help-13

[cedar kiln]

@quasi plover Has your question been resolved?

[quasi plover]

whats that

[livid hound]

look up ambiguity since law

[quasi plover]

0 help ngl

to answer my question

[cedar kiln]

@quasi plover Has your question been resolved?

[vagrant hamlet]

do u still require help?

[quasi plover]

i still require help to answer my question yes

[vagrant hamlet]

okay

so basically

u have already found the essential

try using the alternate angle rule

[quasi plover]

i still didnt find the B angle

[vagrant hamlet]

u dont need to find angle B

do u?

u dont

[quasi plover]

oh

what do you mean then, where do i use the alternate angle rule

[vagrant hamlet]

try this

since u already found angle ABC

[quasi plover]

i still cannot manage to find that straight line

[quasi plover]

but abc is wrong

because it said obtuse angle

[vagrant hamlet]

oh?

[quasi plover]

but the angle that i got is acute

[vagrant hamlet]

let me calculate

[quasi plover]

ok

take your time

[vagrant hamlet]

try this

ignore the words ontop

oh wait

nop

NOP

[quasi plover]

how do you get the 68

;-;

[vagrant hamlet]

wait

68

180-112

[quasi plover]

ohhhhhhh

[vagrant hamlet]

ye

[quasi plover]

how does that help now when we know that

[vagrant hamlet]

i dont know, im just trying-

tbh

i always do that when i attempt this kind of question

[quasi plover]

lol ok understandable 😄

[vagrant hamlet]

okay

let me share a video with u

https://youtu.be/G2Riehij1x0

[quasi plover]

ok...

[vagrant hamlet]

this is based on the ambiguous case

[quasi plover]

so this is actually gonna help right?

[vagrant hamlet]

ur question is an ambiguous case

yep

try it

[quasi plover]

after i watch it what do i do?

[vagrant hamlet]

u try to follow what the question does

i mean

the video

it give steps

[quasi plover]

ok ill do it

[vagrant hamlet]

ye

[quasi plover]

after i finished doing things in the tolet

[vagrant hamlet]

omg

what the answer given to u for the question btw

[quasi plover]

I dont have the markscheme

i just finished with smth

boutta do the question]

[cedar kiln]

@quasi plover Has your question been resolved?

Channel closed

Closed by @quasi plover

Use .reopen if this was a mistake.

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.

[near rapids]

HELP

.close

[cedar kiln]

Channel closed

Closed by @near rapids

Use .reopen if this was a mistake.

[near rapids]

.reopen

[cedar kiln]

✅

[near rapids]

.close

[cedar kiln]

Channel closed

Closed by @near rapids

Use .reopen if this was a mistake.

[near rapids]

.reopen

[cedar kiln]

✅

[near rapids]

.close

[cedar kiln]

Channel closed

Closed by @near rapids

Use .reopen if this was a mistake.

[near rapids]

.reopen

[cedar kiln]

✅

[near rapids]

.close

[cedar kiln]

Channel closed

Closed by @near rapids

Use .reopen if this was a mistake.

[cerulean sail]

[tropic oxide]

@near rapids what are you doing

[cedar kiln]

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.

[near rapids]

No one was replying so I tried to get attention 🥲

[near rapids]

.close

[cedar kiln]

Channel closed

Closed by @near rapids

Use .reopen if this was a mistake.

[latent bloom]

.reopen

[cedar kiln]

✅

[latent bloom]

What

Post your question again

[tropic oxide]

good way to divert attention from yourself, tbh.

[latent bloom]

Aaaaand they leave

Sad

Wanted to actually help

Oh well

[dusk finch]

.close?

[cedar kiln]

@near rapids Has your question been resolved?

[latent bloom]

Yep

.close

[cedar kiln]

Channel closed

Closed by @latent bloom

Use .reopen if this was a mistake.

[latent bloom]

For good this time

[cedar kiln]

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.

[silver cradle]

AB = AD CB=CD BD: y = -1/3x+3. How do I work out C?

[obsidian pine]

The gradient of line AC must be the inverted gradient of BD because it is orthogonal to the line BD

This means the gradient of line AC is 3

ABD appears to be an equilateral triangle so AB=BD=DA

B has co-ordinates (0,3)

[silver cradle]

D (9,0)

[obsidian pine]

That helps now

The line AC cuts through the middle of the line BD

[silver cradle]

this doesn't matter

[obsidian pine]

um ok

The co-ordinates of this intersection must be half way across the line BD

This can be found by finding the difference between the co-ordinates of those points

so co-ordinates of D subtracted by co-ordinates of B

Then divide by 2

Understand so far?

[silver cradle]

not this

[obsidian pine]

Ok

[silver cradle]

didn't quite understand

[obsidian pine]

Two co-ordinate points (x,y) and (a,b) where a>x

The change in coordinates for x is a-x

and the change in coordinates for y is b-y

Then divide these changes by 2

and add them on to the coordinates (x,y)

so then the coordinate point half way between these points is ((x+a)/2,(y+b)/2)

[silver cradle]

ok thanks, I need to go so I'll try to figure this out later

.close

[cedar kiln]

Channel closed

Closed by @silver cradle

Use .reopen if this was a mistake.

[obsidian pine]

do u want the answer for reference?

[cedar kiln]

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.

[rough adder]

ok so

[rough adder]

I understand that when you have something like this

you can switch the derivatives into:

my question is

what happens when one of those derivatives is a second derivative?

for example:

are you able to swap them still? like would the above equation become this:

?

[south tundra]

Yes, that's perfectly fine

[rough adder]

slay, thank you so much

.close

[cedar kiln]

Channel closed

Closed by @rough adder

Use .reopen if this was a mistake.

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.

[shy coyote]

Could anyone guide me on how to annotate this? P(X ≤ 6) = Σi=0 to 6 (36 choose i)

[tropic oxide]

define "annotate"?

[cerulean valley]

Do you mean 0 to x?

[tropic oxide]

did you forget something in that sum?

[shy coyote]

I probably used the wrong word, just to like actually "use" the functions instead of explaining them

like for example

idk if that makes sense

[cerulean valley]

I don't think it does

Ohhhhhh

[shy coyote]

like

[cerulean valley]

I get it

[shy coyote]

i dont know how to write it in maths terms

instead of word/english terms lmao

[cerulean valley]

You write it like a vector

[mighty drift]

$\binom{36}{i}$

[wraith dagger]

mateo713

[cerulean valley]

Yeah

[shy coyote]

thats it

tysm guys <33

.close

[cedar kiln]

Channel closed

Closed by @shy coyote

Use .reopen if this was a mistake.

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.

[half quiver]

Hi

[half quiver]

Could anyone help me find what i did wrong / what i could have done to solve it easier for this question

Determine the real value(s) of the parameter m in the equation 2z5 − z4 + (12 + m)z3 − (4 + m)z2 + 8mz − 16 = 0 such that this equation has a double root that is purely imaginary. Also, determine the roots of the equation for the found parameter value(s).

[cedar kiln]

@half quiver Has your question been resolved?

[half quiver]

<@&286206848099549185>

[cedar kiln]

@half quiver Has your question been resolved?

[cedar kiln]

@half quiver Has your question been resolved?

[half quiver]

<@&286206848099549185>

[half quiver]

<@&286206848099549185>

[burnt vapor]

If the imaginary root is double you could make z3 = ai and z4 = -ai

Because both ai and -ai must appear two times

[half quiver]

couldng z3 and z4 be real values instead of imaginary?

[burnt vapor]

There is a double imaginary solution (ai), so -ai is also a double solution, therefore there are at least four imaginary roots

So no, z3 and z4 can't be real

Except for the case a = 0, of course

But 0 is not a solution in this case

[half quiver]

im confused.. i get that ai and -ai have to be solutions but why 4?:( so we have z1= ai z2= -ai and then z3 = ai and z4 = -ai again?

[burnt vapor]

Well

ai is a solution

So you can write z1 = ai

But it is a double solution (it says that in the question)

So you can write z2 = ai

Then, since z1 = ai, -ai must be a solution too

So write z3 = -ai

And similarly z4 = -ai

[half quiver]

ohhh i get it now thanks!

could you possibly help me with this one too?

answer shld be m>=1

question : for what values of m does A only have real eigenvalues

[burnt vapor]

You made a mistake from line 3 to line 4

The second determinant is -m+λ

But you wrote (-m)(+λ)

[half quiver]

last q (im so sorry): what did i do wrong here? the question is : Show that the equation
z3 + 2z + 2 = 0
has only one real root and that the other two roots are complex and lie outside the circle |z| = √2.

[storm mural]

rolles theorem for the first one
you forgot +a^2 here

[half quiver]

oh bruh.. thanks. so what i did wouldve been correct too?

[storm mural]

not sure

[half quiver]

Determine an invertible matrix P and a matrix C of the form
C =[a −b; b a]
such that the given matrices from exercise 1 have the form A = P CP^-1.

I dont understand how they got that matrix for C ..

[half quiver]

nvm!

.close

[cedar kiln]

Channel closed

Closed by @half quiver

Use .reopen if this was a mistake.

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.

[crimson sedge]

tell me your approach and then I can help you out

yep

good

that's correct

yep it's needed

you need to sort and then find the middle value if total terms are odd or else for even no. of terms you have to take average of middle 2 terms

It's definition

[crimson sedge]

this is definition of median

[cedar kiln]

@crimson sedge Has your question been resolved?

Channel closed

Closed by @smoky moat

Use .reopen if this was a mistake.

[cedar kiln]

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.

[lethal bison]

answer is 912 anyone knows how?

[lethal bison]

.close

[cedar kiln]

Channel closed

Closed by @lethal bison

Use .reopen if this was a mistake.

[cedar kiln]

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.

[smoky stump]

But the answer is 7

what did. I do wrong?

[flint plinth]

what is the question? part of it is missing

[smoky stump]

that is the entire question

Oh

it is to find the shortest distane

[flint plinth]

one way is to observe that you want (a-x) to be orthogonal to the line

[smoky stump]

yeah

[flint plinth]

i just tried that and got lambda = -1 (which results in a distance of 7 as claimed)

what did you get for lambda?

[smoky stump]

I didn't get a lambda,

All I did was plug into this calculator

[flint plinth]

oh, i see

[smoky stump]

These were my vectors

And I got th3 normal of the projection of L-A onto v

which results in a 9

L is the point on the line

[flint plinth]

does Projection do the right thing if v is not a unit vector, or do you have to scale v ?

[smoky stump]

it is the same thing

it is just the (a dot b)/ (a^2) * vector a

[flint plinth]

what if you made it a-L instead of L-a

[smoky stump]

what do you think?

[flint plinth]

oops no scratch that

hang on

[smoky stump]

Have you got it?

[flint plinth]

ah here we go

Projection(a-L, v) gives you a vector

you need to add L to that to get the point on the line, the one that is closest to a

[smoky stump]

yes

yeah, but what is the point of getting on the line

[flint plinth]

you want to compute the distance from a to the closest point on the line

a is a point

[smoky stump]

yeah

[flint plinth]

you need a second point

[smoky stump]

yeah

I did with a-L

[flint plinth]

a-L is a vector from L to a

you project that onto the vector v

so far so good

the result is a vector

[smoky stump]

yeah

[flint plinth]

the result is not a point on the line

[smoky stump]

sure it is not

[flint plinth]

to get the point on the line you need to add L

so the right formula would be:

norm(a - Projection(a-L, v) - L)

or if you prefer:

just let w = a-L, then the formula is simply norm(w - Projection(w, v))

[smoky stump]

that gets you 14

I also don't see how a-L would matter since you are orientating back to the 0th axis

[flint plinth]

>> L = [1, 2, -5];
a = [-2, 1, 5];
v = [6, 3, -4];
>> w = a - L
w =
    -3    -1    10
>> % project w onto v
>> v_unit = v / norm(v)
v_unit =
         0.768221279597376         0.384110639798688        -0.512147519731584
>> proj = (w*v_unit')*v_unit
proj =
    -6    -3     4
>> offset = w - proj
offset =
     3     2     6
>> norm(offset)
ans =
     7
>> 

^matlab

[smoky stump]

dammit, I don't understand it. screw it

I got it, it was just the stupid calculator broken

.clos

thanks

.close

[cedar kiln]

Channel closed

Closed by @smoky stump

Use .reopen if this was a mistake.

[cedar kiln]

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.

[plush glacier]

Is this correct?

[buoyant latch]

Missing bracket on the last one

f’(x) = (6x-1)cos(3x² - x)

[cedar kiln]

@plush glacier Has your question been resolved?

[plush glacier]

Ohh alright, thanks

But in the 2nd one, i dont need to seperate them by parenthesis right?

.close

[cedar kiln]

Channel closed

Closed by @plush glacier

Use .reopen if this was a mistake.

[cedar kiln]

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.

[acoustic plinth]

how do i find the derivative of this?

[uneven iron]

First, take the derivative of 2^x, and then, take the derivative of 2tIn(2), and subtract both results

[acoustic plinth]

what’s the derivative of ln(2)

so like 2^t is 2^t * ln(2) right

[uneven iron]

$\frac{d}{dx} \ln(a) = \frac{1}{a}$

[wraith dagger]

Rub05

[uneven iron]

Yes

[acoustic plinth]

so like that?

[lyric narwhal]

No

[uneven iron]

Yes, but that's not the final result

[lyric narwhal]

ln(2) is just a constant

You treat it like a constant

[acoustic plinth]

yes someone also told me that

[uneven iron]

Oh, that's correct, my bad

[acoustic plinth]

so like what will it be

ln(2)

[uneven iron]

0

[acoustic plinth]

1 ?

[lyric narwhal]

No

[uneven iron]

It's a constant

[lyric narwhal]

You have 2ln(2)t

[acoustic plinth]

ohh okayy i got itt

[lyric narwhal]

So it's a constant * your differentiating variable

[acoustic plinth]

like that?

[lyric narwhal]

No

[acoustic plinth]

the derivative of 2^t is 2^t * ln(2) right ?

[lyric narwhal]

Yes it is

[lyric narwhal]

.

[acoustic plinth]

that’s what u mean?

[lyric narwhal]

No.

How would you differentiate something like 2t

[acoustic plinth]

2

for example 2x = 2

[lyric narwhal]

So in your case, you have 2ln(2)t

But it's still the same

[lyric narwhal]

Bad notation

[acoustic plinth]

can u like tell me what’s wrong here

[lyric narwhal]

Why are you differentiating that as aa product

Firstly, that's not how you differentiate products anyway

Secondly, this is not a product, it is a constant multiple of t

[acoustic plinth]

product rule?

[lyric narwhal]

I just said you don't need to use the product rule

[acoustic plinth]

sorry i’m just confused

can u tell me what should i do

[lyric narwhal]

How would you differentiate, say 9x

[worldly iron]

Well, if you apply the differentiation formula, it is 9 only

you can actually verify it by yourself using the first principal method

you know, the limit defintion of a derivative

[lyric narwhal]

?

I know what the answer is

And I didn't ask yoj

[worldly iron]

this is not a place for personal help, see #❓how-to-get-help

[lyric narwhal]

I'm helping op..

Pls stop

[acoustic plinth]

it’s going to be like that right

[lyric narwhal]

Yes

[acoustic plinth]

okayy

thanks alot

[cedar kiln]

@acoustic plinth Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[crimson sedge]

given l1 & l2 are perpendicular

[crimson sedge]

can anyone tel me how did they do it?

i mean how beta= alpha + 90

[gritty viper]

just start writing angles in terms of alpha and beta

maybe just alpha

every single angle in this picture can be found in terms of alpha

so just find beta in terms of alpha

[crimson sedge]

i don't really understand

[gritty viper]

Find angles in terms of alpha

[crimson sedge]

so is beta + alpha = 180?

[gritty viper]

No

[storm mural]

the angle next to beta is inside the triangle

the triangle has that, alpha, and a 90d egree angle

inside it

[crimson sedge]

then is it like this ??

[gritty viper]

yeah that's correct

[crimson sedge]

oh ok

thx

[gritty viper]

np

[crimson sedge]

.close

[cedar kiln]

Channel closed

Closed by @safe flicker

Use .reopen if this was a mistake.

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.

Channel closed

Closed due to the original message being deleted

[woven moon]

.close

[cedar kiln]

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.

[cerulean bloom]

If x belongs to the set (-2,5) then find the interval in which 1/x lies.

[dull oxide]

[cerulean bloom]

The answer would be (-∞, -0.5) U (0.2, ∞), I think but what about the value x=0? Is it's reciprocal to be considered here as well?

[cerulean bloom]

nice cat btw

[dull oxide]

How'd you get this?

[proven current]

yea that dont make sense

[gritty galleon]

x=0 is intermediate no

[cerulean bloom]

by thinking i guess

cuz like

[dull oxide]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[steep umbra]

X = 0 doesnt lie in (2,5)

[cerulean bloom]

-2,5 i meant

sorry for the typo

[dull oxide]

x is in the interval (-2, 5)?

[cerulean bloom]

yes

[dull oxide]

Then this is correct. You don't consider x=0

[cerulean bloom]

Oh

dumb me ig

[dull oxide]

Nw

[cerulean bloom]

my teacher told that the reciprocal of 0 should be infinite

[proven current]

thats not a number though

[steep umbra]

Noooo

[dull oxide]

Your teacher is making bad assumptions

[cerulean bloom]

now i was confused whether it was negative or positive infinite

[cerulean bloom]

wym

[dull oxide]

1/0 is not infinite

[steep umbra]

1/0 is not equal to infinity

[dull oxide]

It's undefined

[cerulean bloom]

like

error 404

i guesss

[steep umbra]

Yup

[cerulean bloom]

thanks guys

[proven current]

403 forbidden

[dull oxide]

If you get into limits, you can consider infinite. But in regular arithmetic, no. It's simply not defined

[cerulean bloom]

how to close this question thing

[cerulean bloom]

okay thanks

[dull oxide]

.coose

[cerulean bloom]

.coose

.close

[cedar kiln]

Channel closed

Closed by @cerulean bloom

Use .reopen if this was a mistake.

[cerulean bloom]

.close

why does it still show it in the occupied things

.close

[crystal raptor]

Takes time for bot to move it

.coose

[cerulean bloom]

oh okay

.coose

[cedar kiln]

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.

[vagrant hamlet]

im stuck

im not sure how to continue

[tropic oxide]

this step is illegal

[vagrant hamlet]

wha

[tropic oxide]

ln(a+b) ≠ ln(a) + ln(b)

[vagrant hamlet]

?

i dont see it

[tropic oxide]

you tried to take the logarithm of both sides, yes?

[vagrant hamlet]

yea

[tropic oxide]

you cannot simplify $\ln(e^t + 30e^{-t})$ into $\ln(e^t) + \ln(30e^{-t})$

[wraith dagger]

Ann

[tropic oxide]

(and later on you cannot simplify ln(30 e^(-t)) into ln(30) * ln(e^-t)

[vagrant hamlet]

they are together?

what

[tropic oxide]

"they"? "together"?

[vagrant hamlet]

yes

i thought they are single term

no?

[tropic oxide]

who are "they"

[vagrant hamlet]

e^t+30e^-t

[tropic oxide]

these are two things which are added

they are not one term

and my point still stands, the logarithm of a sum does NOT equal the sum of the logarithms

[vagrant hamlet]

ok

so

i ln

then what

[tropic oxide]

i don't think taking the logarithm of both sides is helpful at this point.

[vagrant hamlet]

i can see

[tropic oxide]

you're better off substituting x := e^t and then solving the resulting almost-quadratic equation

[vagrant hamlet]

ok

i didnt get the answer

the answer at the back is 1.70

[tropic oxide]

what is x^-?

[vagrant hamlet]

30/x

no?

[tropic oxide]

yeah but why do you write it this way

also i just realized.

[vagrant hamlet]

sub x

uh

[tropic oxide]

what's "sub x"?

also i just realized you have in fact been solving the wrong equation all along

you want ds/dt = 0, not s = 0.

[vagrant hamlet]

let e^t be x

oh

its a

dirivative thingy

[tropic oxide]

you have

when s = 0
this is an incorrect reading of the problem

yes, it's the "dirivative thingy" that you want.

[vagrant hamlet]

but

[tropic oxide]

you screwed up your algebra, which drew my attention first.

[vagrant hamlet]

how do u know whether it is

derivative

[tropic oxide]

you know whether "it" "is" "derivative" by reading "it"

[vagrant hamlet]

the question doesnt seem to give clues

[tropic oxide]

yeah it does.

[vagrant hamlet]

how do u know

[tropic oxide]

"at rest"

do you know what this means

[vagrant hamlet]

the particle is not moving

[tropic oxide]

exactly

and what does it mean that the particle isn't moving, in terms of its equation of motion?

[vagrant hamlet]

s metres is 0

[tropic oxide]

no

[vagrant hamlet]

oh

[tropic oxide]

s=0 means the particle is located at position 0

[vagrant hamlet]

oh

[tropic oxide]

says nothing about where it's going

[vagrant hamlet]

Im finding the speed?

somethinbg like that?

[tropic oxide]

no

you're finding when the speed equals 0

[vagrant hamlet]

oh

s/t

does displacement do a thing here to tell me whether it is a derivative quesiton

oh

so its the ds/dt

[tropic oxide]

.

as i was saying!

[vagrant hamlet]

are there similar questions like that but itsnt about derivatives

if the question change to when the particle is at 0 metres it means the question will not be derivative anymore?

[tropic oxide]

it means that THAT question won't involve derivatives anymore yes

[vagrant hamlet]

yes

okay i understand

i cant seem to get the right answer now

[hidden pike]

if s is correct, there is a problem in the derivative

[vagrant hamlet]

hm?

31 give 0

e^t also =0

[hidden pike]

oh nvm

[vagrant hamlet]

ye

[hidden pike]

haven't read that ds/dt = 0

[vagrant hamlet]

oh

did i do something wrong

[hidden pike]

normally

d/dx(31 - e^x - 30 e^(-x)) = 30 e^(-x) - e^x

[vagrant hamlet]

hm

i dont think so-

[hidden pike]

why

[vagrant hamlet]

e is a number

all number = 0

[hidden pike]

hm

but if you mean e = euler's constant, then e=2,71

but the derivative of e^x is different than 0

cuz is not just a number

is a number to the power of x

[vagrant hamlet]

oh

i see

[hidden pike]

so

when you do the derivative

you get 1e^x

[vagrant hamlet]

yes

[hidden pike]

imagine now you have e^-x

the derivative is -e^-x

[vagrant hamlet]

yes

i see

[hidden pike]

so that's it

the result can't be just 30e^-x

try it again, and when you have the result ping me

ill check

[vagrant hamlet]

ok

t= 1.70

3sf

@hidden pike

[hidden pike]

show how you did

[hidden pike]

hmm

are you sure?

[vagrant hamlet]

i think so

[hidden pike]

yep it's that

[vagrant hamlet]

yay

[hidden pike]

well done buddy

[vagrant hamlet]

thanmk

[cedar kiln]

@vagrant hamlet Has your question been resolved?

Channel closed

Closed by @vagrant hamlet

Use .reopen if this was a mistake.

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.

[smoky trout]

Hi, can someone exlpain to me how this:

[smoky trout]

can be written in this form:

[cedar kiln]

@smoky trout Has your question been resolved?

[cedar kiln]

@smoky trout Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[cerulean bloom]

When the product of four consecutive odd positive integers divided by 4, what are the set of remainders?

[cerulean bloom]

How do we even start solving this

[tropic oxide]

you know that remainders mod n behave nicely wrt addition and multiplication, yes?

[cerulean bloom]

what

I think I couldn't understand what you mean

[crimson delta]

(a+b) mod n = [(a mod n) + (b mod n) ] mod n

i.e. take remainders first, then add (and then maybe take remainder again)

is the same as add first, then take remainders

[cerulean bloom]

by mod n do you mean |n|?

[radiant topaz]

Ni

[crimson delta]

a mod n is the remainder of a when divided by n

[violet night]

modular arithmetic

[cerulean bloom]

oh

i didn't know that

I have to learn some concepts i guess

which concept exactly tho?

[radiant topaz]

.

[cerulean bloom]

Oh okay

thanks

[violet night]

you could probably get the answer without it

[cerulean bloom]

how tho

[violet night]

ahhh well I don't know how 'proper' it is, if you expand everything , take a 4 out, obviously any 4n has a 0 remainder when divided by 4 so you just have to consider the constant you're left with, but I wouldn't know how to explain going from that to the remainder without modular arithmetic tbf , I guess write your constant as 4n+k and that k would be your remainder

[tropic oxide]

let the lowest of your integers be 2n+1

then the next one up will be 2n+3

and so on

then write down the product and expand it as far as is necessary

[radiant topaz]

Do you think (2n-1)(2n-3)(2n+1)(2n+3) choices are better?

[violet night]

steps in getting the correct answer are the same but it does simplify nicer

[cedar kiln]

@cerulean bloom Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[sly hemlock]

I'm not sure where to start here...

[south tundra]

Rewrite e^(x + 3) as e^3 * e^x

And bring 6e^3 behind the sign of integration

[sly hemlock]

ok. i'm just barely starting with integration so the methods haven't clicked yet

can you explain what you did there? i'm still lost

[south tundra]

$\int_1^36e^{x+3}\dd{x}=\int_1^36e^xe^3\dd{x}=6e^3\int_1^3e^x\dd{x}$

[wraith dagger]

A Lonely Bean

[sly hemlock]

ok. i misread your previous message

now i have to evaluate using that last step

so i would get 6 * 1^3 * (2/n)

because dx = delta x

but i don't know the value of n

to use given the formula delta x = b-a/n

[south tundra]

Huh? You are given that the integral of e^x from 1 to 3 is e^3 - e

[south tundra]

So it's $6e^3(e^3 - e)$

[wraith dagger]

A Lonely Bean

[sly hemlock]

ok

for some reason i didn't see the constant e at first

that makes sense.

i need to find a similar example problem. i see how this problem works out, but i don't understand the concept

[south tundra]

Basically you need to get the form that's given

[sly hemlock]

ok

what are the forms themselves called so i can do some googling

[south tundra]

Whatever is given

[sly hemlock]

ok. thanks

.close

[cedar kiln]

Channel closed

Closed by @sly hemlock

Use .reopen if this was a mistake.

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.

[red pumice]

How come you can say

[red pumice]

that Y - Y^ = (I - H)Y

since we hace Y = Xbeta + epsilon

and Y^ = Xbeta^

so Y - Y^ = Xbeta - Xbeta^ + epsilon

with Xbeta^ = H

or alternatively, can i say that cov(Y, Y^) = 0?

[cedar kiln]

@red pumice Has your question been resolved?

[red pumice]

.close

[cedar kiln]

Channel closed

Closed by @red pumice

Use .reopen if this was a mistake.

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.

[rain light]

Does anyone know how to write out the lambert W function in a calculator (precisely the FX-991 ES Casio Calculator)

[glossy fossil]

I need help with pedal equation

[rain light]

make ur own help channel

[glossy fossil]

?

[south tundra]

This channel is occupied

[south tundra]

You want to graph it?

[rain light]

no no no, like i wanna find out the value of it

[south tundra]

Ah, then not sure if regular calculators can do that

[rain light]

is there no way to somehow alternatively write out the lambert W function?

[south tundra]

I guess you could use its Taylor series

[rain light]

Sorry, I don't know whats a Taylor series

[south tundra]

It's a polynomial function serving as an approximation of another function

By including infinitely many terms of it you get the function itself

[rain light]

interesting, sorry am still at the basics of calculus

[south tundra]

E.g. sinx = x - x^3/3! + x^5/5! - ...

[rain light]

so how do I exactly find out the value of a certain W function using "taylor series"?

[south tundra]

So, this is the general formula for taylor series

[rain light]

Ok but what does this have to do with the W function?

[south tundra]

Using it yields that the cubic approximation for W function is x - x^2 + 1.5x^3

[south tundra]

[rain light]

what? are you saying thats the equation for the W function or something? lmao sorry im dumb as shit

[south tundra]

x - x^2 + 1.5x^3 is the (cubic) approximation for W(x) around 0

The more terms you involve the better the approximation will be

[rain light]

[rain light]

yeah no I am absolutely clueless, I guess I will just wait till College till I learn this shit.

[cedar kiln]

@rain light Has your question been resolved?

[cedar kiln]

@rain light Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[peak void]

how do I go about doing this problem?

i know the first step is to write it in integral form

so then its the int from 0 to 1 of e^3x^2

then from there i evaluated that at 0, which turned out to be 1

then the derivative of that evaluated at 0 also turned out to be 1

i don't know if i was doing it wrong, but then the second derivative turned out to be 0, which I know isn't right

[cedar kiln]

@peak void Has your question been resolved?

[peak void]

<@&286206848099549185>

[peak void]

<@&286206848099549185> ?

[swift field]

hi

can you wait like 20 mins i can come hlep you

[peak void]

👋

[peak void]

sure

[cedar kiln]

@peak void Has your question been resolved?

[peak void]

😢

<@&286206848099549185>

i really need some help here

🙂

[storm mural]

you gotta write the maclaurin series for this inside the integral and then integrate from 0 to 1

term by term

[peak void]

ok let me try that thank you

[swift field]

I COMPLETELY FORGOT

[swift field]

yes do that

[peak void]

you forgot about me 😦

[swift field]

my baddd

but yea just use

[peak void]

how do i use that in this problem?

[storm mural]

that's the series for e^x

you can adapt it here though

[peak void]

so do i use substitution then?

[storm mural]

ya

[peak void]

and instead of x^n in the numerator

i would have 3x^2?

^n

[storm mural]

yes

[peak void]

then how would i take the integral of that

im sorry this is rlly confusing for me

[swift field]

use the series expansion

to get the first 5 terms

and integrate each term seperatuely

so

[peak void]

im working it out now

[swift field]

integratre that

[peak void]

thank you so much

big help

.close

[cedar kiln]

Channel closed

Closed by @peak void

Use .reopen if this was a mistake.

[cedar kiln]

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.

[peak void]

so i am using this formula

[radiant topaz]

Good

[peak void]

so the first term is 0

the second term is 0x

and the third term should be (2x^2) /2

where am i going wrong here

[radiant topaz]

Try this again

[peak void]

ok the deriative of sin(x^2) is 2xcos(x^2)

[radiant topaz]

Ok

[peak void]

oh ok so then it should be 2x^3/2

but then the twos cancel out

and you just get x^3

with no denominator?

[radiant topaz]

What?

[dusk finch]

This should be evaluated at x=0, giving you 2*0*cos(0)

and anything times 0 is 0

[radiant topaz]

That's already done

King is moving to second derivative

[dusk finch]

this is second derivative?

[peak void]

i am getting more confused now lmao

[radiant topaz]

I mean third

[radiant topaz]

Find the derivative of 2xcos x²

[peak void]

that would be for the fourth term right?

[radiant topaz]

Third

Derivative

And fouth term yes

[peak void]

ok so for the third term

is it 0?

i dont think it should be

i keep getting x^3

[radiant topaz]

It's not 0

Although get f'''(0) first then divide it by 3!

[peak void]

would that impact the third term?

[radiant topaz]

No

[peak void]

where is the three in the denominator coming from?

[radiant topaz]

You will get coefficient of x³

[radiant topaz]

(f'''(0)/3!) x³

[peak void]

ok for the third derivative

i got

2cos(x^2) + sin(x^2)(4x^2)

[radiant topaz]

At x=0

[peak void]

then the term w sin cancels out

and im left w 2

[radiant topaz]

Yes

[peak void]

ok...

so then

[radiant topaz]

Divide by 3! To get coefficient of x³

[peak void]

1/3

i get that it got the answer but it like doesn't make sense

[radiant topaz]

You know the series for sin x?

[peak void]

(-1)^n/(2n+1)! * x^2n+1

[radiant topaz]

Where's x

[peak void]

like that?

[wraith dagger]

fäf

[radiant topaz]

You didn't

[peak void]

wait im sorry how did you get from the first line to the second

[radiant topaz]

Integration

[peak void]

oh you just integrated

why do you do that here

[radiant topaz]

f'(x) = sin(x²)

So f(x) = integration of sin(x²) + c

[peak void]

OOOOOOH

i see it now

you use substitution

put in the x^2 into the sin series

then since that is f'

you have to find the series for f

so you take the integral of that

then you just start plugging in values for n starting with 0 and taking steps of 1

[radiant topaz]

Yes

[peak void]

thank you so much

sorry that took so long for me to understand

[radiant topaz]

The first method we did wasn't really working here

[peak void]

i just had to see it done out like that

[radiant topaz]

I thought we would eliminate options

But that didn't happen

[peak void]

this helped a ton

[radiant topaz]

Alright

[peak void]

it gave what you were saying context

have a good day

[radiant topaz]

Nite

You too

[peak void]

🙂

.close

[cedar kiln]

Channel closed

Closed by @peak void

Use .reopen if this was a mistake.

[radiant topaz]

Wish me good night and sweet dreams

I miss them

[peak void]

oh yea how could i forget

gn, sweet dreams

dont let the bed bugs bite

[cedar kiln]

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.

[eternal knot]

Calculus with parametric equations (original question below)

[ionic finch]

@eternal knot

[eternal knot]

hello

[ionic finch]

How did you find (0,2)?

[eternal knot]

I plugged in t= -1 for x and y

oh shoot i messed up

for y I did (-1)^4 +1 instead of (-1)^4 -1

[ionic finch]

right

[eternal knot]

got it thanks

[ionic finch]

Great

[eternal knot]

.close

[cedar kiln]

Channel closed

Closed by @eternal knot

Use .reopen if this was a mistake.

[cedar kiln]

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.

[ebon island]

Can someone recommend me a yt video about direct proofs?
Prove using the direct method that a product of two integers is divided by 7, if
one of its factors is divided by 7.

[surreal aurora]

I cannot recommend a yt video but I can help you understand direct proof.

[ebon island]

we can try, english is not my first language

can u help me with that problem?

[cedar kiln]

@ebon island Has your question been resolved?

[cedar kiln]

@ebon island Has your question been resolved?

[midnight zealot]

Lol is this webassign

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[outer sierra]

anybody here do logic prolog?

[frosty vapor]

i know a small bit, i might be able to help

[cedar kiln]

@outer sierra Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[ebon smelt]

I dont understand this but can anyone show me how to do a couple

<@&286206848099549185>

[dull oxide]

$\sqrt[n]{a}=a^{\frac{1}{n}}$

[wraith dagger]

SWR

[dull oxide]

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

[wraith dagger]

SWR

[dull oxide]

$(a\cdot b)^c=a^c\cdot b^c$

[wraith dagger]

SWR

[ebon smelt]

👍

[dull oxide]

That's all you need for all of them

[ebon smelt]

and these are the answers after I conver them

<@&286206848099549185>

I still dont get it with the formulas

[nimble veldt]

what is your question?

[ebon smelt]

.

[ebon smelt]

.

I dont get the fractions were the exponent it

is*

I am on number 3

[nimble veldt]

use this ....

[ebon smelt]

I know but I just get a exponent inside the squareroot

[nimble veldt]

and this ...

[nimble veldt]

and what do you expect?

[ebon smelt]

How do I get it out

[nimble veldt]

i do not understand, what is your expected result?

[ebon smelt]

But after that formula I get

[nimble veldt]

2^5 = 2^3*2^2

[ebon smelt]

what about number 10

[nimble veldt]

hmmm, you have all rules you need posted by swr. if you have troubes with them it would be nice to tell where you stuck,

[ebon smelt]

what do I do since I have to pharentesses

do I multiply the whole thing

and how

[nimble veldt]

this

[ebon smelt]

k

[nimble veldt]

and this

[ebon smelt]

okay I think I got it

I was confused on this one thing

the rest is now easy

ty

[nimble veldt]

youre welcome.

[cedar kiln]

@ebon smelt Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[lethal bison]

what's this symbol called?

[latent bloom]

Product

But it's the capital version of Pi

Π

But in this form, you use it to denote a product of a group of numbers

[lethal bison]

alright thank you

.close

[cedar kiln]

Channel closed

Closed by @lethal bison

Use .reopen if this was a mistake.

[latent bloom]

[cedar kiln]

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.

Channel closed

Closed due to the original message being deleted

[cedar kiln]

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.

[crimson sedge]

I really don't know where to start...

I've tried multiplying out

that doesn't work

[inland ocean]

Can you show your work

[crimson sedge]

It's on paper so I'll just tell you what I got:
So I multipleid out by the denominators and got:
(2 + cotx + tanx)(cosx - sinx) = (cotx + 1) (cosx) - (1+tanx)(sinx)

it didn't really seem to add up to anything

[inland ocean]

For the first step rather than multiplying convert the tan and cot in terms of sine and cosine

First

[crimson sedge]

ah gotcha

I was thinking that but decided against because of the + 1

[crimson sedge]

I did that but it looks like a mess

cosx / 1 + (sinx / cosx)

how would I proceed with that

[inland ocean]

Yeah it does ngl

[crimson sedge]

yup plus it's a part a question

which leads to another part b

so I'd lose out on 8 marks if i couldn't solve it

[inland ocean]

Okay wait a min

$\frac{cosx}{1 + tanx} - \frac{sinx}{1 + cotx}$

[wraith dagger]

ColdTee

[inland ocean]

Yeah converting is better option here

Easier as well

[crimson sedge]

how would it work

[inland ocean]

$\frac{cos^2x}{cosx + sinx} - \frac{sin^2x}{cosx + sinx}$

[wraith dagger]

ColdTee

[inland ocean]

If you simplify

Try it out

Use (a^2 - b^2) = (a+b)(a-b) for numerator

[crimson sedge]

well then I have to begin rooting cosx + sinx doesn't that just become a mess

[inland ocean]

Rooting?

I dont understand

Why do you need to root

$\frac{cos^2x - sin^2x}{cosx + sinx}$

[wraith dagger]

ColdTee

[crimson sedge]

oh

I thought you meant turning that entire thing and rooting it to put it into the for of a-b a+b