help-13

[granite eagle]

.close

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[worldly iron]

Where I did wrong?

[worldly iron]

Can anyone help me?

[granite eagle]

think abt what u did on case 1 once again

i just saw that

maybe something wrong before that too

[still gust]

need help

[worldly iron]

#❓how-to-get-help

[granite eagle]

so look at the point u do +/- for the first time

i think u should seperate the cases there

bring sin^4 x to the lhs in both cases

[worldly iron]

@granite eagle I did that

still not helps

[granite eagle]

well ur stuff isnt wrong either

tho ur cocclusion in case one 1 rubbish it

case one give udefined answer so discard it

cosx=0 gives ur solutions

whatever they maybe

[worldly iron]

dude, it doesn't

I checked the calculator

the difference in LHS and RHS is nonzero

[granite eagle]

at which step exactly?

the last one?

[worldly iron]

At second step

I plugged the x=pi/2 value into the equation at first to check if it's the solution

<@&286206848099549185> help

[granite eagle]

no bounds?

[worldly iron]

wdym?

[granite eagle]

range of x

also i dont really see a mistake, if u r very stuck u can just plug that value into every epression from the bottom before the cases split; u'll find where u went off

[worldly iron]

@granite eagle holy sex dude!!!!

I fucking forgot 1 thing!

still not helping

dammit!!

<@&286206848099549185> help!!

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[tough gazelle]

.rotate

Someone help

[mental trail]

,rwcr

[tough gazelle]

Pardon?

[mental trail]

,rw

Trying to rotate

[tough gazelle]

Ok

[random shale]

,rotate

[mental trail]

Bruh xd

[random shale]

what have you tried? @tough gazelle

[tough gazelle]

I don’t know what to do

If I find the volume of the water

And the volume of a cube

After I add them togetherr how do I find the level increase

[mental trail]

Find the volume of water displaced

[tough gazelle]

🤔

[tough gazelle]

Wdym

[mental trail]

First, what is the volume introduced in the water ?

[tough gazelle]

80

M squared

[mental trail]

Volume

Not area

Also how did you get 80 ?

[tough gazelle]

4 x 5 x 6

The dimensions of the water

In the cistern

[mental trail]

Yes

ButI was asking about the volume added to the water

What is the volume of the cube added to the water ?

[tough gazelle]

89

Sorry my mom called

[dim obsidian]

please noir

[random shale]

#❓how-to-get-help

[tough gazelle]

I’m asking for help myself

Go to another channel

[tough gazelle]

89

?

Ok so this is what I tried

Since it is a rectangular prism

Of dimensions 4 5 and unspecified height

But the water filled up is 6

So volume of water

Is 80

So then the cube

Which is 3 3 3

Is 27

80 plus 27 is 107

So then 107 in terms of the prism would bring

L times w times x(variable) equals 107

So 20x is 107

107 divided by 20

5.35

But that’s not the answer

,rotate

[tough gazelle]

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[crimson sedge]

need help with c. pls

[livid hound]

identify the horizontal lines that intersect the curve only twice

[crimson sedge]

I don't understand

it intersects 3 times

[long arrow]

it depends on which line you choose

for example this one intersects f(x) three times

try to find it such that it intersects the graph at only two points

[crimson sedge]

does the line have to go across the whole curve

[crimson sedge]

I can't find it...

[long arrow]

one of them

[crimson sedge]

So there are multiple answers?

[crimson sedge]

but if you put the line up it could still intersect the graph at 2 points?

[long arrow]

yes

this is what I was thinking about

[crimson sedge]

this is also correct?

Okay I understand now

Thank you

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[turbid sail]

How could i can get the ordered in lineal funtion?

[turbid sail]

,close

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[radiant portal]

In an office​ building, 66 offices are currently being rented. This represents ​60% of the total units. How many offices are there in the​ building?

[hoary panther]

Yes

@radiant portal

[radiant portal]

yes what

[queen palm]

Hello

@radiant portal

[radiant portal]

hi

[queen palm]

Can you tell me the answer to this question

What is x

[hoary panther]

60% is 66

[radiant portal]

i need help with mine

[queen palm]

Oh

[hoary panther]

/3

20% is 22

22 * 5 is 100%

[radiant portal]

thanks

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[quick raft]

I am on part C. I dont understand what A is. Is it a set of vectors? Is it the transformation?

[quick raft]

I think if i do the RREF on transpose(a), I would find that one of these is dependent? Im not positive if maybe the Image would be transpose(a) and only looking at the independent ones

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[devout star]

f(x) = x(x-a)(x-b)
0<a<b
solve for integers, a, b and c, where f(x) is divisible by (x-c)

[cedar kiln]

@devout star Has your question been resolved?

[cedar kiln]

@devout star Has your question been resolved?

[cedar kiln]

@devout star Has your question been resolved?

[thick cipher]

What are you looking for ? x? a? b?

[devout star]

I'm not sure what it means where it says p(x) is divisible by (x-c)

a, b, c, are all integers

[thick cipher]

Oh I see

More logical now

You know that c is a root of P

Thus P(c)=0

[cedar kiln]

@devout star Has your question been resolved?

[devout star]

how do you know c is a root of p

WAIT P(x) / (x-c), substitute c for x, and if c is a root of P then P(c)=0 and c-c=0, so you get 0/0 which is undefined

so if c is a root of P then P(x) is not divisible by (x-c)

[cedar kiln]

@devout star Has your question been resolved?

[worldly iron]

Is this asking for a relation equation of a,b,c?

[devout star]

it's just asking to find a, b , and c, so that p(x) is divisible by (x-c)

I'm not sure what it means by divisible

maybe there's more than 1 set of answers

for a, b, and c

[worldly iron]

that's that

you maybe need a relation equation for a,b and c

[livid tusk]

if (x-c) divides P(x) then P(x)=(x-c)q(x). thus, P(c)=(c-c)q(c)=0

when we say that $a$ divides $b$" or $b$ is divisible by $a$", what we mean is that

$b=a\cdot$something.

[livid tusk]

there should be infinitely many i think
actually no, there might only be a finite amount.

[wraith dagger]

nix

[livid tusk]

but the important thing to take away from P(x) is divisible by x-c is that P(c)=0

[devout star]

what does q(x) mean?

[livid tusk]

just some polynomial. it doesn't really matter what it is

the point is that P(x)=(x-c) times something

[devout star]

yea

[devout star]

c is the root of P, so if you plug in c for x, 0 =0 times something

but p(x) / (x-c) if you plug in c, you get 0/0, undefined

[livid tusk]

have you taken calculus

[devout star]

I'm taking pre calc, this question should be pre calc

[devout star]

so that (times something) cannot be undefined in order to make p(x) divisible by (x-c) true

[livid tusk]

ok well the thing is that 0/0 can sometimes "have" a proper value. usually when it comes to polynomials, it's fine. so don't worry about dividing by zero until you take calculus. you don't need to do any dividing for this problem

[devout star]

ok

so does that mean a, b and , c can be anything

[livid tusk]

but as an example, $x^2-1=(x-1)(x+1)$. so $x-1$ divides $x^2-1$. that's really all we're saying.

[wraith dagger]

nix

[livid tusk]

absolutely not

what if i do a=b=c=0. then we just get that 13=0 which doesn't work at all

[devout star]

wait but 0<a<b, a, b, c are integers

[livid tusk]

just use that P(c)=0 and then you'll see that there are only a few possible integers c can be

[devout star]

ok

so
P(c) = c(c-a)(c-b) = 0

so either c = 0 or (c-a) =0 or (c-b) = 0

[livid tusk]

check the question again

you lost the 13 somewhere

[devout star]

oh yea

oops

so the function part has to equal 13 so that it cancels -13 and get 0

[livid tusk]

yeah you get c(c-a)(c-b)=13

and since a,b,c are integers, and 13 is prime, you only have so many options.

because notice it's in the form 13=c*something. that means c divides 13.

[devout star]

yea that's what I was thinking

so it would be 1 * 1 * 13

c cannot be 13 because then a=b and 0<a<b

so c=1

a=2

b=14

1*(-1)*(-13) = 13

wow that's so satisfying

I solved it

[livid tusk]

very nice 😎

i counted 12 total solutions if a,b,c can be any integers. but with the condition that 0<a<b, then there might be only one.

[devout star]

thanks for the guidance

[worldly iron]

if you are done, close this channel

[devout star]

how

[worldly iron]

.close

[worldly iron]

@devout star type this

[devout star]

.close

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[hoary charm]

how does

become

Also this is describing the linear transformation of rotation through an angle theta

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[heady salmon]

How would i find these two angles?

[heady salmon]

would i do sec-1(-2)

nvm found out how

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[high spruce]

huh

[zenith sail]

Big O notation https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation

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[dawn moth]

Hi!
Let's say I've a circle with a radius r and a center C (x,y) that is the centre of a square in a grid.
What are the squares in that grid that touch the border of the circle ?
Is there any algorithm to compute it ?

[dawn moth]

for example :

[paper cobalt]

hard question

midpoint circle theorem could be useful but not sure

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

[dawn laurel]

maybe you could start with some smaller circles and see if different radii have anything in common

[paper cobalt]

i feel like this is a comp sci question

but id say you redraw the circle with the that algorithm i linked

if it matches perfectly you know how many iterations it takes to complete one quadrant and then u times by 4 to get amount of squares

(Also lookin at the circle u linked, its not done efficiently so that theorem would skip a few)

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@dawn moth Has your question been resolved?

[dawn moth]

okay thanks!

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[light pine]

how would u do 10^10 mod 13

[light pine]

without doing 10^10 / 13 via long division

[dire geode]

repeated squaring

10 = -3 mod 13. 10^2 = (-3)^2 = 9 mod 13

repeat until you get 10^8

[light pine]

why 10^8?

also for 10^8 i get 6561 mod 13 is that right?

[dire geode]

no point in squaring 10^8

you should be modding by 13 at every step

until your remainder is in the range -12 to 0 or 0 to 12

what's 10^2 mod 13?

[light pine]

9

[dire geode]

use the same process for 10^4

[light pine]

3 mod 13?

@dire geode

[dire geode]

how'd you get this step?

[light pine]

10000 / 13 gives 769 remainder 3

wait

[dire geode]

just square the previous value

100 = 9 mod 13

[light pine]

eh?

im confused

[dire geode]

if a = b mod n, then a^2 = b^2 mod n

have you learned that yet?

[fair geyser]

X = 10^2 mod 13
Y = (X * X) mod 13
Z = (Y * 10) mod 13
finally (Z * Z) mod 13

[light pine]

@dire geode

is the working out correct

[dire geode]

you should be reducing at every step

apply the same logic that you used to get 9 here

[dire geode]

re-read

[dire geode]

and re-read

[light pine]

@dire geode

ive worked hard on this

lol

hows it now?

[remote condor]

I love notability

[light pine]

what have i done wrong here?

<@&286206848099549185>

[random shale]

You haven't done anything wrong?

That is correct

[light pine]

Excuse me I was being thick

But anyway

LETSSSGOOOOO

I DIDNT GO TO THE LECTURE AND I LEARNT HOW TO DO THIS

LETSGOOOOOOOO

[random shale]

😂😂😂😂😂

Op bro

Best way to learn this stuff is lecture notes

[light pine]

Bro icl

My lecture notes r filled with too much text and it’s long

1. YouTube is sick. 2. Ppl on this discord help a lot

Idk what id do without this discord icl it’s helping me learn more than I do in lectures

[crimson sedge]

bro youre here 😭

[light pine]

Brev

Why u stalking me

[crimson sedge]

nah i was just vibing here and saw you in this channel

[light pine]

Nah I’m joking

As u can see

I’m on the grind

[crimson sedge]

just like the average imperial student innit

[cedar kiln]

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[hoary charm]

need urgent help

[hoary charm]

this is a matrix transformation where B is the standard basis in R3

I want to rewrite the transformation in respect to a basis {(1,0,1) , (0,1,1), (1,1,0)}

but I got the solution

which is apparently wrong

can someone tell me what mistake I made

nvm I figured it

out I just keep mixing up Q and Q^-1

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[untold hornet]

i thought it would be

,w ((26 choose 2) * (26 choose 1))/(52 choose 3)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by sytez#9920

[untold hornet]

but there's only 4 options

3/8, 2/3, 1/2, 1/8

[cedar kiln]

@untold hornet Has your question been resolved?

[untold hornet]

<@&286206848099549185>

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[untold hornet]

.closed

.close

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[vapid jolt]

Where did they get 40.8114 and 0.9750 from?
In my calculator I have Listed the years by difference in L1, and the N02(ppb) in L2

When calculating with LinReg y=ax+b

a=-0.98387
b=41.77013
r^2=0.9997905998
r=--09998952944

I need to use this to make a equation for exponential decay. Ao(1 - r)^t

Not sure if this makes any sense, any idea what I did wrong?

[broken mist]

You'll want to use ExpReg not LinReg

[vapid jolt]

Ohh, let me try that rq

I about going to double check my tables rq, B seems right

still not sure where 40.8114 came from

L1 is years and L2 is the N02(ppb)

They all seem right

solved i

solved it

I should have started L1 at 0 not 1

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[obtuse zephyr]

https://cdn.discordapp.com/attachments/490557019623915520/1035721886761504848/unknown.png

[cedar kiln]

@obtuse zephyr Has your question been resolved?

[tough valve]

what did u try?

[cedar kiln]

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[balmy yoke]

is this the final factorization

[balmy yoke]

like no more simplification possible?

[rain drift]

doesn't look like it. Looks like it's as simplified as possible

[cold briar]

,w factor 16w^3x^2 + 24wx^4 - 8wx^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by MarveI#4096

[cold briar]

yeah, i think that's it

[balmy yoke]

and how do you solve this

factor it*

I don't find a GCF

[livid hound]

apply techniques for factorising quadratics

note that this is very similar to
$$x^2 - 5x - 66$$

[wraith dagger]

ℝamonov

[balmy yoke]

oh

one sec

[balmy yoke]

no I don't

[worn wave]

it is

[balmy yoke]

its 6 AM, can barely focus fr

[worn wave]

the variables are just confusing you

[livid hound]

that wasn't a yes/no question

[balmy yoke]

ikr

[worn wave]

try rearranging the terms

and youll see it resembles a quadratic equation

the only thing that makes this “Hard” is the variables

they are just making you confused

but it’s the same thing

[balmy yoke]

wbt the GCF

there ain't one

[worn wave]

GCF is?

sorry. My syllabus doesn’t teach that

[balmy yoke]

greatest common factor

[worn wave]

you don’t need to do that

you just rearrange the terms

In into a quadratic equation

Hint: try rearranging the y variables

[livid hound]

well gcf is the first thing to look for, if there isn't one (other than 1) and/or you've already factored it out,
you continue with other methods if applciable

also expression not equation

[balmy yoke]

other methods like what 😭

[livid hound]

cross method, ac method etc

note that this is very similar to
$$x^2 - 5x - 66$$
would you have any issue factorising this simplified example?

[wraith dagger]

ℝamonov

[balmy yoke]

mhm

no I can't solve

[worn wave]

rynite

solve that quadratic express

or factor it

tell us what you got

[balmy yoke]

idk I'm used on getting the gcf, and simplifying, but nothin here

[livid hound]

have you factored any quadratics before?

if not i'd recommend looking up some vids on that first

[balmy yoke]

yes I did

[worn wave]

then it shouldn’t be a problem

[balmy yoke]

like this

[worn wave]

yes it’s like this

[balmy yoke]

but all this has a GCF

that's my problem

[worn wave]

can you show one working

using GCF

solve any questions from the picture

i wanna see how you solve using this method

since I never heard of this method, unless I did and I heard another term of it

[livid hound]

none of the questions above involve gcf

[balmy yoke]

wait

shti

I know now

sorry lmfao

[worn wave]

so what’s your answer

[balmy yoke]

two numbers that multiply to give 66 and add to become -5 right?

[worn wave]

yes

[livid hound]

not quite

[balmy yoke]

x(

[livid hound]

multiply to **-**66

[balmy yoke]

oh yeah

[worn wave]

-66 in the expression js the product of the roots

-5 is the sum of roots

[livid hound]

no

[worn wave]

what

that’s what I learnt from my textbook?

[livid hound]

sum of roots is actually -(-5) = 5

[balmy yoke]

so 6, and -11

[livid hound]

so 6, and -11
yes

hence the factorisation of
$$x^2 - 5x - 66$$
will be $(x+6)(x-11)$

[wraith dagger]

ℝamonov

[worn wave]

now try doing it with y variable

[livid hound]

the process for factoring $x^2 - 5xy - 66y^2$ is pretty much the same. \
instead of finding the part that multiply to -66 and sum to -5, \
you'd look for the pair that multiply to $-66y^2$ and sum to $-5y$

[wraith dagger]

ℝamonov

[worn wave]

so you can say the roots of this quadratic expression has y variables in it

This graph is rather unique

[balmy yoke]

so, add a y to each number?

root*

[worn wave]

what do you think? Try adding it

and see if it yields the expression

[balmy yoke]

yeah it looks correct to me

[worn wave]

now factorise it

[balmy yoke]

(x+6y)(x-11y)

?

[worn wave]

correct

[balmy yoke]

sweet

thank y'all

🫶

[worn wave]

close when you’re done

@balmy yoke type .close

[balmy yoke]

aight

.close

[cedar kiln]

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[pure grotto]

Anyone know how to save transparent images onto iPhone library?

[pure grotto]

Or iPad, but same issue

[worn wave]

this

is not

where you ask non-math questions

[pure grotto]

h e l p

[worn wave]

no

[pure grotto]

I’ll do a n y t h i n g

[worn wave]

maybe you can close this channel and ask it in #chill

instead of misusing the purpose of help channels

just type .close

[pure grotto]

No one’s gonna reply there tho

[rain drift]

g o o g l e ( or #chill )

[pure grotto]

I already googled

[worn wave]

did I say I’m not going to reply there

[pure grotto]

I’ve been trying for 1 hour

[worn wave]

either you do it now

[pure grotto]

OK

[worn wave]

or ask someone else for help

[pure grotto]

.close

[cedar kiln]

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[cinder blaze]

probabilty question stuck on the top two

<@&286206848099549185>

[rain drift]

hi there. Do you have your prob tree available?

[cinder blaze]

yeah my hand writting a bit messy but ill send one through

[rain drift]

tree looks good

so for a, it looks like there are is two scenarios where we draw out no more than 1, and that's drawing CH and HC. So I believe we just add those probabilities together

(aka P(CH) + P(HC)

[cinder blaze]

ok i will give it a go

[rain drift]

||(4/15) + (14/45) = (12 + 14)/45 = 28/45||

the above is what I got so you can double check your work

[cinder blaze]

thats 26/45 and yeah its not right sadly

[rain drift]

oh whoops yeah your right it is 26/45 lol

basic math am i rite?

wait 26/45 wasn't right?

dang maybe I just don't remember my prob lol

[cinder blaze]

the phrase no more means limeted to 1 so even pulling p(hh) counts i think

[rain drift]

ugh damnit you're right

there's an implied 0 there

so yeah P(HH) + P(CH) + P(HC)

(these always used to trip me up in prob and stats)

[cinder blaze]

yeah this is in the engnering maths course

[rain drift]

well the good news is that b will be similar

[cinder blaze]

dam still not right ahaha

[rain drift]

one or more implies P(CC) + P(CH) + P(HC) or 1 - P(HH)

[rain drift]

...wut?

does it give you the correct answer?

[cinder blaze]

yeah i dont think u can add down the end of the trees since they are conditional

[rain drift]

welp

I think I need to swallow my pride and revisit prob and stats lol

[cinder blaze]

thanks for ur help though

no dont its the worst math topic

[rain drift]

yeah sure, sorry I couldn't help

hmmm, well if you get it wrong does it let you know what the correct answer was? Mabye we can reverse engineer the answer lmao

[cinder blaze]

nah it dsont just that u got it wrong

[rain drift]

BOO

BOOOOO

[cinder blaze]

<@&286206848099549185>

.close

[cedar kiln]

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[cinder blaze]

stuck on probability question

<@&286206848099549185>

[cedar kiln]

@cinder blaze Has your question been resolved?

[cedar kiln]

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[heady salmon]

what do you mean greater then or equal to zero?

[heady salmon]

k can equal 1,2,3,4,5
but why does this stop only at k=2 and not go on to k=5

the largest Radian is 7pi over 3 which is 420 degrees

[cedar kiln]

@heady salmon Has your question been resolved?

[heady salmon]

.close

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[cedar kiln]

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[native flint]

I'm trying to find all the roots. This is what I'm doing.
$-x^3 + 6x^2 - 12x + 8 = 0 \rightarrow x^2(-x + 6) + -4(3x -2) = 0 \rightarrow (x^2 - 4)(-x + 6)(3x-2) \rightarrow (x - 2)(x + 2)(-x + 6)(3x-2)$

[wraith dagger]

OceanBro

[native flint]

jeez

$-x^3 + 6x^2 - 12x + 8 = 0 \rightarrow x^2(-x + 6) + -4(3x -2) = 0 \rightarrow (x^2 - 4)(-x + 6)(3x-2) \rightarrow (x - 2)(x + 2)(-x + 6)(3x-2)$

[wraith dagger]

OceanBro

[native flint]

But this doesn't seem right. Considering the fact that the polynomial has a degree of 3. But with my factoring, I'd have a degree 4 polynomial

[misty rain]

Your third step seems a bit wrong

check it again

like second to third step

[native flint]

Is it?

I shouldn't add the x^2 - 4 ?

[misty rain]

factor out x^2(-x + 6) + -4(3x -2) = 0
then factor out (x^2 - 4)(-x + 6)(3x-2)

see what happens

[native flint]

yeah it's (x-2)(x+2)(-x+6)(3x-2) no?

[misty rain]

sorry not factor i meant expand

mbmb

[native flint]

what should I expand

[misty rain]

both

then compare

[native flint]

I'm...getting the same thing

x^2(-x + 6) + -4(3x -2) = 0

becomes into the original equation

Is my perception skewed?

[misty rain]

what about the second equation

you can tell just from looking at it that (x^2 - 4)(-x + 6)(3x-2) expands to a polynomial with a degree of 4

[native flint]

Right

and it shouldn't

since the degree is 3 of the original one

But if I were to factor out only a single x

then I'd have x(-x^2+6x)

[misty rain]

yes

so you know your third step is wrong

[native flint]

what is wrong tho

[misty rain]

look at your third step

[native flint]

I'm checking but I don't know what is wrong

[misty rain]

you go from x^2(-x + 6) + -4(3x -2) to (x^2 - 4)(-x + 6)(3x-2)

[native flint]

yes

Is that wrong

I'm so confused

[misty rain]

the whole factoring thing has to be equal to itself

[native flint]

Nathaan, question, would you give it a try and solve it?

[misty rain]

so any step in your proof (if factored) has to be equivalent to the original equation

[native flint]

And tell me what you get as your factored answer

[misty rain]

that defeats the purpose

ill try to explain it though

[native flint]

I mean, I did like 5 of these questions and this is the only one that's not getting to my head

I know the concept, I just dont know why this particular one is like this

or where I'm messing up, since all the steps seem to add up in my head

and on the paper

[misty rain]

what math level are you

algebra?

precalc?

[native flint]

precalc,yes

[misty rain]

are you aware of difference of cubes?

[native flint]

What is that

a³ - b³ = (a - b) (a² + ab + b²)

This?

[misty rain]

and when you expand it more?

[native flint]

(a-b)(a+b)^2

[misty rain]

expand

not factor

[native flint]

The equation is (x-2)^3 ?

[misty rain]

(a - b) (a² + ab + b²) expand this

[native flint]

a^3 - b^3

Um

You know what

[misty rain]

wait fk-

[native flint]

I'll just skip this question, thanks man!

[misty rain]

im tripping

[native flint]

One question wont kill the teacher

[misty rain]

i gave you the wrong formula

[native flint]

no tha'ts me

[misty rain]

not difference of cubes

[native flint]

[misty rain]

cube of differences

[native flint]

It's fine man, I'll go get some coffee

[misty rain]

like (a-b)^3

[native flint]

yes

that's (a-2)^3

which is the answer, thank god

.close

[cedar kiln]

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[simple flare]

Can anyone tell me where I’m wrong for this?

The final answer should be the RHS upper top

[misty rain]

What is the context

and what are you trying to find

[simple flare]

I need to find the general solution from (1)

<@&286206848099549185>

[cedar kiln]

@simple flare Has your question been resolved?

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[crimson sedge]

i dont get sine waves can some1 give me a quick crash course about it

[ivory mauve]

YO

so

[toxic moat]

,w graph sin

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Circle#0025

[toxic moat]

quick crash course done

[crimson sedge]

thank you

.close

[cedar kiln]

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[sacred grail]

ah yes

[zenith sail]

wow i guess that was very effective

[sacred grail]

chapter 4J graphs of trigonometric functions

[crimson sedge]

ye lmao

[sacred grail]

cambridge university press

page 341?

[crimson sedge]

we use it for personal study

[crimson sedge]

ye

[cedar kiln]

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[oblique lynx]

I have this linear map between two linear spaces and I have to find map matrix with reagards to B and C, how to do thath?

[crimson sedge]

what does a general vector (a,b)^T in B get mapped to?

[cedar kiln]

@oblique lynx Has your question been resolved?

[oblique lynx]

ehm, whats T?

[crimson sedge]

transpose

was too lazy to write out a column vector

so basically, what is $f(\begin{pmatrix}a\b\end{pmatrix})$?

[oblique lynx]

like (b,a)?

ive never done anything with transpose matrices

[crimson sedge]

dont worry about the transpose

i'll write it out in full from now on

[oblique lynx]

okeyy

[wraith dagger]

Toby

[oblique lynx]

well I can tell that it maps to R^3

and it should be (0,x,y), right?

[crimson sedge]

yup

$f(\begin{pmatrix}a\b\end{pmatrix})= \begin{pmatrix}0\a\b\end{pmatrix}$

[wraith dagger]

Toby

[crimson sedge]

now what size matrix is required to make this transformation

[oblique lynx]

yes

also, can i have quick question before we continue?

[crimson sedge]

sure

[oblique lynx]

what does actually Lin(R,R^2) means? like is it that I take something from field R and make it a 1x2 matrix? We havent really done this too much, so its my first time working with space of linear maps

[crimson sedge]

I've never seen that notation before, so not sure

but I think it means R^2 as an R vector space

[oblique lynx]

oh yes, we use alot of different notation here so ig that its something like that xd

[crimson sedge]

so yes, its the space of 1x2 matrices (or length 2 column vectors)

[oblique lynx]

alright

so no I have to find the maps matrix with rexpect to basis B and C, how to approach that

[crimson sedge]

yeah, firstly, what is the size of this matrix?

[oblique lynx]

i think that $\mathbb{R}^{2x3}$

[wraith dagger]

marejak023

[oblique lynx]

oh god, nvm, you got the point hopefullly

[crimson sedge]

I think its 3x2, but yeah

[oblique lynx]

i thouth that first are rows then columns

also

ive drawn a commutation diagram, which helps a bit, ill send it here

[crimson sedge]

yeah, number of rows then number of columns

so we have $\begin{pmatrix}\cdot&\cdot\\cdot&\cdot\\cdot&\cdot\end{pmatrix}\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}0\a\b\end{pmatrix}$

[wraith dagger]

Toby

[oblique lynx]

here is the diagram

[oblique lynx]

yes

[crimson sedge]

nice diag!

[oblique lynx]

thanks ^-^

my prof uses them alot and now they actually makes sense to me, it is a useful tool

[crimson sedge]

yeah, its a good visualisation to figure out what goes where

what must the two circled dots be here?

[oblique lynx]

1 0

noo

wait yes

[crimson sedge]

yeah 1,0

how about the third row?

[oblique lynx]

alright

0 1

[crimson sedge]

yup

first row?

[oblique lynx]

and first 0 0

[crimson sedge]

exactly: $\begin{pmatrix}0&0\1&0\0&1\end{pmatrix}\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}0\a\b\end{pmatrix}$

[wraith dagger]

Toby

[crimson sedge]

thats the answer now :)

[oblique lynx]

cool, so now we have map matrix

[crimson sedge]

note that the matrix is very familiar

[oblique lynx]

wait

[crimson sedge]

it looks just like the two vectors in the q

[oblique lynx]

shouldnt we calculate the coordiantes in respect to the basis?

or is this it

[crimson sedge]

yeah, since (1,0) and (0,1) is a basis, we could have jumped straight to the answer by putting these two together:

[oblique lynx]

cool

and if the basis were something non-canonical

[crimson sedge]

if we dont have (1,0) and (0,1) then we have to do some row reductions to make it canonical

[oblique lynx]

then i would caluclates coords of the images

okeyy

cool, thanks for your help

.close

[cedar kiln]

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@light pine Has your question been resolved?

[cedar kiln]

@light pine Has your question been resolved?

[neon moon]

Start by finding the cycle of 9 in modulo 7.

For example
9^1 mod 7 = a = …
9^2 mod 7 = b = 9a mod 7 = …
9^3 mod 7 = 9b mod 7 = …

See if the pattern repeats, and locate the cycle of such pattern to go for 9th power mod 7.

Ohhh shoot this is for (ii)…

(iii) is also “easy”. Try consider 10 mod 11 = -1, then you can obtain those powers of -1

[cedar kiln]

@light pine Has your question been resolved?

[cedar kiln]

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[opaque bane]

Suppose I draw a unit circle on a piece of paper located on the plane (x,y), then using the plane (y,z) as the axis of reflection, I fold this piece of paper in half. Now, if I fold it completely in half, the length of the projection of the circumfirence on the (x,y) plane would become 2; if I don't fold it at all, the projected circumference is pi.
Let's call this angle of folding θ, and the projected circumference C.
How can I find the relationship between θ and C?

[opaque bane]

Let's try the forum

.close

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[worn wave]

I got absolutely NO CLUE where to start.

[reef venture]

You should start with y and simplify that log expression

then convert to same base

[worn wave]

Alright hold on

[worn wave]

Simplify even further?

Never mind, someone helped me, thanks still

.close

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[pulsar crag]

guys help

[jade yew]

Hi, i thought of a weird math question that I can;t solve till now. Can you help me solve it? question:
0x2=0
5x2=10

[pulsar crag]

duh u multiply

[jade yew]

sorry i accidentally sent

its not finish yet

[pulsar crag]

can u help

composition and decomposition

[jade yew]

soli im also here to ask

[pulsar crag]

😭

[jade yew]

good luck to find a master to help you

[cedar kiln]

@pulsar crag Has your question been resolved?

[pulsar crag]

pleas ehelp

[cedar kiln]

@pulsar crag Has your question been resolved?

[pulsar crag]

guys please help

composition and decomposition

[cedar kiln]

@pulsar crag Has your question been resolved?

[cedar kiln]

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[bright fractal]

hello, can someone pls tell me how to factor this ?

[sullen gull]

Have you found the roots?

[bright fractal]

the what

[sullen gull]

A root is a value of x that makes the equation = 0

[bright fractal]

ye

1

[random shale]

so what is a factor of the polynomial?

[sullen gull]

You know how to use complex numbers?

[bright fractal]

how are complex numbers related to this

[sullen gull]

factoring through roots -> all roots are needed
and this only has 2 complex roots

(if you want to factor it fully, that is)

[bright fractal]

......what

I just want to factor it so that one of the factors is x-1

[sullen gull]

This is one way to factor it

[bright fractal]

how did u do that

[sullen gull]

This is another valid factoring of the polynomial

[bright fractal]

what is the rule for this

[sullen gull]

Polynomial division

[bright fractal]

what are the steps ;-;

[sullen gull]

https://www.youtube.com/watch?v=_FSXJmESFmQ

you divide out one factor and then use the fact that

$g(z)*\frac{p(z)}{g(z)}=p(z)$

[wraith dagger]

Tech support

[inland widget]

noo use syntehtic division @bright fractal

[bright fractal]

I don't understand anything

[sullen gull]

What level is this?
(do you know what an integral and derivative is)

[crimson sedge]

synthetic division will be more detrimental for you to rely on in the future, because it is more limited

so just use long division

[inland widget]

why

[upbeat lotus]

Find the zeros of the polynomial

Then divide polynomial by the zeros of the polynomial f(x)

Then factor the new polynomial

Another method
Middle term

[inland widget]

are integrals and derivatives relevant to the question or are you just asking

[crimson sedge]

Everything you can do with synthetic division can be done with regular long division. The reverse is not true, though. you can’t find the quotient and remainder by synthetic division when dividing by a quadratic polynomial

[inland widget]

yea so use long division when you need

[bright fractal]

please, I just wanna see a step-by-step version of how this

[bright fractal]

turned into this

[inland widget]

it takes wayyy longer to use long division theres 0 reason not use synthetic

[sullen gull]

Depending on what level this person is on different concepts can be used to explain

[upbeat lotus]

Yupp but synthetic division would be easy here

[inland widget]

can you explain using derivatives and integrals

[inland widget]

exactly

[bright fractal]

;-;

[sullen gull]

nope but I need to ask if they know some math concept

if you know what an integral is you probably can use factor theorem

[upbeat lotus]

${4x^3}+{2x^2}+{x}-{3}$

[inland widget]

yea @sullen gull whats the best way to solve this

using calc

[upbeat lotus]

Where is tex?

[wraith dagger]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[sullen gull]

$if p(a)=0 then (z-a) is a factor in p(z)

wtf what did I just write

[upbeat lotus]

@bright fractal can you find the zeros of polynomial means which f(x) should we use to get zero

[bright fractal]

......

[crimson sedge]

okay @bright fractal

so here is the thing, i dont understand why this channel became such a mess but thats fine

[upbeat lotus]

Sorry 😅

[bright fractal]

I know that one of the factors is x-1 since th elimit is x->1

[bright fractal]

I just wanna see a step-by-step image of how the function got factored into this

[crimson sedge]

when you are factoring cubic polynomials, the first thing you want to look for is to try and factor by grouping. if we had something like $4x^{3}-8x^{2}+6x-12$ then you can factor it into the form $4x^{2}(x-2)+6(x-2)$

oop

[bright fractal]

ye i tried factoring by grouping and it didn't work

[crimson sedge]

okay, just saying it now for the future

anywyas, since you cannot do factoring by grouping

you have to do trial and error

[wraith dagger]

♡LexQa♡

[sullen gull]

1. Find the root (x=1)
2. Factor theorem gives (x-1) as factor
3. Divide out this factor
4. Multiply the new thing you got by (x-1)
5. Factored

[crimson sedge]

the best way to go about it is to look at the constant term, -3, and look at its factors of +-1 or +-3

that way, you would find the root, which is your first goal

[bright fractal]

....wat

[crimson sedge]

taking x = 1, does indeed give you a root, after you test it. Next step, you have to either use long division or synthetic division for the root x = 1 or x-1 to get the root factored out

[crimson sedge]

You seem really confused about the fundamentals of the concept, so I would recommend watching this

https://www.youtube.com/watch?v=rnAKWPM-YBc

it will explain how to factor the question you showed

[bright fractal]

ohhhhhh

tysm

[crimson sedge]

Also

research about "Rational Root theorem"

[bright fractal]

Just knew abt synthetic division today

[crimson sedge]

that's okay

[cedar kiln]

@bright fractal Has your question been resolved?

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[cedar kiln]

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[errant belfry]

If someone can solve solve this I'll be surprised

[errant belfry]

√(bcx(b+c+x))+√(cax(c+a+x))+√(abx(a+b+x))=√(abc(a+b+c))

for x

[chrome tulip]

roots are confusing

√+ what does this mean ?

[errant belfry]

ignore the +

[chrome tulip]

also √.bcx means ?? root(bcm), where does it ends

[errant belfry]

ignore the .

[lilac rampart]

the answer is 2

[errant belfry]

?

[lilac rampart]

yeah

[gritty tide]

its 2i can confirm

[zenith sail]

I mean, it's possible to manipulate the equation to get rid of all the roots

You'll be left with a polynomial in terms of x, but it'll be 4th degree I think

[errant belfry]

But you'll get octic in terms of x

[zenith sail]

is it octic? That's even worse

[errant belfry]

not easily solvable unless obvious

[zenith sail]

And it'll be a giant pain to even get to that point in the first place

I think it's "possible" "maybe"

[errant belfry]

yah I did that and it wasn't worth it

there's a mountain hiding behind those roots

[gritty tide]

where did you even get this from and what is the equation even called or related to

[zenith sail]

It's octic but all the powers are even, right?

[errant belfry]

You'll need a whale sized brain to eat all the terms on the LHS

[errant belfry]

no

[zenith sail]

oh wait, no you're right

[errant belfry]

Heron's on Soddy descartes

[errant belfry]

There's a solution to the problem

but it's not easily solvable backwards with a sanity check either

[cedar kiln]

@errant belfry Has your question been resolved?

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[analog osprey]

It appears to me that they are skipping some steps here, can anyone help me figure out how they are getting to 1/6?

[long arrow]

-1 + 3 = 2

2/2 = 1

and 1^(-2/3) is 1

[wraith dagger]

Modus

[analog osprey]

oh I see. I just missed the negative in front of the 1. So I have been working with (1+3/2) on the inside. Was driving me nuts......I need a break

Thank you again @long arrow, you're saving my ass left and right today lol

[cedar kiln]

@analog osprey Has your question been resolved?

[cedar kiln]

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[fossil spire]

how tf do u prove transitive prop. I've really tried everything... Transative prop is = If angles are congrount to congrent angles then they are congrent to each other

[fossil spire]

This is geomatry btw

[gaunt hamlet]

Are you asking how to prove that if angle A = angle B and angle B = angle C, then angle A = angle C?

[fossil spire]

yea

I mean no one really knows

so imma assume its impossiable

not impossiable

but impossiable in theorem format

[gaunt hamlet]

Two angles are congruent if they have the same measure, right?

[fossil spire]

yea...

[gaunt hamlet]

So write your statement in terms of the measures of the angles

Since the measure is just a number

[fossil spire]

so you want me to assume m = x?

[gaunt hamlet]

You don't have to assume anything

If angle A = angle B, what does that tell you about the measures?

[fossil spire]

they are congrent

[gaunt hamlet]

Well, I think "equal" is the better word

Since we're comparing numbers, not shapes

In other words, mA = mB, right?

[fossil spire]

yea

[gaunt hamlet]

What about the statement "angle B = angle C?"

[fossil spire]

so I would do
a congrent to b | given
m angle A is equal to m angle B | if congrent angles then equal mesurments

given?

[gaunt hamlet]

Idk, that's just the definition of congruent angles

[fossil spire]

mhm okay

Ill try to ask my math teacher

.close

[cedar kiln]

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[rain drift]

hi there how did you get that answer?

[kind matrix]

INF

[haughty wraith]

might not just wanna give the answer lol

[rain drift]

please don't just give the answer.

[haughty wraith]

numerator has higher power than denom

im guessing thats where u went wrong

[fiery berry]

@rain drift i divide x^2 numereator and denom

so i should divide x^3

[rain drift]

yep. The trick is to divide by the highest power in the numerator and denominator

for these rational limit problems

[fiery berry]

ok i will try to find result

[crimson sedge]

there is a rule that if the exponent of x at the up part bigger than the exponent of x at the down part

[gaunt hamlet]

I think it's fine to divide x²

By the top and bottom

[crimson sedge]

if one (-),other one (+), the answer is - infinite

if both (-) or both (+), the answer is + infinite

[fiery berry]

i find denom 5/(x)^3 - 2/x

still undefined

[gaunt hamlet]

What did you get when you divided by x²