help-10

[fathom tangle]

i get it now

[obtuse pebble]

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[mighty sequoia]

I'm really confused about number b - how would you get x =/= 2? i get x =/= 1, but i have no idea where the 2 came from

[timid silo]

Because x = 2 isn't in the domain of f(x)

[mighty sequoia]

well yeah, but how do you figure that out from the problem?

[timid silo]

Because the denominator of f(x) is x-2

Therefore x != 2

[mighty sequoia]

OHHH

oh wait yeah, don't domain restrictions carry over from the equations you're dividing? just wanna know for future reference

[timid silo]

Yes

You got it

[mighty sequoia]

thanks a bunch!

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[timid silo]

Hello
What rules are there for the mod sign?

[timid silo]

as in, congruence?

for example, a is congruent to b mod n

a-b equals n ( times x)

7 is congruent to 27 mod 5
27 - 7 equals 20, 5 times 4 equals 20

so it works

are there any other rules like that?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

Hello, I can't help you with the problem because I'm not super familiar with Modulo math, but I can link you to a nice pdf that might be what you're looking for

https://people.eecs.berkeley.edu/~oholtz/H90/congruence.pdf

[timid silo]

Ah unfortunate, but thank you nevertheless! :D

No problem!

[obtuse pebble]

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@signal cobalt Has your question been resolved?

[signal cobalt]

<@&286206848099549185>

[lyric pebble]

Since unit vectors have magnitude of 1, you can find a unit vector by first finding the magnitude of those vectors (<2,-5> and <1,-2,4>), and then just divide each component by its magnitude.

[signal cobalt]

oh

so?

ans = (2sqrt(29)/29,-5sqrt(29)/29)?

[obtuse pebble]

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[sand root]

Real quick does anyone know what this topic is called? My instructor just kinda jumped into it. Want to know what to look up after to find more info on it. Something about triangles inside of a circle with reference angles 😅

[obtuse pebble]

@sand root Has your question been resolved?

[fickle shell]

you should look up unit circle ig

[obtuse pebble]

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[sand root]

Ty will look into it

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[toxic elm]

can someone help me with this question please?

[obtuse pebble]

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[fringe dome]

Hello I need help with this question

[fringe dome]

1(e)

My working is this but the answer is supposed to be 1 or -2/3 I can’t figure out where went wrong

[obtuse pebble]

@fringe dome Has your question been resolved?

[fringe dome]

<@&286206848099549185>

I checked from photomath but i can’t understand

Step 4 and 5 how do i suddenly get the “2x-3x”

.close

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[clever condor]

um

<@&286206848099549185> ?

https://tenor.com/view/bigsmile-monkey-smiling-gif-22687322

@zinc stump xD?

@ me if anyone answers

its my hw xD

[spiral maple]

Dont auto ping helpers

[clever condor]

sorry ig

[spiral maple]

Or random users.

But anyway.... what have you tried?

[clever condor]

coolio

[spiral maple]

Ok, wont help you then

[clever condor]

no i just dont understand how to do it

wow

bruh

ffs

[zinc stump]

For, say $b$, plug in $x = -1$ in the formula, and compute $y$

[warm shale]

polikuj2

[obtuse pebble]

@clever condor Has your question been resolved?

[naive fossil]

You just take the values in the left most column and plug into x and see what it equals
Ex:
For -3 : y = 2-5(-3) <- replaced x with the left most column value
y = 2 - (-15)
y = 2 + 15
y = 17
a = 17

[obtuse pebble]

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[mortal jolt]

Hi, so im am doing derivatives but first I must clear a parenthysis and im having trouble with it. It is: ( x + 2 )^2

[mortal jolt]

I cant remember how to this, because i know i cant apply the distribuitive.

[undone forge]

Well that's a hard one

[oak locust]

$$y=(x+2)^2$$
$$y=x^2+4x+4$$

[naive fossil]

You're...trying to find the derivative?

[warm shale]

bakasta

[mortal jolt]

Well first i got to clear the parenthesis

[naive fossil]

Not if you use the power rule.

[mortal jolt]

what is it ?

[undone forge]

I saw this question before

but I don't remember how to solve it

[naive fossil]

You can use either method - power question or use bakasta's breakdown and take the derivative.

*Power rule not power question - brain fart

[mortal jolt]

Yeah ok, i think i got it.

Thanks so much it really helped and i have a exam tomorrow.

.close

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[naive fossil]

Good luck!

[obtuse pebble]

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[timid silo]

can someone explain how to do y = ln10^x

[naive fossil]

Are you solving for x? Taking the derivative? What?

[timid silo]

taking the derivative

[hybrid gull]

@timid silo what have you tried?

And is the function $y = \ln\left(10^x\right)$

[warm shale]

Shen

[timid silo]

yeah, so at first, I got 10^x(10)/10^x

but i thought that it wasn't right

[unique solstice]

You can simplify it a bit and make it easier on yourself

[naive fossil]

Use those ln rules for exponents, so true

[timid silo]

<@&286206848099549185>

[naive fossil]

Use those hints - try simplifying ln(10^x) using ln rules for exponents.

[timid silo]

i already did that

it's really 10^x part that I do not understand

because it's inside of the ln()

[naive fossil]

y = x ln 10
Product rule

[timid silo]

ln(10) ?

[obtuse pebble]

@timid silo Has your question been resolved?

[shut trellis]

Yeah that would be the derivative

[timid silo]

thanks

.close

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[molten hazel]

.I miss class again and don't understand this question

[unique solstice]

Do you know soh cah toa

[molten hazel]

Yes

[unique solstice]

Is x opposite or adjacent

To the angle z

[molten hazel]

Adj

[zinc flax]

is it sin cos and tan in some other language ??

[unique solstice]

Its an acronym

[zinc flax]

ohk

[unique solstice]

SOH = Sin opp/hyp

CAH = Cos adj/hyp

[zinc flax]

oh k

[unique solstice]

Etc

[unique solstice]

So plug in the proper stuff to generate a formula

[molten hazel]

Is it sin= x/28

[obtuse pebble]

@molten hazel Has your question been resolved?

[unique solstice]

Sin of what? @molten hazel

Sorry I took a really long time

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[woeful talon]

How would I do this?

[woeful talon]

With a calculator

Teacher said we can use calculators

easy???

[quaint sequoia]

integral of 1/x = ln |x|

[woeful talon]

K

I know this goes against the point but what's the answer?

@quaint sequoia

[quaint sequoia]

nvm you do use u sub
u = x-1 du = 1
then plug in 3 and 2 for x into the u function to make your new bounds

[woeful talon]

k

[quaint sequoia]

then you should have this $\\int_{(2)-1}^{(3)-1} \frac{1}{u}du$

[warm shale]

Dogecode

[quaint sequoia]

NOW you can simply use the int 1/x = ln|x| identity to solve it from that form
sorry for the confusion

[woeful talon]

Okay

So what to do now?

[quaint sequoia]

$\int \frac{1}{x}dx = \ln(\abs{x})$

[warm shale]

Dogecode

[quaint sequoia]

then just plug in the bounds using the fundamental theorem of calculus

you know what that is right

[woeful talon]

K i'll be honest I don't know any of this shit and haven't been taught any of this

But, i just kinda need the answer

Not for school

it's cause a friend is being annoying in the slightest

and shoved this up my face

LOL

[quaint sequoia]

$\int_a^b f'(x) = f(b)-f(a)$

[warm shale]

Dogecode

[woeful talon]

wait

so what now

[quaint sequoia]

plugging in the bounds gets ln(2) - ln(1)
ln 1 is zero so the answer is just ln(2)

[woeful talon]

yo tysm

[quaint sequoia]

np

[obtuse pebble]

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[distant cave]

Fredrik runs and uses an app to measure how far he runs over time. The app has registered data according to the task graph, where the x-axis is the time in minutes from start and the y-axis is how many meters he has run. What is Fredrik's average speed during the last three minutes of the run?

[timid silo]

Rise/run of the start and end

Just a line between the end of the graph and the origin

And calc the slope of that

[plain owl]

*average speed during the last three minutes

[obtuse pebble]

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[distant cave]

Wait

[distant cave]

Fredrik runs and uses an app to measure how far he runs over time. The app has registered data according to the task graph, where the x-axis is the time in minutes from start and the y-axis is how many meters he has run. What is Fredrik's average speed during the last three minutes of the run?

[distant cave]

I don't know what that means

Why do we take from x = 4 to x = 7

[high lily]

those are the last 3 minutes of his run

[distant cave]

Wouldn't it be x=5 til x = 7?

5, 6, 7

[high lily]

no

the time from the 5th minute to the 6th minute is 1 minute
similarly the time from the 6th minute tot he 7th minute is 1 minute
for total of 2 minutes from the 5th minute to the 7th minute

[distant cave]

i dont know how to draw a line with geogebra tbh

any advice?

do i just do $m=\frac{1200-800}{7-4}$?

[warm shale]

Theophania

[lunar plank]

i think so

[warm shale]

Theophania

[distant cave]

Ok let's hope it's accurate

[lunar plank]

133.33 m/min

[distant cave]

In a physics experiment, a tomato was thrown straight into the air. The students created a model for a function where f (x) is the number of meters above the ground after x seconds. The function they developed is $f\left(x\right)=1.5+4x-x^2.$
\newline
\newline
What will be the highest height of the tomato?

[warm shale]

Theophania

[lunar plank]

i think can be view from the graph ?

[distant cave]

Isn't it just f'(x) = 0?

4-2x = 0
-2x = -4
x = 2

Hmm yeah...

[lunar plank]

yea something like this

but im not sure whether im correct or wrong

[distant cave]

.close

[obtuse pebble]

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[fervent valley]

Can someone help me, how do I find the volume bounded by y=ln(x),y=1,and x=5 revolved around x-axis using the Washer method.

[fervent valley]

This is what i have: $V= \pi \int_{e}^{5}{[(\ln x)^2 - (1)^2]}dx$

[warm shale]

MKG

[fervent valley]

but integrating (ln x)^2 will need By-parts i think, which I am not allowed to use for this problem

[timid silo]

my b

[obtuse pebble]

@fervent valley Has your question been resolved?

[high lily]

also your equation for volume isn't quite correct

[mild ocean]

is this even possible without integration by parts

yeah it’s possible without ibp

wait no it isn’t

there’s an alternate way to do it but it also requires ibp

[obtuse pebble]

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[sharp yacht]

If 52 men can do a piece of work in 27 days how long will 39 men take to complete the same work

[sharp yacht]

<@&286206848099549185>

[drifting wraith]

think of them having working speed

52 men in total have 1/27 work/day of speed

find speed of one man

by dividing

etc.

[obtuse pebble]

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[obtuse pebble]

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[haughty shore]

Could someone help me with L’opital Rule question

[jolly ginkgo]

Hint : lim e^(f(x)) = e^(lim f(x))

[haughty shore]

Yeah I did that

Also I think you forgot there is an ln when you take it to power of e

But I don’t really get how that gives us the proper answer

[jolly ginkgo]

No I didn't

[hybrid gull]

If you want to leave the original limit unchanged, then this would be the approach you take

$e^{\lim_{x\to0} \ln\left(x^{\sin x}\right)$

[warm shale]

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

[jolly ginkgo]

Use \displaystyle

[hybrid gull]

Never used that

[jolly ginkgo]

$e^{\displaystyle\lim_{x\to 0} \ln\left(x^{\sin x}\right)}$

[hybrid gull]

Regardless, you get the idea - that's what it's supposed to look like

Idk why the limit thing isn't working

[warm shale]

it's Sam

[haughty shore]

So yeah that’s what I had but I wasn’t sure what to do after
I took the sin(x) out and then changed it to csc and moved it to bottom to make it in quotient form

[hybrid gull]

And then you can apply L'Hospital's Rule

$e^\left({\displaystyle\lim_{x\to 0} \ln\left(x^{\sin x}\right)}\right)$

[warm shale]

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

[haughty shore]

But it doesn’t really help it gives
1/xcsc(x)cot(x)

Negative of that

[hybrid gull]

What if you tried to clear the embedded complex fraction

$\frac{1}{x\csc{x}\cot{x}} = \frac{\sin^2{x}}{x\cos{x}}$

[haughty shore]

Meaning moving sin and tan up

[warm shale]

Shen

[hybrid gull]

ye

[haughty shore]

But that’s fells like what was the purpose of moving them down in first place so that’s why I was struggling to understand but if thats how you do it then thanks

[hybrid gull]

You're "moving them down" to apply l'hopital's - and then you "move them back up" to clear the complex fraction

[haughty shore]

I see ok let me finish the solve

[haughty shore]

This also gives 0/0 so then we take it again and then we get 0/1 which means final answer is 1?

[hybrid gull]

You don't necessarily have do LHopital's again

You can split the limit into (sin x)/x • tan x = 1 * 0 = 0

[haughty shore]

oh makes sense thanks

.close

[obtuse pebble]

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[pine glen]

How would I answer this question?

[unique solstice]

Well if x is less than 1 the function is x+2

And if it's greater than or equal to 1 it's 5

That should be enough to solve it

[dim patio]

Why you didn't answer my reply...

[pine glen]

I was taking a shower

And didn't see it

[dim patio]

Do you know which one of the two diagram is f(x)=5?

[pine glen]

uhhh

A?

[dim patio]

I mean

Ahh

F(x)= numbers always parallel with horizontal

Got i

It?

[pine glen]

Hmmm

[dim patio]

Then it is a or b

[pine glen]

would it be b?

[dim patio]

Just give two value of x( <1) and draw the f(x)=x+2 diagram then you will get the answer.

[pine glen]

Ok gotcha

[dim patio]

Wait

Oh nothing it is right.

[obtuse pebble]

@pine glen Has your question been resolved?

[obtuse pebble]

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[maiden glade]

I've set up a triangle with angle bisector AD

to divide angle A in half

then we have BD = ac/(b+c) and CD = ab/(b+c)

and I'm stuck

[zinc flax]

I've another method idk this method sorry

if u need that I can tell u that

[maiden glade]

sure

[zinc flax]

area of the triangle is 1/2 cbsinA ?

[maiden glade]

yes

[zinc flax]

write area in terms of heron's formula sqrt(s(s-a).........

shit shit leave this

u hv cosine rule right ?? @maiden glade

[maiden glade]

law of cosines?

[zinc flax]

ya

[maiden glade]

yea

[zinc flax]

write that cosA as 1-2sin^2(A/2)

find sin(A/2)

got it ? @maiden glade

[maiden glade]

let me try it

thank you I solved it now

@zinc flax

[zinc flax]

np

any doubts abt that u can dm too

[maiden glade]

ok

.close

[obtuse pebble]

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[pale tusk]

hey there

[pale tusk]

i need crypto math

help

i mean that is maybe for a professor

[obtuse pebble]

@pale tusk Has your question been resolved?

[zinc flax]

sorry to ask but isn't crypto math kind of a code lang ?

encrypt and decrypt type

[pale tusk]

yes

i need that sort of help

[obtuse pebble]

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[pale tusk]

.reopen

[obtuse pebble]

✅

[zinc flax]

oh k
we can make code language in many ways right ?

you want a coded one ?

[pale tusk]

no wtf

this is not how a lang is made

i want to decode smth

[zinc flax]

there should be a decryption key to decode right ?

there should be a key to make a code language right ??

[pale tusk]

idk about the lang

but i dont have to make a new lang xD

i have to decrypt smth

and get the data

like if we have

dafjnfnajufdubddnsjk

i have to decrypt

and

get the data

from it

[zinc flax]

do u have the key ?

[pale tusk]

wait a second

https://medium.com/@alessandro_mazzeo/how-the-covid-19-digital-certificate-works-eb0523b8c1ac

check this

then this

https://github.com/jojo2234/Human-Method

[zinc flax]

do uk the algorithm ?

[pale tusk]

here is

[zinc flax]

ya checking

[pale tusk]

but everything is in java

and i am nodejs dev

but thats my friend project

so i can ask him things

[zinc flax]

it is beyond my knowledge sorry I'm just in class 12
I'll ask my friends and let uk if possible

so u need the key for decrypting ??

[pale tusk]

yes

i have to get the private key

[pale tusk]

thanks a lot bro

[zinc flax]

np np

I'm sorry infact

[runic obsidian]

what do you need to decode and how was it encrypted?

if you don't know how it was encrypted, what is it for?

[zinc flax]

it is there in the articles which he/she sent

[pale tusk]

he 😄

[pale tusk]

its private key

as zachkaupp said

u can check the links above

thanks

[runic obsidian]

so you are confident that the links above relate to your problem?

[pale tusk]

this explains in detail this https://github.com/jojo2234/Human-Method

[runic obsidian]

I don't think that answers my question...

[pale tusk]

yes? my problem is in this code:

https://github.com/jojo2234/Human-Method

just read the articles

[obtuse pebble]

@pale tusk Has your question been resolved?

[obtuse pebble]

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[timid silo]

so i had the equation (1/((x-3)^2)) + (4/(x(x-3))) - (2/(x-3)) = 0
and someone helped me to factor out x-3 and get to this, now i am not sure how to go about this

[timid silo]

like what do i do now to solve it

[pale tusk]

@runic obsidian the ticket was closed

now what

will u help me pls?

[timid silo]

please go to an other channel

#❓how-to-get-help

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

no

lol

[timid silo]

..

[timid silo]

@royal locust

[royal locust]

Ok

[timid silo]

yo

[royal locust]

So

[timid silo]

ye

[royal locust]

We know that the $\frac{1}{x-3}$ can’t be 0

[warm shale]

SilverTongue

[timid silo]

wait

[royal locust]

?

[timid silo]

doesnt it have to be 0?

1/x-3 (everything else) = 0

[royal locust]

Well 1/x-3 can’t possibly be 0

[timid silo]

but like

the whole thing = 0

and the only way for that

is that something * 0 is happening

[royal locust]

Oh

Ok so

Let’s say I have like

x(x-4) = 0

[timid silo]

ye

[royal locust]

What values of x make this true

[timid silo]

0

or 4

[royal locust]

Ok yeah

So let’s say we have

$\frac{x-4}{x}$

[warm shale]

SilverTongue

[royal locust]

What values of x make this 0

[timid silo]

hm

0

and 4

i think

[royal locust]

No just 4

0 would make this undefined

[timid silo]

oh yeah

[royal locust]

Remember you can’t divide by 0

[timid silo]

i confused the two

[royal locust]

Which is why this is true

[timid silo]

i see

x cant be 0 bc u dividing with x

[royal locust]

Exactly

So

[timid silo]

alright

wait wait

it means that

x cant be 3

if its 0 its fine bc its -3

[royal locust]

Yes

That’s right

Well

For that specific term

[timid silo]

ye

[royal locust]

Ok so

Now

Because the 1/x-3 can’t be 0

The only part that can be 0 is the right side

[timid silo]

alright

[royal locust]

So

How can you solve it for 0

The right part

[timid silo]

i mean

i was thinking to just solve it like a normal equation

[royal locust]

Yep

[timid silo]

and multiply the bottom part by the rest of the brackets

so do that?

[royal locust]

Wait wym

[timid silo]

like

multiply this by the bottom part of the fractions

like a normal equation

so x + 4(x-3) - 2(x-3)x

wait no

then i end up with like x^3

i dont understand what to do

[royal locust]

Ok

Sorry so

You have

Is equal to 0 right

[timid silo]

yes

wait so if im not wrong

= 2?

[mild ocean]

yes

because you added 2 to both sides

[hushed star]

how do i do question 15?

[mild ocean]

go to a different unoccupied channel and i’ll help

[royal locust]

Yes so now what can you do

Also sorry if I’m slow to reply I’m driving home rn

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[final lichen]

Alright so I'm supposed to find A^(16)th.. Honestly not sure what I can do here

[final lichen]

I tried it out up until A^(4)

and the 6's stay consistent, they just keep being multiplied by 6

but the 1's and the 0 on the top right are not consistent

[true rain]

have you tried decomposing the matrix?

[final lichen]

perhaps row reduce to like
1, 1/6,0
0,1,1/6
0,0,1

[final lichen]

you mean row reduce?

oh nvm

[true rain]

Like LU

er

[final lichen]

alright i see, how would i decompose this then?

[true rain]

er wait

let me think about this

[final lichen]

alrighty

[true rain]

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

so you get $A=Q\Lambda Q^{-1}$

[warm shale]

KurtDee

[true rain]

$A^2=Q\Lambda Q^{-1}Q\Lambda Q^{-1} = Q\Lambda^2 Q^{-1}$

[warm shale]

KurtDee

[final lichen]

mhmm

[true rain]

$A^{n+1}=Q\Lambda^n Q^{-1}Q\Lambda Q^{-1} = Q\Lambda^{n+1} Q^{-1}$

[warm shale]

KurtDee

[true rain]

boom proved

[final lichen]

whats Q and lambda in this situation?

[true rain]

that sounds like your homework? :^)

[final lichen]

frick

well thank you Kurt

[true rain]

that should give you the answer pretty easily

since lambda is a diagonal matrix

only 2 "difficult" matrix multiplications at the nd

[final lichen]

can i ask another question or would i need another channel?

[true rain]

uh sure

why not

[final lichen]

this one is easier, so im supposed to find a linear transformation for the vector u being mapped on vxu

where x is cross product

but how do i map a vector on another vector?

projection?

[true rain]

You can use a matrix :^)

[final lichen]

yeah but..

how

[true rain]

okay

$u \in \mathbb R ^n,, v \in \mathbb R ^m$

[warm shale]

KurtDee

[true rain]

$A \in \mathbb R^{n,m},, u \cdot A \in \mathbb R^m$

[warm shale]

KurtDee

[true rain]

make sense?

[final lichen]

ohhh

[true rain]

The exact dynamics depend on your problem

[final lichen]

i think so

[true rain]

of what "A" is

[final lichen]

so i write the matrix for the cross product

and then multiple u by that matrix

[true rain]

are u and v the same size?

[final lichen]

yeah

[true rain]

So you're really just asking for a matrix A such that

$u\cdot A \cdot v = u \cross v$ ?

[warm shale]

KurtDee

[true rain]

or just u dot A = u cross v

can you post the problem?

[final lichen]

Determine the matrix A for the linear mapping T:R^3-> R^ 3 which is defined by the vector u first being mapped at v cross u where v = (- 5,2,5)

[true rain]

yeah so it's $u\cdot A = v \cross u$

[warm shale]

KurtDee

[final lichen]

where A is the cross product right?

[true rain]

A is the linear transformation you're solving for

[final lichen]

ohh

how would I find A then lol

[true rain]

one step at a time baby

That's the generic equation of the cross product

plugin in $v$ since it's given

[warm shale]

KurtDee

[true rain]

and then solve for A

[final lichen]

ohhhhh

that makes so much sense

alright alright, i promise this is the last one, this convo is rlly helping me grasp lin alg better, why is it that a mirror on x=z axis gives us a matrix which looks like
(0,0,1
0,1,0
1,0,0)

[true rain]

I mean like lets think of this intuitively

You dont even need to think about this graphically

so we're flipping x and z, y remains the same

That's what a matrix really is

so lets refer back to the identity matrix

okay well, what does this really mean

in terms of vector matrix multiplication

you're really turning each row into a new value at that row

using the components as an index

so like row 1 only uses column 1 to influence

and etc

Basically it's saying "x is mapped to x, y is mapped to y, z is mapped to z"

and if you have like $2*I_3$ it's basically saying $x is mapped to 2x etc$

[warm shale]

KurtDee

[true rain]

so the sentiment here is to map z to x when you reflect against the z=x

literally x=z

so you map x to z, y to y, z to x

which gives you that matrix

[final lichen]

mhmmmm

wow

man

honestly ive been feeling kinda lost in our last few lectures, you just helped deviate away from that like a crap ton

thank you @true rain

[true rain]

yeah no problem

I fundamentally believe math isn't taught right

[final lichen]

bro

i live in Sweden and its sad that we do no theoretical stuff

its all computational

euclidean math

[lofty quiver]

Go watch some three blues one brown

[true rain]

It's not even the content, math teachign is based off a lot of assumptions that knowing the math is understanding the theory

[final lichen]

already seen the series

[true rain]

anything else

[final lichen]

ok i promise

last thing

has been bugging me for a while cause i cant find any content on it

write the vector AB as a linear combination of AC and AD

[meager raptor]

Should I live ,

[true rain]

should you

????

[final lichen]

yes please

[meager raptor]

Guys I'm serious

[true rain]

this is not the appropriate place to discuss this

is this a perfect hexagon?

can you tell me about AD

[final lichen]

yeah

[final lichen]

well its a straight line

[true rain]

AC-AD = CD, invert to get BA

if AD is a straight line then

Really it depends on if it's a perfect hexagon or not

[final lichen]

it is

[true rain]

then yeah

AC-AD= CD invert x to get BA

or you could uhh

cut AD in half

that gives you an equal triangle

[final lichen]

so -AC+AD?

cause its inverted?

[true rain]

By invert x I mean like

$CD \cdot \begin{bmatrix}-1 & 0 \ 0 & 1 \end{bmatrix}

[warm shale]

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

[true rain]

But if you cut the vector AD in half

that's the same thing as BC

So you could do AC - AD/2

[final lichen]

ok i found the same thing

i think i fucked it up on the other one then

i wrote AE is -AC+AD

[true rain]

if it's a perfect hexagon there are only like 3 unique vectors

you should be able to solve for them given two of them

and then any line is a linear combination

so I would just take that method

[final lichen]

shouldnt AB and AE be the same then

[true rain]

inverted on y

[final lichen]

so -AC-AD/2

[true rain]

is what

[final lichen]

AE

[true rain]

You can't just like

negative vector

it inverts on both axis

not just one

[final lichen]

im sorry but they havent taught us this matrix method for finding linear combinations

[true rain]

But adding a reflection definitely seems like BM and not the point on the exercise

actually it's not even reflected on y

it would depend on the orientation of the hexagon to beginwith

so it's definitely BM

you could just do a 60\deg rotation anyways

[final lichen]

man i get what ur saying

but the answer sheet is literally

only coordinate vectors

[true rain]

Show me?

[timid silo]

meth

[true rain]

I mean like yeah you could do it like that

I wouldn't even worry about it then

[final lichen]

what you mean

[true rain]

Just solve the question

[final lichen]

thats the thing idk how to 💀

and i keep trying to find examples of this but theyre all different from mine

ok tell me at least AF is correct right?

cause AC+AD make the same vector as AF

[true rain]

That is not true

[final lichen]

why not

[true rain]

I mean like look at it

that would give you a vector outside of the hexagon

[final lichen]

AC+AD give me CD no?

[true rain]

No

[final lichen]

ohh

ok i think im realizing

my sillyness now

i think ill write a point at the center as AD/2

that should make it easier to find

[true rain]

It should

AD/2 is a good vector

[final lichen]

ur a good vector

fr tho thank you kurt

that was like everything i had

[true rain]

you're welcome

[final lichen]

have a good day

[true rain]

I'll try

[final lichen]

.close

[obtuse pebble]

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[final lichen]

YOU BETTAH

[obtuse pebble]

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[thick torrent]

Halp! I'm trying to do this integral by parts but it's just getting worse and worse. Both functions tan^-1(t) and tanh(t) are difficult to work with and have complicated antiderivatives...

[mild ocean]

@thick torrent you don’t actually do the integrals in these problems

d/dx F(x)

probably not

[thick torrent]

Thanks, I used the fundamental theorem of calculus.

[obtuse pebble]

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[obtuse pebble]

@celest gorge Has your question been resolved?

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[fervent axle]

How would I differentiate e^(10x+1)

[fervent axle]

Very cool new system

Although I guess it's tough to run without enough helpers

[forest sinew]

youll need the chain rule

and the power rule, i guess

$f(x)=10x+1 \ g(y) = e^y \ g(f(x)) = e^{10x+1}$

[warm shale]

jan Niku

[forest sinew]

so $g(f(x))' = f'(x)\cdot g'(f(x))$

[warm shale]

jan Niku

[forest sinew]

@fervent axle is that enough information to sort though your problem?

all the pieces should be here, youll just need to assemble them

[obtuse pebble]

@fervent axle Has your question been resolved?

[obtuse pebble]

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[fervent axle]

ty

.close

o.

[obtuse pebble]

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[celest gorge]

https://cdn.discordapp.com/attachments/486841085210132490/911406119874867210/unknown.png

[celest gorge]

can someone help me find the two posssible tangents from point (20,2) to the circle

[obtuse pebble]

@celest gorge Has your question been resolved?

[celest gorge]

.close

[obtuse pebble]

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[livid dust]

Pls help I'm so damn confused

[livid dust]

The other ones looked like this

Yea pretty much

OHHHH

THAT makes so much sense

Thank you! And yes, that works

.close

[obtuse pebble]

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[warm shale]

fall

[obtuse pebble]

@long roost Has your question been resolved?

[obtuse pebble]

@long roost Has your question been resolved?

[obtuse pebble]

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[junior inlet]

you know eigenvectors?

[strong vale]

That topic appears later in the chapter. I don't think that's method they want used to solve this problem

[junior inlet]

uh
i don't know how else to solve it...
what else have you learned?

[junior inlet]

uh
i don't know how to solve it just using those...

[strong vale]

is there a parametric form for equation of plane?

then general parametric can be subbed in and equated to itself

would that work?

[junior inlet]

yeah i think so

(why haven't i figured this out-i'm dumb)

[spiral maple]

Quick to note that the vector [1,0,0]^T maps to itself

[obtuse pebble]

@strong vale Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[small rain]

$$\text{Could anyone help me out with this problem?}$$
$$\text{ABC is a right - angled triangle at A, O is the midpoint of BC}$$
$$AH \bot BC = H; HP \bot AB = P$$
$$\text{Let J be the circumcenter of triangle BPC, prove that AH = 2OJ}$$

[warm shale]

ind1v

[obtuse pebble]

@small rain Has your question been resolved?

[obtuse pebble]

@small rain Has your question been resolved?

[obtuse pebble]

@small rain Has your question been resolved?

[jolly ginkgo]

Multiply by (x,y,z) and equate to (x,y,z)

[obtuse pebble]

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[obtuse pebble]

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[stone parrot]

Simplify. Express using positive exponents.

6^5 over 6^2

A. 6^10
B. 6^7
C. 6^3
D. 6^-3

[stone parrot]

Um, i dont really remember how to find this simplified

BTW, this is homework

<@&286206848099549185> ?

[hollow jungle]

6^5 / 6^2

[stone parrot]

hey

[hollow jungle]

so when u devide exponents

[stone parrot]

divide exponents?

[hollow jungle]

you either subtract or add onto the exponent

[stone parrot]

oh ok

[hollow jungle]

so

thing of it like this

*think

x^n / x^y = x^(n-y)

[stone parrot]

ohhh

yeah i get it now thanks

lemme answer it and check it

[hollow jungle]

k

[stone parrot]

yup thanks

[hollow jungle]

np

[stone parrot]

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[obtuse pebble]

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[silent imp]

why did he make it 2.1 it has to be less then 2 right?

x is approaching 2 ???

[obtuse pebble]

@silent imp Has your question been resolved?

[zinc flax]

they can approach from the other side too right ?

--> 2 <--

[silent imp]

i guess?

[zinc flax]

that y they put 2.1

u need answer for that limit ??

[silent imp]

i’m assuming when they put lim there asking for the limit as x approaches that number? not actually x though ?

[zinc flax]

the ans for the 1st line is 4

[silent imp]

yes ik but i’m trying to say is that wording correct

what i just wrote above

[zinc flax]

oh kk

[silent imp]

is it??

[zinc flax]

we usually take some values which are way closer like 2.0001

[silent imp]

ok i see

[zinc flax]

so that higher terms can be neglected

[obtuse pebble]

@silent imp Has your question been resolved?

[runic obsidian]

what they are trying to do is plug in an x value very close to 2 (which they can't plug in because it would divide by 0) to find what plugging in 2 should give them. the number needs to be very close to 2, like 2.1, to give a good approximation (it will not be exact because it is meant to be an approximation)

[high lily]

considering limit properties and factorisation is usually better

[obtuse pebble]

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[willow needle]

how do i answer this

[high lily]

consider formulae for areas of basic shapes and area addition/subtraction

[obtuse pebble]

@willow needle Has your question been resolved?

[willow needle]

i tried but i couldnt figure it out

[plush dew]

What did you try?

[obtuse pebble]

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[marsh fable]

i dont get whats going on here

[marsh fable]

the inductive step

mainly

[topaz bridge]

$2^k<k! (inductive hypothesis)\
2^k2<2k!\
2^{k+1}<2*k!\
2^{k+1}<(k+1)*k! (because 2<k+1, so inequality is still true)\
2^{k+1}<(k+1)!$

[topaz bridge]

i don't know how to make space in latex, but i think you should understand

[warm shale]

tomzizek

[marsh fable]

i dont get the 2nd step

[topaz bridge]

multiplying both sides by 2

[marsh fable]

why

why dont you just substitue k+1 into n

[topaz bridge]

$2^k*2<(k+1)*k!$

[warm shale]

tomzizek

[topaz bridge]

now what?

[marsh fable]

🤔 😓

[topaz bridge]

you can't change k+1 into 2, bc we can't be sure if this doesn't change anything

maybe there is some answer, but i can't find it

[marsh fable]

oh ok

thanks anyway

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[remote crypt]

How many 6-digit odd numbers can be formed from the digits 1,2,3,4,5 and 6 (with repetition) such that they are divisible by 3? I know that the sum of the digits should be divisible by 3 too but other than that I'm lost as to how to approach this problem

[runic obsidian]

are you allowed to write a program to brute force this? or does it need to be solved mathematically

[obtuse pebble]

@remote crypt Has your question been resolved?

[remote crypt]

Mathematically. It's for a discrete math class

[unique solstice]

I mean you could define 6 ways to express the numbers divisibility by 3 and solve a system of 6 equations

[clever bloom]

It looks like 6^5 to me.

[runic obsidian]

why? remember they all need to be divisible by 3...

[clever bloom]

Choose any combination for the first five digits, then the last one can be chosen to make the digit sum divisible by three and the number an odd number.

And the last digit must be unique.

For 12345. 1+2+3+4+5 = 15. Last digit has to be 1.

So 123451

Etc.

[runic obsidian]

@remote crypt is this what you were looking for?

[remote crypt]

Yeah i believe it is. Tyvm

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[minor lichen]

anyone recognise how to do this. it is to do with vecots

[obtuse pebble]

@minor lichen Has your question been resolved?

[quaint glen]

find the $x$ and $y$ coordinates(the $\textbf{i}$ and $\textbf{j}$ components in other words) of $\lambda \textbf{a}+ \mu \textbf{b}$ in terms of $\mu$ and $\lambda$ and equate them to the $x$ and $y$ coordinates of $5\textbf{i}+3\textbf{j}$ to get a system of equations you can solve to find $\mu$ and $\lambda$

[warm shale]

Sneaky

[obtuse pebble]

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[grave shadow]

Hello guys

[grave shadow]

I need some help

relate your age to one of your family members, form a polynomial equation that will represents your ages and solve. Supposed that the sum of your age with your father is 60 years. six years ago, your father's age was five times your age. What will be your age six years from now?

just give me the representation and the equation

[dim patio]

First equation: Y+F=60

Second equation: F-6=5.(Y-6)

[obtuse pebble]

@grave shadow Has your question been resolved?

[grave shadow]

What does F stands for?

[high lily]

F for Father's present age

[grave shadow]

I just happen to figure it out btw thanks

[obtuse pebble]

@grave shadow Has your question been resolved?

[grand zenith]

what's so interesting about factoring simple trinomials

can anyone help?

[obtuse pebble]

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[safe shadow]

How does Wolfram Alpha get complex y from $y=(x+1)\cdot(-2)^x$?

[warm shale]

wolex9

[safe shadow]

<@&286206848099549185>

<@&286206848099549185>

Oh, I see now

Rational exponents

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[sweet perch]

Example numbers: 150 rocks in a bag. 75 red, 75 blue. What is the probability that you get 5 red rocks if you take out 25 rocks in total. What's a formula for this?

[obtuse pebble]

@sweet perch Has your question been resolved?

[obtuse pebble]

@sweet perch Has your question been resolved?

[lyric pebble]

This sounds like a hypergeometric distribution problem (we're taking out the rocks without putting it back) and not binomial (replacing rocks per draw). The formula below can be used. n is the number of trials/draws, x is the number of successes in the trials/draws, N is the number of elements in the population, and r is the number of successes in the population.

$\frac{\binom{r,x}\binom{N-r}{n-x}}{\binom{N}{n}}$

$\frac{\binom{r}{x}\binom{N-r}{n-x}}{\binom{N}{n}}$

[warm shale]

SafeDolphin

[sweet perch]

Awesome mate. Will wrap my head around that and code it. Thanks 😄

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[vague stump]

Could somebody explain how we use Cos/sin in such a question?

[spiral maple]

right angle trig

[vague stump]

Could you draw it please

[spiral maple]

you have the drawing

just draw the vector components of tension tip to tail

[vague stump]

i still don't get what determines the use of em

[spiral maple]

cause... you need to decompose the vector

what

[vague stump]

mhm i don't it get dammit

[spiral maple]

Draw your tension vector
Make your rectangular component vectors
The top angle will be theta, and thus you can find the lengths of the component vectors

[obtuse pebble]

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[spiral maple]

$\lim_{x\to a}g(x^2)=-1\implies\lim_{x\to a}g(x)=-1$

I feel like it's false but I'm struggling to find a counterexample.

[warm shale]

Mosh

[spiral maple]

It's a T or F w/ justification question btw, so either justify its truth or prove it's falsehood

[jolly ginkgo]

g(x)= -x/4 , a=2

[spiral maple]

Yeah, that'd do it

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[jolly ginkgo]

👍

[obtuse pebble]

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[timid silo]

what would be the set W1∩W2 ? I need it to find a basis of W1∩W2. Please if anyone could help

[timid silo]

W are subspaces in a finite-dimensional v.s.

[paper quest]

y = -x and t = -z

then $W_1 = {(x,-x,z,-z); x,z \in \bR} \ = {(x,-x,0,0) + (0,0,z,-z); x,z \in \bR} \ = {x(1,-1,0,0)+z(0,0,1,-1); x,z \in \bR} \ = span{(1,-1,0,0),(0,0,1,-1)}$

[warm shale]

Salah

[paper quest]

@timid silo yooo

is that clear?

[timid silo]

this is the span of W_1. I was questioning what would be the set W1∩W2 in order to find the basis and then the dimension.

[paper quest]

this is the first step to find the intersection

[timid silo]

yeah I already found out the span of w1,w2 and their bases and their dimension

idk how to work from there on to W1∩W2

[paper quest]

well see if there is common vector in W1 and W2

if there is no one then the intersection is the 0element

[timid silo]

I do not find a common vector.. but in the case the intersection is 0 then the dimension formula of dim(W1+W2) would not work as it would be 2+2-0 = 4

[paper quest]

what you did there?

[timid silo]

for W_1 exactly what u did and for W_2 I used ( 0 0 1 0 ) as e3 and ( 0 0 0 1) as e4 (as my coordinate vectors) and find out that the basis is {( 0 0 1 0), ( 0 0 0 1)}. Also ofc dim(W1)=dim(W2)=2