help-0

[wicked spear]

@earnest panther

[earnest panther]

As CST correctly mentioned, similarity of triangles could help you here.
Are you familiar with it?
If so, which triangles do you think would be similar in this case?

[wicked spear]

ya im familiar with similar triangles

while taking d as the top

ABD AND CBD

@earnest panther

[earnest panther]

Correct, but I believe you have to rethink about the order of the vertices.
Think about it this way: What vertex in triangle CBD would have an angle equal to that of vertex A in triangle ABD?

[wicked spear]

c?

[earnest panther]

Not quite. Try labeling $A$ as $\alpha$ and computing the angles in $CBD$ using $\alpha$. You'll find out that $C$ is $90^{\circ} - \alpha$, not $\alpha$.

[ocean seal]

RoiKadmon

[wicked spear]

what

[earnest panther]

I'll try to draw it. Give me a minute.

[wicked spear]

thank you

[earnest panther]

Here, we denote $A$ as $\alpha$, and we try to fill in the other angles in the diagram. We find out that $C$ is $90^{\circ} - \alpha$, and not $\alpha$. We'd like to find some angle in triangle $CBD$ whose value is $\alpha$.

[ocean seal]

RoiKadmon

[wicked spear]

so how do u find the values

is it possible for u to vc by any chance

@earnest panther

[earnest panther]

You can do that by using the fact that the sum of the angles in a triangle is $180^{\circ}$. We used it in triangle $ADC$, for example.
Also, is the VC in the server?

[ocean seal]

RoiKadmon

[earnest panther]

(Either way, I think I should be able to.)

[wicked spear]

ill add u and we can do it on dms

[scarlet quarry]

hi

[lone heart]

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[primal galleon]

I have a question regarding the sum of the exterior angles of an n-sided polygon (n-gon). I understand that for a cyclic n-gon, the sum of its exterior angles is always 360 degrees. However, I'm not sure how to prove this for a general n-gon.

I came across a solution that involves drawing n triangles within the n-gon and using the fact that each triangle has an interior angle sum of 180 degrees to find the sum of all the interior angles in the n-gon. To do this, you can start by drawing an n-gon and connecting each vertex to all other vertices to form n triangles. However, I'm having trouble visualizing how to draw the n-gon.

The solution suggests starting with a circle and marking its center point, and then drawing lines from the center to the edges of the circle to form the n-gon. Can anyone provide a more detailed explanation or another approach to solving this problem?

[quasi vector]

To do this, you can start by drawing an n-gon and connecting each vertex to all other vertices to form n triangles.

wrong

you connect each vertex to the center

or any point within the polygon

it shouldn't matter

The solution suggests starting with a circle and marking its center point, and then drawing lines from the center to the edges of the circle to form the n-gon.

that only allows you to prove this fact for cyclic polygons, which not all polygons are

[primal galleon]

So how do we prove it for more than cyclic polygons?

[quasi vector]

you can just choose any point within the polygon

it shouldn't matter I believe

[primal galleon]

Okay so I drew a 4-gon with vertices labeled 1, 2, 3, and 4. To find the sum of the exterior angles, you would draw lines from each vertex to a point within the polygon, say point P.

Then, it shouldn't matter where point P is located within the polygon, as long as it's not on the boundary. We could even choose point P to be the center of the polygon, but that's not necessary. The important thing is to draw lines from each vertex to a common point within the polygon.

[quasi vector]

correct

[primal galleon]

Picture 1)
Picture 2)

[quasi vector]

how is that a quadrilateral

in the second picture

[primal galleon]

Sorry, in the second diagram, the quadrilateral is formed by the lines connecting vertices 1, 2, 3, and P. The lines from vertex 4 are not shown in the diagram, but they would connect vertex 4 to point P as well.

So imagine a line connecting point P and 4.

[quasi vector]

but 1234 don't even form a quadrilateral

three points cant be collinear

i mean

they can

but

that complicates things

[primal galleon]

So how would you draw

[quasi vector]

something like this

[inner path]

so to have only one exterior angle of a n-gone it's 180 minus the interior angle ?

[quasi vector]

correc

[inner path]

ok so the sum of all exterior angles of a n-gone are $180*n-x$ where x is the sum of all angles in a n-gone

[ocean seal]

phoestaclies

[inner path]

and you know that for a n-gone the sum of it's interior angles are $180*(n-2)$

[ocean seal]

phoestaclies

[inner path]

so we have our 360

[primal galleon]

Okay after that what do we do?

@quasi vectorYou there?

[quasi vector]

find the sum of angles in each triangle

[primal galleon]

Can you show me with a drawing please?

[quasi vector]

here: <ABP + <BPA + <PAB = 180

do the same for all 4 triangles that are there

and add all the equations

[primal galleon]

Okay thanks.

.close

[lone heart]

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[alpine sable]

hi, i have this equation $(1+k)(k+1)^2=0$

[ocean seal]

ryuzaki16

[alpine sable]

my book says it's equivalent to $(1+k)(1+k+1)=0$

[ocean seal]

ryuzaki16

[alpine sable]

but why?

[modern sedge]

you sure?

[lilac stratus]

The hell.

[modern sedge]

is there any more context?

[alpine sable]

i can write all the equation if you want, this is one of the passage

[modern sedge]

you can send photo/screenshot

[modern sedge]

for general k, thhis is false

[alpine sable]

the exercise's explanation is not in english tho

i have to find the value of k

this is the exercise

i have one solution of the equation

and with the substitution i have to find k

this is the equation

$x^2 + kx + (k+1)^2 = 0$

[ocean seal]

ryuzaki16

[alpine sable]

i have one solution, given by the book, which is x = 1

so the next steps (in the book) are

$(1)^2 + k(1)+(k+1)^2 =0$
$1 + k+(k+1)^2 =0$

[ocean seal]

ryuzaki16

[alpine sable]

$(1+k)(k+1)^2=0$

[ocean seal]

ryuzaki16

[alpine sable]

$(1+k)(1+k+1)=0$

[ocean seal]

ryuzaki16

[alpine sable]

and finally $(1+k)(2+k)=0$

[ocean seal]

ryuzaki16

[alpine sable]

i don't know why they did this $(1+k)(1+k+1)=0$

i'm confused

[ocean seal]

ryuzaki16

[modern sedge]

the result is correct

[alpine sable]

i thought the power of 2 should be $k^2+2k+1$

[modern sedge]

yeah, by doing this you would arrive at the same result

I dont get what they did there

[ocean seal]

ryuzaki16

[alpine sable]

i'm glad that i'm not the only one

[modern sedge]

the only solution of this is k=-1

it MUST be incorrect

Oh isnt it actually (1+k)+(k+1)^2?

then it would make sense

then they just factored out 1+k

leaving (k+1)(1+(k+1))

[modern sedge]

yeah, must be this one

[alpine sable]

yes, this is what i wrote

[modern sedge]

this is multiplication

not addition

[alpine sable]

i don't understand. how they factored out (1+k) with the power?

[alpine sable]

damn you right, i wrote it in the wrong way really sorry

[modern sedge]

well so
(k+1)+(k+1)^2=0
(k+1)+(k+1)(k+1)=0
(k+1)(1+(k+1))=0

which is (k+1)(k+2)=0

[alpine sable]

so they factored out what? the factor or the first (k+1)?

[modern sedge]

for simplicity, let's replace k+1 with u for now

so u+u*u=0

now they factored out u

leaving u(1+u)=0

and then you can again substitute k+1 for u

[alpine sable]

yes it makes sense, god i have to spot this things automatically. i'll have a hard time when i'll take the test

thank you

.close

[lone heart]

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[chilly dove]

How do I solve a question like this?

[sudden hinge]

i'd first convert the mixed fraction into a normal fraction of the form a/b. then to go from % to fraction you simply divide by 100. example:
6 3/8 = 51/8, divide by 100 and you get 51/800

[vale wigeon]

do you know in general how to convert percentages to fractions

[chilly dove]

yes

[chilly dove]

alright

[vale wigeon]

then what was troubling you with these in particular?

[chilly dove]

idk what to do if its fractions and percentages im just confusd

[sage gull]

When its a mixed fraction as a percentage I suggest converting them into improper fractions as they are easier to work with. Then you already know how to convert a fraction into a percentage right? just multiply by 100. To convert it back to a fraction just divide by 100.

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[exotic locust]

I have the following equation. And I want to solve for dy/dx.

[vale wigeon]

you fucked up and deleted your original message

now you must abandon this channel and open a new one

[exotic locust]

wait what

why

If I use implicit differentiation and use implicit differentitation I get

But If I first multiply the original equation by y on both sides and distribute and then use the product rule and solve for dy/dx I get

So why am I get different answers?

.close

did i break this channel

[lone heart]

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[brave shard]

WHO pinged me

[plain flame]

i did

[wary stream]

And me

[languid bolt]

not me

[lone heart]

@brave shard Has your question been resolved?

[lone heart]

@brave shard Has your question been resolved?

[lone heart]

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[chilly dove]

How do I solve this problem

[chrome ibex]

ii dont understand

[subtle birch]

You doubt has been closed. If any further doubts please open a new channel

[subtle birch]

Since ratio of boys and girls is given and total strength is given, we can easily find respective number of boys and girls

[chilly dove]

how

nvm i got it

[subtle birch]

Let the ratio be 3x:2x

[subtle birch]

Oh, that's nice

[chilly dove]

you get 24:16

[queen smelt]

mhm

[subtle birch]

Yeah

[chilly dove]

i dont understand what to do for B C and D tho

[queen smelt]

you got 3 people that answered wrong what bout the remaining

[chilly dove]

nvm i think i got it, i just understood it wrong

sorry

[chrome ibex]

i need help with this problem... Given that3

ok i understand now

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[haughty sorrel]

If you call 0123456 as a 7 digit number then the same number 00123456 can also be called as a 8 digit number by this logic

So you ignore the Zeros in start

As it doesn't really make any sense

The contribution of these zeros to the value is nothing

[languid bolt]

why can't we call 00123456 an 8 digit number though

[haughty sorrel]

I am not talking about telephone numbers

[languid bolt]

are you reading my mind?

😨

[haughty sorrel]

In case of telephone numbers you may take zeros as digits

[languid bolt]

as an arithmetical value, then no

[haughty sorrel]

Yea

[lone heart]

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[vagrant falcon]

What conclusions can we make using a chi-square value without giving the alpha? For example, if I got 3,2,23,2, is there a high probability that you can reject the null hypothesis since 23 is statisically significant?

[gusty gorge]

why can't you give the alpha

the alpha is a personal choice

[vagrant falcon]

i know that, but im just wondering it is possible to make conclusions just from a chi sqaure componnet

[gusty gorge]

you're talking about rejecting a null hypothesis without talking about alpha?

[vagrant falcon]

yea

with just a chi sqaure componnet

that has a big difference

[gusty gorge]

the alpha that you give defines what you consider as big

the alpha is literally "how far does the test statistic have to be before I call it statistically significant"

and you're basically asking "can I call something statistically significant without talking about alpha"

[vagrant falcon]

ok, so essentially you can't come up with a conclusion just based of a chi-square component

[gusty gorge]

no, your question just makes very little sense

let me use an analogy

you see a building

you ask "can I call this building tall, because it looks tall, but without telling you how tall a building has to be to considered tall?"

[vagrant falcon]

oh ok

it just came across my head because i was doing problems

[gusty gorge]

and in general, you shouldn't be designing your testing procedure after seeing the data

you should set your significance level + statistical tests before seeing the data at all or computing stuff from it

because otherwise you introduce biases into designing the tests

[vagrant falcon]

and alpha was given, but I noticed a pattern that when I reject the null hypothesis, there is usually a value that sticks out

for example, 5,3,2,5,32

[gusty gorge]

I'm not sure what you mean by that

the chi-square test is based off a single statistic

[vagrant falcon]

im referring to the components

[gusty gorge]

wdym components? it's a single number

[vagrant falcon]

in a GOF, your calculating observed - expected squared divided by expected for all the data points

and you add up all those to get your test statistic

so when i refer to my example, 5+3+2+5+32 = 47, 47 is the statistic

5,3,2,5,32 are the components

[gusty gorge]

it's not always the case that one particular entry in the contingency table makes the test statistic really high

sometimes it's a built-up set of irregularities

[vagrant falcon]

ojk

.close

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[finite trout]

(a ) please

[swift musk]

divide both sides by sin θ tan θ, to start

(motivated perhaps by the fact that doing so will give you an isolated "2", like in the target equation)

[cinder tundra]

express all trigonometric functions in terms of sine and cosine and then simplify the equation.

[limpid spade]

yes

[cinder tundra]

Like this:
4cos(a) - 1 = 2sin(a) * tan(a)
4cos(a) - 1 = 2sin(a) * (sin(a)/cos(a))

[limpid spade]

yes

[cinder tundra]

You only need to use the Pythagorean identity and you're done

[finite trout]

oh alright fair enough

tysm

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[lone heart]

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[limpid turret]

5+(2*3)

[alpine sable]

sorry i solved it by accident

.close'

.close

[lone heart]

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[limpid turret]

Question:

[last ether]

🙂

[limpid turret]

🙂

[limpid turret]

I didn't ask if I could ask or if anyone knew

[limpid spade]

I didn't ask for you to ask if you asked if you could ask or if anyone knew.

[limpid turret]

I didn't ask to be told to ask

[limpid spade]

Skills issue

[limpid turret]

https://tenor.com/view/1v1me-fight-me-duel-duel-me-come-at-me-bro-gif-19835961

[hidden fable]

what even is the problem here

[limpid spade]

his problem is my problem with his problem

[limpid turret]

I wanna know how you can swing that R rated word around this PG13 server

[limpid spade]

whats the word

[limpid turret]

I can't say it. I'll be banned.

It's not the talogist part

[limpid spade]

u may say fuck

[white lichen]

hi what's 1+1

i need help its so hard

hello help here plz

[limpid turret]

!help

[lone heart]

Please read #❓how-to-get-help

[white lichen]

umm

!help

[lone heart]

Please read #❓how-to-get-help

@limpid turret Has your question been resolved?

[keen pasture]

I have a problem with people having a problem with Fucktalogist having a problem with their problem

[lone heart]

@limpid turret Has your question been resolved?

[lone heart]

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[sick gulch]

I am trying the statistics question Q14(i) from the 2007 STEP 3 paper.

Firstly, I got $\frac {n\pi} {n+1}$ as they did (see answer image), but I want to know what their $P(x<X<x+\delta x)$ notation means. If I had to guess, it's a way of writing the PDF as the CDF as when $\delta x \rightarrow 0$, we get $P(X=x)$ but I'm not sure about that as the right hand side would become 0, I think.

Moreover, I didn't know how to approach the second question "Find also the expected area of the smallest circle, with centre O, that encloses all the
(n-1) holes nearest to O.". I can express the probability of (n-1) holes being in the smallest circle $Y$, it's just that I don't know how to consider the last point furthest from O that is outside $Y$.

[ocean seal]

bostock

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@sick gulch Has your question been resolved?

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@sick gulch Has your question been resolved?

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[zealous sequoia]

i need help

[zealous sequoia]

i think this is how its solved

i have no idea where to begin

how to do it

or what to even do

i thought by finding both ark lenghs

i can get the total P

by using the formula

for P of a circle

and adding them both together

minusing

the ark lebgh

of both

anyways i cant do that obv

cuz there is no theta

to find the ark lengh

if anyone at all is not busy please respond got math paper 2 mocks tmrw

and am struggling

[gray isle]

you are told that the centres are 5cm apart

[zealous sequoia]

what should i do with that info?

[gray isle]

you'll have 3-4-5 triangles

[zealous sequoia]

OH

is there a trig ratio for dat?

[gray isle]

you should recognise what's special about a triangle with those lengths

[zealous sequoia]

it should be 90 degrees

am guessing

[gray isle]

3,4,5 is a Pythagorean triple and you have a right triangle if that's what you mean

[zealous sequoia]

wait nvm i can use cosine rule

i have no idea how i didnt see that

💀

my bad

[gray isle]

with right triangle, basic right triangle trig is enough

[zealous sequoia]

oh

@gray isle may i ask

u

to draw the right angle triangle

on the circle

i kinda got lost

like using the windows scrible tingy

nvm found it

YEAAAA NOW I CAN JUST SUE

WOOOH

THANKS M8

[mystic vortex]

How do you know that's a right angle?

[zealous sequoia]

3-4-5

side lenghs

[mystic vortex]

Ohh I see cause that can make a right triangle and there's no way to change the angles of you know all the sides

Cool

And then you just use an inverse trig function?

[zealous sequoia]

yup

[mystic vortex]

Nice

Good luck on the paper

[zealous sequoia]

thx m8

@gray isle is this the correct answer?

[gray isle]

did you skip steps?

you didn't seem to multiply the top angle by 2

also you're rounding too early

[lone heart]

@zealous sequoia Has your question been resolved?

[zealous sequoia]

wait

yes

my bad

thanks for pointing it out

[zealous sequoia]

when should i typically round?

[gray isle]

at the end

[zealous sequoia]

alr alr thanks alot

really understood the question

cheers!

.close

[lone heart]

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[rotund grove]

So I’m supposed to find the area between the functions in terms of both x and y and they gave different answers which is not normal??? Anyone know what I did wrong?

[lone heart]

@rotund grove Has your question been resolved?

[gray isle]

y=sqrt(x) isn't equivalent to x=y^2
so you'll need to adjust how you set up your integrals

you messed up signs in the integration you did as well

sqrt(x) - (x-2) isn't the same as sqrt(x) - x - 2

[rotund grove]

wait was is it equal to

[rotund grove]

rightt

[gray isle]

y = sqrt(x) for x>0

[rotund grove]

ohh right okk ty

[gray isle]

integrating wrt x you'd consider those regions seprately

as for integrating wrt y,
you'd consider the y-coordinates of the intersection points,
-1 and 2
and integrate with those bounds

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[mossy scroll]

i need help

[mossy scroll]

i need to turn a 3D prism and tilt it

so i tried this on geogebra

but its not 90 DEGREES

it must be a right prism

how can i do this?

im using this as a base

where C and B are -1 on the z axis

which i did to try and tilt it

but C and B can be any z value i do not mind

[rocky grove]

I think you can construct a perpendicular plane to the plane you have on the base

[mossy scroll]

and then what?

[rocky grove]

Then you can copy the base plane to some level you want and start copying your base up to there

[mossy scroll]

i dont really get it

could u try it on geogebra

i did this

[lone heart]

@mossy scroll Has your question been resolved?

[last ether]

<@&268886789983436800> HenriM seems to be a bit triggerhappy

_ _

[mossy scroll]

@crisp iron

what calculations do i need to make this a right prism?

where the base is tilted

[lone heart]

@mossy scroll Has your question been resolved?

[lone heart]

@mossy scroll Has your question been resolved?

[mossy scroll]

hello?

[mossy scroll]

i need to make a right prism with this base, how can I do so? <@&286206848099549185>

[lone heart]

@mossy scroll Has your question been resolved?

[mossy scroll]

im gonna kill this bot

[lone heart]

@mossy scroll Has your question been resolved?

[proud swift]

Maybe you need to find the mid points of any one opposite side pair

And then just project it upwards or downwards

This won't do

Maybe just rotate the given figure 60 degrees

You would get a prism kind of stucture

[lone heart]

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[tulip oasis]

is this the right set up?

[merry depot]

yes

[long axle]

Me when a helpee gets the difference quotient right

[tulip oasis]

nice thanks😄

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[thick lynx]

"Compute the taylor polynomial $T_2(f, x, 0)$."

[unreal kiln]

7

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@thick lynx Has your question been resolved?

[marble onyx]

I don't think this is standard notation so you might have to check the lecture notes or something. I guess going to the second derivative is probably safe

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@thick lynx Has your question been resolved?

[thick lynx]

Thanks

.close

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[lone ocean]

helppp

[pseudo ice]

Can you find the length of AC? Using sohcahtoa would do for that

[lone ocean]

uh

what

[pseudo ice]

ADC is a right angle triangle

[lone ocean]

yea but idk sohcahtoa

we hvnt learnt it yet

[pseudo ice]

So you haven’t learned the sin = opp/hyp type stuff yet? Peak

[lone ocean]

nope

i googled the q n i got it

thank u tho

.close

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[balmy quiver]

I have a linear subspace V of 3x3 matrices with following properties:

1. middle column and middle row are same vectors.
2. sum of corner values equals to 0

I'm tasked to find a set of 6 generators for matrices with the given properties above.

I'm trying really hard but I just get to 7.

I'm working with this general formula for the matrices:

| a x b |
| x y z |
| c z d |

where a + b + c + d = 0

In the generator I have 4 "canonic" matrices for each corner value. Their sum would be

101
000
101

Then I have three for the middle row and column.

010
100
000

000
001
010

000
010
000

[lone heart]

@balmy quiver Has your question been resolved?

[mortal trellis]

well the sum of the 4 corner matrices isn't in the space

and neither is each individual one if they are what I am assuming

[balmy quiver]

oh right, but this could (and its variation) be

| -1 0 1 |
| 0 0 0 |
| 0 0 0 |

[mortal trellis]

ok

how many of those can you choose linearly independently

[balmy quiver]

I think it's still four

| -1 0 0 |
| 0 0 0 |
| 1 0 0 |

| 0 0 1 |
| 0 0 0 |
| 0 0 -1 |

| 0 0 0 |
| 0 0 0 |
| -1 0 1 |

[mortal trellis]

those are not linearly independent

lets do it as vectors

how many vectors (a,b,c,d) can you choose with a+b+c+d=0

should it really be all R^4?

which it would be if you could choose 4 linearly independent ones

[balmy quiver]

I feel like I'm not even sure whether I understand your question rn

[balmy quiver]

(1,0,-1,0), (1,-1,0,0), (1,0,0,-1) hmm. Here I can't figure out the fourth linearly independent vector

[mortal trellis]

there isnt one

[balmy quiver]

yep

I don't even need it, since this should span the matrices?

[mortal trellis]

well the space of those matrices but yes

I guess true it just says generators

although wait

it does specifically say 6

so you cant just put too many

[balmy quiver]

yeah, but If I create the three new matrices and the three original ones (see above - they should cover 1) condition) it could be it, or am I missing something?

[mortal trellis]

if you put those 6 then thats correct

[balmy quiver]

yeah that's what I meant

if there is no other issue in the original vectors

[mortal trellis]

the other three are fine

[balmy quiver]

Ok, thank you I think I can continue from here. Thank you!

.close

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[agile grove]

Hi I am wondering about the importance of rational functions in the real world? Just simple explanations that would be easy to remember would be appreciated.

[agile grove]

I tried doing some research, but am not finding a satisfactory answer, and have a few queries.

[mortal trellis]

questions like these are hard to answer

[marsh rapids]

Just note that sometimes it's also a rational function in another variable, which makes it much more general

[mortal trellis]

rational functions just turn up from time to time

but explicit examples "from the real world" will always be too complicated

somewhere deep down they will include rational functions in some way

[marsh rapids]

For instance transfer functions, used in signal processing and electronics, for instance to remove ambient noise, are rational functions in jw/w0

And so the knowledge of these classes of functions lets you quickly understand their practical behavior

[high rapids]

So many YouTube notifications bro what you watch

[agile grove]

My apologies, but I see it as if I can understand some applications of it within my knowledge

It would not only help me retain the info better, but be more excited about what I am doing.

[marsh rapids]

I have a horrible tendency to not unsubscribe, even to people I haven't watched at all in over a year

[high rapids]

I always turn off notifications and unsubscribe

[agile grove]

think you'd be able to make a more simplified explanation?

I heard that rational functions

can be used in rates

so if you perhaps delve on that would be nice.

[marsh rapids]

"rates" has got to be one of the least precise things lol. Could you be more precise ? Cause rn idk

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[agile grove]

Thanks found a video on khan academy that helped a lot.

[lone heart]

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[clear yacht]

Hey gyze! Does the outlier include the upper and lower fence. Thanks!

[vapid shuttle]

Could you clarify your question more

[clear yacht]

Lets say that I calculate the upper and lower fence and then the question asks if a value is an outlier and it is exactly on the lower fence

Would it be an outlier or no

[vapid shuttle]

the upper and lower fence being the bounds of the IQR or being the top and bottom ends of the range?

[limpid spade]

If a data point falls below the lower fence or above the upper fence, it is considered an outlier.

[clear yacht]

Ok so not inclusive

Thanks!

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[amber plover]

can someone please explain to me how this function is not odd

[amber plover]

f(-x)= different sign for every number

[atomic acorn]

for a function to be odd, f(x) = -f(-x) for every x in the function's domain

here the equality isn't obeyed so

[amber plover]

isn't a function odd if the entire equation has a different sign if it is multiplied by -1

[atomic acorn]

the function is odd if you multiply f(-x) with -1 and end up with the original function

in your soln they've compared f(-x) and -f(x) and they both aren't equal

[atomic acorn]

and this is not true

[amber plover]

i am so confused hold one let me show you something

[amber plover]

they're not equal

[atomic acorn]

you need to replace x with -x

And see if the entire function once multiplied by -1 gives you the original function

[outer lark]

might help to think graphically if you are getting confused

[amber plover]

no i got it now it's just that the teacher worded and formatted this so weird

[amber plover]

thanks for help btw

.close

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[glacial goblet]

I have the the equations of the forces in the x and y directions that are in equilibrium, im just kinda stuck here because the angles aren't exact can someone tell me where to go from here?

[atomic acorn]

what's the problem if your angles aren't exact

[glacial goblet]

because i have simultaneous equations right

im just not sure if i divide by cos52, cos 38, sin 52 or sin38

because i have to find the tension right?

[atomic acorn]

2 eqns 2 variables

[atomic acorn]

Oh you can't use a calculator?

[glacial goblet]

ik but i cant just sub willy nilly

[atomic acorn]

Either way it's not too much labor

[glacial goblet]

because i have t1cos52 = t2cos38
t1sin52 + t2sin38 = 50

[atomic acorn]

that's it, solve for t1, t2

[glacial goblet]

should i just make one of them the subject then?

that would over complicate the question i think

[atomic acorn]

it's your wish man

[glacial goblet]

alr ill close this the

then*

.close

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[vapid shuttle]

!status

[lone heart]

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

[lone heart]

@amber hemlock Has your question been resolved?

[pseudo ice]

Are you familiar with how you do partial fractions?

[amber hemlock]

no...

[pseudo ice]

Well that makes stuff a bit hard

You haven't done partial fractions before at all?

[amber hemlock]

I do it but get it incorrectly a lot.

[pseudo ice]

Hmm, alright, well at least for the form here they give,
[
\frac{6x^{3} + 2x^{2} - x + 12}{(2x^{2} - 1)(x^{2} + 1)}
]
Do you know the general partial fractions form you would go for?

[ocean seal]

@pseudo ice

[lone heart]

@amber hemlock Has your question been resolved?

[amber hemlock]

I figured it out. Thank you sm/

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[solid furnace]

when using the chain rule for questions such as 3cos(x^3) how come i only take the derivative of 3cos (-3sin) and then leave the x^3 as is, why doesnt the x^3 become (x^3)^0 and cancel out. the answer the this question iirc is -3sin(x^3)(3x^2) [pls correct me if im wrong] but im confused as to why it isnt -3sin(3x^2)

[merry depot]

chain rule says d/dx( f(g(x)) ) = f'(g(x))g'(x)
So you take the derivative of the outside function (sine) and leave the inside (x^3) alone.
Then multiply by the derivative of the inside function (3x^2)

[solid furnace]

for the question 4(sinx)^3 i take away from the power and get 12(sinx)^2(cosx), i still dont get why i dont take from the power in the first :/

[alpine sable]

$4sin^3x$

[ocean seal]

pulse

[solid furnace]

i dont understand

<@&286206848099549185>

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@solid furnace Has your question been resolved?

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@solid furnace Has your question been resolved?

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[ionic wyvern]

Does anyone know how to find an unknown matrix when A, the unknown and B are all 4x4 matrices. Without using the inverse of matrix A

[limpid turret]

Post the actual question please

[chrome salmon]

Post the actual question please

[kindred lodge]

Post the actual question please

[ocean seal]

BusterKeaton15
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lone heart]

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@ionic wyvern Has your question been resolved?

[ionic wyvern]

.close

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[alpine sable]

can someone please help me with these problems for my math homework?

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[merry herald]

explain process pls

[lone heart]

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[smoky oxide]

yo

[lone heart]

@merry herald Has your question been resolved?

[fickle ravine]

Have you drawn a picture of this said pyramid? It may help you to understand what you are doing

[merry herald]

Oh I figured it out

Forgot to close this

.close

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[meager kestrel]

How do you do these questions without a calculator is there any tricks(I'm Joining a math competition)

[marsh rapids]

Yeah these are specifically about having some understanding of what you're doing to do them by hand easily

For the first one set x = 2019.625

That should make it clearer

[meager kestrel]

make it a algebra? ok

[marsh rapids]

Now your expression is 6 times shorter

So it's much easier to see what's going on

[meager kestrel]

ah i see never thought about using algebra in these kinds of pquestions

[marsh rapids]

It's classic that when they take a large, random number, its value doesn't matter and is only here to confuse you

[meager kestrel]

i see

so i make it x and then jsut calculate it normally

and after that turn the x to normal again

right?

[marsh rapids]

Yes

Since you found it's x - 0.625

[meager kestrel]

ye

so much easier

[marsh rapids]

All 3 are easily done by abstracting a bit and then looking at the values only at the end

[meager kestrel]

ok

tysm

[marsh rapids]

2 is a classic method

Knowing how to count factors is useful

[meager kestrel]

ok

can you tell me

cause factor my weakest topic

[marsh rapids]

so x = 5^n 11^m

How many factors ?

What does a factor of x look like ?

[meager kestrel]

????

[marsh rapids]

You want to count them, you might as well figure out what they are

[meager kestrel]

ok

[marsh rapids]

We want to find what (n,m) couples lead to precisely 6 factors

Then find the smallest corresponding x

[marsh rapids]

Any progress yet ?

[meager kestrel]

gimme 1 sec

using my brain

is the answer D?

[marsh rapids]

Yes

[meager kestrel]

i understand it abit

since i did the tree method

and i got 275 through trial and error

[marsh rapids]

It's all the 5^i 11^j, with 0<= i <= n and 0 <= j <= m
Which leads to (n+1)(m+1) factors (i.e. (i,j) combinations, as they're all distinct factors)
6 = 2*3 or 6*1 so you have 5^5, 5^2 11^1, 5^1 11^2, 11^5
25 * 11 is the smallest of these

[marsh rapids]

It's all theory until the final bit

[meager kestrel]

confused with that method

your comparing everything basically

[marsh rapids]

1. figure out what a factor looks like
2. find how many there are
3. find the x that have 6 factors
4. find the smallest one

I also split my message in 4 lines

[marsh rapids]

Does the method make sense ?

[meager kestrel]

yes

[marsh rapids]

What point confuses you in practice?

[meager kestrel]

idk

sometimes ye

[marsh rapids]

So is it clear?

[meager kestrel]

yep

sry for late response

[marsh rapids]

Don't worry, 3) is much easier

Hint: ||look at the values y takes on||

[meager kestrel]

ok

[lone heart]

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[tall trout]

I've got the answer a and b

I'm currently stuck on c

[prime badge]

it's transpose

reflection in x = y

[tall trout]

How did you get that?

[prime badge]

it's a thing

i don't know

[tall trout]

oh.

[prime badge]

now you don't know too

[tall trout]

i thought reflection y = x is a reflection line passing through (0,0) going upper right direction

[prime badge]

actually wait

you're right, it's probably the other one

so, we take y = -x

and move it up

[worn fox]

You were right the first time

[prime badge]

got it

i was rotating it anti-clockwise

so yeah, y = x does the job

[tall trout]

Oh right.

Thanks

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[lost glen]

when deriving a trig function to a power, do you do it like this?

[chrome salmon]

Ok

For derivative?

[lost glen]

yes

[chrome salmon]

Yes you can use chain rule like this

[lost glen]

oh ok, lemme try can I show you my answer to see if its correct

nvm

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[tall trout]

couldn't find answer

[lone heart]

@tall trout Has your question been resolved?

[tall trout]

<@&286206848099549185>

[jovial mason]

part a or b

[tall trout]

i think i got answer for part a, but idk if I'm correct

[jovial mason]

what did u get

[tall trout]

(5;1)

[jovial mason]

oh wait its vectors

[tall trout]

yeah

i try to put it the best i can

[jovial mason]

ngl if no one else answers here just chuck it in to chat gpt

its right like 90% of the time

and its usually obvious if its wrong

[tall trout]

didnt think about that

[jovial mason]

ye

[tall trout]

the thing is

how do i actually give it enough information to work it out

as this is a cord question

[jovial mason]

you could probably just tell it the coordinates of each corner on the shape and say it is filled in for the area

[tall trout]

cant

[jovial mason]

🗿

[tall trout]

ok this is the last question on my homework

still cant figure it out

[jovial mason]

its been a while since ive done vectors

they hella cringe

is there like an example u can look at?

that walks you through a similar question

[tall trout]

there's a video

I have no idea what that crazy lady in the video is on about tho 💀

[jovial mason]

is it a yt link

[tall trout]

no.

[tall trout]

.close

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[lone heart]

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[fierce delta]

what is happening in the second line of working out?

[limpid spade]

which

(0,8) etcN*

[fierce delta]

(0, 8) -> 8 = 9a - 10

yup

[limpid spade]

THEY PLUGGED IN THE VALUES

Caps

[gray isle]

they substituted that point into the equation
i.e. subbed x=0 and y = 8

[fierce delta]

ive got 8 = a(0)^2 + b(0) - 10, clearly missing the 9a. could you tell me why i'm missing the 9?

i tried putting it in the form of y = ax^2 + bx + c

[limpid spade]

plug it in the equation above

[fierce delta]

the -3 from the turning point

[limpid spade]

y=a(x+3)²-10

[gray isle]

c isn't -10

you're conflating general form with vertex form

[fierce delta]

yeah i just realised

with the first line of working out, the 3 and 10 are from the turning point right?

wait, in order to solve this question was I first required to use the vertex/transformation format?

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[severe ruin]

in ABC triangle AB=6 AC=8 <A/(<B+<C)=1/5 Sabc=?

[mellow grail]

what..?

[rocky grove]

So you want the perimeter of the triangle?

[severe ruin]

area

[rocky grove]

Okay

[severe ruin]

can do area with Heron but cant find other side

[rocky grove]

You have a relationship between the angles

Use it

Find A

[severe ruin]

A=30

6x=180 x=30

[rocky grove]

Okay

And you have AB = 6 and AC = 8

And the angle between them is 30°

[severe ruin]

AC=8

[rocky grove]

Oh ye

Okay

There is a relationship for Side-Angle-Side to get the area

Do you know of it?

[severe ruin]

nope

[rocky grove]

Okay

Are you aware of trigonometry?

[severe ruin]

yes

[rocky grove]

Okay

$A = \frac 12 ab \sin{\theta}$

[ocean seal]

VulcanOne

[rocky grove]

Where a and b are sides that share the angle θ

[severe ruin]

so

30= 1/2 ab sin theta?

[rocky grove]

Nope

Theta is your angle

That you just got

A

Angle A

[severe ruin]

A=1/2ab sin 30

[rocky grove]

Yep

[severe ruin]

oh i get it now

[rocky grove]

[severe ruin]

thanks

[rocky grove]

My pleasure :)

[severe ruin]

.close

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[dense shadow]

Can someone explain how to solve cubic equations?

[gray isle]

depends on the cubic

[dense shadow]

I have this : 8^x-7 * 4^x+14 * 2^x-8=0
I make it into : 2^3x-7 * 2^2x+14 * 2^x-8=0
2^x=t
t^3-7t^2+14t-8=0

and what now

[gray isle]

do you mean 14*t

or just 14t

[dense shadow]

Oh 14t sorry

[gray isle]

approach with rational root theorem in an attempt to factorise

[dense shadow]

Isnt there an easier method? Im pretty sure we did learn in school but i already forgot

[gray isle]

not really no

[dense shadow]

hmm strange, maybe we didnt learn yet

ok lemme check how that theorem works

[gray isle]

its a systematic way to find nice roots if they exist

[dense shadow]

what about the Cardan formula?

is it harder?

to do that way

[gray isle]

yes, much more tedious

you want to wish you don't have to resort to that

[dense shadow]

oh i see

[gray isle]

once you find one nice root, factorisation will make it easier to find others

this is applicable to higher degree polynomials as well

[dense shadow]

ok ty

.close

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[paper sage]

$sin 3x=cos 2x$

[paper sage]

how do i begin¨?

can i use sin x=cos(x-pi/2)?

can someone help me solve this please

[ocean seal]

Yousssef

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@paper sage Has your question been resolved?

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[obsidian summit]

how do you calculate beginning inventory in a production budget? i'm trying to wrap my head around it but can't seem to understand

here is my progress so far

[obsidian summit]

please @ me when respond

<@&286206848099549185> \

i can send my assignment if need be

<@&286206848099549185>

<@&268886789983436800>

[next brook]

Sigh.

[obsidian summit]

.close

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[native nebula]

help

[native nebula]

i can't find the inequality

[gray isle]

there are more efficient ways to approach this

[native nebula]

what are they

[gray isle]

note that |x-3| is positive for x != 3

so considering when x-3 is positive will tell you the solutions to this inequaltiy

[native nebula]

can't understand 😅 🥲

ok can you tell me more about what i have done

@gray isle

[gray isle]

firstly, which part of what I said don't you understand

[native nebula]

this

[gray isle]

did you have any issue with

note that |x-3| is positive for x != 3

[native nebula]

no

[gray isle]

so if you have
$$\frac{\text{positive value}}{\red{\text{something}}}$$
that will only be positive if that $\red{\text{something}}$ is positive right?

[ocean seal]

ℝamonov

[gray isle]

(from principle of products of positive and negative numbers)

[native nebula]

yeah true

[gray isle]

so applying that, determining when the denominator is greater than 0 is enough

[native nebula]

yeah

is that the only thing I have to do

?

[gray isle]

pretty much

[native nebula]

can you point out the mistake in my method?

[gray isle]

poor explanation of work, incomplete

of doing case work, you should explicitly state what your doing
and clearly state the conclusions

[native nebula]

can I ask you a question?

[gray isle]

about

[native nebula]

maths

[gray isle]

what about maths

just jump straight to the question you want to ask

[native nebula]

why i don't know the things you know? are you in high school?

if you wish to answer

[gray isle]

past high school already, depends on your teachers and willingness/dedication to self learn

and for the second case, you'd need to find the intersection between x>3 and x!=3

which leads to the same result of the simpler approach

[native nebula]

1>0 doesn't lead to (3, infinity)?

[gray isle]

1>0 is outright true,
but you'd need to consider the conditions under when that's happening
i.e. the case you're taking

which was for x>3

and the conclusion from that is that the original inequality is true for x>3

[native nebula]

ok ic

[gray isle]

1<0 is false indicating there are no solutions for the case x<=3

[native nebula]

one simple question, how many hours everyday should I spend on maths? like now you have seen my level of understanding...

[gray isle]

depends on how well you soak in info

manage time for other subjects as well

[native nebula]

on an average?

7-8 hrs?

.close

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[zinc snow]

I have some questions on how to simply reflect a vector on a matrix. (Precalculus)

[soft saffron]

Ask them instead of stating that

[zinc snow]

That relatively is the question

[lone heart]

@zinc snow Has your question been resolved?

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[neon bronze]

Hello. How do I continue?

[neon bronze]

I am trying to find the value of the limit at the end

<@&286206848099549185>

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@neon bronze Has your question been resolved?

[neon bronze]

<@&286206848099549185>

[delicate jacinth]

$\lim_{x\to 0} (x^2-x+1)^{\frac{n}{x}}$?

[ocean seal]

eotales

[delicate jacinth]

is this the limit?

[neon bronze]

Yeah

[delicate jacinth]

okay

[neon bronze]

Do you have any idea?

[delicate jacinth]

I'm trying to somehow use euler's limit here

[neon bronze]

I feel like you have to take ln but I am not sure

Oh, what's that?

Oh

[delicate jacinth]

$\lim_{x\to 0} (x+1)^{\frac{1}{x}} = e$

[neon bronze]

Ok I searched it up I didn't know it had that name

Yeah

[ocean seal]

eotales

[delicate jacinth]

create a channel for u

!help

[lone heart]

Please read #❓how-to-get-help

[delicate jacinth]

<@&268886789983436800>

[neon bronze]

Can you read #❓how-to-get-help

Create your own help channel and ask your question there

@delicate jacinth I got it, I had to use the expansion for ln(1+x)

[delicate jacinth]

yeah

I'm not satisfied

I will send this to my teacher XD

this should be possible without L'hopital or series

[neon bronze]

Oh ok sure, let me know if your teacher sends the solution

@delicate jacinth feel free to DM me the solution
I am closing the channel for now
Thanks for your help

.close

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[paper sage]

how do i sketch this graph

[wintry birch]

First you need to find the zero points and asymptotes

[paper sage]

it dosen't have any vertical asymtot

[wintry birch]

Yes so that means

[paper sage]

since the denominator dosen't have any roots

they i should use PFD

[wintry birch]

And what about horizontal/diagonal

(I am sorry english isn't my first language 😭 so the terminology might be off)

[paper sage]

horizontal/diagonal can i found by maybe take the derivative?

[paper sage]

no problem at all, im so glad you can help

[unique imp]

Yes, you should differentiate! Maybe even twice to find possible inflexions

[tight wind]

just plot it in GDC

[paper sage]

i should first do polynimail division

[tight wind]

hehe ik its not helpful sorry ill get out of here

[paper sage]

because the nominator is bigger than denomitator right?

[wintry birch]

With the diagonal can't you just do a euclidean sharing (this /)