help-10

[boreal sleet]

du=18x+8 dx

dx=du/18x+8

[high lily]

um missing dx and how are you getting +8

[boreal sleet]

Aaaaaaa

du should be 18x+18

[high lily]

missing dx

its not optional

[boreal sleet]

ok

How does this become 1/18

[high lily]

factorise the denominator

you shouldn't be having issues with this sort of algebra when doing integral calculus
and i would highly recommend that you do a review asap

[boreal sleet]

lmao

thank you very much

.close

[obtuse pebble]

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[foggy flame]

I don’t know how to do bearing at all. Is there someone who can explain as well as help me complete this whole question?

[foggy flame]

There is also b

[unreal musk]

very tl;dr bearings are the clockwise angle from north [and written with leading zeros, e.g. 60 degrees clockwise from north would be written as 060 degrees]

From the information they've given, you can form a triangle

[foggy flame]

Ooh

Okay

I’ll try that, so tell me if it’s correct or not okay?

Like this?

[high lily]

no

that is way off from what you're supposed to be doing

start with a compass (NESW) axis centred at A

then draw a 200° arc clockwise from the north of A

can you show us what you have after doing that

[foggy flame]

Can you tell me what that is? I don’t know bearing terms too well

Do you need a compass?

[high lily]

draw a giant + shape,
with N at the top, E to the right, S to the bottom, W to the left

and the point A at the centre of that, where the lines intersect

[foggy flame]

Like this?

[high lily]

no

the NESW axis itself is fine,
but why did you draw a full circle

when i requested that your arc should only be 200°

[foggy flame]

How much is 200 degrees

[high lily]

do you know how many degrees there are in a right angle?

[foggy flame]

90 degrees

[high lily]

yes

use that to estimate where 200° will be

[foggy flame]

I’ll try my best

I think this? But it looks wrong to me

[high lily]

its good

as mentioned earlier, you should label the centre point (where the NS and EW axis intersect) as A

and draw a line starting from A in the direction of where your arc stopped

[foggy flame]

Like this?

[high lily]

yes

and for the part from B to C, same idea

draw another on of these axis centred at B,
make a 110° arc clockwise from north of B, etc...

[foggy flame]

Alright

Is that the whole answer? Or is there written equations?

[high lily]

this is just the diagram stage

and a decent diagram will give you an idea of where everything is
and what can be used to determine what's being asked

[foggy flame]

Okay

Is my other drawing correct?

[obtuse pebble]

@foggy flame Has your question been resolved?

[foggy flame]

Still doing it

Though I’m not really sure how

[sonic anchor]

draw it on the same diagram not two different ones

where you just 'pretend' that B is now the center

[foggy flame]

Ooh

How do you draw the line for C then?

[sonic anchor]

you draw the 110 degree arc around B then draw the line in the direction the arc stops

[sonic anchor]

or are you talking about the last line? that's just connecting A to C

[foggy flame]

Hmm

I’m not sure what you mean so I’ll send

[sonic anchor]

alr

[sonic anchor]

yea so what you're doing wrong is your still treating A as the center

like just pretend theres another compass with B as its center or w.e

[foggy flame]

Smol???

Sorry that just took me by suprise. I didn’t expect that

[sonic anchor]

it doesnt matter if you make it bigger

lmao

its just for direction purposes

so can you draw the arc from there and then the line?

and then the last line to connect A and C

[obtuse pebble]

@foggy flame Has your question been resolved?

[sonic anchor]

are you still doing it or are you stuck?

[foggy flame]

Stuck

Very stuck

So I went and did other problems but I’m back here now lMAO

[sonic anchor]

lol ok

[foggy flame]

No cause help

Idk what I’m doing wrong

But tbh I think

I’ll just ask my teacher

[foggy flame]

My friend who has done this proble said that there was a protracter involved

[sonic anchor]

yes to measure the angles

[unreal musk]

[you don't actually need one unless you want a very accurate diagram, but one isn't really necessary if the aim is for you to use trigonometry etc]

[sonic anchor]

trig not even necessary for this

is just pythagoras

or is that considered trig idk

[unreal musk]

[not really, hence my "etc" (meaning to "just work it out") - though there may be a part asking to find the bearing from C to A or something(!)
just that I've been doing problems like this haha]

[sonic anchor]

@foggy flame do you understand the diagram?

[sonic anchor]

oh i see, ngl in my country these problems dont exist

never heard of a bearing

[foggy flame]

Not too much

[sonic anchor]

hmm ill use the 2 different diagrams you did before

idk its hard to explain

[unreal musk]

@foggy flame may I ask where you're getting this question from (as in which qualification/level this is)? I'm thinking maybe a video might help to explain how to do these?

[foggy flame]

Grade 10 Past Paper

2018 P1R

[sonic anchor]

here i put your two diagrams together

[unreal musk]

Cool cool

[sonic anchor]

does that diagram make sense? like just the construction of it atleast

except the fact that 15km looks way longer than 30km 💀

[foggy flame]

Lmaoooo

It does

Okay but

I don’t think bearings are supposed to look like that

Like it’s sorta just like actaul ship bearings

But made into a Maths equation

[sonic anchor]

its just a visual representation

[foggy flame]

I just don’t know whot o do it

[sonic anchor]

its like having a protractor

putting the center on the point

and having 0 point upwards

[foggy flame]

Okay so how do you do the written equation? O-o

[sonic anchor]

then measure 200 and 110 degrees respectively

[foggy flame]

Sorry I’m not thinking straight anymore, it’s 2am

Actaully

I’ll ask my teacher for bearings like I said

[foggy flame]

Can you help me with something else?

[sonic anchor]

connect AC and you have a triangle

[foggy flame]

Wow traingle

[sonic anchor]

15km and 30km sides

and the trick is to find this angle here

110-x

[foggy flame]

Ooh

[sonic anchor]

x is the big green angle

so what is x equal?

[foggy flame]

100?

180 - 110 + 30

[sonic anchor]

hmm no

its this triangle btw

you have one 90 degree

and the other you can find out from the 200 degree bearing

[foggy flame]

Oh

You were asking about the traingle

And not the degrees

[sonic anchor]

yes the red angle is 110-x right?

[foggy flame]

How?

[foggy flame]

Yep

[sonic anchor]

so we need to find x

and the angles in the green triangle add up to 180

we have 90 and x and the last angle up there at A

lets call it y

can you tell what y is from the 200 degree bearing?

i put a blue angle there to help

[foggy flame]

Hmm um

I’m actaully not sure

Actaully wait brb

Nvm

[sonic anchor]

so 200 degrees you went 90 + 90 + 20 right

[foggy flame]

Yes

I think

I’m a little confused

[sonic anchor]

on what specifically

ok nvm so can you agree the blue angle i drew should be 20 degrees?

becuase the bearing is 200, so it goes 90 to the east axis, another 90 to reach the south, then 20 more

hmm maybe this is too hard

[unreal musk]

You could notice that the north-south lines are parallel, and so that 20 degrees [the (barely visible!) blue one] and the angle marked x are "Z - angles" so must be equal, if that helps any more

And then notice that 20 degrees is within that 110 degrees

[obtuse pebble]

@foggy flame Has your question been resolved?

[obtuse pebble]

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[fringe carbon]

not sure how to do this one

[gilded topaz]

first, what is the area of the circle?

[fringe carbon]

area is 25 pie

then what i gotta do

[gilded topaz]

then, note that $10 \pi$ is $\frac{2}{5}$ of $25 \pi$, so the angle of the sector should be...?

[warm shale]

FireBlazer

[fringe carbon]

still not really sure

gimme a sec maybe im overthinking it

yea idk wyha to do from there

[gilded topaz]

a sector whose area is 2/5 of the intire circle will have an angle that is 2/5 of 360

[fringe carbon]

i got it thx bro

[gilded topaz]

no problem

[obtuse pebble]

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[dense hornet]

-3*what=-b

[tardy epoch]

is what a variable?

or is b the variable?

[timber fox]

$-3 \cdot a = -b$ ???

[warm shale]

Clarkie

[static beacon]

,rotate

[static beacon]

wtf

[dense hornet]

What

[static beacon]

how is that even a question

[dense hornet]

the missing information

On top ofb

[static beacon]

do you know what b is?

[timber fox]

it has an answer ?

[dense hornet]

Idk

[timber fox]

im saying it does

[dense hornet]

12a-b

[timber fox]

$-3 * x = -b$

[warm shale]

Clarkie

[timber fox]

solve for x

[obtuse pebble]

@dense hornet Has your question been resolved?

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Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
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[shadow lava]

what is the rule here? composite function needs to keep the brackets? even after square and root cancel each other?

[frosty spoke]

why do you need a rule

[shadow lava]

for composite functions

[frosty spoke]

this follows pretty simply from what the first thing means

it has very little to do with composite functions

[shadow lava]

alright

so this answer is correct

(-Inf, 3] is the domain

[nocturne minnow]

You simplified that wrong

[shadow lava]

simplified what?

[nocturne minnow]

Wait, the circled part is wrong because of the lack of brackets

[shadow lava]

that's what I'm asking about

when you cancel a root like this

do you keep brackets?

not just as a composite function, just question in general

say if there was no radical, it was exponentials

[frosty spoke]

why are you memorizing stuff like this

[warm canopy]

it reads "3 minus the quantity (sqrt(9-3x))^2" which is the same as "3 minus the quantity 9-3x", i.e. what you wrote at the bottom

[shadow lava]

lol

[frosty spoke]

if you substitute something into it

[nocturne minnow]

Because it's $(\sqrt{9 - 3x})^2$

[warm shale]

dldh06

[shadow lava]

yes

[frosty spoke]

if you write something like a - b

[nocturne minnow]

You substituted into a function of a power

[frosty spoke]

and then substitute b = 3 - c

what do you think you should do

[shadow lava]

so I need to keep the brackets?

[warm canopy]

"keep"?

[shadow lava]

"i.e. what you wrote at the bottom"

[nocturne minnow]

.

[warm canopy]

right but there were no brackets before to "keep"

[shadow lava]

so this is what it should be

[warm canopy]

yes

[shadow lava]

OK

thank you

for being straight forward with the help

[nocturne minnow]

You're over complicating it. Because the function you plugged in was more than one term long, you needed brackets

[shadow lava]

yes I see that now.. I don't think I am over complicating, I am just forgetting the brackets when I cancel the radical

the radical implies brackets, that's how it is written with exponent

[nocturne minnow]

No, you're over complicating it

[shadow lava]

how am I over complicating it?

[frosty spoke]

yeah see no that's just gross to remember

[nocturne minnow]

If f(x) = -3x, you don't need brackets

[shadow lava]

of course, but this is 9-3x

[frosty spoke]

the root of all of your problems with brackets is that you're just memorizing a bunch of rules to put them in

and you don't actually know why they're true or anything like that

[shadow lava]

but it needs brackets

[frosty spoke]

it's actually very simple if you think about things in relation to values that you're substituting in

[nocturne minnow]

Hence

Because the function you plugged in was more than one term long, you needed brackets

[frosty spoke]

what, you think everyone memorizes all 10000 situations where they need to throw brackets in?

[timber fox]

just got flash banged

[shadow lava]

huh?

[shadow lava]

Because the function you plugged in was more than one term long, you needed brackets

what are the other 99,999 situations you are referring to?

[frosty spoke]

how many times have you asked about brackets

[shadow lava]

lol

[frosty spoke]

they're all literally the same rule

[nocturne minnow]

You realize that was also exaggeration

[frosty spoke]

if you change one value to the other, that value is meant to be in brackets

sometimes they're removable

[shadow lava]

why we acting like Terence Fletcher here.. I'm just asking simple questions!

[frosty spoke]

because you keep asking the same question and memorizing special cases

and making this FAR more complicated than it actually is

this literally comes down to if you have something = something + something else

if you substitute something else2 for that something else

[neon eagle]

Just think about the order in which u do things

[frosty spoke]

it's assumed that the something else2 is done first

[nocturne minnow]

And memorizing isn't a good idea either

[frosty spoke]

like think about why substitution works

it's because you're plugging something else that's EQUAL

[nocturne minnow]

You're trying to come up with shortcuts

[frosty spoke]

you're like "oh I'm doing derivatives I need brackets"

"oh I'm doing radicals, I need or don't need brackets?"

"oh I'm doing exponentials"

like it has nothing to do with ANY of those

[neon eagle]

^

it’s literally just that the order in which you do binary operations can matter sometimes

[frosty spoke]

it's a simple matter of the order of operations, and on a more fundamental level, you need to think about what substitution means

[neon eagle]

And when it matters u just put in brackets

[shadow lava]

I feel like we are misunderstanding my question, or taking it out of context. I am just asking if the brackets remain when you cancel a radical with multiple terms for the radicand. It's a pretty big deal for the final answer. Now I know the answer and these comments are crapping all over me for being so stupid to ask it in the first place

[frosty spoke]

you're proving my point right now

"when you cancel a radical with multiple terms for the radicand"

it has NOTHING to do with those special cases

[shadow lava]

Because the function you plugged in was more than one term long, you needed brackets

[frosty spoke]

NOTHING to do with radicals

[shadow lava]

oh

[frosty spoke]

NOTHING to do with derivatives

[shadow lava]

it's to do with the g(x) composition?

[frosty spoke]

NOTHING to do with composition

bruh

I give up

I think I've tried to explain this like 4 times to you already

but you just ignore it and keep doing this

[shadow lava]

then why does 3 - 9 - 3x become 3 - 9 + 3x

[frosty spoke]

IT DOESN'T

those two things are not the same

[neon eagle]

??

[neon eagle]

idek what the problem is

[warm canopy]

i said the thing you wrote at the bottom i.e. not the thing you have circled

[nocturne minnow]

It's 3 - (9 - 3x) = 3 - 9 + 3x

Brackets are needed

[shadow lava]

@frosty spoke needs to cool his jets here

[frosty spoke]

ok bye

I can teach u how to get some free microSD cards if u need them to store all of the rules ur going to memorize

[shadow lava]

it's not helping just getting angry

[nocturne minnow]

Not really, he has a valid point. You're practically memorizing and asking the same question over and over again tbh

[latent walrus]

you do have a habit of over complicating stuff - not everything is necessarily a unique rule

[neon eagle]

just think about it

[fathom flicker]

Avid, I hope you realize no one thinks you are stupid for asking these questions. Perhaps the channel is getting a little heated but that is because some of the helpers think that you are overcomplicating things for yourself by memorizing lots of rules about function notation/parentheticals that if you instead just understood the way they work in general would be unnecessary to memorize at all. You do ask the same types of questions a lot, which is fine if that is what you need, but I think the point a few of the helpers are trying to get across to you is that sometimes you choose to ignore the general behavior of how functions/order of operations work, which needlessly complicates a lot of your work. @shadow lava

[nocturne minnow]

Also @rocky goblet are you typing like an essay? I just see you typing for ages

[rocky goblet]

i'm just thinking a lot about what to type

[neon eagle]

its hard to explain how to know when you should put brackets

its something noone thinks about because ig its kinda intuitive

[balmy mortar]

If you're not sure, then put every pair in brackets, and don't let anything be implicit

1 + 2 + 3 - 4 = 1 + (2 + (3 - 4))

On the right, there is nothing unclear about the order operations should be performed.

[nocturne minnow]

Unless you have way too many terms and then you end up with like 10 sets of brackets

[balmy mortar]

No ambiguity

[frosty spoke]

write with postfix or prefix notation

get that stack ready

[balmy mortar]

unfortunately, the problem comes when you need to understand what someone else (or the problem) has written, in which case you do need to remember conventions that occur where brackets are missing

[rocky goblet]

tbh brackets don't even really "exist"
they're part of the notation, they're not part of the actual expression

[fathom flicker]

what's your question about that Avid

[shadow lava]

well.. just based on what Symbolab is saying (if we can trust them...)

• sqrt(x^2 + 100) is equal to just that, and there is no distribute law being applied

but - (sqrt(x^2 + 100)^2 uses the distribute law

[balmy mortar]

symbolab is a bot

so its explanations might be... off...
Anyways, I agree in this case

[fathom flicker]

that's because in one of them the square root is no longer there

the square root kind of acts as its own parenthesis

doesn't it?

[balmy mortar]

yeah...

its sqrt(1 + 1) for example

as opposed to sqrt(1) + 1

The distributive law being applied is from observing:

,,-(a+b) = -1\times(a+b) = (-1\times a) + (-1\times b) \ = (-a) + (-b) = -a-b

that middle step is the distributive law

symbolab didn't write it all out

[balmy mortar]

I've put in 'unnecessary' brackets in some places, but it makes it clearer.

[shadow lava]

OK

[sonic anchor]

the distributive law does not change the outcome anyways. you do the thing inside the squareroot first (PEMDAS) then multiply -1

[shadow lava]

OK

[sonic anchor]

in the other case you would do whatever is in the parenthesis first (PEMDAS) and multiply with -1

distributing just helps you algebraically sometimes

[balmy mortar]

Well back to the original question --- whenever you substitute into functions, yes you need brackets

,,f(x) = x^2 = (x)^2

[balmy mortar]

Why? Because let x = 2 = 1+1

[balmy mortar]

clearly f(1+1) should equal f(2), but it fails to be if you miss out brackets

1 + 1^2 is interpreted as 1 + (1^2), not (1 + 1)^2

[shadow lava]

thank you for the help, I will start applying brackets to my composite functions from now on, so that I don't forget the distribution

[warm canopy]

why 9x?

[shadow lava]

oops

[rocky goblet]

...that's such an overly specific conclusion to draw from this

[sonic anchor]

i mean intuitively its just you want to subtract all of x^2

and not only parts of it

[rocky goblet]

part of the point of what people were trying to say here is like

[balmy mortar]

its this.

[rocky goblet]

brackets don't have different rules based on what particular thing you're doing

[balmy mortar]

its 'all' of x being squared, not just 'part' of it

[sonic anchor]

and then later also subtracted

[shadow lava]

I don't agree with this statement, and I don't understand the point trying to be made

[rocky goblet]

here's a question: what actually is an expression?

[shadow lava]

it can be any math that doesn't contain an equals sign, or >, < signs, I think?

[rocky goblet]

ok well more specifically

is it just a sequence of symbols

or is it something else and the symbols are just a representation of that

[balmy mortar]

thats a hmm

I think of an expression as 'something to be evaluated'

(even though you might actually not evaluate it at any point)

Maybe 'something that could be evaluated'

[shadow lava]

that's true too

[balmy mortar]

So you can substitute any numbers for your letters

and you end up with just a number.

[rocky goblet]

...hm
this isn't the type of thing i was looking for and i don't know how to specify the distinction

[shadow lava]

according to cuemath.com An expression in math is a sentence with a minimum of two numbers/variables and at least one math operation in it.

[balmy mortar]

The importance of brackets is not just about function composition - it's fundamentally about how we express our... expressions

[shadow lava]

https://www.cuemath.com/algebra/expression-definition/

[rocky goblet]

i guess i'll just say what i consider an expression to be

[balmy mortar]

We need to make it clear what we mean when we write down a bunch of symbols

[latent walrus]

we still talking about brackets?

[rocky goblet]

ignoring square roots and stuff and taking just the most basic stuff, there are six ways to construct an expression:

1. a number
2. a variable
3. add two expressions together
4. subtract an expression from another expression
5. multiply two expressions together
6. divide an expression by another expression

[balmy mortar]

Thats the channel topic

[shadow lava]

i don't think so.. i'm not sure

[frosty spoke]

I always thought about it as a sequence of calculations you could do on the variables

and the written forms of the expressions are the same if they result in the same sequence of calculations, like on a computational graph

and obviously some expressions are almost the same as the others if you can just use some basic properties to transform them into that

[shadow lava]

only 6 ways?

[frosty spoke]

but like under that line of thinking, a + b + c is not the same as a + (b+c), but if you associativity, then you can transform it into a+b+c

[rocky goblet]

well there are way more if you consider more complicated things that i was ignoring there

[shadow lava]

OK.. what is the point with all of this?

[frosty spoke]

bee actually constructing a context-free grammar on expressions LOL

[rocky goblet]

well if you look at my definition of what expressions are

expressions are made out of other expressions

they're not strings of symbols, they're trees

[sonic anchor]

this is getting deep

[warm canopy]

avid just stick with this honestly

[rocky goblet]

i promise i am going somewhere with this

so

expressions are made out of other expressions, they're not just strings of symbols

so, the rules for manipulating expressions don't operate on symbols, they operate on expressions

[nocturne minnow]

.... That's where your over thinking comes, you're trying to take everything we say to heart and only focus on those instead of getting the full understanding. Like out of that message, your first instinct was to question if there was only 6 ways. Math isn't a set form, you have hundreds of different variations. What bee listed was probably the first 6 that came to mind

[shadow lava]

sounds to me like bee was over simplifying things

[rocky goblet]

yes, and i also said that i was oversimplifying

[sonic anchor]

because is an analogy

i think

[rocky goblet]

i could list enough types of expressions to construct everything, but that would be ridiculous and you wouldn't understand it

[warm canopy]

i think youre oversimplifying to the point of confusion honestly, avid knows what an expression is

[frosty spoke]

no the tree thing is actually a thing used in computer science lol

[balmy mortar]

The way we are trying to get at this is from a programming perspective

[rocky goblet]

i'm not sure they do though
an expression isn't a sequence of symbols and they're acting like it is

[balmy mortar]

yes, expressions should be interpreted in a 'fixed' way, as a machine does it.

[rocky goblet]

the brackets are just in the string of symbols, they're not part of the expression

so if you take something like addition being commutative

this says that if you add two expressions together, it doesn't matter which way round they are, the result you get is equal either way

so if you call one expression "x" and one expression "y", this is "x + y = y + x" (in standard mathematical notation)

or "x y + = y x +" in reverse polish notation, it's the same rule however you write it

what brackets do is just disambiguate how to interpret a string of symbols as an expression

[fading quest]

@shadow lava $$\sqrt{ax+b}\equiv(ax+b)^{\f12}$$

[warm shale]

Duh Hello

[rocky goblet]

if you have "x + y * z", it's not clear whether this means
multiply y and z, then add x
add x and y, then multiply by z

[fading quest]

thats all you need to know, it already has brackets

its in the notation

[nocturne minnow]

I like to think about that math is like chess. Probably a really terrible analogy, but the end is going to be the same, there are hundreds of ways to start, and as you get closer to the end goal, you have fewer possibilities.

[fading quest]

you should know what the notation you are using means

[balmy mortar]

i think this is info overload x10

[rocky goblet]

yeah i'm starting to feel like i'm losing track of which way i'm going with this

[sonic anchor]

at this point just never use brackets and trust your intuition

[nocturne minnow]

Same, I don't know what even happened

[balmy mortar]

more like always use brackets

[sonic anchor]

your intuition will guide the way

[fading quest]

i believe the only reason that we use the root notation over brackets is because it helps us remember when to put the $\pm$ sign in front

[warm shale]

Duh Hello

[fading quest]

but then again thats only true for even roots

[shadow lava]

yeah, the radical notation and exponential notation. it's when you have square root, for whatever reason symbolab is wrong about this..

[rocky goblet]

ok, attempt 2 at explaining:

brackets just disambiguate which order to apply things in
they don't actually interact with rules for manipulating expressions in any meaningful way, they are only part of how you write expressions

[warm canopy]

the point is avid doesnt yet have that intuition, they're trying to acquire it

[fading quest]

what is wrong about it?

[sonic anchor]

ye im just shitposting ill stop

[shadow lava]

why wouldn't they both follow the same rule?

[frosty spoke]

they do

[fading quest]

they are following the same rule

[shadow lava]

Oh

[fading quest]

just 1 has been simplified and the other cant be simplified

[balmy mortar]

The 2nd can be left as -(a+b)

[sonic anchor]

would you say -1*(a+b)=-a-b?

if so they're the same

[balmy mortar]

and sometimes -(a+b) is a better looking form to leave it at

[sonic anchor]

its what i said before. just because one is distributed it doesnt change the meaning of it

thats the whole point

[fading quest]

-a-b is 1 less symbol

[sonic anchor]

its just algebraic manipulation

or w.e you want to call it

[shadow lava]

alright, so brackets are optional, and it doesn't change the result

[rocky goblet]

well

[balmy mortar]

it certainly can !

[frosty spoke]

I guess what bee is trying to convey is that you do your algebraic manipulation rules on the underlying expressions

not on the list of symbols

I call it "not randomly jumbling the symbols around and hoping it works"

[balmy mortar]

you probably mean something different from what you are saying here

[fading quest]

this is not the takeaway you should be having from this conversation

[rocky goblet]

brackets disambiguate which expression you're talking about in cases where it's unclear

[shadow lava]

just in this case, I mean

[balmy mortar]

(1+1)-1 = 1
1+(1-1) = -1

1+1-1 is interpreted as (1+1)-1

[nocturne minnow]

Not quite because -(9 - 3x) and -9 - 3x are two different expressions

[sonic anchor]

no the bracket is still important

[balmy mortar]

but they are optional nowhere for the problem at hand

[fading quest]

i would reword it to "you can have 2 equivalent expressions even if one has brackets and the other doesnt"

[shadow lava]

right

[rocky goblet]

look at this: 1 + 2 * 3
is the result of this 1 + 6, or 3 * 3?

[shadow lava]

yeah

like how we can say radical is the same thing as brackets with exponent

[rocky goblet]

so we use brackets
1 + (2 * 3) is 1 + 6, (1 + 2) * 3 is 3 * 3

[shadow lava]

fractions are the same thing as (numerator) / (denominator)

even though we don't see the brackets with fractions, they are there, surrounding the numerator and denominator seperately

[frosty spoke]

not really there

it's more just that the fraction bar means you compute the numerator and denominator first

this is what bee was trying to say wrt expressions and trees

wait I said I was giving up

byebye

[rocky goblet]

yeah there aren't really "hidden brackets" there, it's just, the notation represents an expression

[fading quest]

id say its more true in this case that there are hidden brackets than the fraction case

[sonic anchor]

well its one way to think of it atleast

[fading quest]

the fraction case is just what people tell themselves to remember how they can manipulate expressions

[nocturne minnow]

Because $a^{\frac{m}{n}} = \sqrt[n]{a^m} = a^m \cdot a^{\frac{1}{n}}$

[shadow lava]

like if this problem was presented as a rational expression, it would make complete sense how to solve it

[warm shale]

dldh06

[fading quest]

a square root is by definition the half exponent of the expression in a root, aka parenthesis

[shadow lava]

but for this example, it's kind of an abuse of math notation to write it this way

[sonic anchor]

its not really is it?

i mean its still clear what the answer is no?

[rocky goblet]

yeah the entire point of this "problem" is that it's not actually valid

[shadow lava]

everyone agrees on the answer?

[fading quest]

you keep showing that example, but you know that there is a correct way to solve it by just left to right rules and bodmas/pedmas

[rocky goblet]

it's not a mathematical question, it's a notational question
any expression that that could represent has a trivial answer, the debate is over the ultimately meaningless question of which expression that is

[balmy mortar]

the answer is "brackets needed"

[sonic anchor]

no, but that's because people are more used to fractions

[balmy mortar]

or 'what did you mean when you wrote that'

[sonic anchor]

because its more clear

[nocturne minnow]

.... Let's not bring this up, overall it's ambiguous with it written like that

[shadow lava]

well, if we follow BEDMAS, it's brackets first. so 6 / 2(3)

[fading quest]

people who dont know that the answer is 9 to that are oblivious to the order of operations

although the expression can still be written badly

[rocky goblet]

i say that doesn't have an answer

[fading quest]

even tho it technically follows the rules

[balmy mortar]

[nocturne minnow]

Also 1 + 2 = 3

[balmy mortar]

This is about as bad as insisting the natural numbers have 0 in it (or dont)

[shadow lava]

yes, and then multiplication, division in whatever order comes first (left to right)
so 6 / 2 = 3
3(3) = 9

[rocky goblet]

here's a question: have more people been to russia than i have?

[balmy mortar]

What it evaluates to is down to the convention you are following for evaluating such an expression

[shadow lava]

probably not recently

[rocky goblet]

what does this question actually mean? which two things are being compared?

[nocturne minnow]

Here's a fun fact: did you know that there are more airplanes in the sea then submarines in the sky?

Unrelated but fun fact

[rocky goblet]

...that's not very surprising
it's not that hard to imagine how an aeroplane could unintentionally end up in the sea, while it's far harder to accidentally put a submarine in the sky (and i don't think anyone is intentionally putting submarines in the sky)
but also that's kind of irrelevant

[latent walrus]

[sonic anchor]

so what did you take away from this avid 💀

[balmy mortar]

[fading quest]

i dont know, there were quite a few nuclear detonation tests on submarines. one would have at least been airborne for a little while

[rocky goblet]

well it's not in the sky anymore is it

[fading quest]

perhaps some shrapnel landed in a tree

[rocky goblet]

is that really "a submarine" "in the sky"

[fading quest]

i dont know, maybe we need to define that

[balmy mortar]

this is a good example of a terribly asked question if you think about it carefully

what the heck the rest of the chat doing

[frosty spoke]

who says that planes aren't submarines though

they're just submarines that sink

[rocky goblet]

yeah it's a completely nonsense question, it doesn't actually mean anything
it just kind of looks valid until you actually think about it

[obtuse pebble]

@shadow lava Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[lone bolt]

why 1 =1 here

[lone bolt]

f and f' i mean = each other

[daring rock]

they're saying e^x is its own derivative

for the function f(x) =e^x, it's true that f'(x) = f(x)

[lone bolt]

but

y

shouldnt it be 0

if e is constant

c = 0

[daring rock]

e is constant. e^x is not constant

[lone bolt]

when derivative

whhats x

[rocky goblet]

the derivative of e is 0, but the derivative of e^x isn't zero

[lone bolt]

then

[torn wave]

x is the variable

[lone bolt]

if var in expo

by itself

then it = itself

but if its e^2x

then what

2x?

2?

[latent walrus]

2e^2x, would be the derivative

[lone bolt]

o?

[rocky goblet]

only if the base is e

[lone bolt]

2 stays same

in expo?

[rocky goblet]

the derivative of 2^x isn't 2^x

for something like e^2x just use the chain rule

[lone bolt]

chain rule would be

if prob:

e^2x^3

= 6e^2x^3?

[sonic anchor]

6 is not the derivative of 2x^3

[rocky goblet]

to clarify: is that $e^{2x^3}$?

[warm shale]

bee

[lone bolt]

yes

[rocky goblet]

so what's the derivative of $2x^3$?

[warm shale]

bee

[lone bolt]

6x^2

good?

[lone bolt]

= 6x^2 or = 6e^2x^3?

[torn wave]

Yes the derivative of 2x^3 is indeed 6x^2

[sonic anchor]

you know the chain rule

and know the derivative of 2x^3

[lone bolt]

ye

same sht

ye?

just

use

extended chain

but when

is it appropriate

to use ext chain

vs not

[sonic anchor]

idk what that means

[torn wave]

I think he's confused on whether chain rule is valid to use on certain equations and to what extent

[lone bolt]

^

[torn wave]

when the answer is that any valid function can use the chain rule, it's just redundant past a certain point

[rocky goblet]

what's the "extended" chain rule?

[blazing lark]

A chain rule within a chain rule maybe?

[rocky goblet]

as in if we're differentiating $f(g(h(x)))$?

[warm shale]

bee

[sonic anchor]

i think maybe if your are compositing into something that is already considered an inner function?

so you need to do it twice or w.e

idk i nver heard of it

[blazing lark]

Like cos(sin(x^2))

“Extended” use of chain, idk

[lone bolt]

i think

[sonic anchor]

ok so back to the question its d/dx e^u du/dx

[lone bolt]

if its

[sonic anchor]

u=2x^3

[lone bolt]

e^x^3

then u whip out the ext chain rule

[rocky goblet]

what is the extended chain rule though?

[lone bolt]

=ex^3 * 2x^2?

is this valid

[sonic anchor]

no

[rocky goblet]

"ex^3"?

[sonic anchor]

oh is it not to my question?

[rocky goblet]

is this meant to be $e^{x^3}$?

[warm shale]

bee

[lone bolt]

ext chain

[lone bolt]

here

[rocky goblet]

that's just the chain rule, i don't see what's extended about it

[lone bolt]

= e^x^3?

[rocky goblet]

then yes that's correct, the derivative of $e^{x^3}$ is $e^{x^3} \times 2x^2$

[warm shale]

bee

[lone bolt]

ty

[sonic anchor]

yea so using the notation you sent us f(x)=e^x and g(x)=x^3

so you use the chain rule

[lone bolt]

ty

guys

.close

[obtuse pebble]

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[crude pawn]

How do i solve this if f(t) as t approaches infinity is k

[fierce lagoon]

Use end behavior

[crude pawn]

So just k?

[fierce lagoon]

Since your evaluating t to infinity, you only care about the functions that grow real fast and much more

If f(t) as t -> inf is k, then f(t) might as well be k

2t > k as t -> inf

So you can literally not care about f(t)

5t > arctan(t) as t-> inf

[crude pawn]

Mk

[fierce lagoon]

So really your limit is equivalent to $\limit{\frac{2t}{5t}}{t}{\infty}$

[warm shale]

Umbraleviathan

[crude pawn]

Oh okay

.close

[obtuse pebble]

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[timid silo]

$-^3\sqrt{x^9}$

[warm shale]

okokok

[timid silo]

evaluate the expression

[heavy depot]

$-\sqrt[3]{x^9}$

[warm shale]

CrEpasPmkinPie

[timid silo]

ye

[latent walrus]

do you know how to express a cube root as an exponent?

[timid silo]

x^3

[neon eagle]

a cube root not a cube

[timid silo]

no

[latent walrus]

$\sqrt[3]{x}=x^{\frac{1}{3}}$

[warm shale]

AℤØ

[latent walrus]

the same applies to other roots, sqrt(x)=x^1/2, fourth root of x =x^1/4, etc

[timid silo]

ok

where did the negative go

[latent walrus]

this isnt the question i was just showing how to express roots

[neon eagle]

the stuff hes writing isn't related to the question

well it is but its not the question

[timid silo]

three thirds is one right

[latent walrus]

yeah?

[timid silo]

ok so its x^1/3 * x^9

[latent walrus]

no

its (x^9)^1/3

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[hollow sonnet]

How does this look?

[obtuse pebble]

@hollow sonnet Has your question been resolved?

[hollow sonnet]

<@&286206848099549185>

.close

[obtuse pebble]

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[teal turret]

@hollow sonnet

[hollow sonnet]

Yes

Figured it wasn’t gonna be seen lol

So I closed

[obtuse pebble]

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[slate breach]

I am stuck trying to formulate a specific algebraic sequence:
I am trying to calculate a specific Magic the Gathering theoretical question where if I have (Y) of something, then going 2(y+1)= x, then 2(x+1) = z, then on and on and on for 54 sequences. How would I formulate this?
I’m currently on n=14 and we’re in the billions already so if there’s a way to formulate this instead of manually solving for each sequence, that’d be good.

TLDR; trying to formulate a sequence that summates the result and the prior result together. I’m probably massively overthinking this.

[grizzled shore]

So n₀ = n₀
n₁ = 2 * (n₀ + 1)
n₂ = 2 * (n₁ + 1)
.
.
.
n_(k) = 2 * (n_(k-1) + 1)

[slate breach]

What would k represent?

[grizzled shore]

See how you have y then x then z

[slate breach]

Some number?

[grizzled shore]

k is just the number of these

So y = n₀, x = n₁, z = n₂

[slate breach]

Ok. Thanks! Haven’t done true math in what feels like ages 🤣

[grizzled shore]

If we sub in n₂, n₂ = 2 * ( 2 * (n₀ + 1) + 1)

So we get 2 * ((2n₀ + 2) + 1)

So the n₀ term goes up by 2^k

The constant term goes from 0 to 2 to 6 to 14

So +1 * 2

U₀ = 0
U₁ = 2U₀ + 2
U₂ = 2²U₀ + 2² + 2
U₃ = 2³U₀ + 2³ + 2² + 2
U₄ = 2⁴U₀ + 2⁴ + 2³ + 2² + 2

U_k being the constant term

Now rewrite it as $n_k = 2^kn_0 + U_k$

[warm shale]

Frosst

[grizzled shore]

Now sub in what we have here

$n_k = 2^kn_0 + \sum_{i=1}^k 2^i$

[warm shale]

Frosst

[slate breach]

I knew we had a summation in here 🤣 holy cow. I’m gonna attempt to plug things in here and see if it lines up with doing it the slow and hard way.

[grizzled shore]

Now let k = 54 and boom

There’s a cool thing you can do to the sum

$\sum_{i=0}^n 2^i = 2^{n+1}-1$

[warm shale]

Frosst

[grizzled shore]

Use this to simplify this

I believe you can do it

[slate breach]

Hmmm. I haven’t used summations in a very long time. I can give it a shot.

[grizzled shore]

You don’t have to prove it, but you can use it to simplify this

[slate breach]

Ok thanks!

So I tried plugging values in and it’s not going where I’ve already calculated so far.

Maybe I’ve possibly grown so old that I’m unable to understand expressions anymore 🤣

[grizzled shore]

Show me

[slate breach]

Sure thing. Image is downloading

Sorry, I’m in healthcare and a sticky note is the largest thing around me that won’t get me yoinked lol. Those are the first few numbers in the sequence, and I’m trying to find a way to simplify or jump to later numbers in the sequence.

[tardy epoch]

don't recommend getting arrested

[slate breach]

Gotta watch my t’s and I’s 😉

Wait, I figured it out. Thanks!!

.close

[obtuse pebble]

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[grizzled shore]

,calc 2^2 * 53 + 2^3- 2

[warm shale]

Result:

218

[obtuse pebble]

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[hidden sky]

State the degree and the value of the lead coefficient if the fourth differences are constant and equal to 12. type the full solution please

[frosty spoke]

sounds like something you'd say to chatgpt

[oak prawn]

chat gpt sucks for solutions like this

if you want a resource use homeworkify.net and paste a chegg link in there to unlock it

[frosty spoke]

I mean the prompt literally "type the full solution please" LOL

[oak prawn]

yeah no this is beyond me

[royal basin]

we are not chatGPT and we do not give out answers. @hidden sky

[frosty spoke]

I think that this is something that has an analogy with derivatives in calculus

so for example, if the second differences are constant and equal to 2

the degree is 2, and the lead coefficient is 1

because d/dx x^2 = 2x

[hidden sky]

State the degree and the value of the lead coefficient for the table

[oak prawn]

do you have any work attempting this thus far

[hidden sky]

The degree = 2
lead coefficient = 1/2.

this is what i got

is it correct?

[frosty spoke]

no

[obtuse pebble]

@hidden sky Has your question been resolved?

[obtuse pebble]

@hidden sky Has your question been resolved?

[obtuse pebble]

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[upper plover]

can anyone solve this problem?

[upper plover]

eliminate the arbitrary constant
x^2(3b-x)=(x+b)y^2

i just have a fight with my proffesor

[obtuse pebble]

@upper plover Has your question been resolved?

[high lily]

is this the original problem?

question doesn't make much sense

[obtuse pebble]

@upper plover Has your question been resolved?

[obtuse pebble]

@upper plover Has your question been resolved?

[obtuse pebble]

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[distant schooner]

f and g are functions defined by f(x) = sqrt(x-1), x(greater or equal to)1 and g(x) = sqrt(5-x), x(less than or equal to)5.
Find (f+g)(x) and state the domain

[distant schooner]

i have no idea

what f+g(x) is

[zenith vale]

well

you can write it as

f(x)+g(x)

its a composite functions

[distant schooner]

what does composite mean

so

sqrt(x-1) + sqrt(5-x)

?

[zenith vale]

yes

now find the domain

of it

as u know

[distant schooner]

how do i find the domain of thaaat

[zenith vale]

for f(x) its x >=1

[distant schooner]

yes

[zenith vale]

for g(x) <=5

so x belongs to

[distant schooner]

domain will be between 1,5

[zenith vale]

nope

see it again

[distant schooner]

whaat

for f(x)

its above 1

and g(x) is below 5

[zenith vale]

x is greater than 1 and less than 5 is it possible or not]

[distant schooner]

yes

[zenith vale]

yes

than your correct

but

remember

the functions are sqrt

and everything written under sqrt is must be positive

right

?

so answer is 1<=x<=5

$1<=x<=5$

[warm shale]

KiriSavage

[distant schooner]

uh

yes.

[zenith vale]

@distant schooner

you got it

?

[distant schooner]

isnt that the general rule where if there is a sqrt the number must be positive

[zenith vale]

so always check your domain by putting in the question

[zenith vale]

yes

it is

i sended you a msg

see it plss

do u got your answer buddy

you can close

now

[distant schooner]

thanks brother

.close

[obtuse pebble]

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[pliant epoch]

ive a 2nd order ODE here, but idk how i can break it into 2 1st order ODE, so that i can use euler's method to approach

[pliant epoch]

i understand what its going on here, but idk how to use it onto the equation above

[obtuse pebble]

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[obtuse pebble]

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[compact raptor]

okay for #88, can someone explain this part?

like they i dont get it

like how did they get that equation from "integrating"

[sonic anchor]

do you understand why integrating both sides gives us ln(y)=ln(cx^2)?

[compact raptor]

no

[sonic anchor]

what is the integral of 1/y?

[compact raptor]

lny

oh

[sonic anchor]

yes so that part is ok

[compact raptor]

lmao

[sonic anchor]

what is the integral of 1/x

[compact raptor]

lnx

[sonic anchor]

yes and a constant added to both sides btw

cause you integrate

[compact raptor]

yeah

but why is it only on x side

is the constant in this case like dy and dx?

[sonic anchor]

becase you can call the constant on the left side c_1 and constant on right side c_2

then subtract c_1 on both sides which gives you (c_2-c_1) on right hand side only

and just 'rename' that constant to c

[compact raptor]

but i thought c would be adding

and not multiple?

like

integral of dx would be x+C

why is it like cx in this case

[sonic anchor]

instead of calling it c lets call it c_3

[compact raptor]

why not lny+C=lnx^2 + C

[compact raptor]

ok

[sonic anchor]

you can then rename it and say c=ln(e^c_3)

ln and e are inverses they cancel eachother btw

so you have 2ln(x)+ln(c)

and use your logarithm rules

[compact raptor]

what

whyh do we have e now

ur confusing me

[sonic anchor]

what is ln(e^c_3)=?

[compact raptor]

c_3 but why does that matter

[sonic anchor]

yea so we havnt changed the value we just rewrote it

[compact raptor]

explain

we dont have e

so idk how this would make a difference

[sonic anchor]

oh sorry we rewrite c=e^c_3

[compact raptor]

i dont ge tit

[sonic anchor]

so we end up with ln(c)=ln(e^c_3)=c_3

[compact raptor]

dont get it

i get lny

and ln(x^2)

but where does the c come from

[sonic anchor]

ok, did you understand the first part of getting the constants to the right hand side?

[compact raptor]

i mean

wouldnt that be like

lny=lnx^2 + C

why is the c in front

[sonic anchor]

yea, just follow me step by step

dont rush to the ending

[compact raptor]

i dont get ur e process though

since there is no e

[sonic anchor]

yes please man just listen and follow step by step

[compact raptor]

ok

[sonic anchor]

we have $ln(y)+c_1=2ln(x)+c_2$ correct?

[warm shale]

Køter

[sonic anchor]

we rewrite this $ln(y)=2ln(x)+(c_2-c_1)$

[warm shale]

Køter

[compact raptor]

yes

[compact raptor]

yes

[sonic anchor]

c_2-c_1 is just a constant and we will call this new constant c_3 so $ln(y)=2ln(x)+c_3$

[warm shale]

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

[compact raptor]

yes

[sonic anchor]

ok all good so far

now if we take the constant c_3

and do this ln(e^c_3) it will not change the value right?

its still the same constant

[compact raptor]

i mean

yeah

[sonic anchor]

ok good

e^c_3 is also just a constant

we write this to c

so we have ln(c)

[compact raptor]

ok

[sonic anchor]

we end up with ln(y)=2ln(x)+ln(c)

still the same constant because ln(c)=ln(e^c_3)=c_3

right?

[compact raptor]

yeah

[sonic anchor]

so now we just use logarithm rules

$ln(a^b)=b * ln(a)$ and $ln(a)+ln(b)=ln(a * b)$

[warm shale]

Køter

[compact raptor]

oh ok

i get it now

thank u

[sonic anchor]

no problem

and then they e^everything=e^everything after

and yea solve for the constant c but oyu prob know that

[compact raptor]

wait

but how do u get the parametric equations

like where did these come from

[compact raptor]

wait are u still here

[sonic anchor]

ye im trying to look at the whole problem to figure it out

[compact raptor]

kk

thx

[sonic anchor]

@compact raptor

you solve for k

[compact raptor]

but what is k

[sonic anchor]

from this we know that x(t)=ke^(-2t)

and y(t)=ke^(-4t)

[compact raptor]

also one more thing

how does that become ln(c) /

doesnt it just become c_3

[sonic anchor]

becuase we call the constant c=e^c_3

[compact raptor]

u can do that?

[sonic anchor]

yea e^c_3 is still just a constant

like we did with (c_2-c_1)

[compact raptor]

ok

ohh

[sonic anchor]

so back to solving k

you agree with my x(t) and y(t)?

[compact raptor]

uh

let me see

[compact raptor]

how

[sonic anchor]

y'=ky is a common differential equation

solution being y=ke^(kt)

[compact raptor]

how do we know

[sonic anchor]

know what

[compact raptor]

like

[sonic anchor]

that its a solution?

[compact raptor]

how did u get x(t) and y(t)

[sonic anchor]

if you see a differential equation of this form: y'=ky

[compact raptor]

okay let me see

[sonic anchor]

its very common and thats the solution. i rather not go into the proof of why thats the case

[compact raptor]

so do we just isolate k?

[compact raptor]

wait i have a question

[sonic anchor]

but if you want more trust you can just differentiate it and see for yourself

[compact raptor]

why couldnt we just find y in terms of x

and then parametrize that?