help-49

[wary thorn]

You haven’t seen the epsilon delta definition of continuity yet?

[plain phoenix]

that I did see

[runic hamlet]

what does the notation U_ei(xi) mean

[wary thorn]

In R you can mostly think of open balls as open intervals

There’s some caveats but that’s about it

[plain phoenix]

alright!

[runic hamlet]

you should probably first stick with convergence of sequences before you worry about continuity of functions

[plain phoenix]

the vicinity (I'm not so sure of the english word for it) of xi with the radius ei, I think

[runic hamlet]

ok thats words. how about some symbols

set notation

[plain phoenix]

What are you asking?

To rewrite U_ei(xi)?

[runic hamlet]

to give the definition of U_ei(xi)

U_ei(xi) = {...}

fill in the dots

[plain phoenix]

Maybe U_ei(xi) = {x_1, x_2, x_3...}. If I got it somewhat right, the set should consist of all the elements past a certain point that converge against x_i

[runic hamlet]

the definition of U_ei(xi) should be completely separate from any sequence or anything

[runic hamlet]

this, but in set notation

[plain phoenix]

ah

I think I know

U_ei(xi) = {| a_n - x_i| < eps}

i hope 😅

[runic hamlet]

well, closer

U_ei(xi) = {x in R: |x-xi| < ei}

so in your case U_1(3) would be the open interval from 2 to 4

so (2,4)

and the intersection of that with D is just {3}

[plain phoenix]

why? I'm so confused about this

your D was 1,2,3,4,3,4,5,3,1,2,3,4,3,3,3,3,3,3,3,...

[runic hamlet]

no

D was {1,2,3,4,5}

the sequence (an) which converges to 3 was 1,2,3,4,3,4,5,3,1,2,3,4,3,3,3,3,3,3,3,...

[plain phoenix]

ah yes, but still!

oh wait, open interval

that means it doesn't contain 2 and 4 right?

[runic hamlet]

yes

[plain phoenix]

ohhh I think I'm starting to get it a bit

let me reread everything

[plain phoenix]

could you explain this to me a little more? I'm scared I'm coming up with my own wild explanations

[runic hamlet]

U_1(3) are the numbers with |x-3|<1

so less than 1 away from 3

so between 2 and 4

[plain phoenix]

that makes.... a lot of sense

thanks!

and why would that always be the case, even if we have a different sequence and different epsilon?

[runic hamlet]

well we can always take the minimum distance to the other points

there are only finitely many

one of them will be the closest

[plain phoenix]

so if I tried to apply our example to the general case

argh I'm still lost 😦 I did understand your example, though

So we have U_ei(xi) = {x in R: |x-xi| < ei > 0} where ei is the minmimal distance between x_i and x_j from D

and then there is D, which is anything, but it's finite

so if the intersection is supposedly just x_i

then the only x, for which U_ei(xi) = {x in R: |x-xi| < ei > 0} is true, must be x_i

xi - xi is 0

which is indeed smaller than something that is bigger than 0

but why can't there be some xj, where xj - xi is smaller than ei

[runic hamlet]

well that xj would have to be in D

but ei was specifically chosen so that this cant happen

[plain phoenix]

AH YES

I got it 😭

thank you so much

is it important that D is finite?

[runic hamlet]

yes

otherwise e_i wouldnt have to exist

[plain phoenix]

true

Could you please also explain to me why then it follows that the sequence (f(a_n)) has an end piece (f(x_i))?

[runic hamlet]

well if the a_n are all equal to x_i after some point

then also all the f(a_n) are equal to f(x_i)

[plain phoenix]

right!

I really need to famliarise myself with convergence more. It's all buzzing in my head

Thank you for the help! That's all I was wondering about

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[slate sigil]

A plane travelled for 600 kilometres from airport A, on a bearing of 210°, before changing direction and travelling a further 250 kilometres on a bearing of 300° to land at airport B.

Calculate the three-figure bearing of airport B from airport A.

[spare ore]

To find the bearing of airport B from airport A, we need to find the angle between the two paths.
The plane travelled from airport A on a bearing of 210° and then travelled on a bearing of 300° to airport B.

The angle between these two paths is the difference between the two bearings:
[300° - 210° = ?°.]

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[twilit field]

$f\left(x\right)=\sqrt{\log\left(\frac{\left(5x-x^2\right)}{6}\right)}$

[grand pond]

Why am. I here

[twilit field]

Find the domain

My attempt:-

$\frac{\left(\left(5x\right)-x^2\right)}{6}>1$

[grand pond]

Why am. I here

[twilit field]

so we have

$5x-x^2>6\ $

[honest kelp]

the expresison can be equal to 1, too

[twilit field]

got it. Thanks

so we have $(x-6)(x+1)>0$

[grand pond]

Why am. I here

[twilit field]

or the domain is R-(-1,6)

however my book says the domain is [2,3]

how?

[honest kelp]

x^2 - 5x +6 <= 0

:

[2,3]

[twilit field]

ah, OK, got it. Tysm!

[honest kelp]

yw)

[twilit field]

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[rocky quest]

help

[rocky quest]

if x and y cannot be 0

if i put 1 + 1 = 0 it cannot be

[midnight plank]

@rocky quest Has your question been resolved?

[lyric patrol]

how i would solve it is by graphing it if u know how to graph in slope intercept u can convert that equation to slope intercept

[rocky quest]

ill try it

[rocky quest]

so quadrant 2 and 4?

[lyric patrol]

yep

hope that helps

[rocky quest]

ill keep this in mind

thank you

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[twilit field]

Find the range of $e^x-e^{-x}$

[grand pond]

Why am. I here

[twilit field]

First I found the limit at $+\infty$

[grand pond]

Why am. I here

[twilit field]

which is $+\infty$

[grand pond]

Why am. I here

[twilit field]

then what?

Do I check if it's monotonically increasing now

Ok, so on differentiating I get $e^x+e^{-x}$ , which is greater than 0 for all x

[grand pond]

Why am. I here

[twilit field]

so does this prove that the range is $R$?

[grand pond]

Why am. I here

[civic wedge]

The limit as x to infty is +infty, and as x to -infty, -infty. Then yes, MVT will get that the range is $\mathbb{R}$

[grand pond]

pramana

[twilit field]

hmm, I don't know the MVT, so is my reasoning correct?

[civic wedge]

there are increasing functions whose range in not all of $\mathbb{R}$

[grand pond]

pramana

[twilit field]

yes, but I've proved it's increasing everywhere

[civic wedge]

$e^x$ is increasing everywhere and its range is $(0,\infty)$

[grand pond]

pramana

[twilit field]

oh, so I can do this by finding the limit at $-\infty$ and then whatever I said follows even without MVT right?

Thanks

[grand pond]

Why am. I here

[twilit field]

Thanks

[civic wedge]

I'm pretty sure you need MVT to prove that it hits all $y\in\mathbb{R}$, but maybe there's a way to do it w/o that

[grand pond]

pramana

[twilit field]

Hmm, OK. Thanks!

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[signal sleet]

what would the arc equal

someone please help

confused

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[cerulean heron]

i cant solve this

[eager oxide]

have you tried exchanging the order of integration?

[cerulean heron]

yes

[eager oxide]

did you adjust the bounds when you changed it?

[cerulean heron]

yes

[eager oxide]

what step are you stuck at

looks good to me, now you can evaluate the left integral with substitution and the right with power rule and it should get you the right answer

[cerulean heron]

what should i substitute u for ?

[eager oxide]

u = 1+(y^3)/2

[cerulean heron]

hmm

ok thanks let me try that

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[arctic forum]

Find the coordinates of the circumcenter of the triangle with the given vertices. 6. T(-6, -5), U(0, -1), V(0, -5)

[arctic forum]

I am confused on how to find the circumcenter

[proud fog]

It’s the point of intersection of perpendicular bisectors of the sides

[arctic forum]

um

can you give me an example

or a picture

[proud fog]

Like this

Circumcentre would be pt. O

[arctic forum]

Like this?

[proud fog]

The perpendicular bisector of the bottom side is off

The circumcenter should be located at the mid point of the hypotenuse in this case

[arctic forum]

Oh

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[static elbow]

help prove

i have theta=x and whatever the other one is =y

simlified right hand side and got sin^2(x)+cos^2(y) / cos^(x)+sin^(y)

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[twilit field]

If $ f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ and $f(1)=7$ Then $\sum_{r=1}^nf\left(r\right)$ is?

[grand pond]

Why am. I here

[twilit field]

My attempt:-

Let's define $f'\left(x\right)=\lim_{h\ \rightarrow0}\left(\frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}\right)$

[grand pond]

Why am. I here

[twilit field]

or

$f'\left(x\right)=\lim_{h\ \rightarrow0}\left(\frac{\left(f\left(x\right)+f\left(h\right)-f\left(x\right)\right)}{h}\right)$

[grand pond]

Why am. I here

[runic hamlet]

I wanna stop you right there. there is no reason to believe that f is continuous or even differentiable

[twilit field]

hmm, OK

[runic hamlet]

the problem is much simpler than that

[twilit field]

yeah, thanks!

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[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

OK, I've tried something, want to makesure I'm on the right track

$f(1+0)=f(1)+f(0)$

[grand pond]

Why am. I here

[runic hamlet]

yes

[twilit field]

so $f(0)=0$

[grand pond]

Why am. I here

[last slate]

yea thats good

[runic hamlet]

now you are on the right track. play around with this

[twilit field]

and then $f(2)=14; f(3)=21$ and so on

[grand pond]

Why am. I here

[runic hamlet]

yes

[last slate]

yep!

[twilit field]

so it's simply $\frac{7(n)(n+1)}{2}$

[grand pond]

Why am. I here

[last slate]

thats true

[twilit field]

Thanks!

[last slate]

I mean, the general stratergy at a low level involving f.eq. is to just play around

[twilit field]

Yeah, I know I've solved such problems before, I'm just too used to calculus . Sorry. Thanks for all the help Arch and Denascite !

Can I close this now?

[runic hamlet]

unless you have further questions

[last slate]

lol, that's fine, if you've seen a lot of calc related questions, then that seems the natural way to think

[twilit field]

.close

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[deft quarry]

Apparently the answer here was 26.9, but I don't understand why? I can't see any issues in my math or sig figs? Did I just horribly misunderstand cross multiplication ? (The answer I'm talking about is the last number, in red)

[sinful trout]

seems correct

@deft quarry i think they just wanted you to round it off...

[deft quarry]

I did, 26.897 is rounded from an even larger amnt of decimals

and the only number in there with limited sig figs is 453.59 (5)

[sinful trout]

well, the sig figures come after the decimal point-

[deft quarry]

this one is going badly too. I'm doing something weird mathematically and in regards to sig figs T~T

[sinful trout]

if we have 3 sig figs that means three 3 dps. i can't write 1,001.3234 as 100 with 3 sf--it'll be 1,001.323

@deft quarry use 3 d.p.s for this one too

[deft quarry]

I thought that was only for adittion/subtraction, and multiplication was all the sig figs

[sinful trout]

...

no-

[deft quarry]

wha

[sinful trout]

why would you think that?

[deft quarry]

I can show you what our teacher gave us to read?

[sinful trout]

if someone asks you to write 5748.948 with 2 sf. they won't want you to write 58... they'll mean 5748.95

[deft quarry]

but in that case, wouldn't the 26.897 then be even longer?

[sinful trout]

what?

no it'd be 26.90

[deft quarry]

but that's 4 sig figs D:?

[sinful trout]

no...

sig figs is after the decimal point-

so 2 sig figs

[deft quarry]

oh I see what you're saying-

[sinful trout]

yea...

[deft quarry]

I was following the rules as I understood them in class, but maybe there's just a weird disconnect, hrm

[sinful trout]

!done

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[deft quarry]

.close

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[deft quarry]

ty :)

[sinful trout]

np

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[twilit field]

Find the nature of $\ln(\ln(\ln(x)))$

[grand pond]

Why am. I here

[twilit field]

i.e is it surjective, injective etc

Other than calculus, how else can I solve this

Oh , and the map is from $(e,\infty$ to R

[grand pond]

Why am. I here

[cedar coral]

well, start by the definitions

what does it mean that it's injective?

what does it mean that it's surjective?

can you prove either?

[twilit field]

One-One

[cedar coral]

that's not a definition tho. That's another way to call it

[strong lava]

By applying elementary properties of functions

[twilit field]

Every image has at least one pre-image

[twilit field]

such as?

[cedar coral]

well then, can you prove that every image has at least one pre-image?

[strong lava]

This is ln applied 3 times
Properties may carry over
Study that

For instance, the composition of 2 increasing functions is increasing

[twilit field]

hmm, OK. Thanks

[twilit field]

I guess, ln(x) when given an input between e and infty outputs a positive number in R+ ,so the output would be negative, and though the log of a negative number isn't defined in R, every image in R will have a pre-image.

[twilit field]

As for proving it's one-one I'd probably use this fact.

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[hallow mauve]

Can someone help me with this question pls

[muted steeple]

!status

[midnight plank]

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

[hallow mauve]

2

[muted steeple]

then tell us what you've got

[hallow mauve]

I know it’s 40% of men as men was the last subject mentioned

[muted steeple]

what is 40% of men? i dont understand what you mean

[hallow mauve]

40% of men were unmarried

[muted steeple]

okay anything else you can draw out from the question as its stated to you?

[hallow mauve]

Lemme try

Total no of women = 4/5 x men

I don’t know how to set up an equation from the second sentence

After unmarried

[midnight plank]

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

[hallow mauve]

<@&286206848099549185>

[midnight plank]

@hallow mauve Has your question been resolved?

[hallow mauve]

<@&286206848099549185>

[midnight plank]

@hallow mauve Has your question been resolved?

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[twilit field]

let $f(x)$ be a polynomial function satisfying $f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x}).$ and $f(2)=9$. then

[grand pond]

Why am. I here

[twilit field]

which is true

A) 2f(4)=3f(6)

2. 14f(1)=f(3)

3. 9f(3)=f(5)

f(10)=f(11)

where do I start, Would differentitating both sides help?

or do I say $f(x)=a_0+a_1x+a_2x^2...a_nx^n$?

[grand pond]

Why am. I here

[twilit field]

would that help more?

[twilit field]

Following on this idea, we get $(a_0+a_1x+a_2x^2...a_nx^n)(a_0 +\frac{a_1}{x}+\frac{a_2}{x^2}...\frac{a_n}{x^n})$= a_0+a_1x+a_2x^2...a_nx^n + a_0 +\frac{a_1}{x}+\frac{a_2}{x^2}..$

[grand pond]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lethal lava]

Try to collect terms one side

It looks like a special format

You have ab=a+b

Take one side then by adding one both sides

You will get

No by subtracting one ig you will get something like (a-1)(b-1)=1

[glass dome]

with these sorts of functional equations, often it's good to just throw some numbers in

let x = 1, then f(1)^2 = 2f(1)

[twilit field]

hmm, i did think of that, let me try. Thanks!

[glass dome]

like after that'll you know f(1) and f(2) which is probably enough to get a whole bunch of things

[lethal lava]

You see (a-1) and (b-1) should be reciprocal of each other and a=f(x) , b=f(1/x) so anything of form 1+x power n or 1-x power n would work

Here f(2)=9 therefore 1+2 powern=9 therefore n=3

So f(x)= 1+x³

[twilit field]

ah yeah, it;s a polynomial function. Got it . Thanks!

[lethal lava]

It could not be 1-x power n form because 2 power n cannot be equal to -8

[twilit field]

Thanks a lot!

I'll close this channel now if that's fine

[lethal lava]

Yes and that was the derivation

[lethal lava]

Sure

[twilit field]

.close

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[twilit field]

How many functions satisfy lcm(f(n),n)-hcf(f(n),n)<5

[twilit field]

how would I even start this

hint please

[steep hinge]

wdym with hcf

[twilit field]

highest common factor

and Lcm is least common multiple

[steep hinge]

oh the gcd

[twilit field]

oh, forgot to mention the function is from N to N

[feral sedge]

to visualize, i would try see what happens for various values of f(n) and n

[twilit field]

hmm, let me try, thanks

[feral sedge]

for questions that involve weird functions, this is always the go-to first step

[twilit field]

hmm, so for instance let's consider f(2) and n=2

let f(2) be say 6

[twilit field]

so that would be 4

I don't see how this helps

[feral sedge]

it helps because <5 is an extremely restrictive constraint

for any given value of n, there are very few values of f(n) that satisfy the provided inequality

this is because lcm grows much faster than gcd (not a good way of putting it but)

[twilit field]

ah, thanks! How do I think like this😢

[harsh marten]

Is it for every n?

[twilit field]

every n belonging to N

[feral sedge]

i believe probably primarily just messing around with problems until you see something
and over time, you build your intuition

in this case, like if n = 2, then we can see that f(n) absolutely cannot be larger than 6 otherwise we fail to satisfy the inequality

[twilit field]

wait, n can't be larger than 6?

can't f(n) be a decreasing function?

[feral sedge]

i mean f(n) cannot be larger than 6

[twilit field]

huh, why?

[feral sedge]

let's see what happens when f(n) = 7 or something

lcm - gcd = 14 - 1 = 13, greater than 5

what about f(n) = 8 ? then,

lcm - gcd = 8 - 2 = 6, also greater than 5

and from here, you can see that you won't get any other values where the inequality is satisfied for n = 2

[feral sedge]

so

i think the most effecive way to do this problem would be to start by finding all possible pairs of (n, f(n)) that satisfy the given inequality

[twilit field]

the question doesn't ask for the soln, just the number of solutions (0,1, infinite or finitely many )

So I think this would be enough, thanks!

[feral sedge]

oh that's the question?

well that would have been slightly useful to know beforehand

[twilit field]

yup, should have mentioned that, sorry

[feral sedge]

😅

[twilit field]

Thanks again

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[tropic marten]

could i screen share to someone and they can help me find stem and leaf plot of whole numbers and decimals? im confused on how to write it down

[last slate]

hello

what do u need

[tropic marten]

could we go into a call so i could show you?

[midnight plank]

@tropic marten Has your question been resolved?

[last slate]

no i cannot

send ss

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[normal citrus]

how do we derivate a function like that

[visual tiger]

use FTC

[normal citrus]

you mean i calculate the integral and then i derivate it?

[visual tiger]

if you don't have theorems that allow you to manipulate this directly, maybe try to make something like $\int_x^1\frac{e^{-t}}{t}dt$ appear

[grand pond]

rafilou2003

[visual tiger]

no no need

just say that an antiderivative exists

[normal citrus]

ok but i don't know its expression

if it would be an integral a to x i would say that the derivate is what inside the integral but here it's not the case

[visual tiger]

no need

name the antiderivative you picked "F"

[normal citrus]

ok then it's equal to F'(+infinite) - F'(x)

[visual tiger]

no

so $\varphi(x) = \lim_{a\to\infty}F(a) - F(x)$

[grand pond]

rafilou2003

[normal citrus]

yes

then i derivate it?

[visual tiger]

yes

[normal citrus]

then it's equal to lim f(a) - f(x) = -f(x)

[visual tiger]

? what is f

[normal citrus]

what's inside the integral

[visual tiger]

ok but the first thing you wrote is not true

$\lim_{a\to\infty}F(a)$ is a constant

[grand pond]

rafilou2003

[visual tiger]

so when you differentiate it

you get 0

no matter what lim(f(a)) is

[normal citrus]

ok i see

thx

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[plain phoenix]

Hello! 👋 Unfortunately, I do not understand the bold statements in the proof of the following proposition. Could someone please explain them to me?

[plain phoenix]

Proposition: Let f : D → R be continuous in a. Let t < f(a) < z. Then there is a δ-environment Uδ(a) of a such that for all x ∈ D ∩ Uδ(a), t < f(x) < z.

The proof: Assume that such an environment U does not exist. Then for every δ-environment of a there is an "exception x" with f(x) ≥ z or f(x) ≤ t. In particular, for every U_1/n(a), n ∈ N, there is an xn ∈ U_1/n (a) ∩ D with f(xn) ≥ z or f(xn) ≤ t. Then (xn) converges to a, and since f is continuous in a, (f(xn)) must converge to f(a). However, since all f(xn) are not contained in (t, z), f(a) = limn→∞ (f(xn)) cannot be contained in (t, z) either. This is a contradiction to our assumption that t < f(a) < z.

I'm currently reading an introduction to calculus book, and although I found the chapter on convergence comprehensible and was able to do the exercises which I revised today, I'm struggling with the current chapter on continuity of functions, where convergence is constantly used as a tool. I find it very frustrating 😦

[steep hinge]

allright what we have

do you know the defintion of a function continuous at a point z?

[plain phoenix]

no

[steep hinge]

mhh maybe this is the problem

however

[plain phoenix]

I don't think I've read about it

[steep hinge]

so that assumption is totally dark to you ?

[plain phoenix]

ohhh, no

I do know the definiton of that

but I don't see how it's relevant in my highlighted parts

I discovered it being used here: f is continuous in a, (f(xn)) must converge to f(a)

but not elsewhere

[steep hinge]

okk

"since all f(xn) are not contained in (t, z)" here it is used that f(x_n)>=z or f(x_n)<=t hence f(x_n) is in (-infinity, t] or [z, infinity)

hence f(x_n) is not in (t,z)

is this clear to you?

[plain phoenix]

embarrassingly clear 😅 how did that not make sense to me before

I think I really led myself astray with the examples that I came up with for clarity

[steep hinge]

[still sluice]

Calculus introduction books lol

[plain phoenix]

How about this: Then (xn) converges to a?

[still sluice]

Is it linear

Or invertable

[steep hinge]

xn converges to a by xn's construction ...you pick every xn closer to x

[plain phoenix]

hmmm

[steep hinge]

what is the problem?

[plain phoenix]

maybe there is none. I'm just slow to read this mathy language. One moment!

[steep hinge]

you can insult me if that help your understanding

[plain phoenix]

what? why would I want to insult you?

[still sluice]

So you could then
lim n-> infinity f(x_n)=f(x)?
You said f(a) = f(x_n) so if it’s invert-able, then f^-1(f(a)) = f^-1(f(x_n)) so a = x_n in your thoughts right?
Also you could use an example to make it clear.

[still sluice]

And always use an example to bring the mathy language from abstract world

[steep hinge]

i don t know maybe that can help people to think better ..i reapet i dont really know

[still sluice]

You would want to define sets and functions and variables first

[plain phoenix]

so to me, the red part reads as: there is xn in the interval (a + 1/n, a - 1/n)

[steep hinge]

yes exaclty

hence |a-xn|<1/n

and goes to zero when n goes to infinity i.e. $\lim_{n\to\infty}x_n=a$

[grand pond]

everg

[still sluice]

Go on with example, it’s a good one

[plain phoenix]

this looks familiar. I recognise the epislon definition there

but I can't see immediatelyhow xn ∈ (a + 1/n, a - 1/n) is the same as |a-xn|<1/n. Let me try to grasp it

[plain phoenix]

so to do this

if a = 3

and 1/n, let's say 2

then xn ∈ (5, 1)

and this |a-xn|<1/n would translate to |3 - xn| < 2

[still sluice]

Notice that absolute value makes things weirder, 1/n must be positive and if n-> infinity, it’ll be 0 and Archimedean Principle works

[plain phoenix]

okay, i got it

[still sluice]

I was talking about Archimedes today so I saw it directly

0 <= x_n <= lim n-> infinity

[steep hinge]

look at this: xn in (a - 1/n, a + 1/n) implies xn>a - 1/n AND xn<a + 1/n implies xn-a>-1/n AND xn-a<1/n IMPLIES |xn-a|<1/n

since if -1/n<c<1/n then |c|<1/n

[still sluice]

So we were limiting the whole time and we could figure out that it’ll be 0 on limit

Or undefined as output

[plain phoenix]

this is great 😭 thank you so much for this. I think not being able to look at intervals that way before was largely what caused me so much trouble!

[steep hinge]

chill it is just get used to this sh*t

[still sluice]

I just constructed the worst function to visualise undefinity here, I just wanted to visualise

[plain phoenix]

alright, I think I get the proof now. I just went over it again and it seems clear. Thanks!

[steep hinge]

np ;D

[plain phoenix]

yeah, thanks 😁

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[ocean token]

Im stuck on the 1st one, Ima also need help with the 2nd one too

[midnight plank]

@ocean token Has your question been resolved?

[harsh marten]

<@&268886789983436800> there is a mess

[normal zenith]

good

what why

lmao

[midnight plank]

@ocean token Has your question been resolved?

[solid iris]

.close

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[rose badger]

Hello, can I get some help?

[surreal moon]

!da2a

[midnight plank]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[rose badger]

The city of Austin is erecting a radio tower to boost cell phone coverage. They must install guy wires to support the tower in the wind. The guy-wires must attach to the tower at a point 56 feet above the base of the tower, and must form a 50° angle with the ground. Assuming that the tower is on level ground, how far from the base of the tower will the guy-wires be secured to the ground?

This

How do I began?

[wary thorn]

I'm not doing that

[surreal moon]

City of @wary thorn

[rose badger]

?

[surreal moon]

Just pinging a friend for some good laughs

[rose badger]

I agree, it isn't a great city

[wary thorn]

Wow

[signal sigil]

lol

[wary thorn]

Catching strays now

[rose badger]

OOOOHHHHHH, I see

[wary thorn]

In all seriousness, draw a picture

[rose badger]

Bam

[surreal moon]

drawing a diagram may help. Sounds like it would be a right triangle

[wary thorn]

No. Draw your own picture

and label relevant angles

and lengths

[rose badger]

Done

[glass dome]

so now you can do trig right

[rose badger]

If I knew it

That is why I am here

[surreal moon]

Do you know SohCahToa?

[rose badger]

Yes

Sin Opposite x Hypotenuse / Cosin Adjacent x Hypotenuse / Tangent Opposite x Adjacent

?

[rose badger]

Yes

[rose badger]

Hello?

[surreal moon]

You know the angle, and the opposite side, and you want the adjacent side. So that's all you really need

[rose badger]

How do I turn that into the bottom side though?

[surreal moon]

You know this angle

[rose badger]

Yes

[surreal moon]

What is the length of the side opposite this angle?

[rose badger]

56

[surreal moon]

and you want to find the length of the side adjacent to the angle, do you not?

[rose badger]

yes

[surreal moon]

so then which of soh, cah, toa do you think you should use?

[midnight plank]

@rose badger Has your question been resolved?

[rose badger]

TOA

[rose badger]

But the problem is, I don't know how to use the opposite side and the angle, to find the adjacent side

[grave marsh]

I think in this case you need to use Sine and when you come to the division part, take the sine inverse of the division

That’s if you wanna find the bottom angle

[nocturne yew]

tan (θ) = opposite/adjacent

then rearrange

so adjacent = opposite/tan(θ)

that's assuming u have all those values

dk what the question is lol

[rose badger]

How do I find TAN if I don't know adjacent?

[rose badger]

The bottom side, and can you lay that out into an equation?

[grave marsh]

Sine 📐 = opposite/hypotenuse

[rose badger]

Right, sohcahtoa

[grave marsh]

Eg. Sine📐= 2/4
📐= sine-1 (2/4)

[grave marsh]

Yeah

[rose badger]

Thanks!

[midnight plank]

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[twilit field]

$g(f(x))=x^2$ for $x\geq0$ and $g(f(x))=e^x-1$ for $x<0$. Then find the nature of $g$ and $f$

[grand pond]

Why am. I here

[twilit field]

(one-one , onto etc)

Could I have a hint

so my first thought was to differentitate the function

so $f'(x)g'(f(x))=2x$

[grand pond]

Why am. I here

[twilit field]

for x>0

and $f'(x)g'(f(x))=e^x$

[grand pond]

Why am. I here

[twilit field]

for x<0

<@&286206848099549185>

Just a hint please, no answers

[placid spoke]

is this to show that g(f) is 1-1?

you taking the derivative

[twilit field]

This is a MCQ

the options are g is onto

f is onto

g is one-one

and f is one-one

[placid spoke]

find out of g(f) is 1-1, onto, etc. there is a Theorem that says if g(f) is one of those two, you can say something about g and or f without knowing what g or f are

I can spoil what the theorem says if you want

[twilit field]

hmm, let me think about this , I'll reopen this channell if I don;t get it. Thanks!

.close

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[vocal trail]

!close

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[orchid gazelle]

The last here how did he get this ?

[orchid gazelle]

How is it (alpha - beta)^2 - 4 alpha beta

[neon rune]

must be a typo

[orchid gazelle]

Owhh

[neon rune]

should be (α+β)^2-4αβ

[orchid gazelle]

Owh i see

How do we get that doe

From the question we have (alpha^2 - beta^2) = 144

[neon rune]

$\left(α+β\right)^{2}-4αβ=α^{2}+β^{2}+2αβ-4αβ=α^{2}+β^{2}-2αβ=\left(α-β\right)^{2}$

[grand pond]

B-eard

[neon rune]

no.
notice the wording:
squared difference; not difference of squares

[orchid gazelle]

Owhhh yeah

I just noticed lmao

What should i do if it was difference of squares ?

[neon rune]

before proceeding into the process, you know what? I solved the exact question with misreading it as difference of squares about an year ago, when I was in 10th

[orchid gazelle]

Owhh lol

[neon rune]

Solvable but long ass method

[orchid gazelle]

I see i see

Anyways thank u for the help !

[neon rune]

$α^{2}-β^{2}=\left(α+β\right)\left(α-β\right)=\left(α+β\right)\left(\sqrt{\left(α+β\right)^{2}-4αβ}\right)$

[grand pond]

B-eard

[neon rune]

This is how you would go on by solving it if it were "DIFFERENCE OF SQUARES"

[orchid gazelle]

Hmm i see

How is that um (alpha - beta) that though

[neon rune]

difference of squares formula

[orchid gazelle]

Owh right right

I'll note that down

[neon rune]

lemme guess

You are in 10th preparing for boards whilst watching PW lectures right?

[orchid gazelle]

Yup but not the Pw lectures lol

Im doin problems from rd sharma

[neon rune]

Oh okay yeah I remember this was from RD sharma

yeah

[orchid gazelle]

Yeaa

Thank u for the help

.close

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[neon rune]

welcome

[midnight plank]

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[hot zinc]

Hi, how do I find the range for a parabola with a maximum value of 9?

[mellow plinth]

if the maximum value is 9, can the range include values more than 9?

[forest thorn]

Want to know how to solve these, the repeating decimal question I didn’t really understand the guys method in the video

[hot zinc]

!occupied

[midnight plank]

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).

[hot zinc]

No. I'm just not too sure how to write it. I wrote (-inf, 9)

ahh omg

I just realized my mistake

it is (-inf, 9]

[mellow plinth]

yeah lol

[hot zinc]

.close

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[orchid gazelle]

is there a relation between sum of zeroes and product of zeroes ?

[orchid gazelle]

.close

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[ebon parrot]

is this just {1}?

since any a/1 can be just a

i mean g is a function that we dont know so is it just Z or?

[lethal path]

All rational number inputs can map to the same number

It doesn't matter which number that is, 0, pi, sqrt2 etc

[ebon parrot]

ye I mean q can be a codomain but is it the smallest co domain?

[lethal path]

Well the function can't map to an empty set

Like I don't think that makes sense for a function

Idk

[placid spoke]

keep in mind that the function can take in ANY rational number in lowest terms. what are the possible numerators for a rational number?

[ebon parrot]

any integer

[lethal path]

Ah okay I think I misread the q

[placid spoke]

yep

that's the answer lol

[ebon parrot]

so just z

okok bet ty that makes sense

[placid spoke]

pretty much

[ebon parrot]

😄

!close

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[bitter tartan]

what rule should i use to solve

[lusty elm]

chain rule is fastest maybe

[sharp coral]

rewrite the fraction as a power, then use power rule & chain rule

[bitter tartan]

why the chain rule looking confusing

[fallow scarab]

Do power rule first then chain rule will be obvious

[slender walrus]

various ways. apply what you know.

with a fraction, you can apply quotient rule if you want

its actually less conversion / switching between exponential forms than the method above

[midnight plank]

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[frail heart]

I know how to get the denominator but can't get the numerator

This is the step I'm stuck on

[azure furnace]

Simplify numerator further

Use identity

[frail heart]

How?

[azure furnace]

(a+b)(a-b) = a^2 - b^2

Use this identity

[frail heart]

Like this?

[vague eagle]

wait sorry aha

yes, 4^2 expands to 16

3^2 expands to 9

[frail heart]

Yea so now do we get y²-14?

[vague eagle]

okay heres how id do

u just cross multiply

i.e.

a/b x c/d = (ad + bc)/bd

[frail heart]

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[midnight flame]

if x > y > 0 and z =/ (does not equal) 0, the inequality which is NOT always correct is

a) x-z > y-z
b) x^2 > y^2
c) x/z^2 > y/z^2
d) xz > yz

[proven slate]

with a),b),c) the value of z doens't affect the inequality
with d), if z is negative, the inequality is not true

[midnight plank]

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.

[proven slate]

bruh i explained to them

i didn't give the ans

[nimble copper]

The two aren't mutually exclusive

You did give the answer, you just explained why it was the answer

[proven slate]

so how could i explain that such simple inequality

[midnight flame]

.close

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[proven slate]

if u could, try to do it in a better way

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[lament fox]

How can I evaluate this infinite sum?

[midnight plank]

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[fair meadow]

Hello guys, do añyone know a trick of choosing a scale for a graph work.

[slender walrus]

choose something that'll allow you to fit what you want to graph on the page

[twilit field]

You just chose a scale that's convenient , the skill comes with experience . For instance if the x axis has data in multiples of 100, as does the y axis, chose 100 as the scale

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[low tiger]

Pls help me solve this question

[wanton shore]

|x| + |y| >= |x+y|

And equality holds when x and y are of same signs

[tardy wyvern]

Thats something triangle inequality right?

[wanton shore]

More general form?

[midnight plank]

@low tiger Has your question been resolved?

[low tiger]

I'm sorry could u elaborate pls

[low tiger]

How did u come to this

[wanton shore]

Like you had to remember this

There is an easy proof for this

[midnight plank]

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[low tiger]

Oh

.close

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[last slate]

How can i obtain the "yellow line" with a function

[last slate]

on this graph

ive tried several functions cant seem to find

[twilit jetty]

immediate idea is to graph something like x = my + b + a/(y + c)

you can test this by typing that into desmos then finding out an m, b, a, and c that work

[last slate]

yes i'll use varying values

for those

let me give it a try

[twilit jetty]

alr

[last slate]

also what is the name of this function?

[twilit jetty]

its usual shape is y = (polynomial in x) / (a different polynomial in x) where it's called a "rational function"
then chose the numerator polynomial to be (mx + b)(x + c) + a and the denominator polynomial to be x + c
so y = ((mx + b)(x + c) + a)/(x + c) simplifies to y = mx + b + a/(x+c)
this is a particular rational function that looks like 1/x but moved around and slanted

then I swapped x and y to get it sideways

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[last slate]

i cannot graph it on desmos

it gives error

even though i am varying the values

[midnight plank]

@last slate Has your question been resolved?

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[oblique torrent]

f: R[X] $\longrightarrow$ A \
P $\mapsto$ (P(0), P'(0))

where A is the ring $(R^2, +, \ast)$

[grand pond]

lilisworld.

[steep hinge]

hence $P(x)=a_0+a_1x+...+a_nx^n\to (a_0,a_1)$

[grand pond]

everg

[oblique torrent]

ok

so how do i show it's surjective? because (a0, a1) in R² and R² in (a0, a1)?

[steep hinge]

for every $a,b\in R$, the polynomial p(x)=a+bx map to (a,b)

[grand pond]

everg

[steep hinge]

so it is surjective ...but this map is not a homomorphism

[oblique torrent]

thank you 👍

.close

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[twilit field]

Is there any way to do this than manually computing the maxima and minima ?

[steep hinge]

lol you are still helpful

do you know what is $\lim_{+\infty}f(x)$?

[grand pond]

everg

[twilit field]

yes . 0, I think

[steep hinge]

however no there is no method to compute manually maxima and minima

[twilit field]

hmm, I was actually talking about $g(x)$

[grand pond]

Why am. I here

[steep hinge]

but it is easy to prove that if $a_n, b_n\to0 \implies \max(a_n,b_n)\to 0$

[grand pond]

everg

[steep hinge]

i asked for $f$ at first

[grand pond]

everg

[twilit field]

ah, got it. TYSM

I'll close this now, if that's fine with you

[steep hinge]

yes

.close

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[glossy sleet]

i don't understand the solution here

in the second step they get the original equation

and they multiply the original eqn by the conjugate of the denominator, and also by some other fraction

shouldn't these be addition?

ik there's probably a really easy way to solve this but i just started learning calculus a week ago so i don't know anything

[midnight plank]

@glossy sleet Has your question been resolved?

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@glossy sleet Has your question been resolved?

[glossy sleet]

<@&286206848099549185>

[slender walrus]

no

they're applying conjugates twice

what's highlighted is (conj of numerator)/(conj of numerator)

the expression in the middle is the original fraction

and the thing below is (conj of denom)/(conj of denom)

[glossy sleet]

ok i got it now ty

.close

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[young wing]

if i need to use the chain rule for this, what's the inside and outside function?

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[violet whale]

can sm1 help

[rose kayak]

use the fact that for absolute inequalities have the following property:

|ax - b| < c

ax - b < c
-(ax-b) < c = -ax + b < c

In both cases, you solve for x and then combine the inequalities into one statment

[violet whale]

i still do not understand

[rose kayak]

https://www.youtube.com/watch?v=ZJvOGkykJ44

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[drifting mason]

im having a hard time understanding this—why is potential energy being used as the “output force”? The Pe isnt what is making the mass move. I dont get it

the kinetic energy should be 3 Joules, no? Because the kinetic energy = (1/2) * (6 kg) * (1 ms^-1)^2, which is 3 J

additionally, if W = Fs, how do objects that are moving at constant velocity have work??? F = ma and if acceleration is 0, then that would make the W = Fs equation equal 0

additionally, why does the force carrying the plate HAVE to be Fn??? Wouldnt there be an Fa thats parallel to the displacement of the plate????

[midnight plank]

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@drifting mason Has your question been resolved?

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@drifting mason Has your question been resolved?

[tulip token]

if you carry a plate around at a constant velocity, you will see that there is no horizontal force required to keep it moving

in any case, it's asking about how much is done to support it

[tulip token]

the kinetic energy is already in the mass that's moving vertically; the extra energy added is to the gravitational potential energy of the system

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[wanton shore]

Find no. Of terms in the expansion of
(1+x+x⁴)^n

[wary thorn]

!status

[midnight plank]

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

[wanton shore]

I observed the pattern and my answer is
4n-2, n>=2

4

@wary thorn

[wary thorn]

!show

[midnight plank]

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

[wary thorn]

, rotate

[wanton shore]

??

[wispy sonnet]

lol

[midnight plank]

@wanton shore Has your question been resolved?

[placid spoke]

gonna be honest I can't read this like at all

[placid spoke]

this is correct, but are you asked to prove it?

[wanton shore]

Nah i was just told to give answer by observation

Now that you mention it is there any proof?

[placid spoke]

you'd use some sort of counting argument, this is a combinatorics kinda problem

or induction if that doesn't work

but it would be a little icky

[wanton shore]

If i do same for (1+x+x⁵)^n
I get 5(n-1), n>=3

Can you check this

[placid spoke]

sure im bored

[wanton shore]

Induction would work i think

[placid spoke]

seems correct to me

induction would be hard bc predicting what powers show up for a general case would be super hard

[wanton shore]

Ok thanks

.close

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[tranquil kite]

I think I'm supposed to use a hypergeometric probability distribution to solve this. The way I know how to do this requires that we know the number of successes in the population, r. Where I'm stuck is finding r. I can't think of how to get the number of successes in 15 restaurants from the average price, $48.60, and the threshold at which a result is considered a success, $50.

[midnight plank]

@tranquil kite Has your question been resolved?

[tranquil kite]

<@&286206848099549185>

[mighty flame]

i think you have to multiply the chance of each step

so the first restaurant probability that it exceeds 50 is like

1/5 yeah

then after you go to a different restaurant

the chance is 2/14

or wait

[tranquil kite]

From what my textbook says:

Mean = n(r/N)

So solving for r I do:

u = n(r/N)
u/n = r/N
(u*N)/n = r

But when I plug in the values:

r = (48.6 * 15) / 3
r = 243

I don't think that's possible because the population is only 15.

[mighty flame]

Ok question is that average cost of $48.60 part of the solution

Like if you're gonna need that value then I'm not gonna be of much help at all

[tranquil kite]

I assumed we need 48.6 in order to determine r.

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[last slate]

I need help with some factoring.

[last slate]

How would I factor with negative exponents

I am not really sure how to

[slender walrus]

factor based on the lowest power

[last slate]

what do you mean by that?

[slender walrus]

consider how you'd factorise something like
p^5 + 3p^3 + 2p^2
walk me through your thought process

[midnight plank]

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[twilit field]

I would like a hint here

[twilit field]

I started by finding the vertices of both the parabolas

the max of f(x) is $1-\frac{b^2}{4}$

[grand pond]

Why am. I here

[twilit field]

and the min of $g(x)=c-1$

[grand pond]

Why am. I here

[peak walrus]

Try equate the two parabolas

2x^2+(2-b)x+c-1

And then I’m pretty sure the solution would be when the discriminant = 0 although haven’t tried it myself

[twilit field]

hmm, so $-x^2+bx+1=x^2+2x+c$. ?

[peak walrus]

Lemme try it first actually

[grand pond]

Why am. I here

[twilit field]

and 1-b^2/4=c-1

[peak walrus]

Yes

I believe so

Just solve that for c

[twilit field]

ok, thanks!

.close

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[mossy nacelle]

Does anyone know how to solve for part b? The answer key states there's a second answer where 2cosx + 1 = 0, but I can't figure out how to solve for the equation in any other way where these two don't cancel out.

[brittle shoal]

it is correct

next you can write 1 as tan45°

and you get the answer for x

[mossy nacelle]

Yeah, but there's supposed to be two cases

This is the first

The other is 2cosx + 1 = 0

Which I guess would be cosx = -1/2

[fallen galleon]

instead of dividing you subtract

[brittle shoal]

wait you did it wrong

[fallen galleon]

so have sinx(2cos(x)-1) = cosx(2cos(x)-1)

[brittle shoal]

cos 2x can be written as $2cos^2x -1$

[grand pond]

斯韋裡

[brittle shoal]

not +1

[azure furnace]

Dude there's problem in your both solution 💀

[mossy nacelle]

Oh, that was the old thing I was trying. I solved for a and I guess I erased that by accident

You factor out sinx and cosx and cancel them out to get tanx = 1, don't you?

[fallen galleon]

but what if 2cos(x)-1 is 0

you're effectively dividing by 0

[azure furnace]

Yeah but how did you reach on the result that
both numerator and denominator has 2cosx +1
Cuz the cos2x identity is 2cos^x -1
Not +1

[mossy nacelle]

I'mma redo the math on this rq

[azure furnace]

If it is so then this problem will become undefined if that soln in the picture is correct
Cuz it'll result in 0/0 form

[fallen galleon]

no you're correct up to the last part

look at what i wrote on my picture

[azure furnace]

How???

[fallen galleon]

you can subtract the RHS from both sides

then since you have a common factor of 2cos(x)-1 you can factor that out too

and you're left with (sinx - cosx)(2cos(x)-1) = 0

and thats where your cases come in

@mossy nacelle ^

[mossy nacelle]

Sorry I'm here

I just redid part a to double check it

It should be 2cosx + 1

[azure furnace]

Ohh yeah you're right
But look at the first step
How is cos2x + cosx = cosx (2cosx +1)

[mossy nacelle]

Sending the new proof in a sec

That's part a

[fallen galleon]

ur missing the +1

cos(2x) + cos(x) + 1

[fallen galleon]

yes the proof is correct

and then for part b instead of cancelling them out you factor them like i said before