help-36

[magic rampart]

so now that u have the side length, solve for the perimeter first

[wind bough]

ohh

[magic rampart]

then input it into the formula

[wind bough]

its 253.76 right

the perimeter

[magic rampart]

yes

actually

i got 261.78

u type into calculator wrong maybe?

[wind bough]

i did 8 * 31.72

[magic rampart]

oh sorry

32.72

[wind bough]

ohh

[magic rampart]

mb

[wind bough]

so its 5170 right

[magic rampart]

yes perfect

its a bit of a pain

since only given apothem

but for questions like that use the apothem to work out the side length

then apply ap/2

[wind bough]

okay thank you so much for your help 🙂

[magic rampart]

all good

if you don't have anymore questions .close

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[sacred coral]

There's no way to add to 6 and multiply to -5

I'm trying to get my x intercepts

This is for graphing parabolas

[onyx peak]

note the -x^2

[final tangle]

note that the coefficient of -x^2 is -1 rather than 1

[onyx peak]

also this equation should be 0 = -x^2 + 6x - 5

[sacred coral]

Yea I see that, do I just factor out the negative for the x intercepts?

[onyx peak]

sure, or you can directly divide the equation by -1

0 = -x^2 + 6x - 5

this equation

[sacred coral]

And this doesn't effect the vertex part or y intercept?

This is only for the x intercepts?

I feel like that'd be too far fetched

[onyx peak]

it would make the y coordinate of vertex different

but for x intercepts, you are just solving equation 0 = -x^2 + 6x - 5

and you can multiply both sides of the equation by constant -1

doing that, you get -1 * 0 = x^2 - 6x + 5

but -1 * 0 is just 0

so it becomes simply 0 = x^2 - 6x + 5

[sacred coral]

Oh I see

[sacred coral]

How would it effect them? Like I already have the vertex as (3, -4)

But that was before I was solving for the x intercepts

[onyx peak]

yes, thats correct

it would only affect it if you did the multiplication by -1 during the process of finding vertex

but since you're only using the modified equation to find x intercepts, its fine

[sacred coral]

So would this be the correct way to find the x intercepts?

And then does y intercept remain -5? Or change to 5 after the thing to find the x intercepts

[onyx peak]

not quite

3 + -2 isn't 5

[sacred coral]

Oh yea

[onyx peak]

Also it should add to -6 and multiply to 5

[sacred coral]

So that'd be -5, -1 then

And the x intercepts would be (5, 1)

Sorry this is taking so long to get 😞I have a test on this in a few hours

But would this be the correct final result?

[final saddle]

@sacred coral Has your question been resolved?

[sacred coral]

@onyx peak bud

@helper

<@&286206848099549185>

[fathom walrus]

@sacred coral

what are u trying to find

just intercepts?

,w y = -x^2 + 6x - 5

[soft zealot]

Results provided by WolframAlpha

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[fathom walrus]

there u go

[sacred coral]

Thanks

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[fiery prism]

just making sure am correct

[fiery prism]

i just need to find the "cable"'s hypotenuse correct?

[sleek quartz]

yes

[fiery prism]

thank you

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[ancient monolith]

how do u do part a

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[wise light]

this not the same as -sqrt2 - 4 , sqrt 2 -4 ??

[silver roost]

You forgot the plus sign

[wise light]

where?

[whole pecan]

-4+sqrt2

[wise light]

thers no -4 + sqrt2

[whole pecan]

That’s the same

[topaz elbow]

the different answers ^

you gave -4√2 as your second root

[topaz elbow]

√2 is the same as +√2

[wise light]

if it was

-4 - sqrt 2, -4sqrt2 its the same right?

[solar glade]

No

Sqrt2 -4 is equal to -4+sqrt2

Not -4sqrt2

[wise light]

oh okay

i get it

yeah okay

how about this one

i did 4x^2 +10x = -15 so far is that right

[solar glade]

You should start by dividing by 4 both sides

[silver roost]

No 😭

[solar glade]

You want to complete the square so, unless you do this everyday u better let the coefficient of x^2 as 1

[silver roost]

Easier to let it be 4 imo but u do u

[wise light]

0/4 = 0

[solar glade]

Correct

[wise light]

u saying divide just the x^2 by 4?

[solar glade]

And the other side?

[wise light]

becuse the 10x and -15 is not /4

divisible

[solar glade]

You let as a fraction

[wise light]

okay

[wise light]

x^2 +10x/4 = 15/4

x^2 + 5x +25 =

[solar glade]

What did u do?

[wise light]

10x/4 x 2/1

[solar glade]

You have to do like this. (x+10/8)^2 - 100/64 = 15/4

[wise light]

the way u do it is complicated

cant we just take out 4x^2

[solar glade]

Try your way and show how do u complete the square

[silver roost]

Bru

[silver roost]

You can’t multiply things unless you do the same thing to both sides

You’ll be changing the equation

[solar glade]

Correct

Whatever you do in one side of the equation you have to do in the other

I recommend you do my way, it is very easy once you know how to do it

[wise light]

okay

x^2 + 5x/2 = 15/4

[solar glade]

Ok

Now complete the square in the left side

[wise light]

okay wait

[solar glade]

No wait

Wait

Ok do it

[wise light]

i want to divide 5x/2 by 2

and square it right

[solar glade]

Do it because the way u explain is incorrect

But maybe u mean the correct

So do it and i see

[wise light]

take half of 5x/2

and square

?

[solar glade]

I wanna see how u do it

Write it

[wise light]

25/16

[solar glade]

Write the whole thing

[flint kettle]

Lol just do the quadratic eq ur not gonna be able to complete the square most of the time if ur doing anything except curated problems

I mean u will do it anytime

But its slower than the quadratic eq

[solar glade]

?

If you cant read dont talk here please

[wise light]

i need to learn comp the square because theres some problems like graphing rational equations

that need comp the square

[solar glade]

Keep on this raileu

Write the whole thing

After completing the square

[wise light]

wait im lost

x^2 + 5x/2 + _______ =15/4

idk how u get 25/16

[solar glade]

Check this

Look at this for a second

Forget about your equation

x^2 +ax = (x+a/2)^2-(a/2)^2

a can be any number

Now use in your left side

x+(5/2)x

a=5/2

[solar glade]

I am not sure i cannot read that

[wise light]

x^2 + 5/2 = (x + 5/2 )^2 - (5/2)^2

[solar glade]

Not exactly

x^2 + (5/2)x

[wise light]

i dont get this

[solar glade]

U forgot the x multiplying

That is how u complete any square

[wise light]

can we agree that this so far is right

[solar glade]

Yes

[wise light]

okay from how i understand it lets say i have this x2 + 8x + 14 = 0

[wise light]

i have to move 14 to right side right

then after that

i divide 8 by 2 to get 4

then square the thing to get 16

[wise light]

i wanna do this but

[wise light]

for this one

[solar glade]

(x+8/2)^2 -16=-14

Like this

Do you know why?

Can you expand this? (x+4)^2

[wise light]

ur doing it differently from what im doing

[solar glade]

Doesnt matter, the important thing is u understand why u add these numbers

[wise light]

x^2 + 8x + 16

[solar glade]

Ok

No

8x

Ok

But u have x^2+8x

U dont have the 16

Right?

[wise light]

u said expand (x+4)^2

[solar glade]

Yes

[wise light]

so i expandded

[solar glade]

I want u to answer my question

In the original equation

We dont have that +16

Right?

[wise light]

x^2 + 8x +16 = -14 + 16

i dont have the 16

[solar glade]

U add it

[wise light]

but i will eventually

[solar glade]

To make it true

[wise light]

x^2 + 8x +16 = 2

(x+4)^2 = 2

i get this question

[solar glade]

Ok if you want to do like that

U can do it

Divide 5/2 by 2

[wise light]

becomes 5/2 x 2/1 right

[solar glade]

No i said divide

U divided 8/2

Now u do the same here

U divide 5/2 by 2

[wise light]

dont u have to take reciprocal of 2

[solar glade]

Imagine this number

20/2

What is that

[wise light]

10

[solar glade]

Now divide 10/2

[wise light]

5

[solar glade]

Ok now divide 20/2 by 2

(20/2)/2

Isnt that 5?

[wise light]

yes

[solar glade]

Which is 20/4

[wise light]

5

[solar glade]

So when u have 5/2 divided by 2

U have 5/4

5/2 divided by 2/1

You can see like this

[wise light]

isnt it supposed to be

5x / 2 times 2/ 1

[solar glade]

Why, u have to divide by 2 not multiply by 2

[wise light]

dividing fractions is just multiplying it by the reciprocal

no

i did it wrong

the reciprocal of 2 is 1/2

[solar glade]

Yes

[wise light]

i was doing 2/1 instead of 1/2

[solar glade]

Now u are doing correctly

So now u have 5/4 right?

[wise light]

ok now i get why i got 25/16

[solar glade]

Nice

[wise light]

that was so frustrating

ok so far

[wise light]

yes?

[solar glade]

Yes

[wise light]

i know the right side is 85/16

can u go over how to get the

right side in the form (x+ __) ^2

[solar glade]

U already have it dont

Done

[wise light]

(x+5/4)^2

[solar glade]

Yeah

[wise light]

so the half of the b?

[solar glade]

Because when u expand that

U will get twice

[wise light]

its just (x+b)^2

?

[solar glade]

Thats why u do half

[solar glade]

This is correct

[wise light]

x = -5/4 +- sqrt85/4

its same as this?

[solar glade]

Unless im missing something it looks correct to me

[wise light]

great

thank you

completing the square sucks

[solar glade]

I recommend u doing my way

It is easier

I will write here right quick

[wise light]

okay

[solar glade]

You have this

[wise light]

i dont understand the way u do it but i want to understand

[solar glade]

x^2+bx+c=0

(x+b/2)^2 - (b/2)^2 + c =0

This way u complete the square in one step

And u dont have to calculate anything before completing it

Now yoi just rearrange the terms

[wise light]

like this

?

[solar glade]

Make the correction

I did a typo

I corrected it

[solar glade]

Yea like that

[wise light]

so u dont move c to other side>

[solar glade]

They are numbers, u do now

You can even do the math of c and -(b/2)^2

And affer that put in the right side

Because u already completed the square

In the first step

I will show u with an example

[wise light]

x^2 +8x +14 = 0

(x+16)^2 - 50 = 0

[solar glade]

No no

b/2 not 2b

[wise light]

(x+4)^2 -16 + 14 = 0

[solar glade]

Yes

Like that

[wise light]

(x+4)^2 -2=0

[solar glade]

Perfect

[wise light]

so i just move 2 then sqrt?

[solar glade]

Correct

[wise light]

to get the x 's

[solar glade]

Isnt this faster?

[wise light]

ok let me try with the oen with fraction

(x + 5/2 ) ^2 - (5/2)^2 -15 = 0?

[solar glade]

No

b is already 5/2

You have to do b/2

It should be (x+5/4)^2-(5/4)^2-15/4=0

Remember one thing

What i showed u was for x^2+bx+c

In that exercise u have 4x^2 in the beginning

[wise light]

is this right

[solar glade]

Yes

[wise light]

so now i can do the other way

since a is 1

yes?

[solar glade]

Yes now just b=5/2 and c =-15/4

Do b/2 remmeber

[solar glade]

Yes

[wise light]

and why is the first one (x+ 5/4)^2 again?

and not (x + 5/2)^2

[solar glade]

Because u had 5x/2

So b = 5/2

[wise light]

the half of it

[solar glade]

And we want b/2

[wise light]

yeahhhhhhh

[solar glade]

Nice

Now i have to leave, i hope u understood the method

[wise light]

okay thank you

[solar glade]

Practice with this method and u will solve quadratics in 10 seconds

[wise light]

.clsoe

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[faint rock]

Hi! I need to find the antiderivative to f(x)=ln(x^2+8x+25)

[hybrid heath]

!status

[final saddle]

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

[faint rock]

2 I have tried using 1*ln(x^2+8x+25) and doing partial integration from that but I can't seem to get it to work

[amber holly]

Show your work

[faint rock]

How? With a picture?

[amber holly]

Yeah, or at least just type the new thing that you need to integrate

[faint rock]

So i integrated 1ln(x^2+8x+25) and got xln(x^2+8x+25) - int(x/(x^2+8x+25)) but I kind of got into a loop

[amber holly]

I don't see any loops, you now only need to integrate a rational function; Also, are you sure the new integrand shouldn't be (2x^2 + 8x)/(x^2 + 8x + 25)?

[faint rock]

I don't see how the 1 would become (2x^2 + 8x)

[amber holly]

Tell me the derivative of ln(x^2 + 8x + 25)

[faint rock]

Ah alright i see now

[amber holly]

Anyway, start by performing the polynomial division

[final saddle]

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[rugged fiber]

I dont understand how they got the highlighted part

[solemn umbra]

the x-intercept is where the graph intersects the x-axis, this is where y = z = 0. if you set y = z = 0, you get what they say

similarly for y and z intercepts

[final saddle]

@rugged fiber Has your question been resolved?

[rugged fiber]

Aaaaah

Thankyou so much

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[bold badge]

What does it mean for the right figure to be according to left cosets of K in relation to the left figure?

[bold badge]

Like, I don't get where the inside structure went

[zenith pollen]

K is like ±1

[bold badge]

so it's just included in the "nodes" themselves?

[zenith pollen]

yea you just absorb them together

[bold badge]

I see

[solemn umbra]

i think they are saying the right diagram is the cayley diagram for the group Q_8 / K, with generators {i,j,-1}

[bold badge]

so for D3 mod <s> where s is the reflections it would just be like this?

the arrows are multiplication by r ofc

actually, they should be one directional, since rrH is Hr not H

so pretend I drew it like H to rH to Hr to H

[zenith pollen]

yea directed

[bold badge]

I'll fix that in my drawing rn actually

Ah, so it's fine then

[solemn umbra]

you could also write them all as left cosets, since usually the quotient group is the collection of left cosets. but it doesn't really change anything i guess. looks good

[bold badge]

but the problem states that this diagram cannot be the diagram for a group

how come?

to me this looks like D3 mod s would be isomorphic to the group of just rotations of a triangle

[solemn umbra]

isn't this the cayley graph of the cyclic group of order 3, where the generating set is the generator?

[bold badge]

that's what I believe

but I guess there is an issue in here somewhere?

but I can't seem to find it

[solemn umbra]

the book says this directed graph can't be the cayley graph of any group?

[bold badge]

I'll post what it states in it's entirety, perhaps I'm misunderstanding

The stuff below the problem seems to be an explanation of the issue yet I can't seem to find out why that's the case

[solemn umbra]

oh, i forgot {1,s} is not a normal subgroup

hmm

[bold badge]

is that important?

must we have that the left and right cosets are the same?

[solemn umbra]

well, now D_3 / <s> isn't going to be a group, its just a set of left cosets

[bold badge]

well according to the diagram, it looks like a group

it's associative, has an "identity" H, and is closed

inverses would be r^2 and r

[solemn umbra]

well okay, you can define a group operation on {H, rH, r^2H} that makes it into a group. but the one induced by the quotients doesn't work out: its supposed to be defined by aH * bH = (ab)H whenever a,b in D_3.

so, you want rH * r^2H = r^3H = H, for example.

but
rH = {r, rs} = rsH = {rs, rs^2}
and
r^2H = {r^2, r^2s} = srH = {sr, r^2}.
and you would want (rs)H * (srH) = (rssr)H = r^2H. but r^2H is not r^3H = H

so you get weird things like rH times srH goes to r^2H, but rH times r^2H goes to H, even thought srH and r^2H are equal

and i think this is going to give you a weird cayley graph, since its supposed to be with respect to the generators r and s, not just r

(sorry for calling them cayley graphs, that's just the terminology i'm familiar with)

[bold badge]

I saw Cayley graph and Cayley diagram, that part is fine, idc either way

[solemn umbra]

yay

[bold badge]

but hmm

I'm gonna think about what you wrote a bit more

I also fcked it up since r^2s is not sr

sr^2 is rs

wait

[solemn umbra]

i think r^2s is sr

[bold badge]

r^2s is sr

ye

[solemn umbra]

[bold badge]

I wrote something else weird

oh I had..

well wait, nvm, I think It is fine rn

ohh I had sr^2 is rs

UGH WAIT

FUCK

let me share my diagram

I think I'm tripping

[solemn umbra]

i don't even know how i'd try to draw the cayley graph. you could write H, rH and r^2H, but now multiplying by srH and r^2H does different things to rH depending on if you write it as rH or rsH. lol

[bold badge]

I guess that's the point of the problem then

[solemn umbra]

so maybe they want you to write tons of cosets like rH, rsH, srH, r^2H, H, r^2sH, etc

i don't know

[bold badge]

I see

I only did the cosets Hr and rH

but also I didn't verify that r^2H is Hr

idk why I did that part

[solemn umbra]

yea, D_3 / <s> in this case is supposed to mean the set {H, rH, r^2H}, i think. its just that the cosets don't form a group (under the group law induced from D_3)

and rH has multiple representatives, like sr^2H or rsH

[bold badge]

oh i excluded multiplication by s

bc if it was just r, it'd be fine

well

wait..

omg

fuck this triangle bullshit, I wish the problem was with D4 instead

sigh

[solemn umbra]

lol but then it would work because {1,s} would be normal

[bold badge]

yeah

[solemn umbra]

[bold badge]

😭

I'm gonna try this again later tbh

thanks for the help though

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[glad basalt]

Why is this 2/3

I do F(b) - F(a)

So you get 2/3 - 0 = 2/3

But then there's a 2 outside

so 4/3?

[mellow axle]

2$\int u^2 = \frac{2}{3}u^3$ not $\frac{4}{3}u^3$

[soft zealot]

chebyshev's infinite pee norm

[glad basalt]

Yes thats what it says in the picture?

It is bounded

that is why I am confused

0 - 1

I did F(b) - F(a) and I get 2/3

But then theres a 2 outside the integral that would make it 4/3

But the professor says its just 2/3 not 4/3

[solar glade]

What is the integral of u^2

@glad basalt

[glad basalt]

u^3/3

[solar glade]

Then where is your confusion

[solar glade]

I mean

[glad basalt]

This is the antiderivative

2/3 * u^3

to solve we do F(b) - F(a)

F(1) - F(0)

[solar glade]

That is 2 int u^2

Why do u want to multiply by 2 twice

[glad basalt]

Oh

So they just brought the 2 in

I gotch you

[solar glade]

U do this

2(F(b)-F(a))

2(1/3-0)

2/3-0

2/3

[glad basalt]

Ok

I thought 2/3 u^3 was the integral and the 2 was outside

thanks

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[sacred lynx]

AB and CD lines are parallel CD=5AB need to find BC if OB=3

[tranquil pine]

Do you understand the parity between the angles of the two triangles?

Because the lines are vertical and you have two vertically opposed angles (shitty translation)?

So if CD = 5AB implies that you can downsize the bigger triangle into the smaller one diving the sides by 5

Or if you want to amplify, just multiply by 5

Thus BC = OB x 5 = 15

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[jagged stump]

how to get started on this one?

[celest crane]

$f_x \cdot (x - x_1) + f_y \cdot (y - y_1) + f_z \cdot (z - z_1) = 0$

[soft zealot]

Kookiemon

[jagged stump]

ah same equation as always then. one moment

so dh/dy would be f_x and dh/dz would be f_y ?

[celest crane]

You have the general equation x = h(y,z). If you put that into standard form, you would get h(y,z) - x = 0.

From there, you would evaluate the partial derivatives, of which you are given two by the problem.

[jagged stump]

ohoh ok

[celest crane]

Looks good.

[jagged stump]

yippeee ill try it

it worked ! thank you

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[gentle trout]

no clue

any theorem i can use that i can just research

[hybrid heath]

this is just a system of linear equations

[warm python]

use matrices to solve the system of linear equations

[gentle trout]

.close

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[fathom walrus]

dang bruh

i was just about to share a nice explanation

oh well

[gentle trout]

my fault

.reopen

[final saddle]

✅

[gentle trout]

show me how you dd it

always nic to see different ways

[fathom walrus]

nah its the same method, just a textbook explanation lol

[gentle trout]

.close(cool)

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[fathom walrus]

lol thanks for the (cool)

🦦

[vital surge]

no

do not use linear system

do not use polynomial interpolation

use method of finite differences

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[vital surge]

k

[vital surge]

notice ur given x-values -2, 0, 2, 4, 6 satisfy an arithmetic progression
then, u may apply method of finite differences

so, set up figure that looks like this:
(each successive row is determined by finding the "difference" between the two terms above)

x-values -2     0     2     4     6
y-values  ?     1     15    133   427
             ?     14    118   294
                ?     104   176
                   ?     72

then, this method tells us that the bottom question mark is a 72. by finite differences, this is because the 4th row from the y-values must be a constant as we are dealing with a cubic

then, complete the figure as follows:

x-values -2     0     2     4     6
y-values  19    1     15    133   427
            -18    14    118   294
                32    104   176
                   72    72

hence, p(-2) = 19

.close

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[vital surge]

(if this explanation does not not make perfect sense internet will help you probably)

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[final saddle]

@tranquil pine Has your question been resolved?

[zenith pollen]

all the answers are true so we need an image of those postulates

also I need the skullcry emoji

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@tranquil pine Has your question been resolved?

[brave forge]

@zenith pollen how genius

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[plucky obsidian]

4.8/2.5=9.7/MN MNx4.8=24.25  MN=5.05

are my ans

corect

guys pls i need help fast

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[tidal idol]

Need help with (b). Thanks in advance

[final saddle]

@tidal idol Has your question been resolved?

[final saddle]

@tidal idol Has your question been resolved?

[tranquil pine]

ok what have u tried?

you should first use partial fractions to expand out f(z)

[tidal idol]

Yes, I did the partial fraction for (a). After that should I assign u = z-2 and u+1 = z-1?

[tranquil pine]

oh wait i misread

[tranquil pine]

i think so yes

[tidal idol]

So it becomes 1/(u+1) + 2/u, and for the first term can be expanded into (1 - u + u^2 - u^3...) I think. And that's the final result?

[tranquil pine]

hmm what is ur final result in terms of z?

[tidal idol]

2(z-2)^-1 + (1 - (z-2)^-1 + (z-2)^-2) + ...)

[tranquil pine]

that doesnt feel like it would make sense

[tidal idol]

How so?

[tranquil pine]

one sec ill try writing it out

[tidal idol]

Alright

Did you find the solution? @tranquil pine

[tranquil pine]

yeah sorry i am not so sure. Although u might be better off asking in #real-complex-analysis

as the help channels are usually meant for highschool level content

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[frosty swift]

Using the double angle formula find the exact values of cos 15

[frosty swift]

cos 30 = 2 cos ^2 15 - 1

root3/2 = 2 cos^2 15-1

cos 15 = root(root3 - 1 / 4)

where'd I go wrong? or is it the wrong formula

i know we can use compound angle identities, cos (45-30) = sin45sin30+cos45cos30

[wintry kindle]

how did you get root(root3 - 1/4)

[frosty swift]

but is compound angle identities the same as double angle formula?

[frosty swift]

i bet i did something stupid okay so

cos 30 = root 3/2

[wintry kindle]

solve stepwise

[frosty swift]

root3/2 = 2 cos^2 15 - 1

okay

[wintry kindle]

yop continue

[frosty swift]

root3/2 +1 = 2 cos^2 15

(root3/2 +1) / 2 = cos^2 15

root ((root3/2 +1) / 2) = cos 15

[wintry kindle]

yop

[frosty swift]

root ((root3 +1) / 4) = cos 15

is that right?

[wintry kindle]

no

[frosty swift]

what can I write root ((root3/2 +1) / 2) as?

[wintry kindle]

do you need to rewrite it?

[frosty swift]

cant I do root ((root3 +1) / 2 * 2)

[frosty swift]

well not really i just supposed i had to simplify it

[wintry kindle]

is it root(3)/2 or root(3/2)

[frosty swift]

former

[wintry kindle]

k well then the +1 becomes +2

since you exclude /2 from the brackets

root(3)/2 + 1 = 1/2 * (root(3) + 2)

[frosty swift]

HUH

[wintry kindle]

what's confusing you :D

[frosty swift]

wait

OH

that makes sense

[wintry kindle]

so you get root((root(3)+2)/4)

[frosty swift]

i had to write it down on paper

[wintry kindle]

but in essence the root(3)+2 term will hinder you from further simplifying I think

[frosty swift]

oh

so after root ((root3/2 +1) / 2) = cos 15

what should I do?

[wintry kindle]

prob nothing

[frosty swift]

sweet

also, this is called the double angle formula right?

and compound identities is (A+-B)?

[wintry kindle]

Ys, since you used cos(2A) = 2cos²(A) - 1

[frosty swift]

sweet, but then how do we do tan 15?

[wintry kindle]

note tan = sin/cos

[frosty swift]

sin 15/cos 15

okay so ill like try it

oh wait

[wintry kindle]

yop

[frosty swift]

i feel like im gonna go horribly wrong

[wintry kindle]

In case the fraction isn't easy to simplify you can also try tan(15) = tan(45-30)

[frosty swift]

but thats compound identities right?

im trying to solve it through double angle

[wintry kindle]

k then the before method

[frosty swift]

sin 30 = 2 sin 15 cos 15
1/2 = 2 sin 15 cos 15
1/4 = sin 15 cos 15
1/ 4 cos 15 = sin 15

is that right?

cos 15 = root ((root3/2 +1) / 2)

[wintry kindle]

neat

[frosty swift]

okay so, do I replace cos 15 into sin 15?

1/4 ( root ((root3/2 +1) / 2)) ?

(1/4 ( root ((root3/2 +1) / 2))) / root ((root3/2 +1) / 2) = tan 15

this looks disgusting omdifnm

[wintry kindle]

and therefore sin15 / cos15 = 1/(2*(root(3)/2+1) after you insert

don't worry since you multiply roots

You can combine them

rootX * rootX = X

[frosty swift]

sweet okay wait ill write it on paper

[wintry kindle]

This can be further simplified, but it's a little tricky

I'll hint that you can get rid of the fraction entirely

[frosty swift]

THIS IS so ew

[wintry kindle]

:]

[frosty swift]

woah

um

[wintry kindle]

= 1/(root(3)+2)

= [magic...]

= 2 - root(3)

[frosty swift]

WAOAH WHAT

[wintry kindle]

experiment with it 🦑

gtg but wish thee gl

[frosty swift]

THANK YOUUU

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@frosty swift Has your question been resolved?

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[frosty swift]

find tan 15 using double angle

[frosty swift]

tan 30 = 2 tan 15/1-tan^2 15

1/root3 = 2 tan 15 / 1-tan^2 15

what next tho

[final saddle]

@frosty swift Has your question been resolved?

[frosty swift]

GOT THE answer for this one

another question

Given that cos A = 1/2 and A lies in 4th quadrant, find value of cos of A/2, sin of A/2 and determine in which quadrant the < A/2 lies.
Cos A/2
cos A = 2cos^2 (A/2) -1
1 + 1/2 = 2 cos^2 (A/2)
3/4 = cos^2 (A/2)
+- root3/4 = cos (A/2)

Sin A/2
Cos A = cos^2 (A/2) - sin^2 (A/2)
1/2 = cos^2 (A/2) - sin^2 (A/2)
1/2 = root 3/4 - sin^2 (A/2)
-root1/2 + root 3/4 = sin^2 (A/2)

is this okay? and where do I go after this?

[midnight sparrow]

So what IS A here

[frosty swift]

im not suree

[midnight sparrow]

Try

[frosty swift]

A is the angle in 4th quad?

[midnight sparrow]

Yes so what IS the angle A

[frosty swift]

UM

Cos A = 1/2

so 60?

[midnight sparrow]

Not exactly

[frosty swift]

UM

i dont get it

[midnight sparrow]

So with cos A = 1/2 WE have A = pi/3 or A = -pi/3 right ?

[frosty swift]

ya

[midnight sparrow]

Knowing that A is in 4th quadrant what IS A ?

[frosty swift]

UM

pi - pi/3?

[midnight sparrow]

?

[frosty swift]

2 pi / 3?

[midnight sparrow]

Why

[frosty swift]

IS IT NOT

[midnight sparrow]

Do you understand this ?

[frosty swift]

yeah

[midnight sparrow]

So A can't be 2pi/3

[frosty swift]

okay

[midnight sparrow]

So what IS A ?

[frosty swift]

A IS - pi/3

[midnight sparrow]

👍

[frosty swift]

YAYO MG

[midnight sparrow]

What is A/2 ?

[frosty swift]

-pi/3 / 2?

[midnight sparrow]

👍

Which IS ?

[frosty swift]

1/2

HELP

[midnight sparrow]

?

[frosty swift]

I didnt write that

[midnight sparrow]

-pi/6 right ?

[frosty swift]

-pi/6

[midnight sparrow]

👍

Now you need to know cos -pi/6 and sin -pi/6 and it's done

[frosty swift]

WHAT IS THIS METHOD

I WAS TRYING TO USE THE OFRMULA 😭 😭

ive just been blindly following the formula lmao

[midnight sparrow]

It's good to use formula sometimes but when you don't know what to do draw a picture Can help sometimes

[frosty swift]

i got it like conceptually

but i still dont know how to apply the formula

im pretty sure its wrong but idk why

[midnight sparrow]

Do exercises and you will know what to use don't worries

[frosty swift]

OK BUT IN TERMS OF FORMULA

im doing something terribly wrong

and idk what

Given that cos A = 1/2 and A lies in 4th quadrant, find value of cos of A/2, sin of A/2 and determine in which quadrant the < A/2 lies.
Cos A/2
cos A = 2cos^2 (A/2) -1
1 + 1/2 = 2 cos^2 (A/2)
3/4 = cos^2 (A/2)
+- root3/4 = cos (A/2)

Sin A/2
Cos A = cos^2 (A/2) - sin^2 (A/2)
1/2 = cos^2 (A/2) - sin^2 (A/2)
1/2 = - root 3/4 ^2 - sin^2 (A/2)
1/2 = 3/4 - sin^2 (A/2)
sin^2 (A/2) = 3/4 - 1/2

[final saddle]

@frosty swift Has your question been resolved?

[frosty swift]

<@&286206848099549185>

[tranquil pine]

is A in 0 to 2pi

[frosty swift]

ya?

i mean i understood A = 60/-60

but idk how to do through formula, where am i fucking up

[tranquil pine]

why not 300

[frosty swift]

OKAY ya could be that too

BUT LIKE LOOK AT THE FORMULA

why cant i solve through formula

actually no, it is 300 cus its 4th quadrant

which is -60

[tranquil pine]

what about 660

[frosty swift]

but where am i going wrong in formula

HELP

OKAY

[tranquil pine]

i dont have the answer lmao, im more conufused than you B)

[frosty swift]

😭 😭 😭

Given that cos A = 1/2 and A lies in 4th quadrant, find value of cos of A/2, sin of A/2 and determine in which quadrant the < A/2 lies.
Cos A/2
cos A = 2cos^2 (A/2) -1
1 + 1/2 = 2 cos^2 (A/2)
3/4 = cos^2 (A/2)
+- root3/4 = cos (A/2)

Sin A/2
Cos A = cos^2 (A/2) - sin^2 (A/2)
1/2 = cos^2 (A/2) - sin^2 (A/2)
1/2 = - root 3/4 ^2 - sin^2 (A/2)
1/2 = 3/4 - sin^2 (A/2)
sin^2 (A/2) = 3/4 - 1/2

<@&286206848099549185>

[midnight sparrow]

What's the problem WE already solve that

[frosty swift]

THE FORMULA

how do i solve it

through the formula

[midnight sparrow]

K first Can you decide whereas it's plus or minus Root 3/4 for cos A/2

[tranquil pine]

acc to his eqn, it should be positive

but it can be negative too :v

[midnight sparrow]

No

Remember A lies in 4th quadrant

WE talk earlier of what IS A but he don't remember

I can't help if you make no efforts

[tranquil pine]

yea

why does it matter tho

it can be negative

A/2

[midnight sparrow]

So what IS A ?

WE have litteraly thé value of A

[tranquil pine]

A isnt a single value

[midnight sparrow]

Yeah sure but it can't be 20 for example

[tranquil pine]

its infinite

[midnight sparrow]

So A = x mod 2pi if you want to bé rigorous

What is x

[tranquil pine]

A = x |2pi| ?

[midnight sparrow]

Maybe you don't know this notation

Hum just Say 1 value of A so

You are in high school ?

[tranquil pine]

..

[tranquil pine]

why 1 value

what value are you taking of A?

[midnight sparrow]

What is A pls 😂

[tranquil pine]

its 5pi/3

+2pi

-2pi

[midnight sparrow]

Yes finally 😂

[tranquil pine]

uh.

?

i dont get how this proves the value cant be negative

[midnight sparrow]

Cause when WE divide this angle WE got also in the 4 th quadrant so -root 3/4

Hum plus sry

Calculation seems ok I think

Ye all good

[tranquil pine]

what..

[final saddle]

@frosty swift Has your question been resolved?

[tranquil pine]

dont close due to timeout

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[hard mist]

$ln(p^2+1)=2y$

[soft zealot]

rainy

[hard mist]

How do I solve for p here?

i can do:

$p^2+1 = e^{2y}$

but that doesnt help me i think

[tranquil pine]

no.....

shouldnt it be e^(ln(p^2+1)) = e^(2y)

holy hell so many brackets

[soft zealot]

rainy

[hard mist]

fixed

So like this then?

[tranquil pine]

no

you need p

u have p^2

[hard mist]

yeah i mean ^ is my starting point

from there i can solve

just needed to get that ln out

[tranquil pine]

yea thats correct

[hard mist]

Ok thanks

.close

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[hard mist]

I dont know how to continue with this

[hard mist]

usually what i do is substitute p' for dy/dx

but idk if its dy/dx, dx/dy, ...

ah, apperantly i have to replace it by dp/dy

.close

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[shut helm]

Hi! How can I prove/disprove that an integer n is a sum of 8 integer cubes? So far all I have tried is trying out different values and since 1 and 0 are part of the cubes list, it seems like almost any integer can be represented as a sum of 8 integer cubes.

[versed crater]

I’m not sure but perhaps something about base 3?

[mellow axle]

Well 2^3 is 8

[sturdy cypress]

or base 2

[mellow axle]

And you can represent any number k between 0 and 7 with k * 1^3

[shut helm]

I see yes

I am sorry I am a little dense . Then how would I go on with the problem?

[mellow axle]

Well how can you express numbers between 8 and 15?

16 and 23?

[sturdy cypress]

what about 23

maybe you have to use negatives

[mellow axle]

no no

er

I see I confused my idea because I was thinking in base 8 after you guys said base 2 lmao

[versed crater]

I was more so thinking to partition the base 3 number into 8 parts that all have 2’s in every digit

No that’s not quite right

2 in one particular digit?

I’m not too sure where I was going, it just sounded like a good idea

[sturdy cypress]

you can't do 23 unless it's 27 − 1 − 1 − 1 −1

[versed crater]

I guess the other 3 are 0’s?

[sturdy cypress]

yeah

or 8 8 8 −1

[mellow axle]

So it's known that you can express n as the sum of 5 cubes, but I'm trying to think of there's a nice way to express it as 8 without just using the decomposition for 5 cubes

[versed crater]

Hold up

2220 is a cube number in base 3 is it not?

,w 2220 base 3 to base 10

[soft zealot]

Results provided by WolframAlpha

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

Requested by frosst

[versed crater]

Or not

But 2222 is

[mellow axle]

2222 is 80?

If 2220 is 78

[versed crater]

No that’s a 4th power

[versed crater]

It’s 81

I think

Wait no it’s 80

[mellow axle]

3000 is a cube though

Cause it's 1000 cubed

[versed crater]

1000 you mean

[mellow axle]

er 10 cubed not 1000 cubed

[versed crater]

No 10³ is 1000

Even in base 3

[mellow axle]

Yeah

yea

[versed crater]

What about 11³

,w 11^3 base 3

[soft zealot]

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Requested by frosst

[versed crater]

,calc cbrt(1331)

[soft zealot]

Result:

11

[versed crater]

Lmao no shit dumbass

,w (11 base 3)^3

[soft zealot]

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Requested by frosst

[mellow axle]

Ok well my idea was between consecutive cubes, n^3 and (n+1)^3, there are 3n^2 + 3n = 3n(n+1) terms which is divisible by 6

so find an expression for each 6k + r

And induct

[final saddle]

@shut helm Has your question been resolved?

[shut helm]

Interesting! Will try. Thank you for the idea

.close

[final saddle]

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[cloud portal]

Why can you always do multiplication in any order?

[nova iron]

because multiplication is commutative on the set of real numbers

[marble agate]

its also associative

[cloud portal]

But why? What is the proof

[hybrid heath]

there's a number of ways to get there

at the root, you want to look at what a Peano System is

and you'll need to know proof by induction

ultimately though, everything is based on a set of axioms

[cloud portal]

.close

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[unique phoenix]

Hi! I have the following linear algebra exercise to solve:
$A = \begin{pmatrix}
1 & a & b\
0 & 1 & c\
0 & 0 & 2
\end{pmatrix}$. Show A is diagonisable if and only if the polynomial $x^2 -3x + 2$ annuls A.\

I immediately thought about using the characteristic polynomial which should always be annuled by A so I calculated
$\chi (x) = (x-1) \cdot (x-1)(x-2) = (x-1)(x^2 -3x + 2)$ and this got me thinking. Normally we'd only care about the unique roots when looking for roots i.e. since $ (x^2 -3x +2) $ is equal to 0 exactly whenever $\chi (x)$ is equal to 0 we don't care about the extra factor of $(x-1)$. However this only holds for numbers. If you put A into the polynomial, we can still have $\chi (A) = 0$ without necessarily having $(A^2 - 3A + 2I)$ if I am understanding correctly, so how can I prove that $A$ annuls $(x^2 -3x +2)$ if and only if it annuls $\chi (x)$?

[soft zealot]

mippen

[unique phoenix]

.close

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[bold needle]

how do i approach?

[terse siren]

the sequence converges

100 seems sufficiently large to just set n to infinite and calculate the limit

[bold needle]

idk anything about sequence convergence and divergence

can you speak in layman terms

[terse siren]

it means as n gets bigger the sequence gets closer and closer to one value

[terse siren]

alternatively you can just calculate the first few terms

[bold needle]

oh yes i get that

but

[terse siren]

and stop when the 1st decimal place is constant

[bold needle]

but my original idea was to get a bound in terms of n

there was another idea i had
generating functions! but i have never used gen functions in recursive relations when its in denominators

[terse siren]

oh

you can just calculate the limit though

[bold needle]

can yyou walk me through it

i have never calculated limits of recursive relation

[terse siren]

1st step is realizing that it converges

[bold needle]

okay ! then

[terse siren]

when n goes to infinity

a_(n) = a_(n-1) = l

then solve for l

[pseudo knot]

I can do it with python, not sure about mathematical solution

[terse siren]

oh wait

that doesn't work here

[bold needle]

that would give 0 = 4

which is false

[terse siren]

yeah

where did I go wrong

[bold needle]

i found a similar problem on mathstack but i didnt get the cruxx of it

https://math.stackexchange.com/questions/1401132/solve-the-sequences-inequality

also idk about the converges part

[terse siren]

oh

[bold needle]

but the sequence sure is increasing

[terse siren]

it doesn't converge

it's a harmonic sum situation

bruv

[bold needle]

what do i do with it now

[terse siren]

uhh

approximate with integral?

[bold needle]

💀
how do we do that

[terse siren]

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

happy reading

[bold needle]

https://tenor.com/view/american-psycho-patrick-bateman-sweating-sweat-sweaty-gif-21722513

[terse siren]

we need to write the sequence as a sum first

[bold needle]

how do you even do that?

the mathstack had diff approach of bounding idk

[terse siren]

recursion

a_n = a1 + 4S + 4/(a1 + S)

if my math isn't off

where S is the sum of 1/a_k from 1 to k

ok this is already complicated

okay

that's not helpful either

[pseudo knot]

Ok I got it