help-10

[obtuse pebble]

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[rotund vector]

Could anyone help me out with this question?

[devout yarrow]

r u allowed to use a graphing calculator?

thatll help u visualise

if not then thats ok

[obtuse pebble]

@rotund vector Has your question been resolved?

[rotund vector]

Nope, not allowed.

[devout yarrow]

oh ok

u have rough paper?

[rotund vector]

Yup

[devout yarrow]

or graph paper

[devout yarrow]

sketch

[rotund vector]

So I've roughly sketched the vertical lines and the quadratic

[devout yarrow]

show it

[rotund vector]

I'll just send it through desmos

When it is rotated about the x-axis it looks like it would form a spherical ellipsoid-like shape

[devout yarrow]

do u know how to intgrate?

or r u expected to solve it without calculus

[rotund vector]

Yeah

[devout yarrow]

ok

so u can find the area of the shape contained by the four graphs

[rotund vector]

Using the washer method?

[devout yarrow]

?

just normal

[rotund vector]

Or disc method

[devout yarrow]

dont care abt 3d first

just do 2d

[rotund vector]

Alright

[devout yarrow]

$$\int^{0.5}_{-0.5} -2x^2 +1 dx$$

[warm shale]

everything_addict

[devout yarrow]

sry abt trash formatting

[rotund vector]

All good

[rotund vector]

5/6

What would I do now?

@devout yarrow

[devout yarrow]

show me ur working

i got 1/4

maybe im wrong

[rotund vector]

Yes, you are wrong.

[devout yarrow]

oh

yeah

sry

calculation error

[rotund vector]

$$\int {-0.5}^{0.5}-2x^2+1\cdot dx=\left[-\frac{2}{3}x^3+x\right]{{{-0.5}}}^{^{0.5}}$$

[warm shale]

Lex1729

[devout yarrow]

yep

i got that

but i calculated it incorrectly

sry abt that

[rotund vector]

$$\left[-\frac{2}{3}\left(0.5\right)^3+0.5\right]-\left[-\frac{2}{3}\left(-0.5\right)^3-0.5\right]=\frac{5}{6}$$

[warm shale]

Lex1729

[rotund vector]

All good

[devout yarrow]

so now u know the 2d integral

u know how to calculate 3d?

[rotund vector]

Well, I can use the formula V = pi * r^2 right?

Although I'm not really too familliar with 3D integrals

Only recently started studying it

[devout yarrow]

yes

however

https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-volumes-2009-1.pdf

part 5

we have to express x as a function of y

so similarly

[rotund vector]

Why do we need to express x as a function of y?

[devout yarrow]

because we r rotating abt the y-axis

not x axis

[rotund vector]

But is says x-axis?

[rotund vector]

.

[devout yarrow]

ah shoot

yep

sry im doing more harm than good

so yeah

[rotund vector]

So I'm doing dx?

[devout yarrow]

$\int^{0.5}_{-0.5} \pi (-2x^2 + 1)^2 dx$

[warm shale]

everything_addict

[devout yarrow]

yes

[rotund vector]

Alright, so now how would I find the 3D integral?

[devout yarrow]

gimme a bit

lemme work that

[rotund vector]

I know I'm supposed to use V = pi * r^2

I'm just not sure how I would know what the radius is

[devout yarrow]

we r technically doing that

rght

[devout yarrow]

we r squaring the function

[rotund vector]

Oh ok, so the function is the radius?

Is there any way to picture that? because I'm confused on how that would look like on a graph/visually

[devout yarrow]

not really

ok lets solve this first

then ill explain

[rotund vector]

Okay

I get the answer to be 43/60

(43/60)pi

[devout yarrow]

show ur working

ill check for u

[rotund vector]

I'm pretty sure it is correct

I just checked it on a calculator

[devout yarrow]

yes

i did it by hand

its correct

[rotund vector]

Alright, what next?

gtg soon could we finish this off?

[devout yarrow]

so thats ur answer

but its not in the options

and idk why

[rotund vector]

Yeah, that's what I thought.

I have a question though

Why is -2x^(2) + 1 the radius?

Because we just substituted it for r in the equation V = pi * r^2

[devout yarrow]

no

we also did a integral

i cant really explain that

im sorry

[rotund vector]

Yeah, but why is r the function?

[devout yarrow]

could u find someone else

[rotund vector]

Oh ok

[devout yarrow]

im not very proficient

in calc

[rotund vector]

No worries, thanks for helping out 👍

[devout yarrow]

no rpoblem

[rotund vector]

.close

[obtuse pebble]

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[obtuse pebble]

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[polar mountain]

so i need help again

[polar mountain]

its the same thing with area

maybe its me but

i justdont understand why the square doesnt add up

[hollow island]

isnt it the area of the square minus the area of the 4 right angled triangles?

[polar mountain]

i guess

but for me it kind alooks like a trapezium

[hollow island]

yes, but it might be more difficult to calculate the area of the trapezium

[polar mountain]

so how would i split it into 4

[high lily]

wdym by

square doesn't add up

[polar mountain]

what would the base be n stuff

@high lily usually isnt the squares side like equal

[honest trellis]

Find area of the four white triangles

Add them

[polar mountain]

oh i didnt even realise

[high lily]

always, not usually

the sides of squares are always of equal length

[honest trellis]

Then subtract that sum from 10^2

[polar mountain]

how did u get that?

[honest trellis]

The answer is $10^2-(86)/2 - (62)/2 - (43)/2 - (74)/2$

[warm shale]

Rebag9

[honest trellis]

You want the purple area right

[polar mountain]

oh

yeah

[honest trellis]

Do u understand

[polar mountain]

yes

@honest trellis im just wondering where u got 10 squared frm

[honest trellis]

The area of the dotted square

[polar mountain]

oooooh

i see it

thanks

[obtuse pebble]

@polar mountain Has your question been resolved?

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[obtuse pebble]

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[steel prawn]

Prove for any positive integer k, that either k | (k − 1)!, k is a perfect square, or k is prime.

[steel prawn]

@ruby fulcrum

<@&286206848099549185>

[wise tangle]

ok ok

If k is a positive integer, then k - 1 is also a positive integer and k is a factor of (k - 1)!, so k | (k - 1)!.
If k is a perfect square, then k = n × n, where n is any positive integer. This means that k can be written as a product of two identical numbers n, which is the definition of a perfect square.
If k is prime, then k cannot be divided by any other positive integers other than 1 and itself.

[steel prawn]

ok

[wise tangle]

alright now u can close it

[steel prawn]

but how does it proce anything, it is 9makr assignment question

prove*

@ruby fulcrum

[wise tangle]

1. when a number is a factor of another number, it means that if you divide the first number by the second, you get a whole number without any decimals.

2. a perfect square is a certain type of number that can be written as the product of two identical numbers. For example, if you take the number 36, it can be written as 36 = 6 × 6, which means it's a perfect square.

[steel prawn]

but who do you know thet k is a factor of (k-1)!

but how do you know that k is a factor of (k-1)!

@ruby fulcrum

[wise tangle]

example : take k = 7. you can write the factorial of 7 as:
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
now, we can divide 7 by each number in the equation one by one, and we can see that for each number, we get a whole number without decimal points, which means it's a factor.

[steel prawn]

so u saying if 7 is divided by 6 we do not decimals

[wise tangle]

yes

[steel prawn]

u dumb

7/6 is 1.166667

u literally get decimals

[wise tangle]

When we divide 7 by 6, we get a whole number without any decimal points. So then 7 is a factor of (7-1)! Which means 7 | (7-1)!

[steel prawn]

did you even go to school man

[wise tangle]

You're right, 7/6 does have decimals. But in mathematics, when we divide 7 by 6, we get 1 and 1/4 as the fraction because we can't have a decimal point in a fraction. We can't have a decimal point in a fraction because the definition of a fraction is that it's a division of two numbers. And when we divide two numbers, we can't have a decimal point either.

[steel prawn]

then you proved nothing form the original question

[wise tangle]

Yes, you're right, I haven't yet fully proven that k | (k-1)!. I just showed that for a specific number k, it will be a factor of (k-1)!
But I can give another example as well. Take k = 10.
10 | (10-1)!
To prove this, we first need to expand the right-hand side:
(10-1)! = (10(9) 8 (7) 6 (5) 4 (3) 2 (1).

[steel prawn]

.close

[obtuse pebble]

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[obtuse pebble]

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[rough bridge]

How would I find the slope? Do I put it into a graphic calculator ?

[obtuse pebble]

@rough bridge Has your question been resolved?

[tardy tusk]

y=mx+b where m is a slope ig

Lemme check

so m=(y2-y1)/(x2-x1)

[hollow goblet]

yes the differences of y over x

[rough bridge]

m is slope right ? how do I gind the y2 and y1

and x 1 x2 are those the intercepts

[hollow goblet]

y2 y1 and x2 x1 are any points on the line

you need 2 points to find slope

of a line

[tardy tusk]

You literally have big dots

On the line

[hollow goblet]

then use those

generally any two points on a line can be used for (x1, y1) and (x2, y2)

this case just gives you points on each line

[tardy tusk]

Ig we take out from bigger values less values
Or what?

[hollow goblet]

doesn't matter

y2 -y1

then divide byy x2 - x1

[tardy tusk]

Okk

[hollow goblet]

the order of points is not important

[obtuse pebble]

@rough bridge Has your question been resolved?

[obtuse pebble]

@rough bridge Has your question been resolved?

[rough bridge]

Do I put the whole point while doing the equation

so for examplex1 x2 (3,2) - (2,1) or 3-2

[rough bridge]

I used points and got 1/3 and it was wrong for the first line

[obtuse pebble]

@rough bridge Has your question been resolved?

[obtuse pebble]

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[warm shale]

random

[brisk matrix]

this is an ok start, but that integral might be divergent

why not take int_0^x

as for an example, im not really sure how to give a hint on it without spoiling the result
i had to look it up myself but there are examples out there

maybe a good first step is to start with a continuous function that is nowhere differentiable, and mess with it a bit

im not sure what the generalization is meant to be
if we are looking at curves g : R -> R^m, then theorem 10 tells you than it suffices to look at the components rather than the whole vector

[warm shale]

random

[brisk matrix]

that might also be the case, and you'd probably want to generate some function who's hessian only exists at 1 point

also thinking about it for a bit, there are very simple examples R^n -> R^m with hessians only at a single point, provided you showed that there is a C^1 function f : R -> R with f'' only at one point

[obtuse pebble]

@abstract hazel Has your question been resolved?

[brisk matrix]

think even simpler than that

btw your example of cont. everywhere and differentiable at 0 is good

so now you have a function f : R -> R that satisfies the desired conditions

let's call that f
can you define an explicit map g : R -> R^2 that has a second derivative only at 0?

yes, but i also want to point out

we could have used

g = (f, 0)

so that was my attempt at hinting how some g : R^2 -> R could also be constructed

or in general, some R^n -> R^m map

be more explicit

g(x,y) = ...

just f(x)

but yes, that is the idea

yup

although

that is not sufficient

you'll also need to multiply by f(y)

you are basically trying to force each variable to be 0 in order to get a second derivative

if you just let g(x,y) = f(x), then g will have a total second derivative for all y as long as x = 0

so over a whole line (rather than a point)

but multiplying by f(y) fixes that. ok rant over

[obtuse pebble]

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[obtuse pebble]

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[fast berry]

Any ideas on how to give a counterexample/proof of this?

[obtuse pebble]

@fast berry Has your question been resolved?

[fast berry]

<@&286206848099549185>

[twin sandal]

I'd assume this is true bc in order to prove it's not, you'd need to show that all binary operations fail

which would be like impossible

[bronze mica]

I'm leaning towards the opposite answer, because I feel like there should be ternary operations that would force the binary one to be too nice

[fast berry]

I imagine there is a counterexample, but I'm not sure how I could make one

[twin sandal]

wouldn't you need to show that every binary operation fails? how would you even try to construct a counterexample

[fast berry]

yeah, that's why I'm kinda lost

I think there's an alternative to testing all operations

[bronze mica]

not if A is finite

yes

finite A obviously has counterexamples

there are more ternary operations than there are binary ones, so there can't be as surjection from binary operations to ternary operations, let alone one specifically given by left-induction

[twin sandal]

visualizing that would be hard lol, binary operations at least have tables

but good point

[bronze mica]

so do ternary ones

[twin sandal]

oh please god no

[fast berry]

that's right, and I could give a counterexample but how can I make sure there isn't a very specific binary operation that would work for that case?

[twin sandal]

if A is finite then it has a finite amount of binary operations

[fast berry]

yes, n^n^2

[bronze mica]

honestly other than picking the base case scenario of n=2 (small correction, if |A|=1 there is a unique ternary operation left-induced by the unique binary one so there are no counterexamples) and going through the list of all 16 ternary operations left-induced by binary ones I don't know how to find it

good news is there probably is a nice way of desrcribing it

but idk what it is

this does feel too painful

[fast berry]

yeah, I'm sure there's a trick to solve this but I can't see it

[bronze mica]

oh

just try force a ternary operation such that the binary one that would induce it has to take at least three values

so e.g. if you want star(a,a,a) = a but star(a,b,a) = b, then you force a . a =/= a . b

idk if you can actually manage to also get e.g. a.b =/= b.b and a.a =/= b.b though

ok it is doable

[fast berry]

( a . a =/= a . b) here what are you referring to when using . and =/=

[bronze mica]

sorry by =/= I meant not equal to

and . was the desired binary operation because idk how to latex a diamond like that

[fast berry]

ah, got it

[obtuse pebble]

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[dapper wing]

can someone help with this, I am confused what the last two questions are asking

[dapper wing]

this is the example form the book

i doont really get where they got 90 from in the example?

.close

[obtuse pebble]

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[mint arrow]

Would this be 2.08

[mint arrow]

the formula i used was wavelength = velocity/frequency

x 1 for each i got 2.08 meters

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mint arrow]

Four

[safe haven]

yes

[mint arrow]

Thanks

This would be E

cause of the amplitude right?

amplitude being the most affected variable if energy of wave was doubled

@safe haven

?

[safe haven]

wont it be D tho

[mint arrow]

but D isn't the amplitude

[safe haven]

true

[mint arrow]

idrk what D is

.close

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[obtuse pebble]

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[timid silo]

lim x-> 9 (\sqrt(x) - 3)/(x-9)

[timid silo]

the denominator is 0 so its undefined.

[shy vigil]

factor it

[timid silo]

so is the numerator

so it is indeterminate rather

how do you factor this?

[timid silo]

not relevent to this question

but sure

you can multiply by the conjugate rather

[shy vigil]

it is very relevant

[timid silo]

it is. You saying it is undefined is incorrect

[timid silo]

the book mentions the conjugate solution, so that is probably more like it

so, multiply by (x+9)/(x+9)

thanks

.solved

[obtuse pebble]

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[arctic wadi]

how do I solve this?

[latent walrus]

call the number 'x'

try write an equation

[arctic wadi]

(1/8)x = (1/2)x-2?

[latent walrus]

seems alright

[arctic wadi]

what next?

[latent walrus]

solve the equation for x

should be quite standard algebra

[obtuse pebble]

@arctic wadi Has your question been resolved?

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[fierce elbow]

Find out the elongation in block (in cm). If mass, area of criss-section and young modulus of block are m,A and y respectively. (where $A = 1mm^2 , Y = 210^10 N/m^2 , l = 1m, F = 32 10^2 N$ )

[warm shale]

❄ѕησωƒℓαкє❄

[rich vault]

Hello there, got a pretty basic finance question if someone could help me

[latent walrus]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[latent walrus]

any ideas? probably would be good to start at the definition of the youngs modulus

[fierce elbow]

I got an answer

But Im not sure if its right

cuz the question wasnt mcq type

and I cant find the answer on internet either

[latent walrus]

what did you get, ill check it in a mo

[fierce elbow]

16

Its most probably wrong

still

[latent walrus]

i got similar but im a factor of 100 out

0.16

[fierce elbow]

did u convert $1 mm^2 to 10^-6 m$

[warm shale]

❄ѕησωƒℓαкє❄

[latent walrus]

i did

[fierce elbow]

then how

Can I send the pic of my solving

[latent walrus]

,w (32*10^2 *1)/(10^(-6) 210^10)

[fierce elbow]

tho its very badly written

[latent walrus]

sure

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aldrnari_

[fierce elbow]

yo

The answer is in m

U gotta convert into cm

[latent walrus]

aha, right i neglected that

yeah 16 is fine then

[fierce elbow]

That means Ive got it right

lets goo

4 marks

[latent walrus]

nicely done

[fierce elbow]

Am I supposed to ask another question here

or make a new chanel

maybe later

[fierce elbow]

thx anyway

.close

[obtuse pebble]

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[obtuse pebble]

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[open bronze]

help

[open bronze]

is this correct

[obtuse pebble]

@open bronze Has your question been resolved?

[obtuse pebble]

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[visual heart]

is this proof correct? or is there any changes I should make?

[obtuse pebble]

@visual heart Has your question been resolved?

[brave bramble]

It's correct

[visual heart]

thanks

[obtuse pebble]

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[civic bone]

check it out

[civic bone]

I started just putting my proofs in latex so its easier for you guys to read em lol

https://i.imgur.com/LM7uthz.png

can someone validate it pls?

[green shuttle]

why so complicated with the contradiction

[civic bone]

hoh

[dark stirrup]

yeah, the contradiction assumption is pointless

[civic bone]

yeah I guess it's constructive without that lol

that's just how I started thinking about it

[dark stirrup]

You're basically saying "Assume Q is false, <proves Q is true>, therefore it is contradictory that Q is false and it must be true"

[civic bone]

lool

[green shuttle]

like just show this

you never used that the tail of (b_n) is unbounded

[civic bone]

yeah I just thought contradiction would be easier before I worked it out and didn't really think about it

[dark stirrup]

It happens a lot.

I wish we had a !nocontradiction bot message

[civic bone]

https://i.imgur.com/QCLh33I.png

okay so now its solid?

[green shuttle]

there's something i don't love about the wording at the end but yea sure

[dark stirrup]

Why is it required that $M\ge 0$?

[warm shale]

SWR

[civic bone]

the arbitrary pair part?

idk how to say it better

[green shuttle]

no, why the iff holds

[dark stirrup]

oh bounded, not bounded above

[green shuttle]

like it's ok

[civic bone]

^

[civic bone]

how would you say it more clearly

[hidden laurel]

What maths course are you doing

[green shuttle]

idk i might write a without loss of generality at the start

[wild swallow]

you appear to be using the align* environment without knowing how it works, resulting in all your equations being aligned on the right

[dark stirrup]

Proof is fine.

[green shuttle]

but yea it's fine

[dark stirrup]

"arbitrary pair" is an odd wording I don't see often

[civic bone]

yeah I just installed it, how do I fix this

was bothering me too lol

[green shuttle]

& moment

[civic bone]

  |b_n| - |a_n| \leq |a_n - b_n| \leq \epsilon\\
  |b_n| \leq |a_n - b_n| + |a_n| \leq \epsilon + |a_n| \leq \epsilon + M
\end{flalign*}

[dark stirrup]

I myself like to say "Proving the converse is done in a similar way"

[wild swallow]

replace flalign* with gather*

[civic bone]

now they're just center aligned 😄

[wild swallow]

i dont think aligning the equations in some other way makes sense

[civic bone]

I guessthe epsilon at the end aligning

but yeah there are 2

[zenith raft]

&\implies

[wild swallow]

theyre just two separate inequalities that you're grouping together, theres no need align them as that would be you trying to emphasise some relationship between the two equations

[zenith raft]

hi snow

[civic bone]

oki

while the gang is all here

what do yall think of this one

https://i.imgur.com/2onM3pJ.png

[zenith raft]

here comes snow to critique the latex

[wild swallow]

i hate it

[green shuttle]

that block of inequalities is poorly written

[civic bone]

-_-

chill yo this was the first one i typed up

i mean the proof itself

quit roasting my shit D:

[green shuttle]

both the math and the tex lol

[civic bone]

whats wrong with the math 😦

[green shuttle]

it lacks logical flow

[wild swallow]

i have no idea what you did here

[civic bone]

subtract 3rd inequality from 2nd

[kind hawk]

thats not how inequalities work

[civic bone]

o?

waaait this is wrong?

[wild swallow]

this conclusion is surely not correct

[civic bone]

they're epsilon-close for each epsilon so they're equivalent no?

[wild swallow]

(a_n) and (b_n) are not guaranteed to be the same

[kind hawk]

equivalent and equal are very different things

[zenith raft]

if the sequences were equal the proof would be pretty simple

[civic bone]

o nio

welp

problem for tomorrow

thanks bois

.close

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[shrewd heart]

im not sure how to do this im supposed to find the value of x

[twilit meteor]

which question

[shrewd heart]

im just confused on all of it cause i zoned out in class

[twilit meteor]

which grade

[shrewd heart]

11

[twilit meteor]

factorize each of the fractions in lhs

[shrewd heart]

whats lhs

[twilit meteor]

left hand side

[latent walrus]

factorise the left side? Do you mean re-express the rhs

[shrewd heart]

i think i get it

[latent walrus]

alrighty, what do you get for a

sorry i mean b

[shrewd heart]

nvm bro i dont get it at all

[obtuse pebble]

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[obtuse pebble]

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[hollow anvil]

Suppose from a group of 5 students with same abilities, 3 are boys. If three students are chosen to be their representatives, construct a table for the probability distribution of the number of boys chosen.
its about probablity

[hollow anvil]

Suppose from a group of 5 students with same abilities, 3 are boys. If three students are chosen to be their representatives, construct a table for the probability distribution of the number of boys chosen.

its about probablity

hello

@analog vault

😦

@teal ledge

.close

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[obtuse pebble]

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[serene elk]

how should i represent this data

I used a bar graph but it doesn't include the totals

[obtuse pebble]

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[runic grotto]

anyone help me with this question? "given a normal distribution with μ = 76.4, σ = 2.9 cm, what value of Height corresponds to a Z score of -0.84?"

[alpine verge]

do you know what the formula for z-score is?

[runic grotto]

uh i believe its z= x-μ/σ

[alpine verge]

yep

we know what z, mu, and sigma are yes?

we just use this formula to solve for x

[runic grotto]

yes

so then is it gonna be x = 76.4-0.84 (2.9)?

[alpine verge]

yep

[runic grotto]

is x = 73.96 the final answer for it

[alpine verge]

yep

looks good

[runic grotto]

im still getting confused, is the left side of the graph always negative?

because I have this question "The height of 12 month old boys is normally distributed with μ = 76.4, σ = 2.9 cm. What is the 20th percentile for height?"

i got -0.52 for the z value of the 20%

[alpine verge]

yeah

anything that is less than 50% would have a negative z score

since the mean is the 50th percentile

anything less than the mean would be negative

[runic grotto]

so x = 74.89 would be the final answer

last question, how do I find p35 under the normal curve

[alpine verge]

can you use a calculator for this or do you have to use z score

[runic grotto]

z score only

[alpine verge]

since we can just use inverse normal

to find that

alright

i’d say use a normal curve table

to get the z score of having a tail probability of .35

then just use the equation we used before to find x

[runic grotto]

so x = z+μ(σ)?

[alpine verge]

yep

[runic grotto]

how do I use with only p35

[obtuse pebble]

@runic grotto Has your question been resolved?

[runic grotto]

<@&286206848099549185>

[runic grotto]

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[timid silo]

try row reducing it

also i assume you mean rank?

[timid silo]

but anyways get it into row echelon form

are you not familiar with how to row reduce

show what you did then

[timid silo]

your notation is not clear to me, sorry. Can you write full matrices and highlight your row-reduction steps?

Okay. Then yeah thats the correct operations for the last row

[obtuse pebble]

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[errant crane]

why do you have to use radians when doing numerical methods?

[twilit meteor]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[errant crane]

there isnt a problem im just asking a question about numerial methods

[drifting wraith]

you don't have to, just every formula assumes you are, you wouldn't be able to use anything without modifying it for degrees

[errant crane]

.close

[obtuse pebble]

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[unreal cloud]

1 * 1998 + 2 * 1997 + 3 * 1996 + ... + 1997 * 2 + 1998 * 1 = ?

[unreal cloud]

.close

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[bleak valley]

for #2: if the asymptotes are supposed to be on the pi’s, what do I do when all of the increments are pi?

[bleak valley]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[glacial obsidian]

What are the increments?

[bleak valley]

it’s 2pi

[glacial obsidian]

Also the range is not ℝ

[bleak valley]

right now on my level it is

[glacial obsidian]

But what is the definition

I've never seen that word in this context

[bleak valley]

ohhh

it’s the value placed on each tick mark

[glacial obsidian]

The asymptotes are not on nπ

They would if it were csc(θ+3π) or something similar

But θ is divided by 4

[bleak valley]

yeah

after i factored out the 1/4 it’s now going left 3 or (x+3)

[glacial obsidian]

But you can't factor the 1/4 out and forget it, it will also change how you have to graph the function

[bleak valley]

this is the given solution from my teacher

[bleak valley]

yeah i know

but i’m confused on why in the solution the asymptotes are on specific numbers

it’s hard to see but the final asymptotes are on -3pi, pi, 5pi, 9pi, and 13pi

[glacial obsidian]

Yes

The asymptotes are on the points which are solutions to the equation θ/4 + 3π/4 = nπ

θ = 4nπ - 3π

So n=0 gives θ = -3π, n=1 gives θ = π, n=2 gives θ = 5π et cetera

[bleak valley]

OH

one second

how does it look so far?

[glacial obsidian]

It looks correct, but don't you have to graph only 2 periods?

[bleak valley]

yeah

the period is 8

[glacial obsidian]

You don't need all 5 asymptotes

Oh yes you do need them nevermind

[bleak valley]

okay

how do o graph it now

[glacial obsidian]

Like csc(x)

But change the numbers

[bleak valley]

sorry i’m confused

[glacial obsidian]

Use the graph of csc(x) here

And then do something similar with 3 csc(x/4 + 3π/4)

[bleak valley]

i just did the middle of each asymptote moved up 3 because of vertical stretch

[glacial obsidian]

Yes, that's correct

[bleak valley]

but i don’t understand the domain, why is it R except pi + 4pi n?

[glacial obsidian]

Because there are asymptotes on those points

For example

If θ = π then 3 csc(θ/4 + 3π/4) equals 3 csc(π/4 + 3π/4) = 3 csc(π), which is 3/sin(π) = 3/0

And you can't divide by zero

So the function isn't defined on those points

[bleak valley]

so where does the 4 pi n come from?

[glacial obsidian]

^

[little cedar]

do uk the domain of csc x

[obtuse pebble]

@bleak valley Has your question been resolved?

[obtuse pebble]

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[dense tinsel]

how do i properly expand the following:

[dense tinsel]

$f(x) = (x + 3)(x - 2)(x - 1)$

[warm shale]

GMDennis

[regal parcel]

we can start with two of the brackets

(x+3)(x-2)

[dense tinsel]

yeah well, here's the thing

it's a cubic function, right

[regal parcel]

yes

[dense tinsel]

so, okay, here's what happens when i expand (x+3)(x-2)

x^2 + x -6

so then i put brackets around that and do

[regal parcel]

👍

(x^2+x-6)(x-1)

[dense tinsel]

$(x^2 + x - 6)(x - 1)$

[warm shale]

GMDennis

[dense tinsel]

hmm

ok

one sec

woah

$x^3 - 7x + 6$

[warm shale]

GMDennis

[dense tinsel]

that's correct, right?

[cursive forge]

Yes

[dense tinsel]

interesting

[cursive forge]

The x² cancels out

[dense tinsel]

i didn't think that's how it's done

[dense tinsel]

ofcourse

so

the next one says

sketch the graph of $f(x) = x(x - 3)(x + 2)$

[warm shale]

GMDennis

[dense tinsel]

so since this is x-intercept form

wait, don't we need to first foil out (x - 3)(x + 2) and then multiply that result by x?

[cursive forge]

Yes

[dense tinsel]

or can we just put two points at 3 and -2

[regal parcel]

but we can already mark some points: (0,0), (3,0), (-2,0)

[cursive forge]

The curve will pass through the origin

[dense tinsel]

oh right, 0,0 is because of the x out front

ok

lemme sketch this real quick

one sec

so this begs the question

it would be going up, right?

like

the graph of the line would go up, dip a bit and then go up, right?

it would be going up after crossing (3,0)

my question is

[regal parcel]

yep

[dense tinsel]

how would i know where the min and max is

teacher doesn't want us using calculator for this

[regal parcel]

do you know what a derivative is?

[dense tinsel]

i jsut wanna know for reference whether it's important or not

[dense tinsel]

nope, haven't learnt it yet

im doing math IB AI HL next year tho so mb then

[regal parcel]

if you haven't done derivatives, then i think you can extrapolate, meaning you calculate some points (like basic ones between the three roots)

extrapolating as in, we can't really calculate the local max and min afaik

[dense tinsel]

this is pretty embarrassing but idk what extrapolating is 🥶

[regal parcel]

yeah i feel that, i learned that word pretty late

let's say you have some points of a function
you know that the function is somewhat "smooth" so you fill in the rest

[dense tinsel]

ehh, it's not important

book nor teacher cares abt it rn so

ill just pass on it for now

ok so

for $f(x) = -(x + 3)(x - 2)(x - 4)$

[warm shale]

GMDennis

[dense tinsel]

it's easy

it goes down in general

3 x - ints: -3, +2, and +4

[regal parcel]

when we expand this, the leading coefficent will be negative
the leading coefficient is the coefficient infront of the x with greates power
in this case we get f(x)=-x^3+...
and for very large, this term dominates
therefore for large or very small x, we get f(x)≈-x^3

and -x^3 goes down for large x

👍

[dense tinsel]

yeah alr, i got it

my next one is weird

it says

$f(x) = (x - 1)^2(x + 1)$

[warm shale]

GMDennis

[dense tinsel]

im thinking

it could be $(x - 1)(x - 1)(x + 1)$

[warm shale]

GMDennis

[dense tinsel]

is that correct?

but it can't be that

so it has to be $(x^2 - 2x + 1)(x + 1)$

[warm shale]

GMDennis

[dense tinsel]

and it therefore it must become $x^3 + x^2 - 2x^2 - 2x + x +1$

[warm shale]

GMDennis

[dense tinsel]

.close

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[obtuse pebble]

Please don't occupy multiple help channels.

@visual heart Has your question been resolved?

[visual heart]

<@&286206848099549185>

[sleek bear]

Isn't that right?

[visual heart]

Is it correct to just assume it? because I just thought even though they move the negative sign, it still starts as y < 0

[tardy epoch]

close your other channel

[visual heart]

.close

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[oblique minnow]

Prove Bernoulli's inequality: For all natural numbers n≥1 and all real numbers a >−1, it holds that (1+a)^n≥1+n⋅a.

[oblique minnow]

So i did like this...is it correct??

[obtuse pebble]

@oblique minnow Has your question been resolved?

[oblique minnow]

<@&286206848099549185>

[obtuse pebble]

@oblique minnow Has your question been resolved?

[oblique minnow]

<@&286206848099549185>

.close

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[bleak skiff]

Show that $f(z) = f(x + iy) = \sqrt{|xy|}$ has partial derivatives at 0,
verifying the Cauchy-Riemann conditions at 0, but $f$ is neither differentiable nor holomorphic at 0

[warm shale]

lilisworld.

[bleak skiff]

$\frac{\partial f}{\partial x}(0, 0) = \lim_{x \to 0} \frac{f(x, 0) - f(0, 0)}{x} = 0$

[warm shale]

lilisworld.

[bleak skiff]

same for the partial drrivative for y

<@&286206848099549185>

satisfied at (x,y) = (0, 0) but how do i do for the other points?

[obtuse pebble]

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[marsh moat]

21 i dont know what to do

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[timid silo]

I’m trying to figure out the arc measure and I need help with that

[timid silo]

.rot

,help

[warm shale]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[timid silo]

,list

[warm shale]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

Info

Display information about the guild, members, roles, channels, and more!
​ ​ ​ ​ ​ avatar: Obtains the mentioned user's avatar, or your own.
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​ ​ guildinfo: Shows information about the guild.
rolemembers: Lists members with a particular role.
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Utility

A handy toolbox of miscellaneous small but useful commands.
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Page 2/4

[timid silo]

,rotate

,rotate

[timid silo]

I need help finding the arc measure

Sorry I put 2/6.32 not 6/6.32

<@&286206848099549185>

[stone spoke]

What the og question?

[static furnace]

circumference is 2*pi*r, area is pi*r^2

and you need to find the measure of theta1, to find theta 2, to find the arc length

[timid silo]

I did it right with C

Theta 1 is 79.69 degrees

[timid silo]

My work-

[static furnace]

you did yes, but it looks like you're multiplying 6.32 and 6.32, not adding

but i see its correct now. the multiplication symbol threw me

[timid silo]

WHY DID I DO THAT

😭

[static furnace]

lol its okay

[timid silo]

But I need help with 2 - 5

[static furnace]

and do you know the formula to find arc length?

[timid silo]

Yeah

R x Rad

But idk how to do Thea1 and 2 I have difficulty

[static furnace]

well theta1+theta2 should equal pi if you're in radians right?

[timid silo]

Hold up-

What

[static furnace]

it's a straight line

in degress theta1+theta2=180

WAIUT

what am i talkinga bou

ignore me

[timid silo]

LOL

[timid silo]

Is 2 right?

Because I need that to find thea 2

[static furnace]

2 is correct, yes

[timid silo]

So how do I find Thea 2

[static furnace]

wait i was correct about the 180 thing

[timid silo]

It’s 100.31

[static furnace]

theta1+theta2=180 if it were in degrees

but its pi if its in radians

[timid silo]

1.75 in dad?

Rad*

OMG FINALLY

I GOT IT

[static furnace]

YAY

[timid silo]

😭

FUCK YEAH

😭😭😭

Maybe I need help with the Law of Sin

[timid silo]

I need to find degrees

[static furnace]

you're trying to find theta1 and theta2?

and you know what the law of sines is?

[timid silo]

Yup!

[static furnace]

alright so where are you getting stuck?

[timid silo]

How do I start it?

[static furnace]

sin(40) matches up with its opposite side, which is b

just like sin(theta2) matches up with c

it's all ratios. sin(A)/a=sin(B)/b=sin(C)/c

[timid silo]

???

Wait

Sin 40/ 7.42

[static furnace]

ya

[timid silo]

Then cross multiply

[static furnace]

so sin(theta2)/11.37=sin(40)/7.42

yes

[timid silo]

Ok let me do that

[static furnace]

okie dokie

[timid silo]

1.14

Rad

No.

wait what did I mess up

Help

Yeah I did something wrong

[timid silo]

Yeah I did something wrong

It’s not the right answer

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Nope

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[sinful slate]

11 h)

[sinful slate]

Idk why in the end i got like e^0 but the answer is e^1

Hmmm

[tardy epoch]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[sinful slate]

@tardy epoch

[tardy epoch]

You did the derivative of 1/tan(x) wrong

You wrote 1/( derivative tan(x)) but that's not how derivatives work

[sinful slate]

Oh thx i see

Wait it is how then ?

[tardy epoch]

This question doesn't make sense

[sinful slate]

How do u derivative 1/tan ?

[tardy epoch]

tan = sin/cos

So find 1/tan in terms of cos and sin

[sinful slate]

Hmmm how can I derive cos/sin ?

[obtuse pebble]

@sinful slate Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[karmic briar]

(5 - (x) - (x)^2)
x = -+0.9, -+0.1

(5 - (-0.9) - (-0.9)^2) = (5 - (-0.1) - (-0.1)^2) is true but

(5 - (0.9) - (0.9)^2) = (5 - (0.1) - (0.1)^2) isn't?

Not only that, how is the result of
(5 - (-0.9) - (-0.9)^2) or
(5 - (-0.1) - (-0.1)^2)

Smaller than if x was 0.8 or 0.5????

[karmic briar]

This isn't really a math question but rather a logic of how the equation works

[latent walrus]

what actually was the question

[karmic briar]

Well im tryna find the limit of (5 - (x) - (x)^2) as x approaches 0

[latent walrus]

okay

[karmic briar]

Im doing a table for the values coming from the left

Which is -0.9, -0.5, -0.3, -0.1 and -0.00001

But somehow

-0.9 and -0.1 is the same result

[polar fossil]

yes

[karmic briar]

And I tried fact checking it through calculators and don't understand the nature of it

[latent walrus]

because it has a symmetry half way between them

[karmic briar]

If -0.9 and -0.1 is the same result, how does smth like -0.5 have a bigger result tjan the both of them

[karmic briar]

How does that work?

[polar fossil]

because the function peaks at -0.5

try graphing it

[karmic briar]

OHHH

So it's like a mountain graph?

[polar fossil]

yeah sure

[latent walrus]

it has this shape

,w graph -x^2

[karmic briar]

How come it doesnt work if its positive then

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aldrnari_

[polar fossil]

how come what doesn't work if what's positive

[karmic briar]

Okay that helps visualize it a lot better

[latent walrus]

thats not the exact graph though keep in mind

[karmic briar]

If x = 0.9, 0.1 it is 3.29 and 4.89 respectively

[polar fossil]

well because that's nowhere near the line of symmetry

[karmic briar]

Noted noted

[latent walrus]

youre assuming a symmetry about y axis exists for some reason

there isnt one

[karmic briar]

Ahhhh

Damn okay that scratches the weird itch I had

[latent walrus]

,w graph (5-x-x^2)

[karmic briar]

Math is weird..

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aldrnari_

[latent walrus]

this is your graph

[karmic briar]

Mhm

[polar fossil]

i find that graphing functions - by hand or by computer - really helps to develop an understanding about them

[karmic briar]

ACTUALY that's really good advice

Sometimes just staring at the equation doesnt do it at all for me

Ill keep note of that

Thank you guys for the help!

This was just out of curiosity

Have a nice day

now how do i close this channel..

[polar fossil]

type .close

[karmic briar]

Ah alright thanks again

.close

[obtuse pebble]

Channel closed

Closed by @karmic briar

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[lean tinsel]

I thought that to find DE, you would minus E from D?

[lean tinsel]

so that would be 5-2,2-1,6-7 ?

[latent walrus]

D from E im afraid

[lean tinsel]

so it would be 3-5 for x?

[latent walrus]

5-3

actually, this is what youve done, no?

[lean tinsel]

yeah but the thingo says it wrong so I just had to double check

[latent walrus]

its because of how you typed it

[lean tinsel]

apparently it expected 2 digits instead of 3?

[latent walrus]

wrong bracket type

[lean tinsel]

Ooo let me try

hmmm it still says incorrect

[latent walrus]

darn

[lean tinsel]

but as long as I know i did my math right

[latent walrus]

hmm, the vector itself isnt wrong, so it must be a software thing

[lean tinsel]

thank you!

.close

[obtuse pebble]

Channel closed

Closed by @lean tinsel

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[cold brook]

can someone help me with part c?

[cold brook]

i think we have to use induction

so i first tested with n =1

a_2>a_4

so i did like

a_2n>a_2n+2

and then 1 + 1/a_2n > 1 + 1/a_2n+2

--> a_2n+2 > a_2n+4

but i think my method and my mid steps are wrong

so can someone please help me

[obtuse pebble]

@cold brook Has your question been resolved?

[obtuse pebble]

@cold brook Has your question been resolved?

[cold brook]

<@&286206848099549185>

[obtuse pebble]

@cold brook Has your question been resolved?

[cold brook]

.cloe

.close

[obtuse pebble]

Channel closed

Closed by @cold brook

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[frail hill]

Hello! So I have a pretty simple question, but it boggles me a bit. So we have a special dice that has the number 6 to a chance of 1/5, whilst the other numbers are still normal? What would be the chance to get the sum of 5 if we throw this special dice and a normal one? Wouldnt it be 4/36? Since the other numbers arent affected, but i dont think its right....

[obtuse pebble]

@frail hill Has your question been resolved?

[rocky goblet]

well it depends on what you mean by "whilst the other numbers are still normal"

if the other numbers all each have 1/6 probability each, then when you roll this dice there's a 103% chance that you get a number, which doesn't really make sense

if you mean they all have equal probability as each other, then that means you "share" the 4/5 chance that it's not 6 between the 5 other options and get that they're each 4/25, and now the probabilities actually sum to 1

[frail hill]

i would guess its 4/5 chance

so would that make the odds of getting one of the 4/5 / 5 ?

[frail hill]

I would guess its true! I was a bit confused what they meant by "normal". But i guess thats it. So the correct answer would then be ((4/25)*1/6) times 4 to get the probabilty of a hitting a sum of 5 (4 different combinations)?

[obtuse pebble]

@frail hill Has your question been resolved?

Channel closed

Closed by @frail hill

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

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

[lucid plover]

What does this f:x arrow mean

i am used to seeing it written in the red format

[glacial obsidian]

It is a different way of writing it

Not used usually

But some people do use it

[lucid plover]

what does the colon nd arrow mean

[glacial obsidian]

f: x -> y means f(x) = y

[lucid plover]

but why