocal cobalt]

yea

[trim lichen]

yeah so

ok

slightly sidestepped but related question, have you solved quadratic inequalities or perhaps polynomial or rational inequalities before, in general?

[vocal cobalt]

? if its part of precalc or algebra 2 then i might have done it

[trim lichen]

it should be a precalc thing

i mean like

ok, lemme just give you an example question of the kind im asking about

$\frac{x(x-7)}{(x-2)(x-4)(x+5)} < 0$

[solid kiln]

Ann

[trim lichen]

would you be able to solve an inequality like this?

[vocal cobalt]

i might immediately crash out

[trim lichen]

💀

[lunar stirrup]

this is exactly the sort of thing u use wavy curve for

[trim lichen]

yeah i was hoping to introduce either wavy curve or sign tables

a sign table would be more convenient for our usecase

[lunar stirrup]

ohk but wavy curve is more fun

[trim lichen]

ok how about something simpler then

$\frac{(x+2)(x-7)}{x-12} > 0$

[solid kiln]

Ann

[trim lichen]

what about this

does this still crash you @vocal cobalt

[vocal cobalt]

i might be able to give a try

i just pick a number in the interval and plug and chug right?

[trim lichen]

mmmm nope.

i would like to avoid plug and chug, and also i don't want to encourage it at all.

do the words "wavy curve" or "sign table" ring any bells at all for you

[vocal cobalt]

no 😭

sorry

[trim jolt]

@vocal cobalt Has your question been resolved?

[trim jolt]

@vocal cobalt Has your question been resolved?

[jolly saffron]

casework will be insane with this stuff

so you should look up the wavy curve method

[vocal cobalt]

ok lol

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[dull pilot]

real quick before i ask a less obvious question $\lim_{x \to 2} \frac{\sqrt{1-\cos(2(x-2))}}{x-2}$ is not defined simply because $\sqrt{\sin^2(x-2)} \neq \sin{x-2}$, right?

[solid kiln]

Percy

[jolly saffron]

yes

[dull pilot]

i.e. the left-right handed limits are defined, but this isn't because one would be $\sqrt2$ and the other would be $-\sqrt2$

[solid kiln]

Percy

[dull pilot]

yeah okay i figured

thanks

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[dull pilot]

whoops forgot parantheses lol

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[unkempt oriole]

No

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[bright seal]

$|\frac{3-4i}{4+3i}|^2$

[solid kiln]

Task Bot

[bright seal]

This exercise comes to me -1, but wolfram says does 1

so wolfram was wrong

[bright quarry]

,w |\frac{3-4i}{4+3i}|^2

[solid kiln]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by kniefy

[bright seal]

I think it's wrong

[soft pendant]

why not you ?

[bright quarry]

cool

[bright seal]

$| -i |^2 = i^2 =-1$

[solid kiln]

Task Bot

[bright seal]

But on Photomath it does 1

So strange

[simple panther]

Dude, its the magnitude squared

[soft pendant]

you are doing |a|^2 = a^2

[bright quarry]

[simple panther]

$|-i|^2=1^2=1$

[solid kiln]

Suika

[bright seal]

🤦

Can you explain?

[simple panther]

What website is that?

[soft pendant]

its doing $|a|^2 = a^2$

[solid kiln]

jandro0103

[soft pendant]

$|-i|^2 = i^2 = -1$

[solid kiln]

jandro0103

[soft pendant]

but you have to do

[simple panther]

$|a+bi|=\sqrt{a^2+b^2}$

[solid kiln]

Suika

[soft pendant]

$|-i|^2 = \sqrt{(-1)^2}^2 = 1$

[solid kiln]

jandro0103

[soft pendant]

this is correct!

[simple panther]

Yes

[bright seal]

Oh well they are making an easy thing in difficult

|-3|=3

Right?

so 3^2=9

|-i|=i

I^2 = -1

Easy

[simple panther]

that's not how mod works for complex numbers

[simple panther]

see this

[soft pendant]

|-3| = $\sqrt{(-3)^2} = 3$

[solid kiln]

jandro0103

[soft pendant]

ez

[simple panther]

$|0-1i|=\sqrt{0^2+(-1)^2}=1$

[solid kiln]

Suika

[soft pendant]

these are basics things !

[bright seal]

Can you explain this pls

[soft pendant]

@simple panther

[bright seal]

But here I have drawn the absolute value in another way

[soft pendant]

but why in the (2) they writed the same thing ?

[simple panther]

That website probably doesn't treat |...| the same way

So u mean 1 website is right while every other source is wrong?

[soft pendant]

guys sorry but in the (2) of my screenshot is |z1/z2| = |z1|/|z2| if z2 is not 0

right ?

because they writed the same thing twice

can you confirm ?

[bright seal]

I understand now that there are more use than ||

[simple panther]

I don't thnk so

[bright seal]

Ei

@simple panther !occupied

!occupied

[trim jolt]

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).

[soft pendant]

im gonna open an help

[empty orchid]

Absolute values produce positive outputs, so no.

[bright seal]

.

[empty orchid]

What platform is this?

[bright seal]

Photomath

[empty orchid]

Yeah, probably wrong.

|-i|=1.

And obviously, 1^2=1.

[bright seal]

|-3|=3

3^2=9

[empty orchid]

Also notice that $z\cdot\bar{z}=|z|^2$.

[solid kiln]

;(

[bright seal]

|-i|=i

i^2=-1

[empty orchid]

Huh?

No.

[bright seal]

,w |-i|

[solid kiln]

Results provided by WolframAlpha

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

Requested by sofficino9188

[empty orchid]

The absolute value defines a distance from the origin of a complex number.

[bright seal]

Mmm

[empty orchid]

Hence, it can never be complex.

Do you understand that?

[bright seal]

$\rho=1$

[solid kiln]

Task Bot

[bright seal]

So you have to find Rho?

The radius?

[empty orchid]

Yeah.

[bright seal]

👍

Clear

[empty orchid]

Like, the polar definition is $a+bi=|z|(\cos(\theta)+i\sin(\theta))$, for $\theta=\arctan(\frac{b}{a})$.

[solid kiln]

;(

[empty orchid]

z=a+bi in this case.

[bright seal]

Thanks!

[empty orchid]

(Sorry about the other day.)

[bright seal]

.close

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[wraith hinge]

Can someone please explain how to do the first one?

[wraith hinge]

I know that we have to use some identity here but I don't know what the first step is supposed to be...

[simple panther]

Try squaring the given equation

[wraith hinge]

You mean x - 1/x = 3?

[jolly saffron]

for the second one find 1/x, and factor out x^2 from the quartic

you'll get something similar

(a quadratic in x + 1/x)

oh you didn't ask for the second one mb

[hollow cypress]

id just solve for x....

👀

[trim lichen]

laborious!

[hollow cypress]

tbf in an exam setting where I cant figure out what to do

and this would show up

it's not THAT bad

[wraith hinge]

WAIT

I think I got it.

[surreal bone]

m²+y² = (m-y)²+2my

m^3+y^3 = (m+y)(m²+y²-my)

[frail heron]

Phenomenal

[wraith hinge]

Alright, thanks everyone! My answer matches. Can't believe I got stuck on such a simple problem after doing so many hard ones. 😭

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[upper owl]

Would the local minimums be 1 and 2?

[jolly saffron]

why?

[upper owl]

I’m not sure if the sharp turn is a local

[jolly saffron]

why 1 specifically

it is

[queen beacon]

Well, you should also say that there’s a local minimum at this point

[queen beacon]

Yes

[jolly saffron]

the only requirement for a minima is f(x - h) < f(x) < f(x + h)

[queen beacon]

The only thing that matters is that everything around it is bigger yes

[upper owl]

Ok thanks

[jolly saffron]

are my inequalities right lol

also don't forget x = 7 in case you did

[upper owl]

It’s asking for the values and x=7 is the same value

As the dot

[queen beacon]

Yes

[upper owl]

So the values are just 1,2

[formal jasper]

am i rememberign it wrong

[queen beacon]

Don’t forget that’s a local minimum too but ye

No

[formal jasper]

oh

[upper owl]

they only giving the graph

But that’s a way to find it

[formal jasper]

oh i think i remember it
f′ on minimums-maksimums were 0 right

[upper owl]

Yes

[formal jasper]

thanks :3

[queen beacon]

Yes

[upper owl]

.clise

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[unborn fossil]

need a bit help 🙏

[unborn fossil]

im just lost in general

[split chasm]

first determine the:
current total from the first 10 reviews
required total from all 20 reviews

[unborn fossil]

would it be 75 * 10?

aka 750

[split chasm]

currently, yes

[unborn fossil]

alright, but thats just an average 😭

i get it tho, now what?

[split chasm]

wdym

what's just an average

[unborn fossil]

the 750?

[split chasm]

750 is the total from the first 10

[unborn fossil]

or thats the addition considering teh average

right

alright, so what would we do from here?

[split chasm]

required total from all 20 reviews

what's the minimum total rating needed to have an average of at least 85

[unborn fossil]

would that be 85 * 10? because we already have the first 10.

[split chasm]

no

forget about the 10 for this part

85+ is the average from 20 reviews

[unborn fossil]

yes, but you cannot change the average of the first 10 reviews

[split chasm]

yes

[unborn fossil]

but it would be 20 * 85 so 1700

[split chasm]

but right now, we first want to know to total

[unborn fossil]

alrighty

[split chasm]

and now you can subtract the total of the first 10 to determine how many points you'd need to gain

[unborn fossil]

so 1700 - 750?

[split chasm]

yes

[unborn fossil]

aka 950

[split chasm]

and now consider the best case scenario

[unborn fossil]

all 100's?

for 9 tests

[split chasm]

yes

[unborn fossil]

not 10 because

we want to find the minum

so 9 * 100

900

and then 950 - 900?

so 50?

[split chasm]

yes

[unborn fossil]

ahh ty! that helps

have a good day

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[soft pendant]

how can i prove that

i think it has to do with binomial formula

[surreal bone]

what about using sin(x+y) ?

sin(x+y) = sin(x)cos(y)+sin(y)cos(x)

[soft pendant]

oh wait i need to go eat

ill back soon

[lunar stirrup]

substitute theta = pi/2 :D

jokes

first factorize the rhs

,w factorize 16 cos^4 theta - 12 cos^2 theta + 1

[solid kiln]

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[soft pendant]

im back

[lunar stirrup]

nvm dont factorize the rhs

yeah split sin 5theta as sin (2 theta + 3theta)

[soft pendant]

i cant touch the rhs

i need to use the binomial formula

i have to use this

<@&286206848099549185>

[hard musk]

is this further maths becuase you would have to use the imaginearly of the expansion of sin5x i guess?

[soft pendant]

yes

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[hard musk]

did u get it

[soft pendant]

yes it was connected to the previous exercise

sorry

[hard musk]

would you mind showing lol

[soft pendant]

yes wait

[daring viper]

oh with complex numbers

[hard musk]

yes de movires i think

[soft pendant]

In the end I skipped the conversions because I was bored, so I wrote the result right away

[hard musk]

yh thankyou

[soft pendant]

If you are interested you can do it yourself from the start

[daring viper]

very nice method.

[soft pendant]

i started from (a) and then do (b)

but you can start from (b) and also do (a)

this is what the book does

[daring viper]

what book is this?

[soft pendant]

its complex - variables - schaum - 2 edition

[daring viper]

ic thanks

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[soft pendant]

hello

[soft pendant]

z = -15 -8i

find the square root

i found r = 17

[solid kiln]

jandro0103

[soft pendant]

but i noticed that the point x = -15 , y = -8 is in the third quadrant

but theta is in the fourth qudrant

how can i fix ?

[trim lichen]

0.49 rad is not in the fourth quadrant

[soft pendant]

i always confuse when i have to fix the angle using pi,-pi etc

[trim lichen]

0.49 rad is in the FIRST quadrant

did you find it as arctan(Im/Re)?

[soft pendant]

oh yes sorry

why i said fourth 😢

anyway

z is in the third quadrant

[trim lichen]

did you find 0.49 rad via arctan(Im/Re)

[soft pendant]

yes

[trim lichen]

right then you are off by half a turn

do you like your arguments -pi to pi or 0 to 2pi?

or do you not care

[soft pendant]

i like use 0 to 2pi

[trim lichen]

ok in that case you have to fix it by adding pi

i was briefly considering sending this but that might have gone badly

(with the joke being that you need to adjust by pi or -pi but it does not matter much which one you go for)

[soft pendant]

could you explain to me why? I mean otherwise next time I wouldn't know how to do it again

[soft pendant]

.

[trim lichen]

do you understand why you are off by a half turn

[soft pendant]

i sneezed 0 time

[soft pendant]

yes

[trim lichen]

well then, (-1)^0 = ?

:P

[soft pendant]

because im in the first quadrant so to reach the third i need to do hald turn

isnt it pi/2 ?

[trim lichen]

yeah so because of your principal argument range of [0,2pi) you need to add pi to stay in range

[trim lichen]

no, a full turn is 2pi

[soft pendant]

i mean

since i need to do half turn

isnt it pi/2?

this

that is, to go from the first to the third quadrant I don't rotate 90 degrees?

oh maybe with pi/2 im going in the second quadrant

not in the third ?

@trim lichen is this the reason ?

[trim lichen]

sure

• 2pi is a full turn
• 1pi is a half turn
• ½pi is a quarter turn

[soft pendant]

okay thanks

.close

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[bronze jetty]

anyone plz help in this ques i think only A and C should be answers ... It is said that ABC are answers anyone plz confirm

[bronze jetty]

<@&286206848099549185>

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@bronze jetty Has your question been resolved?

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[harsh siren]

a friend sent it to me

[trim lichen]

looks very ugly

[trim jolt]

@harsh siren Has your question been resolved?

[undone gorge]

It's a lengthy method but you know F is the origin, so you can find G's cordinate.
Then find H's cordinate using polar method:
x = x0 + r * cos(theta) and
y = y0 + r * sin(theta)
And do the same for J and I.

You can then find the parabola using 3 points

And you have the line passing through FG and cuts the parabola at A and E

Find A and E by substituting the line in parabola's eqn.

Then use distance formula

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[sweet turret]

the answer is not polynomial 3

[sweet turret]

is it describing polynomial order 2?

[fringe thunder]

ts pmo!

[clear cloud]

!occupied

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[fringe thunder]

pleasee tell the ai to speed things up a bit

[fringe thunder]

okay thank you Bot

or maybe you can hurry ybau a little

[clear cloud]

!occupied

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[fringe thunder]

!occupied

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[fringe thunder]

Never stop learning. He who stops learning, stops living.

!occupied

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[clear cloud]

Why make you think of a polynomial ?

[sweet turret]

it goes up, reach peak, and go down

kind of describes -x^2

[split pulsar]

Depends on if you want to model a single day or multiple. If single, then a quartic fits. If multiple, you need a sinusoid, which is a complex exponential.

[clear cloud]

Its generally exponential in physics

[sweet turret]

the answer is polynomial 2 💀

this question doesnt specify clearly

[clear cloud]

Yeah ok why not

[sweet turret]

how would it be exponential

when theres a fall

[clear cloud]

Exponential can decrease

Just get a negative thing in it

[sweet turret]

yea but the question wants increase and decrese

[clear cloud]

e^-x²/2

Gaussian

,w graph e^(-x²/2)

[sweet turret]

wtf ok im not that deep

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[west sleet]

I think the intended answer is just order 2

[clear cloud]

Anyway

[lunar stirrup]

isnt that just downwards facing parabola

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[full dock]

i need help with starting

[full dock]

do i have to put this into parametric form?

[dusky thunder]

what do you know about parallel planes and their normal vectors

[full dock]

the normal vector n is orthogonal to the vector x-p

[dull pilot]

thats a start

[full dock]

amd a direction vector is a nonzero vector in R3 is parallel to the plane in R3

[dusky thunder]

is there a relation between two parallel planes and their normal vectors is what i meant

[full dock]

ohh

wait

if two planes are parallel

their normal vectors can be expressed in terms of some scalar?

[dull pilot]

[dusky thunder]

not exactly

[dull pilot]

vector in terms of a scalar?

[full dock]

like

n = kn or something

where k is real number

[dusky thunder]

i think they meant one normal vector can be expressed as a scalar multiple of the other normal vector

the thing about normal vectors is that they are directional so saying they can be scaled is obvious. one useful fact though is

parallel planes have the same normal vectors

[full dock]

yeah thats what i meant

[full dock]

ohhh okay

[dusky thunder]

it is akin to saying two parallel lines have the same slope

now moving on, do you know how to express a plane equation using the normal vector

[full dock]

i think so using the normal form

We got taught general and parametric vector form of a plane

But I’m not sure when to use which

[trim lichen]

whichever is more convenient in context

there's kinda no way to teach that exactly, you just gotta practice a bunch and develop a feel for it

[trim jolt]

@full dock Has your question been resolved?

[full dock]

so what would be my next step? the normal from is n dot (x-p) = 0 right

[trim lichen]

your 1st step for this question, you mean?

[full dock]

yes

[trim lichen]

you don't actually need to do any fancy form conversion here at all

you need to know two things:

• from an equation of a plane in the form ax + by + cz + d = 0, equations of planes parallel to it can be produced by messing with the constant term d

(and possibly multiplication of the entire eq by a constant)

• the plane you're looking for needs to both arise in this way and pass thru P

clearly the ones that begin with -15x - 3y ... could not have arisen from multiplying -x-3y-3z+21=0 by anything

so those are out

and you now need to decide between option 1 and option 2

do you understand thus far

[full dock]

so my goal is to make the equation im given look like one of the options?

[trim lichen]

mm no

i think you're kinda overthinking it rn

or maybe im not exactly helping it by yapping about answer option elimination

the upshot is like

ok, we want a plane parallel to -x - 3y - 3z + 21 = 0 which also passes thru P = (2, -2, 3)

yes?

[full dock]

yes

[trim lichen]

the way we do this is we replace the constant term (and JUST the constant term! keep everything else as-is) with a blank space:

-x - 3y - 3z + _____ = 0

and now ask ourselves the following question:

what needs to go into the blank to make the point P satisfy this equation?

[full dock]

ohh okay

would we substitute in our point?

[trim lichen]

yes

[full dock]

is it a good idea if i assign the blank space a variable or smth

[trim lichen]

yes it is

[full dock]

so i get d=5

well the blank spot

so -x - 3y - 3z + 5 = 0

ah which is the first option

[full dock]

also one more question, im not sure how to start this i know a unit vector e1 for example has <1,0,0> for R3 but is it talking about that here?

[trim lichen]

busy sorry

[full dock]

allg

.close

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[chilly bobcat]

hi

is this still alive

by chance

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[tame hull]

"Casper owns three cars of the same brand and drives to work with one of them. Each morning, he randomly picks one car and one key. Unfortunately, the cars don't always start even if he picks the correct key. Car 1 starts with a probability of 0.7, Car 2 with a probability of 0.5, and Car 3 with a probability of 0.3. What is the probability that the car starts on the first attempt?"

[exotic pine]

i think you can split it into the chance of correct car+key combo, and the average chance of the car starting

the average wihld be 0.7+0.5+0.3/3=0.5, so 1/2 chance of starting first try assuming correct car+key

[tame hull]

in the solution book it says that the answer is 1/6

[faint stream]

Imma try this.

[tame hull]

I solved it

[tribal juniper]

you forgot the key

But nice u solved it, close it with .close then

[tame hull]

my bad

[exotic pine]

?

[tribal juniper]

there are three keys

and three cars

[main sigil]

the message above it says it

[tame hull]

U needed to get the amount of kombinations of cars and keys, 1/3 x 1/3

[main sigil]

you'd just multiply that average by 1/3

[exotic pine]

the average wihld be 0.7+0.5+0.3/3=0.5, so 1/2 chance of starting first try assuming correct car+key? then i wanted him to calculate the probability of getting the same car and key, which would be 1/3, so 1/3×1/2=1/6?

[tame hull]

and then 1/9 x the average

.close

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[tame hull]

right my bad

[faint stream]

@tame hull just in case, this might help.

It's still the same as his, just a little longer.

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[broken cave]

Can a 7 degree equation be a straight line?

[broken cave]

Is this 7 degree equation a straight line ?

[fleet bear]

One root doesn't imply straight line

[broken cave]

,w plot x^7+14x^5+16x^3+30x-560=0

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[broken cave]

I think I zoomed too much

[fleet bear]

Mhmm

[clear cloud]

Its probably why

[fleet bear]

Wrong axis

[broken cave]

Bro what

[fleet bear]

-1500 < y < 1500

-5 < x < 5

Based off wolfram

[broken cave]

Okay got it

Thanks

It's not a straight line

It just doesn't have any turning points so yeah

It's not a string line because it has different values for diff x

Straight*

Yeah

.close

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[vocal cobalt]

what method do I use to differentiate this? do i first do ln on both function to simplify?

[lunar stirrup]

product rule?

oh nevermind

just saw the x^2 term

[lunar stirrup]

e^ln x^2 is just x^2

[vocal cobalt]

yea i just noticed that

thanks

.close

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[short raft]

all info is shown on diagram, the radius were looking for is smaller than the given one (r1 < r2, r2 = 9)
ive tried making a perpendicular from O2 to the other circle (indicated with red lines) but it doesnt get me anywhere
-# ...i swear if its something plain obvious again

[trim lichen]

do you have a name for the bottom-most vertex of the red triangle

if not i will just call it X

[short raft]

x is fine yeah

[trim lichen]

is $\angle O_1XO_2$ meant to be right?

[solid kiln]

Ann

[short raft]

yep

[trim lichen]

ok, then $XO_2 = 12$, $XO_1 = 9-r$ and $O_1O_2 = 9+r$

[solid kiln]

Ann

[short raft]

yeah

[trim lichen]

and XO1O2 is a right triangle

[short raft]

yep

ive gotten to that point but i dont know what to do, pythegoream doesnt seem to work as theres simply only 1 side we know about

(XO2)

[trim lichen]

what do you mean

pythagoras works just fine

you get an equation in r

write it down and solve it

[short raft]

ima trust you and try

[trim lichen]

do it

[short raft]

WHATTTT

THATS SO CLEVER

anyways i get sqrt root of 72

[trim lichen]

sqrt root is a bit silly

it's like ATM machine

it already stands for square root :P

... but show your work?

[short raft]

oh yeah xD

[trim lichen]

i think you've gotten something wrong here but i wanna see where and what

[short raft]

oh

i uhhh forgot that i need to use (a - b)^2 instead of just doing a^2 - b^2