help-23

[quasi bison]

that is a good strategy.

[zenith glen]

I'll do that first then

But I've no idea how to do the second one

[versed vector]

wdym

you can use the sin^-1/cos^-1/tan^-1 function on your calculator to find the angle

[zenith glen]

Wait

You mean the angle for BFC?

Hmm

[quasi bison]

BFC yes

[zenith glen]

Ohh

Sorry, I didn't read the question well enough

Thanks for clarifying

This was embarrassing xD

Thanks a lot

.close

[safe radish]

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[opaque sorrel]

Log n! = log 1 +... + log n ~ integrate (1, n) log x dx

can someone briefly explain this to me?

[upbeat adder]

the first step is Log n! = log 1 +... + log n

do you understand that part?

[opaque sorrel]

yes

its just expanding n! right

[upbeat adder]

yep

[opaque sorrel]

okay

whats next

[upbeat adder]

ok, now we'd like to get a rough idea of what that is equal to

and essentially, we're taking a sum of values of the logarithm function

we don't have a great way to know exactly what its equal to without just calculating

but we using the trick of integration to find an approximation

so usually we think of integration as summing up thinner and thinner rectangles under the curve of the function we want to integrate

[opaque sorrel]

right

[upbeat adder]

and so to approximate the integral, one might use rectangles of a certain width

say 0.25

[opaque sorrel]

okay

[upbeat adder]

so if I want to integrate a function f from 1 to 2, I'd say it's about f(1)*0.25 + f(1.25)*0.25 + f(1.5)*0.25 + f(1.75)*0.25

have you seen this type of stuff before?

[opaque sorrel]

not quite but i think i roughly follow

[upbeat adder]

if your step is of size i, then the area of the k-th rectangle under the curve is f(a + ki)*i where a is the start of the interval

f(a+ki) is the value of the function at the start of the rectangle, and i is the width

so by multiplying them you get the area of that rectangle

is that clear?

[opaque sorrel]

one sec im trying to sum up what uve been saying to see whether im understanding u properly

[upbeat adder]

I might be explaining it very badly but its the basic idea of integration, sum up lots of rectangles under the curve and the step size i tends to 0 to give you the value of the integral

[opaque sorrel]

okay so to sum up wat ur tryna say:
we are trying to find log(n!), aka finding the sum of values of the logarithm function

integration is a bunch of rectangles below a curve, so now were using this tactic to find the approximated value of log(n!)

we expand it to log1 + log2 + log3 + ...
so the rectangles under the graph would be like f(1) * a + f(2) * a + ..., where a is the area of a rectangle under the graph

[opaque sorrel]

yes

[upbeat adder]

I should have been more clear, I'm doing a bit of the detour with the integration stuff

what I'm trying to get at is that usually, you use rectangles to approximate an integral

[opaque sorrel]

right

[upbeat adder]

but in our case here, we will be doing the exact opposite, we'll be using the integral which we know the value of (since we know the integral of log x) and we'll notice that our log(n!) is actually summing up rectangles in such a way as to sort of approximate that integral

[opaque sorrel]

right

how does this imply log n! ~= n log n tho?

[opaque sorrel]

or this

[upbeat adder]

because notice that you're summing up values of logarithm with a step size of 1, each time you evaluate it 1 further

[opaque sorrel]

yeah

[upbeat adder]

so log 1 + log 2 + log 3 + ....

that's like setting i = 1

the width is 1 and the step size is 1

so whatever it is equal to should be somewhat close to the integral of log x, since its an approximation of log x

[opaque sorrel]

i dont quite get the i = 1

sorry, but do u mind drawing it out lol

i think itll be easier to understand on a graph

[upbeat adder]

I can try, yeah

[opaque sorrel]

tysm

[upbeat adder]

not sure how I'll send it

[opaque sorrel]

hmm

[upbeat adder]

I'll take a picture

[opaque sorrel]

maybe just a really brief sketch

yes sure

[upbeat adder]

sorry its taking a while

[opaque sorrel]

yep sure nw

ur the one helping XD

[upbeat adder]

darn

how do I rotate

[opaque sorrel]

umm

[upbeat adder]

there's a command

[opaque sorrel]

im not sure

yea

someone help

.rotate

,rotate

[opaque sorrel]

oh dam

,rotate 180

[opaque sorrel]

,rotate 90

[opaque sorrel]

aye

[upbeat adder]

there ya go

[opaque sorrel]

lemem take a look

[upbeat adder]

so the integral of log x is the area under the curve

and I drew these boxes here that seem to, if I were to sum up their area, be about equal to the area under the curve

and how do I calculate their area? well their width is always 1, and their height is just the value of log at the start of the box

so for the k-th box, it's log(k)

so we sum up their area by doing the sum of log(k) for k going from 1 to n

[opaque sorrel]

how did u know that their width is always 1

is it just something that we decided on

[upbeat adder]

yeah

I chose the box to be that way

[opaque sorrel]

ohh right

i follow yes

[upbeat adder]

so you agree that the area of all those boxes combined is our log 1 + log 2 + log 3 + ... ?

[opaque sorrel]

yepp

[upbeat adder]

ok, but now if we know the value of the integral, we get an approximation for that sum right?

since they roughly represent the same area

[opaque sorrel]

wait one momenet

so now i understood this: log n! = log 1 + log2 + ... logn

and that this is just the area under curve

for logn!

[upbeat adder]

not quite no

[opaque sorrel]

then now we are going to try to approximate it

oh

[upbeat adder]

log n! is just a number, its not a function

so log n! has no curve

instead, we know that log n! is exactly equal to the area of all the boxes I drew

[opaque sorrel]

ah so its a number that represents the approximated area under curve

[upbeat adder]

yes

[opaque sorrel]

under the log curve consisting of log1 + ... logn

okay

so now we are trying to find the approximation for it

which can be represented in
log n! ~= n log n or (the pic)

[upbeat adder]

sterling's approximation is actually quite hard to get to

[opaque sorrel]

oh i see

how about nlogn?

[upbeat adder]

the approximation we get is much simpler

n logn is what we get, yeah

[opaque sorrel]

ohh i see

[upbeat adder]

its a worse approximation, but much easier to establish

[opaque sorrel]

so how do we get to it o.o

[upbeat adder]

so logn! is the exact area of all the boxes combined

but the area of all the boxes combined is an approximation for the exact area under the curve of log x

therefore, log n! is an approximation for the exact area under the curve of log x

[opaque sorrel]

right

[upbeat adder]

but that goes both ways, the exact area under the curve of log x is an approximation for log n!

[opaque sorrel]

right

[upbeat adder]

so now we just need to calculate that exact area, and we get our approximation for log n!

and how do we calculate the exact area under the curve of log x?

[opaque sorrel]

its not by integration right? since we established that integration is an approximation

[upbeat adder]

integration is not an approximation

if you integrate the function, you get the exact area under the curve

I might have confused you with my previous things

the integral itself is exact

[opaque sorrel]

oh but the one we did is that we made the "boxes" larger and made the height log x

so the one we did is an approximation

[upbeat adder]

yeah

[opaque sorrel]

cus the real integration has much smaller widths

oh

[upbeat adder]

the boxes are an approximation of the integral

yes, exactly

[opaque sorrel]

ah i see

so we can integrate the log curve

[upbeat adder]

yep

[opaque sorrel]

to get the exact value

[upbeat adder]

of the area, yeah

there is a little detail that might be easy to miss, but its about the interval on which we integrate log x

[opaque sorrel]

wait i still dont quite see how we can get nlogn, sorry

i get everything weve discussed but not nlogn

[upbeat adder]

we'll get to it

[opaque sorrel]

oh

right

[upbeat adder]

so notice on my little drawing firstly that the very first box is actually just a line with no height, since log 1 = 0

[opaque sorrel]

right

[upbeat adder]

but also that if you count only the first box, then in theory you get an approximation of the area under the curve of log x between 1 and 2

[opaque sorrel]

which is 0?

cus theres no height

[upbeat adder]

yeah that's why I say in theory, obviously its a bad approximation but I'm making a point about the interval of integration

[opaque sorrel]

ah i see

[upbeat adder]

if I take the first 2 boxes, they go from 1 to 3, first box goes from 1 to 2 then second from 2 to 3

so if I take n boxes, I'm actually approximating the integral of log x from 1 to n + 1 instead of from 1 to n as you might expect initially

[opaque sorrel]

right

[upbeat adder]

so we get log n! ~= integral log x dx from 1 to n+1

[opaque sorrel]

right

[upbeat adder]

have you ever integrated the log function?

[opaque sorrel]

uh lemme think

nope

im looking at it online now

[upbeat adder]

ok, so there's a little trick

you can do it by parts, if you take one of the parts to be u = 1

log x = 1 * log x

it sort of comes out of nowhere but this is an integral that is just good to know

[opaque sorrel]

alright

ill try some later

[upbeat adder]

you can work it out yourself if you want, or you can just believe me that you get x*log x - x as the answer

you can check by differentiating that, it gives you log x + 1 - 1 = log x

[opaque sorrel]

x*logx -x is after integrating logx right?

[upbeat adder]

yep

[opaque sorrel]

right

[upbeat adder]

and since this integral has bounds, you just evaluate it at those bounds

[opaque sorrel]

right

okay wait so

log(n!) can be expressed as log1 + ... logn

then in the rectangles drawn

the first rectangle has height 0 so the area under curve is 0 which is theoretically right

then since to find exact area under curve for logn from range 1 ... n + 1, we integrate log(n!)

and we will get x *logx - x

then we have to sub the 1 and n + 1 inside

these are the "bounds"

then im assuming after this we would get xlogx?

[upbeat adder]

integrate the log function, not log(n!), but yes

[opaque sorrel]

ah right

[opaque sorrel]

is this right?

[upbeat adder]

well we can check

[opaque sorrel]

thing is, isnt this the exact value under curve

wait no

its because of the 0 - 1 rectangle

so its inaccurate but approximately accurate

i think

[upbeat adder]

yeah

not just the first rectangle

[opaque sorrel]

ahhh i see

[upbeat adder]

all the rectangles have a small mistake

[opaque sorrel]

the tiny gaps between all recntagles and the curve?

[upbeat adder]

but it gets smaller and smaller

yes

thats the approximation

[opaque sorrel]

ah okay i see

interesting

thanks so much dude

[upbeat adder]

no problem

[opaque sorrel]

when did u learn this

[upbeat adder]

I kind of learned it several times

but in discrete maths I did an exercise which made me kind of figure it out by myself

but I had seen very similar things before

[opaque sorrel]

ah i see..

do u mind if i ask u these types of qns in the future lol

cus ur explanation is really clear compared to alot of other people here ngl

but if u dont want me disturbing its fine haha

[upbeat adder]

I might not be available super often, but I don't mind the idea

[opaque sorrel]

ill close this

.close

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[feral linden]

Ask

Strange person

[trim venture]

Dr.Strange Multiverse of Madness

jk

see above

[upbeat adder]

woops

how did that happen

.close

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[lucid tartan]

Hey guys I've got a question. Was anyone in grade 10/ year 11 and planning to go to the ib or ap track? If anyone is now in thd ib or ap track can you tell me the topics you guys studied in that year 11 / grade 10 not only for math but other subjects please? Thank you.

[warm cosmos]

well im in 11 now in calc ab

so i dont think i can really help you

[lucid tartan]

Ohhh alright thanks tho. Tho do u have any advice that could be helpful for me this year?

[warm cosmos]

what year are you in, and what are your classes right now

[lucid tartan]

I'm turning year 11/ grade 10 my classes r math physics bio chem English drama moral education history i think that's pretty much it the rest is related to languages of my homecountry and stuff

Tho this year I will be taking calculus and that other stuff

[safe radish]

@lucid tartan Has your question been resolved?

[warm cosmos]

pretty good classes

I would just say organize yourself and dont fall behind on anything

[safe radish]

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[smoky aspen]

is this correct?

[cyan mantle]

4x4 matrix

[smoky aspen]

would you be able to explain

[safe radish]

@smoky aspen Has your question been resolved?

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[bold aurora]

I’m trying to do this with polar coordinates up top. Where am I going wrong? Is it wrong for me to cancel the derivative with the integral there somehow?

[bold aurora]

Nvm I was integrating by parts wrong at the end

.close

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[plucky elk]

Is this just spam?

Yea definitely self promoting spam

[severe laurel]

<@&268886789983436800>

[warped roost]

.close

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[supple coral]

Hello guys,

I have to learn about implication at the moment. I have a problem with the third case and hope someone can finally clear it out for me.

Let's say p->q.

Why is it that if p is wrong and q true the whole statement is true?

If I say it's raining is p and q the street is wet, the statement says its not raining but the street is wet. Why is this statement true if the forth statement says when its not raining the street is not wet. Doesn't the third and forth statement contradict themselfs?

And how can the street be wet when it's not raining?

I understand that the original statement when it's raining the street is wet does not have any relation with the third and fourth case cause the p is not fulfilled. But why do mathematicians still say that the third and fourth case are still true although we can't verify any of our original statement with it.

Isn't it better to write a question mark there cause we can't tell?

[severe laurel]

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

[open birch]

@supple coral p -> q means that 'p implies q'. Which, translated, means that if p is true then q must be true. If p is not true, the statement says nothing about q. q might be false, it might be true. Therefore, the whole statement is true because p was not true -- therefore the statement isn't wrong

[supple coral]

So you guys say that because we can't prove our statement in those cases we can't say it's wrong and therefor have to say it's true?

But doesn't that implication in logic bring a lot of problems with it?

If any of those cases in logic is used, it is not the truth, but it is the vacuous truth, which means it can't be proven. If it can't be proven logic that uses those cases is unusable in a practical matter?

@clr @hsp

[misty bay]

no

you could technically assign other truth values to such statements, but they would make a lot of dealing with it very inconvenient (you'd have to special case a lot of things in general statements)

for example, one pretty intuitive notion is that if something is true for all members of set B, then it's true for all members of set A if A is a subset of B

what if A is the empty set?

are you going to have to qualify "except the empty set" after each time you say that?

people choose to define p->q as "not p or q" because it makes everything much easier

we want things like q -> q to be intuitively true without considering the truth value of q

if you make the comment that "p -> q" should be false or indeterminate when p is false, then suddenly you have a whole lot of inconvenience

another example deals with the laws of arithmetic: maybe we like to formalize the fact that if x = y, then x + 1 = y + 1, perhaps as an implication (x = y) -> (x+1 = y+1). Under schemes where you don't accept vacuous truth, then are you going to say that the implication based on the laws of arithmetic is invalid or semi-valid when I say something like 0 = 1 and derive 1 = 2 from that?

[safe radish]

@supple coral Has your question been resolved?

[safe radish]

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[lean otter]

Find all f:R->R that satisfies

[lean otter]

I'm pretty much stuck on this, I tried plugging in a few numbers but I couldn't find anything useful

Can someone give me an idea on where to start on this?

[safe radish]

@lean otter Has your question been resolved?

[jagged bluff]

Not sure what the final answer will be, but definitely useful to plug in x=0.

[safe radish]

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[flint nest]

x2+y2= 25

[flint nest]

how do I isolate Y

I'm literally a 4.0 student, I just saved math for the last year of my degree because it gives me nightmares

whatever I don't have to defend myself

help

well I need to figure out if this is a function. the prof says to isolate Y

[dreamy comet]

x²+y²=25
y²=25-x²
y=+-√(25-x²)

[flint nest]

okay so not a function because of the square root

[dense token]

cos(cos x)

what rule do i apply?

how do i dfferntiate this?

[cedar rover]

#❓how-to-get-help

[dense token]

oh shit someonese channel mb

[cedar rover]

what do you mean by "not a function because of the square root"?

[safe radish]

@flint nest Has your question been resolved?

[cedar rover]

Here's what I mean, y = √(x) is still a function, even with the square root
As a rule of thumb, when y is raised to something even, it's usually not a function

[safe radish]

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[cedar rover]

f

[safe radish]

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[vital dirge]

What does z sub k mean?

[vital dirge]

I was watching ta video about De Moivres theorem

[safe radish]

@vital dirge Has your question been resolved?

[little geyser]

can you show the complete expression of z subscript k?

might be able to help than

[quasi bison]

@vital dirge there is not one n'th root but n

in this video they are named z_1, z_2, ..., z_n

[safe radish]

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[hidden oriole]

why when I move a function y=kx+b left horizontally for a units, the the function will be y=k(x+a)+b and not y=k(x-a)+b

[dense tiger]

Think about it graphically

[hidden oriole]

I thought about it graphically

still didn't work

[lean otter]

you're moving it to the left

[tranquil sentinel]

Instead of thinking of it as moving your function to the left by an amount a, think of it as moving your graph to the right by an amount a

[lean otter]

so you're moving it a negative amount

so say you move it 4 units left

then it would be y=k(x-(-4))+b

[hidden oriole]

how can it benefit me in anyways

I still don't get it

[tranquil sentinel]

It's how that works...

[whole acorn]

just think of it as a graph going up and down, if you add, the graph goes up and the function will look like it shifts to the left

[lean otter]

wait really?

you're moving the graph?

i thought it was moving the function

[tranquil sentinel]

It's the same thing seen in different ways

It's just to understand where the seemingly wrong sign comes from

[silver condor]

Think of a line y=x
y is 1 when x is 1 right

But if you give the x side a bonus one.
y=x+1
Then you can get the y value of 1 while x is zero

[whole acorn]

it'll look like it shifted to the left when in fact, it was moved up (+) and vice versa

[silver condor]

Plus one to x, is like saying x needs one less to get the same y value

[lean otter]

i always learned it like inside is opposite, outside is the same

[hidden oriole]

wait wait I get it now thanks

[lean otter]

so if it's inside parenthesis then you change the signs

and if it's outside then it stays the same

[hidden oriole]

ok

.close

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[soft pine]

Hi, I am stuck here in the "explain your reasoning" part I have done the calculations and have gotten K = 0,1 but I do not know how to explain the reasoning

[thin token]

the condition is

lhl=rhl= f(0)

[hexed night]

Exactly

You need to calculate the limits for x->0 for RHL and LHL

And then equate the two and find K

[soft pine]

i see

how would you explain that after you get the limits and find K

[hexed night]

What do you mean

We don't need a further explanation

[soft pine]

it asks you to explain the reasoning

so I thought we needed extra explainations

[hexed night]

The reasoning is as @thin token mentioned

[thin token]

yea basically write it in a full sentence form

[hexed night]

That's the definition of a continuous function at a point

[thin token]

yes exactly

[soft pine]

ohhh okay

I get it

thank you guys

[thin token]

.close

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[amber kelp]

Find the probability of picking 3 vowels and 2 consonants when 5 letters are picked (without replacement) from a set of alphabet tiles

[amber kelp]

So far I have 5/26 *4/25 * 3/24 for the probability of the vowels

But I think that might be off

[dense tiger]

Are you familiar with the choose function

Like 3 choose 2 =3

[amber kelp]

No

[dense tiger]

Are you familiar with factorials

[amber kelp]

Yeah

[dense tiger]

What class is this may I ask?

[amber kelp]

Algebra 2

[dense tiger]

Gotcha

[amber kelp]

So I’m learning P(ab)=P(a)*P(b following a )

For dependent probability

Or

P(ab)=P(a)*P(b)

For independent probability

[amber kelp]

Does that make sense

[hexed night]

@amber kelp have you been taught permutations and combinations?

Combinatorics?

[amber kelp]

Yeah

When order matters or doesn’t matter in statistics

[hexed night]

So you are familiar with nCr?

[amber kelp]

Yep

[hexed night]

Cool

[dense tiger]

n c r

is n choose r

[hexed night]

So the total ways of choosing 5 letters from the alphabet is 26C5

Do you understand why?

[amber kelp]

Ohhh that makes sense

Yeah

Except since every time you take the letter out and it stays out, do both numbers just drop one??

[hexed night]

No we are just talking about the number of ways for now

Not the replacement

[amber kelp]

Ok

[hexed night]

That calculation will be as follows: There are 26C5 choices for you to choose a 5 letter word. The choices to choose 3 vowels are 5C3

And the choices to select the 2 consonants are 21C2

So the total number of ways of selecting a word with 3 vowels and 2 consonants is 5C3 × 21C2

[amber kelp]

So 2100?

[hexed night]

Now to get the probability you need to divide this by 26C5, the total ways

[amber kelp]

Oh that makes sense

So 2100/65780

[hexed night]

Yes

[amber kelp]

Or 105/3289

[signal osprey]

damn i like this solution

nice KD

[amber kelp]

.close

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[amber kelp]

.reopen

[safe radish]

✅

[amber kelp]

How would you do this if you replace the letter

So you always have all 26 letters each time

.close

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[little sorrel]

idk how to do

[foggy salmon]

initial velocity is when t=0

acceleration is rate of change of v so differentiate v wrt t

when u get an expression for acceleration, solve for when it equals to zero

the last one u can take the time obtained in part c and sub it in to get max velocity

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[sand condor]

Need help

[sand condor]

Got stuck here

[upbeat swan]

,rotate

[safe radish]

@sand condor Has your question been resolved?

[sand condor]

<@&286206848099549185>

[safe radish]

@sand condor Has your question been resolved?

[molten crest]

What according to you comes next?

@sand condor

(2^1/3)^1/2 gives us what

?

[sand condor]

No idea

Because 1/3 doesent have same numerator

[molten crest]

(a^n)^m is the first question, let's tackle this step by step

And also

You've nearly solved the problem\

[sand condor]

Easy that’s a^nm

[molten crest]

LCM

Exactly

Now

Do the same with

(2^1/3)^1/2

[sand condor]

Dont we need same numerator

[molten crest]

Ye

We'll make them same

[sand condor]

Nearest is 6

[molten crest]

Again, first, what's this (2^1/3)^1/2?

Like you've said before

With the a^mn example

[sand condor]

1*.5 = 0.5

[molten crest]

The fractional form, please

Huh?

(2^1/3)^1/2 = 2^(1/3 * 1/2)

Got it until here?

[sand condor]

2^1/3 * 1/2 = 2/6 *3/6?

5/6?

[molten crest]

It's 2 raised to 1/2, remember?

[sand condor]

Yes

2^5/6?

[molten crest]

How did you arrive at 5/6?

[sand condor]

Cus we need same numerator

[molten crest]

Have you learnt how to multiply two fractions? Could you help me out with a/b * d/c

[sand condor]

Ad/bc

[molten crest]

Exactly

So, when we do 1/2 * 1/3, we get...

[sand condor]

1/6

[molten crest]

Yes

Good job

1/6

Now, we have 2^(1/6) x 2^(-4/3)

[sand condor]

Ok

[molten crest]

Now, we have to make the denominators equal

Can you tell me why?

[sand condor]

Because we have Base 2

[molten crest]

Yes

Because we have the base = 2 in both the terms

What happens when we multiply them?

What is a^b x a^c

[sand condor]

A^b+c

[molten crest]

Yes

So, now

We have 2^(1/6) x 2^(-4/3) = 2^{1/6 + (-4/3)}

[sand condor]

(-8/6)

2^-7/6

[molten crest]

Yes

Good job

[sand condor]

My answers are giving me diffrent

[molten crest]

Did you the calculations right until here?

[sand condor]

Is ours wrong or this

[molten crest]

Ours is wrong

[sand condor]

Why is ours wrong

[molten crest]

Wait...

[sand condor]

You got it?

[molten crest]

Yes

[sand condor]

What was it

[molten crest]

You missed a 2^1/3

[sand condor]

How

Idk How they got that answer

[molten crest]

ignore the arrow mark

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[novel magnet]

I don’t understand how they got the second step

[tawny bone]

Factor out 1/2 first

now you are only integrating

$2\int{u^{-3}}$

wellll

and then you can factor out the 4 as awell

[novel magnet]

Oh

Ok

[flat frigate]

Rαιη

[tawny bone]

so the 4 and $\frac{1}{2}$ cancels out to get the 2

[flat frigate]

Rαιη

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[pliant zephyr]

a

[worthy hemlock]

b?

[pliant zephyr]

sorry was just trying to claim haha

sorry

[cedar rover]

ahah it's ok, what's the question?

[pliant zephyr]

basically im doing some u-substitution but im stuck after getting u and du let me get the question typed up rq and ill walk though what i've done so far

[cedar rover]

If you don't want to type you can send a photo of your work

[pliant zephyr]

well actually its not much work ill just paste a pic of the question and explain what i have

so I have
u = e^2x
du = 2e^2x dx

but now I am confused if I even used the right term to sub cause i believe I could sub everything if I make e^2x u and the composition in the sin du and have 5/2 outside the integral

but i dont know if thats even correct

[cedar rover]

How about u = 5e^(2x)? That will make stuff easier

So we don't leave that 5 inside the sin

[pliant zephyr]

ah word so
u = 5e^2x
du = 10e^2x dx

yeah i just did that and it was 10x easier lol

[cedar rover]

Did you get it?

[pliant zephyr]

yes i got -1/10 cos (5e^2x)

tysm

[cedar rover]

Nice, well done

[pliant zephyr]

have a great day thanks again

[cedar rover]

u too

[pliant zephyr]

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[pliant zephyr]

$\int{4sin^{3}(t)cos(t) dt$

[flat frigate]

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

[pliant zephyr]

if i wanna find the intergral i could move the 4 out of the intergral and just do substitution to solve it right?

[proper crypt]

Yes, 4 multiplied by the integral

[pliant zephyr]

bet and if I wanted sin^3(t) to be U the deriative would have to do the chain rule twice for the exponent and the trig function right?

[final halo]

Think of a more simple sub

[pliant zephyr]

could i do just sin(t) and have u^3 du?

[final halo]

Hell yeah

[pliant zephyr]

word I just get confused with the format of trig functions and exponents lol

also since this is a definite integral so if I have sin^4 (x) I would find the value of sin(x) and take that to the power of 4 right?

[final halo]

Why do you need to do that?

[pliant zephyr]

well i didnt show it but its the definite integral from pi/2 and 0

[final halo]

Right so change your bounds according to your usub

And do the integral as you normally would but with u

[pliant zephyr]

for sure thanks

should i get into the habit of changing the bounds to my usub because what i typically do is just sub u with the original value so I can use the original bounds

[final halo]

As in, you replace u back with sin(x) at the end and use the original bounds?

[pliant zephyr]

yeah

i know it works but is it a bad habit or is it fine?

[final halo]

You can but you need to be careful in the intermediate steps

Something like this works

$\int_{x=0}^{x=\frac{\pi}{2}}u^3, \dd u$

[flat frigate]

ΣAC

[final halo]

So people know which bounds are for which variable

[pliant zephyr]

ah ok should I do it like that cause I just typically dont write the bounds for the U steps until i change back

[final halo]

If you don't put "variable=bounds" ppl will assume its the bounds for the variable at the end of the integral

[pliant zephyr]

bet thanks

[final halo]

Good practice to do that yeah

Good stuff keep at it

[pliant zephyr]

thank you have a good night man

.close

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[novel magnet]

I

[novel magnet]

I’m kinda stuck after this step

Cuz

It keeps on repeating

[lean otter]

what is that

polynomial division?

[novel magnet]

Yes

[lean otter]

what are you dividing by

$1+ux$?

[flat frigate]

jswatj

[novel magnet]

1 + 4x

[lean otter]

ohhh

I see

write it as $4x+1$

[flat frigate]

jswatj

[lean otter]

and your goal should be to get $4x\rightarrow 12x$ not $1\rightarrow 12x$

[flat frigate]

jswatj

[novel magnet]

Oh

So I divide 12x by 4x and not 1?

[lean otter]

yes

[novel magnet]

Wait how come.. this one video I watched he did the leftward digit first

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[solemn vault]

Hi

[solemn vault]

Q.28 Is the sketch of the img correct??

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[solemn vault]

<@&286206848099549185>

.cloa3

.clos3

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[unkempt solstice]

Wtf bruh

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[eternal sparrow]

Need help with solving log equations please?

[rose swallow]

Ok.

[eternal sparrow]

Here it is

[foggy salmon]

hustlers university is good

when u add logs of the same base, u can multiply what is inside aka
log a+log b=log ab

then from there take 6^ left and right side

[balmy sky]

lemme try

wait

[eternal sparrow]

I’ll try

[eternal sparrow]

Is that right?

[foggy salmon]

yep now take 6^ left and right side

[eternal sparrow]

Like that?

[foggy salmon]

6^ is 6 to the power of

alternatively just take the definition of log

eg
log a(b)=c means
a^c=b

[eternal sparrow]

Is that correct?

@foggy salmon

[foggy salmon]

^ means to the power of, not multiply

it should be 6^2 not 6*2

[foggy salmon]

which gets u 36

[eternal sparrow]

Ohhh wait I’ll fix it, sorry

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[eternal sparrow]

Thanks so much @foggy salmon

[foggy salmon]

np

[eternal sparrow]

I really appreciate it

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[foggy salmon]

yea seems right

[lean otter]

both?

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[olive terrace]

Hi everyone, quick question: what is the name of the rule/formula to get the 2nd equation?

[hybrid ermine]

its just factorization

[olive terrace]

But wouldn't it be:

[hybrid ermine]

Well you can write it that way, but you want to write the whole thing as a product of 2 factors

So you split the 2x as -3x + x

[olive terrace]

I'm not sure how you get -3x

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[karmic coral]

I have to prove the circled proposition by induction

Any tips ?

It’s easy when n = 0

[foggy salmon]

assume it is true for when it is k for some k in N

and try to use the statement u got from that to show that that would imply that Pk+1 is true as well

[karmic coral]

yeah that what I’m trying to do

how can I go from t^n+1 to t^n+2 in the integral ?

[foggy salmon]

so

wait sry whats the proposition

i cant really understand

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[inner urchin]

ok so i got everything in therms of x and i got: x=lny, x= -(lny)/3 and x = ln5 i am having trouble as to where i go from here

[inner urchin]

here is the graph

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@inner urchin Has your question been resolved?

[inner urchin]

close

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[wind mason]

.ask need to find a combination that equals 75

omg how i use this lmao

[tame charm]

Combination that equals 75? Wdym

[wind mason]

.show

first 3 are the inputs, the coloumn farthest right is output

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@wind mason Has your question been resolved?

[cedar rover]

it's unclear what you (or they) mean by "combination", but the three numbers of the input always give 58 if you sum them

[wind mason]

Input 1,2,3 = 75

The numbers I’ve posted is what I’ve tried

I can’t find the right combo to get to 75

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@wind mason Has your question been resolved?

[wind mason]

.close

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[dire fjord]

Hi

[dire fjord]

Why is x1 = a + delta x

[fiery prawn]

Each width is equidistant

i.e change in x

[foggy salmon]

they r saying a is kinda x0, x1 is y after x increases some interval, triangle x is the rate of change od y

[dire fjord]

I mean why is it a+ ndelta x

xn should probably be a+n delta x why is that

[foggy salmon]

oh wait the trangle is the delta x oops

no y's

[dire fjord]

Also why are we adding a

[foggy salmon]

they r just saying the width of one S is triangle x

[dire fjord]

a is the left endpoint of the interval [a;b]

That's exactly why they are adding a

Bc it's the left endpoint of the interval [a;b]

[foggy salmon]

sry im doing smt rn

[dire fjord]

It's ok np

Don't mind do what you are doing

Have a nice day in this time I'll be thinking about it

Ok now I understand the case for x1

x1 is a+delta x bc the right endpoint is x0+ width of the strip

Which is a+delta x

And that's also the case for x2,x3....xn

[foggy salmon]

wait so u kinda figured out ur question i guess?

[dire fjord]

Bc as we go from x1 to xn ,the width increases from delta x to ndeltax

[dire fjord]

Yes

I don't surrender to the question

I keep trying till I find a way to solve it

[foggy salmon]

this isn't exactly a question tho lol

[dire fjord]

I try to think in a different way if I see that the way of thinking I am using is useless here

Yea ik

It's not an exercise or something

It's an idea

But I should understand it to continue reading

Bc if I skip this without understanding this idea I may face problems later on

Anyways tysm for your time and have a nice day

.close

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[tranquil sentinel]

Where are you stuck?

[cobalt bough]

b) Idk where to start

[tranquil sentinel]

Think about what it's asking: what are the values of x that make y=5x-x² bigger than y=4?

[tranquil sentinel]

sorry for the typo y=4, not x=4

[cobalt bough]

2

[tranquil sentinel]

It's not going to be a single value, but an interval

Try to look at the graph to find your answer. I which interval does the graph of y=5x-x² goes above the graph of y=4?

[cobalt bough]

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[vast obsidian]

<@&268886789983436800>

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[warped roost]

.close

[vast obsidian]

Thanks

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[lean otter]

I actually have no idea how to go about this question?

[lean otter]

i can only guess that 16x^(2)+4x+1 is the answer because they are all perfect squares

or am i suppose to factor every equation before finding the answer?

it says incorrect

4 and 16 are also perfect squares so why isnt that the answer?

[glass pivot]

perfect square is when you can factorize and have (ax+b)^2

[lean otter]

wait is it saying the middle number must be 2ab

and a is 5 and b is 6

[glass pivot]

Because (ax+b)^2 = a^2x^2 + 2abx + b^2 when we expand

[worthy hemlock]

2abx

[glass pivot]

Ah yes sorry

[lean otter]

so to solve that i need to remember that equation

a^2x^2 + 2abx + c

and see which of the options fit that

[glass pivot]

Yes

[worthy hemlock]

Also, it's $(ax+b)^2 = a^2x^2 + 2abx + b^2$

[flat frigate]

dldh06

[lean otter]

so it factor a perfect square trinomial i just convert it from the second to the first equation above?

(5x+6)^2?

[glass pivot]

Thanks dldh06

[worthy hemlock]

What you need to do, for each answer choice, because you know it's a perfect square trinomial, you need to square root the a term and c term of each choice, then do 2ab, to find the middle term

So for example, 25x^2 + 60x + 36

What's the square root of a and c?

[lean otter]

5 6

[worthy hemlock]

So then what's 2ab?

[lean otter]

60

[worthy hemlock]

Is there multiple answers?

[lean otter]

no

i had another question here also

[worthy hemlock]

Then ask

[lean otter]

my teacher told me to factor a trinomial where a doesnt equal 1

then i have to find 2 numbers whose product is a times c and sum is b

do i apply that here to this equation to factor it

so 900 and 30 times 30

is 900 and 30 plus 30 is 60

i get (5x+6)^2

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[neat kiln]

For first multiply it by π/180

And second 180/π

Yeah coz π radians is 180 degrees

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[hollow siren]

Hello, This is messing with me. I don’t know how to start

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[thick sphinx]

Not sure if it is considered maths, but what exactly is the difference between the proposition form (rough translation from my language) and the boolean form in bool algebra?

[thick sphinx]

Let me know if its the wrong place to ask that

[polar lynx]

What do you mean by "proposition form" and "boolean form"?

@thick sphinx

[thick sphinx]

Erm hold on, I'll try to get examples

(A) A blue car
(B) Bill Gates is the president of the USA
(C) x * 4 => 25
(D) The nile river is the longest river in the world.

Rough translation of the question is something like:
Which one(s) are a proposition (or aren't), or a boolean form?

its... weird

[polar lynx]

Generally speaking a proposition can be true or false.

[thick sphinx]

yea, isnt that the same thing for bool?

[polar lynx]

I don't know what you mean by "boolean form"?

[thick sphinx]

hmmm

one sec

[polar lynx]

These things might end up depending on your specific definitions for "proposition form" and "boolean form" fwiw.

[thick sphinx]

Wait so

If we check the examples of earlier

x * 4 => 25

that cant be a proposition right?

because we dont know x? / can be both true or false

[polar lynx]

I'd say it depends on x. Like, if they mean x is a real number then x * 4 => 25 has to have some truth value, we just don't know what it is.

If not then it could mean anything so I'd say it doesn't have a truth value.

Whatever source you're using though should tell you via defns or examples whether they consider statements with variables in then to count as propositions I guess.

[thick sphinx]

Yeah, I'll ask the teacher next time I see him

Thanks though

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[lean otter]

how would i say "if x is smaller than r, y isnt a part of real numbers"

[eager ravine]

$$x < r \implies y \notin \mathbb R$$

[flat frigate]

jnmwn

[eager ravine]

@lean otter

[lean otter]

are there other ways?

[eager ravine]

maybe

[keen jolt]

can u help me too lenny

[lean otter]

just go in a free help channel :)

[lean otter]

using what

[keen jolt]

okay

[lean otter]

like what is x,r?

like sqrt(r-x) = y

??

if x is bigger than r its complex

you could say that

you could say either

but one is a statement the other is more of an equation

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[muted imp]

Let a, b be positive integers such that b^n+n is a multiple of a^n+n
for all positive integers n. Prove that a=b

[inner tapir]

is this true?

a = 1, b = 3, n = 1 holds...

i mean to say 3^1 + 1 is a multiple of 1^1 + 1, but clearly a ≠ b

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[peak estuary]

but it's not a multiple for every n. take n=2 and 3^2+2=11 is not a multiple of 1^2+2=3

[inner tapir]

🤦 of course mb

[muted imp]

I found this on the stackexchange, I just don't quite understand it

https://math.stackexchange.com/questions/988167/if-for-all-n-in-bbbn-an-n-divides-bn-n-then-a-b

[peak estuary]

which part of it do you not understand

[muted imp]

Yes

[peak estuary]

not helpful

[muted imp]

I mean as you said

Doesn't this demonstrate this is false?

[peak estuary]

no

because the statement says that a^n+n divides b^n+n for all n

[muted imp]

Ah I see

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[vague light]

Why is the ratio test usually used to find the radius of convergence?

[stray socket]

Quandale Dingle here. It's because the ratio test tests whether it converges absolutely, and within a certain domain because we have x, if that makes sense

Here

https://en.m.wikipedia.org/wiki/Radius_of_convergence

[vague light]

wikipedia is actually cracked out

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[river granite]

For a linear regression to be a good estimator of the relationship between two variables, it needs to satisfy a few assumptions. One of these is linearity in the relationship. As far as I understand, this assumption can be satisfied even if the relationship is quadratic, like this:

[river granite]

$$y_{i}=\alpha +\beta_{1} x_{i}+\beta_{2} (x_{i}^{2})+\epsilon $$

[flat frigate]

AlexanderJ

[river granite]

...because the linear model still 'sees' the variables linearly; the independent variables are still all only summing to create the dependent variable, and the model 'doesn't care' whether the x² term is actually "x squared."
(I got this from this StEx answer: https://stats.stackexchange.com/questions/92065/why-is-polynomial-regression-considered-a-special-case-of-multiple-linear-regres)

Another separate assumption is that variables aren't multicolinear. As an example: a video I'm watching is explaining that if the number of car/pedestrian accidents in a neighborhood Y depended on the number of people who live in the neighborhood X₁ and the number of cars that drive through the neighborhood X₂, then this would violate the assumption because X₂ is clearly affected by X₁.

But, if we can have a regression that features both x and x² in it, like the equation I wrote above, and it's a valid linear regression, then why doesn't featuring both x and x² in a regression break the "no multicolinearity" assumption?

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[river granite]

<@&286206848099549185> ? 🙏

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[mossy remnant]

Hello sorry for interrupting but I need help ;-;

[lean chasm]

• Ask your math question in a clear, concise manner.
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[mossy remnant]

So how do I solve that question, because I need to get the common monomial factor.

I've tried searching how to solve it but end up getting the same answer but without the solution

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[chrome pasture]

e²

[mossy remnant]

thank youu

[chrome pasture]

say .close when you’re finished 👍

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[warm cosmos]

please help me understand this

ik that the I is for the x value and the J is fot the y

but i get different answers when i use sin and cos

i got 28 for sin and 151 for cos

so idk what i did wrong

[worthy hemlock]

Post work

[warm cosmos]

its mostly on my calc

but ill put it on paper

you can plug in the values in your calc just to make sure

Do you see anything i did wrong

[worthy hemlock]

So I can't see anything wrong, but what I recall from problems like this you need to use tangent for the angle

[warm cosmos]

Ok lemme see what I get

I got negative 28

[worthy hemlock]

Well, next thing is, you know it's (-15, 8) meaning it's in the second quadrant, so you need to adjust the value of -28 to be in quad 2, by adding 180 or 360

[warm cosmos]

150 it is

so it would be either 1 or 2

and i would say 1 because x comes before the y coordinate

[worthy hemlock]

How would you know if cos comes before the sin?

[warm cosmos]

well the cosine function deals with the x coordinate while the sine function is with the y value and if we were to put that as a ordered pair it would be (x,y)

but idk

that just what i was thinking

[worthy hemlock]

Well, you can check that, because you are given 17<cos(151), sin(151)> so you can then do 17cos(151), 17sin(151)

[warm cosmos]

so it would be answer 1?

[worthy hemlock]

Did you check it?

[warm cosmos]

lemme see

i have to finish the other problems in the section to check it

but they are all the same just with different numbers

[warm cosmos]

yes!

it was right

i got wrong this question though

ohhh i see what i did wrong

i got 32

but since they were both negative

that menas it should of been on quadrant 3

ok anyways thank you

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