help-10

[silent thorn]

you can do that as well

[woeful folio]

Never learnt differentiate

[silent thorn]

ok show me how youll complete the square

I never heard of that

[woeful folio]

I can’t

It’s x^3

[silent thorn]

obviously

its a cubic functiom

if you dont know calculus, the easiest way is to plot points kn the graph

on*

[woeful folio]

Alright I’ll do that if it comes in the exam

Help with a

<@&286206848099549185>

Cosine Rule?

[cosmic quest]

tan = Opp./ Adj. tan(38deg) * 6.4cm = CB , isnt it?

[woeful folio]

Cos = adj/hyp

[cosmic quest]

i meant tan lol mb

[woeful folio]

But what’s the Adjacent?

Opposite is CB

[cosmic quest]

6.4cm, no?

[woeful folio]

Won’t it be 10.9??

Nah 10.9 is hypotenuse right?

Is DCB = 90??

[cosmic quest]

Make DCB its own triangle, 10.9 is longest so its hypotenuse, 6.4 is adjacent

[woeful folio]

Yep

So whatever is longest is always hypotenuse?

How will you find Angle ADB??

Isn’t it 90?

Cause it’s right angled

[cosmic quest]

Unless it has a square to mark it as a 90° then it isnt 90°

[woeful folio]

135??

[cosmic quest]

180 - (38+45) = 97

[woeful folio]

That’s true

So how do I get D??

[cosmic quest]

Angle D, u add 45 and 35, then since u have to add angle D to make it 180° u just subtract 45+35 from 180

[woeful folio]

Alright

AD is 10.9

So it’s a isosceles?

If it’s a isosceles then it’ll be 180-45-45 which is 90

So it’s a right angle

And using that I can find AB

But bro @cosmic quest @keen garnet

How do I find the area of this trapezium?

Is 1/2(a+b)h

I don’t have height

@cosmic quest @keen garnet

<@&286206848099549185>

[cosmic quest]

Disregard what i said before, use only trig

[woeful folio]

Ok

How do I find area?

Do you think it’s possible if I find Area of both Triangles using 1/2abSinC and add the two together to get the area?

Damn no help bro

But thanks

.close

[obtuse pebble]

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[rigid lintel]

somehow i am stuck on proving NH is closed under the operation

i have proved the rest

suppose x, y in NH with x = nh and y = n'h'

then nhn'h' .... therefore nhn'h' is in NH

i guess i could show hn'h' is in H

ah

.close

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[modest pine]

anyone know about p/q

[modest pine]

when solving polynomials

ive been looking for resources online but can't get back to it

i just know you can divide all factors of p by all factors of q to get all possible factors of a polynomial with biggest degree term p and smallest degree q

but cant remember exactly if it was p/q or q/p

or if the biggest degree was p or q

and vice verse

a

nvm found one

.close

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[pure walrus]

I created an operation that is relevant to some work I'm doing and i'm trying to figure out how to express it in terms of absolute values. The way it works is that for two positive numbers or one positive and one negative, the output is the sum of the absolute values, but for two negative numbers, the result is the positive difference of the absolute values.
For example:
1◇2 = |1|+|2| = |1+2| = 3
-1◇2 = |-1|+|2| = |(-1)-(2)| = 3
-1◇-2 = |(-1)-(-2)| = 1

[tardy epoch]

Do you have a question

[pure walrus]

How do I express the operation in terms of addition, subtraction, and absolute values

without having to define it case by case

[tardy epoch]

If you're gonna define an operation, you have to cover all the cases

[pure walrus]

It's an operation over all real numbers, and I did cover all the cases

It's a commutative operation

and -a◇b = a◇-b = a◇b

[tardy epoch]

Then you did do this already?

[pure walrus]

I did it case by case, but I'd like a universal expression

[tardy epoch]

What do you mean universal expression

Absolute value is already case by case

,tex .abs def

[warm shale]

rie.mann

[pure walrus]

I'd like a single expression in terms of addition, subtrraction, and absolute value where I can plug the numbers in and get the result of the operation

I guess if other operations are necessary like multiplication and division that's ok, but I assumed that they wouldnt be required

[tardy epoch]

Doesn't sound possible

[polar fossil]

abs value can also be defined as $|x| = \sqrt {x^2}$ if you want it to, so maybe you can do something similar

[warm shale]

Hayley

[polar fossil]

but it's going to be painful

[obtuse pebble]

@pure walrus Has your question been resolved?

[pure walrus]

You could actually think about the case of a◇b where a and b are both positive as the exception

because for all other cases a◇b = |a-b|

[novel grotto]

max{x,y}+sgn(max{x,y})*min{x,y}?

i don't think it can be expressed elementarily

seeing as the ++ and -- cases are different

probably have to use sgn or max min

uh it doesn't really work for 0 issit?

[pure walrus]

a◇0 = 0◇a = |a|

0◇0 = 0

[obtuse pebble]

@pure walrus Has your question been resolved?

[obtuse pebble]

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[sharp tundra]

the text book instructs taht I use the ratio test to find convergence or divergence for this series. when i use the ratio test i use l'hopitals to get a convergent answer, however the correct answer is that its divergent.

[ruby path]

$\lim_{n \to \infty} \frac{(-1)^n \left ( \frac 32 \right )^{n + 1} \frac{1}{(n + 1)^3}}{(-1)^{n - 1} \left ( \frac 32 \right )^n \frac{1}{n^3}}$

Honestly I give up

[warm shale]

neonperseus

[ruby path]

This is what you have?

There's really no need to LH this

[brisk matrix]

!show

[obtuse pebble]

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

@sharp tundra Has your question been resolved?

[sharp tundra]

this is what i have

oh i forgot the abs but i still regarded that in my work

@ruby path @brisk matrix (sorry for the ping)

[shrewd wasp]

$\lim {n \rightarrow \infty}\left|\frac{\frac{(-1)^{(n+1)-1} 3^{n+1}}{2^{n+1}(n+1)^3}}{\frac{(-1)^{n-1} 3^n}{2^n n^3}}\right| = \lim{n \rightarrow \infty} \frac{3}{2}\left| \frac{n^3}{(n+1)^3}\right| = \frac{3\left(\left|\lim _{n \rightarrow \infty} \frac{n^3}{(n+1)^3}\right|\right)}{2} = \frac{3\left|\lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}{ }^3\right|}{2} = \frac{3}{2} > 1$ thus the series divergent. You can also use limit or root test.

[warm shale]

adzetto

[sharp tundra]

thanks

.close

[obtuse pebble]

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[opaque galleon]

(multivariable calculus) when you can solve a limit by just plugging in, do you still have to prove using epsilon delta definition?

[opaque galleon]

$
\lim_{(x,y) \to (1,0)} \frac{4x-y}{\sin y -1}
$

[warm shale]

casiofx991exz

[opaque galleon]

I got -4

do I have to do anything else? other than pluggin in

[tardy epoch]

Seems fine

[opaque galleon]

hmm ok

$
\lim_{(x,y) \to (0,0)} \frac{y \ln(x^2+1)}{e^{xy} +2}
$

[warm shale]

casiofx991exz

[opaque galleon]

So I can't just plugin

what am I supposed to do?

oh wait I can

nvm

do all techniques from calc 1 apply?

like multiplying conjugate

etc

factoring

[tardy epoch]

Sure

[obtuse pebble]

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[outer idol]

So apparently for problem c (subquestion ii ))is wrong

[outer idol]

my final answer I got was 17.4 (2 d.p.)

[timid silo]

Can anyone help me with this question
p x 26 = 208

What will be the value of p?

[outer idol]

should be p = 208/26

[timid silo]

Ok

I got the answer

[outer idol]

is it correct?

[timid silo]

Its 8 thank you

[outer idol]

np

[timid silo]

What?

[outer idol]

huh

it means no problem?

[timid silo]

The value of p is 8 ?

[outer idol]

sorry

wait

[timid silo]

Am I correct?

[outer idol]

yeah it should be

but

could you go grab another help channel please?

[timid silo]

Ok man sorry

[outer idol]

then I'll try to help you from there

this is occupied

[outer idol]

it's fine

[sour jacinth]

Whats the question?

[outer idol]

sorry

so

[outer idol]

here

my working out is

\ 1/3 * π ( 2^2 * 5 - 1^2 * 2/3 * 5) /

[sour jacinth]

I think you need to find r2 also

[outer idol]

I did

wait

it's not 1 right?

[sour jacinth]

Whats the value of r2 then?

You see in the figure r2 is the base radius

And probably bigger than the upper one

Sorry

[outer idol]

wait

I thought r1 is the base radius?

[sour jacinth]

How did you get 1 btw

[outer idol]

so

[sour jacinth]

Yes you are right

[outer idol]

when I tried to solve for problem c (subquestion i)

[sour jacinth]

Okay

Do you know about trigonometry?

[outer idol]

Yes

[sour jacinth]

Or you have proven part a right??

Then you just put the values and find r2

R2 is not 1 ig

[outer idol]

The answer is correct

[sour jacinth]

Okay

[outer idol]

But when I used the 1 for subquestion (or call it subproblem) ii I got it wrong

[sour jacinth]

Find r2 bro

For both cases respectively

Like for case 1 you get r2 as 0.2 i think

[outer idol]

;-;

Uh

I don’t get what your saying

[sour jacinth]

Okay

[outer idol]

And how am I suppose to find the R2 exactly?

Like

[sour jacinth]

Using the a part or trigonometry

[outer idol]

I tried to find it by doing scale factor

Oh

[sour jacinth]

You have already proven r1/r2= h1/h2

[outer idol]

Yes

[sour jacinth]

So just substitute the values and find r2

Did you get it now??

[outer idol]

I got it!

That’s for subquestion i

I think I can find it for subquestion ii

[sour jacinth]

Yes

U are right

I did something wrong back there

So you are right

[outer idol]

Got the correct answer!

The r2 for subquestion ii was 4/3

[obtuse pebble]

@outer idol Has your question been resolved?

[obtuse pebble]

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[mossy bridge]

Why is $\int_0^{\infty } te^(-2022t)$ divergent

[sage geode]

{\infty}

And {-2022t}

[mossy bridge]

approches infinity right

[sage geode]

I am talking about the latex code, put {} instead of ()

[brisk matrix]

did you try computing it

[mossy bridge]

o thx

yeaaa

[sage geode]

And who told you that it diverges

[warm shale]

Snow

[mossy bridge]

The problem

[tardy epoch]

Show the original question

Screenshot or pic

[mossy bridge]

one sec

Wait sry

I figured it out!

But ty for ppl who helped

[tardy epoch]

What did you figure out

[royal basin]

let the record state $\int_0^{+\infty} te^{-2022t} \dd{t}$ is in fact NOT divergent.

[warm shale]

Ann

[mossy bridge]

The problem asked to either evaluate or show that it's divergent

I just evaluated

.close

[obtuse pebble]

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[calm dust]

Part (b)…help 😨

[plain stag]

What's the definition of unit step function you've been given?

[obtuse pebble]

@calm dust Has your question been resolved?

[dapper wren]

Why is the function not continuous at x=0?

[plain stag]

#❓how-to-get-help

[dapper wren]

Oh thanks!

[obtuse pebble]

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[timid silo]

Find all pairs (a,b) such that
$GCD(a,b) > \frac{a+b}{3}$

[warm shale]

remington123

[chrome crypt]

there is no discontinuity.

[timid silo]

Can someone help?

You can WLOG assume that a\leq b. Now cpnsider two cases:
Case 1: b is a multiple of a . This will mean that the GCD(a,b)=a. You can move some things around and get an inequality involving a and b. Because b is a multiple of a there are only a few possibilities.
Case 2: b is not a multiple of a. In this case GCD(a, b) is at most a/2 because it cant be a and it has to divide a. Now you can make a stronger innequality and move some things around to get an inequality involving a and b.

If you want further help ping me

[warm shale]

alonelybean

[timid silo]

Thats also a smart way of doing it

[obtuse pebble]

@fierce mango Has your question been resolved?

[obtuse pebble]

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[warm shale]

1shero1

[vapid hamlet]

A more verbose way to write it is y'(x) = g(y(x)) h(x), as y depends on x.

[warm shale]

1shero1

[vapid hamlet]

No

It's [(g \circ y)(x) = g(y(x))]

[warm shale]

zanarcane

[vapid hamlet]

No, what you wrote doesn't make sense

because you would be concatenating two real numbers g(x) and f(x) instead of two functions g and f

anyways, there are plenty of examples for separable ordinary differential equations

y'(x) = x y(x)

[obtuse pebble]

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[granite notch]

I want to transform x (or x + d) such that this limit is equal to 2/pi

[granite notch]

for context, this limit models the density of points on the perimeter of a parametric shape

I want to change x, the input variable for my parametric equation so that the points have a fixed density around the perimeter while keeping the range of x that completes the shape the same

this is what I want to fix

[obtuse pebble]

@granite notch Has your question been resolved?

[obtuse pebble]

@granite notch Has your question been resolved?

[obtuse pebble]

@granite notch Has your question been resolved?

[obtuse pebble]

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[frosty fractal]

How are you supposed to figure out the angle from due north here? I look in the guide and see that it is tangent

and tan = 1/4

seems like one of those questions I am genuinely not supposed to be able to answer without a calculator

.close

[obtuse pebble]

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[twin oracle]

Hello

[twin oracle]

I need help with 123

It says to find an equation for plane that contains line p and touches sphere S

Sry line l

I checked to see if line touches the sphere but i get no real solutions there

[royal basin]

there should be two such planes, right?

"jednačine" means equations, plural

i think

anyway hm

so your line l goes through the point (0,1,1) and its direction vector is (5,6,-2)

[royal basin]

yeah that makes sense. if you got any real solutions it would mean the line crosses the sphere, and so any plane going through said line could never be tangent...

[twin oracle]

So there are 2 planes

[royal basin]

trying to think of the best way to approach this from a computational standpoint, and drawing a blank...

[twin oracle]

I was thinking maybe becouse of rotation of line i cant know for sure that there are 2

[royal basin]

no there are 2 definitely

it's visible from geometry

[twin oracle]

I had the idea that line in one plane that touches the sphere is paralel to the line l

And their difference should be some constant

[royal basin]

hmm...

i think that'll do us no good

ok so here is the plan i came up with

which is certainly very time and energy consuming

and i think there are better ways to do it which i simply cannot see right now

ok fuck i got lost in it myself

1. denote the center of the sphere with C = (2,-2,2), and the direction vector of line l with d = (5,6,-2).
2. express line l in parametric form as x = 5t, y = 1+6t, z = 1-2t
3. find the point on l closest to C, call it A

and at this point i am also stuck

it all feels so close yet so far

i am getting some mental images here but getting lost on how to calculate them

[twin oracle]

Soo if you find an equation of line on from a to c, is it to complicated to use that line as radius for a circle that will intercept the sphere at 2 points we need?

[royal basin]

it's a circle in R^3...

i mean sure you can find the equation of line AC but how will that help lol

[twin oracle]

So vector of that line perpendicular to vector d, and distance from AC, is it the same distance from line to those points?

If it is, then i have onenpoint, its distance and that it is on sphere

Distance from line

Forget the vector thing its irrelevant

[obtuse pebble]

@twin oracle Has your question been resolved?

[obtuse pebble]

@twin oracle Has your question been resolved?

[twin oracle]

So anyone got an idea?

[obtuse pebble]

@twin oracle Has your question been resolved?

[obtuse pebble]

@twin oracle Has your question been resolved?

[civic socket]

can you translate it as english?

[twin oracle]

.

[civic socket]

I just saw it

I'm not completely sure about this

[shrewd wasp]

-The line $(l)$ can be written in parametric form as $x = 5t$, $y = 6t + 1$, and $z = -2t + 1$.

-The sphere $(S)$ has center at $(2, -2, 2)$ and radius $r = 3$.

-The equation of a plane can be written in the form $Ax + By + Cz + D = 0$, where $A$, $B$, $C$, and $D$ are constants, and $A^2 + B^2 + C^2 \neq 0$.

The plane we're looking for is tangent to the sphere, which means it intersects the sphere at exactly one point. The distance from a point $(x_0, y_0, z_0)$ to the line $(l)$ is given by the formula:

$$
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
$$

where $A$, $B$, and $C$ are the direction numbers of the line, and $D$ is the y-intercept. In this case, $A = 5$, $B = 6$, and $C = -2$ (direction vector of line), and the point $(x_0, y_0, z_0)$ is the center of the sphere, $(2, -2, 2)$.

We can solve for $D$ by setting the distance equal to the radius of the sphere, $3$, and solving for $D$. Then, the equation of the plane will be $5x + 6y - 2z + D = 0$.

[warm shale]

adzetto

[shrewd wasp]

,w solve |52 + 6(-2) - 2*2 + D| / sqrt(5^2 + 6^2 + (-2)^2) = 3 for D over real numbers

[warm shale]

Results provided by WolframAlpha

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Requested by adzetto#0

[shrewd wasp]

So, the equations of the planes are: $5x + 6y - 2z + 6 - 3\sqrt{65} = 0$ and $5x + 6y - 2z + 6 + 3\sqrt{65} = 0$.

[warm shale]

adzetto

[twin oracle]

So im confused on the part how can we use componenlts of direction vector of line as coeficients for the plane

Difference of those 2 planes is a constant, which implies that they are parallel. So how can 2 parallel planes contain the same line

I think they can only contain a point from the line

[shrewd wasp]

Sorry for the late answer. I think I misunderstood the question, but if the plane contains the line it can never be tangent to the sphere. So the line is perpendicular to the plane, and we need to change the question or change the line equation.

[twin oracle]

im not sure about that

i think that picture in mind for question is that given line is the intercept of the to planes we seek

and we know that those planes are tangent to the sphere and their intercept is the lihne

now taht also could be the solution but im not sure that is the right intrepetation

[shrewd wasp]

Wow. If one were to consider the statement you made, it presents a highly thought-provoking inquiry. I have not yet discovered a refined solution; however, I will present a more algebraic approach. The procedure I employ involves determining the normal line for a given point on the sphere and subsequently normalising it. This can be achieved by utilising the coordinates of the point on the sphere, as well as the coordinates of two points lying on the line. By doing so, it becomes possible to ascertain the normal vector to the plane formed by these points, with respect to the points on the sphere. Subsequently, the point can be made distinct by intersecting the plane and the sphere and equating its discriminant to zero, or by formulating an additional equation for the distance between the point $(2,-2,2)$ and the centre point, and subsequently solving the system of equations involving three unknowns. I have discovered two planes as follows:

$$2x-2y-z+3=0, \quad -2x+y-2z+1=0$$

[warm shale]

adzetto

[obtuse pebble]

@twin oracle Has your question been resolved?

[tall tusk]

oh fuck!

[obtuse pebble]

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[spice path]

can explain for me how to do

.close

[obtuse pebble]

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[trim pumice]

mean value theorem for integrals to find c
f(x) = 16 sec^2(x), [-pi/4, pi/4]

[trim pumice]

so 16sec^2(pi/4)-16sec^2(-pi/4) = 32-32 = 0

[hollow schooner]

where u getting stuck?

[trim pumice]

so 0 = (pi/2)sec^2(x)

but secant can't be 0?

[urban patrol]

?? how does this imply that 0 = (pi/2)sec^2(x)

you just found the secant line

[trim pumice]

via the equation that i at least learned

hold on

[urban patrol]

MVT for integrals is finding the average value

[trim pumice]

i can't format it
integral [b,a] f(x)dx = (b-a)(f(c))

integral [b,a] f(x)dx = 16sec^2(pi/4)-16sec^2(-pi/4) = 32-32 = 0

[urban patrol]

$\int^{b}_{a} f(x)dx=(b-a)(f(c))$

[warm shale]

Arctic

[trim pumice]

(b-a)(f(c)) = (pi/4-(-pi/4))(sec^2(x) = pi/2(sec^2(x)

[urban patrol]

$\int^b_a f(x)dx = 16sec^2(pi/4)-16sec^2(-pi/4) = 32-32 = 0$

[warm shale]

Arctic

[urban patrol]

but f(x) = 16sec^2(x)

youre saying the integral is the same?

[trim pumice]

well 8pi(sec^2(x) doesn't make a difference but my mistake

[urban patrol]

,w integrate 16sec^2(x)

[warm shale]

Results provided by WolframAlpha

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

Requested by Arctic#1647

[urban patrol]

[trim pumice]

but we aren't integrating that side of the equation

[urban patrol]

what

wdym that side of the equation

you dont know f(c) thats what the integral is for

[trim pumice]

so 16tan(pi/4)-16tan(-pi/4)=8pi(sec^2(x))?

[urban patrol]

$f(c)\neq f(x)$

[warm shale]

Arctic

[trim pumice]

so 16tan(pi/4)-16tan(-pi/4)=8pi(sec^2(c))

[urban patrol]

yes

[obtuse pebble]

@trim pumice Has your question been resolved?

[obtuse pebble]

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[urban patrol]

would i just set A equal to the zero vector?

[brisk matrix]

well not just that

but that's a start yes

[urban patrol]

and then solve ofc

Ann interjects

[royal basin]

Ax = 0, not A = 0.

A isn't 0

[brisk matrix]

oh haha yeah

[urban patrol]

yeah Ax=0 right

i always factor out the x vector

[brisk matrix]

what does that mean

[urban patrol]

like

i dont think about that step

nvm

thank you guys

.close

[obtuse pebble]

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[plain halo]

is it possible to do this q with lhospital? i can do it using limit chain rule but i cant find much info/practice qs
i managed to get to $e^{\frac{6}{({ln{1+\frac{2}{x}})^2}(x^2+2x)}$ but not sure how to cancel the bottom of the fraction in the power

[warm shale]

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

[timid silo]

Can you not use the limit definition of e?

[
\lim_{x\to \infty}\parens{\parens{1+\f2 x}^x}^3 \
\parens{\lim_{x\to \infty} \parens{1+\f2 x}^x}^3
]

[timid silo]

Intuitive from there I gues

[plain halo]

oh i did not know about this lol, im still not quite sure how that simplifies to a 6? do u just multiply the 2 by the 3

[plain halo]

ah okay that makes sense, thank you :)

[obtuse pebble]

@plain halo Has your question been resolved?

[obtuse pebble]

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[steady patrol]

12 v

and 13 b

literally cant even grasp te concept

[grizzled shore]

Literally cannot see anything

Try take a better picture

[balmy nymph]

What does it say?

It's all blurred

[steady patrol]

oh fck

ight wait

12 (v) and 13(b)

is it still blurry?

[severe depot]

is 12 v: TRIANGLES, beginning and ending with a consonant?

[steady patrol]

yep

[severe depot]

for the first letter: 6 possibilities (T, R, N, G, L or S)

[steady patrol]

ok so 6!

[severe depot]

for the last one, 5 (same but you already have one at the beginning)

bruh

no

and then you can choose from the others for the remaining letters

[steady patrol]

6 × 7! × 5 = 151,200

thats the answer in the bookj

[steady patrol]

wat

[severe depot]

yes

[severe depot]

last one has to be a consonant, so either T, R, N, G, L or S, but you already have one of those at the beginning, so one of those isn't available anymore, therefore not 6 but 5

[steady patrol]

ah

wait they cant be the same letter?

the first and last?

[severe depot]

there's no repeating consonant in TRIANGLES

so 2 can't be the same

[timid silo]

what did you use to take this photo 😭

[severe depot]

if you have T at the front, you can't have a T at the back, TRIANGLES only has one T

[steady patrol]

dawg i had to take a pic on my ohone send it to myself on messanger (no discord on phone) download it on my pc and sednd it

[steady patrol]

but like

it said arranged

[severe depot]

arranging is just using the letters of TRIANGLES here

[steady patrol]

like in my mind im thinking one of the possible arrangments can be ttttttttt

[severe depot]

no we're talking about words that use the letters of TRIANGLES

so one T, one R, etc

[steady patrol]

next is tttttttr

aha

wait

is this question stupid

compared to othera asked on this server

others*

[severe depot]

don't compare yourself, i've seen people ask what y * y is and i've seen the most complicated differential equations

everything is allowed and your question is definitely not stupid

[steady patrol]

jonas

ily

❤️

[severe depot]

lmao

[steady patrol]

sexually btw

[severe depot]

welcome 😌

[steady patrol]

opk anyway

ok*

[severe depot]

i

[steady patrol]

13 b

[steady patrol]

shh

just accept ot

it*

[severe depot]

okay okay

i uh i can't read it

could u try for a better pic or type it out

[steady patrol]

ok lemme do this all over again

there

b

[severe depot]

i'm sorry lmao

okay

[steady patrol]

i was joking dude its fine

anyway

[severe depot]

all of them?

like i, ii, iii?

[steady patrol]

yea

[severe depot]

i and ii are pretty similar

[steady patrol]

iii looks hard

[severe depot]

do you know what n! means

[steady patrol]

yes

if n is 8 then n! equals 8x7x6x5x4x3x2x1

[severe depot]

yes

[steady patrol]

factorial basically

[severe depot]

so it's n * (n-1) * (n-2)...

[steady patrol]

yep

till u get to 0

[severe depot]

to 1 but yes

[steady patrol]

but here were gonna consider it a variable tho\

right?

[severe depot]

if you have n!, you can also write it as n * (n-1)!

nah

[severe depot]

can you understand

[steady patrol]

rlly?

[severe depot]

for example

7! = 7*6!

[steady patrol]

OH YEA

[severe depot]

LHS: = 7 * 6 * 5 * 4 * 3 * 2 * 1
RHS = 7 * 6 * 5 * 4 * 3 * 2 * 1

bruh

me when *

[steady patrol]

what

[steady patrol]

wait what

im so confused

[severe depot]

i fixed it

it's discord and it's italics

[steady patrol]

ohh

ok ok

[severe depot]

so do you understand that n! can also be written as n * (n-1)!

?

[steady patrol]

yes

they didnt say that in the book

[severe depot]

could you try (i) now

[steady patrol]

ok

[severe depot]

oh yeah that's uh shitty

[steady patrol]

probs wrong

wait no

we can furthur simplify it

[severe depot]

it's n(n-1)!

[steady patrol]

wait

its n?

final answer

[severe depot]

the solution to (i) yes

[steady patrol]

im too smart smh

[severe depot]

fr

[steady patrol]

what abt iii

ik we supposed to get rid of the fraction first

rght?

[severe depot]

uhm first you can fill in n!

in the two fractions

distributivity

[steady patrol]

so subsitutue n! wsith (n-1)!

n(n-1)!***

[severe depot]

nono first u replace the 1s with n!

[steady patrol]

the wat

[severe depot]

it's a distributive property lol

[steady patrol]

wtf

[severe depot]

it's n!/(n-1)! - n!/n!

i'm explaining it badly

[steady patrol]

oh so jus

replace 1

with n!

[severe depot]

yea just the multiplication

yes

it's just written weirdly to confuse u

[steady patrol]

cuz 1-1

is the same as

[severe depot]

this is very doable right

[steady patrol]

n!-n!

yea

it should be

[severe depot]

lmk your solution

[steady patrol]

ight wait 30 seconds

n! (n-1)

final answer

[severe depot]

no

[steady patrol]

THW FYUDCK

WHY

[severe depot]

so you have n!/(n-1)! - n!/n!

right

[steady patrol]

n!/n! equals 1

first of all

[severe depot]

yes that's 1

[steady patrol]

ok

then

[severe depot]

so we have n!/(n-1)! - 1

[steady patrol]

no

its still a fraction right?

[severe depot]

yes i'm just trash at typing

[steady patrol]

oh

ok

[severe depot]

n!/(n-1)! - 1

right

[steady patrol]

yep

[severe depot]

remember what you could write n! as

[steady patrol]

n!/(n-1)! is just n

[severe depot]

yes

[steady patrol]

ik ik

i simplified it

[severe depot]

so n-1

[steady patrol]

wat

dude

its gonna be

n*(n-1)! / *(n-1)!

which equals n

[severe depot]

yes

[steady patrol]

ok

so

n-1

[severe depot]

indeed

[steady patrol]

is the brackets

okj so now

we have

n! * (n-1)

right?

[severe depot]

huh? no

[steady patrol]

i wanna kms

[severe depot]

wait lemme work it out for u on paper brb

[steady patrol]

ok

.

[severe depot]

nonono

you put the n! in the brackets

[steady patrol]

OOOHOOHHHHHHHHHHHHHHHHH

[severe depot]

kdkfjdkfj

[steady patrol]

i thought u literally just spawned n! out of nowhere and repolaced it with the 1s

[severe depot]

LMAO

u cannot do that

omg jxksjfk no 😭

[steady patrol]

yea i was confused lolll

[severe depot]

sorry i explained it badly

[steady patrol]

its fine dude ur doing great

[severe depot]

do understand that it's n-1 as the solution now

[steady patrol]

yep

[severe depot]

💅

[steady patrol]

ok done thanks alot

btw

is this server actice 24/7

[severe depot]

yea there's always ppl around to help

[steady patrol]

yea i genuinly love that

anyways how do i close this

[severe depot]

it's chill yea

.close

[steady patrol]

.closed

.close

[obtuse pebble]

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[timid silo]

The stick drops and breaks at a random point distributed uniformly across the length. What is the expected length of the smaller part?

[timid silo]

.close

[obtuse pebble]

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[nimble harbor]

the farthest i've got to writing a proof for this is that i've figured out that an even number n with k amount of divisors can be written as: n = 2((1 x n) + (2 x n/2) + ... + (k/2 x 2n/k))/k, i don't know how to progress beyond that

[jagged falcon]

Okay, think about it like this, what are the possible even divisors of a? Like any even divisor would be of the form 2^n*k, where k is odd, right? What is the maximum possible value of n?

[obtuse pebble]

@nimble harbor Has your question been resolved?

[obtuse pebble]

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[karmic hedge]

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

[karmic hedge]

Not really sure what to do here

y^2 = x^2 + 16x + 65
I tried the quadratic formula but I got a negative in the discriminant

[warm canopy]

I remind you of this #help-0 message

[karmic hedge]

Oh yes!!

Sorry haha

Thx again

❤️

.close

[obtuse pebble]

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[karmic hedge]

.reopen

[obtuse pebble]

✅

[karmic hedge]

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

Not really sure what I did wrong here

I did quadratic formula with a = 1, b = 16 and c = (65 - y^2)

Yep sorry thats what I did

my bad haha

Oh wait no I didnt

I think thats the problem 1 moment

Yes youre right it is haha

Thank you!

❤️

.close

[obtuse pebble]

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[nimble harbor]

sorry for the late respond, if 'a' is divisible by 4, then the maximum value for n is infinite, if 'a' is only divisible by 2, then the maximum value for n is 1

[obtuse pebble]

@nimble harbor Has your question been resolved?

[errant lark]

Am i missing something here?

Since a is only divisible by 2 and not 4 : it will have only one 2 in its factorisation.
Basically,
$a = 2 * ( p1^{k1} * p2^{k2} * ... )$
where pi's are odd primes.

So, every even divisor of a must be of form 2*q where q is odd. Now, since there is a one-one relationship between even and odd divisors - they must be equal in number.

To see it more properly - You take any odd divisor which will be having different powers(not more than pi has in the factorisation) of pi's.
for this divisor - 2 * (this odd divisor) is a factor.

[warm shale]

.enemagneto

[obtuse pebble]

@nimble harbor Has your question been resolved?

[jagged falcon]

Yes, so any even divisor of a must be of the form 2k, where k is odd. If 2k is a divisor, then so is k, which is odd. Hence for every even divisor of a, we have a unique odd divisor of a.

Similary, say k is an odd divisor of a. Then k<=a/2. Now k is an odd divisor, 2 is a divisor, hence 2k is a divisor of a, which is even. So for every odd divisor, we have a unique even divisor of a.

So we have a bijective correspondence between even and odd divisors of a, hence they must be equal in number.

@nimble harbor

[obtuse pebble]

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[merry laurel]

How to integrate both

Possible or not

[novel grotto]

it is possible for both

[merry laurel]

Ok

Thanks

[novel grotto]

,texsp first one is consider ||$\frac{d}{dx}(ln(e^x+1))$||

[warm shale]

orthogonal_1

[novel grotto]

second one is kind of similar

[obtuse pebble]

@merry laurel Has your question been resolved?

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[buoyant bear]

i am getting -34 plz help

[obtuse pebble]

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

[buoyant bear]

man i am not stuck i am getting wrong ans

[timid silo]

show your working

[buoyant bear]

man i don't have phone

i can't send the photo

[timid silo]

sketch out what you've done by typing then

[buoyant bear]

so i considered a fxn x^3+ax^2+bx+c

then i do the diffentiation for the first time

and get

3x^2+2ax+b

=a

and put value of 1

and so on

6x+a=b

value i put was 2

and then

6=c

and the coffecient i got was

a=-9,b=-6,c=6

[timid silo]

why is the first derivative =a?

ah nvm, read your next line

your reasoning looks fine

[timid silo]

whats the actual answer?

[buoyant bear]

8

[timid silo]

,w solve 3+2a+b=a, 12+a=b

[warm shale]

Results provided by WolframAlpha

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

Requested by toby____#0

[buoyant bear]

oh fuck

now i got it

[timid silo]

nice

[buoyant bear]

basic maths problem

thanks man🫡

.close

[obtuse pebble]

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[hearty pendant]

this is all solution and i cant get the ssecond step wand what is done after

[jagged falcon]

You are dividing both sides by (1-x)^n to get (1-x)^2(n-r) on RHS, they just expanded that to (1+x^2-2x)^(n-r)

[hearty pendant]

and next one step?

@jaunty stream

<@&286206848099549185>

[bold carbon]

is this not just binomial expansion or smth

[hearty pendant]

yes

[bold carbon]

so couldnt u just use the expansion formula and apply it here

[hearty pendant]

how

[bold carbon]

here

[hearty pendant]

can u tell what happened in third step

[bold carbon]

ok\

maybe

but here is binom exp

https://www.bing.com/images/search?view=detailV2&ccid=dbJ0Xfmu&id=5D8AA9AB3BE7E5BCC2BE79FB19FFC95B975CAA69&thid=OIP.dbJ0Xfmu2_ZQGeFus4dchgHaFj&mediaurl=https%3A%2F%2Fimage.slideserve.com%2F1266454%2Fbinomial-theorem-example-6-l.jpg&cdnurl=https%3A%2F%2Fth.bing.com%2Fth%2Fid%2FR.75b2745df9aedbf65019e16eb3875c86%3Frik%3Daapcl1vJ%252fxn7eQ%26pid%3DImgRaw%26r%3D0&exph=768&expw=1024&q=binomial+expansion&simid=608010925471128843&form=IRPRST&ck=7A6BE59607D63EFAA5DC0F7B4AB79D15&selectedindex=2&ajaxhist=0&ajaxserp=0&vt=0

please dont question me using bing lol

anyay

also

[granite quartz]

WHAT.

[bold carbon]

im sorry

i cant figure it out

[obtuse pebble]

@hearty pendant Has your question been resolved?

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[obsidian wren]

I need help writing a two column proof for this question, and i have completelyyy no clue how to start this or prove the final result. Please help.

[jagged falcon]

Do you know about similar triangles?

[obsidian wren]

yes i have learned about them

[jagged falcon]

Yeah, so try switching PR to the denominator in RHS

[obsidian wren]

i get the fact that rs and qs and pr and rt are proportional

and the triangles are similair by sas similarity theorem

cause verticle angles theorem also mean angle r is same for both

but i dont know where to go after that

[jagged falcon]

What triangles exactly now?

[obsidian wren]

prq and srt

[jagged falcon]

Is it PRQ and SRT or PRQ and TRS?

Remember, the order is important

[obsidian wren]

oh yea mb

sry

[jagged falcon]

Yeah, so what can you say about angles RPQ and RTS?

[obsidian wren]

ummmm

idk

[jagged falcon]

PRQ and TRS are similar, right?

[obsidian wren]

oh yea cause the triangles are similair

[jagged falcon]

So how are angles RPQ and RTS related?

[obsidian wren]

didnt you just answer that?

they are similair?

[jagged falcon]

I meant the ANGLES

[obsidian wren]

sry for making you angry

[jagged falcon]

I'm not angry

[obsidian wren]

but the angles are what congruent?

[jagged falcon]

Angles are equal

[obsidian wren]

isnt congruent and equal the same thing?

[jagged falcon]

Not exactly.

But they are equal, right?

[obsidian wren]

yes

[jagged falcon]

So can you see why the lines PQ and ST are parallel now?

[obsidian wren]

nope

[jagged falcon]

Line PT intercepts PQ and ST, and subtebds equal angles on the opposite sides of the lines PQ and ST. That means PQ and ST must be parallel

[obsidian wren]

intersects could also mean it just touches it at a point right?

[jagged falcon]

Yes

[obsidian wren]

what is subtebds? or what di dyou mean by that?

[jagged falcon]

Subtends

Subtends means making an angle with the line it intersects

[obsidian wren]

ok the only problem is what would i put for reason in the proof

like what theorem postulate

i wrote the given

the verticle angles theorem

the sas similarity theorem

[jagged falcon]

Angles RPQ and RTS are equal, hence the lines are parallel. If you are using similar triangles, this statement need not be proven. I can give you a proof if you are curious, but you can simply write this statement without proof in the solution.

[obsidian wren]

in the proof would i say: <RPQ=<RTS

or congruent

[jagged falcon]

I don't remember the theorem name exactly, but you don't need to prove it

[jagged falcon]

This

Angles are not congruent, triangles are. Congruency is involved when there's some partition using some relation, there needs to be congruency classes.

[obsidian wren]

ok so for the reason of them equal i write corresponding angles of similair triangles are congruent?

[jagged falcon]

Yes

[obsidian wren]

can you give the proof?

i am curious

[jagged falcon]

Okay, wait.

[obsidian wren]

so ummm any idea

[jagged falcon]

You do not need to prove this in your solution. If you are upto similar triangles, it is assumed that you already know this statement.

[obsidian wren]

so can i just say if both angles are equal the lines need to be parallel?

because i need to provide a reason

[jagged falcon]

Yes

[obsidian wren]

ok

ok thanks

its been lots of help discussing this problem with you

.close

[obtuse pebble]

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[kindred oasis]

I'm working on multivariable calculus, Green's Theorem to be specific. C is a circle of radius 2, centered at the origin, so the textbook solves the integral using polar coordinates. Shouldn't it be r^2 instead of r^3 on the second line? x^2+y^2=r^2, right? It's just the textbook that is wrong or am I missing something? Thanks

[pseudo swift]

it's from the change of coordinates from cartesian to polar

dA = r dr d(theta)

[kindred oasis]

Oh ok, now I understand! Sorry, haven't used polar coordinates in a while... Thanks for the help, really appreciate it

.close

[obtuse pebble]

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[serene falcon]

Can some ecplain what to do

[serene falcon]

I translated it from a assingment in my book

[obtuse pebble]

@serene falcon Has your question been resolved?

[night glacier]

are you sure thats the full question

[serene falcon]

first i had to determine the belances and determine if they are stable or instable

i did that

and now this is the next question

[pseudo swift]

the equilibria you mean

what did you get then ?

[pseudo swift]

sounds right yes

are you able to represent what the equilibria and the the stability means geometrically ?

[serene falcon]

stability says if it goes in to the equilibria or from the equilibria. stable means to the equilibria and unstable means goes away from equilibria

[pseudo swift]

yes

so you have a solution

which starts at y=2 (so above y=1)

y=1 is unstable

[serene falcon]

so it goes to infinity

for example y=-3 than it goes to 1 right?

[pseudo swift]

that's the idea yes

[pseudo swift]

indeed

[serene falcon]

ah alright

thankyou

can you help me with something else?

[pseudo swift]

alr

[serene falcon]

whole different subject, but i was able to ditermine c

but now i have to fill in y(4)

but i'm struggling with that

this is the assignment. it is in dutch but they want me to determine y(4) exactly

but i'm really struggling with the y(4) part as you see in the bottum

[pseudo swift]

you should prolly find y as a function of t first

instead of keeping the y^(-3/2) form

I mean you could

it's a bit of a pain though

[serene falcon]

oke let me try that

yeah i'm really struggling with that

I'm stuck after this

[pseudo swift]

well how do you cancel that ^(-3/2) ?

[serene falcon]

yeah i don't know. because normally i when y^c = x is y = csqrt(x)

but that does not work with this

[pseudo swift]

y^c, you mean y^2 though

sqrt(y^3) is not y

[pseudo swift]

ah you mean $\sqrt[c]{x}$

[warm shale]

_aplatypus

[serene falcon]

yeah that

i mean that

[pseudo swift]

that should work also

the exponent is a bit more exotic I agree

[serene falcon]

yeah and i need to calculate exactly so no decimals

and that gives a real weird answer

if you understand what i mean

[pseudo swift]

it will still be valid as an exact answer though

even if you have that -3/2 th root

[serene falcon]

ah but i can fill it in, and give as answer 4^(-3/2)

[pseudo swift]

yes

[serene falcon]

ahh alright!

thank you!

[pseudo swift]

you can simplify that though

[serene falcon]

as 4*sqrt4

[pseudo swift]

wait 4^(-3/2) ?

[serene falcon]

ow wait

[pseudo swift]

how did you get it ?

the exponent seems weird here

i'd expect ^(-2/3) if anything

[pseudo swift]

y^(-3/2) = 4 alright fine

that doesn't mean y = 4^(-3/2)

[serene falcon]

ah yeah i see, why does it flip?

[pseudo swift]

-3/2 * -2/3 = 1

so (y^(-3/2))^(-2/3) = 4^(-2/3)

therefore $y^{\left(-\frac32\right)\left(-\frac23\right)} = 4^{-\frac23}$

[warm shale]

_aplatypus

[pseudo swift]

so y^1 = 4^(-2/3)

[serene falcon]

ah oke. can we simplify 4^(-2/3) into a fraction or is that not possible

so y=4^(2/3)

[pseudo swift]

it's less pretty than 4^-3/2 unfortunately

[serene falcon]

ah allright

thnx for your help!

this is the answer given, but shouldn't it be y(4)=4^(-2/3)

for the complete answer

[pseudo swift]

yes

typo prolly

[serene falcon]

thnx

you're very helpfull

.close

[obtuse pebble]

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[frozen jolt]

How do I solve this? (the answer needs to be in the form of pi, and radicals if there are any)

[raw hinge]

$sec(x) = \frac{1}{cos(x)}$

[warm shale]

_shachar

[severe reef]

9π=8π+π

[frozen jolt]

so 4 revolutions, 1/2 revolution, so it would be (0,-1)?

oh waity

[severe reef]

9π/4= (8π+π)/4
= 2π+π/4

That's just one complete revolution

And then π/4

[frozen jolt]

so one revolution

[severe reef]

So it's just 1/√2,1/√2

[frozen jolt]

oh gotcha

[severe reef]

Always try to break the numerator in terms of multiples of the denominator in such problems

[frozen jolt]

ok

yea i just learned trig yesterday lok

well

[severe reef]

I mean this isn't something exclusive to trig but yeah

[frozen jolt]

specifically unit of circle

[frozen jolt]

so basically, when doing unit of circle, i noticed there are common coordinates based on the revolutions, so it would be a good idea to memorize the ones on this circle?