help-28

[sweet ingot]

There is no difference

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[gritty remnant]

In a linear function, is the rate of change always the same?

[gritty remnant]

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[keen isle]

yes the rate of change is slope

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[nova trench]

Yo, I need help with Cauchy’s Theorem. Will send the pick now

[ashen elk]

i should the able to help, go ahead

[nova trench]

I am stuck in here

[ashen elk]

what's the question lol? sorry i can't understand your writing at all

[nova trench]

Mm, okay. I will do it again now

I need to prove this limit using Cauchy’s Theorem

[ashen elk]

ohhhhh, we're thinking the wrong cauchy's theorem i think. i thought it was about the complex analysis one (with contour integrals). i'm not too sure about how to prove it, sorry

[nova trench]

Okay, np

Status: Still waiting for help

<@&286206848099549185>

<@&286206848099549185>

[cold orchid]

By "Cauchy's Theorem", do you mean "(ε, δ)-definition of limit"?

[nova trench]

yeap

[cold orchid]

okey

[nova trench]

I did like this

And I got stuck

[cold orchid]

${\displaystyle {}^{\forall }\epsilon >0,;{}^{\exists }\delta >0;;;{}^{\forall }x\in \mathbb {R} ;[0<|x-2|<\delta \Rightarrow |x^3-8|<\varepsilon ]}$

[glossy valve]

とも

[nova trench]

Aha, probably i did it

[nova trench]

Here

A bit different

[cold orchid]

I'm not done yet.

[nova trench]

Ah, okay, sorry

[sweet ingot]

$x^2+4x+4=(x-2)^2+6(x-2)+12$

[glossy valve]

ScapeProf

[sweet ingot]

Then some triangle inequality and you get ||<delta(delta^2+6delta+12)||

(Another way is only look at x values close to the limit, so for example only look at x values in [1,3] so you can bound x^2+4x+4 that way)

Then pick delta=min{…}

[nova trench]

Now I am totally confused xD

Can you please explain once more

[sweet ingot]

1st way I did I just factored into (x-2) terms because we know something about those

$x^2+2x+4=(x-2)^2+6(x-2)+12$

[glossy valve]

ScapeProf

[nova trench]

Is it all i need to write?

[sweet ingot]

Meant 2x ofc

[nova trench]

And the problem is proved?

[sweet ingot]

No?

You need to find a value of delta

This is a way to proceed

You said you were stuck?

You were stuck at |(x-2)(x^2+2x+4)| you said - I assume this was because of the last term, so I showed you a way to proceed

[nova trench]

Sorry, I meant that i do not know what. I want to see how to make this finish

[sweet ingot]

You know |x-2|<delta agree? You now have a bunch of terms involving (x-2). By using triangle inequality you can turn all these terms in to |x-2|

[sweet ingot]

And you end up with this

[nova trench]

Okay,

Now i got it

Thanks

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[short cloud]

in some equations why we need to square both sides and multiply, divide with same number. how i can know when i need to do that

[narrow vault]

because the equation remains true when u do that

if u did something different to each side the equation would no longer be true

so u wanna do it when rearranging equations, like trying to solve for something

[short cloud]

yeah

how m i gonna know what i need to do in equation when nothing left to more simplify

[narrow vault]

it depends what your goal is, if youre trying to solve for an unknown/variable, once you have an equation in which the variable is by itself on one side and there's a number on the other side of the equals sign, youre done

[narrow vault]

Like the steps of solving an equation?

[short cloud]

yeah

[narrow vault]

If youre dealing with a linear equation in one unkown you wanna do something like, clear fractions/multiply out brackets, then move all terms in the unkown to one side and all constant terms to the other, then factor out your unkown and divide by whatever its being multiplied by

[short cloud]

like some peoples do by their own like they multiply and divide by same numbers etc etc, how am i gonna know what to do when and how i can practice

[narrow vault]

are you able to do these?

[quiet nimbus]

They are isolating for the variable (trying to make it alone to find a value for it)

[short cloud]

yeah

[narrow vault]

what kind of ones are u struggling with

[short cloud]

and many where i needd to find relations

[narrow vault]

,racw

,rccw

[short cloud]

mostly variable questions

like in this why u/g goes right so it gets n-1

then in next step why did both side square

and here why u^2 - u^2 didnt get cancelled

and wat happen in next step

[short cloud]

.

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[short cloud]

<@&286206848099549185>

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@short cloud Has your question been resolved?

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@short cloud Has your question been resolved?

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[iron tusk]

where do i start.

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[iron tusk]

anyone?

[bronze fiber]

Yoo one sec bro

[iron tusk]

@bronze fiber

[bronze fiber]

I’m just drawing it out for u

So find equation of the hyperbola

And transverse axis means the distance vertically between the two halves of the curve

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[plush egret]

it ties directly in

youre showing that formula is true generally

then using it to differentiate g

since $g(x,y) = x \cdot y(x) \cdot e^{-x}$

[glossy valve]

jan Niku

[plush egret]

3 functions of x

do you see?

[cursive finch]

oh i kinda see it

[plush egret]

so what do you think? can you answer the question?

or still confused

[cursive finch]

so to find the derivative of it, you would do x'ye^-e + xy'e^-x + xy(-e^-x)?

[plush egret]

yea this seems right to me

[cursive finch]

ok and then that would be the answer since theyre all variables right

[plush egret]

some are some arent

[cursive finch]

?

[plush egret]

theres a 10 in there yea?

oops

sorry getting channels mixed up in my head

[cursive finch]

XD all good

[plush egret]

almost the answer

whats $\dv x (x)$

[cursive finch]

oh what am i missing?

[glossy valve]

jan Niku

[cursive finch]

it would just be 1?

[plush egret]

x' doesnt make sense in the answer

yea

but i think i see what you mean like just writing the form of the answer in there but you have to make sure to evaluate everything you can

[cursive finch]

oh so since its 1 it just cancels out

[plush egret]

yup

well cancel is a weird word but yea

[cursive finch]

ok tysm

[cursive finch]

is there a way to reduce them or should i just leaeve it like this : 1ye^-e + xy'e^-x + xy(-e^-x)

[plush egret]

i mean i think thats fine

you can factor or whatever but this maintains the form of the tool you used to thats nice

[cursive finch]

ok

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[velvet flume]

$3x=\sqrt{2x+14}$

[glossy valve]

nerd19

[velvet flume]

yeah im very dumb

😔

can't find a real solution in this

anyone help pls

[hollow geode]

square both sides, the solve the quadratic

[velvet flume]

that would be

$9x²=2x+14$

[glossy valve]

nerd19

[velvet flume]

i think

[hollow geode]

yes

[velvet flume]

whats the next step?

[hollow geode]

"then solve the quadratic"

[velvet flume]

alright

ok it's a math error on the calculator

i tried quadratic formula

can't factor that too

[hollow geode]

yeah just use quad formula

$x=\frac{2\pm\sqrt{(-2)^2-4(9)(-14)}}{2(9)}$

[glossy valve]

Mosh

[velvet flume]

ok nvm it works i forgot negative sign sry 💀

this is the answer btw

$/frac{-1+/sqrt{127}}{9}$

[glossy valve]

nerd19

[abstract prairie]

\

[velvet flume]

ok wait

wrong slash lol

$\frac{-1+\sqrt{127}}{9}$

[glossy valve]

nerd19

[velvet flume]

is this supposed to be x value?

[velvet flume]

wait why did that 14 become negative tho?

[hollow geode]

cause you move everything over to 1 side..

9x^2-2x-14=0

[velvet flume]

oh oki

$\frac{-1\pm\sqrt{127}}{9}$

[glossy valve]

nerd19

[velvet flume]

@hollow geode this what came up in calculator

[hollow geode]

it's wrong.

[velvet flume]

usually u dont need to do anything with that right?

[hollow geode]

-(-2)=2

[velvet flume]

huh

[hollow geode]

so you get 1+-sqrt(whatever)

[velvet flume]

all i did was copy this to the calc

$\frac{1\pm\sqrt{127}}{9}$

[glossy valve]

nerd19

[velvet flume]

is this the answer?

@hollow geode

[hollow geode]

yes.

then of course only 1 works

[velvet flume]

do i need to do anything with this or?

is this whole thing just the final answer?

[hollow geode]

No

you check if both are actually solns

since squaring introduces solns

[velvet flume]

ok so i just substitute them as x

ok so i just substitute both 3x and √2x+14

used negative didn't work

used positived they both showed same solution

so i guess this is the answer

$\frac{1+\sqrt{127}}{9}$

[glossy valve]

nerd19

[velvet flume]

@hollow geode pls say im right 🙏

[hollow geode]

yes.

[velvet flume]

oki thanks for helping meeee!

tysm

also how to close this channel xd

[hollow geode]

.close

[velvet flume]

.close

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[small pebble]

Hey look a new dopamine equation from me:
How would I find when $$\sqrt{\frac{xy}{2x^{2}+y^{2}}}$$ evaluates to an integer for integer inputs of x and y?
Is there a way to solve this?

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[fluid dragon]

$\sqrt{\frac{xy}{2x^{2}+y^{2}}}$

[terse karma]

Perhaps using y=mx

And when x=0 , y is any non jero integer or vice versa

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[mental kettle]

would it be m<BOC = m<AOB (2x+10=4x-15)?

[small pebble]

Wow my internet really died for an hour right after I asked the question.

[small pebble]

No. It says the diagram is not to scale.

AOC being 85 degrees is crucial.

If you look at the angles, what is AOC in relation to AOB and BOC?

[mental kettle]

this is what i did. i combined the like terms and got 6x - 5 then i did 6x - 5 = 85 and i got 15 then i imputed 15 in each x and got 40 for boc and 45 for aob

so x is 15?

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[mental kettle]

<@&286206848099549185> it’s been over 15 mins sorry for ping but can someone tell me if this is correct. 40 boc ! aob 45 and x = 15

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[sullen cloak]

can someone help me rearrange this equation to make t the subject?
x = PMT*{[(1+r/n)^(nt)–1]/(r/n)}+P(1+r/n)^nt

[sullen cloak]

<@&286206848099549185>

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[charred hinge]

could you just send a picture of the q? @sullen cloak

[sullen cloak]

That is the question

[charred hinge]

its really hard to see in this form

is ahat i mean

[sullen cloak]

Oh right

Idk how else to sent it as it’s part of a program I’m making and this is just something I made up

So there’s not like somewhere written nicer

[charred hinge]

okay i dont think im swag enough to do this, sorry

[sullen cloak]

Oh, 😦

That’s fine dw

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[indigo tide]

can someone help me? 😄

[indigo tide]

with part B)

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@indigo tide Has your question been resolved?

[sullen cloak]

.reopen

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[wanton plaza]

I'm having trouble understanding the Lesbegue measure. Can someone help me and solve this exercise in order for me to check if I'm doing it right?

[versed tendon]

isn't lebuesge measure for a set (a,b) = b-a?

[wanton plaza]

This is the definition I have in my class notes but I'm having trouble applying it

[versed tendon]

at the first part

5.2

it's just explaining how disjoint intervals are calculated

so if you have [1, 3] U [7,15]

it tells you it's (3-1) + (15-7)

your exercise is simpler

you don't have disjoint intervals

@wanton plaza are you still unsure of how to calculate the lebesgue measure of an interval?

[wanton plaza]

my problem is calculating lebesgue measure of an interval with infinity

[versed tendon]

infinity + constant = infinity

[wanton plaza]

Ok thanks you really simplified it to me thanks

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[jade owl]

i need help with this problem

[wind pier]

Have you ever heard about simmillar triangles?

[jade owl]

yes but i dont understand them

[wind pier]

There are two triangles in the picture do you see them?

[jade owl]

ye

[wind pier]

They are simmillar because they have angles of the same size (they share one, one is 90 degrees and thus the third one is same too)

[jade owl]

yes i know that

[wind pier]

And when you have simmillar triangles you know that each side of one of them is x times longer than the same side of the other one.

So first you need to find x

[jade owl]

and how do u do htat

[wind pier]

Take lenghts of the same sides of those two triangles(the pair that you know)

Do you know which ones?

[jade owl]

would i have to take 16.35- 12.1= 4.25
16.35/4.25= 3.84705882*1.55= 5.96294117
so then would the answer be 5.96294117

would this be the correct why of solving it?

[wind pier]

Should be

[jade owl]

ok thx for the help

[wind pier]

But it has one flaw. You have to assume that you and the tree are "standing at the same angle".

[jade owl]

what do u mean

[wind pier]

If the angle tree-ground is different than human-ground than the triangles are not simmillar.

[jade owl]

i think that higher math we are just learnign with a lot of assumptions lol

[wind pier]

Ok

use .close if no other help is needed

[jade owl]

ok i dint know that

.close

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[torn jolt]

Hello

[torn jolt]

How to solve this?

[torn jolt]

Question number 2.

I am here.

(sin(x + pi/3) - (sqrt(3)/2)) / x as x approaches 2.

<@&286206848099549185>

[keen spruce]

@torn jolt Do you have to use limits

[torn jolt]

Yes 🙂

@keen spruce

[keen spruce]

Nevermind

[torn jolt]

What happened? @keen spruce

<@&286206848099549185> Will someone please help me? 🙂

[keen spruce]

I forgot how to do this

[torn jolt]

No worries 🙂

[keen spruce]

Anyways

[full marsh]

LOL

that's a good one

[jaunty chasm]

Calc ab in a nutshell

[torn jolt]

@keen spruce If we don't use limit then how can we solve it?

[keen spruce]

We can solve it using differentiation rules or L'Hôpital's rule

I do not remember how to solve this using the limit definition of a derivative, I even just had an exam over it 🙂

[torn jolt]

I see.

Will someone please help me?

<@&286206848099549185>

.close

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[agile vigil]

Suppose S is a relation on a set X which is reflexive and transitive. Then S intersection S inverse is an equivalence relation on X.

[agile vigil]

any idea how to prove this ?

I know the defintions, but i dont know what to do with it

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[agile vigil]

Suppose S is a relation on a set X which is reflexive and transitive. Then $S\cap S^{-1}$ is an equivalence relation on X.

[glossy valve]

Michal

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[gilded bluff]

im not sure where to start with part d

[gilded bluff]

do i use the am-gm inequality?

[hollow geode]

Use your results from a-c most likely

[gilded bluff]

im not sure if my result for b is correct tho

[hollow geode]

you can still use the result..

Ie you can complete part d w/o having completed a-c

[gilded bluff]

kinda not sure where to go about this

[hollow geode]

$x+y+z-3(xyz)^{1/3}\geq 0$ is equivalent to what you're proving

[glossy valve]

Mosh

[gilded bluff]

yep

[hollow geode]

what you're showing in d is analogous to c

[gilded bluff]

so do i use a different expression for x + y + z

[hollow geode]

x+y+z is analogous to a^3+b^3+c^3

[gilded bluff]

yeah

[hollow geode]

d should follow directly from c.

[gilded bluff]

could you elaborate more on this

[hollow geode]

x=a^3.

[gilded bluff]

ohh okay i understand what u mean by analogous ok ok

oh thank you i understand how to prove it now

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[torn jolt]

Hello

[torn jolt]

May I know how to solve 9th question?

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[torn jolt]

<@&286206848099549185>

[torn jolt]

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[young echo]

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[glossy valve]

CalicoRackham

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[gilded bluff]

i dont know how to go about this

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[gilded bluff]

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[grand vessel]

$a^2 + b^2 \geq 2ab$ and $c^2 + d^2 \geq 2cd$ then $a^2+b^2+c^2+d^2 \geq 2ab + 2cd$. You can apply again in $ab + cd \geq 2 \sqrt{ab} \sqrt{cd}$ and you get $a^2+b^2+c^2+d^2 \geq 2(ab + cd) \geq 2(2 \sqrt{abcd})$ and that is what u want.

[glossy valve]

DanielC

[hollow geode]

dont just give answers.

[grand vessel]

sorry, first time

:c

[gilded bluff]

ya'll i just wanted a hint 😭

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[lyric knot]

hello

[lyric knot]

im currently trying to solve this mathematical problem

its a representation that i made in paint so it may be innacuarate

what im trying to do is given a certain point and a 2nd point

get a 3rd point within the radius of a circunference which centre is 1st point

i've been trying to solve it and cant figure out how

is it possible without the use of angles?

[sharp lagoon]

@lyric knot So, same distance from point 1 to point 2 and from point 1 to point 3?

[lyric knot]

no they are just for example

they can be any point within radius of circunference

[sharp lagoon]

I don't understand the problem yet. You have a point. You have a second point. Is the radius related to the distance between the first and second point?

[lyric knot]

the radius is related to the third point

[sharp lagoon]

How?

[lyric knot]

for example point A

point A is on the radius of the circunference

and is on a straight line with point 1 and point 2

[sharp lagoon]

So, point 1 is the center of a circle, and points 1, A, and 2 are on a line?

[lyric knot]

yes

[sharp lagoon]

OK, is 2 on the circle or inside it?

[lyric knot]

yes it can be any point within the radius of the circle

[sharp lagoon]

Is A on the circle or inside it?

[lyric knot]

on the circle

[sharp lagoon]

OK, so 2 has to be different than both 1 and A?

[lyric knot]

yes

[sharp lagoon]

OK, so point 1 is the center, point A is on the circle, and point 2 is collinear with them but distinct from them.

What about point 3?

Can it be collinear with them?

[lyric knot]

point 3 would not be used in that case

point would be used to get point B

yes they would be collinear

[sharp lagoon]

So, you're trying to draw a line that contains 1 and 3 and where that line hits the circle is B?

[lyric knot]

yes exactly

[sharp lagoon]

So, you know ahead of time where 1 is, what the radius is, and where 3 is, and you're asking where B is?

[lyric knot]

yes

[sharp lagoon]

OK, so you get the slope between 1 and 3.

Let me work it out. Let's see.

OK, so point-slope form says the line from 1 to B is:

y - 1_y = m(x - 1_x)

And the distance between 1 and B is:

(y - 1_y)² + (x - 1_x)² = r²

That's from the Pythagorean theorem.

So, if y - 1_y = m(x - 1_x), we can replace y - 1_y with m(x - 1_x).

m²(x - 1_x)² + (x - 1_x)² = r²

That can be factored to:
(m² + 1)(x - 1_x)² = r²

You can get m (the slope between 3 and 1).

You know 1's x component (1_x).

You know the radius.

Then you just solve for x, which is the x component of B.

So, let's solve for x.

(x - 1_x)² = r²/(m² + 1)

x² - 2 · 1_x · x + 1_x² = r/(m² + 1)

x² - 2 · 1_x · x + 1_x² - r/(m² + 1) = 0

Then, you use the quadratic formula with a = 1, b = 2 · 1_x, and c = 1_x² - r/(m² + 1).

Do you have any questions so far?

[lyric knot]

i think i understand it

thanks

[sharp lagoon]

No problem.

Oh, mistake.

It was r² and I forgot the ².

So, a = 1, b = 2 · 1_x, and c = 1_x² - r²/(m² + 1).

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@lyric knot Has your question been resolved?

[sharp lagoon]

@lyric knot Here's Mathematica's solution. cy and cx are the y and x of the center of the circle.

[lyric knot]

thanks

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[gilded bluff]

could someone help me with ciii?

[gilded bluff]

i just want a hint

dont give me the full answer

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[gilded bluff]

<@&286206848099549185>

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[dusk surge]

my answer is 0

because if we take x as 0

numerator is 0

therefore the lim is 0

ohh

wait but the denominator also is 0

thats why my answer is wrong lmao

[grand vessel]

yep, and if u get 0/0, do you know what to do?

[dusk surge]

im tryna multiply by the conjugate

[meager dew]

just lhopital it

[dusk surge]

i didnt learn that yet

but can u teach me

i know we will learn it

[meager dew]

if the limit is 0/0 or infinity/infinity, you take the derivative of the numerator and denominator separately

and that is your new limit

[dusk surge]

oh that easy

wow

[grand vessel]

and this also works, u cant do it in both ways and u should get the same answer

[dusk surge]

cool

did i do it correct?

[meager dew]

yep

[dusk surge]

thanks both

ill study lhopital in a bit

https://giphy.com/gifs/originals-covid-psa-19-igbomc6YEwCS566BJO

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[night nest]

What did you try doing so far

[torn jolt]

Well I tried putting values

And I really got the answer.

But is there any other methods.

[night nest]

Another method would be to rephrase the question, it is asking what's the ratio between successive terms as the Fibonacci sequence continues, does that help?

[torn jolt]

Yes.

[night nest]

so you can assume that nth and (n+1)th Fibonacci numbers are a and b, for large n.
by given recurrence, (n+2)th term is a+b

Ratio between (n+2)th and (n+1)th term = (a+b)/b
Ratio between (n+1)th and n th term = b/a

And since the sequence a_n converges (you can either test if it converges or go off the fact that none of the options say "none of these"), you can say that

Required limit = (a+b)/b = b/a

[torn jolt]

Yes Makes Sense.

[night nest]

[torn jolt]

Well thank you.

[night nest]

Glad to help

[torn jolt]

In Q.25 z21 is only relatively prime with 221 so

I'm thinking that could be the answer. And it is the answer.

But this dosen't feel legel way

[dusk surge]

are you from turkey bim?

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[torn jolt]

Nope

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[torn jolt]

.reopen

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✅

[torn jolt]

<@&286206848099549185>

[dusk surge]

ah ok

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[wise seal]

Okay so the question says that

If |x|<2 and |y|<3 prove that
|2x•3y|<13

[wise seal]

In this situation aren’t I supposed to do
-13<2x3y<13

But I don’t know what to do next

Do I have to divide ?

[vast fossil]

There's no need in getting rid of the absolute values

Here you can use the fact that $\abs{2x\cdot{3y}} = \abs{2x}\cdot\abs{3y} = 6\abs{x}\abs{y}$

[glossy valve]

Touch Our Beans

[wise seal]

What about 13?

Oop anyways I found it thank youuuu

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[mental plank]

whoes doubt is it ?

[keen quiver]

both lol

[mental plank]

k lol

[worthy rose]

r u the real playmaker

[mental plank]

use this guyzz

[mental plank]

ig I've met u guyss bfr I changed my dp from that swordsmen ( kirito dual sword )

[mental plank]

@worthy rose and @keen quiver use this one

[worthy rose]

thank you playmaker

[keen quiver]

yeah i got it thanks

[mental plank]

np

feel free to dm if u' get on this doubt

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@worthy rose Has your question been resolved?

[keen quiver]

Hey

I got the first part right but not the second part

Can you explain it please?

the a and b values in the quadratic are right but not the c value

[worthy rose]

till 70/8 is correct

<@&286206848099549185>

@mental plank

[mental plank]

it seems correct

[worthy rose]

the 2nd part is supposed to equal 147/8

[mental plank]

oh u want me to check the working ?

[keen quiver]

yes plz

[mental plank]

-45/4

[keen quiver]

omfg

lmao

thanks

[mental plank]

np

[keen quiver]

i cant believe i missed that

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[soft zealot]

@plush egret

[plush egret]

?

[hollow geode]

yes cause .reopen is so hard to do

[soft zealot]

So you actually explained it to where i could understand but there isnt a answer with the answer i got

4x3

is 12

but 12 isnt a answer

[hollow geode]

yeah, cause 4*3 was complete nonsense looking at your question

nowhere does a 3 show up out of 8 and 5

[hollow geode]

yeah, tell me where you see a 3

[soft zealot]

This is the Figure

Half of 5 is 3

[hollow geode]

??????

No it isn't

not at all

[soft zealot]

They told me that

[plush egret]

ah, the half doesnt go to each number

thats an error in more ways than one

you have 8 x 5 x 1/2 is the answer

[hollow geode]

I highly doubt jan would say 5/2=3

[plush egret]

but you dont half each number

its just multiplication, you can do 8 times 5, then take half

[hollow geode]

$A_{\Delta}=\frac{bh}{2}$

[glossy valve]

Mosh

[plush egret]

yea

this is the formula you should have in your head

[soft zealot]

Ohhh i get it

[plush egret]

but realizing that like

$\frac{bh}{2} \neq \frac b 2 \frac h 2$

thats maybe important too

[soft zealot]

Oh ok thxs

[glossy valve]

jan Niku

[plush egret]

multiplication and fractions dont work like this

[soft zealot]

Oh

so 10 ft is the answer

[hollow geode]

No..

[hollow geode]

8*5 isnt 20.

[soft zealot]

Well those are the only answers

[hollow geode]

What is 8*5

[soft zealot]

ohhhhh nvm i was doing 5*4

[hollow geode]

yeah.

[soft zealot]

40

[hollow geode]

and half of 40 is 20.

[plush egret]

you should write more 😄

[soft zealot]

so the answer is 20

[plush egret]

its a bummer but you should write out the steps to the problem like this

or youll make little mistakes

[soft zealot]

K

[plush egret]

or something you can look at on paper

to organize your thoughts

[soft zealot]

Im writing rn

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[agile vigil]

bijection between $\mathbb{N} \times \mathbb{N} \mapsto \mathbb{N}$

[glossy valve]

Michal

[agile vigil]

any ideas ?

i have come up with an injection $(n_1,n_2) \mapsto 3^{n_1 +1 } + 5^{n_2 + 1}$

[glossy valve]

Michal

[fathom saddle]

Have you seen the proof that Q is countable? That diagonal trick works here too

[agile vigil]

Yes i have

Hmm

So it means when I can order this set of ordered pairs, we are able to found a bijection between NxN ->N

?

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@agile vigil Has your question been resolved?

[fathom saddle]

Like:
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)
And draw the diagonal lines. That's your bijection

f(2,2) = 5 with this method

[agile vigil]

Ok thanks

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[hot herald]

which question are you stuck on

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[jaunty chasm]

What are the instructions?

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[dusk inlet]

I came up with a proof that is intuitively true but I can't seem to work it out. Say I have a closed interval [a, b] $\subseteq R$, if the length of $[a, b]$ which is given by $d(a-b) < \epsilon$ for a given $\epsilon > 0$ then for any two point $x_1, x_2 \in [a, b]$ we have $d(x_1, x_2) < \epsilon$.

[glossy valve]

Plegasus

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@dusk inlet Has your question been resolved?

[dusk inlet]

Maybe showing $d(a-x_1) + d(x_1-x_2) + d(x_2-b) = d(a-b)$ when I assume $x_1 < x_2$ would work.

[glossy valve]

Plegasus

[dusk inlet]

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[brittle marsh]

Hi

[brittle marsh]

Can someone help me on this problem, I’m doing for A level maths revision for my mock exam

It’s question 4 on here

[light sonnet]

You're solving for theta

Same concept as (x + 2)(x - 1) = 0, solve for x, but your problem is trig related

[brittle marsh]

Do I start by expanding the bracket?

[light sonnet]

No need to

Zero product property

If you had (x + 2)(x - 1) = 0, how would you find x?

[brittle marsh]

-2 and 1

I’m still rly stuck

I can’t see what the answers r for this trig equation

[light sonnet]

How did you do that for my example?

Not visually, but mathematically

[brittle marsh]

I just looked at

[brittle marsh]

Expand and use the quadratic formula

[light sonnet]

No need to

What if it was $(\frac{3}{4}x + 2)(x - 1) = 0$?

[glossy valve]

dldh06

[light sonnet]

How would you solve for x?

[brittle marsh]

I’m not sure tbh, I’d usually just look at it and know the answer or I’d expand and use the formula

[light sonnet]

As I stated, zero product property, meaning that if you have AB = 0, then that means A = 0 or B = 0

Remember that concept?

[brittle marsh]

I’ve never heard of it b4 tbh

[light sonnet]

So then that means $(\frac{3}{4}x + 2) = 0$ or $(x - 1) = 0$

[glossy valve]

dldh06

[brittle marsh]

Oh ok

[light sonnet]

Then you just find x

[brittle marsh]

So u set each bracket to 0?

[light sonnet]

That concept is called zero product property. You probably never referred to the full name

[light sonnet]

But yes

So do the same for the trig question

[brittle marsh]

I’ve done that

[light sonnet]

You wrote it wrong

[brittle marsh]

So from here would I move the 2 and 1 to the other side to figure the unknown

[light sonnet]

Double check the sin one

[brittle marsh]

Is it -0.4

[light sonnet]

The problem was $(1 + \tan \theta)(5 \sin \theta - 2) = 0$

[glossy valve]

dldh06

[brittle marsh]

So it’s 5sinX-2=0?

[light sonnet]

Yes

[brittle marsh]

Ok, then what would I do?

Do I solve the unknown by re arranging?

[light sonnet]

Solve for theta

[brittle marsh]

I’ll do that real quick

[light sonnet]

,rotate

[light sonnet]

Looks good

[brittle marsh]

So what do I do for part ii?

[light sonnet]

Change tan in terms of sine and cosine

Then in the form of AB = 0

[brittle marsh]

So tan becomes sin/cos?

[light sonnet]

Yes

Also, it's not an exam right? I just noticed there were points

[brittle marsh]

It’s a revision sheet

I’ve got my a level maths mock exam soon

This is just revision

Which I’m struggling with

[light sonnet]

So tan x = sin x /cos x

[brittle marsh]

Done this

Do I just do cos-1(0.75) to get all the answers?

[light sonnet]

No

Back to $4sin x = \frac{sinx}{cosx}$

[glossy valve]

dldh06

[brittle marsh]

This is what I’ve done

[light sonnet]

Multiply by cos on both sides

[brittle marsh]

Oh

So I’ve done it wrong?

[light sonnet]

Yes

[brittle marsh]

Can u show me the correct thing to do?

[light sonnet]

What I would do, multiply by cos on both sides

$4sin x = 3 \frac{sinx}{cosx}$

[glossy valve]

dldh06

[light sonnet]

At that step

[brittle marsh]

Ok

What next

I think it gives the same answers

[light sonnet]

It might

But then I would subtract 3sin x

So you have $4 \sin x \cos x - 3 \sin x = 0$

[glossy valve]

dldh06

[light sonnet]

So the cos term is correct

You just need the sin term

Because if you factor, you have $$(\sin x)(4 \cos x - 3) = 0$$

[glossy valve]

dldh06

[light sonnet]

Then zero product property

[brittle marsh]

Ok thanks

I’ve completed the revision for this topic

I haven’t got a mark scheme

Do u think u could check my answers, it’s only 6 questions

[light sonnet]

Sure

[brittle marsh]

Thanks

[light sonnet]

First q doesn't look right

[brittle marsh]

Oh

Part ii?

[light sonnet]

Both parts

[brittle marsh]

Ur capping right?

No way it’s wrong

[light sonnet]

You have $cos^{-1}(0.75) = x + 70$ which is x + 70 = 41.40

[glossy valve]

dldh06

[light sonnet]

Meaning x = -28.59

[brittle marsh]

Ok

I’ve corrected it

Any other mistakes

[light sonnet]

I'm too tired to process the rest

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@brittle marsh Has your question been resolved?

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@brittle marsh Has your question been resolved?

[brittle marsh]

Can anyone check my work?

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@brittle marsh Has your question been resolved?

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[carmine minnow]

what is this

what have you tried

[south spruce]

isolating tan3x

[carmine minnow]

do you know any trigonometric identities that can help you?

something involving sec^2 and tan^2?

[south spruce]

dude i didnt study them

💀

f

yes i see one that can help me in my notes

[carmine minnow]

okay, what does that give you?

[south spruce]

tan3x = tan^2x-tan^2x+1

tan3x=1

3x =pi/4,5pi/4

x=pi/12,5pi/12

thx man

[carmine minnow]

are you sure that's all the solutions?

[south spruce]

uh

no

@carmine minnowif secx = 0 then cos is undefined

[carmine minnow]

you mean if cos x = 0?

can x=3pi/4?

[south spruce]

the equation involves secant

no

3x = pi/4

then x would be pi/12

pi/4 * 1/3

[carmine minnow]

???I don't get what's going on

why can't x be 3pi/4?

[south spruce]

tan3x = 1

3x = pi/4 and 5pi/4

arctan of 1 cannot include 3pi/4 because it doesnt include -1

it's not plus or minus 1

it's just 1

so now we have 2 solutions

3x = pi/4

3x = 5pi/4

[carmine minnow]

how about 3x=9pi/4

[south spruce]

9pi/4 is greater than 2pi

[carmine minnow]

so it would result in x>2pi?

[south spruce]

thats coterminal with pi/4

[carmine minnow]

but when you divide by 3 you get a different angle

[south spruce]

hold on a minute

ok

i think i got it right now

.close

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[torn jolt]

Ok so

[torn jolt]

Am i doing something wrong with my deltas?

Because say i try to find the angle of -85

I should be using 180* degrees - ( -85* degrees)

Right?

But that equals 265*

Then why the heck does its Cos 85* = 275*

Im trying to find de actual angle

But they dont = each other

So im like

What now?

Like

?????

[north forum]

can you speak english

please

[torn jolt]

Isnt that

What

Im doing?.

[north forum]

i didn't get a single word you were saying

[torn jolt]

Ah...

Damn

Ok so

[torn jolt]

Maybe if you can just do it and explain it (its quick) itd be easier if i showed the uh question itself

Basically if Sin(x) = -0.9961 and cos(x) = 0.0871

I need to find the angle of x

Which is well

Just using this right here in the end right?

Since sin(x) = -0.9961

X = -85*

Ok

But then we dont want negative degrees

So im supposed to use that

And it makes 265*

But then why does Cos in my question = 275*

Its supposed to be 265

Right??

Its supposed to be the same

[north forum]

265 + 85 = 350

[torn jolt]

Ik

Thats why im like

What

Why does 180 - (-85) = 265 then???

[tulip marlin]

Are you asking about elementary arithmetic?

You can verify what you said with a calculator.

[torn jolt]

Thats what i mean

My teacher weird af and idk whats happening

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[torn jolt]

No but

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[tropic tundra]

i was wondering how we can use squeeze theorem on b

[tulip marlin]

does the 0 function and 1/n work?

not much clue ngl

[tropic tundra]

lemme check desmos

looks good ty

[tropic tundra]

wait for n=1 it doesn't work

maybe 1/e^n

[tulip marlin]

???

if u only care about infinity it doesnt matter

wait im confused

why does it not work for n = 1, what

[tropic tundra]

1/2^1 = 1/2

wait it does work, bro fractions are so weird

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[cosmic ember]

just post a picture

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Please don't occupy multiple help channels.

[cosmic ember]

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[tardy ruin]

Hello

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Please don't occupy multiple help channels.

[tardy ruin]

<@&286206848099549185>

Can someone help me solve this correctly

If so would be much appreciated

Thx 🙏

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[tardy ruin]

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✅

[tardy ruin]

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[boreal maple]

Set them equal to each other and solve for x

The parabolic and linear function

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@torn jolt Has your question been resolved?

[fervent burrow]

I'm assuming you know how to find the inverse of a function? If so, take the inverse of the two functions in this piecewise separately. Then the interval would be the range of each of the first functions, since an inverse function means the x/y values have been switched.

Then put those in a piecewise function, just how the original function is defined.

And their domain will be defined by the range of the original functions.

Let me know if something is unclear.

That's the type of function they show there. It's just when its broken up into multiple pieces. You use a curly brace and then write all the functions that make it up.

Yes exactly. That is the inverse of the first part.

Nice

Yep, that's the second one.

Now the interval for the first one is: for 0 <= x <= p
Since you swap the x and y's to find the inverse, this means the new interval for the inverse is the y range of the first one, instead of the x domain.

Yes. Then you would write that as the interval for the first function.

Same notation as they have on your paper.

You can see by looking at the graph that the first one goes from point q to 11 on the y, right?

So the range would be q < y < 11

This becomes the interval for the inverse, so: for q < x < 11

Oh yeah, you already found the values of p and q in the last problem. So just use those.

Then do the same thing to find the interval of the second piece.

Write out the whole piecewise function, and you're good

Close, remember to look at the vertical values. It starts at -5 and goes up to point q

so from -5 to 2

Do you understand the notation? You weren't sure what a piecewise function was, so I just want to make sure you understand it now.

What don't you understand about it?

You understand the concept of functions, right? A piecewise is simply a function that is made of pieces of other functions. The notation is to write the functions that make it up on the left, and the interval on the right.

It is what your textbook has.

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[lofty shuttle]

Hi does anyone know how to solve this without using sine and cosine?

Found this question on Instagram