help-10

[autumn adder]

To get a new equatiom

[high lily]

you put 8x^2 instead of 8x^3 into the wolf

[autumn adder]

Equation.

[forest sinew]

oh 😄

,w x^4 + 8x^3 + 20x^2 + 16x -5 =0

[warm shale]

Results provided by WolframAlpha

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Requested by jan Niku#2119

[forest sinew]

something to work backwards from

[short spruce]

factors into quadratics lol

[forest sinew]

yea

[autumn adder]

The imaginary points of intersection will not be present on the real axes graph

[pliant phoenix]

[forest sinew]

thats fine

[pliant phoenix]

Is this is equaled?

[forest sinew]

@pliant phoenix this is like

taking f(x) = g(x)

then subtracting

so f(x) - g(x) = 0

[short spruce]

the first line should say 4x^2 not 4^2

[forest sinew]

this doesnt make sense

[pliant phoenix]

dang

[forest sinew]

but

process is good

i think you may have tripped yourself up

with that 4^2 thing

you should try that again

[pliant phoenix]

how about now?

[short spruce]

where'd the 4x^2 go

[pliant phoenix]

My mistake

so basically

its going overa and being added to the 16

And the seond line is supposed to be twenty

lemme correct

[forest sinew]

hmm

i have the factored polynomial

trying to think of how youd get here though

[pliant phoenix]

This is what the second line down is supposed to be

[forest sinew]

cool

i think i figured out how to do it

well maybe theres an easier way but i think i figured one out

@short spruce are you still helping?

[short spruce]

i still can't find a way to factor it lol, go ahead

[forest sinew]

okay so heres my thought

@pliant phoenix

you look at a couple things

first you have to recognize that we expect f(x) - g(x) = 0 to have 2 real roots

and theyre gonna be different

we only see two places they cross , so we expect distance to onnly equal 0 twice

and theyre different

okay, second thing:

the coefficient of your x^4 term is 1

third thing: 5 only factors a couple ways, so we can guess the ones

our factored equation has the form then

$f(x) - g(x) = (x^2 +bx + 5)(x^2 +dx -1)$

[warm shale]

jan Niku

[forest sinew]

now, it could be -5 and +1

i think you just have to guess

then, you multiply this out

and you equate coefficients

$x^4 + (b+d)x^2 + (5-1+db)x^2 +(5b-d)x - 5 = x^4 +8x^3 +20x^2 +16x -5$

[warm shale]

jan Niku

[forest sinew]

we may extract a linear system of equations from this

$\left( \begin{tabular}{cc|c} 1 & 1 & 8 \ 5 & -1 & 16 \end{tabular} \right)$

[warm shale]

jan Niku

[forest sinew]

for b and d

there is only one solution

$b=d=4$

[warm shale]

jan Niku

[forest sinew]

then $f(x)-g(x) = (x^2 + 4x - 1)(x^2 + 4x + 5)$

[warm shale]

jan Niku

[forest sinew]

this gives you your intersections

😊

i think if you guess wrong

well idk i havent checked since i guessed right

[pliant phoenix]

So its all on my guess ;.;

[forest sinew]

no

the only guess is whether its -5 and 1

or -1 and 5

lemme see what happens

that was a lot

did you track all that

[pliant phoenix]

Okay

[forest sinew]

or is there a place youre confused

[pliant phoenix]

Im reading it over

and understanding

[forest sinew]

or look it over and

yea

no prob

ill check if you guess wrong

hmm

you dont get integer solutions to the system

thats a bad sign

oh

if you guess wrong on factoring -5, the numbers dont work out for the x^2 term

1-5-(some positive number) = 20

not gonna work out

you can restrict sgn(b) = sgn(d) since 1-5+bd=20

i guess db=16

is more sensible

[obtuse pebble]

@pliant phoenix Has your question been resolved?

[pliant phoenix]

Jan imma be honest I am Dead lost with everything have no clue where to start or what to do lol and its due in 50 mins

[forest sinew]

so

we can go piece by piece

if you want

idk how long thatll take

but ping me if u wanna

its 2/3 the points so worth doing

if you wanna spend the time

its really just uhh

1. do f(x) - g(x) = 0

2. we know the form is ( some quadratic )( some other quadratic )

3. you know the leading and guess (pretend you guessed correctly) the constant coefficient in each

4. you multiply it out and equate coefficients

thats like 90% of the problem

[forest sinew]

if you wanna we can work but u gotta say yea

[pliant phoenix]

I got 15 - 20 mins so what ever we can cover

Its g

even If I miss it

[forest sinew]

you need to write out f(X)-g(x)=0

i think thats a key part and you had it, right?

you wrote it somewhere up there ^

[pliant phoenix]

yea

[forest sinew]

okay now its like this

you look at that picture and see them cross in two places, yea?

and then theyre pretty far apart every where else and dont look like theyre gonna cross again

[pliant phoenix]

mhm

[forest sinew]

do you get why this means f(x)-g(x) will be into two quadratics?

like why we know that f(x)-g(x)=0 only has 2 real roots

[pliant phoenix]

Yes I get that

[forest sinew]

okay, so you assume the form

do you get the uhh

coefficient thing?

like getting to here

$(x^2 + bx -1 )( x^2 + dx +5 )$

[warm shale]

jan Niku

[forest sinew]

from where we are now

[pliant phoenix]

[forest sinew]

thats fine lol

its a small leap

look at your f(x)-g(x)

you see the coefficient for the x^4 term?

whats the number out front

[pliant phoenix]

Wym

[forest sinew]

like you have a x^4 term

[pliant phoenix]

the f (x) term?

[forest sinew]

whats the number infront of x^4

no the whole f(x)-g(x) thing

$x^4+8x^3+20x^2+16x-5=0$

[warm shale]

jan Niku

[forest sinew]

whats the number in front of x^4?

[pliant phoenix]

0?

[forest sinew]

0*x^4 = 0

the x^4 isnt 0

its x^4

x^4 times what equals x^4

[pliant phoenix]

1

[forest sinew]

yea

so the x^4 coefficient is 1

look at this and imagine distributing it

theres only one place where we can pick up a x^4 term

its when the x^2 hits the other x^2

we know that when this happens, that we only get one x^4 out

(if this thing is factored correctly at least)

so both x^2 and the other x^2 have to have a 1 coefficient too

does that make sense?

[pliant phoenix]

yup

[forest sinew]

okay

this is also how we know err

how we make a guess that -1 and 5 are the coefficients at the end

its the only place we pick up a constant

all the other terms have a x attached

[pliant phoenix]

mhm

[forest sinew]

now the fun part

you need to distribute

expand this and group it by powers of x

this is the meat of the algebra for this problem, but its just algebra

you need to distribute it, group it by powers of x, then set it equal to x^4 +8x^3+20x^2+16x-5

ive sketched some of it back there ^

[pliant phoenix]

x^4 + (b+d)x^2 + (5-1+db)x^2 +(5b-d)x - 5 = x^4 +8x^3 +20x^2 +16x -5

This one?

[forest sinew]

this looks about right

theres a typo

(b+d)x^3

the second term

it says x^2 but it should be x^3

[pliant phoenix]

x^4 + (b+d)x^3 + (5-1+db)x^2 +(5b-d)x - 5 = x^4 +8x^3 +20x^2 +16x -5

[forest sinew]

$x^4 + (b+d)x^3 + (5-1+db)x^2 +(5b-d)x - 5 = x^4 +8x^3 +20x^2 +16x -5$

[warm shale]

jan Niku

[forest sinew]

algebra is on you though since if you just write this down your teacher is prolly gonna be like what

anyways

we need to equate

do you see which 3 equations we get out of this

we can get 3 equations from this one

all the 3 new ones involve b and d

i can give you one for example, we know that b+d has to be 8

since we have b+d x cubed's on the left

and 8 x cubed's on the right

if you could sketch out these 3 equations and just say you ran out of time i mean

youre basically at the answer at that point

or just write this

and scribble something about "equate" frantically

this is pretty close already

[pliant phoenix]

So the first one is x^4 + 8x^3?

[forest sinew]

no its like

we know the number of x cubed's on the left

it has to be the same as the number of x cubed's on the right

if its not, then they're not the same equation

but we know theyre the same, they have an equals sign between them (we put it there, but still, were in the process of making it true)

[pliant phoenix]

Can U type the first one as an example?

Sorry if its a bother

[forest sinew]

i mean its just the b+d thing

you zero in on x^3 terms

on the left you look for the number in front of x^3

its a variable

b+d

so on the left is b+d many x-cubed's

on the right, there are 8 many x-cubed's

there has to be the same amount of x-cubed's on each side

so then b+d has to equal 8

all this tells us they add to 8, it might be 1 and 7, or 0 and 8, or -1 and 9

we need more equations

so we repeat the process

[pliant phoenix]

5 ad 3?

So Just guess and fill in?

[forest sinew]

it could be 5 and 3, we dont know

no you dont guess

b+d=8 is a line

so what we do is look for another line

and we hope they intersect somewhere

i guess you could guess but its not efficient since there are infinitely many things they could be

you need to equate something else

just like i looked at the x^3 term

and said b+d=8

look at another x to the power term

[pliant phoenix]

(5-1+db)x^2 = 20 x^2 right?

so I need to get numbers that can fit into every sum without messing it up

or they can be different

[forest sinew]

what you said first

we dont know what d and b are

but were getting pieces of the picture here right

like 5-1+db x^2 = 20x^2

lets divide by x^2

so you 5-1+db = 20

so 4+db=20

so db=16

whats two numbers that sum to 8 and multiply to 16?

if you dont like that you even have another equation you can make from x

we know 5b-d=16

so there are ways you can solve these exactly, or you make the not too crazy assumption that the problem is probably designed to have nice numbers so you guess nice numbers

[forest sinew]

and answer questions like this by guess-and-check some single-digit numbers or so

[pliant phoenix]

Shitt Times up tho

[forest sinew]

turn that shit in

ur really close dude

thats a hard problem

[pliant phoenix]

https://tenor.com/view/daredevil-daredevil-dog-tough-dog-red-mask-dog-gif-23418245

[forest sinew]

like really hard

you turned something in right

if not email your teacher that was a lot of work to get no credit

[pliant phoenix]

Yea I sent in the grapph and unfinished sum

[forest sinew]

good job dude

[pliant phoenix]

Bruh It was a group assign ment and a girl was supposed to do this part

[forest sinew]

lol

you should say that

[pliant phoenix]

And she waited until today the day its due

[forest sinew]

tell your teacher

[pliant phoenix]

to say shes stuck

[forest sinew]

you say

look i was almost there

i just ran out of time

i had to do her part

youll get credit

[pliant phoenix]

Alr

[forest sinew]

im gonna go to bed

but congrats for real

[pliant phoenix]

You were awesome tho

[forest sinew]

thats a wild ass problem

you stuck with it

[pliant phoenix]

[forest sinew]

have a good night

lol

dont sad cat

you did good

[pliant phoenix]

Why we gotta learn this type of math to give people injections

[forest sinew]

reasoning

[pliant phoenix]

https://tenor.com/view/coryxkenshin-youtuber-ugh-dismay-gif-17625420

[forest sinew]

suspension of disbelief and confidence

idk

a doctor might say some day give that patient x^53 + 5x^3+29=98 cc of insulin

u never know

alr, night

[mild ocean]

[forest sinew]

[mild ocean]

ah yes x^53 of insulin

[pliant phoenix]

Imma be that doctor and If ii gotta say all that Ill just give em an asprin

[forest sinew]

dont sully me

where were u when the 2 hours of algebra

[mild ocean]

i will sully you, good night cat man

[forest sinew]

come in last minute to shame me

[mild ocean]

yeah

[forest sinew]

it was a candid insulin joke

and im tired

[mild ocean]

[forest sinew]

[pliant phoenix]

https://tenor.com/view/crying-black-guy-meme-sad-gif-11746329

Alr guys

Thanks a lot

[forest sinew]

👋

[mild ocean]

you’re welcome even though i did nothing

darn that was a bad joke

[pliant phoenix]

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[obtuse pebble]

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[ocean cedar]

I have the rotation of an object on the Z axis represented as sin and cos. I would like to turn this into a single value for the user to edit. I guess what I'm asking is how to go from sin and cos to degrees?

edit: solved with atan2

[ocean cedar]

.close

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[thin vine]

Create 3 variations of Polynomial functions in factored form considering the multiplicities.

[urban totem]

<@&286206848099549185>

[sage geode]

What's confusing you here?

[thin vine]

Basically I just need to create polynomial functions in order to graph it

I was wondering if which kind of polynomial function would be easy enough to determine it's zeroes and multiplicities

[sage geode]

Any kind of linear and quadratic function would be simple enough to graph I guess

[thin vine]

How do you determine the behavior of the graph at the roots/ zeros/x-intercepts with regards to its
multiplicities?

[sage geode]

Well for example if you have a root with multiplicity of 2 in a quadratic function

For example (x - 1)^2

Then basically you need to graph a parabola which touches the x-axis at 1

[flat anvil]

e.g.

,w graph (x-2)(x-1)^2

[warm shale]

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Requested by Yottachad#6969

[flat anvil]

see how it touches the axis at x = 1

and passes through it simliar to a linear function at x = 2 (as (x-2) is a factor of mult. 1)

[thin vine]

To be honest, I forgot how to define what a multiplicity is can you help me?

[sage geode]

Informally, if the polynomial has a factor of (x - a)^m, them a has multiplicity of m

[thin vine]

Wait I honestly still don't get it

[sage geode]

For example if you have a polynomial (x - 1)(x - 3)^4, can you see what the multiplicity of 3 is here?

[thin vine]

4?

[sage geode]

Yes

You can see 3 here as a repeated root

[thin vine]

I think I got 4 because I think it would multiply itself 4 times?

[sage geode]

Yes

Since it's raised to the power of 4

[thin vine]

So that's multiplicity?

If it never had an exponent that would mean that the multiplicity is 1?

[sage geode]

Yes

[thin vine]

So for every term that in the polynomial with an exponent should have its own multiplicity?

[sage geode]

Yes

For example when you're solving quadratic equations

Sometimes you see that quadratic formula yields 2 identical roots

So the roots are repeated in that case

And you say that the root has a multiplicity of 2

[thin vine]

In which case, the a and b values?

[sage geode]

When the sqrt(b^2 - 4ac) is 0

[thin vine]

Ah I see

Wait how do you determine the end behaviour of the graph?

[obtuse pebble]

@thin vine Has your question been resolved?

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[drifting jay]

Is it just me or are these questions identical?

[timid silo]

First 2 and next 2 look same

[drifting jay]

Yes

Thanks

.close

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[chilly idol]

Could I get help understanding how to convert this Riemann sum to a definite integral? I'm not understanding it very well

[chilly idol]

this was the original Riemann sum. How did it turn into f(x) = 1/(1+x^2)

[obtuse pebble]

@chilly idol Has your question been resolved?

[raven spire]

Applying this?

[obtuse pebble]

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[frigid ferry]

https://cdn.discordapp.com/attachments/929545200748077056/933983941847838800/unknown.png

[frigid ferry]

is A 75?

[timid silo]

What’s 10sin(65)

[nocturne minnow]

I don't think Matthew will understand how you got to that

[timid silo]

Lol

[drifting jay]

How did you get that?

[timid silo]

10+65

[drifting jay]

Why?

Oh wait you’re not op 😂

[timid silo]

Cause

[frigid ferry]

is that correct

[timid silo]

No

It’s sin65 = x/10

[frigid ferry]

.close

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[timid silo]

Lol

[drifting jay]

Bro that’s still 4.226

[high moon]

earlier you did 9 * 43 = 387 from x/9 = 43
just apply the same idea

[obtuse pebble]

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[timid silo]

so I have to find eigenvalues of a given f in the course they tought us

[timid silo]

$\chi_{A}(\lambda) = \det(\lambda E_n - A) = 0$

[warm shale]

mhD

[timid silo]

but then when I use "matrixcalc.org" to verify my answer I get this used equation

[nocturne minnow]

It's both the same, just the terms are swapped

[timid silo]

alright so they result to the same thing?

[nocturne minnow]

Yes

[timid silo]

okay thanks i might have done something wrong in my calc then

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

turns out I had missed out a couple of things BUT now i am actually stuck trying to understand why -5\lambda while it has to be +5\lambda

matrixcalc returns \lambda^3 + \lambda^2 + 5 \lambda + 3

that damn - is making me loose my crap..

i think where the mistake is... its in my eyes

.close

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[keen jungle]

Hello

[keen jungle]

Anyone that is good with Matrices/Matrix's? I odnt understand what is a matrix row and finding the number of solutions in an equation thanks to the amount of matrix rows

[obtuse pebble]

@keen jungle Has your question been resolved?

[hazy marlin]

do you gave an example of an exercise?

[keen jungle]

ye

[hazy marlin]

well in this particular one there are no solutions.. can you figure out why ?

you say you dont understand what is a matrix row
I mean its just what it is
a row in the matrix,
1432 is a row
0125 etc...

the amount of matrix rows dont help you find the amount of solutions, not really
I think you should go over your noted for this onr first

[keen jungle]

yeah cause the last one is all 0s but still has a value for the 0000 7 but i dont see the use of knowing how many rows there are

Determination of the number of solutions of a linear system using the rank of its augmented matrix

thats the translated name of what im trying to understand but i dont see the use of it

nvm i found why

[hazy marlin]

now that makes for a fuller question lol

[hazy marlin]

alright great

[keen jungle]

Thanks for the help!

[hazy marlin]

I mean.. haven't done much but sure !

[obtuse pebble]

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[timid silo]

Find the number of ways in which we can choose a committee from 4 men and 6 women so that the committee included at least 2 men and exactly twice as many women as men.

[timid silo]

There are two possibilities.
2 men and 4 women.
3 men and 6 women.

[vast wren]

yep

[timid silo]

Now I'm not sure if it should be 4C2 + 6C4 + 4C3 + 6C6 or multiply pairs?

[brave bramble]

There's (4C2)(6C4) ways to make the smaller committee

That is, choose the men and the women, multiply the number of choices together

[timid silo]

Why multiply? Why not addition?

[brave bramble]

Let's say you kept the men the same, but changed the women. That's a completely different committee

So for ALL of the choices of women, there's that many committees

[timid silo]

But it is stated that number of women should be double than number of men.

[brave bramble]

Then, we changed one of the men out, that's a whole extra set of committees for the women again

[brave bramble]

Take an extra read

[timid silo]

Ohh i got you.

Thanks!

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[grim geyser]

<@&286206848099549185>

please?

[obtuse pebble]

@grim geyser Has your question been resolved?

[obtuse pebble]

@grim geyser Has your question been resolved?

[grim geyser]

hello?

or this one please

andthat oe

[obtuse pebble]

@grim geyser Has your question been resolved?

[pine sail]

Hello, which one exactly?

[obtuse pebble]

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[buoyant bobcat]

Heyo, I've been working on a reliable movement for my game for quite some time now, but this has not yet been solved. This is in 3D, so we're working with Vectors. I figured out only yesterday that there is Vector projecting, and I realized that this was VERY helpful for many calculations that I'm doing.

When projecting vectors I've seen that you can project an arbitrary vector onto a plane, by looking at that plane relative to the normal with an orthographic view. What I want to achieve here though is slightly different.

I want to:
Project a vector onto plane A from the orthogonal view of plane B, resulting in a vector that exists only on plane A.

[buoyant bobcat]

(The colours are a bit misleading, but they have nothing to do with the ususal axis colour scheme)
In this picture, plane A would be the red plane, and plane B would be the green one.
I want to project the blue arrow (that exists in on the blue plane) onto the red plane, using the perspective of the green one.

[obtuse pebble]

@buoyant bobcat Has your question been resolved?

[buoyant bobcat]

<@&286206848099549185>

[scenic sundial]

This might be a good CS question @buoyant bobcat

[buoyant bobcat]

CS?

[obtuse pebble]

@buoyant bobcat Has your question been resolved?

[tardy epoch]

do you have equations for your planes?

e.g., the green one looks like it could be z = 5

[buoyant bobcat]

No, I have the normal

[hot sonnet]

I think that you can see a solution by drawing a 2D diagram of this situation

I'll give it a try

[obtuse pebble]

@buoyant bobcat Has your question been resolved?

[obtuse pebble]

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[true pagoda]

would i square first or make 3 posiive first

[short spruce]

bad notation tbh

a calculator would square it first then negate the sign

[true pagoda]

o

also

for this would i add the exponents or multiply the exponents

so ^7 or ^12

[short spruce]

$(ax^b)^c=a^c\cdot x^{b\cdot c}$

[warm shale]

a disappointing son

[true pagoda]

so multiply?

[obtuse pebble]

@true pagoda Has your question been resolved?

[obtuse pebble]

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[alpine forge]

I am confused here

[alpine forge]

So far I've guessed I need to integrate but I don't know why

[random folio]

So have you met the process of separation by variables yet?

[alpine forge]

oh no wait

hold on I wanna try something

[random folio]

Go on

[alpine forge]

hmmm yeah nvm

So I have seen separation by variables but it's been 2 months since then and I need a refresher

[random folio]

Ok so I would start off like this

[alpine forge]

omg

that is soooooo simple

that's all I needed thank you

idk why I didn't think of that. I saw no t in the equation and I got confused. ahahah

[random folio]

No worries, from there it’s just a matter of integrating both sides and finding C before subbing in the value it gives you

[alpine forge]

but of course it's t^0

yup ok thank you!!!

and the +c is only on one side right?

[random folio]

Yeah we normally just put it on the right for convenience’s sake

Bc you get two constants when you integrate both sides but then you can just move the C from the left to the one on the right to make a single constant

That’s essentially what we do there when we just put it on the right hand side

[alpine forge]

yeah it makes sense

because it's unknown constants we can make it into one constant

[random folio]

Yep

[obtuse pebble]

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[timid silo]

Hey I need help for this simultaneous, elimination question

[timid silo]

i’ve attempted it and it isn’t the right answer

[long peak]

for the right side -12b+-12b = -24b, you should subtract one from the other instead

similar for the left side I think

[timid silo]

am i using the wrong signs?

uhoh

[long peak]

28b - -27b = -28b +27b = -b

[timid silo]

OHH

cause i keep using the calc for the negative

i always thought (-x) + (-x) = 0

[long peak]

working looks right apart from a couple of signs though

[timid silo]

thank you for this

so

a = -2?

[long peak]

If you want to check the answer then the easiest way is to use a=-2 to find b then see if those values work in both equations

I think it's right though

[timid silo]

thank you!

b = -5

thank you for helping me fix my work tom

eternally grateful

[long peak]

np

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

Confused

Henlo

Am I doing this right

Hello

Allow me to introduce myself

O

What’s the exact question

[obsidian isle]

what is the original integral

[timid silo]

The second line

Oh

Sorry

[obsidian isle]

what is the original question

[timid silo]

Umm nope

Here lemme

U made a small mistake

13

[timid silo]

Mhm where

I’ll show one sec

[obsidian isle]

u = (2x+3)^1/2

[timid silo]

U forgot to divide 2u by 2

mhm lemme reword this

Now substitute back for U

@timid silo

Ooo

Ty sir

#❓how-to-get-help

This channel is occupied

[willow gust]

👍

[timid silo]

You add a variable to a constant where integrating correct?

Umm u add a constant at the end

Constant of integration

Mhmmm

Alr

.close

[obtuse pebble]

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[willow gust]

Q91

[timid silo]

Post a picture

[timid silo]

,rotate

[timid silo]

Do u know the identity for cos2(theta)

[willow gust]

yeah

should we just divide each by 2?

[timid silo]

No

[sharp granite]

you can't

[timid silo]

That’s not the identity

[willow gust]

then how can we solve?

[timid silo]

Bruh

I just asked to u know identity for cos2(theta )

U said yes

[willow gust]

yeah ik it

[timid silo]

What is it

[willow gust]

cos^2-sin^2

and there are 2 others

[timid silo]

Yes

[willow gust]

so will cos4theta be the same but we multiply by 2 or something?

[timid silo]

Same

Just the half angle will become 2(theta)

cos(a+b) + cos(a-b) = 2cos(a)cos(b)

sin(a+b) - sin(a-b) = 2cos(a)sin(b)

Use these

No what

Actually yeh

These will do better

[willow gust]

👍

thanks for the help

.close

[obtuse pebble]

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[obtuse pebble]

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[winged aspen]

First three seconds
t = 3
s = 26.5

Last 3 seconds
a = 0
t = 3
s = 21.8

s=vt - 1/2(at^2)

26.5 = 3v - 9a/2
3v = 26.5 + 9a/2
v = 26.5/3 + 3a/2

Last three seconds
s = ut + 1/2at^2
s = ut (a = 0)
21.8 = 3u
u = 21.8/3

Sub in
21.8/3 = 26.5/3 + 3a/2
a = -0.93

But the answer is giving me something different

It gives -1.02

[high lily]

,w solve 21.8/3 = 26.5/3 + 3a/2

[warm shale]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[high lily]

,calc -47/45

[warm shale]

Result:

-1.0444444444444

[winged aspen]

....

wait a dam min

welll

.close

[obtuse pebble]

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[timid silo]

I need help with this question

[timid silo]

I know that A + B + C + D + E is summed up to 180°

But I don’t know if I have to use this info

Or how to use this info

[sharp granite]

there are not angles give

not even radius

?

[timid silo]

yes

Only this much info

[sharp granite]

okay so

[timid silo]

This is how much I did

Ik angle APD is 180-(A+D)

And sine of which will be sin(A+D)

And I can use sin law

I can do same for B and E

Not sure how to relate that to each other

[timid silo]

Yup

[sharp granite]

trying to figure out how to approach

[timid silo]

Me too

[sharp granite]

coz we cant know for sure which angle is 90 so cant even start inserting trigs

[timid silo]

Ye

[sharp granite]

lets try to see if any of the angle are equal

[timid silo]

Yeh

Pretty sure they are not

[sharp granite]

so we can at least try to get to some relationship

[timid silo]

Since this is a irregular pentagon

[sharp granite]

yeah

[timid silo]

Just pick 5 random points

Btw can u label the 5 intersection points

[sharp granite]

sure

[timid silo]

Thank you

[timid silo]

Ok thanks

So one this I noticed

Triangle AQB is similar

To Triangle EQD

[sharp granite]

yeah

[timid silo]

So

[sharp granite]

have they mentioned any line being a mid-line?

[timid silo]

We can establish some relationship

Nope

That’s the fun XD

[royal basin]

wait how are AQB and EQD similar?

i see one pair of matching angles but thats it

[timid silo]

One vertically opposite angles in centre

Since they both are straight lines

[sharp granite]

only angle are same

[timid silo]

AQB equals EQD

[royal basin]

yes, i see those angles but that alone isn't enough to claim similarity for the triangles

[timid silo]

2 angles are enough

[sharp granite]

there is just 1

[timid silo]

And there is angles subtended by common chord

[sharp granite]

name?

[timid silo]

BD subtends BAD and BED

BAD=BED

Dayum

[royal basin]

??

ah.

[timid silo]

Yes

Similarly AE subtends the other pair

[sharp granite]

that will only lead to this

[timid silo]

Yeh

And a similarity on side

[sharp granite]

which side?

[timid silo]

AQ/EQ = BQ/QD

By similarity of triangles

[sharp granite]

but we dont have those triangle similar

[timid silo]

We do…

[sharp granite]

let me highlight

[timid silo]

Sure

[sharp granite]

which triangles?

[timid silo]

AQB similar to EQD

We just proved that all 3 angles of both are equal

[timid silo]

Damn this question sucks

Yes

Nvm the similarity didn’t help

[sharp granite]

but still we need to find a right angle

and we cant do that if none of them are mid line

[timid silo]

I mean we can apply sine law

Regardless of 90°

[sharp granite]

we can

but we are here proving

[timid silo]

Proving what

[timid silo]

That’s not the sine law , sine law is within a triangle

[sharp granite]

no i mean

go ahead

[sharp granite]

okay

[timid silo]

Ahhh got it

[sharp granite]

youre right

[timid silo]

Ptolemy theorem

That’s what we need to use

Notice AD and BE are diagonals of a cyclic quadrilateral

ABDE

So we can use Ptolemys theorem

[sharp granite]

we are stupid

[timid silo]

Why

[sharp granite]

we shouldve started with cyclic quadrilateral properties

[timid silo]

Yeh lmao

[sharp granite]

we were looking at triangles

[timid silo]

XD

I’m dum

[sharp granite]

me too hehe

[timid silo]

Let’s do this

Ok so by Ptolemy theorem , we know AE.BD + AB.ED = AD.BE

We can easily show

we know AB/ED=AE/BD

If u want me to prove this I can

[sharp granite]

AD X BE =

[sharp granite]

no i was just making diagram

go ahead

i am reading

[timid silo]

Ok

We know triangle AQB is similar to EQD

By the same logic BQD is similar to AQE

No wait

[sharp granite]

we dont need to go that way

[timid silo]

Why

[sharp granite]

we know that since the points are on a circle

and the 4 points make a cyclic quadrilateral

[timid silo]

Yes

Dayum boi

[sharp granite]

AD x BE = (AE x BD) + (AB x ED)

[timid silo]

Yes

[sharp granite]

shall we divide BE on both?

so we can get to AD/BE?

[timid silo]

Sure

BE^2

Not BE

[sharp granite]

we dividing

[timid silo]

Ik

To get AD/BE we need to divide by BE^2

Otherwise we just get AD

[sharp granite]

oh yeah correct

$\frac{AD}{BE}x\frac{1}{BE^2} = (\frac{AE x BD}{BE^2}) + (\frac{AB x ED}{BE^2})

$\frac{AD}{BE}x\frac{1}{BE^2} = (\frac{AE x BD}{BE^2}) + (\frac{AB x ED}{BE^2})$

[timid silo]

Ayo

[sharp granite]

$\frac{AD}{BE}*\frac{1}{BE^2} = (\frac{AE * BD}{BE^2}) + (\frac{AB * ED}{BE^2})

sorry brain lag

$\frac{AD}{BE}*\frac{1}{BE^2} = (\frac{AE * BD}{BE^2}) + (\frac{AB * ED}{BE^2})$

$\frac{AD}{BE} = (\frac{AE * BD}{BE^2}) + (\frac{AB * ED}{BE^2})$

[warm shale]

moizmoizmoizmoiz

[sharp granite]

good?>

[timid silo]

Yes

[sharp granite]

sine law on what angles?

[timid silo]

Hmm I am thinking

[sharp granite]

ABD and BED?

so we can get AD and BE in terms of sine?

[timid silo]

I’m doing that

[sharp granite]

okie

[timid silo]

Ok got this

By applying sine law

Oh this is getting simpler

[sharp granite]

oh wow

we at least getting somewhere

[timid silo]

Yes

Now im trying to do a nifty trick here

Lemme see if it’s successful

Damn

This question is stubborn

@sharp granite u there

[sharp granite]

HEY

SORRY

[timid silo]

Np

[sharp granite]

what did we end up with?

[timid silo]

Nothing

[timid silo]

?

[sharp granite]

sorry just copy pasting so i dont havbe tp scroll up

[timid silo]

Oh ok

also figured something out

Dont know if this will be useful

But

[sharp granite]

can you write those angle in A B C D E form?

[timid silo]

ABE + BED + AEB + EBD =180°

I figured this much out

[sharp granite]

$\frac{AD}{BE} = \frac{sin(ABE) * sin(BED) + sin(AEB)sin(EBD)}{sin(BDE) * sin(BAE)}$

[warm shale]

moizmoizmoizmoiz

[timid silo]

Also BDE=180-BAE

Hence there sin will be equal

[sharp granite]

oh wow

smart

[timid silo]

I’m getting a gut feeling to do -2 on both sides

Might be wrong tho

[sharp granite]

what grade is this again?

[timid silo]

Pre-college

[sharp granite]

LMAO

which country?

[timid silo]

India

[sharp granite]

bhecnhod

[timid silo]

What happened

[sharp granite]

JEE aspirant

[timid silo]

No

[sharp granite]

hahaha

[timid silo]

ISI/CMI aspirant

[sharp granite]

WOW

[timid silo]

[sharp granite]

abe and bed also add to 180 btw

[timid silo]

No

[sharp granite]

opp angle of similar triangle?

[timid silo]

Wut

[sharp granite]

AQB and EQD similar triangle right?

[timid silo]

Yes

[sharp granite]

yeah so

BAD and BED same angles (similar triangle + angle by same chord)

[timid silo]

Yes

[sharp granite]

hence

[timid silo]

This question is destroying my brain cells

[sharp granite]

mine too

and i started with like 5 brain cells so imagine lmao

[timid silo]

Oomph

Jokes on u I started with 0

[sharp granite]

can we express angle

[timid silo]

-ve brain cells , -69 IQ

[sharp granite]

in terms of ABC whatever

[timid silo]

I’ll try

[sharp granite]

would it be smart to try and test the MCQ in reverse?

like lets just compare on diagram to see if it even makes sense?

[timid silo]

That’s a good idea

[timid silo]

Ooh

I might have got something

[sharp granite]

share

[timid silo]

First I need to write on paper

[sharp granite]

OMFG

WAIT

[timid silo]

What

[sharp granite]

can we sin law to angles

B and D

[timid silo]

?

[sharp granite]

no

i am gonna crey

[timid silo]

No worries

I’ll close for now

I’ll try it later myself

What grade are u in btw

[sharp granite]

i swear to god if its d

[timid silo]

https://tenor.com/view/gordon-ramsay-hells-kitchen-angry-mad-fucking-donkey-gif-17562703 MFW answer is D

[sharp granite]

i am applying for master's

in Ai and Data Sc

[timid silo]

Oooh

IIT ?

[sharp granite]

germany

[timid silo]

Oh ok

Are u in IIT rn

[sharp granite]

nope

are you?

[timid silo]

No

[sharp granite]

you pursuing?

[timid silo]

I just got out of 12th

I’m pursuing ISI/CMI

[sharp granite]

good luck

where you from

[timid silo]

Delhi

[sharp granite]

oh wow

Mumbai

anyways Good Luck man

[timid silo]

Nice

Thanks

[sharp granite]

i will kill someone if D

[timid silo]

Me too

Lmao

[sharp granite]

Geometry and Trig is not my field. i wish i could help better

but we got farther than i expected 😛

[timid silo]

No worries

Yeh

Anyway cya

[sharp granite]

byee

[timid silo]

.close

[obtuse pebble]

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[sharp granite]

.reopen

@timid silo

[sharp granite]

<@&286206848099549185>

<@&286206848099549185>

[timid silo]

yo

im here

@sharp granite

was having lunch

at 6pm in evening

Anyway

[obtuse pebble]

@sharp granite Has your question been resolved?

[sharp granite]

<@&286206848099549185>

[sharp granite]

i thinks it's D bro

[timid silo]

Yeh

Look at #discussion

Bunch of prodigys

[obtuse pebble]

@sharp granite Has your question been resolved?

[obtuse pebble]

@sharp granite Has your question been resolved?

[raven spire]

Ummm :o you might wanna check #help-16 as it's been discussed there already

[sharp granite]

done

[raven spire]

Great :)

[sharp granite]

.close

[obtuse pebble]

Channel closed

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[sharp granite]

@raven spire smart one. 👍

[obtuse pebble]

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[rich holly]

what are these called?

[slow skiff]

Vieta's formulas for quadratic

[rich holly]

the literal translation of those from my native language to english is "roots"

is that correct? Can you call them that too?

[slow skiff]

Yes

[balmy mortar]

what???

are these not just called

sum and product of roots

[sage geode]

It's Vieta's theorem

[slow skiff]

They have a name though

[rich holly]

now this is even more complete

[sage geode]

(special case of a more general thing)

[slow skiff]

Yeah

[rich holly]

thought you called them radicals for some reason lol