help-26

[coarse tusk]

There can be many reasons

Being condescending doesn’t help anybody

[burnt swift]

according to gooji divisibility rules are not needed tho

[coarse tusk]

The way they’re describing is also just using divisibility rules

But sure trial and error kinda works

What you really need is $4\ |\ A2$, or equivalently $4\ |\ (10A+2)$

[thorny flame]

kheerii

[coarse tusk]

Find out all possible values of A from here and see which divide 63

[old cloak]

the commonly stated divisibility rule for 4 is literally just "try it with 2 digits" because everyone that is learning about divisibility knows how to do that

I'll stop

[burnt swift]

.close

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[neon iron]

How do I know that theta is less than π/2

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[dusk pawn]

I wasn't sure where to ask this question, so I asked it here just to be safe.

Hi, I'd like to be able to explain the way to prove Lemma 1.7 (that for all n, n x 10 = n0) in a way that is educationally sound and easy to implement.

I have tried to prove it myself, but I'm not sure how to word the proof such that it is accessible to adults studying functional skills maths.

[shadow salmon]

The notation n0 is kinda confusing imo

[dusk pawn]

Oh really? Let's say n = 7, then that number would be 70. That's kind of what I mean. I hope that clarifies things.

[shadow salmon]

Yeah well is this clarified somewhere? Because this can interpreted as multiplication by 0

But uh maybe that’s not so weird since you described this was for a certain public of people(?)

[dusk pawn]

People studying Functional Skills Maths.

The reason this lemma exists, is so I can show that multiplying by 100 (which is necessary for converting pounds into pence) is about adding two 0's to the right of the number.

[shadow salmon]

Why not explain it in terms of the number base system? This is more or less a feature of base 10

[dusk pawn]

The number base system... That could work, but I want to make sure that the proof 'answers the question', as it were.

[shadow salmon]

Instead of relying on an induction proof which seems a lot more challenging to explain

[dusk pawn]

I mean, proof is hard. I'll drop the induction proof with immediate effect. After all, base 10 might be a better proof for functional skills students.

[shadow salmon]

Obviously if they have past experience and are comfortable with such proofs I can see it as maybe useful but when it’s more or less just to describe a feature of base 10 it seems sort of overkill unless I’m missing something

[dusk pawn]

I mean, this is important for understanding how to convert £4.56 (for example) into pence only. And I'm going to tell you why.

As you know, £1 is equal to 100 pennies, so you've got to do 4 x 100 (which is 400), then add the 56 that is left over.

[shadow salmon]

Yeah so if the importance here is understanding why this works and they haven’t had past experience with say Induction proofs I’m pretty skeptical you’ll achieve anything other than confusion

[dusk pawn]

I mean, it's a lemma (as opposed to a definition), and the lemma needs proving, but I could always go ahead without a proof, since most people already know that multiplying by 10 involves adding a 0 to the right of the number.

[shadow salmon]

Okay, if you’re adamant about this I won’t try and discuss that part of it.

So going back to the proof, in the second step, I.e the inductive hypothesis, you’ve included that inside the induction step, I think you should separate that, as they are different steps

[dusk pawn]

I mean, okay. How can I argue with that?

[shadow salmon]

When you say associativity, what do you mean by that exactly? It seems you’ve confused it with the distributive law

[dusk pawn]

Let's change k+1 to a+b.

Then (a+b) x 10 = (a x 10) + (b x 10).

That's what associativity means, at least in this case.

[shadow salmon]

Yeah this is not associativity I’m afraid

[dusk pawn]

OK then. I concede.

[shadow salmon]

Associativity is in respect to one operator and not multiple, and it’s when the order in which you can calculate say a + b + c doesn’t matter

,w distributive law

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[shadow salmon]

Is it this what you meant?

[dusk pawn]

Yes, I did mean distributive.

Thank you.

[shadow salmon]

Ok alright, also since they aren’t familiar with an induction step (or so I assume) I think it would help if you conclude that this holds for all natural numbers n due to the induction principle or axiom, in order to make the conclusion you wanted to

That is, at least make reference to what is actually allowing you to do this conclusion

[dusk pawn]

I think you were right the first time... but I'll see if I can reference what is causing me to make the conclusion that "Multiplying a number by 10 is equivalent to adding a 0 to the right of the number."

[shadow salmon]

Oh and yeah specify this proof holds for natural numbers

[shadow salmon]

Sounds good

[dusk pawn]

I've changed it to "Multiplying a whole number by 10 is equivalent to adding a 0 to the right of the number."

I suppose since my target audience knows only basic stuff, I'll forget about the proofs for now.

Thanks for helping me to decide that.

[shadow salmon]

No problem! I think it’s cool that you’re at least trying to include that more rigorous taste of math

Instead of leaving things to be true “just because they are”

[dusk pawn]

Thanks. I'm going to close this help channel.

.close

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[raw zenith]

Hey

[raw zenith]

I don't know the formal definition of continuity cause the teacher was taking too long to explain it

but basically what I gathered was

I mean they said that you should be able to draw a function without lifting your pen but that is not very mathematical lol

So formally

If a function is said to be continuous around a point x if LHL(x) = RHL(x) = F(x)?

This is what the definition should be right

is this correct?

This will only happen when the function has no breaks

Wait this does not cover the end points

assuming a function ends at a point then RHL may not exist, so LHL will not be equal RHL

[odd pagoda]

well only the relevant limit at the endpoints has to equal F(x)

[raw zenith]

alright, so only one of the two limits?

[raw zenith]

and this definition then is not for endpoints?

[odd pagoda]

yes

[raw zenith]

alright, thank you

🙂

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[strong sable]

any hints on (c) ? I literally have no clues about on one...Appreciate any helps

[hollow latch]

$I have x>1, x<1, and x^2<1, and x is not equal to 0, how is the answer -1<x<1?$

[strong sable]

.....? Sry this is my channel

[hollow latch]

oops

Accidentally did that, sorry

[ocean terrace]

whats Gn and An?

[opal vault]

Geometric and arithmetic mean

ok

[opal vault]

so if you apply part (b) as they suggested, what do you get?

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[strong sable]

.reopen

[topaz sinew]

✅

[strong sable]

this

[opal vault]

yeah sure, but replace "G_n" and "A_n" by their values here

[opal vault]

like what is geometric mean of a_1,...,a_n,(A_n) 2^m -n times

and what is arithmetic mean of ...

[strong sable]

yea that's the piont idk

I dont know how to write the GM and AM form of a_1,...,a_n,(A_n) 2^m -n times

[opal vault]

ok

what's the arithmetic mean of a_1,..., a_k for example

[opal vault]

I give this just as an example

[strong sable]

a_1+...+a_k/k

[opal vault]

what do the commas mean?

[strong sable]

edited

[opal vault]

ok so

if I said

[opal vault]

keep a_1.... a_n as they are

and now I'm gonna define

a_(n+1) = A_n

a_(n+2) = A_n

and so on

until a_(2^m) = A_n

[opal vault]

what's the arithmetic mean of a_1,....,a_(2^m)?

[strong sable]

wait....why?

isn't An arithmetic mean here?

[opal vault]

it's the arithmetic mean of a_1 up to a_n

it's NOT the arithmetic mean of a_1,...,a_n,A_n 2^m - n times

[strong sable]

yea, why a_(n+1) = A_n ?

[opal vault]

you want the arithmetic mean of this

as (b) said, when $k = 2^m$, you have $(a_1...,a_k)^{1/k} \leq \frac{a_1+...+a_k}{k}$

[thorny flame]

rafilou2003

[opal vault]

Except

you don't have n = 2^m anymore

but you still want to use this result

so what do you do?

you add more terms to take the mean with

and be able to use (b)

[strong sable]

aah

but why a_(n+1) = A_n can lead this result

[opal vault]

that's the point of the hint

it tells you

use (b)

with a_1,...,a_n

and add to that A_n as many times as you need to reach a total of 2^m terms

so

a_(n+1) = A_n

etc

up to a_(2^m) = A_n

now we have 2^m terms

[strong sable]

i see

so AM of a_1,...,a_n,(A_n) 2^m -n times= (a_1+...+a_n+a_2^m)/2^m
GM of a_1,...,a_n,(A_n) 2^m -n times= \sqrt{a_1......a_2^m}

(a_1+...+a_n+a_2^m)/2^m>=\sqrt{a_1......a_2^m}

[topaz sinew]

@strong sable Has your question been resolved?

[strong sable]

@opal vault does this correct?

[topaz sinew]

@strong sable Has your question been resolved?

[opal vault]

yeah that's it I think

it's missing some ... sometimes

for example AM is (a_1+...+a_n+***...+***a_2^m)/2^m

[strong sable]

aaah yes

[strong sable]

then .... It's proven? Done?

[opal vault]

?

we have proven that the geometric mean of a_1,....,a_2^m (call it G_2^m if you want) is smaller than the arithmetic mean of a_1,...,a_2^m (A_2^m)

but we wanted to prove that the geometric mean of a_1,....,a_n (Gn) is smaller than the arithmetic mean of a_1,...,a_n (An)

so no it's not done

write out what G_2^m and A_2^m are in terms of Gn and An

and manipulate the inequality G_(2^m) <= A_(2^m)

to hopefully get G_n <= A_n

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[regal harbor]

how do you find the range here?

[regal harbor]

i'm super confused

[ocean fjord]

hint: what is (x + 1/x)^2

[regal harbor]

i see

brb

[regal harbor]

the expanded form is x^2 + 1/x^2 + 2 though no?

[ocean fjord]

hm

do you know amgm inequality

[regal harbor]

nope

that problem is from your average algebra book, do you need that kind of stuff?

i have never heard of amgm inequality

[ocean fjord]

no you dont

[regal harbor]

i don't see how can you continue this

i have tried multiplying x^2 by both sides but it didn't really take me anywhere

[vapid sonnet]

We have to find y which has a preimage ,you can solve this equation by quadratic formula and find the value of x^2 in terms of y

Then possible values of y is range

[regal harbor]

i'll try applying the quadratic formula

[regal harbor]

you apply it like this right?

[vapid sonnet]

No coffecint of x^2 is y not 1 and quadratic formula equal to x^2

This is quadratic equation in x^2 not x

[regal harbor]

is there a reason behind setting x to x^2?

[vapid sonnet]

Because this is quadratic in x^2 you can write x^4 (x^2)^2

Which give value of x^2

[regal harbor]

oh i see

[vapid sonnet]

Like quadratic formula in x give value of x

it looks like ax^2+bx+c

[regal harbor]

so this is correct right?

i never had a case where you'd have to use a quadratic in order to find range

[vapid sonnet]

There is -1 in front of root

[regal harbor]

yeah whoops

[vapid sonnet]

No,what you done

Upper one is right

[regal harbor]

isn't y squared?

[prisma coral]

[2,…infinity)

[vapid sonnet]

Oh sorry yes

[regal harbor]

yea, i'm aware but can't tell how to get to the answer

[prisma coral]

AM-GM

[vapid sonnet]

There is -y in front of root

[regal harbor]

is there no avoiding this?

[vapid sonnet]

-b+√b^2-4ac

[regal harbor]

i don't think AM-GM is ever used in hs

[prisma coral]

make a perfect square

[prisma coral]

it is

[regal harbor]

oh wow

[prisma coral]

(x-1/x)^2 +2

it can be written as that

[regal harbor]

it can?

[prisma coral]

(x-1/x)^2 = x^2 +1/x^2 -2

[regal harbor]

this doesn't equal to this expression right

[prisma coral]

simolify it

[regal harbor]

alright i think i'll resort to using something like AM GM

[vapid sonnet]

You can see possible values of y is -2<=y<=2 which is range

[prisma coral]

why?

i literally told you x^2 + 1/x^2 = (x-1/x)^2 +2

and (x-1/x)^2 >=0

[regal harbor]

oh alright

i think i'll continue from here

thanks to you both

.close

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[clever citrus]

So I am trying to find the asymptotic runtime of an algorithm. I am kinda stuck on the last step. I found the formula for total number of operations which is

$T(n) = \sum_{i=1}^{n} \frac{(n-i+1)(n-i+2)}{2}$

Now I am not sure what to do next

[thorny flame]

Calc III Victim

[rigid ivy]

Write T(n) in closed form, then use master theorem

Or is T(n) your total runtime and putting it in closed form is where you are struggling?

[clever citrus]

right

acc give me a minute im going to try something again rq

wait a minute so I when I expand this I get

$\frac{1}{2} \sum_{i=1}^{n} (n^2 - 2ni + 3n + 1 - 3i)$

and $\sum_{i=1}^{n} n^2 = n \cdot n^3$ right so could I ignore eveyrhting else since surely this would be the dominant term

[thorny flame]

Calc III Victim

[clever citrus]

so we would have n^3/2 which would just be O(n^3) ?

or is this not correct

[rigid ivy]

The 2ni sum would also be O(n^3) btw

actually, it might even cancel out the n^3 terms leaving you with O(n^2)

[clever citrus]

how

2n * (n(n+1))/2 = n^2(n+1) = n^3 + n^2

oh ur right

so n^3 - n^3 + n^2

[rigid ivy]

maybe

maybe not

You'd need to work out the sum to be sure

[clever citrus]

n^3 - n^3 + n^2 + 3n^2 + n + (3n^2 + 3n)/2

5n^2 + n + (3n^2 + 3n)/2 = (13n^2 + 5n)/2

multiplied with 1/2 outside we get (13n^2 + 5n)/4

so the dominant term would be 13n^2/4 so its O(n^2) ?

@rigid ivy is this correct?

[rigid ivy]

nah I got something different

[clever citrus]

shit

where did I go wrong

[rigid ivy]

not sure how you got this

Okay, first, what is $\sum n^2$?

[thorny flame]

SWR

[clever citrus]

alr so this is how I got it

n^2 = n^3
2ni = 2n * (n(n+1))/2
3n = 3n^2
1 = n
3i = 3(n(n+1))/2

wait shit

my dumbass

added the last one

smh

[clever citrus]

it should have been

n^3 - n^3 + n^2 + 3n^2 + n - (3n^2 + 3n)/2

= (n(n-1))/2

right

so (n^2 - n)/2. O(n^2)

[rigid ivy]

Where's the i² term?

[clever citrus]

wym

there isnt any right

[rigid ivy]

There's an i term in both expressions, so their product should have an i² term, should it not?

[clever citrus]

oh tbh with you I put it in a calculator so it could simplify it for me faster and I got that

let me try again

[rigid ivy]

it probably did i²=-1

[clever citrus]

oh LMAOOO IT DID

fuck

alr so it should have been

n^2 - 2ni + i^2 + 3n - 3i + 2

right

$T(n) = \sum_{i=1}^{n} \frac{n^2 - 2ni + i^2 + 3n - 3i + 2}{2}$

[thorny flame]

Calc III Victim

[clever citrus]

there

alr lemme try this rq brb

n^2 = n^3
2ni = 2n * (n(n+1))/2
i^2 = (n(n+1)(2n+1))/6
3n = 3n^2
3i = 3(n(n+1))/2
2 = n

so it would be

n^3 - (n^3 + n^2) + (n(n+1)(2n+1))/6 + 3n^2 - 3(n(n+1))/2 + n

(n^3 + 3n^2 - n)/n

so O(n^3)?

@rigid ivy

[rigid ivy]

yes

[clever citrus]

tysm

.close

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[stray wave]

can someone plz help

[stray wave]

@everyone

[neon iron]

How far have you gotten so far?

[stray wave]

Nowhere

I'm at decomposition

And I have no idea what to do

[neon iron]

While I type this wdym decomposition- I believe I know by a different name but just making sure before I go on a tangent lol

[stray wave]

Oh yeah that's fine

Decomposition is just a method of factoring

[neon iron]

Okay what I thought just wanted to be sure lmao

[stray wave]

Yup

[neon iron]

Alright, so we have
126x - 3x^2 - 3x^3
And we know it's factors are
-3x(x - A)(x + B)
A and B being unknown values
Remember, factors can divide into their larger numbers perfectly- does knowing this help any?

[stray wave]

Not really lol

Could u please elaborate

[neon iron]

Lol that's fine, the main thing you need to notice to break this down is that we have a known factor of -3x, meaning we can divide that polynomial by -3x, which in this case will turn it into a quadratic that's much easier to decompose

After that it should just be your normal decomposing, you just had to do extra steps to get there

[stray wave]

Yes I already did that and right now I have -3x(-x^2 - x + 42

But I don't know what to do after this

[neon iron]

Remember to change sign when you divide negatives :p

And after this you can kind of ignore 3x and start finding your (x - A)(x + B)

Whats your first step in decomposing this?

[stray wave]

Is it possible you could tell me what the answer is and I could try reverse engineering it

[neon iron]

I could if you want to but I think that would take away some potential decomposing experience, also you only really got like 2 major steps left lol but since we're this close up to you

[stray wave]

I think I'll take the answer and worky way through it because I just need to understand the decomposing and I think I will be able to reverse engineer from there so I'll take the answer please

[neon iron]

(x - 6) should be your final answer

[stray wave]

Thanks and ur sure righr

[neon iron]

If it helps any cause idk if you could reverse engineer this lol
Your first step in decomposing should be to find 2 numbers that multiply to a*c, and also add to b

[stray wave]

Ok got it

Thanks

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@stray wave Has your question been resolved?

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[burnt swift]

<@&286206848099549185>

[weak karma]

Hi you need help?

[topaz sinew]

@burnt swift Has your question been resolved?

[topaz sinew]

@burnt swift Has your question been resolved?

[fallow igloo]

Have you factorized 32767?

[burnt swift]

,w factors 32767

[thorny flame]

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[fallow igloo]

Alr, so a is completed

[burnt swift]

what about b

@fallow igloo

<@&286206848099549185>

@static lava

[weak karma]

whats up

[burnt swift]

can someone

help my fat ass

[weak karma]

with what

[burnt swift]

fucking B man

I cannot solve it

[weak karma]

can you send it again

[burnt swift]

[weak karma]

you have to simplify the expression then separate the compound inequality into two inequalites

[burnt swift]

we need to use little fernat theorem most probably

[weak karma]

yeah

[burnt swift]

how to do it

<@&286206848099549185>

[weak karma]

I'm trying to

[burnt swift]

well

we need to check if this is a prime

32767

,w is 32767 prime

[thorny flame]

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[weak karma]

To find an integer ( x ) such that ( 1 < x < 2 \times 32767 - 1 ) and ( 2 \times 32767 - 1 ) is divisible by ( x ), let’s break it down:

Calculate ( 2 \times 32767 - 1 ): [ 2 \times 32767 - 1 = 65534 - 1 = 65533 ]
We need to find an integer ( x ) such that ( 1 < x < 65533 ) and ( 65533 ) is divisible by ( x ).
The factors of 65533 are the integers that divide 65533 without leaving a remainder.
To find such an ( x ), we can start by checking the prime factorization of 65533. However, for simplicity, one of the factors of 65533 is 65533 itself, which is the largest possible ( x ) in this range.

So, one possible answer is ( x = 65533 ).

That should help a little more

[burnt swift]

.solve

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[quiet dome]

hello

[quiet dome]

could someone help me with a problem

i dont understand this problem

.close

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[quiet dome]

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✅

[quiet dome]

hello

could someone help me with my math question

its also kinda thinking skills

[weak karma]

I think its A

[quiet dome]

can explain pls

wait no

on my answer sheet its D

but i dont understand it

[weak karma]

Okay give me one minute so I can write all down to help you understand

[quiet dome]

ok

thx

[weak karma]

Let’s break down the problem step by step:

Define Variables:
Let ( p ) be the cost of one pencil.
Let ( s ) be the cost of one sharpener.
Let ( x ) be the number of pencils Elsa bought.
Let ( y ) be the number of sharpeners Elsa bought.
Relationships:
The cost of one sharpener is $2 more than the cost of one pencil: ( s = p + 2 ).
Anna bought twice as many pencils and 10 fewer sharpeners than Elsa:
Anna’s pencils: ( 2x )
Anna’s sharpeners: ( y - 10 )
Total Cost:
Elsa’s total cost: ( xp + ys )
Anna’s total cost: ( 2xp + (y - 10)s )
Equal Spending:
Since Elsa and Anna spent the same amount of money: [ xp + ys = 2xp + (y - 10)s ]
Simplify the Equation: [ xp + ys = 2xp + ys - 10s ] [ xp = 10s ] [ x = \frac{10s}{p} ]
Substitute ( s = p + 2 ): [ x = \frac{10(p + 2)}{p} ] [ x = 10 + \frac{20}{p} ]
Minimum Integer Value:
For ( x ) to be an integer, ( \frac{20}{p} ) must be an integer.
Therefore, ( p ) must be a divisor of 20. The divisors of 20 are 1, 2, 4, 5, 10, and 20.
Calculate ( x ) for each divisor:
If ( p = 1 ), ( x = 10 + 20 = 30 )
If ( p = 2 ), ( x = 10 + 10 = 20 )
If ( p = 4 ), ( x = 10 + 5 = 15 )
If ( p = 5 ), ( x = 10 + 4 = 14 )
If ( p = 10 ), ( x = 10 + 2 = 12 )
If ( p = 20 ), ( x = 10 + 1 = 11 )
Minimum ( x ):
The minimum value of ( x ) is 11 when ( p = 20 ).
Total Pencils:
Elsa bought ( x = 11 ) pencils.
Anna bought ( 2x = 22 ) pencils.
Total pencils: ( 11 + 22 = 33 ).
So, the minimum possible number of pencils bought by Elsa and Anna together is 33.

That should help

[quiet dome]

Ok thx

i get it now

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[lusty cedar]

for 9b the solution says 3y - x = 0

I don't understand though

Because what I did was set the deravative equal to 0

you can't divide by 0 too

so I have no idea how 3y - x = 0

[unkempt viper]

Using derivarives to find this point isnt neccessaey, but notice that 3y-x = 0 doesnt set derivative to 0 it kinda sets it to infinity, which makes sense if you look a the picture. The tangent line at point P is vertical and derivative define its slope, so it makes sense that derivative is equal to +- infinity.

Its hard to explain with words, but imagine that the tangent line is rotatin counter clockwise, which you interpret as the derivative having a bigger value, it can never be completly vertical, but it will get close. So in infinity it will be vertical

[lusty cedar]

so can I say the denominator of the deravative will always equate to 0 in questions like these

cause it tends to inifity

infinity

[unkempt viper]

ummm its a dangerous path tbh

I dont see problems like this very often

[lusty cedar]

ohh wait that tangent line

is an asymptope

isnt it

thats why theres the infinity thing

[unkempt viper]

But for me its pretty intuitive, if derivative is 0 then the tangent is horizontal, a vertical tangent means that the function grows infinitely fast

[unkempt viper]

not exaxtly but thats how i explain this to my self

[halcyon lantern]

It just goes completely vertical, but basically yeah

[unkempt viper]

Im not sure how asymptotes work for multivadiable functions, but in single variable limit must be +-inf

here the limit is P which is not inifinity

[lusty cedar]

I think I'll just go with this

until I'm proven otherwise

[unkempt viper]

hahaha sure

[lusty cedar]

thanks lol

[unkempt viper]

With some practice you will gain some intuitiom on how infinity and stuff like this works.

[lusty cedar]

yeah

[unkempt viper]

Its confusing at first

[lusty cedar]

like reciprocal graphs

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[topaz sinew]

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

[harsh shell]

2

[fair thorn]

for the second integral

set e^-3x dx to dv

and u = 2x

then solve

$C-\int 2xe^{-3x} \dd x \to \int \underbrace{2x}{\text{\scriptsize{u}}} \underbrace{e^{-3x} \dd x}{\text{\scriptsize{\dd v}}}$

[harsh shell]

what method is this

[fair thorn]

IBP

[harsh shell]

??

[thorny flame]

ren

$C-\int 2xe^{-3x} \dd x \to \int \underbrace{2x}_{\text{\scriptsize{u}}} \underbrace{e^{-3x} \dd x}_{\text{\scriptsize{\dd v}}}$
```Compilation error:```! LaTeX Error: \mathrm allowed only in math mode.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.1421 ...^{-3x} \dd x}_{\text{\scriptsize{\dd v}}
                                                  }$
You're in trouble here.  Try typing  <return>  to proceed.
If that doesn't work, type  X <return>  to quit.```

[fair thorn]

integration by parts

[harsh shell]

is there no other way of doing this?

cuz i swear i havent learnt that method

[fair thorn]

i... think that's the most optimal method here?

[harsh shell]

alright thank oyu

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[hollow rose]

if you make the top equation 13 = 2xy

then you can add that to the bottom one

[ionic oar]

smart

[hollow rose]

to get x^2 +2xy + y^2 = 36 + 2xy

then (x+y)^2 =49

so x+y = +- 7

I've got these solutions so far using x+y = 7

[ionic oar]

wait why

[hollow rose]

But when i sub into x+y = -7 i get the wrong answer

[hollow rose]

why what?

[ionic oar]

you found x + y by adding 2xy

subtract 2xy and find x - y

you'll have linear equations

oh wait same thing nvm

[hollow rose]

any idea why this is?

[hollow rose]

these solutions are correct

[ionic oar]

did you plug it back in and check if they satisfy

sometimes you do get extraneous solutions

but since xy = k has two branches you should have 2 sets of solutions

each set with two values for x and two values for y

[topaz sinew]

@hollow rose Has your question been resolved?

[hollow rose]

i only plugged it into x + y = +- 7 because this is supposed to be a non-calc question

[ionic oar]

it's fine it's doable without a calculator

[tulip basin]

u can do it this way

the factors of that equation are x²-14x+13 and x²+14x+13

u can find the x solutions

and then substitute it to get y

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[hearty tundra]

Anyone know how to solve this (x^2) ln(x^2))=2ln(2)?

[lucid lagoon]

do uk any log properties

[torpid sparrow]

What did you try?

Notice you have this

A(ln(A)) = B(ln(B))

[hearty tundra]

Can you do anything with that?

[torpid sparrow]

Yes

You can deduce from here and prove with properties of log if u want that A = B

In your case, A = x^2 and B = 2

[hearty tundra]

really? how

[torpid sparrow]

If I say Aln(A) = Bln(B), using log properties you can say ln(A^A) = ln(B^B)

From here, do the inverse function.

e^(ln(A^A)) = e^(ln(B^B))
A^A = B^B
A=B

[hearty tundra]

oh i get it now Thanks a lot:)

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[coarse tusk]

this doesn't necessarily imply A^A=B^B though

the function f(x)=x^x is not a bijection

you can only conclude that here because B>1, and for x>1 f(x) is indeed a bijection

it's an important conclusion

[torpid sparrow]

Yes, i am using the context of the exercise

[coarse tusk]

well, if you had $x^x=(0.8)^{(0.8)}$ then there would be two distinct solutions and not one

[thorny flame]

kheerii

[hearty tundra]

Then you cant do that?

[coarse tusk]

you can here

.reopen

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✅

[hearty tundra]

Why

[coarse tusk]

the function f(x)=x^x is bijective if x>=1

but if x<1 you can have two different values

[torpid sparrow]

Because we hare working with ln2

[coarse tusk]

no we're working with 2

not ln2

[torpid sparrow]

I mean that

[coarse tusk]

$(x^2)^{(x^2)}=2^2$

[thorny flame]

kheerii

[hearty tundra]

But How can you Realise that function has more solutions If its smaller than 1 with out graphing

[coarse tusk]

I know what the graph of x^x looks like

,w plot x^x for -0.5<=x<=2

[thorny flame]

Results provided by WolframAlpha

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

Requested by shikhargupta1207

[hearty tundra]

What If you don’t know?

[coarse tusk]

well, you can differentiate the function to check

[hearty tundra]

And why negative values are not aceptable?

[coarse tusk]

negative values aren't in the domain of x^x

well technically you can say negative integral points are

but they give us negative values anyway so it doesn't matter

[placid lance]

not necessarily

[coarse tusk]

..?

[hearty tundra]

Okay thanks then is this the only way to do it or are there any other ways to solve that equation?

[coarse tusk]

oh well if we have negative even values

it becomes positive

[placid lance]

there's $(-2)^{-2}=0.25$

[thorny flame]

nameless individual

[coarse tusk]

but that's still less than 1

[coarse tusk]

no that method is correct

[placid lance]

$\left(-\frac23\right)^{-\frac23}$

[thorny flame]

nameless individual

[coarse tusk]

I was just pointing out that saying x^2=2 does not necessarily give you all the solutions

[placid lance]

you should just force $x>0$

[thorny flame]

nameless individual

[coarse tusk]

this isn't real

$\sqrt[3]{x}\ne x^{\frac1{3}}$

[thorny flame]

kheerii

[coarse tusk]

not for negative values anyway

for positive values it is

[placid lance]

welp

[placid lance]

but still

[coarse tusk]

yeah we usually do do that

[hearty tundra]

I didnt say it wasnt corect i just asked If there are any more ways to do it or is this the only way

[coarse tusk]

it's implied here because it's (x^2)^(x^2)

[placid lance]

only if you choose the principal value conventionally

[coarse tusk]

i mean, this is the easiest

and probably the only reasonable method

[placid lance]

there's no "only way"
you can always go on a shopping spree before you solve the problem

[torpid sparrow]

I have read that before 🥔

[placid lance]

from me? or where

[torpid sparrow]

Yes , from u lol

[hearty tundra]

How can you do that?

[hearty tundra]

If you differentiate you get xx^(x-1) right? And that sounds pretty useless

[placid lance]

what's xx^(x-1)

latex

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@hearty tundra Has your question been resolved?

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[inland minnow]

Hey Guys
How do I find the derivative of i(x)=17 * x^2 * ln(32x) ?

[fair pulsar]

$f(x)=17x\ln{(32x)}$

[thorny flame]

sealpup321

[fair pulsar]

sorry just to make it easier for me to read

ok so use the product rule with $u=17x$ and $v=ln(32x)$

[thorny flame]

sealpup321

[fair pulsar]

to differentiate $v=ln(32x)$ use the fact that $ln(x) = \frac{1}{x}$

[thorny flame]

sealpup321

[inland minnow]

I tried to use a calcultor for this part, it says 1/x

[placid lance]

to differentiate $v=\text{ln}(32x)$ use the fact that $\frac{\text{d ln}(x)}{\text{dx}} = \frac{1}{x}$

[thorny flame]

Your LaTeX Helper

[inland minnow]

How do I use the product rule for this??

[placid lance]

its the chain rule, then the product rule

[inland minnow]

Is the chain rule for the ln(32x) ?

[placid lance]

yes

[inland minnow]

For the ln(32x) with the chain rule, I got it to be 1/x

How do I use the product rule for the rest?

[fair pulsar]

this is wrong

the formula you want to use for chain rule is $u^{\prime} f(u)$

[thorny flame]

sealpup321

[fair pulsar]

sorry I dissapeared earlier had dinner

[inland minnow]

how? (haven't really seen this)

[fair pulsar]

have you learnt chain rule?

[inland minnow]

not much of it

[fair pulsar]

oh sorry i did make a mistake

$u^{\prime} f^{\prime}(u)$

[thorny flame]

sealpup321

[fair pulsar]

anyway use this and tell me the answer

https://www.mathsisfun.com/calculus/chain-rule.html

https://www.mathsisfun.com/calculus/product-rule.html

@inland minnow you good?

[inland minnow]

I actually don't quite understand how to use the rule you sent me

[fair pulsar]

thats fine lets take it a little slower then

worst thing you can do is get stuck and start to slow down or procrastinate

ok... wanna vc?

[fair pulsar]

fingers are tired from typing

[fair pulsar]

wait never mind I thought there were vc calls in this server

are you still there @inland minnow

[inland minnow]

yes

[fair pulsar]

coolio

so you will want to use the chain rule when you have something trapped in a function

eg. $(x^2+4x+4)^5 \ cos(3x+2) \ ln(32x)$

[thorny flame]

sealpup321

[fair pulsar]

these are all examples on when you would want to use chain rule

basically if something is trapped in brackets you probably want to use chain rule

[inland minnow]

Every calculator I have used gives me ln(32x) as 1/x (for the derivative)

[fair pulsar]

when you do this you have to recognise whats in the brackets as u and whats outside and f(u)

[fair pulsar]

cool and?

as I was saying

[fair pulsar]

tell me what these are in regards to ln(32x)

[fair pulsar]

awww piss you got it here

I missread sorry about that

anyway you dont seem confident work me through what you did and then I'll help you with product rule

[inland minnow]

it's okay, I may have learned something new from that also

[fair pulsar]

phew 🙂

[inland minnow]

The only thing I have done until now is that I tried differentiating 17 and ln(32x), whereas I got 1/x to be the derivative for ln(32x) and differentiating 17, I got 0

[fair pulsar]

$17x^2$ not 17

[thorny flame]

sealpup321

[fair pulsar]

but what I want you to tell me is how you used the chain rule I want to make sure you aren't just using an online calculator

[inland minnow]

Ohh, that's one thing I was stuck on

[fair pulsar]

do you have the answer?

[inland minnow]

not yet

[inland minnow]

I don't know if it's correct

[fair pulsar]

ok I am a little confused?

first whats g(x) and f(x)?

instead of writing stuff down I have to go soon so just message here

[inland minnow]

ok

[fair pulsar]

and I'll go through it

ok lets split the equation into 2

$17x^2 and ln(32x)$

[thorny flame]

sealpup321

[inland minnow]

yes, that's what I was trying to do

[fair pulsar]

ok now

lets start with ln(32x)

$u = 32x \ f(x) = ln(u)$

[thorny flame]

sealpup321

[fair pulsar]

yes?

do you understand so far?

[inland minnow]

yes

[fair pulsar]

ok so using the formula $u^{\prime}f^{\prime}(u)$

[thorny flame]

sealpup321

[fair pulsar]

what do you get?

actually no stop

whats $u^{\prime}$?

[thorny flame]

sealpup321

[fair pulsar]

hello?

well if you aren't here then Imma go

[inland minnow]

I’m here, just needed To tell you I work with this table

[fair pulsar]

ok thats an awful table but not important

wow

no nevermind I hate that table

it doesn't explain anything and just gives you a ton of rules

and it got sin(x) wrong

alright to differentiate 32x

[inland minnow]

yeah, i know i fixed it to -sin

[fair pulsar]

no i mean if you differentiate sin you get cos not -cos

it got them the wrong way round

[fair pulsar]

.

[inland minnow]

It's 32

[fair pulsar]

good

so $u^{\prime}=32$

[thorny flame]

sealpup321

[fair pulsar]

now differentiate ln(u)

for now pretend that u is an x

[fair pulsar]

its on this sheet

[inland minnow]

ln(32) would be 1/32, right?

[fair pulsar]

just tell me what you get if you differentiate ln(u)

1/u

[inland minnow]

1/u

yes

[fair pulsar]

cool

that is $f^{\prime}(u)$

[thorny flame]

sealpup321

[fair pulsar]

so multiply 32 and 1/u to get $u^{\prime}f^{\prime}(u)$

[thorny flame]

sealpup321

[fair pulsar]

do you have the answer?

[inland minnow]

32/u

[fair pulsar]

yup

now you want it with x not u

so replace u what its equal to

what do you get?

[inland minnow]

and that's 1/x

[fair pulsar]

yup there we go

now lets work on 17x^2

what do you get if you differentiate x^2?

[inland minnow]

2x

[fair pulsar]

now multiply that by 17

and?

[inland minnow]

34x

[fair pulsar]

good

cool so we now have the four parts we need to differentiate using the product rule

$u = ln(32x) \ u^{\prime} = \frac{1}{x} \ v = 17x^2 \ v^{\prime} = 32x$

[thorny flame]

sealpup321

[fair pulsar]

yes?

yes or no?

[inland minnow]

I wan't say yes, but not quite sure (mostly yes)

[fair pulsar]

ok thats good

I am giving them names in a way

u is the abbreviated form of ln(32x)

and the $^\prime$ just means I have differentiated it

[thorny flame]

sealpup321

[fair pulsar]

what confuses you?

mate i need to sleep soon

[inland minnow]

Can I use this?

[fair pulsar]

the product rule is $vu^{\prime} + uv^{\prime}$

[thorny flame]

sealpup321

[inland minnow]

sorry

[fair pulsar]

yes its the exact same

just instead of g(x) and h(x) i have use u and v

but they mean the same

[inland minnow]

ok

I think i can figure out the rest

[fair pulsar]

thats the product rule

[fair pulsar]

alright ill give you a minute

[fair pulsar]

$g(x) = ln(32x) \ g^{\prime}(x) = \frac{1}{x} \h(x) = 17x^2 \ h^{\prime}(x) = 32x$

[thorny flame]

sealpup321

[fair pulsar]

made 1 minor mistake

not 32 its 32x

[inland minnow]

oh yea

[fair pulsar]

but otherwise well done!

you did it

by the way this is quite advanced math to be doing (I imagine you are in the german equivalent of gcse) are you in a gesamtshule?

oder gymnasium?

[inland minnow]

lol, no I am in a danish high school in my first year (we are supposed to do this on our second year)

[fair pulsar]

ah fuck I thought it was german on your sheet earlier

lmao

[fair pulsar]

fair enough

anyway that sheet makes things way more complex then it needs to be

I would recommend looking into differentiating in your own time using youtube or something

as that sheet could easily cause a ton of misunderstandings

anyway goodnight

[inland minnow]

good night

Thank you very much for your time, I really appreciate it

[fair pulsar]

no worries and goodluck o7

[inland minnow]

thanks

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[gaunt thunder]

can i get the b in the y=mx+b without having the y or the x ?

[cinder sequoia]

without more context your question doesn't make too much sense

can you get b given points? a graph? an equation?

[carmine lynx]

b the the intersection of the graph with the y axis

without y ig it would not be techincally possible

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[haughty ember]

I simplified this into vertex form as 4(x - 4)^2 + 9 is this correct?

[cinder sequoia]

if you try expanding 4(x-4)^2 + 9, you get something different from y

so that's not the correct vertex form

try factoring the 4 out and manipulate to get something that looks like a perfect square

y = 4(x^2 - 2x) + 5

if that was x^2 - 2x + 1 in the parentheses, you'd have a perfect square

[haughty ember]

Is it y(x-1)^2 - 11?

[cinder sequoia]

4(x-1)^2 - 11?

[haughty ember]

Yeah

my bad, I typed it badly

[cinder sequoia]

almost, double check the constant

that expands to 4x^2 - 8x + 4 - 11

and you want it to be 4x^2 - 8x + 5

so instead of -11, what should it be?

[haughty ember]

+1

[cinder sequoia]

yes

4(x-1)^2 + 1 = 4x^2 - 8x + 5

[haughty ember]

Thank you neil

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✅

[haughty ember]

Hold up neil, I am significantly confused right now

Could you work the problem out for me so I can see the process???

If anyone could work the problem process out for me that would be great

[placid lance]

$4(x-1)^2+1=4(x^2-2x+1)+1=4x^2-8x+4+1=4x^2-8x+5$

[thorny flame]

Your LaTeX Helper

[placid lance]

Do you need help on expanding $(x-1)^2$?

[thorny flame]

Your LaTeX Helper

[placid lance]

@haughty ember

[haughty ember]

Oh, I don’t get how it works

As in the how to get the constant values

And how to establish the -1 value

Because usually when a = 1 you halve b value and add/subtract

[placid lance]

The $-1$ value can be obtained by the formula $\frac b{2a}$.

[thorny flame]

Your LaTeX Helper

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[haughty ember]

Thankyou

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[shrewd birch]

i couldnt get the last two lines like how did the no. of solns become nine...how did we get pi/24, pi/8 etc

[verbal crater]

read equation (ii) and equation(iii) and their conditions for n

[shrewd birch]

yes

then the 5pi/24 qand 7pi/24

like how do we know we have to take those

[verbal crater]

look at your interval

[neon iron]

it states the interval

[verbal crater]

if its in your interval, take it

if not, dont

[shrewd birch]

like how to we know till what number we can include in the interval

like how many values of n can i assume and which one will be the last value of n in the interval

[verbal crater]

just take as many as you can

then stop once you realise its out of your interval

[oblique moat]

hey can any one help me tomorrow at 8:30 with some Hw it’s an online class pls i would do it but im out of time and my class finishes tm.

[neon iron]

I know it is not my questions but can someone explain how we got the equation (i)

[shrewd birch]

why do we not have pi/12?

[shrewd birch]

its a formula for general solutions of trig eqns

it says if sin x= sin alpha
then x= npi+(-1)^n alpha

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[brittle void]

Explain the distinguishing feature of the distance between the foci and a point on the ellipse.

[brittle void]

does this explanation make sense?

is it accurate i mean

The distinguishing feature of the distance between the foci and a point on the ellipse is that the sum of the distances from the point to each focus remains constant for all points on the ellipse. This constant sum is equal to the major axis length of the ellipse.

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[minor wharf]

hi

[minor wharf]

In the figure above, tanB= 3/4. If side BC=15 and side DA=4, what is the length of DE?

[severe eagle]

@minor wharf 6

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@minor wharf Has your question been resolved?

[minor wharf]

why

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@minor wharf Has your question been resolved?

[severe eagle]

@minor wharf

[minor wharf]

.

[minor wharf]

but why multiplying BA with 4/5 to find DB and BA with 3/4 to find DE

and what is the value we are multiplying with 4/5 to find BA?

[severe eagle]

we find DB by subtracting the length of BA by 4, because length of DA is 4
We don't multiply BA by 3/4 to find DE, but we can multiply BD by 3/4 to find DE

[severe eagle]

Because we know there exists a ratio of 3:4:5 for the sides AC:AB:BC
so we can find the length of side AB (or BA) by multiplying BC by 4/5

does that answer your question?

[minor wharf]

hey

[minor wharf]

Nah

Where can I learn the full context from, what are the rules called?

Bro left the server

Anyway thnx for the help 😭😭?..

Btw why do people keep leaving this server, 😭 it's not the first time to see this happening

[minor wharf]

I think I get it now, but we don't even know BD to multiply it by 3/4 is what I'm telling you

[minor wharf]

Okay so that's to find BD?

Basically this (4/5)=(x/15) no ?

4x15= 60

60/5=12

So BD=12 ?

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[heady goblet]

need help with part c

[rose sierra]

!status

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3. I got an answer but I was told that it's wrong.
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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

[heady goblet]

how did it travel 10 m?

shouldnt it be 5m there?

it went from 15 to 10m

in that ime period

time

[rose sierra]

I think you're right it should be 5

[heady goblet]

yh i think its a mistake

i think the rets of the answer is correct

rest

[rose sierra]

ye

[heady goblet]

thank u

[placid lance]

If that answers your question, please type .close to allow other people to use the help channels.

[heady goblet]

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[alpine dirge]

The line where $$ \alpha (\alpha - 1) ... (\alpha - n + 1) (1 +x)^{\alpha - n} = \alpha^{n} (1 + x)^{\alpha - n} $$ doesn't make sense to me

[thorny flame]

sed

[odd juniper]

why not

the entire line?

It's just generalising what you had above

plug in n=2,1,0 etc

do u see?

[alpine dirge]

I get the generalisation but not the equality here

[odd juniper]

which equlaity?

[alpine dirge]

the one i sent

[odd juniper]

that just seems like the factorization identity

Oh

I just realised

I think $a^{\underline n}$ is the falling factorial

[thorny flame]

normalAtmosphericPa=101,325

[alpine dirge]

Oh I didn't notice the underline

(why did they make it so fucking transparent)

[odd juniper]

neither did I

yeah

[alpine dirge]

Thanks anyway, massively helped

[odd juniper]

but if it's from wikipedia, they generally do shitty notation like that

[alpine dirge]

Proofwiki

[odd juniper]

yeah similar

[alpine dirge]

But i mean for all the other brilliant proof they've done i still love it

❤️

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[shell shard]

This is a very famous problem

https://en.wikipedia.org/wiki/Langley's_Adventitious_Angles

[neon iron]

@shell shard

Acc to the book I have the answer is 45

[shell shard]

Hi

[neon iron]

I don't understand how

[shell shard]

Book is wrong

Many such cases

[neon iron]

Thanks 👍

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[severe eagle]

see https://www.khanacademy.org/math/geometry/hs-geo-trig/xff63fac4:hs-geo-trig-ratios-similarity/a/side-ratios-in-right-triangles-as-a-function-of-the-angles

[severe eagle]

BD is found by subtracting the BA by 4, because the length of DA is 4

[minor wharf]

so BD is 8

okay

[severe eagle]

yeah

and using that info you can find DE

[minor wharf]

I found BD is 12 because I used this equation: (4/5)=(x/15)

[minor wharf]

okay

[severe eagle]

yup you got it

[minor wharf]

dyk how?

[minor wharf]

[severe eagle]

You can multiply BD by 3/4 to get DE
this is because the inner angles of triangle BDE and BAC are the same, therefore the ratio for side lengths are also the same

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[mortal sphinx]

Hi

[mortal sphinx]

Determine parametric equations for the line through (-2, 3) and parallel to the line with vector equation Vector I = (−2, 1) + t(6, 4).

How can i do this

[acoustic pecan]

write the line in vector form, then just write the equations for x and y

[mortal sphinx]

Wait so I do tip - tail of (-2,3) and (-2,1)

The answer i get would be [x,y] = [tip-tail ans] + t[6,4]

Then i get the parametric form

Atleast tell me if i am right or wrong

Why should i care abt the slope all I need is the parametric equation

@acoustic pecan Come back

I need you too

[acoustic pecan]

it would be (-2,3)+t(6,4)

[mortal sphinx]

What abt the (-2,1)

[acoustic pecan]

its irrelevant

[mortal sphinx]

oh

Thank you

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[grim inlet]

How do I find their intersections by solving for either x or y? I have tried but I keep getting quartic equations and I don't know where to go from there

[coarse tusk]

yes it does seem like you get a quartic solution

!xy

[topaz sinew]

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

[grim inlet]

Find the area of the region bounded by x^2=y+1 and y^2+2x+1-x=0

[coarse tusk]

why is it 2x+1-x?

that doesn't seem right

[coarse tusk]

huh, weird

[grim inlet]

yes i've seen the graph and im stuck at y(y^3+2y-1)=0

[coarse tusk]

this really doesn't seem right to me..

[jade thunder]

You can solve depressed cubics like that

[grim inlet]

howww

[coarse tusk]

if you're actually expected to find these then god help you

[grim inlet]

Oml

[coarse tusk]

maybe ask the person who assigned the question to you if the data in the question is wrong

[jade thunder]

This looks bad tho

[grim inlet]

allright, but is there not any other way to solve for the intersections?

[coarse tusk]

any way you do it you have to solve a quartic or a cubic or something like that

or you could parametise both the parabolas and solve it that way maybe

[grim inlet]

that wouldnt be a clean solution though right?

[coarse tusk]

if the numbers are nice then it could be

[jade thunder]

(x² -1)² + x + 1 = (x-1)²(x+1)² + (x+1) = (x+1)((x - 1)²(x+1) + 1) = 0

Does that look any easier

Solving (x-1)²(x+1) + 1

[grim inlet]

oohh but how will I go from there

[jade thunder]

,w expand (x-1)^2(x+1) + 1

[thorny flame]

Results provided by WolframAlpha

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Requested by frosst

[jade thunder]

,w factorise y^3 + 2y - 1

[coarse tusk]

very nice