help-36

[craggy plume]

so, set them = to eachother

[gritty raptor]

explain further?

[craggy plume]

(perimeter of rectangle) = 4x + 4

[gritty raptor]

yes

[craggy plume]

(perimeter of circle) = 8pi

[gritty raptor]

8pi = 4x + 4

[craggy plume]

yeah

and that's an equation you can now solve for x for, as required

[gritty raptor]

so

hm

how do I solve for for x?

[craggy plume]

when solving for x, or another variable, your goal is to make it the 'subject of the equation', if you have come across this term before

this just means rearrange it using algebra to make it ' x = (something) ' where the (something) has no xs in

how would you begin rearranging 8pi = 4x + 4 to begin this?

[gritty raptor]

hm

how would I begin?

[gritty raptor]

can you give me an example?

[craggy plume]

Sure

lets say you have 3x+8=14

[gritty raptor]

mhm

[craggy plume]

sorry, lets make it simpler at first

lets say x+8=14

[gritty raptor]

mhm

[craggy plume]

you rearrange to x = by moving the 8 to the other side, which is minusing 8 from each side

x + 8 - 8 = 14 - 8

x = 14 - 8

x = 6

so, you have solved for x = 6

what you do to one side of the =, you must do to the other

[gritty raptor]

what do you mean by misusing?

[craggy plume]

taking away

4 - 5 means 4 minus 5

minusing 5 from 4

[gritty raptor]

sorry hold on

$you rearrange to x = by moving the 8 to the other side, which is minusing 8 from each side
x + 8 - 8 = 14 - 8
x = 14 - 8
x = 6
so, you have solved for x = 6
what you do to one side of the =, you must do to the other$

[soft zealot]

Techroz

[gritty raptor]

oop

[gritty raptor]

hmm

Can I get another example

sorry

I still don't understand

[craggy plume]

sure

[gritty raptor]

thanks

[craggy plume]

x - 4 = 5

to get rid of the -4, we +4 to both sides

x -4 + 4 = 5 + 4

simplify

x = 9

[gritty raptor]

ohh

so back to the other example

[craggy plume]

x + 8 = 14

[gritty raptor]

x+8=14 = x + 8 + 8 = 14 + 8?

[craggy plume]

this is correct, but this will not get us anywhere

try and simplify x + 8 + 8 = 14 + 8

[gritty raptor]

so we keep simplifying

[craggy plume]

what do you get?

[gritty raptor]

yes

so

x + 16 = 22

[craggy plume]

Yes

but this is basically what we started with again

its just x + something = something else

[gritty raptor]

okay

[craggy plume]

when what we really wanted to get was x = something

[gritty raptor]

so back to our question

[craggy plume]

what you should have done is

[gritty raptor]

yes?

[craggy plume]

x + 8 = 14

you want to 'move' the 8 to the other side, right?

so only x is on the left side

[gritty raptor]

yes

[craggy plume]

let me try and explain it less math-y

[gritty raptor]

so x = 14 + 8?

[craggy plume]

when you move something to the other side, you 'flip its sign'

so if i wanted to move to the 8 to the other side, it becomes -8

x = 14 - 8

and then 8 moves from left to right

[gritty raptor]

so basically we remove the 8 form the left side and move it right?

[craggy plume]

Yes because our goal is 'x ='

because then it just tells us what x is

[gritty raptor]

and whether its positive or negative?

it doesn't change?

[craggy plume]

that time we had x+8 on left side

to revisit an example where it is x- something

lets say x-4=3

[gritty raptor]

ohh

[craggy plume]

we move the -4 to the other side

[gritty raptor]

x = 3 + 4

[craggy plume]

so it becomes - -4

which is +4

yes

[gritty raptor]

so

8pi = 4x + 4

wait but does it have to only move on the left?

[craggy plume]

left or right makes no difference

[gritty raptor]

I mean right

ok ok

[craggy plume]

this is how to move from one side to the other

you can also see

8pi = 4x + 4

means

4x + 4 = 8pi

so it doesnt matter

[gritty raptor]

yes

so

8pi = 4x + 4 = 4x - 8pi = 4?

[craggy plume]

i am unsure of what you have done

can you explain what you mean?

[gritty raptor]

hmm

[craggy plume]

this is not right, but i may have explained something incorrectly

[gritty raptor]

if we move it on one side

lets say an integer is positive on the right side

if we move it to the left side it becomes negative

[craggy plume]

Yes, and disappears from the right

do you understand why this is?

it is important

[gritty raptor]

why so?

[craggy plume]

lets analyse it in the more math-y way i tried to at first

[gritty raptor]

ok

[craggy plume]

x + 3 = 7 as an example

i have told you you can move the 3 by making it negative on the right

like, x = 7 - 3

[gritty raptor]

oh

[craggy plume]

this is because if we wanted to get rid of the 3 using algebra

we would -3

+3 -3 = 0

and we want x + 0 = x

so, we want to do x + 3 - 3

to get just x

are you with me so far?

[gritty raptor]

I'll work on it

back to the question

8pi = 4x + 4

so we do

so if we move 4x

to the left

it is now negative

[craggy plume]

Yes

[gritty raptor]

but does it go on the right or left side of the 8pi?

[craggy plume]

it doesnt matter

you will either have

8pi - 4x = 4

or -4x + 8pi = 4

they are the same

[gritty raptor]

ok ok

[craggy plume]

8pi - 4x is a little easier to read, though

[gritty raptor]

8pi - 4x = 4

so how do we find the value of x?

[craggy plume]

well remember

our goal is to write 'x='

i havent shown you how to do it if you have a number multiply x, like here

but your start should be the same

try and write 4x=

[gritty raptor]

4x = 8pi - 4?

[craggy plume]

Yes

that is the correct beginning step

Now, do you know how to get from here to x=

ill give you a hint

[gritty raptor]

do we simplify?

[craggy plume]

what do you do to go from 4x to 4?

sorry

4x to x

what operation do you do?

[gritty raptor]

divide by 4

ohhhhhhh

[craggy plume]

Yes

[gritty raptor]

well

[craggy plume]

and remember what you do to one side of the equation, you must do the exact same thing to the other

[gritty raptor]

the answers

say 2pi - 1

that was simplified by

8pi = 4x + 4

wait

hmm

[craggy plume]

you are on the right lines here

[gritty raptor]

x = 2pi - 1

[craggy plume]

very good!

that's absolutely right

[gritty raptor]

thanks

[craggy plume]

hopefully that made sense, however

[gritty raptor]

yes that did

thank you so much for helping

that helped so much!

[craggy plume]

i would recommend you go back and try some more practice examples about algebra like the ones i showed you as examples

[gritty raptor]

yes

sure will

[craggy plume]

you should try and make sure you are an expert at them before doing geometry as they require you to be proficient in them

good luck

[gritty raptor]

thank you

enjoy the rest of your day!

[final saddle]

@gritty raptor Has your question been resolved?

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[minor bluff]

This is not a test

Please help me

I just want to resolve it

The conditoon is to find a20

[cedar ivy]

No one asked

Get a good angle on that photo

[minor bluff]

I dont have that paper anymore

[minor bluff]

yesterday they told me that they dont solve my exercise because they thought it was from a test

[cedar ivy]

How rude

Get value of a_9

[minor bluff]

a1+8r

Right?

[cedar ivy]

No

[soft zealot]

fäf

[cedar ivy]

So what is a9

[minor bluff]

2⁹*8

?

[cedar ivy]

No

[minor bluff]

I dont understand

How from 2an to 22an-1

[cedar ivy]

we have $a_{n+1} = 2^n \times 8$ hon

[soft zealot]

fäf

[minor bluff]

A9=2⁸*8

?

[cedar ivy]

Ok

[minor bluff]

I cant understand that

[cedar ivy]

It's given in the equation of sequence

[minor bluff]

And a20 =2¹⁹*8?

[cedar ivy]

Nope

Because after a9 the formula changes

[soft zealot]

fäf

[minor bluff]

Aaaaa

Thank You man

You helped me a lot

.close

[final saddle]

Channel closed

Closed by @minor bluff

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[neat forge]

Hi I'm doing this partial fraction integral but I'm stuck, Firstly i tried to fin a value of X in -5=A 3(3x-2) that could give me 0 but i can't find it. Secondly I tried with a system but i don't know what values i should put in A+9B= and A-6B=

[paper zodiac]

no

You forgot to equat a=0 and b=0

you almost had it

the 4th part from the top is good

Jus equate a first -> 0

To find b

and do the same to b to find a

replace the new found value to the third equation from the top

then you can integrate it

@neat forge

[neat forge]

Mh wait

[neat forge]

What do you mean?

[paper zodiac]

wait I just saw u were wrong at the 4th part of your solution

the denominators are in the wrong place

I mean the thing besides a and b

not the denominator

[neat forge]

Sorry I'm not understanding

[paper zodiac]

wait

you also have a -3 at the denominator at the start?

Ok tell me how you come up with the factorization and why is there a -3 besides it

[neat forge]

Mh wich - 3?

[paper zodiac]

Cause ur factorization is ok

but where did -3 showed up

I mean 3

[neat forge]

Oh u mean 3(3x-2)?

The starting 3?

[paper zodiac]

Yeah

Like the factor is good and all

But not the 3

[neat forge]

Because i have a formula that says a(x-x1) (x-x2)

Is it wrong?

[paper zodiac]

If u know how to use it

[neat forge]

Oh

[paper zodiac]

It basically says that a scales up the factor of the two

[neat forge]

So i shouldn't use that?

[paper zodiac]

You can if you know how to

i can make it in that form

it would be 3(x+2/3)(x-1)

U can check your answer if it's correct

Just multiply the things on the denominator again and see if you can track back to the original equation

(3x-2)(x+1) <- try to foil it again

and multiply the answer by 3

see if you can go back to 3x^2 -x -2

[neat forge]

Ok but even with this i can't go on because i don't know how to put the system

Like in the numerator i have a - 5

[paper zodiac]

you also got the factorization wrong

just the signs

[neat forge]

Yes but if have 2 equations in a system but only one numerator what i should do?

Both of them equal to the numerator?

[paper zodiac]

You just make two fractional expression

It's just like adding and subtraction of fractions

Since u know that integrals of expressions separated by addition is easy to integrate

Let's go first at the first problem that I had with ur sol

Your factorization

[neat forge]

Yes

[paper zodiac]

just multiply the coefficient of the x^2 and the constant

so 3 x -2

u will get -6

[neat forge]

Yes

[paper zodiac]

find two multiples of -6 that can be added to get the coefficient of the x

3x^2 (-1)x -2

so -1

-3 and 2 are multiples of 6 that when added makes -1

[neat forge]

Ok

[paper zodiac]

so 3x^2 -3x +2x - 2

factor out the common multiples via distributive property

3x(x-1 ) +2(x-1)

check if it's true

via distributive property

3x * x - 1 * 3x + 2 * x - 1*2

3x^2 - 3x + 2x -2

[neat forge]

Ok

[paper zodiac]

-3x + 2x = -1x

3x^2 - x -2

The same as from the original problem

That's how we do the checking

now we're done let's move on

[paper zodiac]

U need to do distributive property again here

What is something common that are both being multiplied to 3x and 2?

[neat forge]

X-1

[paper zodiac]

yes

so factor that out via distributive property

[neat forge]

(3x+2)(x-1)?

[paper zodiac]

yes

that's it

so let's move on the a and b

[neat forge]

Ok

[paper zodiac]

we want to have a separate thing for the two so it's easier to integrate

-5/(3x+2)(x-1) = ( some numerator or a )/ (3x + 2 or a factor from the original equation ) +(some numerator again or b)/ (x-1 or a factor from the original equation)

[neat forge]

Ok

[paper zodiac]

we want to find something to substitute at the top

so -5/(3x+2)(x-1) = a/3x+2 +b/x-1

multiply both sides by the denominator of -5 or the original factored equation

it would cancel out from the left side

How about the right side?

[neat forge]

Yea and I would have 5=A(3x+2) + B(x-1)?

-5

[paper zodiac]

no

Ok

Maybe let's start making 3x+2 and x-1 as other variable

let 3x+2 = g

let x-1 = k

sub all of it

-5/gk = a / g + b / k

Multiply gk at both side

simplify @neat forge

[neat forge]

-5=

Ak+bg

[paper zodiac]

yeah

and look at these

see the problem?

See what's wrong with ur answer?

[neat forge]

Oh A(x-1) B(3x+2)?

Like here i shouldn't find 2 value of X that gives me A=0 an B=0?

[paper zodiac]

See

Do that one by one

First let A = 0

[neat forge]

That's the problem

Like i put - 1

But when i have 3x - 2 how do i get 0

[paper zodiac]

no no

Just make A = 0

You just need to find what will happen to b if a =0

do the same thing vice versa

[neat forge]

But if a=0 i have (3x+2) +B(x-1)

[paper zodiac]

Oh wait

Mb

I got it the wrong way

Let x = something that makes a = 0

mb

Forgot about that

Let's focus on A(x-1)

[neat forge]

+1

[paper zodiac]

Yeah

[neat forge]

And I get B=-6?

[paper zodiac]

so let x = 1

if that's what u got

lemme check

[neat forge]

Oh

B=-7

[paper zodiac]

nahj

[neat forge]

Oh nl wait

[paper zodiac]

get our equation earlier

the whole equation

[neat forge]

B(3x+2)

[paper zodiac]

nah the one with a and b and the left side

[neat forge]

-5=A(x-1) + B (3x+2)

[paper zodiac]

let x =1

get B

then do that also with the x beside B

to get a

[neat forge]

B=1?

[paper zodiac]

close

almost

[neat forge]

O 0

Like 0

[paper zodiac]

no

u forgot the sign

[neat forge]

Oh - 1

Yea i have - 5

[paper zodiac]

see

yeah

[neat forge]

Thanks

[paper zodiac]

do that to the b also

[neat forge]

Here is the problem

[paper zodiac]

make b = 0 to get a

[neat forge]

So i just have 3x+2?

[paper zodiac]

make it to 0

Like what we did at a

[neat forge]

I don't know how to make it

That's one of the main problem

[paper zodiac]

if we make that to 0, and we multiply it to b, it would become 0

since anything x 0 = 0

3x+2 = 0?

[neat forge]

X=-2/3

[paper zodiac]

yeah

let x be that

[neat forge]

Ohhh

[paper zodiac]

to get A

Then the values of a and b would be substituted at the following equation we made earlier

the a/(blah blah) + b(something)

Of course I want to know if u really understood what I said

[neat forge]

But with 3(-2/3) - 2 wouldn't i get - 4?

[paper zodiac]

no

that's not ur equation

where did -2 came from?

[neat forge]

Oh it's +2

[paper zodiac]

<-

[neat forge]

Sorry

Ok now it's 0

[paper zodiac]

u forgot the to make the change to all of them

that's is not just the whole equation

[neat forge]

?

With 3(-2/3) +2 i get 0

[paper zodiac]

u forgot the left side, a and b

[neat forge]

Ohh yes

A=3

With x=-2/3

[paper zodiac]

no

A is not 3

try it again, there might be some problems with ur calculations

ey wait

nvm im wrong

[neat forge]

It's impossible?

[paper zodiac]

no

ur correct

I did -3/2

Made a mistake

[neat forge]

I also made a mistake

[paper zodiac]

bro don't say that after we made it this far

[neat forge]

I think that's - 5=x

[neat forge]

Nono i mean the a equation

[paper zodiac]

why

[neat forge]

3*-2/3=-2

-2+2=0

[paper zodiac]

yeah

seems correct

[neat forge]

Oh wait

I need to find B

My bad

[paper zodiac]

no it's a

u already did

a = 3

we already have a and b values

sub it from this one

[neat forge]

B=-1 A=-3

[paper zodiac]

this one but without the g and k

[paper zodiac]

a has the wrong sign

[neat forge]

B=-1 A=3

So

[paper zodiac]

yeah

ah u only need the right side from the g and k

[neat forge]

3/x+1 + (-1/3x+2)

And i have to integrate both of them

[paper zodiac]

yeah

[neat forge]

The first one is Ln(3x+1)

[paper zodiac]

I dunno you decide

I'm about to have my morning breakfast

[neat forge]

3Ln(3x+1)

[paper zodiac]

Bout to go hibernate mode

but it's easy

m,aybe go to symbolab to get the final answer

[neat forge]

Just the last question -1/3x+2 is still a Ln?

[paper zodiac]

lemme check

[neat forge]

Like if i integrate it

-Ln(3x+2)?

[paper zodiac]

nah fam

make 3x+2 = u first

du = 3dx

[neat forge]

Oh i use u

[paper zodiac]

so 3dx at the top is du

so 1/u

du

[neat forge]

Ohh okk

[paper zodiac]

so ln|u|

sub u back

[neat forge]

Yea got it

[paper zodiac]

ln |3x+2|

[neat forge]

Thank you so much for your help

Sorry for taking so much of your time

[paper zodiac]

np gtg

imma eat some pandesal now

goodbye

[neat forge]

Thanks again

.close

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• Show your work, and if possible, explain where you are stuck.
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[tranquil pine]

Okay, So, I need help with this question

[tranquil pine]

And I don’t really understand it at all

Because it doesn’t make sense

Hello? Anyone?

[stable fossil]

,rotate

[tranquil pine]

Hello?

Anyone?

Sorry, I need to know how to do this urgently

Please <@&286206848099549185>

[muted prairie]

2+r 2+s 2+t and 2+u are the roots of a different polynomial related to f(x)

so the first question is how do you transform a polynomial to increase all of it's roots by 2

[tranquil pine]

Multiplying it by 2?

I am guessing that’s incorrect

[muted prairie]

nah, that would keep them all the same

think about it more graphically

[tranquil pine]

Double the Constant?

Since that would shift it

[muted prairie]

that would shift it up

which would change the roots pretty unpredictably

the roots correspond to intersects on the graph, right

so if the roots are increasing what's happening to the intercepts

[tranquil pine]

I am not sure

They are also Increasing?

That’s what I would think

Since it’s intersecting at a higher point

[muted prairie]

the y-value of the intercepts are always 0

[tranquil pine]

Yeah, so the Y - Intercept wouldn’t change

[muted prairie]

?

i'm talking about the x-intercepts

the points where y = 0

[tranquil pine]

Yeah

Those would increase right?

[muted prairie]

yeah

we're trying to increase those

the roots

but they wouldn't get "higher"

they'd stay at y=0

[tranquil pine]

Yeah

I get that

But, How would you transform the polynomial for that?

[muted prairie]

so what happens to the x-intercepts when the roots increase by 2

[tranquil pine]

The X - Intercepts increase by 2

[muted prairie]

how does a point "increase"

[tranquil pine]

You have to shift it

That’s what I don’t know

[muted prairie]

do you know how the x-intercepts correspond to the roots

[tranquil pine]

Yeah, the X - Intercepts are the roots

[muted prairie]

the x-intercepts are points though, right

[tranquil pine]

Yes

Well, it depends

[muted prairie]

and the roots are numbers

[tranquil pine]

Yeah, the Points

Like if I have (2,0) as a root

Then I know that 2 is the root

[muted prairie]

as an intercept

(2,0)

so you take the x-values of the x-intercepts, right

[tranquil pine]

Yes

[muted prairie]

so if i increase a root by 2

the x-value of the x-intercept increases by 2

which means it moves right 2 units

[tranquil pine]

Yes

Exactly

Just show do you shift it 2 units

[muted prairie]

yeah

[tranquil pine]

How do you change the polynomial

[muted prairie]

you move it right 2 units

[tranquil pine]

To do that

Yeah, But how do you do that Algebraically

I don’t mean graphing

And then using those

Roots and multiplying them with the X

[muted prairie]

horizontal and vertical shifts are usually covered somewhere in school

but you'd replace x with (x-2) in the polynomial

[tranquil pine]

Yup

I know

[muted prairie]

then re-expand everything out

[tranquil pine]

Yeah

I know

So, You are saying to get the roots, Add 2 and then Multiply?

Because the problem is it’s not Factorable

[muted prairie]

shift the polynomial right by 2

[tranquil pine]

Ik, HOW

[muted prairie]

replace x with x-2

[tranquil pine]

So, 3(x-2)^4 and then do that for each one

[muted prairie]

yeah

[tranquil pine]

And then solve it?

[muted prairie]

no, it'll still be unfactorable

then you have a polynomial that you want the product of the roots of

so use vieta's

if you don't know vieta's try looking it up

[tranquil pine]

Is they any other way to do it?

[muted prairie]

a much harder way where you expand (2+r)(2+s)(2+t)(2+u) and try to write it in terms of the symmetric polynomials findable through vietas

or using a calculator that can calculate the roots directly

other than that no

[tranquil pine]

Could you graph it?

[muted prairie]

,w plot 3x^4-x^3+2x^2+7x+2

[soft zealot]

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[tranquil pine]

I only see 2 roots

Don’t I need 4?

,w find roots of 3x^4-x^3+2x^2+7x+2

[soft zealot]

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[tranquil pine]

Okay, so it would be complex?

[muted prairie]

the roots would be yeah

[tranquil pine]

Okay

Thanks

[muted prairie]

np

[tranquil pine]

But, how would I multiply them?

Even if I don’t know what the roots

Are

But know what the polynomial is

[muted prairie]

i told you

vieta's rules

,w vieta's rules

[soft zealot]

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[muted prairie]

ah whatever they're everywhere online

[tranquil pine]

I searched it up

But, I don’t understand it

[muted prairie]

then watch a video

or work on the parts you don't understand

is there any notation you don't understand

[final saddle]

@tranquil pine Has your question been resolved?

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[tranquil pine]

.reopen

[final saddle]

✅

[tranquil pine]

Um, @muted prairie I just realized, Couldn’t you just sub in x = -2?

That would work

[muted prairie]

for what?

[tranquil pine]

Into the original equation

[muted prairie]

how would knowing f(-2) help

[tranquil pine]

Because, since it’s (r+2) for each root

[muted prairie]

if it was a root, that would be very helpful for sure

but it's not

[tranquil pine]

No, But you can sub in -2

And get the answer

[muted prairie]

ok continue

[tranquil pine]

Then, Each root you are shifting by -2

So, you might as well put x as -2

[muted prairie]

by plugging in -2?

[tranquil pine]

Yeah

[muted prairie]

why might as well?

what does it tell you

you're not shifting roots by plugging in -2, you're just turning a polynomial into a number

[tranquil pine]

The answer to (r+2) and so on

[muted prairie]

there's no reasoning process here

[tranquil pine]

Give me a Second

[muted prairie]

what if the polynomial is f(x) = x

and I want you to tell me (r+2) where r is the root

by your logic you plug in -2 for x

f(-2) = -2

thus r+2 = -2, which is false

[tranquil pine]

You recognize that
f(x)=(x-r)(x-s)(x-t)(x-u)
So when you sub in x=-2 you get
(-2-r)(-2-s)(-2-t)(-2-u)
Which is equal to the desired product

The negatives cancel out

@muted prairie You get it?

[muted prairie]

f(x)/3 is that

[tranquil pine]

Wdym?

[muted prairie]

f(x) has leading coefficient 3

so it can't be that product

so you'd divide it by 3 first

[tranquil pine]

No, You can plug in -2

It would just cancel out

[muted prairie]

what would cancel out

[tranquil pine]

The negatives

So, You would get left with what you desire

[muted prairie]

oml

[tranquil pine]

I literally checked it

It works

As I explained

[muted prairie]

f(x)=3(x-r)(x-s)(x-t)(x-u)

NOT f(x)=(x-r)(x-s)(x-t)(x-u)

[muted prairie]

This is false

So you need to divide f(x) by 3 first

[tranquil pine]

NVM, Yoy are right, but then you just divide by 3, then sub in - 2

Or vice versa

[muted prairie]

yes that would work

[tranquil pine]

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[robust knoll]

Can someone help me find the taylor series of this function? I understand the general method to find it I think, I get the taylor polynomial and try to find patterns and build off of that. What i got for the polynomial to the 3rd order is 0 - (x-pi/4)/1 + 0 + (x-pi/4)^3/3!

[robust knoll]

im just having trouble getting a pattern off of that

ignore the answer i entered, i used a calculator which gave me the macluarin series of it

and then i thought i could just add the c

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@robust knoll Has your question been resolved?

[robust knoll]

alright i made a major mistake, i plugged in pi not pi/4 so i will redo the problem see if im still stuck

any help still appreciated

okay this is the answer, only trouble was finding what power of (-1) will get the sign to be like + - - + repeating

i had to search it up

hopefully i remember that if i have to do it again in the future

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[tranquil pine]

Hey

[tranquil pine]

.close

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[jaunty fable]

I'm wondering how they expand it to $24\cdot5\cdot6\cdot...\cdot n$ etc

[soft zealot]

gnu

[jaunty fable]

where does the 24 and 16 come from

and why is 4 important such that n-4 factors is significant

[deft ravine]

I guess because 3! < 8

So 1/(3!) > 1/8. But we want to bound above

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[jaunty fable]

Ohhh I think I see

Thank you

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[harsh moat]

Hi, i don't understand the last passage

[harsh moat]

i find this

But why it becomes so?

[tranquil pine]

Because of rationalisation

The fraction is amplified with what is on the denominator.

[fickle crater]

https://youtu.be/EgYlGXnq8z8

@harsh moat

[harsh moat]

oh i see

Thank you very much!

[tranquil pine]

If I remember correctly bprp said that we rationalise for an easier division

[harsh moat]

idk i i forgive it lol

[tranquil pine]

It's easier to divide by a rational number rather than an irrational one.

[harsh moat]

I'm sorry

[fickle crater]

yes sir

[harsh moat]

Yeah

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[tranquil pine]

I need help solving for all values of theta here

[lethal estuary]

use an identity so you can work with only sin

[tranquil pine]

could we change cos^2 theta to

1-sin^2 theta

[lethal estuary]

yes

[coral kindle]

you can

[tranquil pine]

Should I just divide both sides by 1-sin^2 theta

to just get rid of it?

[lethal estuary]

how would that help?

[tranquil pine]

just helps get rid of the squared sine

not too sure haha

[lethal estuary]

you can try it and see that it doesn't get rid of it

[tranquil pine]

oh

ok

what should i do from there then

[tranquil pine]

referring to this stage

[lethal estuary]

what do you think you should do next?

[tranquil pine]

distribute the 4?

[lethal estuary]

yes

[lethal estuary]

ok

[tranquil pine]

i think i factor by grouping from here maybe?

[lethal estuary]

grouping what?

[tranquil pine]

like

not rlly grouping

im essentiually solving for the zeros here

is what i mean

[lethal estuary]

yes

[lethal estuary]

looks right

[tranquil pine]

sin theta equal to -1 and sin theta equal to -1/4

i think i made a mistake here

i dont think it can be negative

[lethal estuary]

why not

[tranquil pine]

oh nvm

nevermind

i suppose we take the inverse of both?

[lethal estuary]

yes

[tranquil pine]

i remember someone told me that you shouldnt put the negative sign when taking inverse

is that true?

[lethal estuary]

what do you mean?

[tranquil pine]

like

when i do sin inverse 1.4

1/4*

should it be sin inverse 1/4 or -1/4

[lethal estuary]

if you have $\sin{\theta} = -1$, then you must have $\theta = \sin^{-1}{-1}$

[soft zealot]

cwatson

[tranquil pine]

ok

makes sense

[tranquil pine]

got -14.37

and

-90

[cinder linden]

?

how on earth did you get -14.37

[tranquil pine]

did sin^-1(-1/4) on my calc

and got -14.37

[lethal estuary]

,w calc arcsin(-0.25)

[soft zealot]

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[tranquil pine]

i did it in degree

[cinder linden]

,w -0.25*180/pi

[soft zealot]

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[cinder linden]

ah

[tranquil pine]

mhm

so uhh

what do i do from there

[lethal estuary]

plug in your solutions to the original equation to check that it's right

[tranquil pine]

its correct

are the answers just -90 and -14.37

[lethal estuary]

yes

well

i suppose for -pi/2 you could figure out all the integer multiples or whatever

[tranquil pine]

it says to find all values of theta to the nearest tenth of a degree in 0 is less than or equal to theta <360 degrees

[lethal estuary]

oh, then it should probably be 270 and 355.63 or whatever

[tranquil pine]

okay i see

thank you

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[tranquil pine]

hey

[tranquil pine]

quick question

when finding the degree values

sin theta is negative

meaning the answers would be in quadrants 3 and 4

and since one value of theta is 90 degrees

for quadrant 4

[lethal estuary]

no, it was -90, or 270

[tranquil pine]

would the value be 270+90

or 360-90

[tranquil pine]

oh i see

so it must be 360-90

or 270

and 360-14

[lethal estuary]

yes, the other would be 345.xxx

[tranquil pine]

i see thx

thanks again for the help

very helpful

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[fiery prism]

can someone help me with d

[strange shore]

solving for n

when P = 2000

[fiery prism]

yea

but idk what to do after that

[strange shore]

$$a^n = b$$
$$\log(a^n) = n\log(a) = \log(b)$$
$$n = \frac{\log(b)}{\log(a)}$$

[soft zealot]

Ｈｅｒｅｌｓ

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[pine oxide]

BAC is a right triangle
length of BD = 10
angle BDC = 80°
angle ACB = 40°

it asks the length of AC

[amber holly]

Do you know the law of sines so far?

[final saddle]

@pine oxide Has your question been resolved?

[pine oxide]

this problem is to be solved without using sines

[amber holly]

Huh, weird

What theorems can you use/What toрic is this?

[pine oxide]

topic is right triangles

I can use hypotenuse

euclidean theorem

[amber holly]

Can you use similarity of triangles?

[pine oxide]

yes

[amber holly]

Ok great

[amber holly]

Have you solved for all the angles so far?

Because ABD is a special triangle

[pine oxide]

how ?

[amber holly]

Consider the angleCBD at first

What is it equal to?

[muted prairie]

Ah the classic 30-50-100 triangle

My favorite

[amber holly]

Hm?

[pine oxide]

40

[amber holly]

Oh wait damn

[muted prairie]

truly nothing if not special

[amber holly]

I forgot we have only one right angle

[muted prairie]

are you allowed to use trig

[amber holly]

They said they can’t use sines

[muted prairie]

like, AB = AC*sin(40)

[amber holly]

Law of sines was my first instinct

[muted prairie]

Not law of sines though

Just basic trig

[amber holly]

Yeah, but still

[muted prairie]

well you could use cosines instead

[amber holly]

@pine oxide can you use any trig?

[muted prairie]

sohcahtoa in particular

[pine oxide]

no

only this one like "ab * bc * 1/2 * sin90"

[muted prairie]

...

well you can derive law of sines from that

[pine oxide]

how do u find sin40 then

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[pine oxide]

.reopen

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✅

[muted prairie]

idk

you can solve it in terms of sines and then cancel everything at the end if the only restriction is no calculator

[final saddle]

@pine oxide Has your question been resolved?

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@pine oxide Has your question been resolved?

[pine oxide]

<@&286206848099549185>

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@pine oxide Has your question been resolved?

[final saddle]

@pine oxide Has your question been resolved?

[pine oxide]

. without using trig

[stuck tide]

I don't think that's possible

At some point it will be needed

The answer is: sqrt(3) * 5 / sin40 + 5 / sin50

But not sure how it should be simplified

[stuck tide]

Got it by using the sine theorem

[pine oxide]

not possible

[stuck tide]

Which part?

Sine 40?

[pine oxide]

there must be a solution using either muhteşem üçlü or
euclidean theorem

sine40 is not an option

[stuck tide]

not sure

[stuck tide]

Here you go!

It was indeed solveable without finding the sin40

You get 20 at the end

Called s = sin(40) and c = cos(40)

Not sure about the solution that uses euclidean theorem

[pine oxide]

how did u use that BD = 10

[stuck tide]

I drew a perpendicular line. Let's call that BF.
BF/BD = sin(80) = 2sin(40)cos(40)

So therefore;
BF/10 = 2sin(40)cos(40)

And then I did tangent on both splitted right triangles

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[hot dove]

I would like help on this problem

[tired walrus]

do you know how to find the area of a rectangle from its base and height

[hot dove]

Yes

[tired walrus]

ok then tell me

how do you find the area of a rectangle from its base and height?

[hot dove]

Ok so the area of a rectangle is base x height

The base being its bottom length I guess

And height being it’s well height

In the example

They give the area and b

Which is 6x^2 + 24x - 6

And 3

[tired walrus]

don't use the letter x as a multiplication symbol.

[hot dove]

Ow my bad

[tired walrus]

anyway

set aside this problem for a moment

if you know the area and base of a rectangle how do you find its height?

[hot dove]

Simple you make an equation to solve for h

The equation being a/b = h

[tired walrus]

uppercase A, to be more concrete.

okay, so why not do that in your problem?

[hot dove]

I did but I got radicals

[tired walrus]

?????

how the FUCK did you get radicals. what did you do.

how does dividing (6x^2 + 24x - 6) by 3 produce radicals

[hot dove]

I think I solved for x

[tired walrus]

planning error!

[hot dove]

Wait so we just divide it and call it a day?

[tired walrus]

why WOULDN'T we

[hot dove]

Idk guess I was overthinking it

[tired walrus]

you definitely were

[tranquil pine]

how would you solve for x lmfao

[hot dove]

Don’t know

Im a dumbass so who knows

[tired walrus]

by setting the result equal to 42069 i guess

[tranquil pine]

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[vapid kernel]

does anybody knows how to do this

[untold swallow]

??

[vapid kernel]

ignore the workings i couldnt the answer with what i did

[untold swallow]

,rotate

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@vapid kernel Has your question been resolved?

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@vapid kernel Has your question been resolved?

[iron cove]

I think if you use what they give you on the remainder thing it should fall out

[vapid kernel]

oh mai gad senpai again

[stuck tide]

If 2x - 1 is a factor then if you make it equal to 0.
2x - 1 = 0
x = 1 / 2
Then,
f(1 / 2) = 0
a/8 + b/2 = 1
a + 4b = 8

The remainder of f(x) / (x - 2) is f(2)
The remainder of f(x) / (x + 1) is f(-1)

The question states that 2f(-1) = f(2)

-a + 4 - b - 2 = 8a + 16 + 2b - 2
-a - b = 12 + 8a + 2b
9a + 3b = -12
3a + b = -4

Now combine.
a + 4b = 8, so multiply by 3, 3a + 12b = 24
3a + b = -4

Subtract top to bottom 11b = 28
b = 28/11

This should be correct

[vapid kernel]

whoa

let me check

[stuck tide]

Might just be wrong

[vapid kernel]

unfortunate

the answer for this is a=-2, b=2.5

[stuck tide]

Then something was wrong

[vapid kernel]

i think you are right

[iron cove]

I think I got the same

[vapid kernel]

for the 2f(-1)=f(2) part

[stuck tide]

Logically it seems to be correct

[iron cove]

Yeah

[vapid kernel]

the answer shouldnt be wrong

im trying to check the working

[iron cove]

Both our and their a and b solutions do give 1/2 as a root

[vapid kernel]

wait so the answer is wrong?

[iron cove]

Their answer gives 1/2 as a root

Idk if I can be bothered checking the remainder thing

[vapid kernel]

so uh after checking it myself