help-10

[obtuse pebble]

@cerulean heart Has your question been resolved?

[cerulean heart]

got it finally

.close

[obtuse pebble]

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[stable hinge]

Hello

[fierce lagoon]

Post your question

[fierce lagoon]

Use your log identity

$$\log_{a^b}{(c^d)} = \frac{d}{b}\log_a{c}$$

So what's another way to represent $25$?

[warm shale]

Umbraleviathan

[stable hinge]

what?

[fierce lagoon]

Use this rule

If you have $log_{25}{(5)}$. You wanna find a way to make this nice. You can start by rewriting $25$.

[warm shale]

Umbraleviathan

[fierce lagoon]

There @stable hinge

[warm shale]

Umbraleviathan

[fierce lagoon]

Okay or I guess you can ask and then like dip

[stable hinge]

my bad i have no idea what to do

[fierce lagoon]

$$\log_{25}{(5)}$$
$$\=\log_{5^2}{(5)}$$

[warm shale]

Umbraleviathan

[fierce lagoon]

Use this after that @stable hinge

[obtuse pebble]

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[shadow estuary]

Help me pls

[shadow estuary]

how do i solve this

[pallid flame]

easy way: OP is radius so it makes a right angle

since OT is 11 radius is also 11

so OP is 11 and PR is 60 with OPR right angle

[brazen saddle]

You don't even have to use Pythagoras

Actually

No joke

[pallid flame]

pythag is the easiest since i don’t remember much else from geometry lmao

[shadow estuary]

wait so then what

[brazen saddle]

Since OR is the hypotenuse it's gonna be definitely greater than PR , and the only option that is greater than PR is option (a)

[pallid flame]

OP and PR are legs with right angle OPR

[brazen saddle]

Screw Pythagoras

[shadow estuary]

you just use pythagorean theorem? or u can also do it with basic logic and reasoning

i think i understand now, thx

[scarlet locust]

,w sqrt(3721)

[warm shale]

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[shadow estuary]

wait

let me try

[brazen saddle]

@scarlet locust

[scarlet locust]

yeah i got it

just checking

[brazen saddle]

Lol

[shadow estuary]

11^2 + 60^2 = OR^2

,121 + 3600

,w 121 + 3600

[warm shale]

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Requested by Tachanka#5115

[shadow estuary]

,w sqrt 3721

[warm shale]

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[shadow estuary]

Wow!

it worked

[brazen saddle]

Yeah no shit wow

[shadow estuary]

u a genius

[brazen saddle]

bruh

DuDE

[shadow estuary]

yall geniuses

[brazen saddle]

Thank yew

[obtuse pebble]

@shadow estuary Has your question been resolved?

[pallid grail]

hello

can someone help mw with my question plz

An exponential function with a base of 3 has been stretched vertically by a factor of 2.5 an then reflected in the y-axis. it's asymptote is the line y=-1. it's y-intercept is (0,1.5) you can use this as a "check". write an equation of the function and it's domain and range.

[obtuse pebble]

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[obtuse pebble]

@dreamy anvil Has your question been resolved?

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[untold trout]

is this series convergent?

[untold trout]

any idea on how one might approach it?

Well, at least I know it's conditionally convergent if there's a (-1)^n term before it,

but this approach won't work if it's not alternating

I tried the ratio test but it looks ugly

the root test returns 1 too

ah nvm i solved it

.close

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[patent grail]

Each altitude divides the triangle into 2 sections. Which altitude divides the triangle into 2 sections with the least difference? A triangle has points A(0,4), B(-3,-2), C(2,-4), Each point has an altitude.

[patent grail]

<@&286206848099549185> Clueless , but something related to the slope , right ?

[summer spire]

yea... i think so

[patent grail]

every altitude forms at a right angle , but how can we check the least difference ?

.close

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[timid silo]

hi

[timid silo]

can someone help me understand the limit of

V(0) is 0 btw

I don't understand where the 2alpha^2/3 comes ffrom

what I did was, cotax=cosax/sinax, then for x small sin ax=ax

and so 2/x^2-2/x^2 =0

and we're only left with the alpha squared

but apparently this is not the case

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> anyone available? should be quick enough

oops

we can separate this limit into several component using the additive property of limits, so let's ignore V(x) as x --> 0 since you've stated this is just 0

also we can neglect h^2/2m for now because it's a constant

so we have

$\lim_{x \to 0} \alpha^2 + \lim_{x \to 0} \frac{2\alpha \cot(\alpha x)}{x} - \lim_{x \to 0} \frac{2}{x^2}$, do you agree?

[warm shale]

Renegade

[timid silo]

now taking the limit of a constant is absurd, so we are simplify left with $\alpha^2$ for the first limit

[warm shale]

Renegade

[timid silo]

then I'd suggest you rewrite cot(ax) as cos(ax)/sin(ax) and attempt L'hopital's rule

hmmm alright, but can l'hopitals rule be used for limits that are not 0/0?

yes, for inf/inf

actually here I believe cot(ax)/x diverges

but here we don't have inf/inf

we have 1/xsinax

which is like 0^2

on the denominator

$\frac{2}{x^2}$ goes to $+\infty$

yeah

Agreed

[warm shale]

Renegade

[timid silo]

but cot(ax) doesn't exist

but it's cot(ax)/x

which is cos(ax)/xsin(x)

it doesn't diverge

bc the denominator acts as an even function

since both x and sinax change signs for negative x

by diverge I mean DNE

DNE?

have you learned taylor series?

yup

yeah so cos(x) taylor series is 1 - x^2/2! + x^4/4! and sin(x) is something like x - x^3/3! + x^5/5!

okay

so as the terms go to 0 the denominator is 0

yeah

dne means don't exist

it goes to infinity

you don't need the taylor expansion to see that no?

could the limits cancel each other out

arithmetic with infinity is pointless, so no

well it clearly has a solution

as written above

since this is the answer sheet to an exam, corrected by uni professors

<@&286206848099549185> sorry to ping again, anyone?

[finite vector]

So I think the problem is with $\lim_{x\to 0}\left[ \frac{2\alpha\cot{\alpha x}}{x}-\frac2{x^2} \right]$?

[warm shale]

newbienoob

[timid silo]

yes

[finite vector]

ok so first can u simplify the expression a bit by calling alpha.x as k?

[timid silo]

alright?

[finite vector]

You should end up at this: $\lim_{k\to 0}-2\alpha^2\left[ \frac{1-k\cot(k)}{k^2} \right]$

[warm shale]

newbienoob

[finite vector]

now multiply and divide by sin(k)

and the sin(k) in the denominator tends to k itself as k->0

then apply l'hopital's rule

@timid silo

[timid silo]

so we're left with 1/k^2 - k/k^3

it's basically the same problem we had above..

[finite vector]

did u split it up?

do it without splitting and apply l'hopital's rule

u end up at 1/3

for the term in the brackets

[timid silo]

you're right

my bad

ty!

.close

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[solid cove]

getting the slope is easy but how to get the y value ?

[solid cove]

and why is it zero

[slim leaf]

@solid cove After the slope, you need the point where the tangent meets the curve. And this point has x-coordinate sqrt(3).

So for the y-coordinate, you plug in x=sqrt(3) in y.

And you get y = int from sqrt(3) to sqrt(3) .....

which is just y = 0

[solid cove]

oh right

[solid cove]

yeah thx ❤️

.close

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[cyan quiver]

help

[cyan quiver]

need help with b

ik a is 240

[obtuse pebble]

@cyan quiver Has your question been resolved?

[obtuse pebble]

@cyan quiver Has your question been resolved?

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@cyan quiver Has your question been resolved?

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[timid silo]

f(x)=x²+4x+6
g(x)=f(x) (x<0)
|f(-x-k)-k| (x>=0)
h(k)=number of intersection points of g(x) & y=k/8
lim h(k) =/= lim h(k)
k->a- k->a+
find the sum of every a

[timid silo]

i quit cuz it had too many ks

<@&286206848099549185>

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@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[tough hawk]

Did I do this right?

[obtuse pebble]

@tough hawk Has your question been resolved?

[tough hawk]

<@&286206848099549185>

[obtuse pebble]

@tough hawk Has your question been resolved?

[tough hawk]

<@&286206848099549185>

[foggy marsh]

Looks about right, could you explain more about your procedure though?

[tough hawk]

My math teacher walked me through it and he was making no sense and I was so confused

So I don’t really know

[foggy marsh]

Well the standard way to fit data points to an equation is to use linear regression

In which we try to fit points to a straight line

Here we want to fit to an exponential, but that's kind of hard to do directly. Luckily, if we take the log of the equation, the exponential gets turned into a linear equation, which is easier to fit to

So like p(t) = p0*e^(kt) becomes log(p(t)) = log(p0) + kt

Which looks a lot like y=mx+c

Taking the log all so changes the data points nothing happens to x, but the y values get logged

So keep that in mind

Anyways, now we have some data points to fit to a line. We need to find the optimal m and c for the formula y=mx+c

There are straightforward formulas to find this, do you have to derive those as well? I'm not sure

[obtuse pebble]

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[toxic harbor]

So I'm going through this pre-calc book and there are some formula that I just don't like understand intuitively. Like I can follow the steps and I know which numbers go where for each variable in the formulas, but I don't understand how they are derived.

[toxic harbor]

Is this okay or am I expected to understand the derivations?

[grizzled shore]

What formulas

[toxic harbor]

Like this for example

[grizzled shore]

Some derivations require more advanced math, like say the equation of the volume of a sphere

[toxic harbor]

I don't understand why k is being multplied by t

[toxic harbor]

It's not really advanced. I just can't wrap my head around it

[grizzled shore]

Because every time period

All the money get interest

[toxic harbor]

yeah but that's what t is

[grizzled shore]

Well

Think of kt as how many times it’s compounding

If I ask in 5 years

And it compounds every 4 months

That’s 3 times a year and so it compounds 15 times

[toxic harbor]

oh I see

why is k being devided by r

oh because you want to find the compounded rate not the annual rate

What about here

a(1 + r)^x why is there a one there

I don't get it

Should you just do starting point * r?

and then raise r to the power of x so that you get exponential growth based on the x-value given

Like I just don't get what 1 is there for

[grizzled shore]

1 is so your number is like 1.3 or something

The r value is in percent

As in if u have a 10% growth

r would be 0.1

[toxic harbor]

yeah so why add 1 to it?

if it's 10%

[grizzled shore]

0.1ⁿ gets smaller as n gets bigger

[toxic harbor]

Right interesting

[grizzled shore]

Because you want your value to go from 100% to 110%

Not from 100% to 10%

[toxic harbor]

Oh since it'

s increasing

[grizzled shore]

Yea

[toxic harbor]

Okay I kinda understand that

If we want to have decaying functions

Would we just remove the 1

or subtract 1

[grizzled shore]

We subtract from 1

[toxic harbor]

How come?

[grizzled shore]

10% decrease is 1-0.1 = going from 100% to 90%

[toxic harbor]

hm

[grizzled shore]

When u decrease it again (increasing kt by 1) you once again multiply by 1-0.9

So your 90% becomes 81%

Again and your 81% turns into 72.9%

[toxic harbor]

why not 80 then 70

oh 20% of 90%

[grizzled shore]

Because you’re depleting 10% of the current value

Not 10% of the original value

[toxic harbor]

right right

[grizzled shore]

If a car depreciates 5% every year

Is the car worth $0 in 20 years?

[toxic harbor]

But like is understanding all of this like crucial or can I just know how to use the formulas without understanding why they are like that?

[grizzled shore]

No the parts are still worth a bit of money

[toxic harbor]

No

Makes sense

[grizzled shore]

U can but without full grasp of how the functions and equations work it can be hard to manipulate them to do what you want

[toxic harbor]

Hm I see

Idk what to do

I'm just like not understanding some thing

s

And I have no teacher to like explain

[grizzled shore]

It’s the difference between being able to explain it to someone and just saying plug the numbers in

[grizzled shore]

You have the internet

[toxic harbor]

Yeah'

idk

Okay thank you again

appreciate the help

[grizzled shore]

👍

[toxic harbor]

.close

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[scenic reef]

Hi I'm looking to create a program for boids (flock minded bots)

I am trying to make a push formula. a formula that if the boid is closer to me push it harder and if it further softer.
I tried to come up with something for quite a while now and have nothing.

any help would be appreciated.

[grizzled shore]

so you're looking for a function of "push strength" in terms of distance

something like 1/distance would give u something that pushes strong when it's close and softly when it's further away

[scenic reef]

yeah thats what i just came up with. i will see if i can implement it properly. also why would it be divided by 1?

sorry im kinda bad with math

oh nvm i get it now.

tysm

[grizzled shore]

use desmos and visualise the curve

[obtuse pebble]

@scenic reef Has your question been resolved?

[scenic reef]

is there a way for me to use multiplication instead? to avoid division by 0

.close

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[sturdy isle]

i thought it would be 1/5040 but that isn't right

any help appreciated

[shadow badge]

I also got 1/5040

[sturdy isle]

hmm maybe my prof put in the wrong answers or something

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[noble berry]

there no numbers??

[noble berry]

(nk•nf)/2 isnt right

[slim lake]

It wants it in terms of h

[noble berry]

wdym

(h•nf)/2?

this is what it's giving for the answer box thing

[slim lake]

Area of trapezoid = average of the top and bottom length × height

So h(NF+MD)/2

You need to simplify it all in terms of

The only piece of non trivial information is the fact that MF and ND is perpendicular

[noble berry]

oh alr, ty

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[timid silo]

I need help with this proof

[timid silo]

z^-1=_(z)/|z|^2

[high lily]

wdym by _(z)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

conjugate

[obtuse pebble]

@timid silo Has your question been resolved?

[warm shale]

Raiqua

[bright pumice]

z=a+ib

[warm shale]

Raiqua

[bright pumice]

Since we don't want any complex numbers in the denominator we multiply by the complex conjugate

so

[warm shale]

Raiqua

Raiqua

Raiqua

[timid silo]

is the first step not a-bi/z(re)?

[bright pumice]

Therefore

[hybrid gull]

Why do you need the real part?

[timid silo]

no clue

[bright pumice]

$z^{-1}=\frac{\overline{z}}{|z|^2}$

[warm shale]

Raiqua

[timid silo]

just need a proof but I dont get this step

[hybrid gull]

The proof is written there for ya

[timid silo]

I know ^^^^^

[bright pumice]

uh

On that step I substitute z with a+ib

[timid silo]

i know how to rationalize it

how is _z(conjugate)=1

[bright pumice]

It's not

[bright pumice]

No

[timid silo]

how did rs change

[bright pumice]

Look what I did

z^1=1/z

Which makes

[timid silo]

oh nvm i am straight braindead

[bright pumice]

Saying z=1 is like saying 2/4 = 4/8 therefore 2=4

don't forget the denominator

Did you get it?

[timid silo]

i understand now something just didnt click in the order of the pictures

all good

[bright pumice]

Oh that's probably because I didn't include z^1 sorry bout that

Alright 🙂

[timid silo]

[bright pumice]

Nw lol

[timid silo]

.close

[obtuse pebble]

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[lament oyster]

Hi, I have a problem that asks me to find the percentage of uncertainty of sin 𝞱 when 𝞱 = 15.0° ± 0.5°.
The answer they give is 3.25%, and they take ((sin 15.5 - sin 15) / sin 15) * 100%. But I think it should be ((sin 15.5 - sin 14.5) / sin 15) * 100%, since the uncertainty is ± 0.5°.
So my answer is 6.5%.
Any idea why they use sin 15.5 and sin 15, instead of sin 15.5 and sin 14.5?
Thank you.

[scarlet locust]

the uncertainty is measured from 15°, its supposed to tell how how much the answer can deviate from sin(15°), not how different are the 2 extremes values of the answer

[lament oyster]

Oh, so why they take 15.5 instead of 14.5?

[scarlet locust]

its the same

the difference from 15 is the same that is

[lament oyster]

You mean I could do (sin 14.5 - sin 15) / sin 15 * 100%?

[scarlet locust]

yes you could

you'll just get a negative answer

take the abs of that

[lament oyster]

Yes.

OH, now I got it.

Thank you very much. Like I've been on this problem for days, thank you once again!

[scarlet locust]

[lament oyster]

.close

[obtuse pebble]

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[obtuse pebble]

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[supple cobalt]

How would I find the ratio of DC to AB?

[tranquil arch]

https://www.youtube.com/watch?v=-dlCX72BdmE

[supple cobalt]

Yeah, but my teacher won't allow me to use that video

[tranquil arch]

do you know the method in the video?

[supple cobalt]

No

I have not seen the video yet

[tranquil arch]

do you have some ideas?

[supple cobalt]

My main idea is to draw a straight line starting at p and extending in both directions, therefore that could maybe make the triangles more manageable.

[tranquil arch]

you mean ⊥CD?

[supple cobalt]

Something like this.

But I don't really know where to go from there.

[tranquil arch]

yes, set two heights as h1 and h2

[supple cobalt]

mhm

[tranquil arch]

then you can express area of PCD, PAB, ABCD

actually this is the method in the video

[supple cobalt]

Ah, I see

ok

Thank you!

.close

[obtuse pebble]

Channel closed

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[pallid goblet]

Complete set of ‘x’ satisfying: |x + 2| > 6. I got ans A but its wrong.

[scarlet locust]

why would it be only positive values?

-10 + 2 is -8 and |-8| > 6

[pallid goblet]

wait the ans i got is not in option*

not A

[scarlet locust]

yeah it seems like the options are wrong

[pallid goblet]

(-∞,-8) U (4,∞)

[pallid goblet]

This is my ans

[scarlet locust]

this seems correct

[charred plume]

Answer C is correct, if you just turn the final 8 on its side to get $(4, \infty)$

[warm shale]

daveamayombo

[pallid goblet]

is it for real?

So options are wrong

[scarlet locust]

yes

[pallid goblet]

thx

i got one last question can i ask?

[tulip haven]

sure

[pallid goblet]

I dont understand the question does it mean (x² – 3)(x + 1)<0

[tulip haven]

exactly

it means values of x for which that equation is always negative, you will again use the same method, by the way is that physicswallah dpp or smth?

[pallid goblet]

Ya pw dpp

[tulip haven]

jee 22 or 23?

or 24 💀 ?

[pallid goblet]

ARJUNA 21

Jee 23

[tulip haven]

hmm

[tulip haven]

so do you need any more help?

[pallid goblet]

the ans i got is wrong again

[tulip haven]

same though xD, @scarlet locust u too ? 😳

[scarlet locust]

ye

[pallid goblet]

the roots of this are √3 and -1 right?

[tulip haven]

use a^2-b^2 formula and then that whatever method it is called

[tulip haven]

+- root 3 yes

[tulip haven]

which coaching 😳

[pallid goblet]

oh man my brain didnt thought that

[scarlet locust]

None

[pallid goblet]

those are in a-b a+b form

[pallid goblet]

I m not in coaching too

[tulip haven]

🙏

[tulip haven]

pw is like coaching I guess? 😳

[pallid goblet]

no

Pirated Lectures*

[scarlet locust]

💀

[pallid goblet]

ya but they are advanced i can't understand

pw is only 1 i can understand

[tulip haven]

hmm, Ic, all the best you two , gotta go and study

[pallid goblet]

btw ans?

[pallid goblet]

I m not getting d

[tulip haven]

ah, +-root 3 and -1 are roots so use that curve method and then see the negative region

• root 3, - root 3, -1, let me try once

[pallid goblet]

ohk

[pallid goblet]

i got all except -√3

[tulip haven]

x^2-3= (x- root3)(x+root3)

[pallid goblet]

what how?

i mean

wait

[tulip haven]

x^2-(root3)^2

plus options are wrong

[pallid goblet]

oh

bro i thought

[tulip haven]

answer should have been (-infinity, -root3) union (-1, root 3)

[pallid goblet]

x²-3 = 0
x² = 3
x = √3

[tulip haven]

when u remove square it becomes +- root 3

[pallid goblet]

oh

[tulip haven]

that is class 9 or 10 basic, maybe study those once

[pallid goblet]

my basics aren't strong

[tulip haven]

just see some videos of fundamental of maths class 11

[pallid goblet]

Oh ohk

can i dm u?

[tulip haven]

not accepting dms 😓

[pallid goblet]

Ohk no prob

.close

[obtuse pebble]

Channel closed

Closed by @pallid goblet

Use .reopen if this was a mistake.

[tulip haven]

have a great day!

[obtuse pebble]

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Send your question here to claim the channel.

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[sour wagon]

Calculate the distance from a point to a segment (Point is P, segment is AB). The following is given P(3,4) A(1,0) B(12,0)

[sour wagon]

I used the following equation x-xA/xB-xA = y-yA/yB-yA but I must do something wrong

[timid silo]

hmmm

ig u meant

shortest distance

also AB is the segment? and P is the point?

[sour wagon]

yup

ah sorry yeah the distance from point P to AB

[timid silo]

ok so

first get the slope of the line segment

passing through A , B

[sour wagon]

ok, but there is not like only one equation to do it, I think it might be the way I translate the problem

I don't remember having to do any slopes for it

[timid silo]

actually nvm there is a problem with the general method coz its a line segment

[sour wagon]

x-xA/xB-xA = y-yA/yB-yA so is there another formula besides this one ?

[timid silo]

ig this is a lin alg problem?

[sour wagon]

yeah honestly idk but I think I am getting to the bottom of it

I will try to do it with another formula and see if it's good and post the answer here to verify

[sour wagon]

yeah actually what happens is the following

I should be able to have aX-bY+c=0

but what happens is ,w (x-1/12-1)=(y-0)/0-0

,w (x-1/12-1)=(y-0)/0-0

[warm shale]

Results provided by WolframAlpha

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

Requested by Talos#7370

[sour wagon]

,w (x-1/12-1)=(y-0/0-0)

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Talos#7370

[sour wagon]

,w (x-1/12-1)=(y-0/0-0)

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Talos#7370

[sour wagon]

.close

[obtuse pebble]

Channel closed

Closed by @sour wagon

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[obtuse pebble]

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[kind hawk]

what do you know about (1-1)^n

[fallow meadow]

?

[kind hawk]

use the binomial theorem on (1-1)^n

a=1, b=-1

[obtuse pebble]

@fallow meadow Has your question been resolved?

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[tall vessel]

A ship sails a distance of 120 km on a bearing of 44. How far north or south will it be from its starting point? how do i figure it out

[obtuse pebble]

@tall vessel Has your question been resolved?

[obtuse pebble]

@tall vessel Has your question been resolved?

[sick pelican]

if im not mistaken should be smth like this, just use cos for this one

if you tryna find how far north or south

[tall vessel]

yes, its on a 44 degree angle so N44E i am using triganomtry and calculus to figure out the answer

[distant schooner]

you definitely don't need to use calculus. Just trigonometry

[tall vessel]

oh ok thanks

[distant schooner]

Do you know how to start the question?

[tall vessel]

eh trying to do it now

[sick pelican]

draw a triangle

[tall vessel]

ye

[sick pelican]

and write the angles and length of the side

[tall vessel]

with pythaguras theorum

[sick pelican]

huh

[distant schooner]

no you don't need that

[sick pelican]

sin cos tan lol

[tall vessel]

oh i dont

[distant schooner]

have you learnt trig functions

[sick pelican]

you only have 1 side, how u use pyth thm

[tall vessel]

no i self taught my self

[distant schooner]

ok then learn that first

[sick pelican]

you need to learn sin cos tan first :d

[tall vessel]

im in yr 7 so i havnt learnt this this is yr 12 work

[distant schooner]

this is yr 7 lmao

not yr 12

well depends on where you are

[tall vessel]

australia

[sick pelican]

think thats yr 8 ish for me but ye similar enough

[distant schooner]

welp same

[tall vessel]

the corriculems are different'

[distant schooner]

no its all the same

[sick pelican]

whatever

[tall vessel]

no in australia pythaguras is yr 9 and someone said they learnt it in yt 4

[distant schooner]

Don't try to skip too much, this is gonna be too hard

[sick pelican]

anyways just try to understand how sin cos tan works (btw the angle theta can be either the angle, and the opposite and adjacent side will switch)

[distant schooner]

well how about focusing on this

[tall vessel]

ik

[distant schooner]

ok wait lemme give u a question

[sick pelican]

thats just impossible lmao

[distant schooner]

name the side names

like

hypotenuse adjacent

[sick pelican]

but sure can be used as example

[tall vessel]

my question isnt a right angle

[distant schooner]

yeah but u needa understand this

to be able to do what ur doing right now

[tall vessel]

theres hypotonuse, opposite and adjacent

[sick pelican]

uh it has one but sure lets focus on knowing tf is sin cos tan

[distant schooner]

yeah which one tho

[tall vessel]

ok

[distant schooner]

is 1 adjacent?

[tall vessel]

idk

wait

i think 1 is opposite

[distant schooner]

nope

[tall vessel]

is 1 adjacent

[distant schooner]

memorise this

hypotenuse is the longest side

[tall vessel]

ok

[distant schooner]

opposite is the opposite to the angle

and adjacent is just pretty much left over

[tall vessel]

ok

[sick pelican]

the angle isnt always at the lowest one tho, it can appear at the other position

[tall vessel]

ye

[distant schooner]

ok so basicaly

going back to your question

you said it doesn't have a right angle but it does

if it doesnt, you can't solve your question

[tall vessel]

ye

[distant schooner]

ok so you're trying to find distance for north

[tall vessel]

ye

[distant schooner]

so which way is north

x or y

[tall vessel]

x is going up north

[distant schooner]

yep so it's x right?

so is x hypotenuse, adjacent or opposite

[tall vessel]

x is adjacent cuz opposite to anggle

[distant schooner]

yes good

so what trig function are you going to use to find x?

sin, cos or tan

wait do you know how trig functions work and how to use them

[tall vessel]

i am not sure

[distant schooner]

ok they are commonly used to find angles

so wait

so now you know all the angles

i mean all the sides

[tall vessel]

ok

[distant schooner]

you have 120 as hypotenuse, and the x you want to find as adjacent

so what function do you want to use?

[tall vessel]

coz?

cos

[sick pelican]

[distant schooner]

ok and then what?

[tall vessel]

so if adjacent is cos

im not sure...

[distant schooner]

ok so

it says

C = A/H

yes?

[tall vessel]

yes

[sick pelican]

you sin/cos/tan the angle and that shit is equal to the ratio of 2 sides (depends on what it is)

[distant schooner]

also that symbol with a line going across 0 is theta

its commonly used as unknown angle

[tall vessel]

so how do i use that to the question

[distant schooner]

so if you plug that in, it should be cos(44) = x/120

you're just putting the information you have into the functions

do you get that bit?

[tall vessel]

i get half

[distant schooner]

whats the other half you don't get

[tall vessel]

the theta thing

[sick pelican]

θ is the angle

[tall vessel]

ok

ohhh

[sick pelican]

you sin cos tan an angle to get a ratio

[tall vessel]

so 0 is the theta

[sick pelican]

well yes that symbol θ = theta

[tall vessel]

what would the ratio be?

[distant schooner]

ratio?

[sick pelican]

u use the angle

[sick pelican]

yes lmao

trigonometry ratio

or function if you want to call it like that

[distant schooner]

look at the information that was given to you

[tall vessel]

ok

[tall vessel]

how do i sin/coz/tan

[sick pelican]

so basically lets say an angle is 20

sin(20) is a value

[tall vessel]

ye

[sick pelican]

and the opposite/hypotenuse is always = sin 20

when the angle is 20

[tall vessel]

ye

[sick pelican]

and now you have 44

you know what is adjacent, opposite and hypotenuse

use the corresponding trigonometry ratio

[tall vessel]

yes

[sick pelican]

if you know cos θ = adjacent/hypotenuse

you know what is θ and hypotenuse

[tall vessel]

0 is theta?

[sick pelican]

eh its the symbol

[distant schooner]

a symbol for angle yes

[sick pelican]

not "0" necessarily but sure you can let the symbol look like that for now

[tall vessel]

alr

[sick pelican]

now basically the triangle shit we showed u rotated 180

and have some values in it instead of all unknowns

[tall vessel]

ok

soooo

what we have so far is

x = ad

y = op

[stark tiger]

hi everyone i am new and i have doubt
can anyone please solve it ?

[tall vessel]

wrong chat sorry\

[stark tiger]

why ?

it says maths help

[tall vessel]

this chat is about helping me do triganomotry

[stark tiger]

oh ok

sorry

[sick pelican]

ye open a new one

[stark tiger]

ok

[sick pelican]

yes but u dont need y

[tall vessel]

i am a little confused about 1 thing you said @sick pelican

[tall vessel]

this

[sick pelican]

sin cos tan yes

they all have different ratios

like sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse

they arent the same thing

[tall vessel]

ok

[sick pelican]

since you are finding adjacent and u have hypotenuse, you are using cos

[tall vessel]

isnt adjacent x

[sick pelican]

yes you are trying to find x

you are trying to find how north

[sick pelican]

y is east lol

[tall vessel]

how north what

how far out in north

[sick pelican]

too lazy to type all that crap out

[tall vessel]

ok

[distant schooner]

ye ik

[sick pelican]

if you are finding how far out is north ofc you be finding x

[sick pelican]

yes, was using the picture as reference, always forgetting to turn off mention

[distant schooner]

dw dw

[tall vessel]

so using this info how do i find how far out

[distant schooner]

ok so you've determined its cos

right?

[tall vessel]

yes

because cos is ad/hy

[distant schooner]

just substitute everything over from that information to cos(theta) = a/h

[tall vessel]

what would that look like

so if a/h = cos

[distant schooner]

a = x, h = 120, cos(theta) = cos(44)

do you get that?

[tall vessel]

where is h

is h the h

oh

hypotonuse

ye i get that

so what do i do now

[distant schooner]

ok so it looks like this

yes?

[tall vessel]

yes

[distant schooner]

ok so we need to isolate (x), so bring 120 to the left side

that makes

[tall vessel]

ye

[distant schooner]

then just put that in your calclator

[tall vessel]

ok

do i put it ina trig calc

[distant schooner]

yeah?

make sure its in degree mode

[tall vessel]

would a normal one work

[distant schooner]

liek what

[sick pelican]

if you have a calculator with sin cos tan yes

[tall vessel]

ok i never used a trig calc

[distant schooner]

just use online

[sick pelican]

ye

[distant schooner]

https://www.desmos.com/scientific

[tall vessel]

is this automatically in degree mode

[distant schooner]

yes

[tall vessel]

ok so 86

.32

thank you so much

im just gona ss this chat so i can keep it

[distant schooner]

@sick pelican r u good with factoring

[sick pelican]

yes

[distant schooner]

yea i need some help

[sick pelican]

unless its some extreme shit that i never saw

[sick pelican]

ah great

[tall vessel]

is there a way to save h=this chat for future reference

[sick pelican]

who the fuck factorizes shit in surds

[distant schooner]

jsut go to youtube

[distant schooner]

ig australia does lmao

[tall vessel]

true i have a good memory so hopefully iy pays off

[distant schooner]

i tried to do completing the square, doesn't quite work to get me those answers

[sick pelican]

imma just use the quadratic formula to calculate what's x

[distant schooner]

dude

they ban us from using

quadratic formula

LMAo

[sick pelican]

ok

[distant schooner]

so idk how tf else to do this

[sick pelican]

tbh you can just be patient and break down all these shit

[distant schooner]

please be my guest

not sure how to start this off

[sick pelican]

this shit gonna take a bit of time

[distant schooner]

so.. how do you start it off

[sick pelican]

think it's D

imma check rq

[tall vessel]

i wana say thank you for helping me

[distant schooner]

but how.

[distant schooner]

yeah np

[tall vessel]

can you guys addd me if you want

[distant schooner]

sry dont add randos

[tall vessel]

ok alg

[sick pelican]

yep im right

so i simply broke down every brackets for what is x if that bracket is 0

[tall vessel]

is it alg if icome to you in the future for questions like this

[sick pelican]

for instance, (x+3-√7)

[distant schooner]

yes

[distant schooner]

*just create another ticket and someone will reach you

[sick pelican]

x would be -3+√7

[distant schooner]

yep

[sick pelican]

and i used sum of root

sum of root = -(-6/1) = 6

so i pretty much broke down 10 brackets

but some of them are the same so i think just 4 combination

then i add their root to see if they match

i think its from A to E for the sequence

if the 2 stuff add up = 6, thats the answer

and the 4th combination = 6

which is 3+√7 and 3-√7

[distant schooner]

hol up lemme try to get this

[sick pelican]

basically u see 2 brackets on each option, i find the value of x in each of them when the bracket = 0

and use the sum of root method to find which one suits the requirement

[distant schooner]

ok so the sum of root is 6, how does the other stuff meet that

[sick pelican]

nothing else meets it

only 1 option does

[distant schooner]

[sick pelican]

e.g. you add -3+√7 and -3-√7 you get -6

[tall vessel]

ik your talking about something else but i have one last question, to figure out if its sin cos or tan is it if the either h, a or o is outside the theta like if the theta is in the middle of a and h its cos?

[sick pelican]

you are adding the stuff together to see if the sum of roots or product of root matches the equation

[distant schooner]

oh shit is ee

i see

[sick pelican]

don't get it much

[distant schooner]

but about this part

[sick pelican]

yes

[distant schooner]

how did you branch them

i get that you're trying to get them to = 0

[sick pelican]

oh i simply broke down the x's

and write the combinations

such as A is -3+√7 and -3-√7

B is -3+√7 and 3-√7 and so on

till E

the -> just shows where it links to

[tall vessel]

ok ill ask a question to see the answer then,
how do i figure out to use ad, hy or op, to find out if its sin, coz or tan

[sick pelican]

depends on what the question is asking and what info you need

[distant schooner]

im still not quite understanding, maybe ill ask my teacher

i do have another q. about algebra(my knowledge to know when to use it is trash)

[tall vessel]

ok thanks

[sick pelican]

kk

[tall vessel]

how do i figure the thing out without using a scientific calc?

[distant schooner]

almost impossible

[sick pelican]

pretty much impossible unless its special angles like 30, 45, 60, 90 180 360

[tall vessel]

alright thank you

[distant schooner]

so with this, i try to get rid of the surd

[tall vessel]

whats this now?

[sick pelican]

so you know √ = ^1/2?

[distant schooner]

uh no

[sick pelican]

well now you know

[distant schooner]

?

[sick pelican]

without the -. - was a typo

[distant schooner]

yeah i know that

law

[sick pelican]

let the giant shit inside be a

[distant schooner]

giant shit?

[sick pelican]

$\sqrt(a^{-2})$

the big thing inside

not going to type that shit out

so i just let it as a

[distant schooner]

so

that becomes

[sick pelican]

💀 bruh

[warm shale]

𝓡𝓸𝓼𝓮

[sick pelican]

then square root becomes 1/2

[distant schooner]

lmao

what does it look like after

[sick pelican]

then you multiply -2 and 1/2

and it becomes a^-1

[distant schooner]

yeah i remember that

but then i also dont remember like half of it

[sick pelican]

and a^-1 = 1/a

so basically you make the fraction upside down

[distant schooner]

yes i get that

[sick pelican]

and a = the big thing you have inside

which is 2x^-4/y^-8

[distant schooner]

oh

yeah?

[sick pelican]

then you swap stuffs

[distant schooner]

so

2y^8/x^4

[sick pelican]

and swap it again

because ^-1

[distant schooner]

x^4/2y^8

[sick pelican]

si

[distant schooner]

oh shittt

thanks man

[sick pelican]

np

[distant schooner]

oh yeah another thing

how did he make 8/4 into 6/3

[sick pelican]

8/4 = 2
6/3 = 2

the values didnt change, he just changed the numbers

same ratio

[distant schooner]

oh u can do that?

[sick pelican]

yes, since the power didnt change

i suppose he used it to common the denominator with 5/3

for easier calculation

[distant schooner]

mm

cause i just multiplied 3 and 4 to both sides but welp okay

[sick pelican]

[distant schooner]

alr last one

[sick pelican]

y = mx+c

[sick pelican]

yes

[distant schooner]

is what ive written

[sick pelican]

that

[distant schooner]

but that isnt on any of the..

answer

[sick pelican]

switch subjects

[sick pelican]

try to switch the subjects to y for every option to see which matches or whatever method that can allow u to get the same thing

[distant schooner]

still isnt any of the answers

[sick pelican]

why is 5*4=4

[distant schooner]

oh is that how it works

wait so do i just make the 4 at the back 20?

[sick pelican]

theres another method, since y-intercept = 4, that means (0,4) lies on the linear graph

hence you can sub x = 0 and see if y = 4 in that equation

[sick pelican]

well didnt u multiply 5 for both sides to make 5y?

[distant schooner]

oh

to both sides?

ok well anothe thing learnt

[sick pelican]

wdym

you went from y = -3x/5 + 4

[distant schooner]

i usually just put things from other side to another

[sick pelican]

into 5y = -3x + 4

so you literally multiplied 5

but you didnt multiply 5 to 4

[distant schooner]

[sick pelican]

[distant schooner]

thanks for carrying my ass

goodbye sir.

.close

,close

how.

[sick pelican]

close

idk

oh i know

we arent the owner of this channel

@tall vessel is

he needs to close it

we cant

[distant schooner]

[tall vessel]

oh sorrry

.close

[obtuse pebble]

Channel closed

Closed by @tall vessel

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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[timid silo]

Heyy! Can anyone please help me in a question??
It says:
Find the complex number “z=1+i” that fulfills the condition Z •(2+i)=1

[obsidian isle]

that makes no sense

you are already given the value of z

what is there to solve for

[timid silo]

IDKJJ

Waitrt

Ill send you the pic

Its in albanian

Its the 4th one

What do these mean? Lmao

So is the correct answer B. Z=1+i

[frosty river]

(1+i)(2+i)=2+i+2i+i²=1+3i≠1

$\frac{1}{2+i}=\frac{2-i}{4-i²}=\frac{2}{5}-\frac{1}{5}i$

[warm shale]

jnkena

[obsidian isle]

it makes no sense because it's not a true statement

[frosty river]

I think the question is wrong haha

[timid silo]

Thats education for you in Kosovo 😍

Anyways ill just leave this and head on to something else

[obsidian isle]

🤣

[timid silo]

In the rectangle ABCD, the side AD is
symmetrical with the rib BC, in relation to the axis
Ox. Calculate the perimeter of the rectangle ABCD

Its the question number 20

By the way thank you SOOO much for helping me may god bless you

Help!

[brazen saddle]

Is it a country or no?