help-39

[tribal bloom]

i did alpha and beta

ok i will type my work and text book's answer

[steel rock]

OH MAN ITS SO EASY

Do you know the formula for making quadratic equations?

It is k(x^2 -(Sum of roots)x + (Product of roots)

[tribal bloom]

1/4 is alpha and -1 is the beta bec they are two are the zeroes of the polynomial alpha + beta = 1/4 +(-1) = 1/4-1=1/4-4/4 = -3/4 alphabeta= 1/4-1 = -1/4 = x2 -(-3/4x)+(-1/4)= x2 +3x-1

this is my work

but

textbook's answer is

[steel rock]

So add and multiply the two roots

[tribal bloom]

4x2-x-4

this is textbook's answer

i cna't understand which is correct

[steel rock]

wait let me find the answer for that

[tribal bloom]

ok

[timid ginkgo]

y=(4x-1)(x+1) is a polynomial with zeroes at x=1/4 and -1

[dire tapir]

4(-1)^2 - (-1) - 4 = 1 so that can't be right

[steel rock]

4x^2 + 3x - 1 is what i am getting

[tribal bloom]

so textbook and that website is wrong

same

[dire tapir]

yeah me to

[tribal bloom]

thats wht i also got

[timid ginkgo]

this is correct

[tribal bloom]

waitt

how u got 4x^2/

it is x^2?

[honest oyster]

Does the book show how to get that answer?

[tribal bloom]

there is no steps just the direct answer

[steel rock]

did you use the polynomial formula for forming poly out of roots?

[timid ginkgo]

If u expand (4x-1)(x+1)

[tribal bloom]

i think i didn't

can u show

[steel rock]

If you like used that you have take the LCM of it too so you will get 4x ^ 2

[honest oyster]

How'd you go fromx^2+(3/4)x -1/4 to x^2+3x-1 in the last line

[tribal bloom]

just common out the 4

[tribal bloom]

it is the fractions

[honest oyster]

You didnt factor out the one infront of x^2 tho?

[timid ginkgo]

there is no 4 on the x squared term though

[steel rock]

Why

[tribal bloom]

noo

[timid ginkgo]

in x^2 + 3/4x - 1/4

[tribal bloom]

hmm ye

[tribal bloom]

common out 4

[timid ginkgo]

Do u mean multiply by 4

[tribal bloom]

so it will become like x^2+3-1

[tribal bloom]

ye

[timid ginkgo]

Then you also need to multiply x^2 by 4 = 4x^2

[tribal bloom]

why?

[timid ginkgo]

Because otherwise it isn’t equivalent

[tribal bloom]

say the reason

[timid ginkgo]

Let’s say x+5=0
2x+10=0
Then x must be -5 and it solves both equations (because they are the same equation multiplied by 2)

[tribal bloom]

4(x^2+4/3-1/3)

[timid ginkgo]

For example

[tribal bloom]

now i understand

i didn;'t done x^2 multiply

i only did

-3/5

.

not -3/

3/4

-1/4

[timid ginkgo]

Yeah when u multiply u have to do all terms

[tribal bloom]

hmm yee

but

i have one doubt

if we mutliply by 4

how 4 got cancel out

ohh

i understood

bec lcm of that fraction is 4

so we need to multiply by 4 to cancel out 4

okk

thx everyone

[timid ginkgo]

Like 3/4 times 4 = 12/4 =3

[tribal bloom]

huh?

i didn't understand thaat tho

[timid ginkgo]

U can either cancel the 4, or multiply then simplify and the result is the same

[tribal bloom]

yee

now I UNDERSTAND

[timid ginkgo]

3x4/4 =3x1=3
And bc 3x4=12, 12/4=3 also

[steel rock]

Man I think we have to LCM it then you will get 4x ^2

[tribal bloom]

MATHS IS SO INTERSTING

thx

[timid ginkgo]

Real

[steel rock]

.

[steel rock]

Yep

[tribal bloom]

i already said that

[tribal bloom]

btw thx

[steel rock]

oh ok

[tribal bloom]

thx yall

.close

[pearl pond]

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[steel rock]

Are you a 10th grader?

[tribal bloom]

how did u know?

.reopen

[pearl pond]

✅ Original question: #help-39 message

[steel rock]

Because I just passed out from class 10th

cbse boards

And I got 96%

[tribal bloom]

omg

i also written cbse boards

actually i am doing second board

[steel rock]

Ohh same

[tribal bloom]

in may 15

i got low marks in maths

i need to fix that

so i am full focused on maths

[steel rock]

Yeah for outside delhi it was tough but in delhi it was very very very easy

[tribal bloom]

lol

i am in kerala

so thatswhy it was tuff

[steel rock]

Ohh how much you got if i may know that

in boards

[tribal bloom]

maaths?

[steel rock]

no

[narrow forge]

hi folks, might wanna take the conversation to a discussion channel?

[tribal bloom]

percentage?

[steel rock]

ok

[steel rock]

come in dm

[narrow forge]

!done please

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[tribal bloom]

.cloe

.close

[pearl pond]

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[pearl pond]

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[royal galleon]

Regarding 1. I tried his test case for 2 and I don't see how my answer is equivalent to what he has. What did I do wrong?

[royal galleon]

Ok so am just multiplying the denominator by u^2

[hazy pilot]

Alright first for the f(1/x) let t = 1/u and solve from there

you should get integral of 1 to x of ln(1/u) /(1/u + 1)) * -1/u^2 du

multiply out we get integral 1 to x of ln(u) / (u + u^2) du

now combinte integrals

we get (t+1) ln(t) / ((t+1) * t ) dt

cancel the t+1

we get integral of 1 to x of ln(t)/t dt

now substitute u = ln(t)

we get as final answer 1/2 ln(x)^2

[royal galleon]

I'm not following your combination. We have $\int_1^x \frac{lnu}{u(u+1)}+\frac{lnu}{u+1} du$

[jolly parrot]

BigBen

[hazy pilot]

Yes, for the right integrand, multiply top and bottom by u

[royal galleon]

Oh wait

[hazy pilot]

combine

[royal galleon]

Yes

[hazy pilot]

devide u + 1

you get ln(u) / u

[royal galleon]

Yes I see

[royal galleon]

If we make this sub we have ln (ln(t)) then

[hazy pilot]

No?

[royal galleon]

How? We have lnu. If we set u =lnt we just have ln(ln(t))

[hazy pilot]

Im sorry, you're working with u instead of t, you should do t = ln(u)

[hazy pilot]

I was using t

[royal galleon]

Ok so then we just have t^2/2 and then ln^2(u)/2 . Thank you

.solved

[pearl pond]

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[chrome mango]

Is this the correct formula for Champnowne's constant

[pearl pond]

@chrome mango Has your question been resolved?

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[supple sky]

help

[supple sky]

can someone explain the marking key here? im really confused

ty

<@&286206848099549185>

[worldly jewel]

which part are you having trouble with

because it's just plugging in the formula

[supple sky]

um tbh like most of it

wait omg

i get it...

.close

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[royal galleon]

Regarding 7c we have $f(x)= 2f(x)f'(x) \implies \frac{1}{2}= f'(x)$ if I just use an indefinite integral I have $1/4x^2+c$ or if I integrate from 0 to x we have $1/4x^2$. I don't see how I can obtain the -1

[jolly parrot]

BigBen

[royal galleon]

NVM I see we have the 1/2x + c integrate that from 0 to x and we see that c=-1

[rugged niche]

I found two solutions for c

[royal galleon]

Are you saying c can be 1 or -1 depending on what x is

[rugged niche]

This is not depending on x, there are two functions that solve that equation.

One with c=1 and one with c=-1

If you plug in f(x) = 1/2 x + c, the equation is: 1/4 x^2 + cx = 1/4 x^2 + cx + c^2 - 1 <=> c^2 = 1

[royal galleon]

What prompted you to have c^2-1

[rugged niche]

f^2(x) -1 = (1/2 x + c)^2 -1 = 1/4 x^2 + cx + c^2 - 1 on the right side of the equation

[royal galleon]

Ok I see. Thank you

.solved

[pearl pond]

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[acoustic tangle]

Im unsure about the answer for c

On one hand the minimum point occurs at 2 so I agree

But I thought the probability of any given value in a pdf = 0

[pearl pond]

@acoustic tangle Has your question been resolved?

[acoustic tangle]

<@&286206848099549185>

[hoary relic]

uh what

nvm

[acoustic tangle]

Do you know which it is?

[hoary relic]

on second thought i am not familiar with this topic

[pearl pond]

@acoustic tangle Has your question been resolved?

[pearl pond]

@acoustic tangle Has your question been resolved?

[dense jasper]

The probability of $X$ taking any exact value is $0$ for a continuous distribution. However, the probability distribution function $f$ represents the \textit{relative likelihood} of a variable falling \textit{near} a particular point. Since $f$ is minimized at $x=2$, its density increases as you move toward the boundaries ($x=0$ and $x=4$). Hence, $X$ is more likely to be found within a small interval centered far from $2$ than in an interval of the same width centered near $2$.

[jolly parrot]

Civil Service Pigeon

[acoustic tangle]

Thank you so much

[dense jasper]

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[acoustic tangle]

I have another statistics problem I’m a bit confused with

[dense jasper]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[acoustic tangle]

Ok😭

I did the highlighted bit

But only got 5/8 idk why it’s wrong

[dense jasper]

Doesn’t the expected mean differ for routes A and B?

So if you use a rank sum of 216 with a mean of 180, your z-score calculation is wrong

[acoustic tangle]

Why?

[acoustic tangle]

Oh

I don’t fully understand sorry

[dense jasper]

Recall that the expected mean of a rank sum depends on the number of items in that group

Specifically $E(R)=\frac{n(N+1)}{2}$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

Routes A and B have different numbers of journeys, so their “fair” average sums differ

So when you used the rank sum for route A but the mean for route B, you were comparing the data to the wrong baseline

Thus making your z-score calculation invalid

[pearl pond]

@acoustic tangle Has your question been resolved?

[pearl pond]

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[harsh folio]

I need help!

[harsh folio]

So its 6x8.9 divided by 3

Im having trouble with the 8.9 part

[light helm]

what specifically about the 8.9 is causing issues for you

[empty ibex]

Have you got to 2*8.9 through commutative property?

If so, multiply 8.9 by ten to get 2*89 and then divide by 10 after you've solved it.

[pearl pond]

@harsh folio Has your question been resolved?

[harsh folio]

No

[regal pendant]

have you learned about decimal numbers yet?

[pearl pond]

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[rustic gale]

I wanted to understand linear operator is one-one

T(w)=0 =>w=0

[rustic gale]

W=0 how does it probe it is one-one

[summer imp]

This specifically means that the kernel of your linear operator is just {0}.

If T(x)=T(y) what can you say about T(x)-T(y)?

[rustic gale]

T(x-y)=0

if we take any two different functions....their difference maps to zero function

How does we say it is one-one

[uncut nacelle]

not the same person and idk this well but I think it's ≈

that one

or ~

[pearl pond]

@rustic gale Has your question been resolved?

[worldly jewel]

You mean why (Tv = 0 ⟹ v = 0) implies T is 1-1

[pearl pond]

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[summer urchin]

need help on these 2 dont know where to atart

[bitter herald]

Context?

[summer urchin]

fuck mb ill ss the others

thats part of the question'

[bitter herald]

Okay

So do u know how to compose a function and how to invert it

[summer urchin]

i remember vaguely about swapping x for y then trying to make x the subject

ive just solved c idk about the other 1

i need help on B

[steep saddle]

is fg supposed to be f(g(1)) or (f ○ g)(1)

[vestal tapir]

that's the same

[summer urchin]

it just says fg(1)

[vestal tapir]

i think

[summer urchin]

on the question paper

[bitter herald]

Its the same thing im p sure

[bitter herald]

Yeah thats for inverting

So lets start with c since you already have some idea on how it works

[summer urchin]

ive finished C

[bitter herald]

Oh

Awesomesauce

[summer urchin]

i realised i was overthinking it

i need help on question B if possible

[bitter herald]

For b, try to evaluate g(1)

Can you tell me what that is

[summer urchin]

wouldnt that be 5-3x1 which is just 2

[bitter herald]

Youre right on the money

So now

You take that 2, and chug it in f

So try to evaluate f(2)

And that's it!

[summer urchin]

so 12/3 which is 4

[bitter herald]

Good job :D

[summer urchin]

oh yay thanks for the help lex i might be back with more questions later

[bitter herald]

Aighty just open up a new help channel when the time comes

For now you can just close this one I suppose

[summer urchin]

ok thank you

.close

[pearl pond]

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[pearl pond]

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[warped violet]

<@&268886789983436800>

[pearl pond]

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[warped violet]

[pearl pond]

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[sterile python]

How would I start this? I have no idea to do this...

[rough forge]

Can you check if v1 and v2 are linearly independent

[sterile python]

Using Gaussian elimination?

[rough forge]

I mean sort of, you can check if v1=kv2 is a multiple

which would lead to a system

but i think you don't need gauss

[sterile python]

Because I just start linear algebra introduction in complex vectors and forgot the concepts of linearly vectors

ah yes let me check

[rough forge]

You will get three equations of k and so you can solve one for k and try to see if the others satisfy it

[sterile python]

Right

They are different

[rough forge]

notice that k1, k2 and k3 are all k

you dont have different k's it's a scalar not vector

but yeah you noticed they aren't the same

[sterile python]

wait the k2 and k3 are -i

but yea still not same

[rough forge]

yes

now you need to find a 3rd one

the easiest would be cross product

[sterile python]

how would i do this?

[rough forge]

if you really feel motivated you can also use gram-schmidt

[rain arch]

Is no 6 wrong cuz y’s the sides a fraction

[sterile python]

Can I do another way by making a matrix and check for determinants?

[rough forge]

cross product is the easiest

[slim vigil]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sterile python]

using v3 = v1xv2

[rough forge]

wdym using

[sterile python]

right

using v1 x v2 to find v3

[rough forge]

yes

[sterile python]

I learned these from year 11 and now i forgot all of them, what a shame

[tropic saddle]

a general approach would be to take the matrix (v1, v2, e1, e2, e3) (or any other basis instead of (e1,e2,e3)) and row reduce that. then take pivot columns

of course three dimensions as usual are easier

[rough forge]

[tropic saddle]

or you could take the matrix A with the two rows conj(v1) and conj(v2) and solve Ax=0

[sterile python]

to see if there exists row of zeros?

[tropic saddle]

cant happen because the matrix has rank 3

because the last three columns are independent

[sterile python]

right, so if there are 4 or more we use matrix instead?

[tropic saddle]

basically

most of linear algebra consists of somehow reducing a question to solving a linear system

[rough forge]

which means linear algebra loves gauss

[sterile python]

-3 3i 0 should be it

Alrighty thanks guys!

.close

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[shadow elm]

is this valid

[warped violet]

Uh I am confused

You state that there is no max or min

But then chose max is 1 and min is 0

Or am I misunderstanding

[shadow elm]

i chose option A and the correct option according to it is D

[warped violet]

Oh

That wasn't your explanation

I am still confused though

Do you not understand the explanation?

Because, yes it is correct

[shadow elm]

doesnt a maximum exist on 1-something

ok nvm

we only take it

if 1 is included

.close

[pearl pond]

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[warped violet]

If 1 was included in the interval, it would be the max

But since it's not

You can't find the max

Since if you say 0.99

What about 0.995

Or 0.99998

Right

[shadow elm]

yeah right

[warped violet]

Always something closer to 1

[pearl pond]

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[north talon]

Let $ABC$ be a scalene triangle, let $H$ be it's orthocenter and let $D$ be the foot of the altitude drawn from $A$. Let $S,T$ be points on the circumcircle of $ABC$ such that $BSH$ and $CTH$ are right angles. $\$
Prove that if $AH = 2HD$, then $D,S,T$ are colinear.

[jolly parrot]

Copter

[north talon]

Ig, very clearly S is the intersection of (BDH) and (ABC) and similarly for T

but i dont know what to do with the length restriction because it doesnt seem relavent at all

[pearl pond]

@north talon Has your question been resolved?

[north talon]

<@&286206848099549185>

[slim vigil]

Omg finally someone asks about geo, i've been waiting for this

[waxen wave]

Same, lol.

[north talon]

i do have the solution key but i do not wanna look at it and it seems messy

[slim vigil]

Just read the question, haven't solved this but i feel like ||Euler line being parallel to BC|| might be important

[north talon]

whys that true -w-

[north talon]

ive reflected H across BC which seems promising

[slim vigil]

Think about the centroid

[north talon]

oh, same ratio?

[slim vigil]

Do you know inversion?

[north talon]

i know what it is but idk how to use it

[slim vigil]

Well it would solve this problem

[north talon]

.-.

[slim vigil]

Let M, N be the midpoints of AB, AC. Can you prove that AMHN is cyclic?

[slim vigil]

The sol with inversion is so nice, i think it's worth learning inversion just for this

I am trying to find a sol without inversion for you though

[north talon]

im thinking of smth interesting

Let X be the midpoint of BH, and Y be the midpoint of HC

both are the center of the circumcircle

[slim vigil]

Yes

[north talon]

BS is a shared chord so OX is the perpendicular bisector of BS

hence OX parallel to HS

[slim vigil]

Correct

[north talon]

it suffices to show <BDS = <TDC

[slim vigil]

Very good

And that is true because...

[north talon]

by angle chasing <BDC = <BHS = <OXH, and <TDC = OYH

so we need to show <OXH = <OYH <=> O,X,Y,H are cyclic?

[slim vigil]

Yeah that's fine too

[slim vigil]

I had another way right here. Note that M,H,T are colinear and so are S, H, N.

But yours work too

Finish the angle chase

[north talon]

hmmmm

[slim vigil]

Use the pairs of parallel lines you found

Btw I hope you will see the inversion sol afterwards, it's so beautiful

[north talon]

why cant i get anything omg

OH//XY//BC

oh nvm i got it

[slim vigil]

Very nice

You know about Euler circle and directed lengths right?

[slim vigil]

Inversion at H with power HA.HD (directed lengths) turn the problem into this, which is very simple angle chasing

[north talon]

i dont really get directed lengths

[slim vigil]

Like the segments have direction

[north talon]

hmmm

i thought of a sol with homothety instead lmao

[slim vigil]

For example you probably know that HA.HD=HB.HE
But if I ask you to construct a point X on HB such that HB.HX=HA.HD there would be 2 options

Which are G and E here

But if i say HB.HX=HA.HD (directed lengths) then E would be the only option

[north talon]

reflect H across BC so AH' is twice AH, well known that ABH'C cyclic

[slim vigil]

Because HA and HD have opposite directions, so must be HB and HX, which means in that case X ≡ E

[slim vigil]

Yeah that's how you do it

HN // H'C

[north talon]

what would the inversion look like

i actually have learnt it a bit and i didnt get any of it

[slim vigil]

You get this?

So the inversion at H with power HA.HD (directed lengths) swap B and E

[north talon]

yo what the freak

[north talon]

i do not get this 😭

[slim vigil]

The image of B through the inversion at H power HA.HD (directed lengths, i will stop specifying from now on) is the point X on HB so that HB.HX=HA.HD

[split void]

um do you know what inversion is

[north talon]

i do, i just suck at it

[slim vigil]

Directed lengths is needed because normal segment lengths would give 2 solutions, E and G (in this pic), that's what directed lengths are for

So that inversion swaps A and D (obviously), swaps B and E, C and F

So (ABC) becomes (DEF), aka the Euler circle

(BFD) becomes the line AEC

[north talon]

oh shit

[slim vigil]

S is on both (ABC) and (BFD) so its image is the intersection of Euler circle and AEC, which is just N, the midpoint of AC

Similarly T swaps with M

[north talon]

i see i see

[slim vigil]

S, D, T being collinear is equivalent to their images (N, A, M correspondingly) being on a circle with the center of inversion (H in this case)

So inversion turns the whole problem into proving AMHN is cyclic, which is just angle chasing (as you figured it out)

[north talon]

T-T

[slim vigil]

Don't worry, i've been through this feeling

Take your time to understand it, please tell me if some parts were confusing

[north talon]

contest is in a month, can he lock in in time ⁉️

[north talon]

i will look into inversion, thanks

[slim vigil]

Out of curiosity, what contest are you taking? Also good luck on that

[north talon]

im one of the respresentatives to take the national olympiad twt

[slim vigil]

So it's the national MO?

[north talon]

thai national olym

yeah

[slim vigil]

That's quite early, mine doesn't start until late December

Best of luck to you

We should be friends :3

[north talon]

they need the national oly to be done before qualifiers which is at the end of august lol

sure!

[slim vigil]

Type .close once you're done reading btw ^^

[north talon]

ok

[north talon]

you were able to spot this cause well known, im guessing?

[slim vigil]

Yeah you develop an instinct for it. This was the first thing i tried when i saw the question

[north talon]

holy aura

.close

[pearl pond]

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[pearl pond]

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[hearty flume]

I need help verifying my answer for this question: Prove that $\lim_{x\to a} \sqrt x = \sqrt a$ if a > 0 and for x and a in $\mathbb{R}$

[jolly parrot]

Toaster

[smoky gull]

Do you know what does limit mean

[hearty flume]

yeah

Ok, so here's my answer:

$\lim_{x\to a} \sqrt x = \sqrt a $ then there's $0<|x-a|<\delta \Rightarrow |\sqrt(x)-\sqrt(a)|<\epsilon$ \ Look at $|\sqrt x -\sqrt a |$. $|\sqrt x -\sqrt a | = \frac{|x-a|}{\sqrt x +\sqrt a }$ \ If $|x-a|<1 \Rightarrow -1<x-a<1 \Rightarrow a-1<x<1+a \Rightarrow \sqrt {a-1} <\sqrt x <\sqrt {1+a} \Rightarrow \sqrt {a-1} +\sqrt {a} <\sqrt {x} +\sqrt {a} <\sqrt {a+1} +\sqrt a $ \ Look at $\frac{|x-a|}{\sqrt x+\sqrt{a}} = \frac{1}{\sqrt x + \sqrt a}|x-a|<\epsilon$ \ We can find a C>0 s.t $|x-a|<C \ \Rightarrow \frac{1}{\sqrt x + \sqrt a}|x-a|<C|x-a| \text{ and we are assuming } x and a \in \mathbb{R}$ so $x>0$ and $a>0$ (otherwise x and a would be undefined). \ So we can have $C = \frac{1}{\sqrt{a-1}+\sqrt(a)}$ (since $\frac{1}{\sqrt x + \sqrt a} < \frac{1}{\sqrt{a-1} + \sqrt a}$) so let $\delta = \min{1, \epsilon(\sqrt{a-1}+\sqrt a )}$ then since $\frac{1}{\sqrt x + \sqrt a}< \frac{1}{\sqrt{a-1}+\sqrt{a}}$ and $|x-a|<\epsilon(\sqrt{a-1}+ \sqrt a)$ then $\frac{|x-a|}{\sqrt x + \sqrt a} < \frac{1}{\sqrt{a-1}+\sqrt{a}}\epsilon(\sqrt{a-1}+ \sqrt a) = \epsilon $

[jolly parrot]

Toaster

[dense jasper]

what if a is less than 1

[autumn fossil]

and btw you dont need to include the whole process of finding delta in your proof. You can just let delta = [whatever you find] and then prove that this delta works

[hearty flume]

Oh ok

[hearty flume]

Yeah, that's why mine is just a partial answer

[honest oyster]

Shouldn't it be for x>0?

[hearty flume]

Oh yeah, that too

[hearty flume]

But it wouldn't be in R if it was < 0

[honest oyster]

You said x in R, never said anything about sqrt(x)

[wet osprey]

you can't say x in R because x is not a variable here lol

[hearty flume]

mb

[autumn fossil]

there is a much easier lower bound on sqrt(x) + sqrt(a) btw

[wet osprey]

$\lim_{x\to a}\sqrt{x}$ is either a number in $\bR$ or it doesn't exist

[jolly parrot]

frosst

[wet osprey]

there is no x

(for a in R, assuming we dont have complex number shenanigans)

[hearty flume]

what

[wet osprey]

it's like saying $\int_{0}^1f(x),dx$ for some $x\in \bR$

[jolly parrot]

frosst

[wet osprey]

it similary doesn't make sense because the x is not an actual variable it's a dummy variable

[hearty flume]

x is just there to denote any real number > 0 right?

[wet osprey]

well in your question you dont really need to say where the x is from

you just need a in R

[hearty flume]

Wait what's the easier lower bound?

[autumn fossil]

how would you put a lower bound on sqrt(x) + sqrt(3)

fisrt thing that comes to your mind

[hearty flume]

my first thought was x + 3

[autumn fossil]

you need sth that doesnt depend on x

ideally

[hearty flume]

Oh wait

I'm dumb

its 0 right?

[autumn fossil]

0 actually doesnt work

[wet osprey]

okay but you can't put 0 in the denominator >.>

[autumn fossil]

and its not like sqrt(x) + sqrt(3) can ever be 0, thats quite bad lower bound

you can do a bit better

[wet osprey]

what's the next best thing that you could actually put in the denominator

[hearty flume]

sqrt(3)

[autumn fossil]

and in general?

[hearty flume]

my brain is not braining today sorry

[wet osprey]

sqrt(x) + sqrt(a)

lower bound for this

[hearty flume]

sqrt(a)

[wet osprey]

yeah

[hearty flume]

Wait but if you take the reciprocal wouldn't the inequality sign flip?

[autumn fossil]

Yeah

you need an upper bound of the fraction and hence a lower bound of the denominator

[hearty flume]

oh wait yeah, so $\delta = \min{(1, \epsilon(\sqrt a))}$?

[jolly parrot]

Toaster

[autumn fossil]

well, you dont need the 1 anymore

[hearty flume]

oh

[autumn fossil]

you can just pick delta = epsilon sqrt(a)

[hearty flume]

ohhhhhhhh

So i'm done here?

[autumn fossil]

yep, once you verify that it works, you're done

you just need to check that whenever |x-a| < delta, |sqrt(x) - sqrt(a)| < eps

[hearty flume]

since $\frac{|x-a|}{\sqrt x + \sqrt a} < \epsilon$ and since $|\sqrt x - \sqrt a| = \frac{|x-a|}{\sqrt x + \sqrt a}$ we good here 🙂

[jolly parrot]

Toaster

[hearty flume]

You know, I hate it when the emoticons turn into emojis

$\text{:)}$

[jolly parrot]

Toaster

[honest oyster]

(:

[hearty flume]

alright

ty

.close

[pearl pond]

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[brisk steeple]

How is the empty set {} contained in every set?

[boreal raptor]

the empty set is literally nothingness

everything has nothingness

wait

[stoic imp]

the empty set {} is a subset of every set, that means that every element in the empty set is contained in every set, since there is no elements

is vacously true

[boreal raptor]

ah

[brisk steeple]

So, does the set A = {1, 2} contain the set {}? So A ∈ {}?

[stoic imp]

no

[brisk steeple]

My teacher said: NO!!!!!!!?????!?!?? (jk)

[brisk steeple]

You just contradicted what you said earlier.

[stoic imp]

the empty set clearly is not an element of this set, however it is a subset of this set

[twin sigil]

stop the cap @stoic imp

[stoic imp]

I am not joking

[rough stream]

Let's use careful terminology here.
X is a subset of Y, denoted X ⊆ Y if every element in X is also in Y.

No matter what Y is, {} is a subset of Y, because every element of {} can be found in Y.

[stoic imp]

since the empty set doesnt contain any elements, I can say whatever I want about his elements

[rough stream]

You might say it's weird to make a statement about "every element of {}". There are none!

[brisk steeple]

Are the elements the entities stated explicitly while writing a set using the Roster method?

[rough stream]

A = {1,2} is defining a set called "A", and that set contains elements "1" and "2"

[maiden pendant]

I'm sorry to interrupt the debate about subsets here with my random question 😭 , perchance are you from Spain or a spanish speaking country? (edit: okay, gotcha, I'm from Spain too, that's why I was asking)

[rough stream]

{} is an empty set

Now, to be very clear, {} is a "subset" of Y.
{} is not an "element" of Y.

[brisk steeple]

Ok. I understand the element thing.
The fact that the empty set is a subset of every set seems semi-true because it's just not incorrect.

[rough stream]

It's vacuously true.
In math, something is true if a counter-example cannot be provided. This is because of the truth table of the implication

[brisk steeple]

Is a vacuously true statement also just true?

[summer imp]

Yes. A statement like $\forall x \in A, P(x)$ is true "by default" (vacuously) if $A$ is empty.

[jolly parrot]

Azyrashacorki

[brisk steeple]

ok. So something is either true or false

got it

thanks everyone

.close

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[radiant jay]

Hello I've been trying to figure out this problem in Karlheinz Spindler's Algebra and I'm not even sure if the hypotheses lead to the conclusion at this point. How do I prove this?

[radiant jay]

Let $R$ be a ring with identity element 1. Show that all elements of $R$ which are no [sic] zero-divisors have the same order in the abelian group $(R, +)$.

[jolly parrot]

SherlockSage

[autumn fossil]

Here is a hint:
Prove that the elements have same order as 1

[pearl pond]

@radiant jay Has your question been resolved?

[pearl pond]

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[royal galleon]

I used the fact the $e^c = e^x + \sqrt{e^{2x}+1}$ and $e^{-c} = \frac{1}{e^x + \sqrt{e^{2x}+1}}. To find $\frac{3}{2} = \frac{e^x+\sqrt{e^{2x}+1}}{e^x+\sqrt{e^{2x}+1}}$

[royal galleon]

one sec

[rough forge]

You just forgot to close

[royal galleon]

ye so I am unsure how to proceed. I looked at the key and he has this. I don't see how he obtains the blue part

[jolly parrot]

I used the fact the $e^c = e^x + \sqrt{e^{2x}+1}$ and $e^{-c} = \frac{1}{e^x + \sqrt{e^{2x}+1}}$. To find $$\frac{3}{2} = \frac{e^x+\sqrt{e^{2x}+1}}{e^x+\sqrt{e^{2x}+1}}$$

[rough forge]

It seems like they merged the fraction into one

[royal galleon]

wait I should have a +1 in my fraction

[royal galleon]

what about our simplification?

[rough forge]

Yeah I am thinking how that results to e^x

[rough forge]

It's just factoring yeah, I thought I was naive

If you collect the terms in the numerator

And then it's just ln(3/4)=ln(3)-ln(4)=ln(3)-ln(2²)=ln(3)-2ln(2)

[royal galleon]

yes I see it now. thank you

.solved

[pearl pond]

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[plain ember]

how do you approach something like this?

[hazy pilot]

Do you need to do this by hand?

[plain ember]

by hand as in? i need to approximate the right expression based on the left expression

[hazy pilot]

No I meant actually using pen and paper or by the use of a computer

[plain ember]

pen and paper

[wooden merlin]

A way is to graph the function and try to use trigonometry to estimate the area under it.

[plain ember]

using trigonometry how?

[pearl pond]

@plain ember Has your question been resolved?

[viscid shale]

Take a few values of you function, construct this sort of diagram and you can relatively easily approximate the area up to some arbitrary precision

You can see you get a combination of rectangles and triangles

On a slight note, this is a well known algorithm for approximating integrals.

[pearl pond]

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[bitter herald]

Translate question

[jolly parrot]

haseeb ♥

[ashen ivy]

how far have u gotten

[stoic imp]

im unsure how to proceed

also this is forall d in N

suppose d | p then d in div+(p)

[ashen ivy]

last you left off, you wanted to prove $$\big( (d < p \wedge d\mid p) \implies d = 1 \big) \implies p \quad\text{ prime}$$

[jolly parrot]

haseeb ♥

[ashen ivy]

realized i said the wrong thing

hopefully you noticed

[stoic imp]

what

this looks correct to me

[pearl pond]

@stoic imp Has your question been resolved?

[ashen ivy]

well that's a problem
one is a => b
one is (a => c) => b

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[woven falcon]

I wrote down the triangle inequality , then tried tightening the bound by using AM GM inequality but ended up back on the triangle inequality

Can somebody give me a hint about the right approach here??

[autumn fossil]

Suppose its not true. Then a > b > c and consequently a^n > b^n > c^n. Prove that triangle inequality will be violated for large n

[pearl pond]

@woven falcon Has your question been resolved?

[pearl pond]

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[woven falcon]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[woven falcon]

How would I go about doing so??

[autumn fossil]

can you just state which triangle inequality you think would get violated?

there are 3, so which one of them is most likely to fail?

[woven falcon]

a power n - c power n is less than or equal to b power n?????

[autumn fossil]

yeah

or in other words
a^n <= b^n + c^n

this one will probably get violated quite quickly, since a^n will just grow faster

if you know limits, you could use them

[woven falcon]

Hmm but then how do we proceed to show that the triangles are isosceles?

[autumn fossil]

well, triangle inequality getting violated would be a contracdition

hence the assumption a > b > c would be wrong

and so at least one of the equalities would have to hold

meaning it would be isosceles

[woven falcon]

So how can we show it gets violated??

I’m sorry, im a little behind on writing proofs

[autumn fossil]

are you familiar with limits?

if not, it can be done without them. Limits just make it feel a bit more natural ig

you need to show that for large n, a^n > b^n + c^n

[pearl pond]

@woven falcon Has your question been resolved?

[woven falcon]

Yes

[autumn fossil]

divide both sides by b^n and take a limit then

[woven falcon]

Ohh lemme try that

Ok im still not getting it

[fleet ridge]

@autumn fossil 's method is 100% correct

But, you dont need to prove a^n > b^n + c^n

From the triangle side formula, we get $$a^n + b^n > c^n$$

[jolly parrot]

-TimeLord-

[fleet ridge]

Rearanging, we get:
$$a^n + b^n - c^n > 0$$

[jolly parrot]

-TimeLord-

-TimeLord-

[fleet ridge]

(I dont know how to make an actual fraction)

But anyways

[fleet ridge]

Ignore this please

Wait never mind. I somehow proves that the triangles are completely valid

Anyways time to switch it into how the triangles are valid

$$1 + b^n/a^n - c^n/a^n > 0$$

[jolly parrot]

-TimeLord-

[fleet ridge]

As @autumn fossil said, we can take the limit as n approaches infinity on the left side. If it is greater than 0, then its good

$$lim n-> infinity (1 + b^n/a^n - c^n/a^n)$$

[jolly parrot]

-TimeLord-

[fleet ridge]

$$lim of n-> infinity (1 + b^n/a^n - c^n/a^n)$$

[jolly parrot]

-TimeLord-

[fleet ridge]

idk how to write it

solving, we get the answer as 1

so it seems the triangles are valid

[woven falcon]

It’s fine, got it , thanksss

.close

[pearl pond]

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[fleet ridge]

ok

whats the proof? I woudl liek to know

[tulip ore]

to write limits with latex, you type \lim_{n\to\infty}

[fleet ridge]

ohh

[tulip ore]

also, its faster to use discord bot if you just start the message with ,, instead of covering with $$s

[fleet ridge]

Do you know the proof of the question tho?

[fleet ridge]

Ok

[fleet ridge]

I would really want to knwo

[autumn fossil]

This was my way: Assume a>b>c, then a^n > 2b^n > b^n + c^n for large n, namely n > log(2) / log(a/b)

And so for large n triangle inequality fails - contradiction

Hence one of the > must actuqlly be >= resulting in isosceles tri

[fleet ridge]

oh

what about for fractions?

[autumn fossil]

Alternative through the limits, you could do
a^n > b^n + c^n
Iff
(a/b)^n > 1 + (c/b)^n
Take lim, you get infty on LHS and 1 on RHS. So for large n it must hold

[autumn fossil]

Wdym

[fleet ridge]

if a=1/2 and b=1/4

[autumn fossil]

Thats still okay

[fleet ridge]

log(a/b)=log(2) right?

so n>1

[autumn fossil]

Indeed.

[fleet ridge]

but i dont see a contradiction

[autumn fossil]

(1/2)^2 > (1/4)^2 + (1/4)^2

Triangle inequality doesnt hold

[fleet ridge]

oh. My bad

Thanks

[pearl pond]

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[austere wraith]

Anyone help with this

[warped violet]

!status

[pearl pond]

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

[pearl pond]

@austere wraith Has your question been resolved?

[pearl pond]

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[empty hemlock]

so say i have 2 3d vector v and w
i want to collapse vector w into v in sense that, if we put a plane that is perpendicular to v at the end of w,
our new vector is at the intersection of that plane and v
how do i calculate this ?

[empty hemlock]

after watching this 3b1b video, i feel like you can do it by taking the dot product of w and normalized v ? and then multiply it by normalized v ?
but im not quite sure, and idk how to verify it

[stuck hatch]

can you draw an image of what you want to achieve?

[empty hemlock]

holup

[summer imp]

Yes, that is the case and aligns with the usual definition of a projection of a vector on another.
The dot product of w with normalized v gives you the component of w in the direction of v. Multiplying that by normalized v gives you specifically the vector for that component.

[summer imp]

https://www.desmos.com/3d/6po3b6wh5s

[summer imp]

If $\vb{v}$ and $\vb{w}$ are linearly independent, then they lie in a common plane. This plane has a basis of orthogonal unit vectors, $\vb{v}{||}$ and $\vb{v}{\perp}$, where $\vb{v}{||} = \hat{\vb{v}}$ and so you get a basis of $\R^3$ : ${\vb{v}{||}, \vb{v}{\perp}, \vb{u}}$. You can take $\vb{u} = \vb{v}{||} \times \vb{v}_{\perp}$, but we don't really care about $\vb{u}$.\

Then this means that $\vb{w} = a\vb{v}{\} + b\vb{v}{\perp} + c\vb{u}$ for some $a,b,c \in \R$. $\vb{w} \cdot \vb{v}{||} = a$, which is the component of $\vb{w}$ in the direction of $\vb{v}{||}$, and therefore in the direction of $\vb{v}$.\
$a\vb{v}_{||}$ is specifically the vector pointing in this direction.\

You can check that $a\vb{v}_{||}$ lies in the plane $\vb{v}\cdot (\vb{x} - \vb{w}) = 0$.

[jolly parrot]

Azyrashacorki

[pearl pond]

@empty hemlock Has your question been resolved?

[summer imp]

The point is that :

• the collapsing you're talking about is really just projection in R^3
• it corresponds to computing the component of w which goes solely in the same direction as v
• you can check that the vector in the direction of v with this component points exactly to the intersection of v and the plane normal to v going through w

[empty hemlock]

is this formula correct ?

wait dotP is wrong, i already move out the normalizing scalar

i think im confusing mysekf from not marking normalized vector

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[empty hemlock]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[summer imp]

The formula is $\text{proj}{\vb{v}}(\vb{w}) = \frac{\vb{w} \cdot \vb{v}}{||\vb{v}||} \frac{\vb{v}}{||\vb{v}||}$.\
In coordinates that means $$\text{proj}{\vb{v}}(\vb{w}) = \frac{v_x w_x + v_y w_y + v_z w_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \cdot \frac{1}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \begin{bmatrix} v_x\ v_y \ v_z\end{bmatrix}=\frac{v_x w_x + v_y w_y + v_z w_z}{v_x^2 + v_y^2 + v_z^2} \begin{bmatrix} v_x\ v_y \ v_z\end{bmatrix}$$.

[jolly parrot]

Azyrashacorki

[empty hemlock]

i got class rn, ill try it later

ty so much

[pearl pond]

@empty hemlock Has your question been resolved?

Channel closed

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[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[proper nova]

you might want to open a new channel

you deleted your original message

[molten kelp]

Yes

[crystal dew]

!msgdel

[pearl pond]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[crystal dew]

please move to another channel.

[molten kelp]

Why..?

[crystal dew]

and please don't delete the initial message when you claim a channel in the future, as a gentle reminder.

[crystal dew]

you deleted the initial message of this channel, thus forcibly closing the channel.

this action is not reversible and the channel cannot be reopened.

[molten kelp]

Oh I got legged

Got it

[hoary relic]

uhh if you are done please type .close

[crystal dew]

it's forcibly closed.

[molten kelp]

.close

[crystal dew]

just leave it be.

[molten kelp]

Sorry guys. Got legged and deleted it by accident

[crystal dew]

nw.

[pearl pond]

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[stoic imp]

ok please can I get some help with this exercise

I kind of understand but at the same time I dont

the idea is for example, if a) is a definition of primality, then
a) holds for p <=> p is prime

so what I have been doing so far is that, I pick random values for p (both prime numbers and composite numbers) and try to see if a) holds when p is prime and if it doesnt hold if p is composite

after this, I get a general idea of whether a) is a possible definition or not (if for example I try many p primes that make a) be true aswell as if I try many composite p that make a) be false)

after this, I get a general idea of whether a) is a definition of primality, suppose every p i try if it is prime it makes a) be true and if p is composite then a) be false, after all this p I try then I would like to prove this a) is a definition of prime number, so I would like to prove
a) <=> p is prime

the left direction <= is simple, suppose p is prime then d | p means d is 1 or d is p, but since d < p then d = 1, so if p is prime a) holds

what I am struggling with is the right direction =>
that is,

a) holds for p => p is prime

.close

[pearl pond]

Channel closed

Closed by @stoic imp

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[pearl pond]

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[winged pulsar]

Stuff like (pi + v) = sin(0) is… called?

[winged pulsar]

So that I can search it on YouTube

[random ermine]

equation

[winged pulsar]

For lessons

Bruh…

Tf that’s not a correct answer

[random ermine]

yes it is

is ur pfp ai?

[dire tapir]

i mean that's what it is

unless you mean trig functions

[winged pulsar]

Troll

[winged pulsar]

I need EXACT thing so i can understand this

[random ermine]

(pi + v) = sin(0) is an equation

maybe you want trigonometric equations

[fresh solar]

Give some more context lol like which chapter, etc

[random ermine]

v = -pi

[winged pulsar]

Nah formula

[random ermine]

no equation

[modest tartan]

...

transformation formula

i think

[fresh solar]

Trigonometric identities

[modest tartan]

put the trigo function before the 1st bracket bruv

my dumb ahh thinking what da fuc is this

[winged pulsar]

Don’t call me dumb

[dire tapir]

complementary angles

[fresh solar]

He didn't call you dumb....

[flint basalt]

these could be called co-function identities, or trig identities

[modest tartan]

i said my dumb ahh u dumbass

now i am calling u dumb

[winged pulsar]

You’re so disrespectful

[modest tartan]

thank you ,dumb

[winged pulsar]

How can these things result into cosinus?

[modest tartan]

the spelling 😭

[random ermine]

(co)sinus

[flint basalt]

if you graph the functions on a graphing calculator you will see they are equal

cosine and sine are the same function, just shifted over a bit.

[winged pulsar]

Nah one is opposite and other adjacent

[flint basalt]

...?

that has nothing to do with this

[random ermine]

if you wanna search these on youtube just type these equations and search

[winged pulsar]

Sin is 0 - 90°
Cos
Tg???

Answer @flint basalt

[flint basalt]

answer what?

[winged pulsar]

What does cos range to

[random ermine]

@winged pulsar what is sin(0)?

[flint basalt]

the range of both sin and cos is -1 to 1

[winged pulsar]

0 is theta

[winged pulsar]

But tg?

[dire tapir]

whats tg

[flint basalt]

what do you mean "but tg"

[winged pulsar]

Tangans

[random ermine]

tan, i'm assuming

[fresh solar]

Tan is called tg I think

[random ermine]

like tg for tangent

[winged pulsar]

Yes tg is called tangens

[flint basalt]

tangent

the range of tan(x) is -inf to +inf

[dire tapir]

the range of tan is all real numbers

[winged pulsar]

In mathematics, "tg" is a notation for the trigonometric tangent function, often used in Eastern European, German, and Russian contexts, or in older mathematical literature. It represents the same function as
tan
t
a
n
. The formula is defined as the ratio of the sine to the cosine, or the opposite side to the adjacent side in a right triangle:
t
g

(

𝜃

)

=
s
i
n
(
𝜃
)
c
o
s
(
𝜃
)

=
o

[flint basalt]

holy ai

[random ermine]

even ai is better than this

[winged pulsar]

Literally tangent

[ebon skiff]

Fr

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[random ermine]

yep

[flint basalt]

i hope you know we cant read this. format it better or take a screenshot of what you are trying to say.

[winged pulsar]

I was searching up the definition.

[winged pulsar]

Not my problem that discord did that

[flint basalt]

do you have an actual question to ask?

[fresh solar]

We get that tangents is called tg

[flint basalt]

sure. whats your question...

[random ermine]

i feel like we answered his original question

[winged pulsar]

Y’all should know better lowkey…

[fresh solar]

Alright, if you are done you can close this then

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[winged pulsar]

Ong don’t test me

[dire tapir]

we're just asking you to close

if you're done

[crystal dew]

well, I've had enough

[winged pulsar]

!close

[crystal dew]

<@&268886789983436800> non-cooperative OP

[flint basalt]

its .close

[dire tapir]

.close

[winged pulsar]

WHAT?

[crystal dew]

on top of stuff like this

[winged pulsar]

.close

[pearl pond]

Channel closed

Closed by @winged pulsar

Use .reopen if this was a mistake.

[winged pulsar]

But it’s true… y’all in college and you should know different namings…

Just like people use • instead of x

[dire tapir]

I never heard of tan being tg

[crystal dew]

doesn't give you the pass to go ham on the helpers.

[flint basalt]

we don't all live in europe.

[dire tapir]

I'm in europe

[winged pulsar]

I’m always now gonna use tg

I’ve been told it’s tg and it’s tg for me

[flint basalt]

and also not all 30k people in this server have gone to the same school as you have

you can use what you want to, but if someone asks you to clarify then you are meant to clarify.

[crystal dew]

the problem isn't with you using tg. the problem is that you are insinuating less of helpers who don't share your notation.

[fresh solar]

OP is probably ragebaiting

[crystal dew]

either way, I'll let the mods look into this

[winged pulsar]

I just simply said what tg is and you all started to bring bitchy about it

[winged pulsar]

I was just saying what is it

[random ermine]

who is bitchy

[crystal dew]

you may want to look at your own attitude as well.

but either way, the mods have been called so

if you want to provide more stuff for the mods to look at by all means

I'm disengaging

[winged pulsar]

There’s no need I’m not breaking any rules

[unborn abyss]

.this seems like a language issue and a... temperament mismatch. maybe next time you'll be matched with helpers that mirror your learning style better.

[pearl pond]

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[quick venture]

is this correct so far

[rough forge]

You are off by a sign

An antiderivative of sin(x) is -cos(x)

[quick venture]

u sure

because ihavethis

[autumn trellis]

your first sign error got cancelled out by yet another sign error giving you the correct answer in the end

and yes the antiderivative of sin is -cos

[rough forge]

Do you want a proof by first principles

[pearl pond]

@quick venture Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

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[pearl pond]

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[autumn trellis]

you can try multiplying the mclaurin series. But i suggest you look a bit more for a pattern. its definitely there. make sure you are evaluating the derivative at 0 at every step. you will find a pattern

[rustic atlas]

use complex numbers

[plush bramble]

the hint was to use complex numbers

[pearl pond]

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Closed by @cosmic fox

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[pearl pond]

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• Be polite and have a nice day!

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[gentle perch]

Hello, may I have some help on my test review? I’m just out of it today and I don’t know a lot about this subject due to personal matters. I’d really appreciate some help.

It’s four pages but we can see how just this first page goes.

[gentle perch]

Here’s the helping paper we can use

Asked for help earlier but got mixed up in something and had to leave

[vestal pier]

is this conversion rates?

[vestal pier]

hi do you want help for all the questions?

or are there any specifics

[gentle perch]

So far, yes. If you have the time. Like I said, I’m not too good at this haha.

[vestal pier]

mind giving me something you want to tackle first

[vestal pier]

anything specific you want to go through first?

[gentle perch]

I guess we can start with d if I have a b c correct?

[vestal pier]

ah conversion of 200km to miles?

[vestal pier]

is it that one?

[gentle perch]

Yep that’s d

[vestal pier]

so if 1 mile is equal to 1.6km

what would you do if you have 200km

[gentle perch]

divide 200 and 1.6?

[vestal pier]

by

you would divide by

1.6

yes

[gentle perch]

yes

[vestal pier]

and what would that get you?

think of 1.6 as a fraction

[gentle perch]

125

[vestal pier]

which is 8/5

yeah

but if 200 is 8/5

you can divide 200 by 8

which is 25

and multiply by 5

for 125

[gentle perch]

i see i see

[vestal pier]

thats if you dont have a calculator

[gentle perch]

we can use calculators but theyre more primitive

ie cant do fractions on em

[vestal pier]

oh

ok moving onto e then

[gentle perch]

theyre like these ones

[vestal pier]

24 degrees c to farenheit

one sec

c equals 5/9 (f-32)

oh wait we are finding f

5/9c +32

is your equation

what would you do in your head

[gentle perch]

and c is 24 right? so am i doing 5/9 x 24 + 32?

[vestal pier]

yea

so you substitute in 24

substitute is a fancy word for replace

[gentle perch]

i got 45.many 3's

[vestal pier]

so you replace certain phrases for numbers

[light helm]

wrong way

[restive solstice]

The conversion for C to F is f = (9/5) . c + 32

[vestal pier]

lile c could be substituted for 30

did he do it wrong

[light helm]

9/5 *
not 5/9 *

[vestal pier]

i was doing it on my phone before