help-10

[fast zodiac]

Id start by finding PR

Which is

[green echo]

But we don't know po or or

[fast zodiac]

You have the perimiter of the rectangle

Qp + Qr is half of that

[green echo]

They add to 26, so po plus Or i'd 13

[fast zodiac]

Which is 13

[green echo]

Yup

What step do I go for now

[fast zodiac]

After you find PR

You can deduce one of the sides

And the subtract that from the radius which is 10

So like

Find OR

And then

10 - OR would be BR

And then do the same on the other side

[green echo]

10 minus PO is AP

[fast zodiac]

And then for BA is just the quarter of the perimiter of the circle

[fast zodiac]

Yep

[fast zodiac]

You have the radius so u can find it too

[green echo]

How do I find OR though

I know it adds to 13 with po

[fast zodiac]

You can find PR right

And you know that the angle PRO is 45

So OR would just be PR times cos(45)

And for OP

Its the same

But

Sin instead of Cos

[green echo]

13 - RO?

As the adjacent

[fast zodiac]

Wait a min

Im dumb

OQ is the radius right

And RP = OQ

So RP is just 10

RP is 10 cos 45

[green echo]

Wait what how is rp the same as the radius

[fast zodiac]

And RO is sin 45

[fast zodiac]

Both diagonals of the rectangle are equal right?

QO can be considered as the radius of the circle

Because the quarter circle is AOB

[green echo]

Yup

[fast zodiac]

And Q is a point on the outer edge

[green echo]

Ohh

I get it

[fast zodiac]

And since RP is also a diagonal

They must have equal length

Since

Its a rectangle

[green echo]

Aight

So we use trig to find the sides

Makes sense

[fast zodiac]

Yea yea but wait

I made a mistake

The angles arent 45

Its a rectangle not a square

So

Both angles must add up to 90 but they cant be equal

[green echo]

Aight

[fast zodiac]

Butt

Now that you know that the diamater is 10

A^2 + B^2 = 100

And

A + B = 13

[green echo]

So RO is 10?

According to my calculator

[fast zodiac]

Nah cause

Wait what

RP is 10

Weve proven that

And RP is the hypothenuse so its the longest side

So RO isnt 10

[green echo]

RO = tan 45 = x/10

Is that right

[fast zodiac]

Im sorry

Its not 45 it cant be

[green echo]

Nbm it's sin

[fast zodiac]

Theyre not 45

[obtuse pebble]

@green echo Has your question been resolved?

[obtuse pebble]

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[pure epoch]

how can i solve this to show that there is a value of n where sn =0

n+3-2cos(4n^2)=0

then

actually hold tht

nvm

stop holding it

help.

actually now im just confused

the graph shows no solutions in positive x

also i understand the answer for the actual question, im just curious

[shut lagoon]

n+3 is an integer

So if n+3 = 2cos(4n^2)

[pure epoch]

yea

[shut lagoon]

Then 2cos(4n^2) is an integer.

[pure epoch]

oh

right

i see

there are no solutions tho, however i do understand waht you mean

[shut lagoon]

You would need 4n^2 to be like pi/3 or pi for this to happen, but that would mean pi is rational :c

[pure epoch]

yeah

wait

i read the question wrong

😂

=/ 0

[shut lagoon]

Bless.

[pure epoch]

thanks u helped i never thought to think of it like that

.close

[obtuse pebble]

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[shut lagoon]

No worries

[obtuse pebble]

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[opaque spindle]

I am not sure that i understand how the proof for the injection works as well as how to prove that it is surjective

[vestal hull]

what is the definition you've got for injective functions?

[vestal hull]

cool, you can consider using the contrapositive too

f(a) = f(b) ⇒ a = b, for all a, b ∈ Domf

So for an injection proof you consider 2 elements in the Domain such that their images are equal, you have to show that those 2 elements are the same

[opaque spindle]

so for this specific example here where they have already proven that the inverse of the function is an injection

its saying

x1 and x2 are part of b
the inverse of the function using x1 and x2 are the same?

[vestal hull]

what do you mean using?

[opaque spindle]

honestly im not sure

i dont understand the proof for the injective part

[vestal hull]

what are you stuck at?

[opaque spindle]

this middle part here

i dont understand how this proof works

[vestal hull]

by definition of the inverse functions, (f ∘ f⁻¹)(x) = x

[opaque spindle]

would that be the bottom part of the definition there ?

or i guess f(f^-1(x)) = x

[vestal hull]

Sorta, but the definition it's used in the proof states that if f⁻¹ is the inverse function of f then (f ∘ f⁻¹)(x) = f(f⁻¹(x)) = x, for all x ∈ B

[opaque spindle]

what is this part

(f ∘ f⁻¹)(x)

[vestal hull]

Alternatively, f⁻¹ is the inverse of f then (f⁻¹ ∘ f)(x) = f⁻¹(f(x)) = x, for all x ∈ A
but since the proof requires to show they are in B, you don't want to use this other part of the definition

[vestal hull]

composition

[opaque spindle]

i have not learnt composition yet

i am supposed to be able to do it without that supposedly

okay let me try to explain maybe how im seeing this

this means that if i take a value x1 and a value x2 from b and pass it through the inverse function the result obtained from the function would be the same?

[dull linden]

yes

[opaque spindle]

okay so this next part here is saying that if i take x2 and put that in the inverse function and then take the obtained value and put it in the original function i would obtain x1

or x2 since they are the same

and since this works then the function is injective

[dull linden]

yep

really, this would be clearer if it said $x_1 = f(f^{-1}(x_1)) = f(f^{-1}(x_2)) = x_2$

[warm shale]

aldo booze

[opaque spindle]

so by what definition is this injective? i thought that injective meant that every value in a could be mapped to a unique value in b

[opaque spindle]

ok i understand this part now

[dull linden]

One definition of injective is that a function $g$ is injective if $x_1 = x_2$ whenever $g(x_1) = g(x_2)$

[warm shale]

aldo booze

[dull linden]

We showed that here, with $g = f^{-1}$

[warm shale]

aldo booze

[opaque spindle]

so when passing two different values in the same function the same result is obtained

[dull linden]

A function is injective if what you said here cannot happen, i.e. two different points always give two different values

[opaque spindle]

maybe im seeing this wrong but im seeing g(x1) = g(x2) as if i pass a value x1 into the function g and i pass a value x2 into the function g then they are equivalent

[dull linden]

Yes, you're right

Injectivity is saying that whenever this happens, $x_1$ and $x_2$ are the same

[opaque spindle]

i dont understand how that shows that every value in a can be mapped to a unique value b

[warm shale]

aldo booze

[opaque spindle]

okay

i see

so x1 and x2 are the same number

[dull linden]

Yes, whenever they map to the same value

[opaque spindle]

okay i understand now

so when passing the same value in the inverse function and then that value into the original function you obtain the same value that you started with

[opaque spindle]

how would it be possible to prove that it is a surjection "by definition" then

i guess i would start with x1 and x2 being part of b

[dull linden]

For surjectivity, you only need to start with one point in B

[opaque spindle]

so i need to prove that every value b can be mapped to by a value a correct?

[dull linden]

Yes

[opaque spindle]

or in this case would it be the other way around since im trying to prove that the inverse is surjective

so every value a can be mapped to by a value b

[dull linden]

Yes, since $f^{-1}$ maps from $B$ to $A$

[warm shale]

aldo booze

[opaque spindle]

could i say something like f^-1(x1) = x2

x1 is part of A

[dull linden]

The general way to show surjectivity is to fix some value $a$ in $A$, and find some value in $B$ which $f^{-1}$ maps to $a$

[warm shale]

aldo booze

[dull linden]

So with what you just said, you would have to establish that there exists an $x_1$ which maps to $x_2$

[warm shale]

aldo booze

[dull linden]

But I suggest using the notation $a$ and $b$ rather than $x_1$ and $x_2$ for this part

[warm shale]

aldo booze

[opaque spindle]

so let a be part of A

f(a) = b
f^-1(b) = a

[dull linden]

Exactly

[opaque spindle]

so i could write a statement like the injective proof provided and say

Let a∈A such that f^-1(b) = a.

Then by definition of inverse we have a = f(f^-1(b))

or i guess the last part wouldnt work

that was just for injective i suppose

[dull linden]

You've almost got it but it's a bit backwards

[dull linden]

You don't want to assume anything about a -- what you wrote here is essentially the whole proof

We want to find b such that f^-1(b) = a

[opaque spindle]

so i could say something like

"let a be part of A such that f(a) = b. By definition of the inverse f^-1(b) = a and so every value b can be mapped by a value a"

or is it still backwards

im not sure

[dull linden]

Essentially, yes, though the last bit is still slightly backwards

But your idea is right: the value f(a) in B maps to a in A

[opaque spindle]

backwards in the sense of that not being the definition?

[dull linden]

You say "every value in b can be mapped by a value a", when I think you mean to say "every a in A can be mapped to by a value b in B"

[opaque spindle]

oh because its the inverse

i see

if it was A -> B then every value in b can be mapped to by a value in a

[dull linden]

Yep

[opaque spindle]

thank you so much!

[obtuse pebble]

@opaque spindle Has your question been resolved?

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[flat geyser]

u sub

[fathom flicker]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mortal thunder]

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[abstract marlin]

can someone help please

[obtuse pebble]

@abstract marlin Has your question been resolved?

[abstract marlin]

no

help?

should i try another room ?

[shell garnet]

hey

what exactly do you need help in @abstract marlin

[abstract marlin]

i dont know how to do this graph

[shell garnet]

you want the equation of the hyperbola right?

[abstract marlin]

finding the equation for a hyperobla

yes

[shell garnet]

the vertex of a hyperbola is (a or b,0)

the foci are points +-(ae,0)

where e is the eccentricity of the hyperbola

as all the points are already given to you

you can find the eccentricity

and then use the fundamental definition of a hyperbola

considering a point (h,k) on it

distance of a point lying on the hyperbola is eccentricity times the distance from its focus

[abstract marlin]

this is what i got but its wrong

would you be able to write out the numbers for me

the step by step?

[shell garnet]

yeah 1 sec

that is wrong because you considered the axis of the hyperbola to be the x axis

which is visually not possible

as the centre of the hyperbola is (1,3)

this is the standard result for a hyperbola

[abstract marlin]

still not getting it

hahah sorry

[shell garnet]

see

the centre of the hyperbola is 1,3 right

and the focus is 1+sqrt 5,3

so the distance between the centre and the focus is sqrt5

which is the eccentricity of the hyperbola in simple terms

the distance between the centre and the vertex is the transverse axis length of the hypebrola

[abstract marlin]

like this?

[shell garnet]

did you understand now?

if you do ae-a

you get the eccentricity

as we know the value of a already

and the formula for its eccentricity when x axis is the transverse axis is as follows

[abstract marlin]

i didnt learn it this way, so i dont know what is ae

[shell garnet]

ae is a times the eccentricity

you can see my image right?

[abstract marlin]

Yes

[shell garnet]

the intersection of the axes there is the centre of the standard hyperbola

marked (0,0)

[abstract marlin]

this is how i learned it

[shell garnet]

the point of the intersection of the bow of the hyperbola with the x axis is the vertex of the hyperbola

and the point marked further ahead is the focus of the hypberola

1 second let me send you the solution

[abstract marlin]

k

[abstract marlin]

intersting

i would of never understood this lol

[shell garnet]

lmao

[abstract marlin]

we didnt learn this with our examples

[shell garnet]

did you now?

yeah just remember this thing

it really helps

if you know any of the two parameters

like a or b or e

you can get the entire equation of the hyperbola

[abstract marlin]

we havent learned e so thats the confusing part

thats probably why i didnt understand you

[shell garnet]

but if the axis of the hyperbola (which is the x axis here) is not parallel to the coordinate axes you have the get hte equation of the curve by its definition

that is a little complex for you maybe right now

[shell garnet]

you dont know what is eccentricity?

how did you learn hyperbola then?

[abstract marlin]

no i dont

[shell garnet]

bruh

[abstract marlin]

lol

[shell garnet]

did you learn parabola?

ellipse?

[abstract marlin]

wasnt in my lessons?

[shell garnet]

where are you from?

[abstract marlin]

i learned ellipse and then hyperbola

[shell garnet]

if i may ask

which country?

[shell garnet]

yeah ellipse and hyperbola are pretty much similar (i like ellipse more though)

[abstract marlin]

do you have time to help me with two other problems

[shell garnet]

uh sure

[abstract marlin]

lol

thanks

so the center is (0,0)

[shell garnet]

well the graph is something like this

ah fuck

no

wait

[abstract marlin]

assuming its (-5,0) amd (5,0)?

[shell garnet]

its like this

my bad

i guess now you can try solving for the equation

[abstract marlin]

ok let me see

[shell garnet]

yeah just dm me ig

imma hop off rn

[abstract marlin]

i dont have ig

for this one i just knopw x^2/25

so far lol

i dont know how you find b if i wasnt given a foci

can anyone else help?

.close

[obtuse pebble]

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[thorn spade]

Could anyone please help me with this?

[thorn spade]

.close

[obtuse pebble]

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[vocal shell]

Any hints on how to solve this? I don't want an answer please. $(4ab - 1) \mid (4a^2-1)^2 \iff a=b$

[warm shale]

7aman

[vocal shell]

I tried substituting x=a-b and didn't find a clean way to divide by x that clearly shows x is 0, and I'm about to try induction rn

[timber island]

Are a and b given to be coprime?

@vocal shell

And just to be sure you need to prove a=b if (4ab-1)|(4a^2-1)^2

Right?

[obtuse pebble]

@vocal shell Has your question been resolved?

[obtuse pebble]

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[nova jolt]

I'm having issues with axis of symmetry I think. I wasn't able to get the right outcome.

[nova jolt]

particularly I am not sure if I am doing it wrong with line 4 where

x=-((-2+4)/2)

I'm not sure if the first minus symbol on the outside of the parenthesis is supposed to be there.

[hollow solstice]

You're finding vertex of parabola?

[nova jolt]

correct

[hollow solstice]

If so then quadratic's 2 roots are 4 and -2 then you casually find it's average

Not extra -ve sign

Which is 1 so vertex's x coordinate is 1

And then you put x=1 in main equation getting 9

So (1,9)

[nova jolt]

ok so the equation for finding the vertex should've been

correct?

[hollow solstice]

Yeah

[oak plover]

You can get the x from -b/2a and the y from plugging x back to the equation

[hollow solstice]

By the way do you know -b/2a, -D/4a format?

Yeah that

[nova jolt]

oh ok

[oak plover]

Where b is x's coefficient and a is the x squared's coefficient

[nova jolt]

I thought that if a was negative you had to put a negative in the axis equation

my materials are a little confusing on this

because they sometimes put a - there

[obtuse pebble]

@nova jolt Has your question been resolved?

[nova jolt]

I'm still having some trouble with it

trying to work it out

[obtuse pebble]

@nova jolt Has your question been resolved?

[obtuse pebble]

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[blissful bane]

Can you please explain what do they mean by discontinuous here?

I didn't get the idea of branch cuts of argz function

These are the figures

[obtuse pebble]

@blissful bane Has your question been resolved?

[obtuse pebble]

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[hollow igloo]

$3^{2tlog_3 4} = 4^{2t}$ this is true yeah?

[warm shale]

Yanek

[hollow igloo]

im not sure if i can use this property of logs like this

here is the equatation im trying to solve:
$x^{2log_3 4} - 7 \cdot 2^{log_3 x^2} + 12 \leq 0$

[warm shale]

Yanek

[hollow igloo]

and i substituted $t = log_3 x \implies 3^t = x$

[warm shale]

Yanek

[hollow igloo]

so i have this
$3^{2tlog_3 4} - 7 \cdot 2^{2t} + 12 \leq 0$

[warm shale]

Yanek

[hollow igloo]

and then: $4^{2t} - 7 \cdot 2^{2t} + 12 \leq 0$

[warm shale]

Yanek

[hollow igloo]

can someone confirm whether im doing this good?

[obtuse pebble]

@hollow igloo Has your question been resolved?

[red ice]

yes that's good and all correct so far

[red ice]

this is true cause a^(bc) = a^(cb) = (a^c)^b

[obtuse pebble]

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[mighty barn]

I need help with this question

<@&286206848099549185>

[obtuse pebble]

@mighty barn Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[plush lava]

hi im kind of confused on what b) is asking?

[plush lava]

i got the maximum number of the population which was 500/e at t=10

what does the max rate of increase and decrease mean?

[candid rock]

at some point the rate of change dN/dt will be at a maximum (or a minimum)

[plush lava]

so am i finding the max?

[candid rock]

when does dN/dt hit its maximum, and what is it

same with the minimum

[plush lava]

but doesnt it only have 1 max/min

[candid rock]

hint: time can't be negatuve

[plush lava]

so like this?

[candid rock]

yeah, but I also don't think your derivative is right

[plush lava]

oop

[candid rock]

e shouldn't be giving you multiple terms in its derivation

[plush lava]

whats the dy/dx of 50te^-0.1t

bc i got smth like 50(e^-0.1t -0.1t^2 * e^-1.1t)

[candid rock]

d/dt of f(t)g(t) = f'(t)g(t) + f(t)g'(t) for f(t) = 50t and g(t) = e^-0.1t

[plush lava]

but online calculator got this thingy

[plush lava]

oh

[plush lava]

thats what i did i think

[candid rock]

and what is d/dt of e^(-0.1t)

[plush lava]

-0.1t*e^oh

ohhhhhhhhhhhhh

e

-0.1t*e^-0.1t

would this be right?

[candid rock]

d/dt of e^(-0.1t) with the chain rule h(t) = f(g(t)) and g(t) = -0.1t and f(t) = e^t gives f'(g(x))g'(x)

f'(x) = e^t (it's just the same)
g'(x) = -0.1
h'(t) = -0.1e^(-0.1t)

[plush lava]

so dy/dx of 50t*e^0.1t
=50 x e^0.1t + 0.1t x e^0.1t x 50t?

[candid rock]

f'(t)g(t) + f(t)g'(t) for f(t) = 50t and g(t) = e^-0.1t
f'(t) = 50
g'(t) = -0.1e^(-0.1t)

derivative of N is then -5e^(-0.1t)) + 50te^(-0.1t) = e^(-0.1t)(50-5t)

[plush lava]

oh

why 5e?

[candid rock]

f'(t)g(t) = 50 * (-0.1e^(-0.1t)) = -0.1 * 50

[plush lava]

ohh

ok got it

thanks!

.close

[obtuse pebble]

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[silver harness]

differentiate 1/(1-10x) ' = the series

when I do derivative I gain one term?

We’re talking taylor and mclauring series

Here

[bright estuary]

@silver harness can u help me with this problem

[obtuse pebble]

@silver harness Has your question been resolved?

[silver harness]

@sonic rapids

<@&286206848099549185>

[cyan sinew]

open a new channel

[silver harness]

<@&286206848099549185>

[obtuse pebble]

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@silver harness Has your question been resolved?

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[pulsar charm]

Find range.

[pulsar charm]

I 1st devided and got y= x²+x+1
It's +ve "a" so open upwards.
Then searched how to find vertex of parabola. Couldn't find how.
So I differentiated it for it's minima. And put it's value to find y in that minima.
Then my range came out to be [7/4, ∞)
Is this technique and answer ok?

i don't have the answers for these questions...

[pulsar charm]

sorry, it's [3/4, ∞)

[polar fossil]

divided

careful... there's a hole in the graph now

bcz there's an invalid value for x

[pulsar charm]

Oh yah...
Thanks for the point.

[pulsar charm]

But do we still need to consider it after we just divided the lower part off—

Like, if I want to minus the y value for when x=1, do I do that for the original function where I dunno how to do it.
Or I put x=1 in the quadratic that came out after the factorization?

[polar fossil]

the point is that we can't divide by 0

so x=1 is a hole in your function

to see what the y value "would have been" you can plug x=1 into your quadrativ

[pulsar charm]

[3/4, ∞) - {3}
Is considered complete answer?

[obtuse pebble]

@pulsar charm Has your question been resolved?

[obtuse pebble]

@pulsar charm Has your question been resolved?

[obtuse pebble]

@pulsar charm Has your question been resolved?

[obtuse pebble]

@pulsar charm Has your question been resolved?

[pulsar charm]

<@&286206848099549185>, I just need confirmation of is my answer correct. I don't have the answer.

[wispy wadi]

no, 3 is in the range too

F(-2) = 3

[3/4, ∞) is correct

[pulsar charm]

oh yes it's quadratic...

[pulsar charm]

Thanks for the help.

.close

[obtuse pebble]

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[somber topaz]

Hi I have a quick question about Real Analisis.

If a metric space X, or subspace of a metric space X, can be entirely contained in a fixed radius ball centered in X, is that space totally bounded?

For example, is a space with 0-1 a metric always totally bounded? and why?

Or the space of all functions R -> R such that sup( | f(x) |) < 1. Is that totally bounded as well? With uniform metric for functions.

[somber topaz]

Srry for english.

[unreal musk]

For the first question, take any infinite set with the discrete metric and choose your ball radius to be strictly less than 1 (e.g. say 0.5). Any ball with radius strictly greater than 1 will contain the whole space

Assuming by 0-1 metric you’re referring to the discrete metric which is 1 when the points are different and 0 when they’re the same, that’s the second para for you

[somber topaz]

So it would be totally bounded?

[somber topaz]

yup!

The definition of totally bounded is that the space is a subset of the union of a finite amount of epsilon balls

[unreal musk]

Nope! Remember total boundedness means that given a radius of your choice, you can cover the whole space with finitely many balls with that radius

[unreal musk]

But you cannot cover an infinite set with the discrete metric with finitely many balls of radius 0.5, because each of those balls contain only one element, their center

[somber topaz]

What about a ball of radius 1 ?

That would contain the whole space would't it?

[unreal musk]

The closed ball would, the open one wouldn’t

[somber topaz]

oh yeah

[unreal musk]

Any ball of radius strictly greater than 1 would contain the whole space though, whether open or closed, which was the point here!

[somber topaz]

Wait is it a radius of my choice?

I thought it had to be an epsilon radius balls

[unreal musk]

(In other words, boundedness doesn’t imply total boundedness in general)

[somber topaz]

And a finite amount of epsilon balls and all that

[unreal musk]

Well as in for whatever radius you choose (you can denote the radius by $\epsilon > 0$), you must be able to cover with finitely many balls that have the radius $\epsilon$

[warm shale]

@unreal musk

[unreal musk]

It isn’t enough that you can find one epsilon that works (similar with convergence etc etc), it has to work for all of them

[somber topaz]

so wait, is this true of not?

Is it enough for total boundness there to find a radius that works?

Such that there is a finite amount of N radius balls?

[unreal musk]

Nope, it isn’t enough, you need to be able to show it for all possible radius choices

For example, any compact set, call it $A$, is totally bounded: choose whatever $\epsilon$ you want, and then the set of open balls ${ B_{\epsilon}(a) : a \in A }$ form an open cover of $A$, compactness implying there’s a finite sub cover of that

[warm shale]

@unreal musk

[somber topaz]

wait, so [0, 1] isn't totally bounded?

[unreal musk]

(A proof you’ve probably seen - and you’re likely to see compactness being equivalent to closure and total boundedness)

[somber topaz]

oh yeahyeah

seen that

[unreal musk]

As an interval of the real numbers with the standard metric, it is, but if you take, for example, the discrete metric, it isn’t

[somber topaz]

Why is it totally bounded with standard metric?

srry haha

But why?

Is it the union of a finite amount of epsilon balls?

[unreal musk]

One thing is that under the standard metric, boundedness does imply total boundedness (a consequence of Bolzano-Weirstrass if I recall correctly)

[somber topaz]

so wait, is [0, 1] totally bounded?

With standard metric

[unreal musk]

Matter of fact, it’s easier to visualise in that case - take an $\epsilon$ of your choice, then consider $k = \ceil{\frac1{\epsilon}}$

[warm shale]

@unreal musk

[unreal musk]

You can cover [0,1] with finitely many intervals (intervals being the same as open balls)

Argh I’d have to write it out but it isn’t too challenging to create the cover of intervals

[somber topaz]

its ok ill look it up

welp thanks!!

[unreal musk]

no worries! Key point overall is that in general boundedness doesn’t [necessarily] imply total boundedness

As for the last question you asked, I’m not too sure and would have to think about it, but I’m in bed atm

[somber topaz]

welp np! I'll ask later

I have to go now anyways

Ty chartbit!! <3

Sleep well!

[unreal musk]

See ya later

[somber topaz]

cya!

!close

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[gilded pumice]

how to write this in a mathematical way?

[gilded pumice]

i mean i got an answer

where x1 = -6, x2 = 3, x3 = 2

i figure there must be others that are possible

[robust raven]

do you need entire solution or ?

[gilded pumice]

not exatly

sure it would be easier, but i kinda want to be in the process of figuring it out

just some kind of pointers or something

idk😅😅

[robust raven]

anythign you know about matrices ?

yoru exercise is typical

for matrices methods

1 method i suggest is Cramer's rule

to us ei tyou nneed know, determinants

eventually, you may use Gaussoan eliimination as well

or inverse matrix method

unless you are pupil in school

then us e school methods

[gilded pumice]

i did one before this

[gilded pumice]

im in school, yeah

[gilded pumice]

not sure for this type of exercise which is recommended by the school, i think we have freedom in what way we solve

[robust raven]

then, for example, determine x2 from the third equation and substitute the expression for the first and second equations, then you will have a system of two equations

$x_{2}=x_{3}+1$

[warm shale]

Joanna Angel

[gilded pumice]

so:
x1+2(x3+1)−2x3=−4
2x1+4(x3+1)+x3=2

now i just need to find a way how to write it as in the other exercise, where it was a lot easier to figure out a relation between the x1,x2,x3 values

[obtuse pebble]

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[midnight steppe]

I don't really know how to start here

[obtuse pebble]

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.close

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[little scroll]

can anyone help me solve this for a family member?

[obtuse pebble]

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[neon dragon]

can someone help me with finding the interquartile range pls? thanks

[azure anchor]

thats the IQR

[neon dragon]

ohh okay i see thank you

[azure anchor]

it wants a distance between the two end points

q3 - q1

[neon dragon]

yeahh ive got it now thankss :)

[obtuse pebble]

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[spring steeple]

5 boys (B$_1$,B$_2$,B$_3$,B$_4$ and B$_5$) and 5 girls (G$_1$,G$_2$,G$_3$,G$_4$ and G$_5$) are to be seated around a round table such that boy and girl sit alternatively and B$_i$ does not sit beside G$_i \forall$ i $\in {1,2,3,4,5}$. Find the total number of such arrangements.

[warm shale]

Normed

[hollow python]

i am confused

[spring steeple]

Now wlog there are 10 options for position 1,4 options for position 2,3 options for position 3,3 options for position 4

[hollow python]

6

[spring steeple]

?

[spring steeple]

But can't say anything about positions 5,6,7,8,9,10 as the number of options would depend on the previous positions

So how do I solve it?

[lone dirge]

Hi

I'd order the boys, and then insert the girls between each pair of boys

[spring steeple]

hmm I see but how do you make sure that B_i and G_i do not sit besides each other?

[lone dirge]

I'd subtract the cases where they do

Oh mm

wait let me think about that

You could find the number of pairings and then order them

[remote skiff]

The graph theoretic way to do this is to consider a bipartite graph between the vertex set of boys and the vertex set of girls then count the number of ways you can form a cycle and remove the duplicates.

[spring steeple]

Nah haven't studied graph theory yet

[lone dirge]

@spring steeple could be a bit cumbersome, but how about inclusion–exclusion?

[obtuse pebble]

@spring steeple Has your question been resolved?

[lone dirge]

@spring steeple are you here?

[obtuse pebble]

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[lavish nymph]

I believe the answer is A can someone check I just found dy/dx and plugged in x=1

[lavish nymph]

<@&286206848099549185>

[finite tide]

hallo

[lavish nymph]

hi can you check my answer

[azure anchor]

can you show any work

[lavish nymph]

I got dy/dx=-2xy-2y/x

and then I plugged in 1 to get -4y

[azure anchor]

and you found y?

[lavish nymph]

yea

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[trail cliff]

I can't seem to figure out how to prove this

[trail cliff]

all work present, happy to elaborate on anything if its not clear!

[latent walrus]

this, the second line is an issue

[trail cliff]

hmm how so, I change sin^2 to 1-cos^2, cancel it out

do the inverse for the bottom

[latent walrus]

the numerator for example is actually, with what youre trying to do,
1+(1-cos^2)*1/cos^2

you only applied the /cos^2 to the cos^2

you completely ignored the 1

[trail cliff]

$1+(1-cos^2)*1/cos^2$

[warm shale]

Bestower

[trail cliff]

ahh I see

alr ill try it now

[obtuse pebble]

@trail cliff Has your question been resolved?

[trail cliff]

Sorry, something came up

[obtuse pebble]

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[trail cliff]

.reopen

[obtuse pebble]

✅

[trail cliff]

but I am still kinda lost, I get 1+ (1-cos^2)/cos^2 all over 1+ (1-sin^x)/sin^2x

not really sure what to do next

I could make 1-cos into sin squared

oh then I could make the top into 1+tan^2

and the bottom into 1+cot^2

when I kept going it seems like that doesn't really help me, still lost 😔

This is what my teacher did for the solution but I don't really understand how she went from the first line to the second

[latent walrus]

they put the numerators and denominators over a common denominator

the numerator for example:

$1+\frac{sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}+\frac{sin^2(x)}{cos^2(x)}$

[warm shale]

AℤØ

[latent walrus]

denominator follows a similar process

[trail cliff]

ahh I see

ya thats prob a much better method then what I did, Ill try to solve using that now

[trail cliff]

would you not have to multiply both terms by cos^2 to put them over cos^2, one becomes cos^2 but how does sin^2x not become sin^2xcos^2x

[trail cliff]

Also how do you even write latex that fast :<

[latent walrus]

just a bit o practice lol

[latent walrus]

im not too sure what you mean

the sin^2 was always over cos^2 already

there was no need to do anything to that term

[trail cliff]

I mean like

would it not have to be $(cos^(2)(1+sin^(2)x))/(cos^(2)x)$

[warm shale]

Bestower

[trail cliff]

good lord that looks bad

[latent walrus]

hmm

$\frac{cos^2(x) (1+sin^2(x))}{cos^2(x)}$

[warm shale]

AℤØ

[latent walrus]

that?

[trail cliff]

$\frac{(cos^(2)(x)(1+sin^(2)x)}{(cos^(2)x)}$

[warm shale]

Bestower

[trail cliff]

good lord

[latent walrus]

1+a/b = b/b + a/b, it wouldnt make much sense to make a/b into ab/b would it

[trail cliff]

you get the idea, if both 1 and sinx are over 1 you would need to multply both terms by cosx^2

[latent walrus]

the sin^2 isnt over 1 though

it was already over cos^2

[trail cliff]

wdym? if we are adressing the numerator by itself then sin^2 is just over 1

[latent walrus]

$sin^2(x)\left(\frac{1}{cos^2(x)}\right)=\frac{sin^2(x)}{cos^2(x)}$

oops

[trail cliff]

😭 welcome to my world

[warm shale]

AℤØ

[latent walrus]

this was in your numerator

then theres a +1

[latent walrus]

and this comes into play

you only needed to change the denominator of the 1

since the sin^2 is already over cos^2

[trail cliff]

oh wait

ya youre right

😭 forgot to multiply the sin^2 to 1 to make it over cos^2

oops

[latent walrus]

multiply the sin^2 to 1?

[trail cliff]

ya, was just blanking

all good now

[latent walrus]

okie dokie, still not too sure what you meant by this

[trail cliff]

like, sin^x(1/cos^2x) is sinx^2/cosx^2

[latent walrus]

aha, yeah alright

[trail cliff]

really simple concept, just forgot it while looking at it 😭

[trail cliff]

look good?

ignore that random floating 2

[latent walrus]

it seems snazzy yes

[trail cliff]

my sheer refusal to put the x needs to stop 😭

snazzy is a great way to describe it thanks

[latent walrus]

no worries

[trail cliff]

don't suppose theres a much snazzier way to prove this

never hurts to know it, even if I don't use the method its good to know

[latent walrus]

hmm, (1+tan^2)/(1+cot^2)=tan^2(1+tan^2)/(tan^2+1)=tan^2=sin^2/cos^2=sin^2sec^2

just randomly threw that one together

[trail cliff]

this is going to look intresting in latex

$(1+tan^2)/(1+cot^2)=tan^2(1+tan^2)/(tan^2+1)=tan^2=sin^2/cos^2=sin^2sec^2$

[warm shale]

Bestower

[latent walrus]

lmao, i can write it if you want

[trail cliff]

that would be great lmfao

❤️

[latent walrus]

$\frac{1+tan^2}{1+cot^2}=\frac{tan^2(1+tan^2)}{tan^2+1}=tan^2=\frac{sin^2}{cos^2}=sin^2sec^2$

[warm shale]

AℤØ

[trail cliff]

thats kinda sick to see just from looking at the question

amazing stuff, thanks man!

[latent walrus]

no worries

[trail cliff]

.close

[obtuse pebble]

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[wanton kestrel]

angle rtc would be 90 degrees because it is a rectangle

90-42=48

48= 2x+6

solve for x

42=2x

x=21

hope that helps 🙂

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[median light]

this helps tyyy

.close

[obtuse pebble]

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[topaz tartan]

i need help

could someone go over step by step how to solve?

[native inlet]

first find a function that tells us how much money Roshaun has at $t$ weeks

[warm shale]

PajamaMamaLlama

[native inlet]

then do the same for Keegan

where these lines meet will be the week they have the saem amount of money :)

[topaz tartan]

ok wait

so r = 150 + 10

and k = 25

well i forgot to add w

with 10 and 25

but im now stuck

[native inlet]

so we have 150+10w and 25w, so now these two lines tell us how much money each person has

so when they have the same amount of money

this will be 150+10w=25w, no? :)

[topaz tartan]

but how does the 25w connect with the other one

or substitution

[native inlet]

wdym?

[topaz tartan]

like how does 150 + 10w = 25w?

[native inlet]

well 150+10w tells us how much money Roshaun has at some number of weeks, right?

[topaz tartan]

yes

[native inlet]

and 25w tells us the same for Keegan

so these two expressions are values of money

[topaz tartan]

wait so are these both slope intercept equations?

[native inlet]

not rreally

[topaz tartan]

alright

[native inlet]

when Keegan's money = Roshuan's money that will be 25w=150+10w

[topaz tartan]

why should we do that tho

[native inlet]

because we are interested when they have the same amount of money :)

[topaz tartan]

oh alr

so now how do we solve our new equation

u there?

[native inlet]

do you know how to solve linear equations?

[topaz tartan]

linear equations

like equations that represent a line?

slope intercept

standard

[native inlet]

like 150+10w=25w

[topaz tartan]

thats sorta like standard

right

[native inlet]

not quite, for standard form of a lien we need y

we have just w

[topaz tartan]

what equation is it then

[native inlet]

there's no line, just a number

[topaz tartan]

alright

wait

so we cant combine like terms

[native inlet]

yep! precisely what you do here

[topaz tartan]

-10w

both sides

[native inlet]

[topaz tartan]

15w

for the one right of =

[native inlet]

yep so we get 15w=150

[topaz tartan]

theres no more 10w

[native inlet]

now, how do we get w to be alone?

[topaz tartan]

ez

15w / 15

and 150/15

[native inlet]

so now, w=? :)

[topaz tartan]

10

[native inlet]

[topaz tartan]

so we found the number of weeks they meet

oh wait its coming together now

now we substitute all the w values with 10

[native inlet]

close, we actualy only need to put w=10 into one of the equations, but yep pretty much :)

[topaz tartan]

yes one of them

10(10) + 150 = 25(10)

correct?

[native inlet]

nope, that will just give 0=0, which is obivous

just 25(10)

[topaz tartan]

so we could choose from the 2 equations from the starts

[native inlet]

yep! and I just chose Teegan's cause his is easier to deal with

[topaz tartan]

dont we do both

since its asking for how much both makes

[native inlet]

well it doesn't matter, because w=10 is where their equal

[topaz tartan]

so both equation would give the same answer anyways

right?

[native inlet]

[topaz tartan]

and now B is

$250

for section B

not what it equals to

you think you could help with another equation

.close

[obtuse pebble]

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[lavish steppe]

can someone explain the steps?

[obtuse pebble]

@lavish steppe Has your question been resolved?

[prisma smelt]

Notice that for all values of y, y= sin x is always 90° ahead of y= cos x

Hence we can deduce that sin x = cos (x-90°)

90° equals to π/2, and what is given in the question is -π/4. If we substitute x by 2alpha, into

sin x = cos (x-π/2)

You will get the first statement

Second line : take out 2 in the 2alpha - π/2 , to use the double angle formula where the angle is (alpha - π/4)

[lavish steppe]

should the second line be -1 instead of +1?

[prisma smelt]

Yes 🤣

[lavish steppe]

ok thank you i get it now

[prisma smelt]

I have another way to solve this, but I don't know what you prefer

[lavish steppe]

do you mind telling me it

[prisma smelt]

It's simpler but maybe longer.. And I'm outside

You can DM me

[lavish steppe]

oh thats fine then

.close

[obtuse pebble]

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[obtuse pebble]

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[latent walrus]

have you attempted putting it into vertex form?

howd it go

nice

but i meant the y=ax^2+bx+7

this isnt entirely correct either

well

factor out a first
$$a[x^2+\frac{b}{a}x]+7$$

[warm shale]

AℤØ

[latent walrus]

not in that way

it would actually be y=a(x-4)^2-5

[latent walrus]

but back to this

what next

i already did

[latent walrus]

here

dont

[latent walrus]

just get this into vertex form

well, its not done

you have to do the rest of the work

what would you normally do

what do you normally do to get something into vertex form

dont let the fact there are letters everywhere make you think its any different to normal

no it isnt

that x shouldnt be there

but its just b^2/4a

quite a wild ping

what? why

why

youre not showing me any work

leads me to believe youre just guessing

why

what makes you think that is the case

https://tenor.com/view/why-persian-room-cat-guardian-but-why-trendizisst-anyaboz-gif-2156104475561014590

well, youre wrong

b/2a=-4

youre welcome

how can i correct you if i dont know how youre even getting stuff

that right

i never gave you an answer to copy down

what do you think the point of my 'random questions' are exactly

the point of my questions is to get you to the next step of the work
when i asked you to show your work its because you pulled some equations from nowhere and i wanted to see how you were doing it

anyway
you got to the point of using ((b/a)/2)^2
so i presume you got to the point of completing the square on a[x^2+b/a x] +7

or at least thought of doing so

plus i dont really appreciate your attitude, im sat in my room helping you at 5:40 in the morning, im not under any obligation to do so

its voluntary

[still belfry]

Yo Azø can you help me out rq in #help-32

I’ve been waiting 40 mins

My question is very short

[latent walrus]

lmao, alr, good luck with your work man

[obtuse pebble]

@shadow canyon Has your question been resolved?

[obtuse pebble]

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[mellow vigil]

help mi

explain why it is non positives only

[latent walrus]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[timid silo]

what is the problem

[mellow vigil]

√x² = -x x=?

[obtuse pebble]

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[latent walrus]

welp

[timid silo]

type .reopen

[mellow vigil]

.reopen

[timid silo]

nvm

just send it in another open help channel

[mellow vigil]

ok

[obtuse pebble]

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[austere whale]

how to deal with exponents where it is powered by a negative fraction, and the numerator is not 1

[austere whale]

so for example,

4^(-3/4)

[narrow vault]

that’s not a negative fraction

[inland matrix]

You could make the numerator 1
For ex if you have
2^(-2/3), you could rewrite it as (2^(-2))^(1/3)

${2^{-2}}^{(1/3)}$

[austere whale]

mb

[warm shale]

Lorentz

[austere whale]

so, (1/4)^(1/3)

[inland matrix]

Yeah

[austere whale]

then cube 1/4

so it will become

[inland matrix]

Cube root*

[austere whale]

yesyes

1/cbr4

cbr is cuberoot

[inland matrix]

Yeah

[austere whale]

then rationalize since radicals cannot be in denominator and all that

[inland matrix]

Yeah you could do that

[austere whale]

thanks

[inland matrix]

Np

[austere whale]

.close

[obtuse pebble]

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[royal sonnet]

not really sure if im doing this right, also dont know where to go from here, any help would be greatly appreciated

[timid silo]

I mean I assume you are done by that point

,w Laplace transform y'' -4y' +4y = t, y(0)=1, y'(0)=0

[warm shale]

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[timid silo]

oh man

your work seems fine to me though

[royal sonnet]

i think im supposed to do the inverse laplace transform like i did iin this question, although im also not sure if this is correct

[timid silo]

no

it's $\ds \m Fs= \cm\LL{\m ft} \tss{and} \m ft = \cm{{\LL}^{-1}}{\m FS}$

[timid silo]

what you did is fine

[royal sonnet]

o okay

thank you

.close

[obtuse pebble]

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[vocal lynx]

guys I need your help. i really dont know how to find this answer :

34 + 46 = 80 million
46 = 10 cash, 36 deposit

is there a way to find out how much is the 34 consists of in cash or deposit with mathematic formula ?

[obtuse pebble]

@vocal lynx Has your question been resolved?

[obtuse pebble]

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[tulip sky]

whats wrong with doing this?

[tulip sky]

$$a^3 + b^3 + c^3 \geq 3abc$$

a^3 + b^3 + c^3 >= 3abc by AM-GM

so if we prove 3abc>= a^2b+b^2c + c^2a

wouldnt that solve the problem?

[brazen gorge]

we don't know the ordering

is it
$$a^3 + b^3 + c^3 \geq 3abc \geq a^2b+b^2c + c^2a$$
or
$$a^3 + b^3 + c^3 \geq a^2b+b^2c + c^2a \geq 3abc$$?

,tex test

huh

anyways

[brazen gorge]

if we try to prove that x>y, and we find that x>z, that does not imply that z>y

[tulip sky]

agreed

but

if we know x>=z

and we find out z>=y

why wouldn't that imply x>=y

[brazen gorge]

it would but that isn't what you have here

[tulip sky]

it says "So you'd be happy if you could show that 3abc>-a^2b + b^2c + c^2a"

and if one manages to show that, wouldn't it prove the inequality?