help-26

[sturdy ingot]

re

[cloud scaffold]

re ?

[sturdy ingot]

آره

yes

[cloud scaffold]

ha fehmtek

for this question I still have the same problem but I think I've got it right Let's consider the following demand function: D(P)=8-2P
At what price does demand become elastic?

[sturdy ingot]

look, there is a better way to understand elasticity

[cloud scaffold]

yes hahahha

[sturdy ingot]

elasticity is defined as the change of demand with respect to change of price

[cloud scaffold]

yes

[sturdy ingot]

say demand changes = 4 and price changes = 2

elasticity is equal to 2

do you know line equation?

a line can be defined as y=mx+b where as m is the rate of change of y with respect to x and b is the gradient (what y equals at x=0)

now the function you gave is D(P)= (-2)P + 8

we can write it just like a line equation

[cloud scaffold]

oe I see

[sturdy ingot]

y = -2x+8

now the rate of change is equal to -2

sometimes the problem will look like this: P(D)=2D+8

in this situation it is x=2y+8

[cloud scaffold]

no2

[sturdy ingot]

idk man im probably dumb

or your paper is wrong

[cloud scaffold]

I think it was just 2

[sturdy ingot]

D(P)=8+2P

?

[cloud scaffold]

no D(P)=8-2P

[sturdy ingot]

give the prices 0 and 1 to the fucntion

demands will be 8 and 6

where 8 is the first demand (demands are decresing)

change of price is 1

change of demand is -2

-2/1=-2

you can try this on any other point on the function

and you will always get the same answer

because it is a line equation

[cloud scaffold]

exactly

you master the social surplus?

[sturdy ingot]

idk. what is the problem

[cloud scaffold]

I have to deal with the social surplus the problem is Prices and quantities demanded at dates t = 0, 1 are as follows:
P0 = 8, P1 = 10 and Q0 = 30, Q1 = 20.
What is the arc elasticity of demand?

[sturdy ingot]

22 i think

sorry gotta go dont have time to explain

[cloud scaffold]

tkt

[topaz sinew]

@cloud scaffold Has your question been resolved?

[cloud scaffold]

I don't think that's it

I have to deal with the social surplus the problem is Prices and quantities demanded at dates t = 0, 1 are as follows:
P0 = 8, P1 = 10 and Q0 = 30, Q1 = 20.
What is the arc elasticity of demand?

I found 3.37

<@&286206848099549185>

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[shadow heart]

does minus infinty square equals to infinty (-∞²=+∞?)

[eternal wing]

That doesn't have any meaning

Are you doing limits maybe?

It's true that if f(x) approaches -infinity, then [f(x)]^2 approaches infinity

[shadow heart]

ofc

[eternal wing]

Then yes

[outer salmon]

You can’t really square infinity because it’s not a number, so multiplication isn’t defined for it. But if you were to extend multiplication to + and - infinity, I suppose that equation would hold true

[eternal wing]

But saying "infinity" as if it was a number is very informal

It should be clear that you mean "infinity" in terms of limit arithmetic

[shadow heart]

im calculating a limit

lim x² (x approaches - infinity)

[outer salmon]

Then yes, that’s infinity

[shadow heart]

aight thanks

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[cinder sphinx]

How do i solve this?

[keen hinge]

Do you need to prove or find solutions

[cinder sphinx]

To prove

[keen hinge]

What are restrictions for x, y ?

[cinder sphinx]

eirher they are in (0,1) or (1,+ infinity)

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[trail yoke]

I have to graph 5-3sin(2x) and I have no clue what to do with the 5-3 part

[trail yoke]

i know that you subtract 3 from y

idk what the 5 does

[mossy tree]

plus 5

sin(x) is normal we know that

is undergoing these changes

period is decreasing

multiplied by -3

and added 5

[trail yoke]

thank you

[mossy tree]

there are two -3sin(2x)

there was a mistake

but doesnt matter

thank you too

[trail yoke]

yah its g but is the final equation just 5-3sin(2x) but rearranged

[mossy tree]

yes

[trail yoke]

that makes a lot of sense ty

[mossy tree]

-3sin(2x)+5 = 5-3sin(2x)

however

you are welcome

[trail yoke]

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[white sandal]

heres my question:

[white sandal]

I get stuck mostly when I get to adding the y1 in the point slope formula. I can never tell when it wants me to add or subtract it and I've done this question for 30+ mins now

I can show some work for other problems but they keep changing the numbers in the table so you can't just keep guessing the question

<@&286206848099549185> pls help :(((

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[white sandal]

<@&286206848099549185>

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[lost sluice]

help 😭

[lost sluice]

how do you rearrange this conic equation for y 😭

-3.9955967133235 x^(2)-0.1070756903598 x y-1.0832535136997 y^(2)+20.8529602534027 x+0.0952418719278 y=13.4514503924045

[gleaming reef]

bring all the terms that don't involve y to the RHS

and factor out y

and then divide

and boom

https://tenor.com/view/ice-spice-umm-judging-gif-27091996

[shrewd flume]

Just out of curiosity what application has you rearranging a hyperspecific equation with that many decimals?

[cursive patrol]

😅 maybe some more context would be nice

[shrewd flume]

My engineering student brain just wants to truncate those numbers to the hundreths place for my sanity

[lost sluice]

😭

im trying to find uhm

the volume of an acorn.

[shrewd flume]

Ah yeah just round all those numbers and slap an error because theres no way you’re actually measuring to subatomic precision

[lost sluice]

😭

[cursive patrol]

ainnoway

[shrewd flume]

So your measurements there are $\pm 0.01$ (units) so use that

[thorny flame]

Sean

[shrewd flume]

No real reason to pain yourself with those insane decimal values you have

Because your dont have precise measurements beyond a hundredth of a unit anyway

So decimal numbers beyond that are just statistical noise fwiw

[lost sluice]

do you know if thats allowed in ib papers lol

[shrewd flume]

Whos grading it

[lost sluice]

mymath teacher 😭

i think he hates me

[shrewd flume]

Ok since it’s a human, the best thing to do is clearly mark your error throughout your calculations

Im assuming units are in inches?

[lost sluice]

yup

[shrewd flume]

Ok so your initial measurements should be thought of as “[measurement] $\pm 0.01$ in.”

[thorny flame]

Sean

[lost sluice]

oakok

how do i justify the equation 💀

like does polynomial regression work for this

or

[shrewd flume]

Now there are specific rules for error propagation but that doesnt rly matter for this

Just round everything to the hundreths for this

[lost sluice]

okaoy epic !!

[shrewd flume]

This shape seems a lot more involved than a polynomial could fit

[lost sluice]

oh 😭

(what does that mean)

[shrewd flume]

Like are you talking about fitting a polynomial to the edge of the acorn to get an equation for its shape?

[lost sluice]

should i split it into smaller sections

[lost sluice]

yeah

the individual sections tho

[shrewd flume]

Yeah each individual section seems simple enough of a curve to work

[lost sluice]

are you familiar with ib requirements by any chance

[shrewd flume]

From there you can use the ol’ rotate around the x axis integration for each section and get a volume

[shrewd flume]

No sorry

[lost sluice]

yeah that was the plan

[shrewd flume]

Didnt take ib in high school

[lost sluice]

but idk how to justify my equations

why they’re the degree that they are

💀

[shrewd flume]

Wdym by justify in this context

[lost sluice]

like why is it quintic

and not cubic

😭

[shrewd flume]

I mean you could compare a cubic regression’s R squared error to that of a quintic

And show how much closer it fits the dataset

And if it’s not a negligible difference that’s justifiable

[lost sluice]

doesnt the R2 just get closer and closer to 1 as you increase the degree

oh

okok

[shrewd flume]

Like yeah more parameters = better tuning but again going back to my error comments earlier that also should be within reason

[lost sluice]

okay 😭

thank you !!

[shrewd flume]

Tbh you could get away with parabolas for the middle sections

Those seem rather parabolic

[lost sluice]

yeah i did parabolas at first

conic fits better but i might just change it back

[shrewd flume]

What was the difference in R2?

[lost sluice]

idk i didnt check

but like i made the parabola degree really high

so the fit was probabkly better

anyways i have to go now thank you for the help

[shrewd flume]

Conics seems like overfitting ngl

Alright nw

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[lone burrow]

Why is this wrong? Both graphs are less then y 6

[scenic chasm]

If I’m reading this right, the range of both of your graphs need to be the set of all numbers less than 6

In both graphs the range indeed includes only numbers less than 6 - but you’re supposed to use every single number that is less than 6

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[twin pollen]

Why is the answer not factored?

[twin pollen]

Cant i factor numurator?

Numerator

[keen venture]

Yep you can!

Are you asked to?

[twin pollen]

My answer key didnt

[twin pollen]

It didnt say factor or anything

[keen venture]

Should it have?

[twin pollen]

Idk

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[nocturne echo]

In (c) how is the range from 0 to pos inf?

we can never get postive values (except 0 and 1) from the f o g equation

when you plug in 0 you get 1, when 1 you get 0, when 4 you get -1

it goes to neg inf from 1

[abstract wadi]

You get what with 4?

[nocturne echo]

oh shit

my bad, thanks

[abstract wadi]

Also, it's \
$(\sqrt{x} -1)^2$

Might make it easier to find the range.

[thorny flame]

! What the hell am I doing here?

[nocturne echo]

got it thanks

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[neon iron]

i don’t know how do i write my answer because it had x10^12

[cursive patrol]

i don't think the intended answer is that large

[neon iron]

you add all the books right?

then 15!

[fair thorn]

i think so

wait-

hang on

are the books the same?

[neon iron]

nope

[fair thorn]

are you sure?

what i mean is, are all the algebra books the same

are all the calc books the same

etc

[neon iron]

yeah

[fair thorn]

well that's smth else altogether

so not 15!

[neon iron]

do i do 4! 5! and 6! ?

there’s an example like this one in my book but the amount of each books are different

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[wanton stone]

Can someone explain me the second part of this exercise?
Why is it just 2 integrals?
Completely bamboozled rn

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[wanton stone]

<@&286206848099549185>

Anyone can give an explanation or maybe see a mistake here ?

[long stirrup]

the answers are obvious

they are maybe like preparing you for more complicated problems where it's not obvious, so when they do integrals it's necessary but familiar

[wanton stone]

Actually it's not obvious for me, cause i do understand the part of P(A ∩ B)/P(B), but i dont understand why it's P(A)*P(B)

Sure if it's A and B are independent you can write P(A ∩ B) = P(A)*P(B) , but where is the "/" P(B)

@long stirrup i just miss this part.

[long stirrup]

the bus arrives uniformly between 10:15 and 10:30

part b is not different from part a

i can't help with this, it's not your fault

[wanton stone]

Thank you in particular, gonna have to rethink about it, maybe i get it somehow.

[long stirrup]

i see what you mean now, they multiply 2 integrals

instead of division by 1/2 it looks like multiplication by 1/2

yeah that's a question mark

[wanton stone]

If i remember correctly P(A|B) is defined as = P(A ∩ B)/P(B) and if P(A ∩ B) = P(A)*P(B) if independently you should get P(A)

Interesting fact:
In the exercises has been already in the past examples some mistakes he made.

[long stirrup]

must have meant this

[wanton stone]

Sure i already considered that, but i dont know. It doesnt really hold with what in here i think. Maybe i miss some important detail.

[long stirrup]

the numerator is not inrependent events, one is completely inside of another

their intersection is the first event

[wanton stone]

AHHHH, i see it now. Omg, it was so easy and direct in front of myself

Thank you so much. Dammit thought too complex.

Didn't considered the intersection of the two events.

[wanton stone]

https://tenor.com/view/stickergiant-you-deserve-a-trophy-trophy-first-place-gold-medal-gif-24960237

[long stirrup]

the gif will do

[wanton stone]

Have a nice day and thank you so much!

.close

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[visual sage]

look at the units, they give you: theta/time

this is correct

@neon iron

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[visual sage]

oh righy i onlky saw the second pic

its limit will be zero

no

it will tend to zero*

so treat it as dx

could be wrong as fucjk but:

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[bright carbon]

I'm just tryin' to solve it, I'm still a beginner in calculus 2

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[vivid mauve]

I need some help

[vivid mauve]

I hope someone can help me today

My problem is

The following information is provided about a function
Dm(f) = {x E Rix -::f:.-61\x-::/:- 6}
f(-1 1) = 0, f(-7) = 0 and f(-3) = 0
f(O) = 2
f(x) > 0 in the intervals]-11; -7[,]-3; 6[ and ]6; oh [
f'(-9) = 0, f'(0) = 0 and f'(9) = 0
fer growing in the intervals ]-oo;-9], ]-6; 6 [ and ] 9; oh [
a) Write down the coordinates of the curves of the graph with the coordinate axes.
b) Sketch the graph of a function that fits the given information.

So we have to draw a graph where the following points should stand

[topaz sinew]

@vivid mauve Has your question been resolved?

[vivid mauve]

Hello...

It's my birthday friends and I hope you will spend some time on this.

<@&286206848099549185>

[dapper furnace]

happy birthday my man! wish you all of best luck!

X axis: -11,0, -7,0, -3, 0
Y axis: 0,2

here you go man

@vivid mauve

[vivid mauve]

Thank you very much!

[vivid mauve]

How do you know?

[dapper furnace]

big brain,

dw, im right

nah, i'd win/

[vivid mauve]

What do I do after that?

[dapper furnace]

well plot the points'

and done

[vivid mauve]

a) Write down the coordinates of the curves of the graph with the coordinate axes.
b) Sketch the graph of a function that fits the given information.

[vivid mauve]

We have task a and b

What kind of one have we made?

[dapper furnace]

ok look

b

take a graph

and plot those poitts

[dapper furnace]

.

[vivid mauve]

Which program should I use?

[vivid mauve]

Do you know geogrobra?

How should I insert them?

[dapper furnace]

your copybook

[vivid mauve]

copybook?

[dapper furnace]

yes, your copybook

[vivid mauve]

do not know it

but should I not use geogrobra

[dapper furnace]

not rly draw a graph in ur notebook

and plot the points

[vivid mauve]

So I have to draw it? What should the points be?

[dapper furnace]

.

[vivid mauve]

So I take a ruler and puts the points on?

[dapper furnace]

first, draw this graph

but make each side max number 14

[vivid mauve]

Do I need tenge for 11?

Ahh okay

14 max right?

[dapper furnace]

and take the cords

yes

here

just the left point draw 11 there

please type .close

@vivid mauve type .close

[vivid mauve]

Bro we are not done

[dapper furnace]

I LITERALLY GAVE YOU THE ANSWERS

[vivid mauve]

In b does it say that we must make a function regulation?

[dapper furnace]

that is the answer.

the rest is on you, i quit.

[vivid mauve]

What

We need to make a function that describes the coordinates?

<@&286206848099549185> is there anyone else who can help.

[radiant tangle]

Can u draw the line on it

Then try it on the GDC calculator idk

Graphic display

I’m actually dumb

I didn’t scroll far up enough

[vivid mauve]

Should I do as he said?

I have the points up there and are they correct?

I think it is task a that has been done further up

Task b we have to create a function

[radiant tangle]

Can u send the initial question again

[vivid mauve]

Yesss

The following information is provided about a function
Dm(f) = {x E Rix -::f:.-61\x-::/:- 6}
f(-1 1) = 0, f(-7) = 0 and f(-3) = 0
f(O) = 2
f(x) > 0 in the intervals]-11; -7[,]-3; 6[ and ]6; oh [
f'(-9) = 0, f'(0) = 0 and f'(9) = 0
fer growing in the intervals ]-oo;-9], ]-6; 6 [ and ] 9; oh [
a) Write down the coordinates of the curves of the graph with the coordinate axes.
b) Sketch the graph of a function that fits the given information.

soooo

Can someone help?

.close

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[neon iron]

.close

.ask

[frail kraken]

Can I ask a question in here

[frail kraken]

Math teacher here trying to teach this to 8th graders for the academic team

.close

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[jolly patio]

i need help with this one

[jolly patio]

i know the total % of Bel, Cherry, and spring is 69.4%

and there's 38 employees scattered into the stores

<@&286206848099549185>

[sleek aurora]

if 69.4% employees work in those places then how much will now work there

[jolly patio]

100-69.3=30.6?

[sleek aurora]

yep so what is the total employee count

[jolly patio]

total employee count is 38

[sleek aurora]

that's just salaried employees

what about hourly employees

[jolly patio]

ok so my question is what is the difference between salaried and hourly

is salary the amount you get per year

and hourly is the amount per hour

a person gets

[sleek aurora]

for salaried they get a fixed amount of money based on their work and hourly workers get their salary based on how many hours they have worked for

[jolly patio]

ok so there was a prior question i had which was this

how do i know if they are talking abotu a salaried or hourly

because here i did 5:38

[sleek aurora]

if they don't mention hourly or salaried then include both if they mention then use the one that's mentioned

[jolly patio]

so for my csse, do i use both salary and hourly, or just salary

[sleek aurora]

it's not mentioned so use both

[jolly patio]

wait i gtg bc i have class but ill come back

.close

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[rustic turtle]

im not really sure how to start/what to do here

i started calculating $\frac{dy}{dx}$ for the $y = \sqrt{ \frac{x}{x+1} }$

[thorny flame]

la garce

[rustic turtle]

but im not really sure if thats the right thing to do

[haughty mural]

afaik, we can say that the gradient of the normal to y will be the inverse of dy/dx

1/(dy/Dx)

and this can be rearranged into the form you are given

[rustic turtle]

ohhh ok

so $\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy} = -1 \sqrt{px(x+1)^q}$

[thorny flame]

la garce

[haughty mural]

yep

[rustic turtle]

is dx/dy the correct notation there?

[haughty mural]

depends

in mathematics, no

because d/dx is not a fraction but an operator

[rustic turtle]

yeah

so is it best to write it $(\frac{dy}{dx})^-1$

[haughty mural]

but often times it functions like a fraction

[thorny flame]

la garce

[haughty mural]

yeah

[rustic turtle]

alright

is it still necessary to calculate dy/dx using the first given equation?

i guess so, and then subsitute that into $(\frac{dy}{dx})^-1$

[thorny flame]

la garce

[haughty mural]

yep

[rustic turtle]

thank you

let me see how it goes before i close it

$\frac {(\frac{x+1}{x})^\frac{1}{2}} {2(x+1)^2}$

[thorny flame]

la garce

[rustic turtle]

this is dy/dx

my arithmetic has been terrible today

[haughty mural]

((x+1)/x)^1/2 = sqrt(x+1)/sqrt(x)

[rustic turtle]

how can i simplify it further

alright, but what after that?

[haughty mural]

we have (x+1)^something in numerator and denominator

a^b/a^c = a^(b-c)

btw

i made a mistake

it is -1/(dy/dx)

[rustic turtle]

$\frac{1}{2x^\frac{1}{2}(x+1)^\frac{3}{2}}$

[thorny flame]

la garce

[rustic turtle]

[ 2 * sqrt(x+1)^3 * sqrt(x) ]^-1

i dont think its possible/necessary to simplify further here

[haughty mural]

there should be a minus

and we can put all that into a sqrt

a^1/2 * b^3/2 = sqrt(a * b^3)

because b^3/2 = (b^3)^1/2 = sqrt(b^3)

[rustic turtle]

$(2\sqrt{x}\sqrt{(x+1)^3})^{-1}$

[thorny flame]

la garce

[rustic turtle]

thats what ive got so far

[haughty mural]

remember that what we got here is dy/dx

[rustic turtle]

yeah

[haughty mural]

we now want -1/(dy/dx)

[rustic turtle]

yep

[haughty mural]

but yeah, looks good i think

[rustic turtle]

alright cool

its an open book test, and its worth 8 marks out of 50

its not that bad if i only get 1/2 of the question right, im quite confident the rest is correct anyway

and the marks dont matter toward a final year mark anyway

alright, so

$-1(2\sqrt{x}\sqrt{(x+1)^3})$

[thorny flame]

la garce

[rustic turtle]

thats the form i need it in

is $-1(\sqrt{\sqrt{2}x(x+1)^3})$ incorrect?

[thorny flame]

la garce

[rustic turtle]

the problem there though is sqrt(2) is not an integer and it specifies p and q are both integers

i followed the same kind of logic as simplifying $\sqrt{9 \times 2} = 3\sqrt{2}$

[thorny flame]

la garce

[rustic turtle]

oh wait, it just has to be rational, i misinterpreted Q as Z

so perfect, p = sqrt(2) and q = 3

@haughty mural thanks for all your help, really appreciate it

[haughty mural]

wait

sqrt(2) is not rational

so we would wanna keep it outside the root

[rustic turtle]

oh right that is true

[haughty mural]

oh wait

we cant

[rustic turtle]

yeah

has to be in this form https://cdn.discordapp.com/attachments/903486479064522752/1204152531371040838/338269a15ad05478ea111bbbf45377bf.png?ex=65d3b16f&is=65c13c6f&hm=0717f4e3a48d1db140e53712a88b60470e7da15ffd676eca92989817af4a2903&

there was probably one minor mistake made somewhere

perhaps instead of 2 it should have been 4

[haughty mural]

wolframalpha gives this for dy/dx

[rustic turtle]

i probably used the wrong rule then

[haughty mural]

ohhhh

[rustic turtle]

i used chain rule with quotient rule

[haughty mural]

im stupid haha

nah, your dy/dx was correct

[rustic turtle]

whew

[haughty mural]

we had a 2 infront of the root

2=sqrt(4)

and 4 is in Q

[rustic turtle]

ohhhhhhhhhh youre right

[haughty mural]

haha

[rustic turtle]

oh my god 💀

[haughty mural]

yeah haha

[rustic turtle]

i was worried my dy/dx was wrong from the start

it would have been an absolute pain to restart

alright, well thanks for all your help, i appreciate it a lot

.close

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[maiden star]

this should be 1 tailed

but it seems like 2 tailed

idk how

shouldnt it be 1 percent

@knotty finch

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@maiden star Has your question been resolved?

[topaz sinew]

@maiden star Has your question been resolved?

[maiden star]

<@&286206848099549185>

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@maiden star Has your question been resolved?

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@maiden star Has your question been resolved?

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[regal peak]

HELP!

[sharp dew]

It is (2w)^3 - 5^3

I mean a^3 - b^3

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[regal peak]

8. I DONT UNDERSTAND

[hollow drum]

What's the goal?

[regal peak]

😢

how do you solve it

is there a formula

[hollow drum]

There's nothing to solve

[stone verge]

factorize it

[regal peak]

yes

[hollow drum]

So the goal is to factor 8w^3 - 125?

[pseudo bear]

That's a perfectly fine polynomial, so there's nothing to solve unless you're given an instruction like factoring it or something like that.

Can you show the instructions above the section that question h is in?

[regal peak]

yes

[pseudo bear]

Oh, OK.

[regal peak]

how do you factor it fully

😢

[thorny flame]

Chai T. Rex

[pseudo bear]

So, let's start there.

What's the cube root of 8w^3?

[regal peak]

hmm

(2w)^3

[pseudo bear]

OK, what's the cube root of 125?

[regal peak]

5

5^3

[pseudo bear]

OK, so we have (2w)^3 - 5^3.

So, with a^3 - b^3, a = 2w and b = 5.

So, one factor will be a - b.

So, one factor is 2w - 5.

[regal peak]

ooooo

[pseudo bear]

Does that make sense so far?

[regal peak]

is it because the formula

[pseudo bear]

It's because whenever you see a^n - b^n with the same exponent on both and a subtraction, one factor will always be a - b.

[regal peak]

what about 4

instead of 3

like 5^4

or 5^6

it works the same right?

ohhhh

[pseudo bear]

Well, if you have a^4 - b^4, one factor will be a - b.

[regal peak]

its because the exponents cancel away

OHHHH

[pseudo bear]

Same with any exponent, as long as you're subtracting the powers and the exponents are the same.

[regal peak]

I GET IT

[pseudo bear]

The exponents don't exactly cancel.

[regal peak]

only for subtracting?

[pseudo bear]

Yes.

[regal peak]

not adding?

okay

[pseudo bear]

And then the other factor will be like (a^2 + ab + b^2).

[regal peak]

how do you know there is only 2 factors

[pseudo bear]

There's more.

This is just the first factoring.

[regal peak]

😢

[pseudo bear]

Then, you factor the (a^2 + ab + b^2).

[regal peak]

okay

then what

then its done?

[pseudo bear]

Do you want to learn how to get the other factor for any exponent?

[regal peak]

yes

[pseudo bear]

Yes, after that, it's done.

[regal peak]

yay

[pseudo bear]

OK, so let's say you have a^5 - b^5.

One factor will be a - b.

The other factor will be a^4 b^0 + a^3 b^1 + a^2 b^2 + a^1 b^3 + a^0 b^4.

See how the exponents on a start at 5 - 1 and go down to zero?

And the exponents on b start at 0 and go up to 5 - 1?

[regal peak]

why

[pseudo bear]

a^4 b^0 + a^3 b^1 + a^2 b^2 + a^1 b^3 + a^0 b^4

Well, first, do you see that pattern?

[regal peak]

yes but how come it wasnt in the other one

[pseudo bear]

What other one?

[regal peak]

this

[pseudo bear]

Oh, well, this is for when you have a^5 - b^5 with a 5 exponent.

You have a 3 exponent.

a^3 - b^3.

So, the other factor will start with 3 - 1 as the exponent on a.

So, that's 2.

So, a^2 b^0 + a^1 b^1 + a^0 b^2.

Which is a^2 + ab + b^2.

So, you subtract one from the matching exponents.

You put that on a.

You put 0 on b.

Then you go down by one each time for a's exponent and up by one each time for b's exponent.

So, if you have a^3 - b^3, you get (a - b)(a^2 b^0 + a^1 b^1 + a^0 b^2) = (a - b)(a^2 + ab + b^2).

If you have a^7 - b^7, you get (a - b)(a^6 b^0 + a^5 b^1 + a^4 b^2 + a^3 b^3 + a^2 b^4 + a^1 b^5 + a^0 b^6) = (a - b)(a^6 + a^5 b + a^4 b^2 + a^3 b^3 + a^2 b^4 + a b^5 + b^6).

Does the process make sense?

[regal peak]

😟

😢

no

its okay

[pseudo bear]

OK, let's try it with you doing it.

That might help.

[regal peak]

i dont need to learn that yet

[pseudo bear]

Oh, OK.

[regal peak]

im in grade 12

yes

[pseudo bear]

Well, back to your problem, you have (a - b)(a^2 + ab + b^2).

So, fill in a and b there. What do you get?

[regal peak]

(3x-5)(3x^2+3x(5)+5^2)

[pseudo bear]

Almost. a = 2w.

[regal peak]

whoops

(2x-5)(2x^2+10x+100)

[pseudo bear]

One thing I recommend when you're substituting is to surround what you put in with parentheses.

So, like ((2w) - (5))((2w)^2 + (2w)(5) + (5)^2).

[regal peak]

okay i will do that next time

[pseudo bear]

The reason why is because (2w)^2 = 4w^2, not 2w^2.

Parentheses help to keep everything straight.

So, you have (2w - 5)(4w^2 + 10w + 25).

And then you can factor 4w^2 + 10w + 25.

Still there?

@regal peak

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@regal peak Has your question been resolved?

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[graceful leaf]

what would be best way to approach this one

Like

30t+50t?

t+2?

i got 3 hours

i hate how i dont have answer key i cant even tell if id di right 😭

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[sharp dew]

what's wrong with my answer

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[wooden widget]

pls

[wooden widget]

idk anything

[mild fractal]

do you know about integration?

[wooden widget]

yes

[mild fractal]

ok

so we have to integrate the velocity

,tex $$s(t) = \int v(t) \dd t$$

[thorny flame]

@mild fractal

[mild fractal]

s(t) is a common notation for position
v(t) is a common notation for velocity

[wooden widget]

is like -1 and 3 the bounds?

[mild fractal]

,tex $$s(t) \int_{-1}^3 \left(t^3 + 29t^2 - 30t\right) \dd t$$

[mild fractal]

i forgot abt that sry

[wooden widget]

no ur all good

[thorny flame]

@mild fractal

[mild fractal]

@wooden widget heres the power rule of integration that will help you

[wooden widget]

do i just evaluate the intergral

[mild fractal]

you do

do you know how to do that?

[wooden widget]

512/3

is the distance the derivative of the intergral

[mild fractal]

yeah thats correct

[wooden widget]

uhh

so

3t^2 + 29 x 2 t - 30?

isnt it 0

[wooden widget]

?

bc its a constant with respect to x

[mild fractal]

the particle's positon is 512/3 units from its starting point

[wooden widget]

isnt it 0

[mild fractal]

nope

[wooden widget]

im confused

[mild fractal]

,tex Net Displacement: $s(t) = \int_{-1}^3 \left( t^3 + 29t^2 - 30t \right) \dd t$

[thorny flame]

@mild fractal

[wooden widget]

Yes

thats ir original intergral

so then what

[wooden widget]

?

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@wooden widget Has your question been resolved?

[wooden widget]

<@&286206848099549185>

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[vapid notch]

confusing me

[vapid notch]

no mention of scale compared to graph

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[analog crown]

How do i prove that if A is the dedekind cut for root(2), then AxA = 2?

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[silk gyro]

How do I solve (i)

[silk gyro]

we're supposed to solve only from these

[silk gyro]

And this is the question

[loud oasis]

you'll need a formula involving arctan

[loud oasis]

yes, use this formula

[silk gyro]

i dont even know where to begin

how do i use it

[loud oasis]

a can be any number

so find the value of a that makes that formula equal to your integral

[silk gyro]

hmm

[topaz sinew]

@silk gyro Has your question been resolved?

[twin canyon]

any helps?

[silk gyro]

yes

.close

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[agile tree]

can someone solve this w me step by step ty

[worthy storm]

for both questions, a reasonable approach would be to first find T(1,0) and T(0,1)

[agile tree]

how do i do this 😭 sorry

[worthy storm]

work out how to express (1,0) as a linear combination of (1, -1) and (2, 1)

then do the same for (0,1)

[topaz sinew]

@agile tree Has your question been resolved?

[topaz sinew]

@agile tree Has your question been resolved?

[vernal matrix]

@agile tree

[vernal matrix]

Are these steps alright? Do you see why we'd want to do them?

[agile tree]

I’m back sorry!

I’m going to try that rn

I don’t understand why tho :0

[vernal matrix]

If you get (1,0) as a linear combination of those two, then by linearity you can find T((1,0)) (same idea for (0,1))

You then would know how T acts on the standard basis vectors, which hopefully should easily get you the matrix

[agile tree]

okay!

[vernal matrix]

(let me know when you have the standard basis vectors in terms of (1, -1) and (2, 1) )

[agile tree]

okk

[agile tree]

is this a matrix?

i got 2 same matrices

[vernal matrix]

Well I mean you could create the basis transition matrices, but more the idea was you find $a$ and $b$ such that $\pmqty{1 \ 0} = a \pmqty{1 \ -1} + b \pmqty{2 \ 1}$ (similar the $c$ and $d$ such that $\pmqty{0 \ 1} = c \pmqty{1 \ -1} + d \pmqty{2 \ 1}$)

[thorny flame]

@vernal matrix

[agile tree]

i turned it into a matrix and solved it, but i dont know how to get a and b from the solved matrix :0

[vernal matrix]

How did you do it?

Hehe, mind reader

[agile tree]

oops i can simplify one step further

ohh

a and b are both 1/3

all a b c d 1/3 right

[vernal matrix]

Yep, sounds good

c and d I think are ever so slightly different

[agile tree]

oh okay

-2/3 ?

d 1/3 ?

wait]

c= -2/3 and d = 1/3

[vernal matrix]

Looks good to me

[agile tree]

okayy

[vernal matrix]

So now, you know that $T$ is linear, and that $\pmqty{1 \ 0} = \frac13 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1}$ and also that $\pmqty{0 \ 1} = -\frac23 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1}$

so can you find $T\qty( \pmqty{1 \ 0} )$ and $T\qty( \pmqty{0 \ 1} )$ from there?

[thorny flame]

@vernal matrix

[agile tree]

i dont know how to use what we just found for the T part

[vernal matrix]

You know what linearity means, right?

[agile tree]

hmm not really now that u say

[vernal matrix]

Naughty

[agile tree]

sorry prof chartbit

[vernal matrix]

Anyways, basically, it means that for any two vectors $\mathbf{v_1, v_2}$ you have $T(\mathbf{v_1} + \mathbf{v_2}) = T(\mathbf{v_1}) + T(\mathbf{v_2})$ , and that if you have a vector $\mathbf{v}$ and scalar $c$ that $T(c\mathbf{v}) = cT(\mathbf{v})$

[agile tree]

linear function straight line

[thorny flame]

@vernal matrix

[agile tree]

uhhhhh

ok!

is T like a function

T(v1) like f(v1)

hm hm

[vernal matrix]

You can write them as "one", so like if you had two vectors $\mathbf{v_1, v_2}$ and two scalars $c_1, c_2$ then $T(c_1 \mathbf{v_1} + c_2 \mathbf{v_2}) = c_1 T(\mathbf{v_1}) + c_2 T( \mathbf{v_2})$

[thorny flame]

@vernal matrix

[agile tree]

ok that makes sense

[vernal matrix]

(and yep, it's a function, a linear one )

[vernal matrix]

Anyways, if we go back up here, we have something that looks like the last message I said, right?

[agile tree]

yes 1/3 being the constants

no

scalars

🧠

wait so to clarify, we know that T is linear because for (1,0) and (0,1), they can be written as a linear combination of (1, -1) and (2,1)

why did we check 1 0 and 0 1 tho?

[vernal matrix]

More that they tell you T is linear, you just take that as is, for a moment the (1,0) and (0,1) don't matter, so assume that it's true

[agile tree]

oh hmm

[vernal matrix]

The idea is that they give you the image of two vectors, and that alongside with linearity allows you to find the image of any other vector, the first step being expressing desired vector in terms of those (1, -1) and (2, 1)

[vernal matrix]

And the idea for the second part of the question is that you want the matrix of the linear map [with respect to the standard basis]

[agile tree]

desired vector is (1 2) ?

[vernal matrix]

So you see where the standard basis vectors go, and then you form the matrix from that (more on that later!)

[vernal matrix]

Well we desire the image of (1, 0) and (0, 1)

[agile tree]

how come those two specifically?

oh

they are standard basis vectors is that why?

[vernal matrix]

Yep, reading my mind again

[agile tree]

[vernal matrix]

Anyways, time to make you do a little bit of work

[agile tree]

i am ready

[vernal matrix]

Back to here

We have $T\qty( {\color{green} \pmqty{1 \ 0}} )$, and know that ${\color{green} \pmqty{1 \ 0} } = \frac13 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1}$, do you have any ideas on what I want you to do now?

[thorny flame]

@vernal matrix

[vernal matrix]

(you better have, seeing you've been reading my mind, this isn't the time to stop )

[agile tree]

T((1 0 )) = (1/3 1/3) ?

wait

uhhhhhhhhhhhh

final answer 1/3 1/3

[vernal matrix]

Wow, you're quite far ahead of me didn't even get the actual answers myself

Give me one second

[agile tree]

oh it should be 2/3

i multipled and added the RHS

[vernal matrix]

2/3 where?

[agile tree]

oh lord lemme do maths again

[vernal matrix]

there should be a 2/3 somewhere sure

Though at least for now, I wanna see your steps

[agile tree]

im gonna send my wrk

nvm idk what im doing 😭

[vernal matrix]

Alright, let's go through this together, bit by bit

[agile tree]

tyty

i tried to apply the a and b to the given vectors but i dont get it

[vernal matrix]

Let's go back to here, we'll do this one step at a time what do you think I want you to do here? (it's a very tiny thing)

[agile tree]

multiply the 1/3 to the (1 -1) and 1/3 (2 1) ?

[vernal matrix]

Nope, don't do that

[agile tree]

hmm

[vernal matrix]

Right, maybe some colours might help

We have $T\qty( {\color{green} \pmqty{1 \ 0}} )$, and know that ${\color{green} \pmqty{1 \ 0} } = \frac13 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1}$, do you have any ideas on what I want you to do now?

[thorny flame]

@vernal matrix

[agile tree]

T(1/3(1 -1) ......)

so plug in the right stuff into the T ?

[vernal matrix]

Yep

That's it! Now, we know that $T$ is linear, and we're at $T\qty( {\color{green} \pmqty{1 \ 0}} ) = T\qty( \frac13 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1} )$ so far, do you have any other ideas of what we can do, given that $T$ is linear?

[thorny flame]

@vernal matrix

[vernal matrix]

Oh, and I think I forgot to say, T is linear

[agile tree]

take out the scalars?

[vernal matrix]

Sounds very good to me, can you show me how?

[agile tree]

yes

[vernal matrix]

Lovely

[agile tree]

ohhh

and we know those

form the question

from

[vernal matrix]

Yep, exactly

Think you know what to do from there!

[vernal matrix]

Perfect great!

Now, to be evil can you do the same idea for T((0, 1)) for me?

[agile tree]

Ok!!

i got $\pmqty{-2/3 \ -1/3}$

[vernal matrix]

Perfect

(oh btw there are hidden backslashes

$T\qty( \pmqty{-2/3 \\ -1/3} )$

gets you $T\qty( \pmqty{-2/3 \ -1/3} )$)

[agile tree]

TY LOL

i am trying to practice latex since i need it for my class XD

[vernal matrix]

Awwww mind you some of these are from the TeXit bot's default preamble which has some of the packages I'm making use of (so in your own you may need to include them!)

[agile tree]

oh i see!

thank you so much !

i realized

my answer should not have T

lol

[vernal matrix]

Heheh it shouldn't, but I'll let you off with that one you've done well and gotten it!

(if you want it as the vector form keep the \pmqty{-2/3 \\ -1/3} there!)

[thorny flame]

risa★

[agile tree]

yippee

[vernal matrix]

Perfect

[agile tree]

what do i do with these two?

vectors...?

images

of

1 0 and 0 1

[vernal matrix]

Well, now, we know how T acts on the standard basis vectors

Do you know how to make the matrix of a linear map?

[agile tree]

one part corresponds to x and the other to y (?) i dunno

[vernal matrix]

Kind of like that, like the first column is the image of the first basis vector, and the second one the image of the second basis vector (as a simplified version)

[agile tree]

okok

so i would combine the two into one matrix basically

as the first and second column

i do remember doing smth similar

how do i know which one to put first?

pls finish typing tho sorry for asking a lot

[vernal matrix]

Put the one corresponding to (1, 0) first, then the (0, 1) second, as that's the order of the basis

[agile tree]

okk

[vernal matrix]

(noooo don't worry! I was trying to figure out how to explain the "extended" version of what I said )

[agile tree]

ty!

[vernal matrix]

You could make a new "backwards standard basis", {(0, 1), (1, 0)} (so the standard basis but in the backwards version, then if you wanted the matrix of T with respect to that new basis, you find what T(0, 1) gets sent to, but in terms of the basis {(0,1), (1, 0)}, and make that your first column

you do similar with the next one, find the coordinates of T(1, 0) with respect to the "backwards basis" and make that your second column

[there is also the idea of taking different bases on each side, in fact, remind me about this at the end, there's something cool!]

[agile tree]

ok i sOrta get it

wait so the standard basis being 1 0 and 0 1 is like a general thing

[vernal matrix]

Yep, that's a general thing so now, we know the matrix B

As I kinda described to you, we have $B = \pmqty{\frac13 & -\frac23 \ \frac23 & -\frac13}$, this matrix is the matrix of $T$ with respect to the standard basis

[thorny flame]

@vernal matrix

[agile tree]

i can use the matrix B to multiply to any vector to get T of that vector?

[vernal matrix]

Yep, so if you have any vector in standard basis, if you multiply by B, you get T of that vector

Now, time to subject you to some more maths

[agile tree]

ok!

[vernal matrix]

Let's go back to the original question, we have that $T\qty( \pmqty{1 \ -1} ) = \pmqty{1 \ 1}$ and $T\qty( \pmqty{2 \ 1} ) = \pmqty{0 \ 1}$, but notice that
[
\pmqty{1 \ 1} = -\frac13 \pmqty{1 \ -1} + \frac23 \pmqty{2 \ 1}
]
and
[
\pmqty{0 \ 1} = -\frac23 \pmqty{1 \ -1} + \frac13 \pmqty{2 \ 1}
]

[agile tree]

yes

[thorny flame]

@vernal matrix

[vernal matrix]

Right, got it sorry for the wait

[agile tree]

noo all good

[vernal matrix]

Anyways, say instead we wanted the matrix of T but with respect to the basis {(1, -1), (2, 1)} (it happens to be a basis )

[agile tree]

okok

[vernal matrix]

We also want to take it in that particular order, so (1, -1) is the first one, and (2, 1) the second basis vector

[agile tree]

as in, instead of (1 0 0 1)

[vernal matrix]

Yep yep

[agile tree]

mhm

[vernal matrix]

Now, anyways, the first column is the coordinates of T(1, -1) but those being with respect to the basis, so those coordinates are (-1/3, 2/3)

And also the second column is similar, the coordinates with T(2, 1) but in the new basis, so those are gonna be (-2/3, 1/3)

Does that at least make sense?

[agile tree]

hmmm

[agile tree]

how come?

[vernal matrix]

Because the coordinates with respect to the new basis are like, for this case, ( [whatever's the coefficient of (1, -1)], [whatever's the coefficient of (2, 1)] )

[agile tree]

oh yes

just like how we solved it previously but now with 1, -1

[vernal matrix]

Yep yep and cause now we have them like here

[agile tree]

ok ok ye

[vernal matrix]

Anyways, cause we have the coordinates of ${\color{green} T\qty( \pmqty{1 \ -1} ) }$ with respect to the new basis is ${\color{green} \pmqty{-\frac13 \ \frac23} }$,

and the coordinates of ${\color{orange} T\qty( \pmqty{2 \ 1} ) }$ with respect to the new basis are ${\color{orange} \pmqty{-\frac23 \ \frac13} }$,

the matrix of $T$, with respect to the basis $\qty{ {\color{green} \pmqty{1 \ -1} }, {\color{orange} \pmqty{2 \ 1} } }$, then becomes
[
\pmqty{ {\color{green} -\frac13 } & {\color{orange} -\frac23 } \ {\color{green} \frac23 } & {\color{orange} \frac13 } }
]

[thorny flame]

@vernal matrix

[vernal matrix]

nice

[agile tree]

yeah!

[vernal matrix]

But hopefully that's somewhat clear how it works?

[agile tree]

how do we know if something is a valid basis ?

[vernal matrix]

You can e.g. check whether they're linearly independent and that you have as many vectors as the dimension of the space (for example here, we have those two vectors in R^2, so just need to check linear independence, which is "easy" because they're not scalar multiples of each other!)

You can check linear independence however you wish

[agile tree]

Free variable -> dependent or independent I forgor :0

[vernal matrix]

Free variables mean you're dependent

[agile tree]

This is how I was trying to solve this part also, but I got something weird so I think it’s wrong, what did I misunderstand? :0

[vernal matrix]

(1, 1) on the RHS!

You wanted the coordinates of T(1, -1) = (1, 1) in terms of the "new basis"!

[agile tree]

i thought new basis was {(1, -1), (2, 1)}?

[vernal matrix]

it is! But you have (1, 1) in standard basis, but want it in "new" basis

[agile tree]

does the 1 1 come from here?

[vernal matrix]

Yep, and the fact that T(1, -1) is (1, 1)

[agile tree]

ohhhh

oooo

aaaa

ooooo

okk

[vernal matrix]

But (1, 1) in standard basis not new basis

[agile tree]

earlier, were we solving for the matrix of T with respect to the basis {(1, 0), (0, 1)} (standard basis)
is my understanding correct :0

[vernal matrix]

Yep, that was the aim before we wanted the matrix of T with respect to the standard basis (on both sides, both domain and image)

The aim now is the matrix of T, but instead the basis {(1, -1), (2, 1)} (again both sides)

[agile tree]

and in that case, we did this
which is why im confused why for the new basis, {(1, -1), (2, 1)} , its not (1,-1) on RHS