help-28

[inland moth]

radius is perpendicular to the tangent line

[crimson saffron]

i just have that the 90 angle and 62 angle from that

what do i do next

[inland moth]

the ? is in what triangle

[crimson saffron]

the smaller triangle inside the circle

[inland moth]

yeah

what properties does that triangle have

[crimson saffron]

ummmmm i’m not sure

[inland moth]

the two sides of the triangle are?

[crimson saffron]

OH SHOOT

BRUH OKAY THANK YOU SO MUCH

i’m tripping out

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[limpid moat]

I can't even read it properly

,zoom

I knew

[granite torrent]

tap/click on the img

[calm flint]

i hope this is any better

[limpid moat]

doesn t change to much ..the elephant in the room is the fact that I cannot zoom

[calm flint]

so anyway

i need help with A. item 6, 7 and 8

actually just 6

according to the module

the answer is

either the intersection of sets A and B is non-empty or A is a subset of B if and only if at least one of the complements of A or B is true

where did non empty come from

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[calm flint]

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[gray spindle]

Hi

[gray spindle]

$$P_{A}(s) = \sum\limits^{n-1}{i=0}a{i}a_{i+s},\quad s=0,1,...,n-1$$

[glossy valve]

cascino

[gray spindle]

$$P_{A}(n-s) = \sum\limits^{n-1}{i=0}a{i}a_{i+s},\quad s=0,1,...,n-1$$

[glossy valve]

cascino

[gray spindle]

I would like some help to prove the symmetric property

that P(s) = P(n-s)

Or to verify that the above is reasonable

Also

$A = {a_0,a_1,...,a_n-1}$

[glossy valve]

cascino

[gray spindle]

we take the modulus of i+s when i+s >= n

its okay

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[vocal sonnet]

How I solve this with algebra

[grave elm]

Why specifically with algebra?

[vocal sonnet]

That's the assignment

[grave elm]

Have you tried substitution?

u = 1/x

[vocal sonnet]

"Find the following limits algebraically."

[vocal sonnet]

yea

[grave elm]

so by that you probably got lim u->+inf arctan(u)

[vocal sonnet]

tan^-1(1/0.000000000001) get's really close to 1.57

[vocal sonnet]

what

[torn jolt]

is the answer pi?

i mean pi/2

[vocal sonnet]

yea

Is there an easy way to know what the relationship a decimal has to pi?

[grave elm]

Have you tried substitution or not?

[vocal sonnet]

yea

[grave elm]

okay, what substituion did you try?

[vocal sonnet]

I got pi/2

tan^-1(1/0.0000000000001)

[grave elm]

ah, so you tried substituting in numbers

[vocal sonnet]

yes

[grave elm]

I meant substituion like u = 1/x

[vocal sonnet]

Ohhh

u get's really close to posive infinity

so

[grave elm]

So it should be this limit I mentioned earlier, right?

[vocal sonnet]

yeah

[glossy valve]

MæthIsAlwaysRight

[grave elm]

alright so

do you know what this should be?

Do you know how does arctan behave as x gets close to infinity?

[vocal sonnet]

not in my head, but using a graph with y scale as pi shows pi/2

I forget all the trig rules :(

[grave elm]

Do you know how does graph of tan(x) look like?

[vocal sonnet]

Lots of vertical asymptopes

[grave elm]

indeed, and those asymptotes happen to be at pi/2 (+- kpi)

[vocal sonnet]

kpi?

[grave elm]

k*pi

where k is any integer

it means like pi/2 + pi, pi/2 + 2pi, pi/2 - 3pi etc

[vocal sonnet]

ooh

Ok I see

[grave elm]

Anyway, we only really care about this part of the graph

[vocal sonnet]

Yes

[grave elm]

this tan(x) is between -pi/2 and pi/2

and as you can see, it approaches +infinity as x approaches pi/2

and so arctan(x) approaches pi/2 as x approaches +infinity

[vocal sonnet]

I see

I need to review precalc trig

Thank you!!!

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[grave elm]

yw

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[urban sedge]

how can i solve this?

[urban sedge]

i think i figured

[limpid moat]

maybe you could use taylor expasion for multivariate function

[urban sedge]

does polar work?

[urban sedge]

didn't learn it

[limpid moat]

yes why not

[urban sedge]

ok ty

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[bold moat]

how to do this

any clues

[grave elm]

looks like you will have to complete the square

[rugged jay]

A will be a constant so we should get rid of all x and x^2s in the -B(x+C)^2 part

[rugged jay]

Yeah basically

[grave elm]

you can factor out 0.04

[bold moat]

so -0.04( (x^2) - (0.8/0.04) ) + 2

[grave elm]

Great start

can you simplify 0.8/0.04

you can rewrite it as (8/10) / (4/100) = 80 / 4

[bold moat]

ah ok

[grave elm]

so it would be...?

[bold moat]

800/40

oh wait

[grave elm]

That's true, can you simplify it even further tho?

as in, calculate it

[bold moat]

80/4 = 20

[grave elm]

-0.04(x^2 - 20x) + 2

so this is what we have now

[grave elm]

Oh, you forgot an x there

[bold moat]

lol

[grave elm]

it should be -(0.8/0.04)x

[bold moat]

-0.04(x - 10)^2 - 10^2) + 2

right?

[grave elm]

Yes, that seems to work

but you forgot (

-0.04((x - 10)^2 - 10^2) + 2

anyway, now you just need to write it as
-0.04(x-10)^2 + 10^2 * 0.04 + 2 and simplify

[bold moat]

• 6

[grave elm]

correct

so 6 - 0.04(x-10)^2

[bold moat]

tysm

[grave elm]

yw

[bold moat]

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[woeful anvil]

I asked a couple chemistry questions here before and got good answers so I’ll ask again.

Hundz Rule states that electron orbitals that are degenerate are first half filled before they are actually filled. How do we know that they are degenerate in the first place?

[torn jolt]

Hund's third rule enables us to minimize the repulsion between electrons - that's why we place them in separate orbitals first

this is maths server]

[woeful anvil]

https://tenor.com/view/的9-gif-27299608

[echo marsh]

simple, you just rote learn which orbitals are degenerate

[woeful anvil]

I see

[torn jolt]

so basically electrons will spin in opposite directions (as denoted by the arrows in the orbital diagrams). Because they have the same negative charge, they'll repel each other. Hund's rule basically minimizes the disorder, if you will

[woeful anvil]

Is there any configuration that can tell us which are degenerate?

[torn jolt]

well degenerate means "of the same energy"

[hallow walrus]

n+l rule

[umbral dome]

degenerate just means they have the same energy level, because the quantum numbers n and l change the energy level but m and s do not

[torn jolt]

yeah

[thorny shuttle]

not really for monoelectronic species

[torn jolt]

so $p_x$ is degenerate to $p_y$ and $p_z$

[glossy valve]

45

[torn jolt]

for example

[woeful anvil]

I see

How do you know which direction to put the arrow in first?

[thorny shuttle]

you need to know

aufbau rule

hunds rule

pauli's exclusion principle

[torn jolt]

by convention we put it pointing upwards first

that's what I've seen in literatures

[woeful anvil]

Alright

[thorny shuttle]

only l gives the answer in this case

[woeful anvil]

So like for 4f you’d put everything for the first 7 orbitals up then for the next 7 down?

[thorny shuttle]

correct

[torn jolt]

yuh

[thorny shuttle]

thre might be exceptions

[woeful anvil]

What about for like 5f?

[torn jolt]

yeah

[thorny shuttle]

for more exchange energy

[torn jolt]

copper and chromium are exceptions to Hund's third rule

[thorny shuttle]

yup

there are a few more radioactive metals too as far as i remember

in the lanthanide and actinide series

[torn jolt]

haven't been taught why that's the case though so I can't help u with that though :p

[thorny shuttle]

more exchange energy

[torn jolt]

well, I understand they're more stable in that case, but what's the underlying reason?

[thorny shuttle]

it has a fully-filled 3d sublevel and a half-filled 4s subshell, which makes it more stable. This is due to the stability of orbitals, which states that completely filled and half-filled orbitals are the most stable.

[torn jolt]

ahhh

[thorny shuttle]

symmetry

and elec repulsion

parallel spin so minimum elec repulsion

[torn jolt]

mmmm

[thorny shuttle]

yep

[woeful anvil]

For Pauli’s exclusion principle, what does it mean no two electrons can have the same set of quantum numbers?

I don’t understand that

[thorny shuttle]

azimuthal quantum number, principal quantum number, spin quantum number and magnetic quantum number

are never the same for any electron

[woeful anvil]

Could you give me an example?

[thorny shuttle]

pick any electron config

and write down its

[woeful anvil]

Ok

[thorny shuttle]

n,l,m,s

it will never be the same

[woeful anvil]

Okok

So like

Here idk if I’m doing this right but

[thorny shuttle]

sure show

[woeful anvil]

4f^14 means n=4, l=3 ml= 14 and ms=+1/2

Right?

And it can’t be replicated

[thorny shuttle]

not really

ms is defined for a particular electron

[woeful anvil]

ok

[thorny shuttle]

not the entire orbital

[woeful anvil]

I see so how would I define 4f^14?

[thorny shuttle]

4th shell
f "3" subshell

[woeful anvil]

Yeah

[thorny shuttle]

has 14 electrons

so completely filled

[woeful anvil]

That means l=3 right

[thorny shuttle]

yes

s=0, p=1, d=2, f=3

[woeful anvil]

14 electrons so ml=14 or no?

Or uh how would you define the ml value

[thorny shuttle]

it means

[woeful anvil]

From 4f^14

[thorny shuttle]

theres 7 possible

magnetic quantum number

[woeful anvil]

I see

And 14 spins right

[thorny shuttle]

which means 7 possible suborbitals division

pxpypz...

[woeful anvil]

ic

[thorny shuttle]

and 14 electrons as each have two possibilities

and yea right

just practice

[woeful anvil]

Is it always the power number divided by half?

[thorny shuttle]

ull get the hang of it

magnetic quantum number is defined as -l to +l

4f^14 means theres 14 electron(filled all orbital dvisions twice). it can also be say 4f^3 it means it has 3 electrons

the ^ notation represents no of electrons at that moment

not the total possible

[woeful anvil]

So ml would equal -3 to 3

[thorny shuttle]

that can be found from m value

yes

thats 7 divisions

in f subshell

which means 14 electron possible

[woeful anvil]

And ml = -14 to 14?

[thorny shuttle]

nope -3 to 3

-3,-2,-1,0,1,2,3

total 7

[woeful anvil]

Oh ok

That’s because of the l value right

[thorny shuttle]

each has 2 electron so 14 possible electrons in f subshell

[thorny shuttle]

m value

which is defined by f

[woeful anvil]

If it was d it woild be -5 to 5

Wait

I mean

-2 to 2

[thorny shuttle]

yes

that would mean

5 possible subshell divisons

or orientation of the orbital

[woeful anvil]

Am I able to write it in interval notation like (-2, 2)

[thorny shuttle]

which means 10 electrons

[thorny shuttle]

dont need to

just use to

it feels like math doing that

then itll be wrong since -2 and 2 are included

[woeful anvil]

okok

And how would I calculate the value for Ms? Just by looking at 4f^14

[thorny shuttle]

magnetic spin?

[woeful anvil]

Correct

[thorny shuttle]

just watch some khan academy

its simple u cant unpaired electrons

[woeful anvil]

oh

[thorny shuttle]

ye

[woeful anvil]

How should I fill this out then since it tells me to list the value for mss

[thorny shuttle]

https://www.youtube.com/watch?v=b4LaNaFQYvc

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[north gust]

can someone check the work on this question please

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[north gust]

no

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[silk rock]

If ( A ) is a proper class, then ( {A} = \emptyset ).

[glossy valve]

Halex

[silk rock]

What does it mean for A to be a proper class and how do I prove this?

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[proven kettle]

So I need to find the roots of the equation on line 1. I plugged everything into the quadratic equation but I get stumped with the sqrt[4(1-sinh^2(theta))], as I am not sure of a nice fitting identity for 1-sinh^2(theta). Cosh^2(theta)=1+sinh^2(theta) not what I have here. Just unsure of where to go from here

[proven kettle]

Alternatively, must I leave it in the form from the step before and take the sqrt[-4sinh^2(theta)+4] = [2isinh(theta)+2] thus the roots would be [(isinh(theta)+- isinh(theta)+1]

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[devout river]

find the condition for this expression to be a multiple of 11 give two posible values for n

[devout river]

12n squared plus 12n plus 11

is there away to do it algebracially if so how

?

[full forum]

@devout river Has your question been resolved?

[devout river]

<@&286206848099549185>

[mental aspen]

is there any condition on n?

[devout river]

no

[mental aspen]

like it has to be a whole number or integer etc?

[devout river]

doesn't say anything about it having to be an integer but considering the level I am at it probaly has to be but if u can find two possible values for n where the equation is a multiple of 11 and u can do it using algebra my teacher won't care

even if not an integer but rahter not

[mental aspen]

okok

[clear reef]

the problem with this is it's actually harder using only integers i think

[mental aspen]

so we have $12n^{2}+12n+11$ right?

[glossy valve]

77²

[devout river]

yes

[mental aspen]

what you can do is that, transform it like $(11n^{2}+11n+11)+(n^{2}+n)$

[glossy valve]

77²

[mental aspen]

seems alright?

[devout river]

yes

you isolate nsquared plus n

[mental aspen]

and what can you do afterwards?

Well without getting in trouble later, we should restrict the domain of our given function to be the set of natural numbers

[devout river]

after u isolate it may sound stupid but if u do over 11 equals 1 then u simply to 11n2 plus 11n equals 11 then u can transofrm to 11n2 plus 11n minus 11 equals 0 then its a quadratic

does that work or no

?

[mental aspen]

erm, not quite

[devout river]

why though?

[mental aspen]

It might work, but it will be trickier

[mental aspen]

it is better to just factor out the 11 from the expression in first parenthesis

[devout river]

wdym

[mental aspen]

like $[11(n^{2}+n+1)]+[n^{2}+n]$

[glossy valve]

77²

[devout river]

like u factor that bit out cause thats always going to be divisble by 11

so u just take n2 plus n

[mental aspen]

seems fine

[devout river]

how did u plan on doing it thouhg

[mental aspen]

,w n^2 + n -11 = 0

[glossy valve]

Results provided by WolframAlpha

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Requested by seventysevensquare

[mental aspen]

lol

[devout river]

not whole number oops

so how do u get it with whole numbers

[mental aspen]

see here we can say that the first thing will be a multiple of 11

do you agree?

[devout river]

yes

[mental aspen]

no just left the n^2 + n thing

[devout river]

yes

[mental aspen]

if we can find some n, such that this thing is divisible by 11
then we are done

[clear reef]

okay wait i just want to ask one clarifying question: for what unit of math is this for?
is this for quadratics or for something like number theory

because that matters a lot here

[devout river]

no just proofs

I think

[mental aspen]

as far as the question is framed
it seems like they just care about any 2 natural values of n

[devout river]

it has to be a whole number earlier on in the question it says n is a whole number but that was 5 parts ago so I did not realise mb

like this is d but it said that on a

sorry

[mental aspen]

okay whatever, we'll be fine

so can you factorise n^2 + n further?

[devout river]

n(n plus 1

[mental aspen]

yeah nice

and also what are the ways you can write 11 as product of 2 whole/natural numbers?

[devout river]

why?

10 1

9 2

8 3

7 4

[mental aspen]

cause we want n(n+1)=11k thing

no

I mean product

not sum

[devout river]

11 x 1

[mental aspen]

yeah

[devout river]

ah

[mental aspen]

I think we should work a bit differently

[devout river]

how

[mental aspen]

what if you assume n to be 11

[devout river]

ah

[mental aspen]

so n(n+1)=11k

[devout river]

working backwords

[mental aspen]

yeah

[devout river]

but is there not a way otherwise

cause working backwords is still a bit of guess work

[mental aspen]

there are possibly infinite number of answers to this problem

Guess and check is like the most basic one imo

and we just want 2 values, so guessing is not a bad idea here

[devout river]

I know 2 values but guessing does not help long run

I want to kknow how to do it withut guessing

[mental aspen]

getting the general solution might involve some other things

involving some reccurence relation type shi

[devout river]

we can do it ur way if its the only way you see but if there is another one u know of I would rahter go down that root

[mental aspen]

that method will take long for me to explain, and I've also an assignment to complete😶

[devout river]

oh

[mental aspen]

I think you can wait for some other helper

,ti

[devout river]

ok thank you

[glossy valve]

The current time for seventysevensquare is 01:58 AM (IST) on Fri, 13/09/2024.

[devout river]

I am sorry

for being a pain

[mental aspen]

nah It's fine

[devout river]

do i ping helper again or no

[mental aspen]

I'm not sure if it is 15 mins or not

maybe check for that

[devout river]

it says 15 mins but I waited I think I ping again

<@&286206848099549185>

[clear reef]

https://www.desmos.com/calculator/gswmixyq80

this is

my best guess

it works out on the original formula

as for my logic, i'm not sure if it's good but it works

[devout river]

wdym

i don't understand it

is there no way to do it using algebra

without graphs

[clear reef]

yeah i didnt use graphs i just

used desmos bc im too lazy to latex it

lmfao

i just have the graph there to verify

[devout river]

but what are the answers then

I am so confused

[clear reef]

okay so

[clear reef]

we want 12n^2 + 12n + 11 = 11b, where n and b are both integers, right

[devout river]

yeah

[clear reef]

so

let's try reshaping this equation a little, and see if it leads us somewhere

isolating a little for n, we get

$12n(n+1) = 11b-11$

[glossy valve]

Serphic

[devout river]

yes

[clear reef]

next: try to get some sort of limitation here

$\frac{11b-11}{12} = n(n+1)$

[glossy valve]

Serphic

[clear reef]

wait

that didnt work

actually no that's fine

notice how we can factor out an 11 on the left

[devout river]

yes

[clear reef]

$\frac{11(b-1)}{12} = n(n+1)$

[glossy valve]

Serphic

[clear reef]

now

this left term can only be an integer if the 12 cancels out

[devout river]

yes

[clear reef]

so that means b has to be some 12k + 1, where k is (again) any integer

you see where im going with that?

[devout river]

yes

[clear reef]

so

we can simplify the left to

$11k = n(n+1)$

[glossy valve]

Serphic

[clear reef]

and finally one last thing to limit n

$k = \frac{n(n+1)}{11}$

now we just said that k has to be an integer, correct

[devout river]

yes

[clear reef]

wait

let me fix that

that's not right lmfao

[glossy valve]

Serphic

[clear reef]

there we go

so

there are two ways that n(n+1)/11 can be an integer

how would we cancel out that denominator?

either n or n+1 has to be a multiple of 11

[devout river]

how did u get to that

oh

I see

very smart

[clear reef]

which means n is either some multiple of 11, or n is some multiple of 11 minus one

boom

[devout river]

so there are infinte answers

[clear reef]

yep

[devout river]

how are u so good

[clear reef]

im going to be honest i thought i fucked up like 9 times here 😭

[devout river]

and they expext my dumbass whos 14 to do and u are in unviersity

[clear reef]

im 17, but these arent problems you should be expected to do at 14

these are problems called diophantine equations

[devout river]

oh

[clear reef]

there's approaches for a lot of problems of this type

but

a lot of them are huge pains in the ass

[devout river]

oh I have no way to thank you enough cause I was stuck

swear everyopne in my class is a genuis cause apparntly they found it easy

[clear reef]

kay wait

this might not be what the teacher wanted you to do though 😭

[devout river]

being the dumbest in top set comes with problems like this

it works

and it uses algebra

[clear reef]

[devout river]

so it works

[devout river]

actually really sorry but why does b-1 equal 12

[devout river]

here how do u get to the nxt step

[clear reef]

b-1 has to equal some multiple of 12, because we want to cancel out the denominator of 12, or else that means n(n+1) is not an integer

and we just said n(n+1) is a whole number at the start

[devout river]

ok

[clear reef]

so if n(n+1) is equal to 11/12 times some number, that number better cancel out the 12 in the denominator

or else that means n(n+1) isnt an integer

[devout river]

ok

thanks for clarifying

[devout river]

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[exotic shoal]

Hiii! I’d just like to check if my sketching of this graph is correct (the limits and points I had to sketch the graph to follow are on the side).

[lime ether]

f(-2) = -1

should the point (-2,-1) be a closed or open circle?

[exotic shoal]

Ohhh yeah closed

Cuz we know it’s a point

[lime ether]

mhm

there isn’t sufficient information for the right side of the graph though

[exotic shoal]

Well it just has to be a possible sketch

[lime ether]

same for the left

yea

it satisfies those

[exotic shoal]

Yay

[lime ether]

i was just wondering if there’s supposed to be more

[exotic shoal]

That’s everything I was given

So I don’t think so

Thanks a lot! @lime ether

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[split wraith]

hello! i was struggling on 8 questions for a review im supposed to do before i take a test and was wondering if i could get help it’s basically all scientific notation stuff

[digital lion]

What do you specifically need help with?

As in what about these questions confuses you

[split wraith]

i’m not entirely sure how to add what’s asked to be added nor can i recall how to express the standard form or scientific notation of a number

[digital lion]

Ok, no worries I can give you a short explanation

The exponent of the $10$ within an expression of scientific notation acts as the amount of decimal places to jump by, positive being to the right and negative to the left. I'll use one of your provided questions as an example:

$ 9.6 \cdot 10^5 $

In the above equation we can see that $10$ has the exponent of $5$, which is a positive number, so we should \emph{move the decimal point to the right five times}.

$ 960000 $

This would be the number expressed in normal form

[glossy valve]

luxzi

[split wraith]

ok so i was right about that. i guess it was the second one

[digital lion]

The second one would be in the opposite direction. i.e. $0.000000071$

[glossy valve]

luxzi

[split wraith]

because it’s a negative?

[digital lion]

Exactly!

[split wraith]

ok

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[rugged narwhal]

How do i find the general maclaurin series of a function? For example e^-5x and 1/(1-x^4)

[gritty rose]

with the definition

[leaden ermine]

with the definition

[rugged narwhal]

What does this mean @leaden ermine

[leaden ermine]

very smart

you replied ping to him but also pinged me

[rugged narwhal]

Yeah i do this often

Ok thanks

[gritty rose]

preferably, you'd learn this from your notes/book in a more familiar notation to you

but if you never learned it "math libre" and "lamar calc" are good search terms to add to your googling

<@&268886789983436800>

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[shrewd holly]

I need help

[dense smelt]

With what

[shrewd holly]

The picture above

I’ll send it again

[dense smelt]

I see it, it doesn’t say anything

[wide grove]

2 squareroot 3 over 3 .- 3 over5

[dense smelt]

What do you need to do with it mf

[shrewd holly]

Oh it says simplify

[dense smelt]

What have you tried

[shrewd holly]

I got the 2root3 over 3 -3over5

[wide grove]

yes

[shrewd holly]

It the answer key is somehow 10root3-9over15

[dense smelt]

2 root 3/ (root 3)^2 seems simplifiable

[dense smelt]

Do not give answers

[wide grove]

oops

[shrewd holly]

lol

[wide grove]

i actually apologize

[shrewd holly]

Can you clarify?

[dense smelt]

Anyway, if you have two fractions a/b - c/d, how do you subtract them

[shrewd holly]

Same denominator?

Make it

[dense smelt]

Do it then

[dense smelt]

With this

[shrewd holly]

Ok

How would I subtract if it isn’t like terms

[dense smelt]

How do you here

[shrewd holly]

Just leave it as is

[dense smelt]

No

how do I subtract a/b - c/d and get one fraction

Do not use ChatGPT

[shrewd holly]

Bruh

Ok mb apologies

But find the common denominator for B and D

[dense smelt]

What’s a common denominator

Give one

[shrewd holly]

let’s say 3/5 - 3/10 the common denominator would be 10 as people say LCM but if you get something else u could simplify it

[dense smelt]

Stop

[shrewd holly]

?

[dense smelt]

If I have you a/b and c/d

What’s a common denominator

Don’t overthink it

[shrewd holly]

E?

[dense smelt]

no

[shrewd holly]

So what r u trying to teach me

[dense smelt]

How to add fractions

Let’s look at our question, a/3 - b/5, what’s a common denominator

[shrewd holly]

15

[dense smelt]

Ok, so what’s a/3 - b/5

[shrewd holly]

5a/15-5b/15 which is an

An

Ab

[dense smelt]

Wrong

[shrewd holly]

Ok

[dense smelt]

Why did you get 5b/15

[shrewd holly]

Oh wait

It’s 3b

[dense smelt]

Ok, make it one fraction

[shrewd holly]

5a3b?

5a-3b

[dense smelt]

Over 15

Replace a = 2 root 3, and b = 3

[shrewd holly]

So if I got 2root 3 over 3-3 over 3 I could get the common denominator which is 5root3 over 15 - 9 over 5

Then

[dense smelt]

No

9 over 15

[shrewd holly]

I said 9 over 15

[dense smelt]

And it’s not 5 root 3

5*2 = 10

[shrewd holly]

Ok can I show my steps

[dense smelt]

Your steps were wrong that led you here

[shrewd holly]

Yes so can I explain it u

[dense smelt]

5 * (2 root 3)

What is this

[shrewd holly]

10root3

[dense smelt]

Ok

So you have 10 root 3 - 9, all over 15

[dense smelt]

Looks like this

[shrewd holly]

Ah ok I get it may I ask a question: when does rationalizing square root denominators come in to play

[dense smelt]

When there’s a square root in the denominator

That’s it

[shrewd holly]

Simplified or not does it matter?

Mixed radical

Let’s say

[dense smelt]

It comes up when there is a square root in the denominator

That’s it

Don’t overthink it

[shrewd holly]

👍

[dense smelt]

There is no deeper meaning

[shrewd holly]

So let’s say 12/root125 I tuned both by root of 125

Times

[dense smelt]

Yep

$\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$

[glossy valve]

Sharp

[dense smelt]

That’s it

[shrewd holly]

I see and thank you for your patience and help

[dense smelt]

Use .close

[shrewd holly]

.close

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[shrewd holly]

.reopen

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✅

[shrewd holly]

Quick question can I simplify 10root3/15 and why

[dense smelt]

You can divide by 5

But that just gets (2 root 3)/3

Where we started from

[shrewd holly]

So r both answers correct

I’m guessing that’s a yes

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[vale solstice]

So im trying to do u substitution to solve this. would the best number for U be 1-4t^3?

[torn fable]

Yes

[vale solstice]

and then that would make du be 12t^2 but what I'm struggling with is do I then turn the equation into sqrt(u) 1/6dt?

[torn fable]

Are you differentiating correctly?

[vale solstice]

I hope so

the 1 would turn into a zero and then the du should be equal to -12t^2 right?

[torn fable]

Correct now

[vale solstice]

so then it would sqrt(u) -1/6dt right

[torn fable]

dt?

[vale solstice]

sorry du

[torn fable]

Yeah

[vale solstice]

and the 1/6 moves to the outside of the integral then?

and then to the anti derivative?

[torn fable]

Sure

[torn fable]

Standard formula

[vale solstice]

and to find the new bounds I plug the old ones into the equation for u and evaluate right?

so I would plug 0 into 1-4t^3 and then 2?

[torn fable]

Yeah

Don't forget your minus

[vale solstice]

so then what im currently thinking is the answer is going to be 1/2sqrt(1-4t^3) evaluated from 1 to -31 and then once i plug the numbers in I then divide it by 1/6?

[torn fable]

No,you don't put the expression back once you change the bounds

[vale solstice]

oh so then it would instead be 1/2sqrt(u) evaulated from 1 to -31 and then once i plug in those values then do the 1/6

so plug in values to u then do the multiplication for 1/6 then subtraction for integral?

[torn fable]

Did you antidifferentaite correctly?

[vale solstice]

oh wait

i dont think i did

[torn fable]

I recommend doing it by writing out the formula

[vale solstice]

it should instead 2/3*u^3/2

i was accidentally doing the derrivative instead of antiderrivative

[torn fable]

Correct

[vale solstice]

so then would it be smarter to do 2/3 and 1/6 and then evaluate the integral?

which gives me 1/9*(1)^3/2 - 1/9(-31)^3/2 but im struggling to do the math for it

[torn fable]

Calculator lol

Also, what

How did you get -31

[vale solstice]

by plugging 2 into 1-4x^3

which was to find the lower bound for u

[torn fable]

It was -2

[vale solstice]

ohhhhh

i didnt see the negative there

so instead it should be 33

[torn fable]

Yeah

The calculation is complex

[vale solstice]

okay I think i got it

Thank you so much!

really appreciate it!

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[torn jolt]

hello

[torn jolt]

y = 3x+1 and x^2 + y^2 -2y - 2 = 0
find the intersection point
^ this is the problem i've been stuck on

I tried solving it by plugging in y into the 2nd equation

i got rt(3/10) but apparently that's wrong

[static bramble]

show your work

[torn jolt]

you want me to take a picture ?

[winged ibex]

it shouldnt be

[torn jolt]

Ill just type it out

[torn jolt]

my teacher said it was

[winged ibex]

have you got the y value correct?

[torn jolt]

didn't go that far, he said this part is wrong

[winged ibex]

i mean there are two intersections

[torn jolt]

ik

[winged ibex]

rt(3/10) and -rt(3/10)

[torn jolt]

it's +-rt(3/10)

yes

then you plug in those into one of the equations and get y

but he said that was wrong?

so idk what im supposed to do from here

[winged ibex]

maybe you gotta rationalize the denominator?

rt(30)/10

[torn jolt]

but

isn't rationalization only for when there's a rt on the denominator

[winged ibex]

well there is!

rt(3)/rt(10)

[torn jolt]

I guess

wouldn't it give the same answer in the end?

for y

[winged ibex]

yeah teachers can be stubborn sometimes

try rt(30)/10 and see

[torn jolt]

alright one second

[torn jolt]

answer is very different

But the according to the method we learned

i should be right

so im just really confused

[winged ibex]

hm

did you copy the equations correctly?

[torn jolt]

yup

i dunno

ill ask my teacher about it tomorrow

but earlier in class he said what i was doing was wrong

its okay anyway

thanks cutie

[winged ibex]

yw :)

[torn jolt]

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[balmy escarp]

im not sure how to start solving for the answer 50, i changed the d to -67 and then im already stuck on what to do next 😓

[tidal citrus]

just plug in 67

where did you get 50 from

[balmy escarp]

idk it says the answer is 50

[tidal citrus]

maybe it is

why are you trying to work backwards

it's so much easier just to plug in d

[balmy escarp]

im not sure what u mean by plugging d in 😓

[tidal citrus]

the problem says the diver wants to dive to a depth (d) of 67 ft

so replace d with 67

and evaluate the expression

[balmy escarp]

oh

okay i try

im stuck, i dont know if i did it right at all

[glossy valve]

faiyrose

[balmy escarp]

34

ah 50

thanks!

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[hollow isle]

Can you someone help me figure out, how to graph 1a?

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@hollow isle Has your question been resolved?

[hollow isle]

<@&286206848099549185>

[pearl ginkgo]

okay so @hollow isle try to draw each of the 3 graphs indivisually

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[muted vigil]

(2/5)x = 82*

solve

[drowsy osprey]

ok whats the answer

[hoary ember]

!noans

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[drowsy osprey]

is the answer 136

[hoary ember]

no

[covert heath]

hmm

[muted vigil]

🤔

[hoary ember]

hoq did you get that?

[drowsy osprey]

82 is 40% of the book
82/2 = 41 so 41 is 20% of the book
41 multiplied by 3 = 60% of the book which is 123

is that how u solve it?

[hoary ember]

yes

[drowsy osprey]

ok ty

[sacred yarrow]

!done

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[covert heath]

lmao

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[glossy valve]

Aya

[rapid rain]

you know how to get g'(y) right

then differentiate that relation

twice

[glossy valve]

Aya

[rapid rain]

yes but you should get the original way to write it

only in terms of one variable

if you want to differentiate wrt y

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[silk sail]

can someone explain this please?

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@silk sail Has your question been resolved?

[undone vector]

Rewrite 8 as 4*2 and remember you can “split” multiplication under square roots

[silk sail]

alr thank you!

what abt this?

like the part where the 8 and 2 was canceled

[undone vector]

It is a bit confusing the way it is written, but they basically just said 8/2=4

[rugged void]

8/2 = 4

[silk sail]

ohhh alright thanks!

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[bright bronze]

that works

[remote zealot]

but the angles arent bordering the side

[bright bronze]

theres 3 angles, which sum up to 180. So if 2 are equal, then the 3rd is too

[remote zealot]

whats this rule called

[bright bronze]

Its the property of any triangle? That the 3 angles of a triangle sum up to 180 degrees

[remote zealot]

im talking abkut the rule of congruence

in triangles

[bright bronze]

AAS becomes ASA

[remote zealot]

sure ig if its a rule

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[waxen steeple]

why arent they considering -1 as root of 1,wouldnt that work.i dont really understand this concept

[errant python]

.reopen

[waxen steeple]

too late sorry

[hot herald]

$\twosqrt{1}{1}$

[glossy valve]

ℝαμOmeganato5

[hot herald]

sqrt(1) is 1 ONLY

[waxen steeple]

wdym -1x-1=1 right?

[errant python]

when you do -1 times -1

[hot herald]

yes.

[errant python]

2 minuses are aplu

[hot herald]

as stated above, -1 is the negative square root of 1

[errant python]

exactly

[hot herald]

sqrt(1) is the positive square root of 1,
which is 1 only

[waxen steeple]

oh whoops sorry

btw whats the point about domain expansion?

dont we usually take all roots.do we always only take roots that lie in domain of equation

[errant python]

if 2 negatives make a positive, that means -1 x -1 = 1

[hot herald]

you only take roots that work in the domain and range of the initial equation

[hot herald]

not the issue they had

certain process like squaring give resultant equations that are easier to manipulate
but may end up generating extraneous/fake solutions that don't actually work in the original equation

[errant python]

theres a negative and positive square root for every square

[waxen steeple]

cause otherwise if rootx is taken to be -1 and x is 1 it will solve right

[errant python]

yes that is true

but theres also a negative square root

so like the harder way

[waxen steeple]

id never heard of ignoring roots that does not lie in domain ,so this was a bit confusing

[errant python]

what grade r u in?

[hot herald]

lets consider something like
x = 1
if you were to square both sides, you'd get
x^2 = 1
the solutions of this new equation are x=-1 or x=1
however its clear that x can't be -1 from what you started with

[errant python]

i agree

but whoever made this math rule also decided that the negative square root is going to be the positive square root times -1

[waxen steeple]

the initial equation does solve for x=1 and rootx=-1 right

[hot herald]

no

[errant python]

both ways r correct

but the positive one is easier

negative is smae but just harder

[waxen steeple]

the squared one supports x as 1

[errant python]

ok look here :

whats -4 times -4?

[waxen steeple]

16

[errant python]

and whats 4 times 4?

[waxen steeple]

but root 16 is +-4

[errant python]

EXACTLY

tahst what i was saying

[waxen steeple]

similarly shoudlnt x root be either -1, or 1

[errant python]

yep

[waxen steeple]

if we take x=1 as soln

the only thing going against it is that its not included in the domain

[errant python]

yeah

[waxen steeple]

which is what i was confused about

[hot herald]

in the initial equation, from the domain and range of the square root,
$$\sqrt{\red{x}} = \blue{x-2}$$
$\red{x} \geq 0, \blue{x-2} \geq 0$

[errant python]

the domain only is using the easier way

[hot herald]

x=1 doesn't work because 1-2 is negative, not a valid output of a (positive) square root function

[errant python]

which is by finding the positive square root

[waxen steeple]

its -1 right root of x can be -1 too

why doesnt it work

[hot herald]

because of what i said at the start

[errant python]

it works

[waxen steeple]

the domain thing?

[errant python]

how does the domain affect it?

[hot herald]

sqrt(1) represents the POSITIVE square root of 1
which is 1 ONLY, and not -1

[hot herald]

the issue here is more about the range

[waxen steeple]

what do you mean?i really dont understand this

sqrt(1) represents the POSITIVE square root of 1
could you please explain this statement

[errant python]

the domain is using the positive square root bc thats easier

[hot herald]

$$\sqrt{1} = 1$$
this only holds one single value

[glossy valve]

ℝαμOmeganato5

[waxen steeple]

wdym easier?

[hot herald]

sqrt(1) isn't -1 or 1,
it's just 1

[errant python]

wait

i dont get it either

arent there supposed to be 2 square roots to a square?

[hot herald]

again refer back to what i initially posted, this is just how the square root is defined

$\twosqrt{1}{1}$

[glossy valve]

ℝαμOmeganato5

[errant python]

just ask the texit app

i dont get it either

[hot herald]

forgot to add something earlier

in the initial equation, from the domain and range of the square root,
$$\sqrt{\red{x}} = \blue{x-2}$$
$\red{x} \geq 0$ and $\blue{x-2} \geq 0 \implies x \geq 2$ \
together you have the restriction that $x \geq 2$, as indicated by their $[2,\infty)$

[glossy valve]

ℝαμOmeganato5

[hot herald]

1 isn't in that interval

[errant python]

i looked up the defintion of extaraneous roots

thsi si what it said

[waxen steeple]

ok so we just have to check the doman before finding roots

[hot herald]

and if you were to plug x=1 in
sqrt(1) = 1 - 2
gives 1 = -1
which doesn't work, indicating that x=1 isn't actually a solution

[errant python]

so this whole thing was SUPPOSED to happen

[waxen steeple]

which is what confused me because i though root of any square gave 2 outputs

[hot herald]

misconception

[waxen steeple]

• and -

[hot herald]

its a misconception from equations like \
$x^2 = k, (k \geq 0) \
\sqrt{x^2} = \sqrt{k} \
|x| = \sqrt{k} \
x = \pm \sqrt{k}$ \
the $\pm$ doesn't come from the first step where you apply the square root

[glossy valve]

ℝαμOmeganato5

[waxen steeple]

ok thanks for the help

misconception i guess

.close

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[tender solar]

could anyone help on where to begin with this

[gritty rose]

!msgdel

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[cold wadi]

how did they get +7 , my answer was -3

[leaden ermine]

So first step is doing -f(x)

-f(x) = -3(2(x+2)/3)²+7

you here?

[cold wadi]

yes

[leaden ermine]

So next step is since it's a stretch you do -f(x/6)

[cold wadi]

yep to 1/9

[leaden ermine]

then you do -f((x-5)/6)

lastly