help-0

[pliant oxide]

i dont even know how to get started lmao

im so bad with log

[abstract fractal]

What have you tried

[pliant oxide]

i dont even know how to get started on this

wanna give me a hint

[abstract fractal]

Maybe try solving for one of the variables on one of the equations

[pliant oxide]

wait so does y = 4(x+1)?

[alpine sable]

raise x to the power of both sides in the first one

and in the second one, raise 3 to the power of both sides

[abstract fractal]

I don't think so

[worn iris]

x^5 = y

[pliant oxide]

oh

that makeks sense

ty

.close

[lone heart]

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[cerulean fjord]

Kate is standing on the top of a 140 metre cliff on a beach and is looking down at 2 boats
on the ocean. Her angles of depression to these two boats are 31° and 43°. How far
apart are the 2 boats?

i just need help on where to put the angles

[cerulean fjord]

would it be at the bottom or the top?

[fair osprey]

?

Try marking in all the angles first?

[cerulean fjord]

the angles of depression

[haughty night]

perhaps think about what angle of depression means

[fair osprey]

Do you know what that is?

[cerulean fjord]

the pov i guess?

[fair osprey]

?

Mark where they would be in the image

[cerulean fjord]

i think i got it

[fair osprey]

Might also want to add the numbers

[fair osprey]

I’m not 100% sure on this but I think now you need to use the sine rule?

[cerulean fjord]

on the triangles on the outside right?

wait are the angles of depression have the same angles on the inside?

[fair osprey]

Yep

[cerulean fjord]

ok thanks

[lone heart]

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[half epoch]

What sort of divine inspiration did the author stumble upon to go from line 1 to line 2 to line 3?

[half epoch]

I’ve verified that $-\tan^{-1}(s) + \frac{\pi}{2} = \tan^{-1}(\frac{1}{s}) \text{ } \forall s > 0$

[ocean seal]

Learath2

[half epoch]

But I have no idea how they came up with this substitution. Nor do I get how the author can just write that inverse laplace transform as it’s completely obvious

[lone heart]

@half epoch Has your question been resolved?

[half epoch]

<@&286206848099549185>

[slender marten]

I think it's an identity.

[half epoch]

Which one? The inverse transform or the initial substitution?

[slender marten]

The -arctan(s) + pi/2 = arctan(s) part.

[half epoch]

Yep it is, found it on wikipedia, thanks.

How I’m expected to know a billion of these is beyond me, but whatever…

Anyone any idea on the inverse transform? Is that also part of some tables?

[slender marten]

I'm so many chapter behind the chapter on Laplace Transform but my book seems to have many mentions of this problem.

I've put some of the material together but I won't be able to add much.

[half epoch]

Hm, I guess it’s just a common occurrence that I need to learn to identify :/

[slender marten]

I'm not sure. 🙂

[wintry parcel]

Whats the question

[half epoch]

How do you go from line 2 to 3

[wintry parcel]

Just integrate use inverse Laplace definition

[half epoch]

The Bromwich integral? Not only is it not thought in this class I don’t even know complex analysis

[wintry parcel]

Ah rip

But u can't find references to this in the book?

They might have explained it in an easier way

[half epoch]

Well you are supposed to use tables and rules like the linearity of the transform e.g. but I don’t see a way to turn this into anything I’ve seen in a table of rules

[half epoch]

Found it in other lecture materials. It’s just a part of the table he hands out at exams… Thanks for the help

.close

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[fierce prairie]

A triangle has a side length 10 and another side length 16. Which of the following(s) could be the area of the triangle?
I. 96 cm^2
II. 80 cm^2
III. 60 cm^2

[fierce prairie]

idk what to do

[high rapids]

Use area of triangle formula

[lone heart]

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[harsh nimbus]

i am being really stupid arentt i

[harsh nimbus]

i am missing something very small right

talking about integrand^

nvm i got it i am stupid af

.close

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[sudden quarry]

I want to know about the general theory of relativity

[tacit arch]

this is not what help channels are for.

[lone heart]

@sudden quarry Has your question been resolved?

[lone heart]

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[plain cipher]

hi guys, i forgot interval notation. is it b or c?

[wanton turtle]

Open brackets mean that number is NOT included (<), closed brackets mean it is included (<=)

[sudden quarry]

Sorry I am new here

[weary wyvern]

first do special relativity

[last ether]

Is this homework

[wanton nova]

Dont all students have summer break?

[last ether]

Some have summer school

[wanton nova]

Ok

[lone heart]

@plain cipher Has your question been resolved?

[plain cipher]

Gotcha, C

[plain cipher]

Summer Homework

[last ether]

What did you get when you solved for x

And yeah, it's C

[plain cipher]

x<9

[lone heart]

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[native blaze]

I need help determining if this maths test is too demanding to be completed in the given time

[native blaze]

https://www.ausbildungspark.com/einstellungstest/konzentrationstest/

The page is in german but you can translate it on your browser

There are 20 questions and 2 minutes, giving you 6 seconds per question

I tried it and couldnt finish in time

[tacit arch]

please help me, but do these inconvenient things first to be able to help me.

[native blaze]

why are you being snarky?

the page is in german and I cannot change it

[worn fox]

the only person who can determine whether it is too demanding for you, is you alone

[frosty lark]

Have you even tried it?

[native blaze]

Yes

I finished 10 questions

I am not sure if this is a good result

[tacit arch]

this isn't a good use of a help channel since you don't have a math question

go to #discussion or something

[native blaze]

alright

.close

[lone heart]

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[lone heart]

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[unreal crow]

Hey guys

[molten pivot]

Hey guys

[unreal crow]

I need your help with a MATLAB problem, please find it attached. I'm just not understanding how we can have 20 charts for the f and r values because they are obviously connected.

[unreal crow]

Hey man, are you going to help me with my problem?

[molten pivot]

4×5 = ?

[unreal crow]

Erm dude, it's not as simple as that.

[worn fox]

get your own help channel if you have a question, this one is occupied

[molten pivot]

I'm answering his question

[worn fox]

no you're not, go elsewhere

[molten pivot]

Yes I am

Did you read what it says

[worn fox]

yeesh im dumb

didnt read

[unreal crow]

We have 4 values for f and 5 values for r. but both f and r are needed to make the equation work.

[worn fox]

mb

[unreal crow]

So we can't just use lets say r1 in the equation without using a value for f unless of course we set it to be 0.

[molten pivot]

Yeah, so for each r value you have to make a chart using each f value

[unreal crow]

So wouldn't that give 5 phase diagrams?

[molten pivot]

It would give 20

(r1,f1), (r1,f2)...
(r2,f1), (r2,f2)...

[unreal crow]

Oh right sorry lol

[molten pivot]

So on and so forth

[unreal crow]

Don't know how I didn't get that before.

So like r1 with f1 then r1 with f2 etc and so on

[molten pivot]

Yes

[unreal crow]

Thanks for your help.

I have another question

[molten pivot]

Sure

[unreal crow]

It feels like I need to combine the phase diagrams together, what's the easiest way to do that?

[molten pivot]

It doesnt say you have to, it just says you can

So perhaps just have one graph for each r value, and plot all 5 f values on the same graph

So that would give you 4 different charts

[unreal crow]

Hmm I could use a for loop for that right?

[molten pivot]

Yeah

A nested loop

[unreal crow]

I don't think that's relevant to my problem man.

No worries my man, maybe you can watch a YouTube video or something instead.

[shell patio]

Lol

Oh sorry. I thought this was general tab

[unreal crow]

It's cool, if you want you could help me with my problem?

[shell patio]

Sry. Idk matlab

[unreal crow]

Okay I used the for loop however I've just gotten one chart, I'm not sure if that's okay.

[molten pivot]

It should be a nested for loop

[unreal crow]

What's the code for that?

[molten pivot]

That produces 20 different charts

Iterate over r
Iterate over f

[unreal crow]

I get that, but how do you do a nested for loop?

[molten pivot]

For loop inside for loop

[unreal crow]

I just wrote:

for f = 0.6
for r = (0.3:0.3:1.5)
% other code
end
end

[molten pivot]

Yeah, but iterate over the f values as well

[unreal crow]

So something like::

for f = 0.5+(0.1:0.1:0.4)
for r = (0.3:0.3:1.5)
% other code
end
end

[unreal crow]

Okay when I did I got the following chart:

[molten pivot]

Yeah that looks messy

Try printing each f value on a different chart

[unreal crow]

So like one chart for r1 and f1 and so on?

[molten pivot]

One chart for f1 and all r values, one chart for f2 and all r values,

So on and so forth

[lone heart]

@unreal crow Has your question been resolved?

[unreal crow]

Okay I'm going to try the subplot command, apparently this is another method of combining charts.

[unreal crow]

So the label would be 2,5- 2 rows and 5 columns.
Let's say I want to plot the 2nd value from row 2 would I write 2:2? Or would that end up plotting values 2 and 1 on row 2?

[lone heart]

@unreal crow Has your question been resolved?

[lone heart]

@unreal crow Has your question been resolved?

[unreal crow]

Can anyone else help please?

[lone heart]

@unreal crow Has your question been resolved?

[remote heron]

what progress have you made again?

i might be able to help

although its been a few months since i was really into matlab

feel free to ping

maybe i will try a bit on my own

lets see if i can remember how to use matlab

https://tenor.com/view/typing-computer-twitter-gif-12385391

[remote heron]

😭

lemme know when ur back around

@unreal crow

[unreal crow]

Hey thanks a lot for replying. I'm feeling sleepy though so I will reply back later. Is it okay if I @ you when I'm awake?

[remote heron]

sure, ill be around all day tomorrow, and fresh in the head too

i was looking at both software (pplane) and just ode45 to accomplish your task

feel free to dm or just open a channel or whatever

[lone heart]

@unreal crow Has your question been resolved?

[serene sapphire]

hasan what's u problem

[unreal crow]

Just not sure how to do the subplots.

[unreal crow]

Are you still there mate?

[unreal crow]

<@&286206848099549185>

[unreal crow]

.closed

[remote heron]

.close

[unreal crow]

.close

[lone heart]

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[edgy flare]

Complex Variables:

This is what I got for part a and b so far:

Not sure if I made a computational error or not.

To determine the image of the real axis, do I just see what the points -1 0 1 translate to?

For the imaginary axis: i, 0, -i ?

[ivory pivot]

you want help for part b ?

[edgy flare]

Assuming my part a is correct.

I’m not sure if I mapped it correctly.

Also, I would guess the real axis maps to the line 1/2x + 1/2

Actually scratch that

It’s confusing because 0 maps to i

[real gazelle]

a looks right to me

recall that circles/lines are mapped to circles/lines

[edgy flare]

Since there isn’t a straight line with the three points it maps to, I can just infer it’s a circle?

[ivory pivot]

it's mpossible because for $z\in \mathbb R \Re{\frac{z+i}{z+1}}=\frac{z}{z+1}$ that is every time <2

[ocean seal]

everg

[real gazelle]

Well

straight lines are really just circles

with infinitely large radius

[edgy flare]

Please don’t do me like this.

😩

[real gazelle]

do you know the riemann sphere

[edgy flare]

Sadly, yes.

[real gazelle]

sadly? :(

[last ether]

And the circumference is the slope of its area

[ivory pivot]

this is an OP argument ...do you know the proof

[real gazelle]

a line in the complex plane is a circle on the riemann sphere

[edgy flare]

He touched on it at the very end of the last lecture and said it’s good to see and know but I won’t be tested on the Riemann Sphere

[real gazelle]

I don't remember but it's standard knowledge for a complex analysis course, look up mobius transformation

you will probably find it

[edgy flare]

My professor has also went through this proof. Lines and circles do go to lines and circles

[ivory pivot]

i find it in my book but is a complement argument

[real gazelle]

yes there is only one line/circle through any three points

[edgy flare]

Does my mapping look correct?

[real gazelle]

a "line" is just a circle with one of the points at infinity

part a looks correct

[edgy flare]

With my mapping the real axis goes to (infinity, i, and 1/2 + i/2)

[ivory pivot]

part a is a text to verify that you are awake

[real gazelle]

yes

[edgy flare]

I’m not sure how I can describe where infinity is.

[real gazelle]

actually you have four points on the real line here

[edgy flare]

Can I make a line going through the two points and assume infinity falls on it at some point?

[edgy flare]

Oh infinity as well?

[real gazelle]

well you know what T(infinity) is

[edgy flare]

Well

My professor draw infinity on both the real and imaginary part

Like imagine an infinity at the ends of my graph

[real gazelle]

yeah you can't really draw infinity on the graph

you need the riemann sphere

the riemann sphere includes the point at infinity

[edgy flare]

So how do I know it is the real line and not the imaginary?

[real gazelle]

(usually the topmost point)

[edgy flare]

Or do I incorporate it with both?

[real gazelle]

it is both

those lines (circles with infinitely large radius) intersect at infinity

you can see this easily if you draw out the sphere

it may help to watch a youtube video about the riemann sphere

so that you can visualize it better

[edgy flare]

Okay so I got [infinity, i, 1/2 + i/2, 1]

For the mapping of the real axis

[real gazelle]

yes

what goes through those points?

[edgy flare]

A line

[real gazelle]

yes!

what line

[edgy flare]

Hm

One sec

-x + i = y

?

[real gazelle]

what are x and y here

[edgy flare]

I’m confused

Like values?

Like one set value?

[ivory pivot]

1/2+i/2 isn t a solution

[edgy flare]

Or

[real gazelle]

like

[edgy flare]

x and y are reals?

[real gazelle]

there is only one variable, z right

yeah

what is their relationship to z

[edgy flare]

Real and imaginary part?

[real gazelle]

yea

so logically

you will not have an i in the equation

because everything is real

lol

[edgy flare]

So y = -x + 1?

[real gazelle]

yes!

[edgy flare]

Also, as a side note before I do the imaginary part, what you said yesterday.

The intuition hit me like a divine power when I first got to work and understood your hints.

[real gazelle]

yay!!!! :))

[edgy flare]

I almost astral projected

[real gazelle]

LOL

[edgy flare]

Anyways,

Okay so for the imaginary axis, I got [i, 0, 1, i+1]

Which is going to need to be a circle with radius sqrt(2)/2?

Centered at 1/2,1/2

[worn fox]

yep!

[real gazelle]

yup

[edgy flare]

Okay now let me figure the unit circle out

Does this look correct so far?

This looks like y = x

And the interior is the left side of it shaded?

Like this?

[lone heart]

@edgy flare Has your question been resolved?

[real gazelle]

yes

[real gazelle]

how do you know

[edgy flare]

My professor said if one interior point of the unit circle maps to a region then all of the interior maps to that region.

Okay, now part c I’m stuck

I pulled all the points from the positive real and imaginary axis and 1st quadrant and plotted the mappings

Not sure how to determine what this is though.

[real gazelle]

yup that's right

[real gazelle]

try plotting where the boundary of the 1st quadrant goes

based on part b

then shade in the 1st quadrant

like you did with the unit disc

[edgy flare]

Would it be this?

The triangle-ish shape I have shaded that is points found in all of the mappings?

[real gazelle]

I am confused, which part is the first quadrant

[edgy flare]

I plotted the things I found in part b

y=-x+1

The circle centered at 1/2,1/2

Radius = (sqrt2)/2

And y=x

Then found the domain of points they all share in the first quadrant

[edgy flare]

Well, I guess this wouldn’t work because the line y=-x + 1 only has points on the line and note below it.

Would it be just the 1st quadrant of the circle? Since it would contain all of the points I have listed?

[real gazelle]

you only want to plot the first quadrant, most of what you did in b is irrelevant to that task, but some pieces of what you did in part b you can use

[edgy flare]

This is the mapping

[real gazelle]

also you don't want the first quadrant of the output, you want where the first quadrant of the input gets mapped to

[edgy flare]

Of the points in the first quadrant

[real gazelle]

don't consider arbitrary points in the first quadrant

[edgy flare]

Why?

[real gazelle]

consider where the boundary of the first quadrant gets mapped to

[real gazelle]

there are infinitely many points in the first quadrant

try plotting all of them..

lol

[edgy flare]

Okay I’ll look after I eat real quick

[edgy flare]

So if I’m looking at the bounds wouldn’t that just be infinity on both the x and y axis?

@real gazelle

[real gazelle]

yes

well

infinity is just one point

consider the entire boundary

[edgy flare]

0 also?

So 0 goes to i

Infinity goes to 1

So the first quadrants maps to y = -x + 1 again?

Well this could also be a quarter of the unit circle.

Hmmm

[real gazelle]

yes 0 is another point on the boundary

but consider where the entire boundary maps to

hint: the boundary is composed of two rays

(idk if you remember from geometry what rays are or not)

[edgy flare]

A Ray from i to 1?

Two rays?

0 to i

0 to 1

?

Do I consider the real axis goes from 0 to infinity and includes 1 so I map from T(0) to T(1) T(inf)?

And the same with T(0) T(i) T(inf) for the imaginary part?

But the imaginary part would be a quarter circle and not a Ray

[real gazelle]

yes

[real gazelle]

yes

[real gazelle]

the ray gets mapped to part of a circle (not a quarter circle)

[edgy flare]

I get the Ray from i to 1 passing through 1/2 + i/2

For the real part

And for the imaginary part, I have to find a Ray that passes through i to i+1 to 1

Yes?

[real gazelle]

wait

no, the rays are in the input space

then they get mapped to shapes in the output space

you already pretty much know what shapes they get mapped to because you basically did that in part b

(do you know what a ray is btw)

[edgy flare]

Maybe

A fixed lower bound with no upper bound?

•——————->

[real gazelle]

yea

exactly

[edgy flare]

Looks like that

[real gazelle]

the boundary of the first quadrant consists of two rays

you know where those two rays get mapped to after the transformation

then, the region that the first quadrant gets mapped to will be between those two shapes

[edgy flare]

This?

With the arc filled in

[real gazelle]

yesss now just check one point in the interior of the first quadrant

to see what area you should shade (inside or outside the semicircle you just drew)

[edgy flare]

Of the input quadrant?

I can test 2?

(2,0)

[real gazelle]

that is on the real line

[edgy flare]

Or no

Yeah

[real gazelle]

I would test something that is inside the quadrant

[edgy flare]

That’s on the boundary

[real gazelle]

yeah, don't test something on the boundary because you already know where the boundary goes

[edgy flare]

I shall test 107,875, 456i

[real gazelle]

LOL

okay

have fun with that

[edgy flare]

Nah

I do 2 + 2i

[real gazelle]

kk

idk why you are making this hard for yourself

look at the work for part a

just use that lol

[edgy flare]

Use T(1+i)

?

[real gazelle]

yea

[edgy flare]

Which goes to 4/5 + 3/5i

Which is inside

[real gazelle]

yup

[edgy flare]

Ouí?

[real gazelle]

yup

there is also this cool site you can use to visualize these

http://davidbau.com/conformal/#z

I don't know how helpful it is, but if you plug in your function, it makes a fun pattern

🙂

[edgy flare]

That looks strange

Looks like a trip

[real gazelle]

it does, but it's a nice tool for visualizing where everything goes

for example you can see the grey unit circle goes to the y=x line

[edgy flare]

Ahhhh

You can!

[real gazelle]

well actually it might be plotting the inverse of that

like

okay wait lemme find a better example

that wasn't a good example lol

[edgy flare]

Okay, I’ll wait before moving on.

[real gazelle]

okay better example, the little white and black circle thing represents z=0 and it is sitting at the point z=-i

that tells you that T(-i) = 0

:)

[edgy flare]

Ahhh I see that.

What’s up with all the colors?

[real gazelle]

the colors give you information about the phase

and then the brightness corresponds to the magnitude

this is a common way to plot complex numbers because you can't really make a normal graph of it (it would be 4D)

another example is that the rainbow colored circle represents the point at infinity

and it is sitting at z = -1

so T(-1) = ∞, just as you would expect

[edgy flare]

Ahh that makes sense.

This is neato

Thanks for this.

I’ll be posting another here in a second.

.close

[lone heart]

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[old pelican]

I need to find the length of the segment OA.

[old pelican]

But idk how.

Info the problem provides: MN = 8, and that H is the middle point between M and N.

O is the centre of the circumference.

I've tried making a rectangle triangle, so I can apply Pythagoras theorem, however, I need the radio to do so.

And anyways, OA is basically the radio so that'd be redundant.

I could use the same idea if I drew triangle HAN, however, I'd need the length of AN to apply the theorem.

[median oar]

Do you see triangle HAN

[old pelican]

ye

[median oar]

Well actually HON is a better triangle to look at

ON is the radius

OH is half the radius

HN is 4

It’s a right angle triangle

So you can pythag for the sides

[old pelican]

Ohh

but, how do I get the radius

[median oar]

That’s what you’re solving for

You want to find OA that’s the radius

Notice that the triangle HON is a right angle triangle

This means its sides have a special relationship (pythag)

[old pelican]

oh hm

But how am I going to find the side lengths? x² + y² = 4?

well um, if i have ON i have OH and viceversa, but I can't find any of those if i just have one side

[median oar]

.

[median oar]

.

Can you construct the pythag equation for triangle HON using ON NH and HO as the sides

[old pelican]

x² = 4² + (x/2)²

?

[median oar]

Yeah

But I’d prefer to use r as the variable

But the equation is right

[old pelican]

But um

how do i find x

or r

whatever

So, r² = 4² + (r/2)²
r² = 16 + (r/2)²
r² = ??

<@&286206848099549185>

[hardy blaze]

You messed up

[old pelican]

oh

where

[hardy blaze]

On r^2 = 16 + (r/2)^2

You need to subtract 16 from both sides

Then square root and then you see 2 options of r

[old pelican]

oh ok

[hardy blaze]

wait no

Look at this photo

[old pelican]

oh so,
r² - 16 = (r² + 16)(r² - 16)?

[hardy blaze]

No (r+4)(r-4)

Since 16= 4^2

[old pelican]

oh yeah right

[hardy blaze]

Yea

[old pelican]

so, um, r² - 16 = (r+4)(r-4), and?

[hardy blaze]

So (r+4)(r-4)= (x/2)^2

Simply the right

[old pelican]

r² = (x/2)²?

nvm

the right

uh

idk
((r+4)(r-4))*2=x²?

[hardy blaze]

No

[old pelican]

idk

[hardy blaze]

Whats a better way of saying(x/2)^2

[old pelican]

uh

idk?

[hardy blaze]

Whats the meaning of a number being squared like 7^2

[old pelican]

the number being multiplied by itself

[hardy blaze]

How many times

For this moment

[old pelican]

once

[median oar]

Expand the power

[hardy blaze]

Yea

[old pelican]

?

[median oar]

Remember your indices rule

$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

[ocean seal]

Frosst

[dreamy widget]

[\left(\frac{x}{2}\right)^2= \frac{x^2}{4}]

[hardy blaze]

HEY

[ocean seal]

Blend

[hardy blaze]

well yea

[median oar]

Don’t just give them the answers lol

[old pelican]

oh right so

(r+4)(r-4)=r²/4

?

[hardy blaze]

Try to make r by itself

@dreamy widget dont you dare

[hardy blaze]

What do you think is the next step?

[old pelican]

simplifying the left?

[hardy blaze]

No its about the right

[old pelican]

hm

uh

idk

[hardy blaze]

I am not gonna lie after you do that i am kinda suck on what to do next

Wait

[hardy blaze]

Oh no

you see r^2=16+ r/2 ^2?

[old pelican]

ye

so r²=16+r²/4

[hardy blaze]

No

Keep it as that

[old pelican]

okie

so?

[hardy blaze]

Subtract a number so its like this

[old pelican]

oh, so
r²-4²=(r+4)(r-4)

nvm

[hardy blaze]

No

[old pelican]

r^2-16=16+ r/2 ^2?

[hardy blaze]

No

[old pelican]

r^2-16=(16+ r/2^2)(16- r/2^2)

[hardy blaze]

Still confused on how geometry turned into algebra but ok

[hardy blaze]

No

Look at the other one

[old pelican]

uhm

(r+4)(r-4)=r²/4

[hardy blaze]

No

[old pelican]

idkkkk

[hardy blaze]

Subtract r/2^2 from both sides

What do you get

[old pelican]

r²-(r/2)²=16

[hardy blaze]

What do you think a and b is

[old pelican]

r and 4?

[hardy blaze]

B is wrong

Look at r/2^2 do you see something

[old pelican]

oh

r² - r/2²=(r+r/2)(r-r/2)?

[hardy blaze]

Yes

[hardy blaze]

Simply this

r - 1/2 r= ??? And r + 1/2r=???

[old pelican]

-r/2 + 2r/2?

[hardy blaze]

No

1-1/2

=??

[old pelican]

1/2

[hardy blaze]

=1/2(r) since r=1

1+1/2?

Not mixed btw

[old pelican]

oh so 1/2(r) * 3/2(r)?

[hardy blaze]

Yea

Simply again

[old pelican]

which would be 3/4(r)?

[hardy blaze]

Yep

[old pelican]

but um, my question was how do i find r

[hardy blaze]

I know how

[hardy blaze]

3/4 r^2

[old pelican]

and i do not

[hardy blaze]

I will show you

So we have 3 r^2/4=16

[old pelican]

ohh

i forgot about the 16 lmaaao

so

[hardy blaze]

Yep

Solve for r

[old pelican]

3r²/4 = 16
3r² = 16*4 (=64)
r² = 64/3 = 21.33333....32?

[hardy blaze]

We can simply

[old pelican]

uh how

[hardy blaze]

Whats sqrt of 64

[old pelican]

oh

yeah right

so

3r²/4 = 16
3r² = 16*4 (=64)
3r = 8

and 8/3 = 2.66666666667

[hardy blaze]

No

sqrt 64/3= 8/sqrt3

[old pelican]

so...?

r=8/sqrt3?

[hardy blaze]

Yea

I had to check my work

And its right

[old pelican]

so uh

r=4.61880215352?

[hardy blaze]

It can also be negative

[old pelican]

wdym

im confused

[hardy blaze]

Square roots can be negative remember?

[old pelican]

oh yeah

so um, now i apply pythag?

[hardy blaze]

Wait

Send the photo again

[hardy blaze]

What are gonna use it on?

[old pelican]

i need to find the length of oa

oh wait

NVM

oa is the radium

but yeah

i need to find the length of oa, mn = 8 and h is the middle point between mn and o and a

[old pelican]

so, oa is 4.61880215352...?

that's uh

[hardy blaze]

Um

[old pelican]

looks off to me

[hardy blaze]

Dont submit it tho

[hardy blaze]

Yea

[old pelican]

btw this is the original pic, i just added the segments ON and NA cos i wanted to apply pythag

not really any difference but yeah

[hardy blaze]

Are you sure?

[hardy blaze]

Ok

You ditched y that's why!

[hardy blaze]

This

I think

[old pelican]

yeah but, isn't y the half of x?

[hardy blaze]

Check

4-64/3

12-64/3

[old pelican]

HN is the hipotenuse, 4
OH is y, and half of x
ON is the radius, x

?

[hardy blaze]

Wait

Bruh

[old pelican]

i gtg, might ask later, thanks for trying to help anyway

/close

.close

[lone heart]

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[hardy blaze]

But i found it out I thought

I found it

@old pelican

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[thin marsh]

im not sure how to do that

<@&286206848099549185>

[drowsy lagoon]

z + 11 > 0

4z-6 > 0

solve for z

[serene sapphire]

yeah

@unreal crow

what was u question

[thin marsh]

which equation do i use

[drowsy lagoon]

solve for both

[thin marsh]

i got 3/2

[drowsy lagoon]

then take the one that can include both answers

z > -11
z > 3/2

[thin marsh]

i got that

[drowsy lagoon]

so it has to be z > 3/2

[thin marsh]

cuz it has to be greater than 0

right

[drowsy lagoon]

it cannot be z > -11 bc it cannot satisfy 4z-6 > 0

[thin marsh]

ok ok got it

ty

[mellow tusk]

i have a doubt, what would be the use of equal sides and a right angle

[drowsy lagoon]

No use actually

[mellow tusk]

alr

@thin marsh u can close

[thin marsh]

.close

[lone heart]

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[dire turret]

<@&286206848099549185>

[vale wigeon]

1. 15-minute rule

2. this is hardly a math problem

[dire turret]

yeah

I just needed help with the potential problems

so
one problem I identified
is that
they only made participants do it during lunch
which means that
they only reviewed what food the restaurant makes at lunch
but the restaurant wants
to know how people like their food overall

[vale wigeon]

that sure is a problem, yes

[dire turret]

to improve it

you can

ask customers at breakfast

and maybe ask just a few number of people

during dinner

selected randomly

ok let me

think of

another problem

do you think there are other problems

with this experiment

other than the timing

because aren't they asking

every customer

since they give them a check and a

survey

[alpine sable]

Hallo guys where can i have othr ppl solve my hw hahaha

Or just hints

[dire turret]

<@&286206848099549185>

[abstract fractal]

Dang, breaking the 15 minute rule twice in one channel

!15m

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[dire turret]

ok

I'm sorry

[tacit arch]

Restate the problem as a math problem with equations

[lone heart]

@dire turret Has your question been resolved?

[dire turret]

nvm I got it

[lone heart]

@dire turret Has your question been resolved?

[lone heart]

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[cinder sundial]

F2 will be equivalent to f1 right

Sorry

.close

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[mellow tusk]

what ik till know is that gx is a linear equation,
g(3)=6

[lone heart]

@mellow tusk Has your question been resolved?

[mellow tusk]

<@&286206848099549185>

[minor needle]

P(3) = 6, let P(x) = Q(x) * (x^2 - 9) + ax + b where g(x) = ax + b

Use P(3) = 6 to create first equation

For 2nd use fact that P(x) is odd

[mellow tusk]

how do i use that

[minor needle]

-P(x) = P(-x)

Hence P(-3) is...

[mellow tusk]

oh

oohk i got 3a+b=6, -3a+b=-6, b=0, a=2

thanks @minor needle

.close

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[faint tusk]

Which online can you recommend Me for making graphs with formulas that is very simple and fast? Thank you.

[chrome plank]

Desmos is pretty nice

Otherwise there's geogebra

[faint tusk]

Used both of them, just to check what else is out there, I prefer not looking so tidy up, colourful, how should I say it.

Okay, I will try again with Desmos, I just need to figure out how to insert formulas. I need a line graph consisting of two lines.

Hey, about Desmos, can you help Me insert a formula: Line A=A*27?

[chrome plank]

Sure

[faint tusk]

I tried, probably would look at some tutorials if you did not accept helping.

[chrome plank]

Type your function in that section

The graph will just pop up

[faint tusk]

Yeah but I do not see where to define variables, Line A in this case.

[chrome plank]

Type it

"add a slider: [variables]" should appear, click that blue button and some sliders should apper

and then you can change the value of the variables

[faint tusk]

Ah, I went A=A*27, that is why I did not find it, sure, I forgot that you have to get an axis... Well, I apologize and thank you.

[chrome plank]

No worries

Let me know if you have more questions

[faint tusk]

Thank you! Yeah, I already said this...

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[frail lantern]

Im just trying to get better understanding of this method
till now when solving, we never had the limitation of the terms being unequal, so we solved it as: separate it into r variables and distribute n amongst r such that one term can take 0 as well i.e n + r - 1 C r - 1
Now that this is unequal, my sir introduces a new term y1 and y2 such that theyre always equal to or greater than 1 this way, x1, x2, x3 will never be equal
but when you say let x2 = x1 + y1
youre suggesting that
x2 > x1 ALWAYS
and thats fine?
because its just division of n not arrangement?
Just, if someone could explain to me what the issue was (the terms not being equal) and how we tackled it, (what we did to solve what problem) would be really nice, Im just trying to get the hang of this all (edited)

[lone heart]

@frail lantern Has your question been resolved?

[frail lantern]

Could someone explain to me why the coefficent of $x^n$ in $n!e^{xr}$ is $r^n$

[ocean seal]

ashllxyy

[frail lantern]

I have 2 questions here, if anyone could answer either, it would really be helpful

[frail lantern]

<@&286206848099549185>

[vale wigeon]

taylor series of e^t

[frail lantern]

what about the series

its right above the steps Ive highlighted

the e^x series

the coefficent of x^2 in that taylor series is (1/2!)^r times, so...

[vale wigeon]

e^(rx) = 1 + rx + r^2 x^2/2 + r^3 x^3/3! + ...

[frail lantern]

I dont unedrstand

the top has (1 + x/1! + x/2! + ...)^r

the terms inside the bracket represent the taylor series for e^x so we substitute that in

and we get e^(xr)

then how does e^(xr) become 1 + rx + r^2x^2/2! +.....

actually no

youre right

yeah

you just replace x with xr

yeah damn, thats a cool way to use it, thanks a ton ❤️

[frail lantern]

if someone could help me out with this as well. Its a new method that they taught in class and I didnt quite get it as welll as the others.

.close

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[wraith marlin]

can someone please explain

I don’t get the question

or why the answer is the answer

T^T

please help

[languid bolt]

use the equation given @wraith marlin

T is time of 400 meters

t is time of 200 meters

T = 2.2t

T = 220% t = 2t + 20% t

so the 400 meters time is twice the 200 meters time plus 20% of the 200 meter time

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[hazy violet]

If the probability of person A passing an exam is 0.05 (5%) and person B passing the same exam is 0.10(10%) then wudnt the probability of both passing be 5/100 * 10/100= 0.5%??

[vale wigeon]

not if B cheats off of A or something to that effect

but if they're at opposite ends of the classroom and pass/fail independently of each other then yes it's 0.5%

[hazy violet]

LOL no..

[hazy violet]

hm..

But in my text book the question says that its 0.02(2%) by using sets

But i dont get it

[vale wigeon]

show the question exactly as stated

[hazy violet]

1 sec

Here u go

[vale wigeon]

ah but yes

the events of their passing or failing are not assumed independent here

[hazy violet]

Oh...

[vale wigeon]

the 2% is part of the data, not derived from each student's individual chance of success

[hazy violet]

Oh...

Also i hve another doubt

[vale wigeon]

yes?

[hazy violet]

OUt of 20 numbers a person chooses 6 numbers

If these numbers match with the 6 numbers chosen by the lottery he wins the lottery

So shudnt the probability of him winning the lottery be nCr squared

Since the numbers chosen shud match withthe winning numbers.

Or is it jus nCr

[vale wigeon]

it's neither nCr nor (nCr)^2

[hazy violet]

Why not?

[vale wigeon]

both of these cannot take values between 0 and 1 other than 1, and the lottery sure is not a guaranteed-win event.

there are 20C6 possible selections of numbers that can be made by the man and by the lottery

[hazy violet]

Yes

[vale wigeon]

the probability that the man picks the same numbers as the lottery is 1/(20C6)

the logic here is the same as in this simpler problem: suppose you and i each roll a dice. what's the probability the two of us get the same number?

[hazy violet]

but for it to be the same combination shudnt we multiply it by itself

1/36

[vale wigeon]

that's the chance that we both get a particular number, say fours

but we could also both get ones

or both get twos

etc.

[hazy violet]

Oh....

Right

Got it

Lets say an airplane is flying and the probability of an engine failing is 1% so the probability of both engines failing at the same time wud be (1/100)* (1/100)

right?

[vale wigeon]

that's if we assume the engines operate independently of one another.

[hazy violet]

Yeah they d

do

[vale wigeon]

if they do then yes

[hazy violet]

Alright

THanx im clear now

.close

[lone heart]

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[grave coral]

hey guys can someone explain to me why we need the middle part , I mean what's the point from it

[vale wigeon]

what middle part

[grave coral]

this one

[keen pasture]

x < y?

If x = y and y was not in Q the statement would be wrong

[grave coral]

but why are they defining z after the implies sign

[mortal trellis]

"x<y implies that there exists a rational z such that x <= z <= y"

you can't define z before the implies because at that point you don't "know" anything about x,y yet. and in general (without x<y) the claim is false

[keen pasture]

Unfortunately it isn't x < z < y 😧 . This would lead to infinitly many rational numbers between x and y 🙁

[grave coral]

how would I know that it means there exists is it the implies sign

[grave coral]

in both cases it's infinite

[mortal trellis]

the backwards E means exists

the => means implies

[grave coral]

I know but they just did a comparison and defined z with no context

am I wrong ?

[mortal trellis]

wdym no context

we take x<y and then we know that there exists a rational number z between them

[keen pasture]

If it was < we could always plug z into the statement. But as we have <= z could always be able to be e.g y, so there could in theory only be 1 z between x and y.

[grave coral]

okkay , I think I get it now

thank you guys !!

It makes a lot of sense

.close

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[mortal river]

Basic algebra question

[keen pasture]

42

[mortal river]

Would this just be

37
X^2, 12x, 2x^2

38
X+8+5+x+x+12+x+2x+x+8+34
or
7x+67?

[last ether]

The meaning of life

[mortal river]

Yeah sorry I’m on my phone it took a second to type that all out

[vale wigeon]

37
X^2, 12x, 2x^2

1. don't capitalize unnecessarily
2. you want the sum of these

[last ether]

just add them and combine like terms

[mortal river]

Capitalization because I’m on my phone

It just did auto caps

[vale wigeon]

38
X+8+5+x+x+12+x+2x+x+8+34
or
7x+67?
yes

[mortal river]

Why would ie ant the sum?

[last ether]

Yeah just make sure to disable it

[vale wigeon]

auto caps can be turned off tbh

[mortal river]

If I’m finding the area?

[last ether]

The question asked for the sum

[vale wigeon]

read question 37 carefully

you want the sum of the areas

[mortal river]

Ohhh

Lol no no you’re right

[last ether]

Yeah so when it comes to variables, make sure if your phone is autocapitalizing things that you manually make it lowercase

Because X ≠ x

[mortal river]

37
x^2+12x+2x^2
3x^2+12x

[last ether]

Yup

[mortal river]

Cool, thanks everyone

.done

[last ether]

Close

Not done

.close

[mortal river]

Oh my b

.close

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[last ether]

Wlater white

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[toxic ferry]

Linear(Vector?) Spaces
How do I go about finding* the intersection between two spans? Sorry if this is a little bit of a broad question, I'm a little rusty on the subject

[mortal trellis]

if $V$ has the basis ${v_1, \ldots, v_n}$ and $W$ has the basis ${w_1, \ldots, w_m}$ and $x\in V\cap W$, that means there exist scalars $a_1, \ldots, a_n, b_1, \ldots, b_m$ such that $x=a_1v_1+\ldots+a_nv_n = b_1w_1+\ldots+b_mw_m$, or equivalently, $a_1v_1+\ldots+a_nv_n-b_1w_1-\ldots-b_mw_m=0$

[ocean seal]

Denascite

[mortal trellis]

so you should study solutions of that last system of equations

[pliant dune]

I'm not certain but you might also be able to consider one of the spanning sets and remove each vector not in the other span

[toxic ferry]

Ah, yeah that makes sense

[toxic ferry]

This would not work for different bases for said spans though

I think.

Consider Sp{(1,0),(0,1)} and Sp({(1,1),(1,2}).
Same spans, different bases

I'll keep this up for just a moment to see that I truly got this though

[mortal trellis]

it might be a good idea to do some dimension arguments first to check how big the intersection can be

[pliant dune]

for this example I mean checking for each vector in {(1,0),(0,1)} if it's in Sp({(1,1),(1,2})

[mortal trellis]

for example if $V, W$ are subspaces of a $k$ dimensional space, then $\dim V\cap W = \dim V + \dim W - \dim (V+W) \geq n+m-k$

[ocean seal]

Denascite

[toxic ferry]

Oh, the question is a little more generalized than that so I'm not sure how that would work - but I think that's what I usually do!

[mortal trellis]

maybe also a good first check to reduce the system of equations

but still not enough

[toxic ferry]

Oh right! The formula looked alien for a moment.
Unfortunately I'm trying to find dim(V + W), which is why I'm trying to find V intersection W 😛

[mortal trellis]

oh...

well reduces to the same problem

find the rank of the matrix (v_1...v_n, w_1,...w_m)

which you do by studying the same system of equations

row reduction etc

[toxic ferry]

It'll take me a little while to type out the question, but basically I have U and W, their dimensions are equal to K, and their bases are:
B of U = {u1,...,uk}
B of W = {u1 - w, ... , uk - w}

So reducing is not an option either I think ^^;

I think the method you mentioned earlier could work though