help-26

[austere sable]

you quite literally move it right or left

if I told you

write the graph of y=x^2

then shift it upwards by 1

y=x^2+1

what would that look like

[neon iron]

I obviously can do that, but these repeating graphs are harder to do that with

[austere sable]

There's no difference here

you're over thinking it

[neon iron]

so alright, say phase shift is pi/2

[austere sable]

It's identical

[neon iron]

what would you do

pi/2 right

[austere sable]

you look at when crosses the x and y acis

then draw from there

same process as you would try to graph x^2+5x+6

find when it crosses the x and y access

same for sine and cosine

[neon iron]

alright, makes sense then

thanks

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[inland oak]

so i get how to solve it by why can we square the D? and have that equal to f(x,y)

is there a reason behind it? though more annoying, can we make D = f(x,y)

[loud ingot]

You square both sides from the previous step

And replace z^2 in terms of x and y that is given

Hence having function parameters of only x and y

[inland oak]

oh okay, thank you

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[bitter pollen]

ive been trying to solve this with a calculator, and it is not working for me, i dont know how to solve for n here:

[bitter pollen]

im not sure how to get isolate n inside the factorial

[keen venture]

Definitely can't

But that's decreasing, so what happens when n = 7? n = 8?

[bitter pollen]

what do you mean

[keen venture]

That's gonna get smaller as n gets larger

[bitter pollen]

Im trying to determine the minimum order of taylory series polynomial it would take to have an absolute error of less than some number, like 10^-6

yes it is

[keen venture]

What happens when n = 7?

[bitter pollen]

the inequality is false

or never satisfied

but yeah false when n=7

[keen venture]

Right, then true when n = 8

[bitter pollen]

so is the technique to plug values until the inequality is satisfied?

[keen venture]

I mean we only plugged in 7 and 8

[bitter pollen]

well at first glance I wouldnt be sure which integer to start with

[keen venture]

So yeah this won't work generally

But it solves this problem!

[bitter pollen]

i guess i could graph it and just eyeball where the function approaches a difference of less near 10^-6

thanks

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[carmine star]

!help

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[carmine star]

how would i find the key points

cuz somehow i am getting them wrong

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[strange loom]

How would you do part b?

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[safe rivet]

hi can anyone help me on this

[safe rivet]

im stuck

thats where i got

idek cause (t) isnt even an angle

im completely lost

[mint crescent]

yeah this is fine

now just calculate sin 0 and cos 0

[safe rivet]

wait what

how would i do that

a and b are different numbers

[mint crescent]

this is irrelevant

it's just calculating standard trig values

[safe rivet]

okay

i see

thanks

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[pale axle]

How can I solve tan(5x) = 0?

[mellow arrow]

the problem is basicall sin(5x)=0

and cos(5x)=\0
but cos(5x) is never zero when sin(5x) is zero, so you basically have that sin(5x)=0

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[topaz tulip]

can anyone help me with this limit : limit ln(sin(x))/e^1/x as x -> 0+

[topaz tulip]

im getting stuck after applying lhoptal rule

once

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[modern rock]

not sure how to solve

should i find all roots first?

[pearl shuttle]

I think you can

[topaz sinew]

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[modern rock]

can anyone verify if the answer is $(z^2-\sqrt{3}z+1)(z^2+1)(z^2+\sqrt{3}z+1)$

[thorny flame]

Galaxy

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[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[random quiver]

@neon iron you still need help ?

[neon iron]

yeah ;-;

[random quiver]

find the first derivative of f(x)

and then determine the critical pointsby setting the dervative to 0

[neon iron]

alr that would be for f(x)

what about g(x)

[random quiver]

do you understand that

since g(x) is f(|x|)

it will have symmetry

in the y axis

meaning that

it would be 5 solutions I think ?

do you have the answers?

[neon iron]

its 5

[random quiver]

yea I meant 5 lol my bad let me explain it

so g(x) is even

there is no extrema at x = 0 unlike some other functions

then you gottak now that

the number of extrema x > 0 will be equalt to the number of extremas for x < 0 for x > 0

see it as this

so number of extremas is 2

total is 5

[neon iron]

i see

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[safe rivet]

how am i supposed to do that

[safe rivet]

i dont understand what they represent

[chilly walrus]

differentiate both sides with respect to t

even though it doesn't appear there, you can still do it

[safe rivet]

wait what im confused how i would go about doing that

[safe rivet]

wdym with respect to t i just do like 3(2) or something? sorry im really confused

[agile harness]

differentiate implicitly

[safe rivet]

where am i supposed to use that

wtv

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[grim parcel]

what did I do wrong?

and why do you need (a-b)^3?

[topaz sinew]

@grim parcel Has your question been resolved?

[grim parcel]

<@&286206848099549185>

[pastel salmon]

hint for you:

$\int_{}^{}\frac{x}{1+\sqrt{x}}\text{ }dx=\int_{}^{}\frac{t^{2}\cdot 2t}{1+t}\text{ }dt$

[thorny flame]

Joanna Angel

[pastel salmon]

then, when you look at the numerator, you will use the formula for the sum of cubes, etc

[grim parcel]

if sqrt(x) = t then x = t^2 instead of T^2 * 2t

[pastel salmon]

?

do not forget about the derivative

i hope it is clear for you 🙂

I quote theorem for you:

$\int_{}^{}f\left[ g\left( x \right) \right]\cdot g'\left( x \right)\text{ }dx=\int_{}^{}f\left( t \right)\text{ }dt$

[thorny flame]

Joanna Angel

[pastel salmon]

so how, koleżko, have you learned this theorem I've written for you?

[grim parcel]

I haven't

I'll just watch a video about it because the explanatation from the teacher's handout is too unclear

[pastel salmon]

$\int_{}^{}\frac{x}{1+\sqrt{x}}\text{ }dx=\int_{}^{}\frac{t^{2}\cdot 2t}{1+t}\text{ }dt=2\int_{}^{}\frac{\left( t^{3}+1 \right)-1}{1+t}\text{ }dt=\\=2\left( \int_{}^{}\frac{\left( 1+t \right)\left( t^{2}-t+1 \right)}{1+t}dt-\int_{{}}^{}\frac{dt}{1+t} \right)=\\=2\left( \int_{}^{}\left( t^{2}-t+1 \right)dt-\ln\left( 1+t \right) \right)=2\left( \frac{t^{3}}{3}-\frac{t^{2}}{2}+t-\ln\left( 1+t \right) \right)+C=\\=2\left( \frac{1}{3}x\sqrt{x}-\frac{1}{2}x+\sqrt{x}-\ln\left( 1+\sqrt{x} \right) \right)+C$

[thorny flame]

Joanna Angel

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@grim parcel Has your question been resolved?

[grim parcel]

Still don't understand this but I finally solved it

thanks

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[slender sonnet]

Need to find the line with the x on it

[topaz sinew]

@slender sonnet Has your question been resolved?

[slender sonnet]

<@&286206848099549185>

[livid ocean]

use tangent secant theorem

[slender sonnet]

like?

how?

[livid ocean]

is length of tangent 20 cm here?

[slender sonnet]

yes

@livid ocean

what do i do?

<@&286206848099549185>

[livid ocean]

this theorem is called tangent secant theorem

use it in your task

[slender sonnet]

tried but the answer is wrong can you try?

[livid ocean]

4x*(4x+5x)=20^2, you need to solve this equation

[slender sonnet]

thanks a lot done it

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[dreamy drum]

it depends on the inequality

easiest way is just to interpret it as you go

just post the question and you can get walkthrough

but as you go higher in math there ususally are less memorizable ways

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[signal abyss]

hi, need help with my homework cause im a bit confused

[signal abyss]

from q1 do i need to account for doubles such as {M, M}, {T, T}, and {A, A}?

or can i simply just do 11C2?

[topaz sinew]

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[signal abyss]

i apologise for pinging <@&286206848099549185>

[long stirrup]

you would overcount {E,A} for example

there are two way to pick it, but it should count once

[signal abyss]

i did a bit more thinking and thought of it as like
MM AA TT I E H C S = 8 groups
so theres 8C2 ways to choose 2 of the 8 groups and then i add 3 to the final result because there will be 3 multisets that are double letters ({A, A}, {M, M}, {T, T})

[long stirrup]

i would do that

[signal abyss]

so its 8C2 + 3 = 28 + 3 = 31?

[long stirrup]

yeah

[signal abyss]

ahh so that means i can apply the same logic to q2 where i group them into 8 right

but instead of doing 8C2 i do 8 * 7 since its ordered

[long stirrup]

sure

[signal abyss]

so (8 * 7) + 3 = 56 + 3 = 59 yeah?

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[carmine pelican]

Sorry but i vaguely remember this , here we are scaling up so wont we divide the determinant' by m and n ? i remeber doing some problem of it was ,matrice or determinant idr, but when i scaled up without dividing i got the wrong answer

[carmine pelican]

Also i certainly remeber wr have to divide sometimes but dont know when the sometimes is

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@carmine pelican Has your question been resolved?

[carmine pelican]

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[carmine pelican]

.reopen

[topaz sinew]

✅

[carmine pelican]

[topaz sinew]

@carmine pelican Has your question been resolved?

[gray ridge]

hihi, to answer your question, adding or subtracting a Row or Column on other rows or columns would not affect the value of the determinant

e.g.
1 3 5
A = 1 3 5
1 3 5
det(A)=
|1 3 5|
|1 3 5|
|1 3 5|

e.g. you can add
-5C1 to C3
-3C1 to C2
and get
| 1 0 0|
| 1 0 0|
| 1 0 0|

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[dusty pelican]

Anyone good with Frieze patterns?

[dusty pelican]

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[drifting swift]

where is the question?

[pulsar patrol]

find vector b

[drifting swift]

i think there might be two different vectors that could be b

is any more info given?

[pulsar patrol]

nah

[drifting swift]

ok, do you know how to calculate scalar projections?

[pulsar patrol]

yeah

[drifting swift]

ok, how do you do it

[pulsar patrol]

proj = u . v / |v|

how did u get 2 ways?

[pulsar patrol]

not sure how to find the individual x and y values for the vector

[drifting swift]

geometric intuition

actually come to think of it

maybe there are more than two ways...

let b = xi + yj

then our projection is (x+4y)/sqrt(x^2+y^2) right

no wait ok. my bad.

[pulsar patrol]

?

[drifting swift]

im gonna need to see the entire exercise that this is from.

i have a feeling we're still missing a lot of info.

[pulsar patrol]

no thats all it is

[drifting swift]

so you need to describe all such vectors, then?

[pulsar patrol]

idk

can you just show how you would get one of the answers

cos im still clueless on how

[drifting swift]

i would probably assume for simplicity that b has unit length

so x^2 + y^2 = 1

then we get 4x + y = 11/sqrt(13) and x^2 + y^2 = 1

suffer through the solution of that system and get your b

idk like

i really do not see any other way lol

[pulsar patrol]

for the 4x+y thing

what happened to the sqrt 17

[drifting swift]

what sqrt(17)?

[pulsar patrol]

magnitude of a

[drifting swift]

isn't b the vector that we project onto?

[pulsar patrol]

oh

[drifting swift]

so only its magnitude would go in the denom

[pulsar patrol]

so magnitude of b is assumed to be 1

[drifting swift]

i would probably assume for simplicity that b has unit length

[pulsar patrol]

ok

thanks

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[modern rock]

i have no clue what to do

[modern rock]

where does x^2-x+1 come into this?

[ruby mural]

Hint: imaginary roots always occur in pairs

[modern rock]

yeah i get that, the conjugate is also a root

[ruby mural]

Oh

[modern rock]

?

[ruby mural]

So you have a quadratic that's a factor of the given polynomial

And a real root as well

You can use long division

[modern rock]

why would i need to do long division?

[ruby mural]

I'm sorry I'm sleep deprived

You can just form the quadratic

Using the imaginary root pair

[modern rock]

aight

[ruby mural]

That would be a factor of the polynomial, since each of the roots of the quadratic is also a root of the given polynomial

[modern rock]

hold up i think messed up the expansion

[ruby mural]

This was to find the 4th root, but not beneficial for part i)

[modern rock]

what does $-\frac{1+i\sqrt{3}x}{2}-\frac{1-i\sqrt{3}x}{2}$ come out to be?

[thorny flame]

Galaxy

[ruby mural]

The denominator is same

So you can get a common numerator
It would be -1 I think

[modern rock]

but that means i got x^2-1+1

should i not be getting x^2-x+1?

[ruby mural]

This is the sum of roots right?

[modern rock]

its when i expanded $(x-\frac{1+i\sqrt{3}}{2})(x-\frac{1-i\sqrt{3}}{2})$

oops

[thorny flame]

Galaxy

[ruby mural]

Yeah

So

[ruby mural]

This is the coefficient of x

(x-a)(x-b) = x² -x(a+b) +ab

That a+b goes with the x

[modern rock]

oh u want me to form the quadratic using the sum and product of roots?

i thought u could form it from the expansion of the roots aswell?

[ruby mural]

I was saying that if you expand, you will get part a
You said we would be gettin x² -1 +1, but it's x² - x +1, since that -1 is a coefficient of x

[ruby mural]

And yes, that's another way
Even this way is fine, but I didn't mean to form using sum and product of roots

[modern rock]

thats why im saying I expanded it wrong, that -1 is from adding the 2nd and 3rd term of the expansion. It should equate to x^2-x+1, but im saying I got x^2-1+1.

[ruby mural]

Oh so you got it now?

[modern rock]

no, i fucked up and i dont know where i went wrong

[modern rock]

these are the middle terms correct?

[ruby mural]

Yes

[modern rock]

then why do i get -1?

[ruby mural]

Coz that's what you get after adding em

And that -1 gets multiplied with x

As I said, it's a coefficient of x

[ruby mural]

What I meant to say by this was....

[modern rock]

yeah i get that its the coefficient when you add the 2 roots, but when you form the quadratic from expansion I should be getting -x but i got -1, the x was losst in translation, but how

[ruby mural]

Let's call the roots a and b for now

So we have (x-a)(x-b)
Which is x(x-b) - a(x-b)

Which is x² -bx -ax +ab

Which is x²-x(a+b)+ab

[modern rock]

right i get that, but how come when i did it, the x was lost

[ruby mural]

Maybe you forgot to multiply

With x

[modern rock]

.

[ruby mural]

Over here you're just addin the two numbers

[modern rock]

right

[ruby mural]

That's y there's no x

[ruby mural]

But this is the thing we need

And when you expand, that sum is a coefficient of x

[modern rock]

this is it

[ruby mural]

I see

[modern rock]

. both 1s should be x's

aight but how do i do b

[topaz sinew]

@modern rock Has your question been resolved?

[topaz sinew]

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[trim wolf]

i have to prove that. Given is that E is the mid point of BC. Thats what i found i may need to some hints.

[flat kindle]

you mean the arc BC?

[trim wolf]

Yep

By knowing that we know that BE = EC

[flat kindle]

how is G constructed?

I only see that G is colinear with A and B

but that's not enough to determine a point

or am I missing something

[trim wolf]

The points C - BA are on a circle.
Point E is the middle of the arc BC, as shown in the video before you.
At point E, a tangent is drawn to the circle.
The tangent cuts the continuation of the chord AB at point G.
The chords BC - AE intersect at point F.

thats the translation. Its not in English

[flat kindle]

oh it's tangent ok

[trim wolf]

and also i have to proove that BC || GE

[flat kindle]

I'm not good enough in geometry

[topaz sinew]

@trim wolf Has your question been resolved?

[topaz sinew]

@trim wolf Has your question been resolved?

[trim wolf]

Question: Is the angle ACE the angle of the arc AE / 2?. And is the angle GEA also the angle of the arc AE / 2?

[topaz sinew]

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[topaz sinew]

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[sick anvil]

Hello! How can I solve this? -(1/x) - 1 - 2X = 0

[drifting swift]

are x and X supposed to be different?

or was that just a typo

[sick anvil]

typo

[drifting swift]

ok

!status

[topaz sinew]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sick anvil]

they are the same

[sick anvil]

1. 😂

[drifting swift]

multiply both sides by x and your equation will become quadratic

maybe also multiply by -1 as well to get rid of the minuses

[sick anvil]

so all would be equal to x?

[drifting swift]

idk what you mean by "all would be equal to x".

[sick anvil]

-1 - 1x -2x^2 = 0

[drifting swift]

what is 0*x?

[sick anvil]

= x

[drifting swift]

what is 0*x?

[sick anvil]

0...

sorry

[drifting swift]

ok so then why put x on the right

when 0*x is not x but 0

-1 - x - 2x^2 = 0

is what you will get

do you see how to continue from here?

[sick anvil]

but here then its not in order?

wait i did something

[vale furnace]

What do you mean by not in order?

[sick anvil]

there are negative

[drifting swift]

so what?

[upbeat crypt]

it dosnt really matter

[sick anvil]

can i show you what ive tryed?

[drifting swift]

sure you can

[upbeat crypt]

go ahead

[drifting swift]

but also if you are so worried about the minus signs, then

maybe also multiply by -1 as well to get rid of the minuses

[sick anvil]

1/x - (2x^2)/x = 1

[drifting swift]

this is incorrect

[upbeat crypt]

did you multiply by x?

if x!=0

[drifting swift]

maybe better if you write it on paper

[sick anvil]

ok 1 minute

this is going to make you go crazy

[drifting swift]

uh oh

thanks for the forewarning, i guess...

[sick anvil]

;')

[drifting swift]

first line is already fucked up

-1/x turned into +1/x

bad

not worth reading afterwards

[sick anvil]

the first one is wrong

[drifting swift]

should i repeat my own suggestion again?

[sick anvil]

ill write it down

so a =-1

[drifting swift]

what is a

[sick anvil]

a b and c

for quadratic equations

[drifting swift]

if you decide to stop at this

-1 - 1x -2x^2 = 0
and REFUSE to follow my other QoL suggestion,

then no, a isn't -1.

a is -2.

however by doing this you're also showing that you are a masochist who likes losing minus signs.

[sick anvil]

i used this

[sick anvil]

oh yeah my bad i just realised

[drifting swift]

i would multiply by -1

get 2x^2 + x + 1 = 0

less minus signs to worry about

less minus signs to lose

[sick anvil]

yess

[drifting swift]

less chances to screw up

[sick anvil]

can i try

and show you

[drifting swift]

sure, why not.

[sick anvil]

so first we calculate delta

delta = b^2 - 4ac

= 8

[drifting swift]

are you sure

[sick anvil]

no

[drifting swift]

1^2 - 4 * 2 * 1 isn't 8

[sick anvil]

i made a mistake

-7

[drifting swift]

are you looking for only real solutions or are complex solutions also expected?

[sick anvil]

i dont know what that is but let me show you the formula that i use

[drifting swift]

is sqrt(-1) a thing?

don't show me any formulas.

tell me only this: does sqrt(-1) exist?

[sick anvil]

its equal to 1

[drifting swift]

[EXTREMELY LOUD INCORRECT BUZZER]

[sick anvil]

square of a negative numbre is positive

[drifting swift]

i said sqrt

SQUARE ROOT

la RACINE

[sick anvil]

oui

[drifting swift]

pas le carré

[sick anvil]

c;est pas posible

[drifting swift]

ok donc on cherche seulement des solutions réelles

[sick anvil]

mais la je fait la formule canonique

[sick anvil]

oui

[drifting swift]

une fois qu'on voit que le discriminant est inférieur à zéro

il faut conclure que l'équation n'a pas de solutions

et c'est TOUT

unless you want to do 1000 times more work than necessary or your teacher is a bureaucrat

[sick anvil]

d'accord !! J'avais 2 autre questions es ce que tu peux toujours m'aider ?

ca fait au moins un mois que j'ai fait dexo sur ca

[drifting swift]

i have to go right now sorry

[sick anvil]

oh ok

byee!!

thanks!!

.close

[topaz sinew]

Channel closed

Closed by @sick anvil

Use .reopen if this was a mistake.

[sick anvil]

.reopen

[topaz sinew]

✅

[sick anvil]

so there is a rectangular triangle

abc in a

we know that ab = 8

ac= 6

and i is the middle of the side [bc]

i am supposed to find the vector -> ai . -> ab

and the vector -> ac . -> ai

and lastly the vector -> ci . -> ba

i have the formula but i dont know the angles

<@&286206848099549185>

[fair socket]

Whats the question

[sick anvil]

i have to find the vectors

ai . ab

ac . ai

ci . ba

[fair socket]

What does . Represents here

Multiplication?

[sick anvil]

let me show you

[fair socket]

Ok

[sick anvil]

scalar products

[fair socket]

Oh

[sick anvil]

so we need to find the cos

[fair socket]

You mean you need to find scalar product of vectors

[sick anvil]

yes but i dont know what the cos are

for each

[fair socket]

Is I mid point of bc

[sick anvil]

i have the values of each side

[sick anvil]

yes

[fair socket]

You can prove triangles BAI and CAI congruent

[sick anvil]

for the first one would be 6.5 * 8 * cos of IAB

[sick anvil]

how ?

[fair socket]

Wait let me prove

[sick anvil]

ok

[fair socket]

Is that right angled triangle isoscales

If it is It can easily be proved

Reply something

[sick anvil]

no

only abc

sorry bathroom\

[fair socket]

Abc isoscales

If yes then I am sending proof

[sick anvil]

nonono

[fair socket]

Oh

[sick anvil]

its a rectangle triangle

soryy

[fair socket]

Wdym by a rectangle triangle means the triangle has one 90 degree angle

[sick anvil]

yes

[fair socket]

Oh and angle ABC is 45 degrees

I can see that on the pic u sent

Is that correct?

[sick anvil]

yes thats what i found

[fair socket]

So means it's isoscales as two of its angle are equal

I am sending proof

Wait

[sick anvil]

ok

[fair socket]

Umm sorry my tablet shutted down wait

This is the proof

[sick anvil]

Yes i get it now!!

[fair socket]

Oh that's nice

👍

[sick anvil]

how can i find the last vector?

ci . ba ?

[fair socket]

K lemme send another pic

[sick anvil]

ok

[fair socket]

My handwriting is bad so if don't understand anything ask me

[sick anvil]

i think the difficlyty is to find the cos

[fair socket]

You have already got the angles

Whats the difficulty in finding cos

[sick anvil]

but here its ci and ba they dont have anithing in common

[fair socket]

You can shift vector parallely

[sick anvil]

?

[fair socket]

So we shift vector ci to ba's starting point

[sick anvil]

ohhh

yessss

i get it

[fair socket]

Got it

[sick anvil]

then the angle is?

[fair socket]

Ok

Angles is 45degrees

[sick anvil]

cia?

[fair socket]

So it's cos is 1/sqrt(2)

[sick anvil]

yep!

thank you!
I have 2more difficult exercices.. would you still like to help?

[fair socket]

Yes I can

But I have to sleep so I don't have much time

[bitter pumice]

yo guys

so I hav to graph from the highest point on the graph. Which equation should I use sin or cos?

[fair socket]

This is not your help channel

[bitter pumice]

Aw man

[sick anvil]

sure, let me write it: we have a flat triangle abc with points abc and m n like this

am = 5ab + ca

and cn = 2ab + ac -3bc

they are all vectors

[fair socket]

So what we have to do

[sick anvil]

we have to show that mn = 0

i have 0 clue how to do this

[fair socket]

You mean to show m and n are the same points

[sick anvil]

its a vector

they are not on the triangle

[fair socket]

Ok but you mean to show that mn =0 means m and n are the same points?

[sick anvil]

yes i think

[fair socket]

Wdym by a flat triangle

[sick anvil]

non flat

do you know french by chance?

[fair socket]

No

[sick anvil]

ok its ok

but that isnt flat

[fair socket]

So what it is

[sick anvil]

the surface isnt flat

[fair socket]

I searched it on Google it showed that the flat triangle is when all the angles are equal

[sick anvil]

ok so its not flat

i dunno 😦

[fair socket]

I am getting confused

[sick anvil]

do you know a french person that is good at maths? to explain this?

[fair socket]

Idk

[sick anvil]

ok forget about this line

[fair socket]

Can you post the pic of question

From where u got it

[sick anvil]

its from a french book 😦

[fair socket]

I can use Google translate

[sick anvil]

ok 1 minute

sent

these two questions

[fair socket]

Ok

Let me solve it

[raven sparrow]

Pour le 25, il s'agit d'appliquer la loi de Chasles : $\vec{AB} + \vec{BC} = \vec{AC}$.

[thorny flame]

Azyrashacorki

[raven sparrow]

Le nerf de la guerre c'est de trouver une façon d'exprimer MN en fonction des autres vecteurs.

[sick anvil]

oui

[raven sparrow]

Alors avec la première equation, on sait se rendre de A à M.

Avec la seconde on sait comment se rendre de C à N

Du coup, si on pouvait savoir se rendre de A à C, on serait bien puisqu'on pourrait suivre le chemin MA + AC + CN.

[sick anvil]

comment on fait pour se rendre a am

[raven sparrow]

La première equation te donne ce "chemin",

[sick anvil]

ah oui!

[raven sparrow]

AM = 5AB + CA

[fair socket]

Bye lune

[sick anvil]

oh ok ! byeee thannks for your exellent help!!

[raven sparrow]

D'une part, remarque que si on a le chemin AM, on a immédiatement le chemin MA en prenant l'inverse comme MA = -AM.

[sick anvil]

ouii

[raven sparrow]

Ensuite il y a plusieurs façon d'ajouter à tout ça le chemin AC : je te recommande de prendre l'équation 2 et d'ajouter AC des deux côtés. De cette façon tu te retrouves avec l'équation AC + CN = 2AB + 2AC - 3BC.

On recherche MA + AC + CN.

On a MA et on a AC + CN, du coup tu peux les additionner et réorganiser le tout avec la loi de Chasles?

[sick anvil]

donc c;est ac + cn = ma?

[sick anvil]

?

[raven sparrow]

Ok je reprends : avec la première équation, on peut prendre MA = -AM = - (5AB + CA). La seconde nous donne AC + CN = 2AB + 2AC - 3BC une fois qu'on additionne AC des deux côtés.

Du coup il ne reste qu'à additionner ces deux expressions pour obtenir MA + AC + CN

Ce qui équivaut à MN

[sick anvil]

7ab + ac + 3bc

[raven sparrow]

Attention au signe devant MA.

C'est -5AB - CA que tu additionnes.

[sick anvil]

ok..

je ne sais pas comment additionner

je vois quil y a 2ab et 5ab

et il y a 2 ac et ca

[raven sparrow]

Oui alors si on additionne avec les signes, on obtient MA + AC + CN = 2AB -5AB + 2AC - CA - 3BC.

Maintenant c'est de la simplification.

Tantôt j'ai dit que MA = -AM. Du coup -CA c'est quoi?

[sick anvil]

ac

[raven sparrow]

Exact

[sick anvil]

alors c'est 3ac

[raven sparrow]

Oui

[sick anvil]

donc -3

[raven sparrow]

Pour AB oui

En somme jusqu'à maintenant : MA + AC + CN = -3AB + 3AC - 3BC.

[sick anvil]

oui!

[raven sparrow]

Bien

Donc

Si tu groupes -3AB et -3BC ensemble, tu remarques quelque chose?

[sick anvil]

non...

[raven sparrow]

Et si on l'écrit sous la forme -3(AB + BC)?

[sick anvil]

toujours pas

[raven sparrow]

Utilise la loi de Chasles.

AB + BC = AC

[sick anvil]

ah oui donc c;est egal a cn

ah non

[raven sparrow]

Bah du coup, on a MA + AC + CN = -3AB + 3AC - 3BC = -3(AB+BC) + 3AC = -3(AC) + 3AC

[sick anvil]

qui est egale a 0 non?

[raven sparrow]

Exact

[sick anvil]

et mn c'est ou?

?

[raven sparrow]

Puisque MN = 0, les points M et N sont en fait le même point n'est-ce pas?

[topaz sinew]

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[topaz sinew]

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[sick nacelle]

I found out that A = 1 with x = 0... how do I find out B and C?

[sonic dawn]

plug in A=1 while subbing in x values other than 0

[sick nacelle]

thank you

.close

[topaz sinew]

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[wet sphinx]

Can anybody help me figure out this question

I think the answer is 13 times 9 and the whole thing over 4 factorial

[sonic dawn]

insufficient information

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@wet sphinx Has your question been resolved?

[old moat]

i am sorry i cannot be of help i am not so good in factorial but if you get it you can please post the workings

[topaz sinew]

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[wet sphinx]

Sorry for posting this question without the full information

Essentially, I want to know how to solve Part c

[topaz sinew]

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@wet sphinx Has your question been resolved?

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[winged tide]

a

[winged tide]

What is the scientific name to describe L? because in arabic its literally same word for "diameter" so what is it called in english?

[sweet shard]

Diagonal

[winged tide]

thanks

[topaz sinew]

@winged tide Has your question been resolved?

[topaz sinew]

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[mild island]

if x = log(base4)5 and y = log(base4)3 then express the following in terms of x and y log(base4)225

[topaz sinew]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[chilly walrus]

factor 225

[topaz sinew]

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[hybrid cedar]

@finite trench

man istg if the ticket closes again

[finite trench]

hi

what's the subscript notation supposed to mean

nth order derivative?

[hybrid cedar]

yes

i know how to solve it but i cant prove it

[finite trench]

i assume you have to use an ode

[hybrid cedar]

leibnits theorem formula

this

[finite trench]

yeah i just searched up the formula

[hybrid cedar]

yess

[hybrid cedar]

if the ticket closes dm or ping me please

it is time for jack to go to bed

[finite trench]

alright

sounds like a fun problem

ill try to solve it meanwhile

[hybrid cedar]

i have solved a similar one

without the square

[finite trench]

imma try to solve it another way

[hybrid cedar]

exactly same

[hybrid cedar]

do it your way lol

[topaz sinew]

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[topaz sinew]

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[real saffron]

Can u help me please

[noble laurel]

!noping

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[upper barn]

hi

[upper barn]

i have a exam soon

so can someone help me out on question 5

my friend wrote this

but i dont get how he got the line of symmertry

[gloomy hare]

a quadratic equation is symmetric on the line that passes through its vertex, so x = -b/2a

[upper barn]

ye i realised that

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[primal ice]

can someone confim if this is the answer for part a?

r(a)=3≠m=4 not onto
r(a)=3=n=3 one-to-one

[topaz sinew]

@primal ice Has your question been resolved?

[primal ice]

no bruh this server is acc doodoo whenver i ask my question no one ever answers it

[topaz sinew]

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[blissful hornet]

Im trying to find the nullclines for a dynamical system

[blissful hornet]

I just cant rearrange the first equation to being k in terms of s

how the hell do i do it

[topaz sinew]

@blissful hornet Has your question been resolved?

[topaz sinew]

@blissful hornet Has your question been resolved?

[topaz sinew]

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[frank dome]

Hello

[frank dome]

How does one make a parabola with points

[wary tulip]

what does this mean

[frank dome]

I gotta make a parabola that fits this

Thingie

Photo is loading

There

[wary tulip]

oh

[frank dome]

I got the points on the side

I just

Mmmmm

[wary tulip]

if those points all lie on a parabola, you only need to use 3 of them to determine the parabola

[left maple]

Can you multiply x and i (imaginary number)

[wary tulip]

!occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[left maple]

Oops

[frank dome]

OH THANK GOD I THOUGHT THATS WHAT I HAD TO DO

I WAS LIKE

WHAT IMAGINARY NUMBER

Anyways

[wary tulip]

a parabola looks like y = ax^2 + bx + c for some real numbers a,b,c

since (0,0) is on the parabola, you have 0 = a*0^2 + b*0 + c

i.e. c = 0

putting some other points in will tell you what a and b are

[frank dome]

So like
0=a(7.28)^2+b(7.28)+c

?

[wary tulip]

yes

[frank dome]

But then how’d I find a b and c

[wary tulip]

pick another point and you'll have two equations in a and b

[frank dome]

I’m sorry I still don’t understand

[wary tulip]

here you have 0 = (7.28^2)a + 7.28b

if you pick another point that will give you another equation

a system of 2 equations

would you know how to find a and b from there?

[frank dome]

I think so…?

Ok so

I used 2 points

To make 2.97=-45.1584a-4.48b

[wary tulip]

wait where did that come from

[frank dome]

Don’t you combine the 2 equations?

Or am I thinking of something else

Hello?

Hello?

:((

[topaz sinew]

@frank dome Has your question been resolved?

[frank dome]

<@&286206848099549185>

[wary tulip]

sorry 😭 i forgot about this channel

what’s the other point you used?

[frank dome]

Ummm

2.8,2.97

[wary tulip]

so how’d you get this?

[frank dome]

Hmmm

Well I multiplied one bye -1, then added them

Please don’t forget about me again

[wary tulip]

IM SORRY 😭

[frank dome]

Your back!!!

That’s all that matters

[wary tulip]

if you do smth like that you want to eliminate one of the variables

[frank dome]

Do..do we not want to

[wary tulip]

0 = (7.28^2)a + 7.28b
2.97 = (2.8^2)a + 2.8b

these are the 2 equations right

[frank dome]

Yeah

Third one would be 2.9=(4.4^2)a+4.4b

+c

[wary tulip]

we don’t need a third one

and we already know c = 0

[frank dome]

Ob

Ok

C=0

[wary tulip]

you could e.g. multiply the second equation by 7.28/2.8 and then have a 7.28b in both

and if you subtract one equation from the other you’ll just have a’s

so you can solve for a

[frank dome]

Wait would I multiply the 2.97 by 7.28/2.8 as well

[wary tulip]

yes

[frank dome]

Ok

Hmmm

I did the equation but it doesn’t quite fit

[wary tulip]

[wary tulip]

,w solve (7.28^2)a + 7.28b, 2.97 = (2.8^2)a + 2.8b

[thorny flame]

Results provided by WolframAlpha

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

Requested by slayyla

[wary tulip]

[wary tulip]

ok so

the parabola looks like it passes through the 3 points we used

and that’s the only parabola that passes through those 3 points

so this problem is fucking stupid

[frank dome]

Hmmmmm yeah

[wary tulip]

there is no parabola that passes through every point

in that list

[frank dome]

Oh the list was what I made bc we had to transport the pic from somewhere else

It’s rough estimate at most

But she said we had to have at least 6 points

I did 8 to see the best ones

[wary tulip]

[frank dome]

If my teacher made a legit impossible task I am rioting

[wary tulip]

what was the original task?

[frank dome]

Make the path of the snowball

I mean….it’s pretty close…?

[wary tulip]

were the points on the screen the ones given?

[frank dome]

Nope

[wary tulip]

exactly

T_T

where did they come from

what information was given

[frank dome]

None

Just the pic

On the back

With the dotted line and igloo

And arrows

DID I MESS UP ON THE POINTS?????

Oh god

[wary tulip]

i doubt any of them other than (0,0) lie on the parabola

[frank dome]

But how would I even make sure that they DO lie on the parabola

Cause

She didn’t give us anything

And the pic of the snowball path is movable and stuff

Expandable

[wary tulip]

idk what to tell ya

[frank dome]

Oh boy

Welp

At least I’ll still know what to do if I have to change it

Thanks for your help!

Imma go cry in the shower lol

/ take a longgggg nap

.close

[topaz sinew]

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[neon iron]

this server is not meant to serve as a medium for people to make financial transactions and requests through. Please respect that and delete what you said

.close

[topaz sinew]

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[neon iron]

<@&268886789983436800> in case as well

[craggy haven]

yeah - what he said ^ and also that would be academic dishonesty so please don't!

[topaz sinew]

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[uneven raven]

!reopen

[uneven raven]

What would be the first steps

Do these 2 look correct?

[topaz sinew]

@uneven raven Has your question been resolved?

[topaz sinew]

@uneven raven Has your question been resolved?

[topaz sinew]

@uneven raven Has your question been resolved?

[wide bone]

,rotate

[wide bone]

you need to use foil

so in 9. multiply 5x to x and 4

then -2 to x and 4

ending up with 5x^2 + 20x -2x +8

same principal for #10 @uneven raven

[uneven raven]

So would that be the anwser?

[wide bone]

yes add 20x and -2x for simplification

use the same method on #10

[uneven raven]

Ngl had no clue how to do it and used Mathway and got this but this doesn’t make a single bit of sense how one thing in this problem got there

Ohhh

I looked up foil

Makes more sense now

What would be the best way to solve this problem

[topaz sinew]

@uneven raven Has your question been resolved?

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[devout depot]

I'm currently experimenting with a 2d particle system using C++ and SFML. I've implemented basic physics, particle collision detection, and other such Variables - but now I'm trying to calculate the actual collision between particles. I've found a video that has this equation that in principle, should work if plugged in. My conundrum comes from how the equation plays with itself and the mathmatical language used.

My questions as follows:

1. What does <> mean in an equation?
2. What does || mean in an equation?
3. How do the division parts get summed up, addition? Multiplication?
4. How do the values get applied to a Vector if a Vector is 2 values, but the beginning of the equation is for calculating a single value?

So for instance
Velocity(10x,5y) = Velocity(10x-5y) - ((2 * 5m^2) - (5 + 5)) (???) etc etc

[devout depot]

Would I have to plug in the x and y individually? Or would the extra additions cause errors in the calculation??

<@&286206848099549185>

[topaz sinew]

@devout depot Has your question been resolved?

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@devout depot Has your question been resolved?

[strange whale]

1. <a , b> is the dot product of a and b
2. ||a|| is the length of a, also ||a||^2 is <a, a>, you only need dot products here in fact
3. yes it's multiplication
4. the result of that equation is a vector yes, v1 is a vector, the result hat v1 is a vector

the only number here is that red thing

the rest are vectors

@devout depot

[devout depot]

Cheers! Ah it makes sense now.

.close

[topaz sinew]

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[light brook]

How do I solve that limit without L'Hopital?

[molten vine]

multiply top and bottom by x-1

then use this: $\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=f'(a)$

[thorny flame]

Adam Chebil

[light brook]

Forgot to add no differential calculus has to be used

[carmine pelican]

then you can substitute t as x-1

X-1 tends to zero

[molten vine]

forgot to add that no substitution has to be used

[light brook]

🤣

Yeah it can be used

[light brook]

Thanks I will try that

.close

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