help-13

[hearty elk]

if current gradiant is 2

for question c

hello?

[cedar kiln]

@hearty elk Has your question been resolved?

[cedar kiln]

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[crimson sedge]

In = \int_0^1 x^{n}\sqrt{1-x}
prove that (3+2n)In=2n.I{n-1}

[crimson sedge]

$In = \int_0^1 x^{n}\sqrt{1-x}
prove that (3+2n)In=2n.I{n-1}$

[wraith dagger]

Seif

[crimson sedge]

$In = \int_0^1 x^{n}\sqrt{1-x}$
prove that $(3+2n)In=2n.I{n-1}$

[wraith dagger]

Seif

[crimson sedge]

@fallen heath i know u love these kind of drugs

😂

[wraith crypt]

that In is cursed

[crimson sedge]

its not ln

[fallen heath]

I_n

[crimson sedge]

I n-1

like the previous one

only ansh understand me

[glad kestrel]

I_{n-1}

[crimson sedge]

yep

prove that $(3+2n)In=2n.I_{n-1}$

[wraith dagger]

Seif

[fallen heath]

$I_n = \int_0^1 x^{n}\sqrt{1-x}$ \

Prove that: $(3+2n)I_n=2nI_{n-1}$

[wraith dagger]

Ansh

[crimson sedge]

yessir

[wraith crypt]

much less cursed XD

[crimson sedge]

hes a magicien

look

[fallen heath]

Look who

[crimson sedge]

i tried ur old method ansh

didnt work

[fallen heath]

XD

[crimson sedge]

its 3 functions

i tried to make them 2 functions

its hard

to do parts

[fallen heath]

Figure: $2n(I_{n-1} - I_n)$

[wraith dagger]

Ansh

[crimson sedge]

im there

[fallen heath]

Should be easy

$= 2n \int_0^1 \overbrace{x^{n-1}}\underbrace{(1-x)^{3/2}} \dd{x}$

[wraith dagger]

Ansh

[crimson sedge]

wtf

[fallen heath]

integrate the first thingy, and differentiate the other one

[crimson sedge]

parts?

[fallen heath]

Yha

[crimson sedge]

ansh can u help me i have problem

with integrate

with parts

idk which one i choose u

and which one i choose v

is it by luck?

[fallen heath]

$=2n\qty[(1-x)^{3/2}\cdot \frac{x^n}{n}]_0^1 + 3I_n$

[wraith dagger]

Ansh

[fallen heath]

the first term is clearly 0 so you're only left with:

$2n(I_{n-1}-I_n) = 3I_n$

[wraith dagger]

Ansh

[fallen heath]

rearrange and derive your result :D

[crimson sedge]

$5^{\frac{a}{b}}$

[wraith dagger]

Seif

[crimson sedge]

what is this equal to

@fallen heath last thing big promiss

and i go to sleep

[fallen heath]

Gd ni8

[cedar kiln]

@crimson sedge Has your question been resolved?

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[bleak ermine]

hello

[bleak ermine]

can someon help me

An electrician charges $15 plus $11 per hour. Another electrician charges $10 plus $15 per hour. For
what amount of time will the cost be the same? What is the cost?

i have 15+11x=10+15x
10+15x+11x+15

[glad kestrel]

how did you go from an equation to an expression

[bleak ermine]

i dont know i thought that was an equation

[glad kestrel]

you went from correct to i don't know what

[bleak ermine]

wait so is that good

[glad kestrel]

going from good to bad? no

10+15x+11x+15
what is this

[bleak ermine]

umm

i dont even know why i put that

i just switched it around

[glad kestrel]

the thing above it is correct

[bleak ermine]

oh

[glad kestrel]

but i don't know how you got to step 2 there

[bleak ermine]

so i just solve that

[glad kestrel]

yes

[bleak ermine]

oh ok

ok

so now i subtracted 11 and im at 15 = 10 + 4x

[glad kestrel]

mk

[bleak ermine]

is that good

i think so

[glad kestrel]

yes

[bleak ermine]

now i have 5= 4x becaus i subrtacted 10

[glad kestrel]

righto

[bleak ermine]

umm

so i divide

now i have 5/4 = 4x/4

[glad kestrel]

mhm

[bleak ermine]

uhhh

x = 1 1/4

i think thats correct

[glad kestrel]

are you instructed to answer as a mixed fraction

[bleak ermine]

i dont think so because i have to find the amount of time

and 1 hour and 1 fourth is like 1h 15 m right

[glad kestrel]

leave it as an improper fraction then

5/4 is good

[bleak ermine]

ok

[glad kestrel]

you can also do that

5/4 hours is also acceptable

[bleak ermine]

now i have to find the cost

so

i have to like replace 5/4 for x right

[glad kestrel]

yes

[bleak ermine]

in one of the equations

[glad kestrel]

mhm

[bleak ermine]

so i subtracted 5/4 from 10

umm idk what that is

[glad kestrel]

why'd you subtract 5/4 from 10

[bleak ermine]

i dont know

i guess ill do 15 - 10

[glad kestrel]

you have the equations
11x+15 and 15x+10

[bleak ermine]

yea

[glad kestrel]

pick one and this is all you do

[bleak ermine]

so i just add it or something

[glad kestrel]

you're subbing 5/4 in for x

you will be multiplying 5/4 by whatever x's coefficient is in whatever expression you choose

[bleak ermine]

oh yea i forgor

my bad

11 times 5/4 is 55/4

[glad kestrel]

indeed it is

[bleak ermine]

thats 13 3/4

so 15 + 13 3/4

thats

28 3/4

so 28.75

[glad kestrel]

yep

[bleak ermine]

oh

so the amount of time is 1 hour 15 minutes and the cost is 28.75

[glad kestrel]

yep

units is $

[bleak ermine]

oh i forgor about that

my bad

ngl

if i had a calculator i would pass all my tests

thanks for helping me

[cedar kiln]

@bleak ermine Has your question been resolved?

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[sour glacier]

I’ve tried everything, please help!

[obsidian coral]

You tried everything but there's nothing inputted?

[cedar kiln]

@sour glacier Has your question been resolved?

[sour glacier]

I reset it, but I put ones in the first row in the columns of each free variable. Then I put 0’s everywhere and made the solution the vector it’s looking for, with 0 on the bottom

I know I’m doing something wrong but I can’t understand, I don’t know what to do with x4 either

[sour glacier]

<@&286206848099549185>

[cedar kiln]

@sour glacier Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[gray vine]

Show by using the definition of derivatives that the function $f\left(x\right)=\left|x-2\right|$ is not able to be derived in $x=2$.

[wraith dagger]

Theophania

[gray vine]

$$f\left(x\right)=\left|x-2\right|$$
$$f'\left(x\right)=\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$$
$$⇒f'\left(x\right)=\lim _{h\to 0}\left(\frac{\left(\left|x-2\right|+h\right)-\left|x-2\right|}{h}\right)$$

[wraith dagger]

Theophania

[gray vine]

Is this right?

$$f:'\left(2\right)=\lim _{h\to 0}\left(\frac{\left(\left|2-2\right|+h\right)-\left|2-2\right|}{h}\right)$$
$$f:'\left(2\right)=\lim _{h\to 0}\left(\frac{h}{h}\right)=1$$

[wraith dagger]

Theophania

[gray vine]

Okay… what do I do next 😅

[gray vine]

I don't know if I did everything correctly

[cinder talon]

simplify that

don't plug in 2 yet

also btw it would be |x+h-2| not |x-2|+h

since the +h is within hte f(x+h)

[gray vine]

hmm ok

is it written like this $\left(\left|x-2+h\right|\right)$ or without the ( and )?

[wraith dagger]

Theophania

[cinder talon]

you don't need the parentheses

thats redundant

[gray vine]

$f:'\left(x\right)=\lim _{h\to 0}\left(\frac{\left|x-2+h\right|-\left|x-2\right|}{h}\right)$

[wraith dagger]

Theophania

[gray vine]

how do I simplify this

x-2?

[cinder talon]

now you can plug in 2

[gray vine]

$f:'\left(2\right)=\lim _{h\to 0}\left(\frac{\left|2-2+h\right|-\left|2-2\right|}{h}\right)$

$f:'\left(2\right)=\lim _{h\to 0}\left(\frac{\left|h\right|}{h}\right)$

[cinder talon]

yep

[gray vine]

= 1

eh wait

[cinder talon]

well it would be f'(2)

no

take the left and right limits

[wraith dagger]

Theophania

Theophania

[gray vine]

how do i do that

[cinder talon]

you haven't learnt left and right limits yet.?

[gray vine]

it could be buried somewhere in my book

but i can't ctrl+f

give me a moment

so here

if h approaches zero from the right, then h>0 and |h| = h

then the second image follows

and if h approaches zero from the left, then h<0 and |h| = -h

[cinder talon]

$\lim_{h \to 0} \frac{\abs{h}}{h}$ exists if and only if $\lim_{h \to 0^+} \frac{\abs{h}}{h} =\lim_{h \to 0^-} \frac{\abs{h}}{h}$

[wraith dagger]

sean

[cinder talon]

exactly

[gray vine]

i must admit this is above my paygrade, i need to watch a video or smth

[cinder talon]

do you know the definition of absolute value?

[gray vine]

it's even more confusing when we throw in absolute value

no, not off the top

[cinder talon]

|x| = x if x≥ 0
      -x if x<0

[gray vine]

yes

why isn't it x≤0?

that part i didn't get

[cinder talon]

well

if |x| = -x when x≤ 0 and |x| = x when x≥ 0, then thats not possible

|x| cannot equal both -x and x

[gray vine]

True

$f'\left(2\right)=\lim _{h\to 0}\left(\frac{\left|h\right|}{h}\right)$

so this is where i should proceed?

[wraith dagger]

Theophania

[cinder talon]

yeah

so first

try take the right limit

[gray vine]

If $h→0$ from the left then $h<0$ and $\left|h\right|=-h$ which gives us $\lim _{h\to :0^-}\left(\frac{|h|}{h}\right)=\lim _{h\to ::0^-}\left(\frac{-h}{h}\right)=-1$

[wraith dagger]

Theophania

[gray vine]

$\lim _{h\to :0^+}\left(\frac{|h|}{h}\right)=\lim _{h\to ::0^+}\left(\frac{h}{h}\right)=1$ if $h>0$ and $h→0$ from the right.

[wraith dagger]

Theophania

[gray vine]

So what does this mean, when the limit is 1 and -1, does that mean the tangent is different at that point and so you can't derive the function at x = 2?

@cinder talon

you or someone else can ping and i'll check later in case you're busy

[cinder talon]

oops sorry

i was doing smth

im back now

uhh

so you cant directly use that yet

[wraith dagger]

sean

[cinder talon]

so what u want to do is use x-2 instead of x within the definition of absolute value

so

|x-2| = x-2 when x-2≥ 0 (which is the same as x≥2)

and

|x-2| = -(x-2) when x-2 < 0 (which is equal to x<2

now you can find the left and right limits

@gray vine

[gray vine]

lol didn't i do that?

from here #help-13 message

down to here #help-13 message

[cinder talon]

oh shit sorry im kinda dumb

[cinder talon]

yeah this is correct

[cinder talon]

this is also correct

now you see tht the left and right limits are not equal

therefore f(x) is not differentiable at f(2)

yeah u got it

[gray vine]

thanks

.close

[cedar kiln]

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[cedar kiln]

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[upper cypress]

let's say you have
P(B) = 0.3
P(A|B) = 0.7
P(A|B') = 0.3

Is the probability of A just 1.0?

[upper cypress]

since it can either be P(A|B) or P(A|B')

actually that doesn't make sense

I think I'm missing something

I'm not sure how P(B) plays a role here

but I get the feeling that it does

[dire geode]

Use the law of total probability

[upper cypress]

oh okay, I'll look that up!

I found a very helpful website based off your suggestion

I think I can solve this now 😄

its these terminology that can help guide my search

Perfect, I got
P(A) = P(A|B)P(B) + P(A|B')P(B')
= 0.7 * 0.3 + 0.3 * 0.7
= .21 + .21
= .42

how do I mark this channel as solved?

[cedar kiln]

@upper cypress Has your question been resolved?

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[plucky summit]

h

[plucky summit]

how would i go about solving this? i could use a calculator but my math teacher doesnt let me

[glad kestrel]

just simplifying it?

[plucky summit]

how would i go about doing that?

if square root of 3 gives a decimal

[glad kestrel]

leave it at sqrt3

[plucky summit]

where did they get the 2 and 3?

from simplifying the original problem?

[glad kestrel]

multiplying by the reciprocal

if given $$\frac{a}{\frac{b}{c}}$$ just multiply both top and bottom by the bottom's reciprocal
$$\frac{a\cdot\textcolor{red}{\frac{c}{b}}}{\frac{b}{c}\cdot\textcolor{red}{\frac{c}{b}}}$$

[wraith dagger]

a disappointing son

[glad kestrel]

then they rationalized

[cedar kiln]

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[candid acorn]

How do I do it ?

[sand ether]

Do you understand that if $x-4$ is a factor of a polynomial $p(x)$ then we can rewrite as

[wraith dagger]

azeem321

[sand ether]

$p(x) = (x-4)f(x)$

[wraith dagger]

azeem321

[sand ether]

Where $f(x)$ is some other polynomial

[wraith dagger]

azeem321

[candid acorn]

Yes

[sand ether]

ok good

[candid acorn]

I am supposed to use synthetic division then factor that

[sand ether]

So $3x^3-20x^2+37x-20=(x-4)(ax^2+bx+c)$. Now you just solve for for $a,b,c$

[wraith dagger]

azeem321

[sand ether]

Oh. you have to use synthetic divison?

[candid acorn]

Yes then factor it

????

[sand ether]

sorry i don't know synthetic division, so maybe someone else will help

[candid acorn]

What about another method?

[sand ether]

$3x^3-20x^2+37x-20=(x-4)(ax^2+bx+c)$. Now you just solve for for $a,b,c$

[wraith dagger]

azeem321

[sand ether]

Now compare coefficients

So $3x^3-20x^2+37x-20 = ax^3-(4a-b)x^2-(4b-c)x - 4c$

[wraith dagger]

azeem321

[cedar kiln]

@candid acorn Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[valid cloud]

I don't quite understand the conditions needed to ensure for N to work.

[fallen heath]

moe-moe cannon

[valid cloud]

What Condition N should have to make it work

Idk if it works if you try case by case

[fallen heath]

I think the mirror reflection chapter from physics would help >_<

[valid cloud]

Yeh I think it involves some distance and things

Like 9 wouldn't work because when distance is 3, It goes 1 4 7 and 1 again, Thats where I don't get

you kinda have to think about what values of n that can ensure, that the "beam" goes through all the spots

[tepid plover]

i'm guessing that n must not be divisible by all integers from 2 to n-1

[fallen heath]

A good way to view it would be taking the moe-moe cannon on one of the midpoints of the edges of a regular polygon

and the mirror- the other midpoints

[valid cloud]

So in that logic 4 doesn't work

[valid cloud]

Hmm how

[tepid plover]

yes, since if distance is 2 it would go 1, 3, 1, 3, and so on

[fallen heath]

oh right, the ceremonial circle

[valid cloud]

How does 9 works there

[fallen heath]

Kuristinaa, have you tried drawing the figure and generalizing it?

Suppose "n" equidistant points on circle

[valid cloud]

I have drawn 10 circles by now, with equal distance but idk what to do

[tepid plover]

the cannon should bounce on all the mirrors

regardless of the distance

[fallen heath]

n/k > 2

[valid cloud]

Yeh Yeh

Can you elaborate?

[fallen heath]

nvm. the beam needs to bounce off all mirrors

[valid cloud]

Hmm so prime numbers?

[tepid plover]

i think so

if n isn't prime, some mirrors would be left if the distance is one of the factors of n (not 1 or n)

[fallen heath]

I think I figured something

[valid cloud]

Hmm

[fallen heath]

Wanna hear me out? (@_@;)

[valid cloud]

Yep

Let's see if the hypothesis that prime numbers work

Do work

[fallen heath]

kek

So basically, we start by assuming "n" equidistant points on the ceremonial circle.

Now, we assign $b_n$ = position of beam after nth reflection

[wraith dagger]

Ansh

[fallen heath]

Let's say we start with "1"

[valid cloud]

Ok Ok

[fallen heath]

We're replacing all the marks $a_1, a_2, \ldots, a_n$ with their subscript value "i"

[wraith dagger]

Ansh

[fallen heath]

So 1 denotes the first mark $a_1$, 2 denotes $a_2$ and so on

[wraith dagger]

Ansh

[fallen heath]

Now, $b_0 = 1$

[wraith dagger]

Ansh

[valid cloud]

Oki Oki

[fallen heath]

Let's say $b_1 = k$ for some $1 < k < \frac{n}{2}$

[wraith dagger]

Ansh

[fallen heath]

n/2 because symmetry argument

Also, if n even then beam goes along the diameter, reflects back, we're done. So n must be odd

[valid cloud]

But not every odd

[fallen heath]

Now, what do you think $b_2$ is?

[wraith dagger]

Ansh

[fallen heath]

in terms of k

[valid cloud]

k^2?

[fallen heath]

Nope

$b_2 = k + (b_1 - b_0) \pmod n$

[wraith dagger]

Ansh

[fallen heath]

which is basically: $b_2 = k + (k - 1) \pmod n$

[wraith dagger]

Ansh

[fallen heath]

what is $b_3$?

[wraith dagger]

Ansh

[valid cloud]

k+k+(k-1) mod n?

[fallen heath]

wait

[valid cloud]

wait

[fallen heath]

$b_3 = k + 2(k-1)$

[wraith dagger]

Ansh

[fallen heath]

note: the distance between each positions is now fixed

since initially we chose b0 = 1, b1 = k, and the distance is now fixed as (k - 1)

and we now have an AP

$b_r = k + (r - 1)(k-1) \pmod n$

[wraith dagger]

Ansh

[valid cloud]

Ooooohhhhh

[fallen heath]

Now, according to given conditions, $b_n = 1 \pmod n \forall k$

[wraith dagger]

Ansh

[fallen heath]

and, you can figure the other condition which denies any of this positions to be "1" for any "r" < n

[valid cloud]

oooohhhh

[fallen heath]

of course there might be a more direct approach I believe but for a crude understanding of the situation, I believe this was the best that came to mind rn.

[valid cloud]

So we test values with them?

[fallen heath]

I dunnoo. restrict "n" somehow maybe?

[valid cloud]

how would it work if lets say we have 16 points

[fallen heath]

Can't you figure something from: $b_n \equiv 1 \pmod n$, , and \ $b_r \neq 1 \pmod n \forall k, , 1<k< \frac{n}{2}$

[wraith dagger]

Ansh

[fallen heath]

Well, $b_n \equiv \pmod n$ always holds for all "k"

[wraith dagger]

Ansh

[fallen heath]

the only restriction should be the second one

[valid cloud]

ok ok

seems complicated

How would it work if N is prime

[fallen heath]

$\forall r \in {1, 2, \dots, n-1}, , \forall k \in {2, \dots, \frac{n-1}{2}}, r(k-1) \not\equiv 0 \pmod n$

[wraith dagger]

Ansh

[fallen heath]

that's the condition T_T

[fallen heath]

does this mean "n" must be prime ;-;

[valid cloud]

16

THANKS

.close

[cedar kiln]

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[fallen heath]

$\forall r \in {1, 2, \dots, n-1}, , \forall k \in {2, \dots, \frac{n-1}{2}}, r(k-1) \not\equiv 0 \pmod n$

[wraith dagger]

Ansh

[fallen heath]

@valid cloud explain

ohhh it's because of some divisor argument

,calc 3 + 2sqrt(2)

[wraith dagger]

Result:

5.8284271247462

[fallen heath]

hmm

Right. We can directly see n = 3, 5 are okay. For n > 5, there isn't any divisor in the choices for "k"

such that correspondingly, there's a choice in "r" for which r. (k-1) is divisible by "n"

hence, all n primes work

[valid cloud]

Yep primes do work

[fallen heath]

I mean, was my reasoning correct?

Ik primes work, and that my reasoning is sound but I feel something loose here lol

[valid cloud]

I Think its a good assumption that its the prime numbers just like that one dude who told me that

[fallen heath]

Or did you figure some simpler soln.

[valid cloud]

Idk i just test values and there should be a pattern

[fallen heath]

Wait, you didn't read my solution?

[valid cloud]

I read it but if you ask whats my thought process, Then i should just test values even if its hard work

[fallen heath]

ooo, hmm! I would agree but how would you test values here?

[valid cloud]

Like 4 wouldnt work if distance is 2 it would be 1 3 1 3

6 if distance is 2 it would go back again

But 9

[fallen heath]

n even automatically removed, what about n = 3k, 5k, 7k? all can be broken into smaller shapes which return after 3, 5, 7 reflections instead of 3kth, 5kth, 7kth

Hence, only primes goes

Woah. this soln. much easier (@_@;)

[valid cloud]

If distance is 3 then 9 would go 1,4,7,1

[fallen heath]

For 9 as well, there's a 3 reflection path,

since 3 | 9

hm, I got it. Checking cases was indeed wise

Ty

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[lilac juniper]

so for a business I'm trying to get the yield. I have the market value which is $8.26 and the par value which is $6.00 and the dividend is 20c per share. How do I calculate the rate of dividend to get the yield and can I even do that with the info given?

TLDR: How do I calculate what the dividend would be to the market value if it was a %. So what % is 20c to $8.26 or how do I calculate that

[lilac juniper]

<@&286206848099549185>

[cedar kiln]

@lilac juniper Has your question been resolved?

[lilac juniper]

Can someone please help me out here <@&286206848099549185>

I got the answer, thanks

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[supple garden]

This is probably a really dumb question but, are you sure you had $x$ anywhere? I make that mistake a lot.

[wraith dagger]

vibeman

[supple garden]

Found the issue? @crimson sedge

What's the value of the line you got?

Yeah, what was the y value of the line?

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[crimson sedge]

Hey guys

[crimson sedge]

I need help with triangle similarities

It's 27th one

I'm gonna translate it

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[hot garden]

Need some help with a geometry\trigonometry proof problem:
"Given a semicircumference of diameter AB = 2r, and center 'O', consider the chords AD = r/2 and BC = r.
Find the surface of the triangle COD"

[hot garden]

I only managed to find out these data using some trig theorems and chord theorems

[fallen heath]

Well, at least you know OD = OC = r, right?

[hot garden]

but the sin and cos aren't nice values, so I don't get how is the solution not including stuff like arcsin or arccos,
I figured out gamma is pi - phi - zeta, but I can't figure out exact value of zeta

[hot garden]

Yeah

[fallen heath]

nvm

[hot garden]

if you mean arcsin, arccos, arctan, arcsec, yes

but they're not rational values in this case , unless i did some horrible errors already

[fallen heath]

so if AB is the diameter

$\angle COD = \pi - \angle AOD - \angle COB$

[wraith dagger]

Ansh

[fallen heath]

and what you need is $CK = OC \sin \angle COK = OC \sin \frac{\angle COD}{2}$

[wraith dagger]

Ansh

[fallen heath]

tbh BC = r, really saves the day

cause then you know $\angle BOC = \frac{\pi}{3}$

[wraith dagger]

Ansh

[hot garden]

yeah BOC is equilateral triangl

i figured that out already

[fallen heath]

$\Delta_{COD} = \frac{1}{2}r^2 \sin \qty(AOD + \frac{\pi}{3})$

[wraith dagger]

Ansh

[fallen heath]

$\cos AOD = \frac{7}{8}$

[wraith dagger]

Ansh

[hot garden]

yeah tho AOD is not a nice angle .-. I wish i could arccos it

[fallen heath]

.-. Can't you just use the sin(A + B) formula?

[hot garden]

I don't see it where should it be used

[hot garden]

so sin(AOD)cos(pi/3) + cos(AOD)sin(pi/3)?

[fallen heath]

Yeah

and cos(AOD) and sin(AOD) are known

so just fill it?

[hot garden]

I don't get where does that surface formula comes from tho,

is it from chord theorem?

[fallen heath]

$\Delta_{COD} = \frac{1}{2} \overline{OC}\cdot \overline{OD} \sin \angle COD$ ?

[wraith dagger]

Ansh

[hot garden]

oh, OD sin(COD) = DC / 2

or not?

[fallen heath]

not

Also, this formula comes from the sine law in triangles

But you'd probably hear of this also when doing circles

[hot garden]

but isn't this true due to chord theorem?

uhhhh thank you tho, I guess I've been assigned an homework but we haven't done sine law yet in school , that's why I wasn't understanding it

[fallen heath]

np

[fallen heath]

After you simplify this using sin(A+B) and use cos(AOD) = 7/8, sin(AOD) = √15 / 8, I'm sure you'll get your answer!

[hot garden]

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[fallen heath]

note that: $\overline{OD} \sin \angle COD$ is actually the height of the triangle COD drawn from vertex D to side OC

[wraith dagger]

Ansh

[cedar kiln]

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[fossil thorn]

Can I get help with the first question please

[fossil thorn]

,rotate

[dire geode]

the shortest distance from a given point to a line intersects the line at a perpendicular

[fossil thorn]

and perpendicular lines have the same slope....?

[dire geode]

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/parallel-and-perpendicular-lines-2/#:~:text=Two lines are perpendicular lines,2 %3D − 1 m 1

you can calculate the original slope from your line's equation, then find the perpendicular slope.

[fossil thorn]

ah okay tysmm

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[gaunt trench]

this is basically asking for a piecewise function right

?

[crimson sedge]

Yeah

[gaunt trench]

ok i so i have this, f(x)= {y = 5 , x =/= 1}

what else would be thgere

[crimson sedge]

Well they also want so if they redefine x its continuous afain

[south tundra]

For example f(x) = x + 1 for x =/= 1 and f(x) = literally any number except 2 for x = 1

This way you can redefine f(x) = x + 1 for x =/= 1 and f(x) = 2 for x = 1, making it continuous

Or should I say removing the discontinuity

[gaunt trench]

ohh

right

okay

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[radiant mountain]

Not sure what 13 is asking

[cedar kiln]

@radiant mountain Has your question been resolved?

[dire geode]

prove there exists a point $1 \le a \le 3$ such that $f(a) = 4$

[wraith dagger]

riemann

[radiant mountain]

ok how do I do that

[dire geode]

"existence" statements usually follow from theorems. look in your chapter for one that can apply

[radiant mountain]

like i think the mean value theorem

[dire geode]

looks good

[cedar kiln]

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[worn mantle]

I keep getting this wrong, need help for an explanation of each

[shrewd brook]

Please can someone help me with this paper

[dawn jetty]

@shrewd brook I think if you were more specific and posted snaps instead of files people would be more responsive, h

but it looks like the channel is occupied already

@shrewd brook #❓how-to-get-help

[rigid ridge]

the first one at the very least is false, a counterexample would be (a,b,c) = (2,1,3) because a^b a^c = 2^1 2^3 = 16 but a^(bc) = 2^3 = 8

and in fact, the actual statement should be the second one, a^b a^c = a^(b+c) by basic properties of exponents: multiplying something b times and then c times is the same as multiplying b+c times!

[worn mantle]

ok!

[rigid ridge]

moreover, a^(-b) = 1/a^b is also true because multiplying something -b times is the same as dividing b times

but a^(-b) = -a^b is then false, a counterexample is (1,1) since 1^-1 = 1 but -1^1 = -1

and a^b = ab is just plain false

https://cdn.discordapp.com/emojis/871537511841816576.webp?size=48&quality=lossless

[worn mantle]

thanks man, really appreciate the explanatons! much clearer to me

[rigid ridge]

np

[worn mantle]

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[inland meadow]

Feels like I need to apply some markov property and bayes rule here, but I don't really know how.

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@inland meadow Has your question been resolved?

[inland meadow]

<@&286206848099549185>

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[wraith crypt]

feels like law of total probability

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[hollow portal]

So I'm having trouble solving for X in the equation y(2-y)=x, I've expanded to x= 2y-y^2 however I am lost on the next step

[dire geode]

that's the answer

x = 2y - y^2

[livid hound]

x was already isolated and you didn't need to do anything at all

[hollow portal]

I mean i need it to be y=x instead of a y-based equation

so like it's currenty f(y) and i need it as f(x)

just a hint in the right direction would be fine I don't necessarily want it done for me

[livid hound]

treat this as a quadratic equation in y

[hollow portal]

ok so leave the x and use quadratic, and then solve? i'll try it thanks!

well i'll have to bring the x over.. hmm ok

ok i think i got, i get 1-(1-x)^(1/2) and/or 1+(1-x)^(1/2), appreciate it guys!

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[fallow yew]

how do i show that they’re similar? i tried look it up on yt for examples and couldn’t find any so if anyone could link a video on how to do so that would be great 😊👍

[crimson sedge]

Hint: what type of triangles are they

[fallow yew]

right?

[crimson sedge]

Yes

[fallow yew]

so do i try to figure the rest to get it to 180

[crimson sedge]

You know they are all right triangles

How can you prove that

Hint: what must every right triangle have

[fallow yew]

90 degrees

[crimson sedge]

Right

And all the triangles have a right angle

That should be a sufficient proof

[fallow yew]

yes but how does the 3 one have a 90 degree

is see the 2

but not the 3

wait

nvm i see it’s the angle marking on top i’m pretty sure

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[fallow yew]

ty btw

[cedar kiln]

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[civic coral]

alright so y=a+bx is the formula for regression analysis

"r" stands for Coefficient of Correlation, S stands for Standard Deviation

[civic coral]

the problem is that the blue underlined equation doesn't looks equivalent to me

the (n-1) at the very bottom have made the equation became illegal

it doesnt equals to Sxy/Sxx

[civic coral]

im not agree to it cuz when i tryin to prove the equation it doesnt looks equivalent

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[civic coral]

<@&286206848099549185>

.reopen

[cedar kiln]

✅

[cedar kiln]

@civic coral Has your question been resolved?

[cedar kiln]

@civic coral Has your question been resolved?

[cedar kiln]

@civic coral Has your question been resolved?

[cedar kiln]

@civic coral Has your question been resolved?

[elfin hemlock]

What is the problem? It works fine

[elfin hemlock]

$\frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2}=\frac{s_{x,y}}{s_x^2}=r_{xy}\frac{s_y}{s_x}$

[wraith dagger]

ScapeProf

[cedar kiln]

@civic coral Has your question been resolved?

[warm bone]

Anyone can help me with Differential Equation?

[cedar kiln]

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[crimson sedge]

why is the answer neg 1?

ln 0 is undefined

why is this pi?

[uncut hedge]

Yes it is. The way to solve it is to let a be the lower bound of your integral, and compute the limit of your integral as a converges to 0+.

[crimson sedge]

wdym

[crimson sedge]

$\lim_{a \longrightarrow 0^+}a(ln(a))$

[uncut hedge]

In your resulting F you have an x ln x term, and you have to compute its value for 0. ln 0 is undefined, ln x and diverges to -inf as x goes to 0+. But does x ln x tends to -inf as x goes to 0+ as well?

[wraith dagger]

Pencil

[crimson sedge]

uhh

[crimson sedge]

Hint: Use L'Hôpital's rule

Ok, nope. I don't follow

ohhhhhhhhhh

ok yeah I remembered.

alright, thank you.

.close

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[crimson sedge]

.reopen

[cedar kiln]

✅

[crimson sedge]

@crimson sedgeWhat about the second one?

oh nvm

I got it

thanks

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[white bough]

About 'A and B are open sets in A∪B', is there any counter-examples？For some topologies?

[cedar kiln]

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[white bough]

No

[obsidian coral]

Don't ask in an occupied channel
Read #❓how-to-get-help

[cedar kiln]

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[dire geode]

I don't really understand this question. What's 'A ?

Are you trying to prove some claim? What is that claim?

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[late pike]

not sure how to get the answer

[dire geode]

there isn't even a question

[late pike]

oh the question is above the answer

if you want the whole questio its

[cedar kiln]

@late pike Has your question been resolved?

[late pike]

<@&286206848099549185>

[dire geode]

can you type this out. it's so poor quality

[late pike]

is that better. sorry i cant type it since idk how to w the proper syntex

[wraith dagger]

Ansh

[fallen heath]

Hey @late pike , I think you're just being too intimidated by the question. It's fairly simple tbh

[late pike]

how would you go about doing it

[fallen heath]

me? I'd say: the question's already at it's second step of solving itself

[late pike]

I need to know how to factor the first question to the second and I should be able to do it from there

[fallen heath]

Indeed, you're right on that

It's the third step of the solution

so? did you make an attempt yourself?

[late pike]

I have the solution just cant understand it

[fallen heath]

$(t\sqrt{x-1})^2 - (x\sqrt{t-1})^2$

[wraith dagger]

Ansh

[fallen heath]

hmmm... how about, since there is a square root in each parenthesis and a square outside each parenthesis, we square both terms within their parentheses?

I think we're capable of doing that-

[late pike]

if that was the case how would you factor t^2(x-1)-x^2(t-1)

since you cant factor (x-1) or (t-1) since they are not the same

[fallen heath]

how about... since we see parenthesis still in play here, we can just go about multiplying?

[late pike]

wouldn't we be back at square 1

[fallen heath]

You think so?

How about .. you write it down instead of thinking?

See what you get

is it same as square 1?

[late pike]

oh I meant the result was the same as the question not the value

[fallen heath]

Ik what you meant.

Did you try getting rid of the parenthesis? by multiplying inside out-

[late pike]

yea you mean like 2t^2 -x^2 +x +2trootx-1 -2xroott-1

[fallen heath]

Oh, you got a square root after multiplying?

[fallen heath]

from here, I thought we would go a step ahead and try not to move backwards

$t^2(x-1) - x^2(t-1)$

[wraith dagger]

Ansh

[late pike]

oh I see what you mean

I thought you were talking about the one before mb mb

[fallen heath]

$= xt^2 - t^2 - x^2t + x^2$

[wraith dagger]

Ansh

[late pike]

yea, that should be it

[fallen heath]

$= (xt^2 - x^2t) - (t^2 - x^2)$

[wraith dagger]

Ansh

[fallen heath]

$=xt(t-x) - (t+x)(t-x)$

[wraith dagger]

Ansh

[fallen heath]

$= (t-x)(xt - (t+x))$

[wraith dagger]

Ansh

[fallen heath]

$= (t-x)(xt - t - x)$

[wraith dagger]

Ansh

[fallen heath]

which gives you your factorisation

[late pike]

yes whohoo

[fallen heath]

but tbh, I feel like you could've put more effort on this ^^

[late pike]

yea I kept trying to factour out t^2 with t^2

[fallen heath]

consider reviewing some basic algebraic factorisation... and don't be intimidated by the question

[late pike]

I should have tried the otehr wway

[fallen heath]

the trick to factorisation is:

if t = x seems to make the whole expression zero, you can be damn sure (t - x) is a factor

whether you use long division or synthetic division method, or a smarter method to divide the polynomial and take out factor (t - x) doesn't matter after then

[late pike]

yes, thanks for the help...I'll be sure to brush up on my skills

.close

[cedar kiln]

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[cedar kiln]

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[wanton sierra]

This is the question I’m struggling on

[vestal copper]

@wanton sierra do you have some approach tot this question?

for the cuboid

[wanton sierra]

I think so yes

[vestal copper]

how would you start to approach it?

remember that your end goal is to find the angle between AG and ABCD

[wanton sierra]

Which angel is that

Is it G

[vestal copper]

angle A

[wanton sierra]

Oh

[vestal copper]

are you okay with reading geometric notation?

[wanton sierra]

To find DG

10 and 15”8

15”8

15.8

Are the sides of the triangle at the bottom

[wanton sierra]

Not sure on that wording

So we do Pythagorean

Phythagros

[vestal copper]

like, do you know how a line can be described by its endpoints?

if I call the left endpoint of a line A and the right endpoint B, then we can write the line as AB

it's just a form of notation we use to reference geometrical objects

similarly, a triangle can be represented as ABC, where each A, B, and C are the corners of the triangle

[wanton sierra]

That’s good yea

I worked out the line

GD

[vestal copper]

which one

ok

wait hold on

[vestal copper]

whats this

inches?

oh, you meant 15.8 cm

[wanton sierra]

Yea

[vestal copper]

no no

you're working with cm

[wanton sierra]

I meant the .

like you said

[vestal copper]

ok

[wanton sierra]

Mind if I send my working

[vestal copper]

go ahead

[wanton sierra]

Well this don’t the working

Isn’t*

But

[vestal copper]

could you label the corners correctly?

[wanton sierra]

The missing one is AG

[vestal copper]

so GD = 18.7 cm?

[wanton sierra]

Rounded

But I have the decimals

Written

[vestal copper]

ok

[wanton sierra]

I’m not sure what to do next

Or do I use trig

[vestal copper]

labeling the triangle will help u

[wanton sierra]

To work out the angle

I’ll try work out the angle I’ll see what you think

[vestal copper]

this is the angle you're trying to find, GAC

[wanton sierra]

Oh

I just worked out the wrong one

[vestal copper]

you will have to at some point

[wanton sierra]

So all I’ve done was pointless?

[vestal copper]

no

okay

it seems you need some outline on how to work this problem

have you practiced finding angles of a right triangle using just its side lengths?

[wanton sierra]

Isn’t that trig

[vestal copper]

(i.e. solving a right triangle)

yes

[wanton sierra]

Let me send you what I did

I think it’s wrong

Angle GAD

[vestal copper]

that's not the right angle

but it's very similar to what you have to do

[wanton sierra]

Ah

[vestal copper]

just focus on your goal

angle GAC

what do you need to find this angle?

[wanton sierra]

It’s a 3D triangle

I have no clue

[vestal copper]

no

no triangle is "3d"

all triangles live in two dimensions, if you look at them from the right perspective

do you see what i did?

[wanton sierra]

Oh

I don’t understand why I’m working out that angle was my problem

[vestal copper]

it's something very similar to what you did here, using trig

[wanton sierra]

To work out AG

[vestal copper]

mm I guess the question assumes you know about projections of lines into planes

[wanton sierra]

I can use trig with this?

For AG

*pythagros

[vestal copper]

yes

the pythagorean theorem

[wanton sierra]

And then I’ll try work out AC

[vestal copper]

yea

you definitely know how to solve these problems

you simply need a push in the right direction

[wanton sierra]

I see

I’ll try get the answer

[vestal copper]

in how to visualize the triangle whose angle they ask to find

[wanton sierra]

How do we know there is a right angle?

[vestal copper]

because the planes BCGF and ABCD are perpendicular

[wanton sierra]

Ohhh

Ok so

[vestal copper]

the planes of a a cuboid are all perpendicular

so any two lines that you can draw in any of those two planes will be perpendicular also

[wanton sierra]

Ok so

That’s my triangle so far

Do I do

20.5 squared - 10 squared

And then square root

[vestal copper]

why

you can find the angle directly using trig

[wanton sierra]

Oh yea

[vestal copper]

what trig function would you use?

[wanton sierra]

Sin

[vestal copper]

yes

[wanton sierra]

I got 29* as my answer

[vestal copper]

,w arcsin(10/sqrt(8.5^2+18.7^2))*180/pi

[wanton sierra]

This next question looks very hard, but you’ve already helped a lot so do you want me to try ask someone else?

My calculation there was

But I used the full decimal version of 20,5

[wraith dagger]

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[vestal copper]

yea that's right

[wanton sierra]

Yay

[vestal copper]

sorry I dont use this bot too often

I need to practice my inquires

[wanton sierra]

It’s fine

Does ^ mean squared

[vestal copper]

it means 'to the power of'

,w 3^3

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by duchat#3932

[wanton sierra]

Oh I see

Ok should I try work out the other one

I habe a choice

Have

It’s either this one

Or the one above

Whichever is easier I can do

Or the orange triangle

[vestal copper]

oops

both are equally easy

the 2nd one requires a bit of a trick

I'm sure you can figure out the 1st one on your own

it's simply pythagoras and trig

I want you to find the angle on that 2nd diagram they're asking for

[wanton sierra]

Which question?

[vestal copper]

like, not numerically, just tell me what angle is it on the sketch

[vestal copper]

^

[wanton sierra]

AEB?

[vestal copper]

no

right now there's no way to represent that angle with letters

you need to draw another point on that figure, call it X

and then you may write the angle

it'd asking for the angle between the base and the slant face

[wanton sierra]

I would try draw it on but I don’t have anything to

[vestal copper]

but can you visualize it more or less?

yes or no?

[wanton sierra]

There’s no letters for the triangle?

[vestal copper]

no

you need to make another letter yourself

[wanton sierra]

I’ll try think

[vestal copper]

ok

[wanton sierra]

I’m rly not sure

Can we try the other one

@vestal copper

I’m rly stuck

And I need to get it right

[vestal copper]

okay @wanton sierra

what do you see when you look at that piramid directly from its face?

like, what would you see of your eyes were leveled with the base of the pyramid and you were staring directly at one of the faces?

what geometrical figure

[cedar kiln]

@wanton sierra Has your question been resolved?

[wanton sierra]

The question refreshed

Nvm I did it

Thanks for your help

[cedar kiln]

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[modern compass]

what have you tried?

[frank nest]

@modern compass sorry for bothering, been here an hour but helpers ignore or dont come. is it possible for you to help mine after this persons question? its just a derivative one. thanks

[crimson sedge]

@frank nest #❓how-to-get-help

[frank nest]

i did for an hour no one helps..

once agains sorry for disrupting but after him, could you?

[gentle shard]

sorry i just don't know how to process the question

[crimson sedge]

This is a great example to work backwords

write out everything

[barren rose]

I made an equation for it it's long af

@gentle shard Ok so

[gentle shard]

yes

[barren rose]

Let's start off with one side of the equation

So we have 3 events of decrease

so there's going to be 3 subtractions of the original x

To get the reamining

And the word problem says 45 is what was remaining after the 3 subtractions

So one side of the equation can be 45

And then we want to subtract from the total(x) 3 times based on what it tells us

So our equation should follow x=x-(s1)-(s2)-(s3)

s is for a subtraction event

So fill in each of those using the information for each event given in the word problem

@gentle shard Understand?

[gentle shard]

will try and let you know

one sec

so will it be x=45-(1/7)-(1/5)-(3*1-1/5)??

@barren rose

[barren rose]

No