help-27

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[jaunty mantle]

Hello friends

[jaunty mantle]

So why is it like that on the right side

Doesn’t the preimage of {z} under f give only the 2 sets in the σ-algebra of M (domain of f) whose measure is z?

Or maybe a better way to phrase it is what does the preimage of {z} under f look like

Here’s what I think: all k ∈ M such that f(k) = z is included in the preimage of {z} under f, ie the union of all such k’s

[supple knot]

Wtf does that even say

[jaunty mantle]

;-;

Let $f$ be a non negative measurable function.
$\$1) Markov Inequality.
$\\int f d\mu \geq z\cdot \mu(\text{preim}_f({z}))$

[woven radish]

Frosst

[jaunty mantle]

What I think it should say:
$\\int f d\mu \geq z\cdot \mu(\text{preim}_f({s\in\mathbb{R}^{+0}| s\geq z}))$

[woven radish]

Frosst

[devout snow]

@jaunty mantle Has your question been resolved?

[arctic field]

[ f \ge z , \chi_{\set {f = z}} ]

[jaunty mantle]

Doesn’t that on his example give only the 2 points where z intersects the function?

[arctic field]

yes lol

[ f \ge z , \chi_{\set {f \ge z}} ]

[arctic field]

this is also true

[jaunty mantle]

But how do we measure just 2 sets

[arctic field]

2 sets?

[jaunty mantle]

Doesn’t the measure function take in a set (an element) of the σ-algebra?

[jaunty mantle]

Well the 2 points are 2 sets aren’t they?

2 sets (elements) of the σ-algebra

[arctic field]

[ \set {f = z} = f^{-1}(z) = \set {x \in \op{dom} f \where f(x) = z} ]

[arctic field]

it's a set

[jaunty mantle]

Ok

Then we can split them up and measure them separately?

Then we add them together?

[arctic field]

well the measure is defined on this set when f is measurable

[jaunty mantle]

But still shouldn’t it be this

Rather than equal

[arctic field]

so there's no need to "split" since μ of that set is well defined

[jaunty mantle]

Cos otherwise you don’t get everything between

[arctic field]

if you want everything in between then yeah

the inequality as stated is still true

[jaunty mantle]

What he’s saying here is that the blue area is less than or equal to the white and blue area

[arctic field]

yes

[jaunty mantle]

Sure but that’s only equality

Which fine you can have non-strict inequality

But that inequality part is meaningless

[arctic field]

i'm not sure i follow what you're saying

[jaunty mantle]

Let’s say the 2 points of intersection is a, b

[arctic field]

sure

[jaunty mantle]

mu({a, b}) and mu({a, a₁, a₂, …, b}) aren’t the same right?

Well, unless the middle bits have measure 0

Then they are equal

[jaunty mantle]

But if they aren’t measure 0 then the right side is bigger than the left side

[arctic field]

yeah

[jaunty mantle]

So what’s the point of saying int f dmu >= z mu({a, b}) when this is trivial, and equality only when the middle bits have measure = mu({a})

Wouldn’t it make more sense to talk about how it’s > when the middle bits have measure > mu({a})

[arctic field]

[jaunty mantle]

Hold up

I wrote bad things

[arctic field]

f could have a part which is constant at z

but i think they meant to write >= z

the usual markov inequality has >=

[jaunty mantle]

Ok if I’m not the only one that thinks there should be a >= so it’s not a trivial statement

Then I’m probably not going crazy yet

I mean I can see that the inequality holds, but is trivial

[arctic field]

i mean

not really trivial

it's still a bit of work

[jaunty mantle]

Well the integral part already measures {a, b}, so with the rest of the integral it must be at least the measure of {a, b}

[arctic field]

it's a consequence of this pointwise inequality of functions

yeah but f^-1(z) could be much more than just two points

and much worse behaved

[jaunty mantle]

Shouldn’t this just have f = z chi_{f=z}

Well it’s not wrong but isn’t it more reasonable

[arctic field]

what

what

it's not equal

[jaunty mantle]

Oh

Oh wait the chi only gives 0 or 1

So when f ≠ z, then f >= 0

Nice

When f = z, then f = z

So in general, f >= z chi_{f=z}

[arctic field]

yeah

[jaunty mantle]

Ok I see

Ok thank you

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[restive river]

Currently, I am working on polynomials. I am comparing the tasks and the content in the books. Could I have made a mistake?

[restive river]

I also watched the tutorial to remind myself

[pseudo basin]

what's the first term of the top one?

x^6?

[restive river]

5

[pseudo basin]

did you maybe mean x^5

[restive river]

yes

[pseudo basin]

okay so you're dividing x^5 + 3x^2 - 4x + 1 by x^2 - 3

[restive river]

-3

[pseudo basin]

and you are confused because terms of different degree ended up under each other

is that what happened

[restive river]

I have the same done task in my notebook and it doesn't say 6x^5

[pseudo basin]

...

this

does not answer my question at all

oh also $3x^2 + 3x^3 \neq 6x^5$ at all if that's what you thought

[woven radish]

Ann

[restive river]

ummm I divided x^2-3 by x^3 and got x^5-3x^3 when I subtract that, what do you get?

frst is 0 but yhen u have 6x^5 and that doesn't match my notebook

[pseudo basin]

you get $3x^3 + 3x^2 - 4x + 1$

[woven radish]

Ann

[pseudo basin]

no you don't have 6x^5.

no you don't have 6x^5.

3x^2 + 3x^3 is not 6x^5.

by your logic x + x is 2x^2.

which it isn't

[restive river]

so tutorial is bad

the signs will change and then we add

[pseudo basin]

no the tutorial isn't bad

YOU are bad for thinking exponents add like this.

do you really honestly think that 3 * x * x + 3 * x * x * x equals (3+3) * x * x * x * x * x?

[restive river]

+3x^2 - (-3x^3) is not 6x^5

[pseudo basin]

you're right it is not!

[restive river]

bro what is this

oh

when I can't just take subtraction i just rewrite them

cus that is +3x^2+3x^3

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[silver anchor]

Whats the best way to simplify this?

[winter torrent]

poly'l division

[silver anchor]

Could you help me with it? Idk how to solve it though

[supple knot]

https://www.youtube.com/watch?v=FxHWoUOq2iQ&ab_channel=TheOrganicChemistryTutor

the video has specific numbers, but you'll have to use variables a, b, and c

[winter torrent]

i would use long division over synthetic

[silver anchor]

Im trying to do it with long division but i dont get it

Ill try it again later thanks

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[supple knot]

https://www.youtube.com/watch?app=desktop&v=_FSXJmESFmQ

[winter torrent]

$\polylongdiv[stage=1]{x^4+x^3-2x^2-5x+6}{x^2-2x+1}$

[woven radish]

hayley!

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[restive river]

$y=2^{2} + 16x + 30$

[woven radish]

Chocolate

[restive river]

how can I turn this to vertex form

[trail eagle]

You're missing an x. Once it's fixed, you find the vertex form by completing the square.

[restive river]

great explanations, Thank you sir

short, using simple words, concise, brief, amazing explanation mate

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[stable atlas]

how do i get the 1/s part

[supple pasture]

Do you know about partial fractions?

[supple pasture]

You mean by how 1/s(s^2+1) is equal to 1/s-s/(s^+1) right?

[stable atlas]

yes

[supple pasture]

Ok

[stable atlas]

like i dont understand how that works

[supple pasture]

If you only want to verify it, you can use the common denominator method for adding and subtracting fractions

[stable atlas]

how do i do that

[stable atlas]

oh ik how to do this

i just dont know how to implement it into my qns

[supple pasture]

In this case a is 1, b is s, c is s, d is s^2+1

[stable atlas]

ahhh okay that makes sense

ty

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[hasty flame]

is my halfassed explanation gud nuf

[devout snow]

@hasty flame Has your question been resolved?

[ornate creek]

I need help

[devout snow]

@hasty flame Has your question been resolved?

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@hasty flame Has your question been resolved?

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@hasty flame Has your question been resolved?

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[short hinge]

I have no idea where to start

Please help me for 7

[winter patrol]

the 2 is on the wrong side, also make sure you write a multiplication symbol between the 2 and the fraction

[short hinge]

Ok

Wait but the big R

Is twice the size

Right

[manic condor]

Yeah. Exactly. It can be confused as a mixed fraction so make sure to use a multiplication symbol or dot.

[prime hollow]

If you multiply it it is 4 times

[short hinge]

Ok I will do that in the future

[short hinge]

But I divide by 4/3

Bc I transpose it

[manic condor]

Yes. So, one with R has more volume. Right?

[short hinge]

Yes

So 2 * formula for R = formula for r

[manic condor]

So shouldn't you be multiplying the smaller one with 2 so that volume of smaller one becomes equal to volume of bigger one?

[short hinge]

wait wtf yeah ur right

Idk what I was thinking thx

So it becomes that

7

Did I do it right? Like I got the same answer but I want to make sure I did it the right way

[manic condor]

Yes.

It's correct.

[short hinge]

Yay

Thx for the help

How to close

[winter patrol]

type .close

[short hinge]

.close

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[languid vessel]

i need help

you gotta work out the area right??

[lucid oracle]

No this graph represents the position of the boat, not the velocity. Remember that the displacement is simply the difference in the position of the object and is independent of the path taken when traveling between those two positions

Therefor the displacement is the distance between where the object is at time t=16 and at time t=0

which is -15 meters from our reference point minus 6 meters from our reference point

[languid vessel]

ohhhh

thx

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[wooden axle]

Can we apply epsilon delta of limit definition to check limit exist?

[wooden axle]

Or just check f(a+h) =f(a-h) existantance

[sour meteor]

You can

[wooden axle]

Can I drop a question?

[sour meteor]

You can

[wooden axle]

Checking limit of it by epsilon delta definition

[sour meteor]

No way it's asking this

[wooden axle]

What?

[sour meteor]

I don't know how to prof $\lim_{x\to 0 } \dfrac{\sin x }{x} = 1$ with eps delta

[woven radish]

tales

[sour meteor]

let the graduate + come

oh

its sin x²

[sour meteor]

Is it specifically written epsilon delta proof?

[wooden axle]

Nope

It's only me who is trying

[sour meteor]

Then you have to prove this with eps delta

[wooden axle]

I know it's exist

[sour meteor]

And prove product rule

[sour meteor]

that's easier

[wooden axle]

I don't know how to do it

[sour meteor]

I searched online and there is no proof

https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1

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[chrome stag]

I'm not sure how to calculate the original price of the car here or how to find the interest paid. The answer to one of them leads to the answer of the other ofc since I know how much Madeline paid for the car. Which do I solve for first and how? (Attachment 1 is the problem and confirmed correct figures I have come up with, Attachment 2 is the work I have done, and Attachment 3 is the given formula sheet)

[devout snow]

@chrome stag Has your question been resolved?

[chrome stag]

<@&286206848099549185> Sorry to bother, just using the ping since I can use it once after 15 minutes

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@chrome stag Has your question been resolved?

[novel totem]

The principal is the original amount of the loan - P by definition

[devout snow]

@chrome stag Has your question been resolved?

[chrome stag]

It was that simple, dang

[chrome stag]

Thanks for the help!

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[dry oxide]

help anyone

and please try to guide me further from the second step i did

because our teacher told this as a hint

i tried using sinC+sinD and inverse in the denominator

but didnt do anything

[red turret]

no need

[dry oxide]

oh yeah

lol

i forgot

[woven radish]

Dyssrupt

[dry oxide]

wait lemme do it this way

[red turret]

you should've noticed that the sum was 90 degrees

[dry oxide]

wait what do you mean, does this identity is apllicable if A+B=90?

[red turret]

no

sin(A+B) will be sin(90) which is 1

so the question wants you to add those

[dry oxide]

wait, i dont know this identity

[woven radish]

Dyssrupt

[red turret]

for denominator

[red turret]

um

use C D formula on where you left

you'll eventually get this

[dry oxide]

wait

this is what i did next

,rotate

[dry oxide]

i think i did something wrong?, didnt i?

[red turret]

combine the pairs to form sin(2x)

[dry oxide]

yeah i thought of that

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[devout snow]

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[supple pasture]

The page: https://en.wikipedia.org/wiki/Non-analytic_smooth_function

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[hardy tapir]

.reopen

what did I do wrong?

[formal bronze]

Show your work

[devout snow]

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[restive river]

I cannot understand these numbers or the concept of Using Quadratic Equations to Represent Motion

[eager lodge]

the quadratic equation is just a curve

the volleyball move along that curve

[pseudo basin]

it is difficult to give good interpretations to the numbers in this problem

[restive river]

thank you

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[restive river]

Hello, I'm needing help with my homework, not necessarily help me get the answer, I'm just a bit unclear what it's asking me. if anybody can let me know it would be great, Like does it want me to list each point on the unit circle or am i supposed to increment

[restive river]

is the problem

[winter patrol]

you're supposed to increment t
as well as write the respective point
(until you get all the red points )

[restive river]

okay so basically filling out each red bullet by incrimenting?

[winter patrol]

yes

[restive river]

Okay, thank you so much lol, I was a bit confused

[winter patrol]

e.g.
what's pi/4 + pi/4
and what are the coordinates of the terminal point there
and continue

[restive river]

oh okay

.close

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[wraith prawn]

i need help simplifying this equation (pardon the handwriting)

[timber pebble]

its just $3 \qty( \frac{9+13}{6} )$?

[woven radish]

jan Niku

[wraith prawn]

yep

[timber pebble]

hmm what have you tried?

[wraith prawn]

well i tried adding 9 + 13

[timber pebble]

whatd you get

[wraith prawn]

22

[timber pebble]

alright

[wraith prawn]

then 22

/ 6

and then just gave me repeating numbers

[timber pebble]

well you have options right

$3 \qty( \frac{22}{6} )$

[woven radish]

jan Niku

[timber pebble]

what else could you do here?

[wraith prawn]

well i dont know how to multiply with the 3 since its outside the quantities

[timber pebble]

hmm what sort of fraction rules do you know

maybe its helpful to write this as

$\frac 31 \cdot \frac{22}{6}$

[woven radish]

jan Niku

[timber pebble]

what can we do here?

how do you deal with these two fractions

[wraith prawn]

oh

3 * 22

[timber pebble]

sure

what do you get

[wraith prawn]

and 1 * 6?

[timber pebble]

yea

[wraith prawn]

wait up

[timber pebble]

$\frac{3 \cdot 22 }{ 1 \cdot 6}$

[woven radish]

jan Niku

[wraith prawn]

66 and 6

[timber pebble]

yea

$\frac{66}{6}$

[woven radish]

jan Niku

[wraith prawn]

so final answer?

[timber pebble]

this simplifies nice

[wraith prawn]

wait no

[timber pebble]

can you see it?

whats 66/6?

60 is 6 times 10

[wraith prawn]

11

[timber pebble]

yup

[wraith prawn]

then

[timber pebble]

cant get much simpler than that

[wraith prawn]

11 * 3

[timber pebble]

❌

[wraith prawn]

oh wait no

[timber pebble]

what makes you think that?

the 3 was uhh

[wraith prawn]

we already did something with the 3

[timber pebble]

whats the right word

well the 66 was created with the 3

[wraith prawn]

yeah a fraction

[timber pebble]

it lives inside 66

[wraith prawn]

yes

[wraith prawn]

what-

anyways

[timber pebble]

idk im just expressing that you already dealt with it

[wraith prawn]

yeah

so could you like make a list of what to do in this situation

like 1st step and last step

then ill be good to go

[timber pebble]

no

i dont know that there is a general process

maybe work outside-in

[wraith prawn]

sorry

[timber pebble]

youll get better at it as you do more

[wraith prawn]

alright

thanks

[timber pebble]

np

[wraith prawn]

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[still radish]

Why my thought was wrong? I had to solve lim x->0 (sin(3x)/x) so I multiplied the number by 3/3, so I could get a (sin(3x)/3x). The problem is that after multiplying by 3/3 I got lim x->0 (3sin(3x)/3x) and the first thing I have tried was split the limits (lim x -> 0 of 3/3x and lim x->0 sin(3x)/3x) I got a 0. According to my text book the answer should be 3, which would make sense if first I find the limit of sin and then multiply by 3, but the limit will make the denominator "disappear"? Probably I'm having an algebraic issue here but I can't see why resolving the sin limit it would make the 3x disappear. I am mistaken? I know that this is a stupid question and I should know the answer

[hybrid elk]

the first thing I have tried was split the limits (lim x -> 0 of 3/3x and lim x->0 sin(3x)/3x)

do you mean you did
$$\lim_{x\to0}\frac{3\sin(3x)}{3x}=\lim_{x\to0}\frac{\sin(3x)}{3x}\cdot\lim_{x\to0}\frac{3}{3x}$$

[woven radish]

light

[still radish]

Yes!

[hybrid elk]

youve applied the limit property incorrectly

[still radish]

Oh, god yes

I see it now, I feel like an idiot

[hybrid elk]

its not a big deal, just a small error to remember in the future

[still radish]

It should have been lim x->0 3sin(3x)?

Thanks light! It makes sense. I really appreciate the help

[hybrid elk]

im not sure how you got rid of the 3x in the denom?

[still radish]

Oh, no, you're right, I can't do that either

[hybrid elk]

$$\lim_{x\to0}\frac{\sin(3x)}{3x}\cdot\lim_{x\to0}\frac{3}{3x}=\lim_{x\to0}\left(\frac{\sin(3x)}{3x}\cdot\frac{3}{3x}\right)$$

[still radish]

Yup! Got it now! Thanks 😄

[woven radish]

light

[hybrid elk]

there we go lol

[still radish]

I appreciate the help

[hybrid elk]

you should split up the 3sin(3x)/3x into two components, (3 * sin(3x)/3x)

[still radish]

Yes, we're talking about this property, right?

[hybrid elk]

yes

you can factor out a 3 from the numerator

and after factoring out that 3, you're left with just limit of sin(3x)/3x as x > 0

[still radish]

Cool, gotcha. Thanks! 😄

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[burnt leaf]

How can I find the reference angle for 100?

It suggests to subtract using multiples of 2pi but I did but I’m not sure if I’m missing something or passed by an answer

[timber pebble]

use the definition

An angle’s reference angle is the measure of the smallest, positive, acute angle t formed by the terminal side of the angle t and the horizontal axis.

is 100 acute?

[timber pebble]

its maybe easier to just think of it as this definition like

make the smallest angle between your angle and the horizontal axis

[burnt leaf]

So it would be anything less than pi/2?

Or the nearest to 2pi

The one in red I tried close to 2pi and close to 0. Still no answer

[devout snow]

@burnt leaf Has your question been resolved?

[burnt leaf]

[burnt leaf]

<@&286206848099549185>

[devout snow]

@burnt leaf Has your question been resolved?

[burnt leaf]

https://tenor.com/view/zerosgifs-gif-20657439

[tall galleon]

The reference angle is the degree or radian angle that would result in you getting back to the horizontal axis the fastest. 100 is in quadrant 2, so figure out how you can get that angle which is in quadrant 2 to the x axis on the left.

That angle is either pi, or 180°.

@burnt leaf

[burnt leaf]

It’s in Radians

100 is not specified as degrees and I already tried it as degrees just in case it was that

[tall galleon]

Try converting 100 radians to degrees and then back to radians, and then subtracting 2pi until you’re within a radius of 2pi.

I find it to be more confusing when the radian angle does not include pi.

[burnt leaf]

https://tenor.com/view/dog-dog-fling-flung-dog-dog-flung-fling-dog-gif-24868321

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[devout snow]

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[eager latch]

I need to express the sequence {...,-3/10,-2/5,-1/2,0,1/2,2/5,3/10,4/17...} using N as my ambient set

[eager latch]

Does it look better to say:

[woven radish]

Invictus

[eager latch]

or,

[woven radish]

Invictus

[eager latch]

typo: ignore the ^2 in the numerators of both sets

[faint zinc]

I would consider the two equivalent

[devout snow]

@eager latch Has your question been resolved?

[eager latch]

yes definitely, but which is more aesthetically pleasing

i prefer the 2nd but i guess the 1st is more "detailed"

[faint zinc]

I prefer the second

And do not think the 1st is more detailed

It's just more verbose

[devout snow]

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[restive river]

Like, $\sum$ do $\prod$ also have property if so what are they can you provide a yt video or explain?

[woven radish]

.fire007

[blissful cloak]

here is a video i found after searching for a bit about sigma
https://www.youtube.com/watch?v=st-6vZESf8A

capital pi is used similarly to sigma, except instead of adding were multiplying

https://www.youtube.com/watch?v=NdXWFvb21Yk

[restive river]

Does it follow the property that $\prod a \pm \prod b = \prod a \pm b$?

[woven radish]

.fire007

[restive river]

With same limit

No right?

[blissful cloak]

nope

[restive river]

Alr kul

Thanks

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[blissful cloak]

np

[devout snow]

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[pastel onyx]

Is it accurate to call this slightly left skewed?

[weak cove]

Left skewed is the opposite

[hard rivet]

Well this distribution doesn't have a very significant skew, so I'd suggest that if you are going to call this a skewed distribution, include a comparison of the mean, median, and mode in your justification.

[hard rivet]

I think @pastel onyx is correct in calling this a (slightly) left-skewed distribution?

[pastel onyx]

okay thanks

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[lusty swift]

Why is the domain of (f∘g)(x) = ((sqrt(x+6)^2)+5 [-6, +inf) and not (-inf, +inf)? (f∘g)(x) = ((sqrt(x+6)^2)+5 simplifies to x+11. Why do we calculate the domain before we square the sqrt(x+6)?

[lusty swift]

To be clear, I understand that the text here says that it is "useful to consider the composition before any simplification", but this doesn't really seem like something that is "useful". It seems like something that is "mandatory".

Do i misunderstand the order of operations? Does squaring (sqrt(x+6)) not counteract the radical? Or are we supposed to take the square root, and then square the result, thereby "undoing" the radical?

[winter patrol]

(f o g)(x) = f(g(x))
g(x) itself needs to be defined at the value of x
to be able to have something to plug into f

e.g taking x=-7 for example
g(-7) = sqrt(-1)
(undefined in the real number system, which is what you're working in)
f(something undefined),
can't compute that

[lusty swift]

I see! Thanks for that!

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[cloud thunder]

im at 2x^2 + 9x -980 = 0

[cloud thunder]

im stuck at (2x+ )(x - )

whats the fast way to find factor for the -980

like im guessing what times what makes 980 but that takes long im sure theres a faster way?

[final storm]

afaik, i cant think of any fast ways for a number that large, what about the quadratic formula?

[cloud thunder]

oh that would b better i suppose

[final storm]

or if calculators are allowed, u can use that

[cloud thunder]

ok lemme try that. ty for suggestion

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[cloud thunder]

.reopen

[devout snow]

✅

[cloud thunder]

no quad would b a bigger number

😭

anyone know any other fast way

[silver crane]

What's ur question?

[silver crane]

solve this?

First combine the like terms

[cloud thunder]

no

[granite tulip]

Hes trynna find the factor for it

[silver crane]

oh

[cloud thunder]

im trying ti find fasyer

way to facotr

[silver crane]

oh

[granite tulip]

Im thinking of sqrt980 or 14sqrt5

Wait....

[silver crane]

Isaac, u want to factorize that thing?

[granite tulip]

Divide allat to 2

To remove the 2x²

[silver crane]

x^2 + 9x/2 -490 = 0?

[granite tulip]

Yes

[cloud thunder]

yheah

[granite tulip]

2x² prevents it

[silver crane]

(x-20)(x+49/2)

done?

[granite tulip]

Looks about right

[silver crane]

For me, I would use quad formla

[cloud thunder]

bro

[granite tulip]

It didnt work for him

[cloud thunder]

it have me like 7k

but i appriciate it thank u

'that hekped a lot

[silver crane]

I use quad to have this...

[silver crane]

how...

[granite tulip]

I dont know ask him lmao

[cloud thunder]

ikd i got a fat sqrt

[granite tulip]

I just backreaded and he said thta

[cloud thunder]

but i appricate it

.close

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[weak cove]

.close

[mental grove]

how do i solve this]

[mental grove]

like it doesnt exactly make sense\

[silver crane]

am here to help u

[mental grove]

so what do i have to do to find it

[silver crane]

Am not sure this is right-angled triangle...

(am gonna have lunch now, lol)

[pseudo basin]

which part doesn't make sense to you?

you have a triangle here in which two sides have been given as 550m and 700m, and the angle between them as 30°.

[mental grove]

idk what to do after that

[pseudo basin]

ok so it's that's you're stumped, and not that the problem itself doesn't make sense.

big difference.

ok now

do you know your basic trig? like soh-cah-toa stuff?

[mental grove]

yh

[pseudo basin]

right

then take a look at this

[mental grove]

so to find the area 0f the traingle i have to find the area of the rectangle

[pseudo basin]

that's what the problem wants you to do, yes.

maybe you know "half-base-times-height" as a canned/memorized formula for the area of a triangle, but this rectangle stuff has always been there as its proof.

anyway, do you catch my drift here with the red line i drew?

[mental grove]

yh

ok so idk how to find the height of the rectangle

[pseudo basin]

well you say you know your basic trig

look at triangle ABH

[mental grove]

yh

would it be cpz

cos

[pseudo basin]

idk what "it" is.

please be more overt in what you are talking about

[mental grove]

would the triangle be coz

coz

cos

[pseudo basin]

i don't know what "a triangle that is cos" or "a triangle that isn't cos" even is tbh.

so either you're confusing me with your wording or you are confused yourself

[mental grove]

what

[pseudo basin]

would the triangle be cos?
i don't have the faintest clue what you might possibly mean by this.

[mental grove]

what

cah

would cah

would it be cah\

[pseudo basin]

again i don't know what "it" is.

are you trying to write out a ratio of sides here?

[mental grove]

yh

[pseudo basin]

ok then write out a ratio of sides

all the points have names so all sides can be talked about without trouble

___(30°) = ___ / ___

fill in the blanks, the first with a trig function, and the other two with sides

[silver crane]

huh

[pseudo basin]

don't give out answers esp. wrong ones!

[silver crane]

ok

[mental grove]

cos 30 =550/700

[pseudo basin]

incorrect

this would only be true if ABC were known to be a right triangle.

which we don't know.

and i am asking you to look at triangle ABH and not at ABC.

and also i drew that red line for a reason... its length is what we want to find.

[mental grove]

idk what to do

[pseudo basin]

ok then let us go through this step by step

i will give you instructions that i expect you to read carefully and follow precisely.

are you willing to do both of those?

[mental grove]

yes

[pseudo basin]

ok

look at triangle ABH in this diagram. of its three sides, only one has a length known to us. which one?

answer in the following format, filling in the blanks:

The only known side is ___, and its length is ___.

[mental grove]

AH and 700\

[pseudo basin]

❌ instruction not followed
❌ info incorrect

when i say "answer in the following format, filling in the blanks" i expect you to copy down the sentence i write and send it back to me with the blanks filled

try again

[mental grove]

The only known side is AH__, and its length is 700 m___.

[pseudo basin]

❌ info incorrect

700 m is NOT the length of AH.
700 m is AC, and AC isn't a side of triangle ABH.

also you fucked up and didn't delete the underscores.

try again

@mental grove

[dim knot]

@mental grove
Formula called 0.5 * a * b * sinA :

[pseudo basin]

1. don't do the number one-half dirty by shoving it into a decimal
2. op still needs to know WHY that formula is what it is

3. don't give out answers

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dim knot]

are u telling me to demonastarte the proof of that formula?

[pseudo basin]

no i'm telling you to back off until OP comes back

[dim knot]

Gotchu

[devout snow]

@mental grove Has your question been resolved?

[devout snow]

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[south rampart]

Hello, I am not sure if this is correct please someone check it. I got everything from YouTube but don’t think it’s correct

<@&286206848099549185>

[rare lagoon]

bruh. Andrew tate

lemme help u

[south rampart]

bro its cool

XD

[rare lagoon]

ik

[south rampart]

alr

[rare lagoon]

I am in the course of the real world rn

[south rampart]

bruh

thas crazy

bro i have been waiting the whole day for someone to help me with this page

i am not joking btw

also if you can can you check my others papers aswell if you have the time

[rare lagoon]

oh

wait

[south rampart]

what

[rare lagoon]

add

me

[south rampart]

alr

[rare lagoon]

so what is the question mate

there are like more than 1 sub question

solving all of them?

[south rampart]

i dont know if what i did is correct

for all of em

[devout snow]

@south rampart Has your question been resolved?

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[tough light]

Bhaskara I sin(x) rational fraction approximation. i dropped out of high school at sophomore year and don’t know the symbols/operations related to alot of symbols especially in respect to the symbols before them. ive read that the brackets/parentheses are solved from the innermost nested to the outer, but ive also read that square brackets are a list or more importantly a range between the zeroth entry and the next so i just need some help understanding the application in this formula. do i divide the former by every integer between zero and the answer to the equation in the brackets or just divide by the solutions on either side of the division symbol.

[tough light]

https://mathshistory.st-andrews.ac.uk/Biographies/Bhaskara_I/

the equation /expression whatever is in that link towards the middle of the oage

page

[pseudo basin]

but ive also read that square brackets are a list or more importantly a range between the zeroth entry and the next so i just need some help understanding the application in this formula.
in this case the square brackets do not mean that

they are just the same as parentheses only they look different

tbh this is better written as a fraction: $\sin(x) = \frac{16x(\pi - x)}{5\pi^2 - 4x(\pi - x)}$

[woven radish]

Ann

[pseudo basin]

@tough light

[tough light]

can yall stick around to let me solve it and check me to make sure i understand please

[pseudo basin]

sure, just tell us exactly what you're solving here

[devout snow]

@tough light Has your question been resolved?

[devout snow]

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[devout snow]

@formal bronze Has your question been resolved?

[formal bronze]

<@&286206848099549185>

[formal bronze]

<@&286206848099549185> 🥲

[devout snow]

@formal bronze Has your question been resolved?

[devout snow]

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[finite briar]

Help

[pseudo basin]

you've determined that the factor count of a^7 is always 1 mod 7

given k ∈ N try to find a number whose 7th power has exactly 7k+1 factors

this number will serve as a witness for the luckiness of 7k+1

[finite briar]

Hi

[finite briar]

I haven't learnt modular arithmetic

[pseudo basin]

ok, then rephrase that as "the factor count of a^7 is always one more than a multiple of 7"

[finite briar]

Help

[pseudo basin]

...

[finite briar]

@finite briar

Etd

Wtf

Sorry

Uhhbmy keyboard got bugged

[finite briar]

K

[finite briar]

How did I figure that out?

[pseudo basin]

the factor count of a^7 is 7 * (SOME SHIT) + 1

where the shit is all made of integers

[strong lava]

help

[finite briar]

If a is prime only then no

[finite briar]

#❓how-to-get-help

[pseudo basin]

please open your own channel. #❓how-to-get-help

[pseudo basin]

au contraire

8 is also of the same form

7*1 + 1

[finite briar]

How did u get this

[pseudo basin]

by looking at this thing you wrote

do you not see that it's all a sum of various assorted multiples of 7 with a one at the end

[finite briar]

I generalized the expansion (7x+1)(7y+1)...

[finite briar]

Oh yes

But is it necessary that- OHHh

Got it

So now I find k?

[finite briar]

Trial and error?

[finite briar]

Meaning N =7k+ 1?

[pseudo basin]

i didn't say to "find k"

N = 7k + 1
sure

[finite briar]

Pls tell me u aren't being sarcastic

[pseudo basin]

see if you can consider something more general than primes but less general than all composites

what's a type of number that's in between those complexity-wise

[finite briar]

[finite briar]

Uhm...what's there except prime and composite and 1 in terms of factorization

[pseudo basin]

i didn't say to discover a new class of numbers disjoint from those

i'll give you one more nudge in that direction and if you don't get it still i'll tell you what i meant

[finite briar]

Okay

[pseudo basin]

i'm talking about composite numbers but only ones whose factorizations are particularly simple

[finite briar]

Hmm so like 2 times 3? I.e. not involving powers > 1? Or only involving 1 number in power?

[pseudo basin]

only involving 1 number in power

shoddy wording but yes that

im talking about prime powers

i.e. numbers of the form p^k

[finite briar]

Okay...

[pseudo basin]

how many divisors does (p^k)^7 have?

[finite briar]

7k + 1

[pseudo basin]

well there you have it don't you

and that works for any natural k

[finite briar]

Ohhh

Got it

[finite briar]

So I just set it less than 100

7k<99

[pseudo basin]

i'd like to acknowledge explicitly that we've proved one important thing:

[finite briar]

Find all multiples of 7 less than 99

[pseudo basin]

a number is lucky IF AND ONLY IF it is of the form 7k+1

[finite briar]

Okay 👍

Tysssm

May I .close?

[pseudo basin]

why do you ask me for permission

it's your channel

you are the only one who knows whether or not you have anything else to ask

[finite briar]

No I don't wanna make u feel u got doorslammed

.close

[devout snow]

Channel closed

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[scenic goblet]

I need help on this math problem, im stuck on what it means by "a sum to infinity of 60", thank you

[manic condor]

Do you know what is a geometric series?

[scenic goblet]

yes

its a series of numbers than goes up by multiplication

[manic condor]

Yes.

The number that they are multiplied by(to get the next term) is called common ratio.

[scenic goblet]

yep

[manic condor]

You might be used to representing common ratio by r, let's say that our common ratio is r.

[scenic goblet]

thats what i usually do

[manic condor]

Since first term is 6, and common ratio is r, what are the terms of the series? Try writing those.

[scenic goblet]

u1=6 u2=6r u3=6r^2?

[manic condor]

Yes and so on.

What they mean is that if you write such infinite terms and sum them all, you get 60.

Using that, you have to find r.

[scenic goblet]

got it, does that mean r is a fraction

less than 1 but greater than -1

[manic condor]

Basically,
$6 + 6r + 6r^2 + \cdots = 60$

[woven radish]

Enemagneto

[manic condor]

Has to be. Otherwise, it won't have a finite sum.

[scenic goblet]

yeah true

[manic condor]

Anyway, you can try to do it now. If you get stuck, i'll be here.

[scenic goblet]

so if the formula is: u1(1-r^n)/1-r, do i put n as infinity or as 60?

[manic condor]

What does n denote?

[scenic goblet]

the number of the sequence

[manic condor]

Number of terms in the sequence.

[scenic goblet]

so it would be infinite

[manic condor]

And, how many terms are there in your summation expression?

[manic condor]

Yes.

But it would be more like taking limit $(n\to \infty)$, as you aren't exactly supposed to substitute infinity.

[woven radish]

Enemagneto

[scenic goblet]

im unsure of what youre saying here

[manic condor]

So, for sum of n terms of Geometric series, you have -
$$S_{n} = u_{1} \cdot \left(\frac{1 - r^n}{1-r}\right)$$

[woven radish]

Enemagneto

[manic condor]

Okay.

[scenic goblet]

👍

[manic condor]

You need to put n as infinity. Right?

[scenic goblet]

Yes i believe so.

[manic condor]

Can you evaluate the expression $S_{n}$ for n = infinity?

[woven radish]

Enemagneto

[scenic goblet]

s-infinity = 6 x (1-r^infinity/1-r)?

[manic condor]

Yes, but how will you calculate value of RHS here?

[scenic goblet]

im not sure

whats RHS?

[manic condor]

Right hand side

I meant this part - 6 x ((1-r^infinity)/(1-r))

[scenic goblet]

i forgot all of this but im not sure if its possible with the infinite power

[manic condor]

Yes. You are supposed to take the limit therefore.

Are you familiar with taking limits?

[scenic goblet]

no not yet

[manic condor]

Okay. Then, we gotta do it analytically.

You know that |r| < 1

Right?

[scenic goblet]

yes

[manic condor]

So, if you take any number such that |r|<1, then if you multiply the number to itself, it becomes smaller(magnitude wise).

For example, r=1/2
r^2 = r * r = 1/4 which is smaller than 1/2.

If you multiply it once more with itself, it's 1/8 which is even smaller.

Good so far?

[scenic goblet]

yes

sounds good

[manic condor]

What if you do this 10 times?

You get $\left(\frac{1}{2}\right)^{10}$.

[woven radish]

Enemagneto

[manic condor]

Which is equal to

[scenic goblet]

exactly

[manic condor]

,w (1/2)^10

[woven radish]

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Requested by .enemagneto

[manic condor]

0.0009

See, it's becoming smaller.

Let's try for hundred times.

,w (1/2)^100

[woven radish]

Results provided by WolframAlpha

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Requested by .enemagneto

[scenic goblet]

small number

[manic condor]

Yes. Point is that as you increase power for a number r, where |r| < 1, magnitude reaches closer to 0.

If power goes to infinity, then it can be taken as just 0.

[scenic goblet]

mhm

[manic condor]

So, you have $r^{\infty} = 0$

[woven radish]

Enemagneto

[scenic goblet]

alright

[manic condor]

Now, you should be able to solve your question, i believe.

[scenic goblet]

s=6(1/1-r)

how do i solve for r

[manic condor]

Well, you know what is S here.

You were given in the question.

After that, just isolate r.

[scenic goblet]

s is 60?

[scenic goblet]

i got that 60=6(1/1-r) so then 10=1/1-r. then i realized that 1/0.1 is 10 so made it so 1-r equals 0.1 which makes r, the common different 0.9. Is this right?

[manic condor]

Yeah, that's correct. You arrived at it little weirdly but it's alright.

[scenic goblet]

how would you have arrived at it

for the future

[scenic goblet]

but then this next one is just as confusing, i get u1=x, u2=x-3 and s/un=75

putting in the formula i got 75=x(1/1-r)

but i have 2 variables

[manic condor]

Well, you can write r in terms of x.

Think about it.

[manic condor]

No, you are good. You probably just put it weirdly. Lol

[scenic goblet]

😄

[scenic goblet]

how, i dont understand

[manic condor]

Well, r is the common ratio.

and you have two consecutive terms.

What will be their ratio?

[scenic goblet]

i dont know

-3

[manic condor]

No. Come on. Just take the ratio of the term.

Since x-3 is the second term, that means that
(x-3) = x * r

So, you get r as ?

[scenic goblet]

i literally dont know

can you give me a hint

[manic condor]

Think about it, buddy.
You have a geometric progression. Where, each term is multiplied by r(common ratio) to get the next term.
That means that since $x$ is the first term, second term would be $x \times r$. Makes sense?

[woven radish]

Enemagneto

[manic condor]

@scenic goblet

[scenic goblet]

yes pal

[manic condor]

But second term is (x-3).

Right?

So, we have $x \times r = x-3$

[woven radish]

Enemagneto

[manic condor]

Now, just isolate r.

You'll get r in terms of x.

Then, it's just one variable. You can solve it.

[scenic goblet]

i literally cant

i dont know how

[manic condor]

Well, how about dividing by x on the both sides?

You get $\frac{x\cdot r}{x} = \frac{x-3}{x}$

[woven radish]

Enemagneto

[brave zodiac]

r=-3

[scenic goblet]

r=-2

[manic condor]

Now, x gets cancelled out, and you are left with
$r = \frac{x-3}{x}$

[woven radish]

Enemagneto

[manic condor]

You solved it already?

Oh wait. You did r = 1-3 ?

[scenic goblet]

yes

[manic condor]

That's wrong.

[scenic goblet]

k

[manic condor]

You get this.

[brave zodiac]

im consfused aswell

me and @scenic goblet are in the same school and math class

could you just give the answer first

and then explain why please 🙏

[manic condor]

I can't.

I can only help you get it.

[brave zodiac]

okay then please help explain

i do get it.

i just dont get the answer

[manic condor]

Well, is this much clear?

[brave zodiac]

wait let me read

[scenic goblet]

can you give us the answer and tell us how you got there, thanks in advance

it woukd be so much more helpful

[brave zodiac]

do we have what xis

[scenic goblet]

as it seems to me as you keep repeating the same thigns

[manic condor]

No. you have what r is, in terms of x.

[manic condor]

Which you can substitute here, and get x.

[brave zodiac]

how would u subsitite that

[scenic goblet]

i havent got r yet so im not really sure as to what to do

[brave zodiac]

think about it, buddy!

[manic condor]

You got r in terms of x.

[scenic goblet]

can i substitte r for x/x-3 in the formula!

OMJEEPERS!

[manic condor]

Not x/(x-3)

(x-3)/x

[brave zodiac]

why

explain.

[manic condor]

.

[brave zodiac]

im puzzled

[scenic goblet]

GOT IT!

[brave zodiac]

oh yeah

tchobs judst forget

silly him!

okok i get

it

do u tchobs

[scenic goblet]

i get it 105%

[brave zodiac]

sorry we are a bit distracted w upping the ratio in the geometrial sequence of our fortnite career

amny thanks talk later

[manic condor]

Yeah.

@scenic goblet Close it if you are done.

[scenic goblet]

@manic condor thank you so extremely much for your help. For the future, could you give alittle hint with each message because i felt you kept repeating the same equation but thanks!

let me quickly plug it into the equation

[brave zodiac]

what do you get

[scenic goblet]

ive hit a roadblock

i got 75=x(1/1-x-3/x), much too confusing

[manic condor]

Well, you learn if you get it on your own. Also, it wasn't tough so i had expected you to get it on your own.
Seemingly, you seem to have too little patience for learning. Try to improve on that.

[manic condor]

try simplifying (x-3)/x a bit more. It would be easier.

Use the fact that $\frac{a-b}{c} = \frac{a}{c} - \frac{b}{c}$

[woven radish]

Enemagneto

[scenic goblet]

you never know, i could be doing matsh way above my level so need more assistance!

[scenic goblet]

will try my best to improve

[manic condor]

Agreed. Regardless, you knew what geometric series is, so it's not too much to expect you to come up with (x-3)/x = r.

[manic condor]

Any progress on this?

[scenic goblet]

i apologize prefusely for my lack of mathematical talent

[brave zodiac]

woudlnt it be 1-3/x?

[scenic goblet]

1-3/x?

[manic condor]

Come on. You know that's not what i want at all. Also, i can't even tell if you are being backhanded or not.
Regardless, let's get back to question.

[manic condor]

Yes

[brave zodiac]

okay

[manic condor]

$\frac{x-3}{x} = \frac{x}{x} - \frac{3}{x} = 1- \frac{3}{x}$

[brave zodiac]

so what to do next tchobs

[woven radish]

Enemagneto

[brave zodiac]

i dont have the formu;a

[manic condor]

Now, look carefully at the denominator.
you have $1 - r$.

[woven radish]

Enemagneto

[scenic goblet]

1-1-3/x

so just -3/x?

[manic condor]

If you substitute this r,
you get $1 - \left(1- \frac{3}{x}\right)$