help-13

[mystic plover]

Say

Sum x=0 to q of |x>|f(x)>

Where q is like the size of the finite field

x is the input and f(x) is the output

I want to know what is the inner product of it with itself and how to do this generally

I guess to normalize

1/q * Sum x=0 to q of |x>|f(x)>

It's not that important to how to do it though

[crimson delta]

you can't divide by q

q=0

so what you want is $\sum_{x\in\mathbb F_q} \langle x, f(x)\rangle$ ?

[wraith dagger]

Denascite

[crimson delta]

how exactly do you define the inner product?

Tr(x*f(x)) or something?

or do you mean that this whole sum is supposed to be the inner product

[cedar kiln]

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[crimson sedge]

How do you simplify this

[gritty viper]

you have to factor

[crimson sedge]

How do you factor the bottom

[velvet mortar]

Factor by grouping

[gleaming cloud]

Yo are you diegoro?

@velvet mortar

[rain drift]

notice that we can re-write $x^3 - x^2 - 4x + 4$ as $x^2(x - 1) - 4(x -1)$ Now you have a common factor of $(x-1)$. Do the same with the numerator and you'll find a common factor

[wraith dagger]

MellowDramaLlama

[velvet mortar]

You have a very creepy description on about me

[gritty viper]

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

You can’t factor by grouping???

.close

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[wary anchor]

what would $2sin(3x) * sin(3x) $ would be

[wary anchor]

what would $2sin(3x) * sin(3x)$ would be

[wraith dagger]

yomiko

[sage forge]

2sin^2(3x)?

[wary anchor]

i guess so right

yeah its right

thanks

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[long swan]

\sin

$\sin x$

[wraith dagger]

Mr. Gamer

[cedar kiln]

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[crimson sedge]

If i question has a number being increased by a percentage that would always be exponential right?

[gleaming cloud]

What's "that"

[crimson sedge]

oh sorry i just realized that question wasn't very specific im talking about like for example you have 1,500 dollars but pay 6.5% interest annually that would be exponential not linear right?

@gleaming cloud

[mystic plover]

You inner product the entire state

The 1/q isn't 1/0

it's 1 over the field size

Because there are q field elements

[crimson delta]

q=0 in F_q

[mystic plover]

That is the normalization

No

The entire thing isn't mod q

And also it won't make sense

[crimson delta]

open a new channel

[mystic plover]

It's just f(x)

And x

Are in mod q

@crimson delta 15

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[midnight cipher]

How to show 1<y<9?

<@&286206848099549185>

[cedar kiln]

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@midnight cipher Has your question been resolved?

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[wild condor]

Not sure how to solve at all

[cedar kiln]

Please don't occupy multiple help channels.

[solemn torrent]

1)angles on a straight line

2. vertically opposite angles

add all the angles in the triangle

=180

and solve for x

[wild condor]

So X=4.7?

I did 3x -6 + 8x + 4 + 130 = 180

[eager sparrow]

@midnight cipher Are you still trying to find out the answer for your question?
I just read about it and I already have a solution

[solemn torrent]

130 isn’t inside the triangle

[wild condor]

Oh

[eager sparrow]

It's a straight line so 130 + the angle beside it = 180

Since they are supplementary angles

Angles that add up to make a straight line

[wild condor]

Not sure how my equation is supposed to look

[eager sparrow]

The inside of a triangle is 180

All the angles inside a triangle adds up to 180

According to vertical angle theorem, opposite angles from intersecting lines are equal
So the angle at top of 8x+4 is also equal to 8x+4

[wild condor]

Hmm I see

[eager sparrow]

Then the right top angle is a supplementary angle for 130
So that angle + 130 =180 or 180-130

Then add all of it and equate to 180

[eager sparrow]

@midnight cipher Try to multiply x-1 on both side

Try to equate for x

The left side would be yx-y = 2x^2 + x - 1
Then transpose the left side of the equation to the right side to create a quadratic equation that equates to 0

0 = 2x^2 -yx+x-1+y

You're "a" would be 2, b would be +1-y, and c would become -1+y

Substituting those in the quadratic equation, we're gonna focus on the square root, because after substitution, inside the square root would become "y^2-10y+9". Equate it to zero since anything negative inside a square root is an imaginary number.

Then you would find that the roots of y is 1 and 9

It means anything between them is an imaginary number and not a real one

[cedar kiln]

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[cedar kiln]

@crimson sedge Has your question been resolved?

[keen elbow]

bro

[cedar kiln]

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[sonic thistle]

$(-1)^{\frac{3}{5}}$

[wraith dagger]

Master Oogway

[sonic thistle]

,w (-1)^(3/5)

[wraith dagger]

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Requested by Master Oogway#0599

[sonic thistle]

Which answer is correct?

[rain drift]

Mathway. -1 raised to an odd power is -1. Similarly, -1 taken to an odd root is also odd. So it never changes sign

[fair geyser]

both are correct, like sqrt(9) has two answers

the one wolfram gives is more default

[cedar kiln]

@sonic thistle Has your question been resolved?

[sonic thistle]

Ann?

Can you help?

I need a third opinion

[tropic oxide]

my opinion is

why do you even care

are you just asking this question for its own sake here

[sonic thistle]

[fair geyser]

the primary goal of the server is to make people seethe via trolling

[tropic oxide]

WA assumes you want exp(3 i pi/5), mathway assumes you want a real number
it is no use to ask "but which is the CORRECT way" - is there a problem you are solving here that requires you to raise negative numbers to fractional powers?

[sonic thistle]

So they're both correct

Thats what i needee to know

[sonic thistle]

So you gave me the wrong answer on purpose?

[fair geyser]

I can't think of a suitable answer at this moment, it's definitely not "yes"

[sonic thistle]

Sure

[fair geyser]

i'll ping if i come up with something

[sonic thistle]

Should i ping mods?

[fair geyser]

why are you asking me, i'm not a mod i don't give pinging permissions

[sonic thistle]

Im just asking

Wether you trolled on purpose

[fair geyser]

i've never been in this scenario, i can;t give advice either

i said i'll ping!

i haven;t come up with an answer yet

have patience etc.

[sonic thistle]

Ok...

[fair geyser]

@sonic thistleI gave you the right answer by mistake, my intention was to give the wrong one

[sonic thistle]

Why was that your intention?

[fair geyser]

i don;t reflect on my intentions, it seems unnecessary

[sonic thistle]

Intentions are still punishable...

[sonic thistle]

What even is that exp thing

[tropic oxide]

exp() is another way to write e^x

exp(z) = e^z

[sonic thistle]

Why would wa assume this

So wa is assuming that i asked e^(3 i pi/5)

Weird

Oh and 1 more thing

It says -1 can only be raised to a fractional exponent with odd numerator

[fair geyser]

this is least correct

because it's not satisfying at all

[sonic thistle]

Therefore the answer is undefined (not real)

Least correct

[sonic thistle]

Or incorrect

?

[fair geyser]

these are all 5 answers

one is even real, imagine that

[sonic thistle]

Ok?

How does this link to the problem

Oh

Wait

Yeah i understand

[sonic thistle]

Isn't the real one -1?

Like Mathway said

[fair geyser]

yeah the leftmost dot

[sonic thistle]

But why are there 5 solutions?

How are there 5 solution?

[fair geyser]

they all give -1 when you do ^(5/3)

[sonic thistle]

And i can not find them using apgebra,

?

[fair geyser]

i don't know

[sonic thistle]

Ok

So you're saying Photomath is wrong

Becuase there IS a real solution

[tropic oxide]

again this bickering over who is right and who is wrong

[sonic thistle]

Well Photomath is saying there is no real solution

But -1 is a solution

So PM is wrong

Imma take that as a yes

Thanks for helping guys

.close

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[pallid spire]

Help

[pallid spire]

I know that triangle inequalities need to be used here

But idk how

[cedar kiln]

@pallid spire Has your question been resolved?

[crimson delta]

$\abs{\abs{a} - b +c} = \abs{(\abs{a} - b)+c}$

[wraith dagger]

Denascite

[crimson delta]

can you apply the triangle inequality to the RHS ?

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[magic prairie]

Ok so, If one plays a video file at 1.5x speed, will this cut the time of the video in half?
I always thought so, but shouldn't 2x speed be halving the time?
How would I go about calculating this?

[cold briar]

time = distance / speed

distance = 1 video
speed = 1x

time = 1/1 = 1 of the video's time

speed2 = 1.5x

time = 1/1.5 = 2/3 of the video's time

speed3 = 2x

time = 1/2 = 1/2 of the video's time

something like this?

[magic prairie]

yes thank you
i dunno why my mind refused to figure this out! my confusion is now gone

.close

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[rich sandal]

Lagrange multipliers are just an alternative to finding the min/max of a function subject to constraints right?

[scenic heath]

Yep

Very useful when it comes to finding mins/maxs

[rich sandal]

And it's usually better to use lagrange multipliers right

Because it's much quicker

[scenic heath]

Yep
As long as you're only under say 1 or 2 contraints. Not sure about how they'd fare against 3+ constraints to your original function though

[rich sandal]

I see thanks

Doing it the normal way takes so incredibly long

[scenic heath]

Depends on what you call the normal way
At some point you may find the lagrangian method the normal way

[rich sandal]

I just learned it so it's new

🙂

for me

thank thought cleared up my question

.close

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[grand heath]

Hi I would like to ask 32

Is the graph right?

[sacred stone]

not right

,w Plot[x^2-3*x+2,{x,0,3}]

[wraith dagger]

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Requested by 秋水#5244

[grand heath]

Ohhhh I got it now thx

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[grand heath]

Is 32 integration that formula right? Can I write it as integrate 3 to 0?

[cedar kiln]

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[still scarab]

https://cdn.discordapp.com/attachments/908078008609431674/1037969786858131556/image.png

[cedar kiln]

@still scarab Has your question been resolved?

[still scarab]

Is this correct?

Or is C correct?

Or am I just wrong

[floral forge]

when f''(x)=0 then f(x) will have an inflection point. check where f''(x)=0 and see if there is an inflection point on f

[still scarab]

So C?

[floral forge]

yes

[still scarab]

👍

[cedar kiln]

@still scarab Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Alright

[crimson sedge]

Im gonna need some serious help

[hushed spoke]

With?

[crimson sedge]

Oh i thought you guys could understand it

My bad

the 2 yellow circles

[hushed spoke]

How to plot?

[crimson sedge]

They have to be put in Y = 3 * x

So yeah

<@&286206848099549185>

[valid yacht]

@crimson sedge Whats your question

[crimson sedge]

This

[valid yacht]

Put it all in one message :/

[valid yacht]

this lacks any problem at all

whats the problem

[crimson sedge]

alright

draw the graph of the function in the coordinate system

@valid yacht that is the question

[valid yacht]

So you have to plot its graph

Note that, graph of a linear equation in two variables is a straight line

which means, you need two points that satisfy $y=3 \cdot x$

[wraith dagger]

n11

[valid yacht]

Now, when I say a point $(x_1, y_1)$ satisfies an equation, it means if I substitute $x=x_1$ and $y=y_1$ in the equation, both L.H.S. and R.H.S. will be equal to each other

[wraith dagger]

n11

[valid yacht]

So then, what values of $x$ and $y$ can you put in $y=3 \cdot x$ such that both LHS and RHS will be equal to each other?

[wraith dagger]

n11

[crimson sedge]

Your making me even more confused 😭

Yeah bro gave up

[valid yacht]

Read this

[crimson sedge]

There is a reason im here yk

[valid yacht]

Lol

[valid yacht]

Alright so, do you understand this?

Alright nvm, lets try something else

Forget it all

[crimson sedge]

Thank you 🙏

[valid yacht]

We have this equation:
$$y=3x$$

[wraith dagger]

n11

n11

[crimson sedge]

Yeah

[valid yacht]

So that means we got one point

Which is (1,3)
1 is the x part, 3 is the y part

[crimson sedge]

v

[valid yacht]

Right?

[crimson sedge]

So this

[valid yacht]

We only have one point yet

A single point is not enough to define a line

why? because a single infinitely many lines can pass through a given point

[crimson sedge]

okay

[valid yacht]

To define a unique line, we need two distinct points

Because if we have two points, then one and only one line can pass through them. We see this by intuition. Do you understand that?

[crimson sedge]

Yeah kind of

[valid yacht]

Alright. We already have one point, namely (1,3)
We need one more

you can think of any value for x, and then find the corresponding value for y

Because for all values of x, there will definitely exist a 3x (which is y)
Remembwr the equation, y=3x

We already have a point (1,3) where x is 1
So think of any value for x, other than 1

[crimson sedge]

2?

[valid yacht]

Grape

If x=2, then y becomes 3(2)=6

So the other point we got is (2,6)

Now you can plot these two points, and join them. There you'll have the line in question defined.

[crimson sedge]

So x=2 Y=6?

[valid yacht]

yes

[crimson sedge]

?

[valid yacht]

Hey thats not (1,3) and (2,6)

Oh wait

One second let me visualise, then

[crimson sedge]

Alright

[valid yacht]

Does this help lol

thats how you plot points

[crimson sedge]

So this

[valid yacht]

Yes

Now the basic set of operations followed to solved such questions is:
(1) Write the equation in terms of one variable.

How you do this, is basically by doing some algebraic manipulation. For example, x+y=3 can be written as y=3-x. Now, for each value of x, you will always have a 3-x, which is y. Thats how you'll get the points.
(2) Find any two points satisfying the equation.
(3) Plot those points.
(4) Extend the line segment to make it a line.

[crimson sedge]

Perfect

I will use this

🙏

[valid yacht]

🤝

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[undone knot]

related rate problem i need help with
i tried finding dh/dt

but did i do it correctly

i know i need dc/dt, but i just wanted to find dh/dt

[drifting marlin]

dh/dt is 0 since you never move away from the launch pad

[undone knot]

so what i did is invalid

[drifting marlin]

Yes

[undone knot]

if we were walking and the rocket was moving at the same time then would my answer for dh/dt be viable

[drifting marlin]

Assuming h is the distance from you to the pad, you would have to be moving yes

[undone knot]

ok thanks

.close

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[cloud egret]

Who can help with me:known the first,solve the second

[cedar kiln]

@cloud egret Has your question been resolved?

[limber canopy]

what are the question?

[cedar kiln]

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[limber canopy]

does any1 know what this is equal to?

[limber canopy]

${ \left({{\lim_{x \to 0}{ {-\ln(1 + 3(e^{-x} - 1))} \over {x} }} - \left({{\lim_{x \to 0}{ {e^{24x} - 1} \over {x} }} \over {(\cos^2x + \sin^2x)}}\right)}\right) \left({{\lim_{x \to 28} {{x^2 - 25} \over {x + 5}}} + {\prod_{k=1}^{4} k}}\right) + {{\prod_{k=1}^{4} k}}{\left({{\lim_{x \to \infty}{{ 23x^{3} - 11x } \over {{ 1x^{3} + 3x^{2} + 11x }}}} - \left({{-3e^{\pi i}} \times (\cos^2x + \sin^2x)}\right) + {\lim_{x \to 27} {{x^2 - 9} \over {x + 3}}}} \right)} }$

[wraith dagger]

Mxrcus

[limber canopy]

69

.close

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[crude socket]

hi I'm trying to understand kernels and images of Linear transformations for linear algebra could someone help me get the concept?

[cedar kiln]

@crude socket Has your question been resolved?

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@crude socket Has your question been resolved?

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[idle elk]

is it true that C ∩ Q is dense in C where C is cantor set?

[cedar kiln]

@idle elk Has your question been resolved?

[modern compass]

yes

[idle elk]

i wrote down that for a,b belong to C ∩ Q and because C is uncountable there exists a x in C that is between a and b,is this correct ?

[cedar kiln]

@idle elk Has your question been resolved?

[cedar kiln]

@idle elk Has your question been resolved?

[cedar kiln]

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[light pine]

say u had a function x^3 >= 2

[light pine]

does this mean its greater than or equal to the y value of 2 on a graph?

[elfin hemlock]

x^3>=2 is not a function

[light pine]

the question asks for me to find the min max sup and inf of x^3 >=2

[elfin hemlock]

yeah and? not a function

[light pine]

ok

forget function then

[elfin hemlock]

but x^3>=2 is the values of x such that x^3>=2

[light pine]

so if u had a graph

of x^3

would everything after y = 2 be x^3 >= 2

where 2 is on the y axis

[elfin hemlock]

wdym after y=2

[light pine]

u have an axis

[elfin hemlock]

if you mean everything above the line y=2

[light pine]

on the y axis

[elfin hemlock]

then yes

[light pine]

yes

i mean that

[light pine]

does that mean in interval notation we can right it as [2, inf)

ie the range

[elfin hemlock]

its more than that

,calc 1.9^3

[wraith dagger]

Result:

6.859

[elfin hemlock]

so x=1.9 also works

[light pine]

confused

nvr mind

im over complicating way too much

.close

[cedar kiln]

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[glass lodge]

Well what’s standard form and what’s vertex

[simple tapir]

i'm currently trying to find it online lol

ax squared plus bx plus c is equal to zero.

that's standard right?

i just can't find anything about the vertix formula

[tame wave]

The equation is in standard form.

To find the vertex you have to find the 2 x intercepts by factoring the equation, then finding the mid point of those two points and plug it in to find the vertex (so the value of y at the axis of symmetry)

[simple tapir]

im gonna save that image

[tame wave]

u know how to factor right?

[simple tapir]

yeah it get...

vertix form would be y = 3(x-20/3)^2+140/3

[tame wave]

Oh sorry, ok I understand the question. So basically the rule for finding the vertex from standard form which is ax^2 + bx + c is that you have to take the b and a value and put them into this equation : -b/2a, and that will give you an x value. So once you get that x value you have to plug it into the main function and then you get your vertex.

[livid hound]

,W expand 3(x-20/3)^2+140/3

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[simple tapir]

yeah that's what i ended up getting

but i can't find and x or y value to answer the question

[livid hound]

i just can't find anything about the vertix formula
there's plenty of stuff on vertex form

[simple tapir]

What do the x and f(x) values you got represent?

isn't that what this is asking?

[tame wave]

you're being asked to get the vertex from standard form. so you have to answer it as v(x,y)

so the coordinates of the vertex

[simple tapir]

x= 5,6,20/3,8,9
y= 55,48,140/3,52,63

[tame wave]

f(x) is your y value. which tells you the point on the y axis where the vertex is. and your x is the point where your vertex is in the x axis

[livid hound]

you can deduce the formula based on how you converted to vertex form

[simple tapir]

imma get in a call with my teacher about this thanks for the help

.close

[cedar kiln]

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[arctic marsh]

Sorry everyone, I'm studying intervals and I can't get why we define "x > n - ε" as a condition for a number to be an "extreme value"

[sleek condor]

an extreme value of an interval?

please express it more formally just to be sure

what would be the extreme value of[a,b]?

[arctic marsh]

Yes, maybe it would be better to say an extreme value of a group

[mighty drift]

Sounds like supremum but he forgot the definition

[sleek condor]

yeah

thats what i thought too

[arctic marsh]

Well they are usually named with sup(A)/inf(A) so it's probably what you're saying

[mighty drift]

Supremum/infimum

[mighty drift]

That's absolutely not the definition

[arctic marsh]

My book says that there are two condition that defines a supremum:

1. x <= sup(A)
2. x > (sup(A) - ε)

[sleek condor]

what is x?

[arctic marsh]

A value of the group
For example, it can be 1/n

[sleek condor]

so it can be anything we want?

[arctic marsh]

Well yes, it depends on the group

[sleek condor]

set*

so its not any number, its a number in the set?

[elfin hemlock]

You are missing some quantifiers clearly

[arctic marsh]

For example, A could be {x|x= 1/n, n contained in N}

[sleek condor]

okay

[arctic marsh]

Oh, sorry, didn't know that

[elfin hemlock]

S is supremum of A if for all x in A we have x<=S

[sleek condor]

there's already something very common called a group in maths

so avoid that

[elfin hemlock]

And there exists an x in A such that x>S-epsilon for any epsilon>0

You can’t just ignore some part of the definition

And only post something about it

[arctic marsh]

Ok that's the part where I get confused

[elfin hemlock]

For all x we have x<=S means its an upper bound

The x>S-epsilon means its the lowest upper bound

[arctic marsh]

And why does it mean that?

[elfin hemlock]

Because if we try a value less than S (S-epsilon)

Its not an upper bound

[sleek condor]

are you familiar with epsilon delta proofs?

[elfin hemlock]

Because there is an x such that its greater than S-epsilon

[sleek condor]

like the definition of a limit?

[arctic marsh]

OOOOH ok I think I got it

[arctic marsh]

Nope, not at all

However, I think I have understood the definition now

Thanks to you all!

.close

[cedar kiln]

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[elfin hemlock]

Next time when asking a question post the full definition

Not just bits of it

[cedar kiln]

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[slow mirage]

How do I solve this?

[slow mirage]

@sonic condor

@fringe socket

@crimson sedge

@sweet stag

[crimson sedge]

tip: formula for the circumference is 2πr

[slow mirage]

I thought it was circumference = Pie x Diameter

[crimson sedge]

it’s the same because of the associative property.

[slow mirage]

whats that

i apologize this is all rocket science to me lmao

[crimson sedge]

to put it in simple terms:
(ab)c=a(bc)
(2x10)5=2(10x5)

[slow mirage]

what?

[crimson sedge]

hate discord’s syntax

[slow mirage]

dannggg that rocket science lookin real rocket science-y

Will I need this for my GED test

[crimson sedge]

I’m not familiar with the US school system so I can’t speak on that.

[crimson sedge]

how this relates to 2πr is that it’s essentially 2xπxr = 2rxπ and 2r is the diameter.

[mighty drift]

Don't name algebraic operation properties to someone who doesn't know the circumference of a circle. Know your audience

[crimson sedge]

you’re right, didn’t think that one through.

[mighty drift]

Also commutativity matters lmao

[slow mirage]

Now you making me feel stupid

💀

yall outta pocket for that one

[cedar kiln]

@slow mirage Has your question been resolved?

[cedar kiln]

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[cedar kiln]

@fathom mural Has your question been resolved?

[cedar kiln]

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[slow badge]

I need helps with this man

[cedar kiln]

@slow badge Has your question been resolved?

[royal wasp]

i cant even tell what it says

[cedar kiln]

@slow badge Has your question been resolved?

[fair geyser]

you have to count objects, it's multiple choice

it's all memorization

when objects are arranged like this: OO the corresponding digit is 2

e.g. the ducks in the last question

and the sand crawlers before the question before the last

[cedar kiln]

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[spice tundra]

Is there a straight forward way to solve 1ci, ii, iii, iv?

[spice tundra]

here are the answers https://media.discordapp.net/attachments/903430333096132618/1038352355760029796/unknown.png

it's not just a simple quadratic though

they want you to answer it in the most funky way

yeah done that

doesn't give the answer though

look at the answers

[drifting marlin]

Seems like some messages were deleted

[spice tundra]

don't think so

i mean i asked this question a while back

but we figured out an incredibly cancerous solution

so i'd like a fresh perspective

[autumn fox]

For (i), (ii) and (iii) you could substitute x² = u. At (ii) this would give you 16u² - 16u + 1 = 0. Here you can solve for values of u with quadratic formula. You should get 2 values for u, where you can put these back into x² = u and get 2 values of x per value of u, leaving you with all 4 values for x where 16x⁴ - 16x² + 1 = 0.

[cedar kiln]

@spice tundra Has your question been resolved?

[cedar kiln]

@spice tundra Has your question been resolved?

[spice tundra]

i tried this but didn't get the rigth answer

could you give it a go?

[autumn fox]

What did you get for the values of u in 16u² - 16u + 1 = 0?

[spice tundra]

i get

$2+\sqrt{3}\ / 4}$

[wraith dagger]

splooze
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spice tundra]

ffs

$ \frac{2 + sqrt(3)}{4}$

$ \frac{2 + sqrt(3)}{4}$

$\frac{2 + sqrt(3)}{4}$

[wraith dagger]

splooze
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spice tundra]

$\frac{2 - sqrt(3)}{4}$

[wraith dagger]

splooze
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spice tundra]

@autumn fox these are my two answers

[autumn fox]

They are correct

[spice tundra]

oh wtf

i get the rigth answer now

[autumn fox]

For LaTeX remember to do \sqrt

[spice tundra]

5pi / 12

[autumn fox]

Si

[spice tundra]

okay

don't know why i didn't get that for i

so i, ii, iii should all be fine with that

what about iv

[autumn fox]

You could use the quartic formula, but umm. I wouldn't do that if I was you

Even the cubic formula is exponentially bigger than the quadratic formula

You could also try factoring into the product of 2 quadratics, but I don't think you can do that nicely with that equation.

[spice tundra]

i mean

question iv

is a binomial expansion of a quartic

which i did in 1 a

so maybe i'm meant to link the fuckers somehow

[autumn fox]

Yeah, that seems to be the general point of the question. I just thought you were looking for an easier way.

But I don't think there is any for (iv)

[spice tundra]

not even sure how to approach iv

don't know how to match the equations

[cedar kiln]

@spice tundra Has your question been resolved?

[lean thunder]

for (iv), divide the entire equation by x^2, and then substitute u = x - 1/x

[spice tundra]

geek

[cedar kiln]

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[open wave]

,calc 4000*1.025^9

[wraith dagger]

Result:

4995.4518797907

[ruby lion]

do yk what rationalising means?

[open wave]

nope

[ruby lion]

hmm

do u know what a rational number is?

[open wave]

yh kinda

[ruby lion]

so what does a rational number mean

[open wave]

can be written as a ratio or fraction with 2 numbers

[ruby lion]

yep

so we have a irrational element in the fractions denominator right

how do you think we can get rid of it

[open wave]

square it?

[ruby lion]

you would have to multiply but something x/x

[ruby lion]

not quite

if u were to square it

you would still have a^2+2ab+b^2

so u would still have irrational element

think about difference of two squares

[open wave]

with numbers could u do 1/(5+ root 3)(5-root 3)

so like put it in brackets and expand

[worn wave]

yes

but first

do you know

what’s an irrational number?

[open wave]

any number that cant be 2 ?

[worn wave]

…. not quite

[open wave]

2 numbers*

[worn wave]

no

wrong

[ruby lion]

it wouldnt just be 1/ whatever.

[worn wave]

irrational numbers are just numbers with decimals that can’t be express in any other way

for example

do you know pi?

do you know what’s the value of pi?

[open wave]

3.14 ect

[worn wave]

yes

that is an irrational number

because you can’t express it in any other way

same goes for certain numbers with certain nth root

[open wave]

wdym what would it be then

[worn wave]

like you can’t express the second root of 2 in any other way than decimals

you can’t express square root of 2 in a fraction

[open wave]

yh

[worn wave]

so now do you understand what it means?

[open wave]

yea

[worn wave]

Okay great

Now rationalise means we make an irrational number rational

for example, let’s say you have a fraction

1 / square root of 2

When a question says rationalise this fraction

it wants you to make the denominator a rational number

[open wave]

*by square root of 2?

[worn wave]

but square root of 2 is irrational

no

[worn wave]

So you have to make it rational

So you need to convert the fraction into another fraction that gives the same result, only difference is that the denominator is not irrational

do you understand?

[open wave]

so change it into a different fraction that gives the same answers, but would the numbers used be the same? or can they be different

[worn wave]

they MUST be different

[open wave]

alright

[worn wave]

the only thing you need to care about is that the denominator is not irrational

you have to the idea now?

[open wave]

yh

[worn wave]

Alright

so let me ask you

[open wave]

how would i do that tho make the denom rational

[worn wave]

that’s what I’m going to teach you

but first

do you understand surd laws?

[open wave]

Explain one and I'll lyk since off top of my head rn idk

[worn wave]

okay

if I take a surd

let’s say surd a order n

and multiply it with itself

what do I get?

[open wave]

Surf a+n

[worn wave]

wrong

actually I’ll rephrase it

imagine

square root of a

and I multiply it with itself

so square root of a x square root of a

what do I get?

[open wave]

U get a

[worn wave]

okay, can you prove it?

[open wave]

√a*√a = √a^2 which squared and root cancel each other out

[worn wave]

great

so if we want to make a root rationalised

[open wave]

Multiply it by it's self ?

[worn wave]

yes, you multiply the fraction with another fraction that’s equal to one, where the nominator and the denominator is the denominator of the first fraction

[open wave]

So with my question itl be like 1/5+√3 x 5+√3/1?

[worn wave]

nope

I’ll teach you slowly

remember the example

1 / sqrt 2

now

can you express 1 in terms of fractions

just express 1 in any fraction

I want a fraction that’s = 1

[open wave]

1/1

[worn wave]

okay

so you notice the denominator and nominator are the same in order to get one

[open wave]

Yea

[worn wave]

so if I make both of it into sqrt of 2, would it be the same?

[open wave]

Yea

[worn wave]

okay

so the idea is

you take the fraction and multiply it with another fraction that’s = 1 in terms of fraction

the fraction must have a condition where the nominator and denominator should be the same

and both of the numbers MUST be the denominator of the first fraction

so it would be

1 / sqrt 2 x sqrt 2 / sqrt 2

[open wave]

Ol

Ok

[worn wave]

so I believe at this point, you know how to multiply fractions

[open wave]

Yea

[worn wave]

so tell me what’s the final fraction

[open wave]

√2/2

[worn wave]

Correct

And now the fraction is rationalise

I’ll give another example

1 / sqrt 3

rationalise it

[open wave]

1/√3 x √3/√3

[worn wave]

yes

now solve it

[open wave]

√3/3

[worn wave]

great

last one

3 / sqrt 5

Rationalise it

[open wave]

3/√5 x √5/√5

[worn wave]

great, solve it

[open wave]

3√5/5

[worn wave]

alright now you can learn about conjugate surfs

so now we can finally start solving your question

[open wave]

Alright

[worn wave]

so this

First tell me what’s the denominator

[open wave]

5+√3

[worn wave]

Alright, now the denominator has 2 terms

how are we going to solve it?

simple

but first let me ask

[open wave]

5√3 x √3/√3

[worn wave]

nope

let me ask

can you express

a^2 - b^2 in another way?

[open wave]

Ab^4 ?

[worn wave]

no

okay never mind let’s ditch that

so, conjugate she’d

surds*

They are surds which “complements” their original surd

it’s hard to explain but I’ll show some examples

[open wave]

Alright

[worn wave]

Let’s say we have 2 - sqrt 3

the conjugate of it is

2 + sqrt 3

we only just negate the operation

next we have 7 + sqrt 2

conjugate is 7 - sqrt 2

[open wave]

so its like the opposite operation?

[worn wave]

yes

that’s the main idea

it’s hard for me to explain but you can look it up later

so essentially

we are going to express 1 in terms of fraction agaian

instead, both nominator and denominator of the fraction is the conjugate surd

So from your question, what’s the conjugate surd for 5 + sqrt 3?

[open wave]

bck sry

5 - sqrt 3

[worn wave]

Great

So create a fraction using that so. It’s = 1

And then multiply with your main question’s fraction

and you should find your answeer

[open wave]

1/(

1/(5+ sqrt 3)(5-sqrt 3)

[worn wave]

what about the 1?

you forgot the 1

the 1 would be 1 x (5-sqrt3)

since

1/(5+sqrt3) x (5-sqrt3)/(5-sqrt3)

[open wave]

ohh ok

[worn wave]

do you understand how everything works now?

[open wave]

ye

[worn wave]

Okay I hope so, now find your answer

[cedar kiln]

@open wave Has your question been resolved?

[open wave]

mb my wifi and shit went off

so all i need to do it solve that and its my answer?

right

.close

[cedar kiln]

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[fair geyser]

makes sense

yeah

[cedar kiln]

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[tender topaz]

I don't understand this step. Anyone who could help?

[tender topaz]

I thought everything got divided by 2^x? but it didn't with the -4^x?

[south tundra]

They divided both sides by 2^x

2^(x - 1) got divided by 2^x so 2^(-1) left

[tender topaz]

But why don't divide the -4^x with 2^x?

[crimson delta]

well they divided the product 4^x*2^(x-1) by 2^x

which results in 4^x*2^-1

you don't divide every factor of the product. just one of them

[tender topaz]

But how? How should you know which one to divide with if you only divide one factor?

[crimson delta]

well the second one is much nicer here

if you lets say divide 9*5 by 3, you could either get 3*5 or 9*(5/3)

clearly the first one is nicer

both are correct tho

[tender topaz]

Oh so you just pick one of your choice, preferably the most easy one.
So it doesn't matter how many factors you have to divide with, as long as you just only pick one? Correct?

[crimson delta]

well depending on situation you might split stuff up

like if you divided 12*30 by 15=3*5, you would do 4*6

where I divided the 12 by the 3 and the 30 by the 5

but yes you take one of the options and preferable the easiest one

[tender topaz]

But in my question it's not possible to split it up right?

Is it maybe also as you can't divide -4^x by 2^x?

[crimson delta]

well in theory it would be possible to split it up but it doesn't help

well you could divide 4^x by 2^x

note that $4^x = (2^2)^x = 2^{2x}$ and then $\frac{4^x}{2^x} = \frac{2^{2x}}{2^x} = 2^x$

[wraith dagger]

Denascite

[crimson delta]

but the other option is easier here

or wait that's overcomplicated, $\frac{4^x}{2^x} = \left(\frac42\right)^x = 2^x$

[wraith dagger]

Denascite

[tender topaz]

Ah thanks! So if you would take that method, the result here would be 2^x . 2^x-1. Correct?

(of the final part)

[tender topaz]

Is this rule only allowed if you have the same amount of X?

[crimson delta]

yes

[crimson delta]

yes

[tender topaz]

And otherwise you use the first one?

[crimson delta]

yes. or some other applications of the various exponent laws

note that $2^x\cdot 2^{x-1} = 2^x\cdot 2^x \cdot 2^{-1} = (2\cdot 2)^x \cdot 2^{-1} = 4^x \cdot 2^{-1}$ so it's the same

[wraith dagger]

Denascite

[tender topaz]

Thank you so much!

Was struggling with this for an hour. Appreciate your help and the useful images and rules!

[crimson delta]

being comfortable with applying all of these rules is very important for all of this stuff

[tender topaz]

Yeah I've found out 😅

Thanks and have a nice day

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[cedar kiln]

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[crimson sedge]

crap i forgot to close the last one

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[simple yoke]

how can i find this limits as x goes to 0

[south tundra]

Are you allowed to use L'hopital's rule?

[simple yoke]

no

[south tundra]

The definition of a derivative?

[delicate patio]

convert the question in this kind of form

and then proceed

@simple yoke did you get any answer?

[cedar kiln]

@simple yoke Has your question been resolved?

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[cold briar]

What's different from linear equations with 2 variables and linear equations with 3 variables other than the number of variables?

[cold briar]

Like... they both have the same solving method

substitution, elimination, (graphing)?

[chrome tulip]

true

[cold briar]

Also other than the fact that you need more equations to solve a 3 variable'd linear equation

Not necessarily, but yeah

Usually you need at least 3 for a 3 variable'd one

and 2 for the 2 variable

[chrome tulip]

yes, n for n is needed ,afaik

[cold briar]

And what's hard about it :/

It's just substitution and elimination right?

Or is there some type of question that I don't know that needs another method to solve

"Involving linear equations with 2 or 3 variables"

[chrome tulip]

afaik, anything can be solved with substitute and eliminate

[cedar kiln]

@cold briar Has your question been resolved?

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[reef mulch]

Im having a hard time trying to prove this identity

[south tundra]

Are you sure this is an identity?

[sacred stone]

I think it's
$$\frac{y^3-x^3}{y-x}$$

[wraith dagger]

秋水

[reef mulch]

i was asked to prove that this is true for all x,y

so im guessing it is

[flint cape]

That's not what the LHS of this is...

[flint cape]

Are you saying that it's a typo and what it should be is the correction here, @sacred stone ?

[sacred stone]

yes

[reef mulch]

oh well

there’s a mistake on the exercise sheet lmfao

[sacred stone]

$\frac{y^3-x^3}{y-x} = x^2+xy+y^2$ is an identity

[wraith dagger]

秋水

[south tundra]

Well wait actually

e^x - e^y = y^3 - x^3
e^x + x^3 = e^y + y^3 ?

I may be doing something wrong

Ah right lol x isn't the same as y

[flint cape]

Assuming so, tho, we can assume (y-x) is not zero, so multiplying through that gives y³ - x³ = (y-x) (x²+xy+y²)

So now we have an alternative statement to prove

which just means to expand the RHS carefully

Interestingly, if it isn't a typo and you're being asked for all solutions, then we're looking for solutions to exp(x) - exp(y) = y³ - x³

And through a little bit of thinking, you're left with y=x being the set of all solutions

[reef mulch]

thats weird

if it’s not true for all x and y then the question is extremely badly formulated

it says (translated): show that that expression is true

[flint cape]

(out of interest, from what language?)

[reef mulch]

french

[flint cape]

ah

Then yh typo, ce doit etre "y³ - x³"

au lieu de "exp(x) - exp(y)"

[reef mulch]

yeah thought so

but even then it would be completely irrelevant to the original problem where we were trying to prove that an exponential function is injective

actually a sum of exponential and polynomial

[flint cape]

What's the whole question here?

[reef mulch]

funnily enough the function is f(x) = x^3 + e^x

one sec

we’re trying to prove bwoc that f is injective

first 2 questions are proving those 2 identities

[flint cape]

Ouais donc c'est une coquille

[reef mulch]

alright then thanks for you help

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[sweet ingot]

Hey.

[sweet ingot]

Prove or disprove:
If A ∪ B is a finite set and A ∪ B is equivalent to A \ B then B = ∅
I tried to solve this but I just can't seen to understand if it's correct or not

[south tundra]

I think I found a simple way to do this

The cardinality of A U B is always bigger than or equal to the cardinality of A

And the cardinality of A \ B is always less than equal to the cardinality of A

Since |A| <= |A U B| = |A \ B| <= |A|

We get that |A U B| = |A|, which is only possible when B is a subset of A

But then, since B is a subset of A, |A \ B| = |A| implies |A| - |B| = |A| and |B| = 0

[sweet ingot]

So the claim is correct

[south tundra]

Yes

[sweet ingot]

Okay, thank you!

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[steep roost]

If on a circle with a radius of 1 , I have 2 unit vectors going to points on the circle. I also have a vector going from vector a to b. I know the distance between a and b. if I went "theta" degrees on the circle from point a to point b, what percent distance of the vector from a to b would be travelled if a line through the center and that distance on vector(a,b) would also go through the distance travelled on the circle?

[steep roost]

Basically I have a known vector a,b. If I went a certain degrees around the circle, how far would I have gone along the direct line from a to b

[crimson sedge]

okay so label the points like this

so we know that the angle BOX is theta1 and angle AOX is theta2

[steep roost]

Yes

[crimson sedge]

now we can let angle BXO be alpha, so AXO is 180-alpha

then do law of sines on triangles BXO and AXO

[steep roost]

Ok I gotta learn law of sines

How do I know BXO and AXO?

[crimson sedge]

you dont

but you know they add up to 180

[steep roost]

true

[crimson sedge]

so you can cancel them out bsaically

[steep roost]

wait I know OBX and OAX tho right?

because they are the same angle no matter what right?

isosceles triangle

[crimson sedge]

yes you do

actually thats probably easier

[steep roost]

ayyy lets go

so I know all angles of the triangles, and I know vector A and B are both length 1

so can't I do some triangle math to get another side length?

[crimson sedge]

you can yes

but it would be in terms of the angle

[steep roost]

The theta angle?

Ok thank you, I think I got it figured out

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[lusty reef]

do u know how to find the area of the circles?

bc it would be the area of the rectangle minuses the area of both circles

no

the total area of the rectangle minuses to area of both circles will give u the shaded area

imagen u have a piece of paper that represent the rectangle, if u cut out (subtract) two circles the remaining paper will be the remaining area (the yellow area)

base times hight

yeah

no

it will be a surd

where did 6 come from?

[cosmic steppe]

The six yellow sections is the area of the rectangle minus the areas of the 2 circles

[lusty reef]

ignore how many sections, its area of rectangle - total area both circles

[cosmic steppe]

What's the formula for the area of a circle

Just pi r^2

So what's the radius of one of the circles

They're congruent circles so you only need to worry about the area of one of them

[lusty reef]

best way i can explain it

[cosmic steppe]

Yooo how much for that NFT

[lusty reef]

six quadrillion Ethereum

[cosmic steppe]

Dude

[lusty reef]

the area of a circle is pi time radius squared

[cosmic steppe]

Don't make me spoon feed you

[lusty reef]

well u can get that by halfling 2, because 2 is both the hight of the rectangle and diameter of the circle