help-17

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[unique jolt]

1

[vocal sleet]

@unique jolt Has your question been resolved?

[lone linden]

Factor $-x(9-x^2)^{-1/2}$ out of the numerator and stuff cancels

[twin meteor]

Civil Service Pigeon

[unique jolt]

could you help me alittle more i dont understand how to

[lone linden]

What don't you get?

[unique jolt]

How would i factor it out when it has a negative exponent

i tried to flip it into the denominator but things got messy and i didnt even know if that was what i was supposde to do

[lone linden]

I assume you're encountering an issue with the second term in the numerator

[unique jolt]

is that not what youre supposed to do? where the (9-x^2)^1/2 goes into the denominator and it can cancel out with the same (9-x^2)^1/2 on the numerator?

[lone linden]

I'm just asking if that's what you're confused on

[unique jolt]

im honestly confused over everything

[lone linden]

so you don't know how to rewrite the first term so you can factor is what you're saying

[unique jolt]

no i do not

[lone linden]

where the (9-x^2)^1/2 goes into the denominator and it can cancel out with the same (9-x^2)^1/2 on the numerator?

I think you've become overly focused on rewriting the negative exponent as a positive exponent here

but would you agree that $$-x^3(9-x^2)^{-1/2}=(-x(9-x^2)^{-1/2})(x^2)$$

[twin meteor]

Civil Service Pigeon

[unique jolt]

yes you are taking out a negative from the -x^3 to flip the term?

[lone linden]

what do you mean by 'flip the term'

[unique jolt]

because the -x times 9-x^2)^-1/2 will make the -1/2 positive?

[lone linden]

are you trying to add the exponents

you can only add the exponents if the bases are the same

[unique jolt]

ok nevermind then ill just keep listening

yes i do see that it would be the same

sorry im just trying to have a clear understanding i wont get ahead of myself

[lone linden]

no no

Idk if you've felt rushed before to understanding something

but we're on discord

it's informal

it's one on one

there's no need to rush

if you want, we can back up and consider some simpler examples

[lone linden]

before coming back to this

[lone linden]

btw, in case it wasn't clear, this is a good thing lol

[unique jolt]

ok haha yeah im pretty new to this topic and i would appreciate simpler examples!

[lone linden]

alright

Ig I'll start by asking you this: Have you been exposed to factoring before in previous algebra courses?

If so, can you list out some examples of factoring?

[unique jolt]

i have a little bit of experience but this one is throwing me off

i was able to do a similiar question to this one but this one is making me confused

let me paste it:

i did this one by facotring out a 4x and a (x-9)^-5

[lone linden]

oh so you've done factoring with negative exponents before

are you confident on how you factored that

or are you a bit shaky with that question

[unique jolt]

all i understand is that when you factor negative exponents, you take the lesser one but what confused me is what was left from the (x-9)^-5 after taking out the ^-4 out

[lone linden]

alright

Yk that $a^b a^c=a^{b+c}$ right

[twin meteor]

Civil Service Pigeon

[unique jolt]

yes

[lone linden]

So we can rewrite $$(x-9)^{-5}=(x-9)^{-4} (x-9)^{-1}$$

[twin meteor]

Civil Service Pigeon

[lone linden]

since (-4)+(-1)=(-5)

[unique jolt]

ahhh ok

[lone linden]

I think the problem is that you've become too focused on memorising rules for specific examples

like this

when you factor negative exponents, you take the lesser one

[lone linden]

but can you see how you would do the question now rewriting it like this?

[unique jolt]

how? i dont see anything different from the original

[lone linden]

Oh I meant 'this' as in what I wrote here

I just repasted the question so you didn't have to scroll

[unique jolt]

oh you could replace (x-9)^-5 with

[lone linden]

exactly

and that would give you $$4x(x-9)^{-4}-4x^2 (x-9)^{-4}(x-9)^{-1}$$

[twin meteor]

Civil Service Pigeon

[unique jolt]

yes, thats the whole thing with it replaced

[lone linden]

mhm

actually I rlly should've said (x-9)^(-4) = (x-9)^(-5) (x-9)^(1) and substituted for (x-9)^(-5)

but you get the idea, right?

[unique jolt]

yes

[lone linden]

👍

ok

wait

frick

wrong picture

copy paste loves to be annoying

ok there we go

let's ignore the denominator for a minute

and just try and factor the numerator $$-x^3 (9-x^2)^{-1/2}-6x(9-x^2)^{1/2}$$

[twin meteor]

Civil Service Pigeon

[lone linden]

So applying the same technique that we used here, what would you get

[unique jolt]

so we are trying to make the two 9-x^2 equal so we can cancel them?

[lone linden]

wdym by 'equal'

are you referring to the exponents

[unique jolt]

yes

[lone linden]

yeah you could think of rewriting the exponent in one of the temrs to achieve that

So what would that give you

[unique jolt]

im not sure how to do it, but would you try to factor out the - in the exponent 1/2?

[lone linden]

take a shot at it and see what happens

[unique jolt]

ok

yeah i dont know how to other than doing the exponent rule of like a^-5 = 1/a^5

[lone linden]

same thing we did here

we want to rewrite the 1/2 so that we also get a -1/2

So you should get that ||1/2 = -1/2 + 1||

[unique jolt]

oh then maybe we could add another term of (9-x^2) multiplied by the (9=x^2)^-1/2 so it becomes (9-x^2)^1/2

since the -1/2 + 1 would make it positive

[lone linden]

yeah you have the right idea

So now we have $$-x^3 (9-x^2)^{-1/2}-6x(9-x^2)(9-x^2)^{-1/2}$$

[twin meteor]

Civil Service Pigeon

[lone linden]

do you see how you would factor it now

It's very similar to the other example you showed me

[unique jolt]

we could take both 9(-x^2)^-1/2 out of the equation?

[lone linden]

yup

factor it out

[unique jolt]

oh okkk

[lone linden]

no need to let me do everything

you're allowed to make mistakes

see how far you can get

[unique jolt]

okok thanks for being so patient >,<

[lone linden]

np

I want to be the person I didn't have growing up 🙂

actually I'm still growing up

I'm only 15 💀

but you get the idea

[unique jolt]

OK DAMN

[lone linden]

?

[unique jolt]

u got it together but thanks so much haha

[lone linden]

np

anyway

back to factoring

[unique jolt]

yeh let me get on it

[lone linden]

[unique jolt]

so then it wouold be -x^3-6x(9-x^2)

[lone linden]

can you write out the whole thing

including what you factored out

[unique jolt]

including like the denomatinor and the pic you sent?

[lone linden]

no just the numerator

ik you factored out (9-x^2)^(-1/2)

so write it with that

[unique jolt]

ok -x^3(9-x^2)^-1/2-6x(9-x^2)(9-x^2)^-1/2

[lone linden]

lemme rephrase

yk how 2x+8=2(x+4)

and that would be factoring out 2

I want you to rewrite the numerator including the part you factored out, in this case, factoring out (9-x^2)^(-1/2)

[unique jolt]

oh ok the way i did the last questions was different formatting and i dont know how i should forat this oen

[lone linden]

no just write it like this

$(9-x^2)^{-1/2}(\cdots)$

[twin meteor]

Civil Service Pigeon

[lone linden]

fill in the ...

[unique jolt]

(9-x^2)^-1/2(-x^3-6x(9-x^2)

?

[lone linden]

yup that looks good, just forgot a parenthesis on the end

[unique jolt]

i just put in whatever was left

[lone linden]

so now we have $$\frac{(9-x^2)^{-1/2}(-x^3-6x(9-x^2))}{x^4}$$

[twin meteor]

Civil Service Pigeon

[lone linden]

any ideas now?

[unique jolt]

should we foil the -x^3-6x(9-x^2)?

[lone linden]

oh 'foil'

It works, but it fails for anything other than multiplying binomials

You can just use the distributive property to say that $$(a+c)(b+d)=a(b+d)+c(b+d)$$

[twin meteor]

Civil Service Pigeon

[lone linden]

no need to memorise an acronym, you can actually understand it

ok that's my little foil speech

[unique jolt]

ok haha that was what i was taught in hs

[lone linden]

but you don't rlly need "foil" here

you can just use distributive

note that it's $-x^3-6x(9-x^2)$ and not $(-x^3-6x)(9-x^2)$

[twin meteor]

Civil Service Pigeon

[lone linden]

if that's what was confusing you

[unique jolt]

yeah i know theres like shortcuts for this stuff but i never really memorized them or was taught to

let me try the distributive

oh i see its like doig the steps but not actually calucating

i got -x^3(9-x^2)-6x(9-x^2)

[lone linden]

oh

[lone linden]

yeah you're getting confused with this

that's why parentheses are important

the first represents only multiplying the -6x, the second represents multiplying the -x^3 and the -6x

Repasting because I'm too lazy to scroll

[unique jolt]

oh are you only supposed to distribute the -6x to the (9-x^2)?

[lone linden]

mhm

[unique jolt]

oh ok i see

then that would be -x^3-54x+6x^3

[lone linden]

yeah

so now we have $$\frac{(9-x^2)^{-1/2}(5x^3-54x)}{x^4}$$

[twin meteor]

Civil Service Pigeon

[lone linden]

what ideas do you have now

[unique jolt]

im not sure all i see is to put the (9-x^2)-1/2 into the bottom and make the exponent positive

[lone linden]

we can do that

$\frac{5x^3-54x}{x^4(9-x^2)^{1/2}}$

[twin meteor]

Civil Service Pigeon

[unique jolt]

oh should we turn the 9-x^2 and put it under a sqrt?

[lone linden]

we could do that too

but focus on the numerator

[unique jolt]

oh we could factor out an x

[lone linden]

[unique jolt]

so x(5x^2 - 54)

[lone linden]

mhm

$\frac{x(5x^2-54)}{x^4 (9-x^2)^{1/2}}$

[twin meteor]

Civil Service Pigeon

[lone linden]

keep going

what else can we simplify

[unique jolt]

well could we factor 5x^2 - 54

i dont think so cause 54 isnt a power of something

maybe we can do the x/x4 = x^-3 or do the power for 9-x^2?

[lone linden]

x/x4 = x^-3

yes we can cancel a factor of x

$\frac{5x^2-54}{x^3 (9-x^2)^{1/2}}$

[twin meteor]

Civil Service Pigeon

[lone linden]

ok now here's where it gets murky

'simplify' is kinda vague

so for your online thingy

does 'simplify' mean no roots in the denominator

[unique jolt]

lemme see

im not sure it just says simplify the expression

[lone linden]

welp

welcome to playing the guessing game

[unique jolt]

what does it mean to have no roots in denominator

haha

i can ty looking at past answers

[lone linden]

do that

[unique jolt]

i got 3 submissions left o~o

[lone linden]

exactly what it says lol

[lone linden]

shoot some over and let's see

[unique jolt]

ok

[lone linden]

just send stuff that involves radicals/fractional exponents

stuff with only polynomials isn't going to help here

[unique jolt]

that latest one seems to be the only one with them

[lone linden]

welp

alright

I'm going to assume this is an american website right

[unique jolt]

yeep

[lone linden]

ok

I'm inclined to go with no radicals in the denominator, all fractional exponents converted to radicals

you mentioned rewriting with the square root

$\frac{5x^2-54}{x^3\sqrt{9-x^2}}$

[twin meteor]

Civil Service Pigeon

[lone linden]

and to "get rid of" the square root, we can multiply the numerator and denominator by $\sqrt{9-x^2}$

[twin meteor]

Civil Service Pigeon

[lone linden]

which gives $\frac{(5x^2-54)\sqrt{9-x^2}}{x^3(9-x^2)}$

[twin meteor]

Civil Service Pigeon

[lone linden]

take a minute to process if you need to

ik I just slammed you with a text wall

[unique jolt]

ok but couldnt you also sqrt it or does the - in x^2 make it not possible

[lone linden]

do you mean $\sqrt{9-x^2}=3-x$

[twin meteor]

Civil Service Pigeon

[unique jolt]

yes

[lone linden]

In general, $\sqrt{a^2-b^2} \neq a-b$

[twin meteor]

Civil Service Pigeon

[lone linden]

A classic counterexample is $a=5, b=4$ because then you have $$\sqrt{5^2-4^2}=5-4 \implies 3=1$$

[twin meteor]

Civil Service Pigeon

[lone linden]

As a nice exercise, you can figure out when $$\sqrt{a^2-b^2}=a-b$$ holds

[unique jolt]

ah right you actually have to square it first then calculate

[twin meteor]

Civil Service Pigeon

[lone linden]

anyway do you understand how we got here

[unique jolt]

yea you cancled the sqrt by multiplyng it self and you have to do it to the numerator as well

[lone linden]

yeah

alright see if that's what it wants

[unique jolt]

ok

it was right!!!

[lone linden]

🥳

[unique jolt]

idk if im pushing my luck, but could you help me with one more question? you are an amazing tutor

[lone linden]

sure why not

[unique jolt]

farthest i got to is factoring the denominators

[lone linden]

ok what did you get from that

[unique jolt]

i got 5/(x+5)(x-6)+ 13/(x-5)(x+3)

[lone linden]

that isn't right

recheck that

[unique jolt]

ok

i see it should the (x-5)(x+6)

[lone linden]

$\frac{5}{(x-5)(x+6)}+\frac{13}{(x-5)(x+3)}$

[twin meteor]

Civil Service Pigeon

[lone linden]

do you remember how you add fractions?

For instance, if I told you to add 3/7 + 2/11, what would you do

[unique jolt]

yeah the denominators should be equal and you have o multiply the numerator wtv u multiply the denom

i would multiply 7 and 3 by 11 and 2 and 11 by 7

thne add the tops

[lone linden]

I'm going to reword that for you

take the least common multiple of the denominators, rewrite both fractions to have said denominator, then add the numerators

well now you have the entire breakdown

so do yk how to find the least common multiple

[unique jolt]

i think i do but not in this case

like with 8 and 6 it would be 24

[lone linden]

In essence, what you do is find the highest exponent of each factor

and multiply all those factors

So if I wanted to find the least common multiple, of say, x(x+2) and (x+2)^2 (x+3)

you have factors of x, (x+2), (x+3)

and the highest exponents are 1, 2, 1, resp.

so you could have x(x+2)^2 (x+3)

omg

copy paste

I hate you

(this is why you don't have two convos at once)

ok here we go

[unique jolt]

haha

[lone linden]

my friends and I are basically ranting about exam questions

it's nothing special lol

[lone linden]

So here, what are the two expressions you want to find the least common multiple of?

[unique jolt]

is it the denomatinors

[lone linden]

mhm

so what would the least common multiple be

[unique jolt]

im not sure

[lone linden]

start by identifying your factors

what are they

[unique jolt]

(x-5)(x+6)(x-5)(x+3)

maybe you could convert it into (x-5)^2(x+6)(x+3)

[lone linden]

the expressions are (x-5)(x+6) and (x-5)(x+3)

the highest exponent of (x-5) that shows up in either is 1

so you would have (x-5)(x+6)(x+3)

it's certainly not wrong to do it with (x-5)^2 instead

but it makes it more complicated

[unique jolt]

oh ok yeah you just use it once cause its the same thing

[lone linden]

mhm

take the least common multiple of the denominators, rewrite both fractions to have said denominator, then add the numerators

continue on

what do you get

[unique jolt]

im a bit confused now

well my guess is x for all of them

i dont underswtand how to take the lcm of experssions

instead of this, would i be able to multiply the first term with (x+3) and the second with (x+6) to make the denominators equal

[lone linden]

that's what I meant lol

yeah do that

[unique jolt]

oh lol

[lone linden]

eh youtube probably has stuff

All I have for you is this

[unique jolt]

ok so then the denominator would be (x-5)(x+6)(x+3)

ad the numerator would be 5(x+3) + 13(x+6)

ok so after doing that i got 18x + 91

[lone linden]

don't think that's right

[unique jolt]

yeah oops

it should be 93 not 91

and i can factor out a 3 so 3(6x+31)

is there anything more i can do or is it completely simplified

[lone linden]

do you think there's anything else to do?

[unique jolt]

it doesnt look like it i cant cancel out anything or factor anymore

[lone linden]

[unique jolt]

eeeeeeee

ok thank you so much!!!! u are amazing

i cant imagine doing 3 more years of this trying to finish my math based degree O_o

[lone linden]

ooh you're in college?

any advice for me lol

[unique jolt]

yeaaaa i just started 2 weeks ago

hmmm well i mean ur amazing in math so dont worry about any math coursees

[lone linden]

being the idiot in my math friend group couldn't be me sobbing

[unique jolt]

well much better than me i suppose

[lone linden]

the prospect of college apps is terrifying 😭

idk where I want to go lol

[unique jolt]

yeah i wanted to do cs at UT austin but saddly didnt get in qq

[lone linden]

aw

[unique jolt]

so now im stuck at tis christian republican school sooo

[lone linden]

I get a lot of mail from religious colleges

I forgot to clean out the mailbox for 2 days and it didn't fit 💀

[unique jolt]

yeah its wtv im barely out of my room anyways but seeing these white men taking abut their favorite bible verse is disturbing

anyways i reccomend doing lots of extra curricular stuff that has to do with ur major/ollege of choice

im a special case tho im only here cuz i was number 2 of this crappy ass school

so i didnt have very much oppurtunites to build my resume

but yeah i reccomend doin stuff like that like personally i should have taken an internship and did morme summer courses\

[lone linden]

Maybe I'm not doing so badly after all

I just need to figure out where to go

[lone linden]

I'm in a specialized so I have the privilege of extra competition and standards 😭

anyway I assume you have more work to do so I'll leave you alone

plus I need to sleep

👋

[unique jolt]

alright bye bye u are amazing and choose wheerever u get that cash money or happiness

bye!!

.close

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[vocal sleet]

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[strange kiln]

I have zero clue what this question is trying to ask and I'm getting a headache trying to think about it

[hidden ermine]

aah

see, 6 * 10^4 = 0.06 * 10^6

i hope uve understood what the qs is asking👍

[strange kiln]

Confusion

[hidden ermine]

?

6 * 10^4=(6 * 10^-2) * 10^6

[strange kiln]

I understand it a bit better now but where does 6 * 10^-2 come from?

This is the only question I'm stuck on I finished everything else

[hidden ermine]

we have to ask ourselves the question: "6 x (what) x 10^6 = 6 x 10^4"

we also know that:
10^a * 10^b = 10^(a+b)

thus, (what) = 10^(-2)

becos then:
-2+6=4

[strange kiln]

I understand it now

[strange kiln]

So putting it into practice for number B wouldn't it be

0.4 x 10-3 = (0.4 x 10^3) x 10-6 ?

[hidden ermine]

yes

[strange kiln]

https://tenor.com/view/yess-yes-gif-25420589

You don't understand man I've been sitting here for an hour or two trying to piece that together

I literally got a headache from it thanks so much

[hidden ermine]

glad to help 😁

• When you’re done, you can close the channel with .close.

[strange kiln]

alright ima work on this one problem and then let you see it

then I'll close it and do the last one myself

[strange kiln]

Is my math mathing? I feel like it might be wrong

[hidden ermine]

the second part and third part are a bit wrong

[strange kiln]

yeah I had a feeling

[hidden ermine]

10^-3 * 10^6 is not 10^5

its 10^3

(-3 + 6)

[strange kiln]

I see where I went wrong

I thought I had to work my way backwards and solve it for 10^3 instead of 10^5

[hidden ermine]

oh

[strange kiln]

I got it now though thanks for everything

.close

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[pliant flint]

very confused on how to go about this problem

[pliant flint]

actually dont understand how a function (which I assume has to have a defined formula here) could have no fixed points but simultaneously when you take fof it would just have them?

[visual oracle]

you just need to find one example

surely you can

[pliant flint]

i mean ive obviously tried googling for an example, moreso just to sort of wrap my head around the idea of a function with no fixed points where fof has fixed points, but havent had much luck

[mild flower]

what about if f swaps points around in some way? then applying it twice would put them back

[pliant flint]

oh wait

so the fixed point is relative to treating fof as a single function

so fof(x) = x counts as a fixed point then?

[gritty sage]

Yes.

[pliant flint]

idk why but i was thinking of it like f(x) =/= x but then you'd need fof(x) = f(x)

guess thats a comprehension problem on my end

.close

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[pliant flint]

thanks for the help btw

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[echo smelt]

can someone confirm if im correct

[echo smelt]

2(3h+2) - 4h = 2h + 4

distributing

also i need helo with another equation

well i think i can solve my other equation but i wanna go step by step with someone

[echo smelt]

noice

but this other equation here

ill send it

[gritty sage]

OK, so do the distribution.

[echo smelt]

so ill call the fraction 0.5

ima just do the whole thing and see if its right

[gritty sage]

OK.

[echo smelt]

-1n - 2

final answer?

[gritty sage]

Yes, except we write -1n as -n (and we write 1n as n).

[echo smelt]

ye forgot

mb

ty

.close

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[gritty sage]

No problem.

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[midnight igloo]

,rotate

[midnight igloo]

What do you know about $\sin(2\theta)$?

[twin meteor]

Luke

[spare blaze]

2sinxcosx?

[midnight igloo]

Yep, so try substituting that into the equation

[spare blaze]

Got this

,rccw

[spare blaze]

I can cancel out 2

[midnight igloo]

You can,

try to make the trig functions the same

[vocal sleet]

@spare blaze Has your question been resolved?

[spare blaze]

Nvm I got the answer

.close

[vocal sleet]

Channel closed

Closed by @spare blaze

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[opal slate]

Hello, I was working with a helper with this one but he has stopped responding and never answered the question if it is correct or not. So is this correct?

[spiral inlet]

the thing you circled? or was that from the previous problem

That expression contains a variable called x, but there is no x in the original equation

[strong grove]

The answer is on right side

[opal slate]

i was aksing if number 4 is correct

the answer is the circled one

[spiral inlet]

where did x come from? and where did n1 and n2 go?

[opal slate]

this is what we ended up with but i dont know if its correct

idrk this is what he was helping me with

so it's wrong?

[opal slate]

no its the one i circled

[spiral inlet]

just to be clear, your answer for number 4 was arcsin(sin(x)) = x?

just because the way the picture is cut, I want to make sure that's not your answer from number 3 overflowing or something

[opal slate]

yes

sorry there was barely any space

[spiral inlet]

yeah you're right lol your teacher didn't give you very much to work with

[opal slate]

yes....

It's usually always like this too.....

[strong grove]

That doesn't seem right tho , u had n1 n2 ¤1 ¤2 but now u only have x

[spiral inlet]

I think I see what happened, I think you misunderstood the last person who helped you

[opal slate]

maybe

[spiral inlet]

Do you understand how you got here?

[opal slate]

not really something about arcsin

[strong grove]

Ah that was general formula

Arcsin cancels sin

So u were left with ¤2

Instead of x

[spiral inlet]

Yeah, I think your last helper told you arcsin(sin(x)) = x, which just means arcsin and sin cancel, and leave x

but in your case it means they leave theta_2 on the right side of that equation

[opal slate]

so what should i do?

[spiral inlet]

cancel the arcsin and sin on the right side of that equation

[strong grove]

Replace x with theta

[opal slate]

can u show me that on the tex thing

[spiral inlet]

The arcsin and sin cancel, that's all

that way you're left with theta_2 on the right side

which is what you want, since the instruction was to solve for theta_2

[opal slate]

wait how do i cancel them out

divide arc sin on both sides

[spiral inlet]

no, they just cancel and disappear

arcsin(sin(theta)) is the same as theta

so just write the same equation again but with theta_2 on the right instead of arcsin(sin(theta_2))

[opal slate]

ok, i will change it

one sec

[spiral inlet]

looks good 👍

[opal slate]

nice

i was wondering do you have time to check another ws i have

same type of questions

[spiral inlet]

Depends on how many questions lol

but I can at least help with one or two

[opal slate]

alr ima send it and u tell me

[spiral inlet]

Did you do these with helpers in this server as well?

[opal slate]

no i did them

[agile wadi]

\

[spiral inlet]

<@&268886789983436800>

ugh wtf

[dire robin]

number 9 is wrong

[opal slate]

omg

omg

omg

omg

omg

<@&268886789983436800>

[agile wadi]

WHY HAVENT I BEEN BANNED YET

[opal slate]

someone please help

[agile wadi]

IM TRYING MY HARDEST

PLEASE JUS BAN ME

[dire robin]

bro

get a life

[agile wadi]

PRETTY SPLEASE

[opal slate]

get out of here

[agile wadi]

when i feel like it

[opal slate]

honestly what are u doing

[agile wadi]

im jus uhh

idk

[spiral inlet]

don't engage with trolls

[agile wadi]

being a dickhead

[opal slate]

get off

[spiral inlet]

I guess there aren't many mods online rn

[opal slate]

ur right

[opal slate]

what do we do

[agile wadi]

yeah listen to your glazer

[agile wadi]

kiss me

[opal slate]

i am reporting you

to discord

[dire robin]

you can block him ig

[flat whale]

<@&268886789983436800>

[opal slate]

block him

[dire robin]

he left the server i think

[opal slate]

i blocked him

omg

[flat whale]

Mods still need to delete history

[opal slate]

that is absolutly disgusting

[spiral inlet]

he did leave

there's only one mod online rn unfortunately

[opal slate]

did yall report it though

like to discord

u can report it

alr

omg

i was never expecting this

[spiral inlet]

anyway

[opal slate]

yes anyway

[spiral inlet]

the last three problems look good

the first three are a little all over the place tho

[opal slate]

alr

do u have time to help me fix them

[spiral inlet]

just curious, you said you did all these yourself? the logic looks a bit different between all of them

[opal slate]

a lot of youtube videos i guess i started doing nonsense

[spiral inlet]

lol ok let's start with number 8, because everything there is right so far, it just isn't done

[opal slate]

whatever to get completion grade

[spiral inlet]

oh wait

[opal slate]

ok

[spiral inlet]

sorry, no it isn't quite right so far

[opal slate]

oh ok

[foggy glacier]

i am dumb

[opal slate]

?

[spiral inlet]

Maybe let's back up to the beginning of number 8

[opal slate]

ok

[spiral inlet]

You've got $T=2\pi\sqrt{\frac{L}{g}}$

[twin meteor]

tatpoj

[opal slate]

yes

[spiral inlet]

and you want to solve for g, which means, you want one side of the equation to be "g"

and the other side to have no g's at all

[opal slate]

yes

[spiral inlet]

so, what can you do to both sides of this equation as a first step, to get rid of some of the stuff that's clogging up the right side where the g is

[opal slate]

is it possible to divide l by both sides

or i guess multiply

[spiral inlet]

no, since the L is under the square root

we'll need to get rid of the square root first before dealing with L

[opal slate]

right

[spiral inlet]

but, even before that, we need to get rid of anything outside the square root

[opal slate]

ok

[spiral inlet]

so, how do we get rid of the stuff outside the square root?

[opal slate]

we square it too?

[spiral inlet]

oh, we could square everything

Not what I was thinking but yeah that works

[opal slate]

whatever is easier you tell me

[spiral inlet]

I think, just try to get the square root part by itself

meaning, get rid of the 2pi

how do we do that?

[opal slate]

ok

divide it

[spiral inlet]

so, if we divide both sides by 2pi, then we have

$\frac{T}{2\pi} = \sqrt{\frac{L}{g}}$

[twin meteor]

tatpoj

[spiral inlet]

(that was actually the same first step you took last time, that part was right)

[opal slate]

yes just about to say that'

niceeeee

[spiral inlet]

but, after that, now we need to get rid of the square root

how do we do that?

[opal slate]

do we square it?

[spiral inlet]

yep

[opal slate]

les gooo

[spiral inlet]

so, can you tell me what the equation will look like after you square both sides?

[opal slate]

wait

You can only put square on num,bers right?

[spiral inlet]

everything we're talking about is a number

L, T, g, pi, and 2

those are all numbers

[opal slate]

oh ok

[spiral inlet]

it's just that, for some of them, we don't know exactly what number we're talking about

like L might be 5 or it might be 5000, but it's definitely a number

[opal slate]

ohhh ok ok

[spiral inlet]

on the right, squaring will undo the square root, but on the left, there's nothing to cancel, so we just square everything

[opal slate]

so would it would be T^2/2pie^2 = L^2/g^2

[spiral inlet]

not quite

so we had $$\frac{T}{2\pi} = \sqrt{\frac{L}{g}}$$ then we square both sides
$$\left(\frac{T}{2\pi}\right)^2 = \left(\sqrt{\frac{L}{g}}\right)^2$$

[twin meteor]

tatpoj

[opal slate]

oh ok

i see

[spiral inlet]

so, on the right, the exponent cancels with the square root and just leave L/g

on the left, there's nothing to cancel with, so we just square it normally

$\frac{T^2}{4\pi^2} = \frac{L}{g}$

[twin meteor]

tatpoj

[spiral inlet]

notice how the T got squared, so did the pi, and so did the 2

[opal slate]

yes and not the right side because squaring cancels out the square root

[spiral inlet]

right, we still squared the right side

but it just canceled with the square root

[opal slate]

yes

[spiral inlet]

so, remember we want to get g by itself

[opal slate]

yes

[spiral inlet]

imo, the easiest thing to do next is get the g on top of the fraction instead of the bottom

and we can do that by just flipping both fractions over

[opal slate]

ok

[spiral inlet]

so, can you show me what that would look like?

[opal slate]

yes

[spiral inlet]

good, now there's just one last step to get g all by itself

[opal slate]

divide L on both sides?

[spiral inlet]

not quite, you already have g/L

which is

g divided by L

you need to do the opposite, to undo that

[opal slate]

multiply?

[spiral inlet]

yes, multiply both sides by L

[opal slate]

ayyyyyy

how do i do that for the left side?

[spiral inlet]

would you be comfortable doing something like

$3 \cdot \frac{2}{5}$

[twin meteor]

tatpoj

[spiral inlet]

it's basically the same

[opal slate]

i am coinfused on where the numbers come from

[spiral inlet]

I just made that up as an example

[opal slate]

oh ok

well yes i am comfortable with that

[spiral inlet]

but if you can do $$3 \cdot \frac{2}{5}$$ then there's no reason you can't do $$L \cdot \frac{4\pi^2}{T^2}$$

[twin meteor]

tatpoj

[spiral inlet]

It's basically the same

[opal slate]

ok i will show you what i get

[spiral inlet]

alright cool

[spiral inlet]

looks like you multiplied by L on the top and bottom

but L is not equal to L/L

L is equal to L/1

just multiply by L on top, not the bottom

[opal slate]

so the T is basically 1

kinda confused why

[spiral inlet]

well, no T isn't necessarily 1

but

$$L \cdot \frac{4\pi}{T^2}$$
$$=\frac{L}{1}\cdot \frac{4\pi}{T^2}$$
$$

[twin meteor]

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

[spiral inlet]

You only want to multiply by L on top

because L is L/1

L is not L/L

[opal slate]

ohhhh

ok

imaginary 1

forgot

[spiral inlet]

well i mean, it's not really imaginary. it's just true that L = L/1

L divided by 1 is L

L divided by L is 1

[opal slate]

yea i gust like to call that

[spiral inlet]

basically you have the right answer, except you shouldn't have L in the denominator

[opal slate]

so this is the correct one

[opal slate]

ok

[opal slate]

nvm

,oh thats the final answer since g is by itself now right?

[spiral inlet]

yep, since g is by itself, and there's no g on the other side

[opal slate]

ayyy

alr ima circle it

u wanna do 7 and 9 aswell?

[spiral inlet]

I'm willing to tackle number 9 but then I'm gonna have to be going

[opal slate]

ok

[spiral inlet]

What you want to do is factor v_f from the right side

do you know what I mean by that?

[opal slate]

no

[spiral inlet]

like how

ab+ac

is equal to

a(b+c)

[opal slate]

ok

[spiral inlet]

do you see how I did that?

[opal slate]

yes

[spiral inlet]

you can do the same thing with the right hand side of number 9

starting from the beginning, I mean

[opal slate]

vf(m1+m2)

[spiral inlet]

good, awesome!

Now, you want to solve for vf

so divide by (m1+m2)

[spiral inlet]

write it like this

before you cancel anything

(m1+m2) should cancel completely

and just leave you with vf

not vf+vf

[spiral inlet]

try factoring out vf first

don't worry about dividing just yet

just rewrite $m_1v_f+m_2v_f$ by factoring out $v_f$

[twin meteor]

tatpoj

[spiral inlet]

then worry about dividing

[opal slate]

omg my bad lemme fix that

[spiral inlet]

yes this looks good

[opal slate]

oh ok

[spiral inlet]

now you want v_f by itself

so you can divide by (m_1+m_2)

which will get v_f by itself on the right side

[opal slate]

thats weird i got Vf + Vf = Vf

but theres also vf on the left side

oh wait

we only divide by the factored side

?

[opal slate]

<@&268886789983436800> I don't know why this person wrote this here for no reason.

[spiral inlet]

what? no, if you have an equation that's true, you MUST do the same thing to both sides, or else it won't be true anymore

[opal slate]

ok

[spiral inlet]

if I have 6=6, I can't just multiply one side by something and not do anything to the other side, and expect the equation to be true anymore

18=6 isn't true

[opal slate]

but i keep getting vf + vf = vf

OH WAIT}
\

i think i wrote it wrong

[spiral inlet]

oh, I see, the m1+m2 doesn't cancel like you think, I think

since it can't be factored out, it isn't multiplied by the same thing

[opal slate]

so what do i do

[spiral inlet]

don't try to simplify it, just show me what you get

[opal slate]

i am not i am just dividinbg m1+m2 and getting vf from both sides

[spiral inlet]

no

$\frac{m_1v_1+m_2v_2}{m_1+m_2}$ is not the same as $(m_1+m_2)$

[twin meteor]

tatpoj

[spiral inlet]

You're trying to cancel the m1 from part of the numerator and part of the denominator

and same with the m2

but you can only cancel something that is a factor of the entire numerator and the entire denominator

[opal slate]

so v1+ V2

[spiral inlet]

hm

no

that's exactly what I'm trying to tell you not to do

[opal slate]

wait is m1v1 like combined

[spiral inlet]

the left side is simply $\frac{m_1v_1 + m_2v_2}{m_1+m_1}$. There just isn't any way to simplify that

[twin meteor]

tatpoj

[spiral inlet]

On the right side, however

[opal slate]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH\

[spiral inlet]

you can cancel $(m_1+m_2)$, on the right side

[twin meteor]

tatpoj

[spiral inlet]

and be left with

[opal slate]

wow thatys just wiw

[spiral inlet]

$\frac{m_1v_1 + m_2v_2}{m_1+m_2} = v_f$

[opal slate]

yes

'

yes

[twin meteor]

tatpoj

[opal slate]

ok i get it now

[spiral inlet]

sorry, I wrote m1 when I meant m2

fixed in that last one

[opal slate]

let me show you my work and u tell me

[spiral inlet]

yeah, just check your subscripts

numerator should be m1v1 + m2v2

not vf or whatever

[opal slate]

the left side for all of them is wrong and should be v1 and v2 rather then vf and vf right?

[spiral inlet]

$\frac{m_1v_1 + m_2v_2}{m_1+m_2} = v_f$

[twin meteor]

tatpoj

[spiral inlet]

I'm curious, what class are you in? this is like algebra or precalc or something?

[opal slate]

This is my summer packet for my upcoming physics class

[spiral inlet]

oh, you wrote vf in all those places, but that's not what the original equation said

[opal slate]

Yea that been the problem the whole time

I was confused cuz of that

[spiral inlet]

that'll get ya

[opal slate]

Yea

Ok I fixed it

Ima show u

[spiral inlet]

I hope this doesn't come off as rude, but I think it will be a big help for you to review some algebra 1 and algebra 2 concepts. In particular solving different sorts of equations

You'll probably be expected to be a bit stronger in this area in a physics class

[opal slate]

yea i should cuz this was diff for me

[opal slate]

yes def

it even said too on the previous page

this like most important

[spiral inlet]

Khan academy is awesome, and free.

Anything in their algebra 1 and 2 sections about solving equations

I think will be a pretty big help

[opal slate]

alr thank you

[spiral inlet]

sure thing 👍

[opal slate]

its correct btw right?

[spiral inlet]

yep, that looks good now

this topic in particular, you can probably find practice problems by searching for "solving literal equations"

[opal slate]

alr Thanks for everything

[opal slate]

alr

[spiral inlet]

literal just means lots of variables

[opal slate]

i will remeber that

right

alr thanks found the videos about it

[spiral inlet]

awesome 👍

[opal slate]

alr cya

.close

[vocal sleet]

Channel closed

Closed by @opal slate

Use .reopen if this was a mistake.

[spiral inlet]

see ya good luck 👍

[opal slate]

thank you

[spiral inlet]

no problem

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[simple urchin]

how..? i solved this and got a completely wrong answer

[simple urchin]

can someone acompany while i solve so they can see what i did wrong

[river minnow]

Sure

[simple urchin]

alright so

waht i did was

replace 2x+2 with the x^2 and the x after 2

and i got

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

after i simplify

i get

[river minnow]

Right

[simple urchin]

4x^2 + 9 - 4x -6 + 5

am i right?

[river minnow]

No, (2x + 3)^2 is not 4x^2 + 9

$(a + b)^2 = a^2 + 2ab + b^2$, use this

[twin meteor]

A Lonely Bean

[simple urchin]

OH WAIT

I FORGOT ABOUT THE REMARKABLE IDENTITY

I'M SO STUPID...

okay so

i would get

2x^2 + 2(2x+3) + 3^2

am i right now?

$2x^2 + 2(2x+3) + 3^2

ion know how to use it

[river minnow]

You meant $(2x)^2 + 2\cdot3\cdot2x + 3^2$ perhaps?

[twin meteor]

A Lonely Bean

[simple urchin]

uhh-

yeah

[river minnow]

Right so this should have been 4x^2 + 12x + 9 - 4x - 6 + 5

[simple urchin]

wait so

4x^2 + 12x + 9

is this not it?

yeah

then

mhm

exactly

so when i simplify

oh

yeah...

so i just need to remember this remarkable identity for hte exam?

(a+b)^2 = a^2 + 2ab + b^2

[river minnow]

Yes

[simple urchin]

ah perfect

now i understsand this question

there are others

how..?

how did we get 19...

i like COMPLETELY dont understand this question

like not at all

[river minnow]

I suppose they meant to say that y is a function of x

So, can you see the black 5 on the table?

[simple urchin]

yeah

oh

wait

y is a function of x yeah

do i just apply the rule that

there is only one unique y for 5

[river minnow]

Yes

[simple urchin]

and is that why hte other 5's y is 19

[river minnow]

Yes

[simple urchin]

oh so i just needed to know that?

thank

you

now

ONE last questio

and im done studying math

finally 🙏

so here

they want x

i haven't studied this so I don't know how to get x

i'm sure its like an equation or smt

how d o iget x?

[river minnow]

Extend the radius BO to a diameter BD and note that the angular measure of arc AD should be 112 degrees

And just use the fact that the angularm easures of BA and AD should add up to 180

[simple urchin]

i dont get it-

[river minnow]

Can you extend the radius BO?

To a diameter

[simple urchin]

oh

So up to angle 74?

BD?

[final zephyr]

sorry for barging in but i feel like you're overcomplicating a bit @river minnow

[river minnow]

No

[river minnow]

Is there a simpler approach?

[final zephyr]

try joining OA

[river minnow]

Ah yeah okay

You are right

I see

[simple urchin]

so in these questions

its only a matter of dividing things

[river minnow]

@simple urchin Forget about what I said and draw the radius OA

[simple urchin]

to get answers?

yup

[river minnow]

The triangle OAB is isosceles, right?

Because OB and OA are radii of the circle

[river minnow]

Right

[simple urchin]

oh

yeah

it is isosceles

[river minnow]

OB = OA so the angles OBA and BAO must be equal

Meaning the angle BAO is 56 degrees

So what should the angle AOB be equal to?

If the other angles in the triangle are 56 and 56

[simple urchin]

56..?

[river minnow]

No, the angle AOB

[simple urchin]

yeah

the triangle aob

wait

angle B is 56 degrees right

[river minnow]

Yes

[simple urchin]

so now

AO = [number] - 56 and then divide by 2

what is that number tho

180?

[river minnow]

AO is not an angle

[simple urchin]

unless im wrong

its a

[river minnow]

The angles should add up to 180, so you have AOB + 56 + 56 = 180

Also notice that AOB has angular measure x so that's just x + 56 + 56 = 180

And solve for x

[simple urchin]

oh

but

why is it

like

2 56's

is it

BO and OB

Or smt

[river minnow]

Because the triangle is isosceles

[simple urchin]

that are congruent

Oh

Oh yeah

okay

So when I have a number

and an iscocles triangle

i'll just do that number x 2 = 180

and then solve for x yh

x + 56 + 56 = 180

yup

okay

so

it'd be..

68?

wait

what-

the answer's 68

[river minnow]

Yes

[simple urchin]

oh okay

i have another example

or question