help-39

[plush bramble]

Doesn't seem likely

[sharp vigil]

a couple of the proofs in the wikipedia page use only elementary calculus (if you agree to brush all the convergence issues under the rug), but they tend to have some slightly nasty algebra along the way

[pearl pond]

@final citrus Has your question been resolved?

[pearl pond]

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[magic ruin]

Can anyone prove this to me..
(I am beginner in coordinate geometry)

[cobalt hinge]

We can use some trigonometry

Let me whip out a diagram

The red is the line we are analyzing, the green is the perpendicular. I included the legs in order to use the trigonometry. Now, what I want you to do is two things:

• Tell me what alpha is in terms of theta
• Tell me the slope of the red and green lines, respectively, in terms of trig functions of theta.
  I gotta go so I hope someone else picks up on this.

[magic ruin]

I have no idea what it is...sorry

[autumn fossil]

there are other ways to prove it

[magic ruin]

please

[autumn fossil]

okay so imagine we have a line

what would the slope be in this case?

[magic ruin]

the hypotenuse?

[autumn fossil]

what's the numerical value of the slope though?

express it using x and y

do you know what does slope mean?

[magic ruin]

a little bit..

it is the line that is crossing x and y(x or y) axis

[autumn fossil]

slope as a number

slope is in some way a property of line

it can be any number

what does that number represent?

would you be able to guess the slope of this line?

yes / no

[magic ruin]

I am completely new to this..

[autumn fossil]

https://www.khanacademy.org/math/algebra-basics/alg-basics-graphing-lines-and-slope/alg-basics-slope/v/introduction-to-slope

then review this

It's a pretty good vid

[magic ruin]

ok

[autumn fossil]

there are several other vids under that one, which you can watch as well

and there is an exercise as well

[magic ruin]

ok i will watch..thanks

.close

[pearl pond]

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[autumn fossil]

https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry

this has also analytic geometry

[pearl pond]

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[woven dove]

أريد أن أسأل عن (التكامل) و (الإقترانات الأسية و اللوغاريتمية)،
سؤالي عنها بشكل عام ، ما هي ، القوانين المتعلقة بها، و أمثلة عنها...

[pearl pond]

@woven dove Has your question been resolved?

[woven dove]

<@&286206848099549185>

[plush bramble]

you might want to translate to english if you want help sooner than later

[robust iris]

积分是微分的逆过程，通过导数求原函数。它计算曲线下的面积，并解决累积问题，例如距离与速度的关系。关键规则包括幂律 (∫xⁿ dx = xⁿ⁺¹/(n+1) + C) 和复函数的代换。

指数函数用于模拟增长/衰减 (f(x)=aˣ)，这在人口增长和放射性衰变中很常见。特殊情况 eˣ 是其自身的导数，这在微积分中至关重要。

对数函数（指数的逆函数）用于解指数方程。自然对数 (ln x) 和常用对数 (log₁₀ x) 将乘法转换为加法，用于分贝和 pH 值标度。

shit wrong lang

1s

[woven dove]

لا أعرف ، إستخدم مترجما

[robust iris]

التكامل يعكس التفاضل، بإيجاد الدوال الأصلية من مشتقاتها. يحسب التكامل المساحات تحت المنحنيات ويحل مسائل التراكم، مثل المسافة من السرعة. تشمل القواعد الأساسية قاعدة القوة (∫xⁿ dx = xⁿ⁺¹/(n+1) + C) والتعويض للدوال المركبة.

تُمثل الدوال الأسية النمو/الاضمحلال (f(x)=aˣ)، والتي تظهر في النمو السكاني والاضمحلال الإشعاعي. الحالة الخاصة eˣ هي مشتقتها الخاصة، وهي أساسية في حساب التفاضل والتكامل.

تحل الدوال اللوغاريتمية (معكوسات الدوال الأسية) المعادلات الأسية. يُحوّل اللوغاريتم الطبيعي (ln x) واللوغاريتم المشترك (log₁₀ x) عملية الضرب إلى جمع، ويُستخدمان في مقاييس الديسيبل والرقم الهيدروجيني.

[woven dove]

أنا لا أفهم

[plush bramble]

neither of those look english. also don't answer for OP

[robust iris]

Wym by "OP"?

[woven dove]

شكرا ، هل لديك صور للقواعد في الرياضيات البحتة؟

[robust iris]

القوانين الأساسية:

• الأسس: aˣ⁺ʸ = aˣaʸ، (aˣ)ʸ = aˣʸ
• اللوغاريتمات: log(xy)=logx+logy، log(xʸ)=ylogx
• التكامل: قاعدة القوى، قاعدة eˣ (∫eˣdx=eˣ+C)، التعويض

أمثلة:

• التمويل: الفائدة المركبة A=Peʳᵗ
• الكيمياء: pH=-log[H⁺]
• الفيزياء: ∫v(t)dt تعطي المسافة

All done?

[desert birch]

this is litteraly something you can study or search up

[woven dove]

I hope

[woven dove]

If that was all thing , why this server?

[desert birch]

wdym lol , who speaks that

[robust iris]

English: Integration reverses differentiation, finding original functions from their derivatives. It calculates areas under curves and solves accumulation problems, like distance from velocity. Key rules include the power rule (∫xⁿ dx = xⁿ⁺¹/(n+1) + C) and substitution for complex functions.

Exponential functions model growth/decay (f(x)=aˣ), appearing in population growth and radioactive decay. The special case eˣ is its own derivative, crucial in calculus.

Logarithmic functions (inverses of exponentials) solve exponential equations. Natural log (ln x) and common log (log₁₀ x) convert multiplication to addition, used in decibel and pH scales.

Key Laws:
- Exponents: aˣ⁺ʸ=aˣaʸ, (aˣ)ʸ=aˣʸ  
- Logs: log(xy)=logx+logy, log(xʸ)=ylogx  
- Integration: Power rule, eˣ rule (∫eˣdx=eˣ+C), substitution  

Examples:  
- Finance: Compound interest A=Peʳᵗ  
- Chemistry: pH=-log[H⁺]  
- Physics: ∫v(t)dt gives distance

[robust iris]

Me

Ppl speak Chinese more than english

[woven dove]

الترجمة مقرفة

[blissful cloak]

native yes, otherwise it actually is English

[woven dove]

ألا يوجد عربي إلا انا؟

[desert birch]

you'd get more luck asking in english here

but you can also ask in your native lang and someone who speaks it may show up

type .close

if your question is answered

[woven dove]

. close

.close

[pearl pond]

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[pearl pond]

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[grim phoenix]

Is this adequate proof

[pearl pond]

@grim phoenix Has your question been resolved?

[grim phoenix]

<@&286206848099549185>

Also B is the answer to this

[hazy bear]

The last step b.c=b.d → c=d isn't correct

[grim phoenix]

can't we divide by b?

[hazy bear]

'I think' because it's assuming that c vector and d vector both are making equal angle with b vector

[hazy bear]

No b is not a number

[toxic lichen]

no thats nonsense

[hazy bear]

It's a vector

[toxic lichen]

scalar product is not multiplication

[grim phoenix]

mb

was it correct up until there

[hazy bear]

Yes there was nothing wrong before that

There should be some other way to prove it.

[grim phoenix]

Is what i've written true, but not usable for the proof?

[hazy bear]

Yes

[grim phoenix]

damn

<@&286206848099549185> how would I prove this

wait

if the OB is the perpendicular bisector

that means angle at vertex C and A are the same

right?

so then let those angles be k

|a| |c| cosk = |a||d|cosk

then dividing |a| and cos k each side we get |c| = |d|?

would that be ok?

[hazy bear]

Yes

But how did you get that angle at A, C is same?

It's same as saying that AO=CO because it looks like that

You are supposed to prove that thing with vectors

You can't use a geometric fact for that

[grim phoenix]

is it not always the case

[hazy bear]

Do you know the theorem that says if two angles of a triangle are same then the opposite sides are also same?

Assuming angles A,C are same is the exact same as assuming AO=CO

You can't assume what you are supposed to prove

[grim phoenix]

it is assumed though that OB is the perpendicular bisector

using that is it possible to show that angle C and angle A are the same

[hazy bear]

But I think there should be some other way because this way we aren't using any Vector to prove anything, it's just geometry

Lol how I am proving your way wrong but myself don't know how to prove it properly

[grim phoenix]

<@&286206848099549185>

[hazy bear]

Ayo i get it

a/2 = b-d = c-b
→ 2b=c+d
→ b=(c+d)/2

Now, b.a=0
[(c+d)/2].(c-d)=0

You were supposed to do this

[grim phoenix]

oh ok

u get difference of two squares

and the square of a vector is the magnitude

( c^2 - d^2 )/2 = 0

[hazy bear]

It is given that OB is perpendicular
→ OB.AC=0
→ b.a=0
→ b(c-d)

And that B is midpoint
→ AB=AC
→ a/2=a/2
→ b-d=c-b

[grim phoenix]

c^2 -d^2 = 0

[hazy bear]

Yes

[grim phoenix]

|c|=|d|

[hazy bear]

yes

[grim phoenix]

isoceles

[hazy bear]

That's the correct way

[grim phoenix]

ight thanks

[hazy bear]

Ok bye gl

[grim phoenix]

thanks

.close

[pearl pond]

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[simple eagle]

Can someone explain the 10th one

[simple eagle]

The ans is c

[versed mica]

which one do you think it should be instead?

[simple eagle]

None of these 😭

[versed mica]

why?

what do you think it should be then

[versed mica]

that says +

[toxic lichen]

you did e^x + e^(-x)

[versed mica]

the question says -

[simple eagle]

Ohhhhhhhhh

I still don't know hwo to do it

[versed mica]

,w 2sinh(x)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by kniefy

[simple eagle]

Huh

[versed mica]

don’t worry about that

[simple eagle]

Okka

[versed mica]

sinh x is defined as (e^x - e^(-x))/2

[simple eagle]

Whattt----

Where

How

[versed mica]

don’t worry about it you can read about it after

hyperbolic trig functions

[simple eagle]

Okay

[karmic rune]

At x->infinity, we have f->infinity, for x->(-infinity) we have f ->(-infinity), by intermediate value theorem, all the values in between are also in the range because f is a continuous function

[versed mica]

do you know IVT?

[simple eagle]

No what's that

Inverse trig?

[versed mica]

no

[simple eagle]

Oh i don't know intermediate value theorem

[versed mica]

intermediate value theorem says that for a continuous function f on an interval [a,b], every value between f(a) and f(b) is hit at least once

[simple eagle]

Ohhh

[versed mica]

so for any y so that f(a) <= y <= f(b), there exists a c in [a, b] such that f(c) = y

[simple eagle]

What do u mean by hit at atleast once

[versed mica]

this

[simple eagle]

Okag so basically every thing that lies between a to b has image

Something like one one function?

[karmic rune]

Not one one more like onto

[simple eagle]

Ohhh okay ic

[versed mica]

no that’s not really the idea this is just true because we’re assuming it’s a function defined on the interval [a, b]

[simple eagle]

Okay okay

[versed mica]

think of it like you have to cross every y value between the the y values at the two endpoints along the journey through the interval

[simple eagle]

Okayy...

[versed mica]

honestly the simplest thing here is process of elimination but you can in fact produce an inverse

[simple eagle]

Can u tell me the elimination part too

[versed mica]

$y = e^x - e^{-x} = \frac{e^{2x} - 1}{e^x} \iff e^{2x} - ye^x - 1 = 0$

[jolly parrot]

knief

[simple eagle]

Can i draw a graph of this?

[simple eagle]

Okayy

Then solve it like a quadratic

[versed mica]

actually i think it’s probably easiest to notice that it’s odd

[simple eagle]

But how will that help me find the range

[versed mica]

so choices a and b are immediately wrong

because f(-x) = -f(x) if we have positive values in the range we must also have negative values and vice versa

immediately eliminates a and b

does this make sense

[versed mica]

yes you can do this

and solve for x

in terms of y

,w arcsinh(x)

[jolly parrot]

Results provided by WolframAlpha

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Requested by kniefy

[versed mica]

annoying

[simple eagle]

Or should i take the vertices of the quad eqn

[simple eagle]

Damnnnn it took me a minute to process that but its actually so smart

[versed mica]

🤝🏻

[simple eagle]

Thankk youu

[versed mica]

you’re welcome sir

[simple eagle]

.close

[pearl pond]

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[cloud maple]

I require some help with verifying my proofs to Linear Algebra questions

[cloud maple]

I shall type them out here:

[jolly parrot]

Klein Bottle

[cloud maple]

seems pretty flipping obvious but oh well

(very similar solution for the second btw)

and then last one is asl follows

[toxic lichen]

that's normal

[vital crescent]

im assuming here -x is defined as the element such that -x+x=0 right?

[cloud maple]

yup yup

I just started linera algebra and some of the proofs just seemmm......trivial

like in high school, we just took this stuff for granted

anyways proof 3:

[vital crescent]

you probably meant $0=0\cdot x+0 \cdot x-0\cdot x=0\cdot x$

[jolly parrot]

qwertytrewq

[vital crescent]

for the second last line

[cloud maple]

$$(-1) \cdot x = -(1) \cdot x = -(1 \cdot x) = -x$$

[jolly parrot]

Klein Bottle

[cloud maple]

and the axiom (1 \cdot x) = x is already given

[vital crescent]

it more or less is a intro to proof as well. just rebuilding things from high school from scratch

[cloud maple]

gotcha

[cloud maple]

Would this one be correct

?

[vital crescent]

$(-1)\cdot x= -(1\cdot x)$ is not clear

[jolly parrot]

qwertytrewq

[cloud maple]

and we move from the second equal to sign to the third equal to sign because of the property:
$$(\lambda \mu) \cdot x = \lambda (\mu \cdot x); \lambda, \mu \in \mathbb{K}, x \in E$$

[jolly parrot]

Klein Bottle

[vital crescent]

$ -(1) \cdot x$ just means $-(1\cdot x)$ unless i misinterpreted what you mean

[jolly parrot]

qwertytrewq

[cloud maple]

I think it does

seems pretty weird though

its just like saying it is cause it is 😅

[vital crescent]

yes that is right, but $\lambda$ is a number, not the "negative sign"

[jolly parrot]

qwertytrewq

[cloud maple]

that is true

how do you reckon I write the second and third step?

[vital crescent]

remember how we define -x. there is only one axiom related to it.

so, if we oughta invoke that axiom somehow

[cloud maple]

$$-x + x = 0$$

this one?

[jolly parrot]

Klein Bottle

[cloud maple]

gotcha

so lemme see

[vital crescent]

the proof (at least the one i have in mind) is less straight forward than the other ones

[cloud maple]

$$(-1) \cdot x$$
by $x + 0 = x$
$$= (-1) \cdot (x + 0)$$
by $-x + x = 0$
$$= (-1) \cdot (x + x - x)$$
and then perhaps something from here?

[jolly parrot]

Klein Bottle

[cloud maple]

and then the distributivity property perhaps?

[vital crescent]

well try it out

it is a good try

[cloud maple]

$$ = -x - x + x = -x$$

[jolly parrot]

Klein Bottle

[cloud maple]

I feel like

the problem with my reasoning is that when I am doing the distributive property

I am RELYING on teh fact that (-1) * x = -x

[vital crescent]

we have not established that (-1)\cdot (x+x-x)=-x-x+x though!

[cloud maple]

exactly!

[vital crescent]

also I kinda recommend at first write minus x as +(-x)

just so it is clear what it means

[cloud maple]

ahh okay

[vital crescent]

that is how minus is defined

[cloud maple]

$$(-1) \cdot (x + x + (-x))$$
$$=-1x -1x + -1(-x)$$

[jolly parrot]

Klein Bottle

[cloud maple]

would this be fine?

AHH THIS SEEMS SO TRIVIAL 😦

[vital crescent]

no we dont have that $(-1)x=-1x$ because the latter means $-(1x)$

[jolly parrot]

qwertytrewq

[vital crescent]

you should add brakets too

because negative of the field element, is different from negative of the vector space element

i would write -(1x)

[cloud maple]

indeed you are correct

ahh then

$$-(1x) - (1x) -1(-x)$$

[jolly parrot]

Klein Bottle

[cloud maple]

would this be correct?

[vital crescent]

but we don't have -(1x)=(-1)x yet 😭

as you can see one is taking negative of the vector space element (1x) and one is taking negative of the field element then multiply by x.

[cloud maple]

that is true 😦

[vital crescent]

its good to go through these, be very careful with every step: question everything you write

[cloud maple]

so do I write it as

and I apologise in advnace if I am wrong : (

$$ = -1 \cdot (x) -1 \cdot (x) - 1 (-x)$$

[jolly parrot]

Klein Bottle

[vital crescent]

well you should be a little bit more precise, write it as + (- something) instead

[vital crescent]

your operators for a vector space, as you can see, is + and *

[frigid lark]

ok klein bottle

look, your name suggests you ask a lot of whys

[cloud maple]

So we can't use negatives since this is a vector space right?

[frigid lark]

lets see.
(-1)(x+x+(-x))

[cloud maple]

$$(-1) \cdot x + (-1) \cdot x + (-1) \cdot (-x)$$

[jolly parrot]

Klein Bottle

[frigid lark]

-(2x-x)

-(x)

@cloud maple in vector space, if we have vector t, then -t is t moved 180 degrees.

[vital crescent]

negative means the additive inverse of an element. in vector space we do have additive inverses. minus sign is sort of different (although we use the same symbol to denote it).

formally, we define $x-y$ to be $x+(-y)$

[jolly parrot]

qwertytrewq

[cloud maple]

noted

[vital crescent]

by that definition what you wrote here is really $$ = -1 \cdot (x) +(-(1 x)) +(- (1 (-x)))$$

[jolly parrot]

qwertytrewq

[cloud maple]

holy moly

alrighty

[vital crescent]

and you still used the $(-1)x=-(1x)$ implicitly in the middle you see

[jolly parrot]

qwertytrewq

[frigid lark]

yes. minus of one is the opposite of one. add it to one, and get zero.

[vital crescent]

we are working in an arbitrary field $\mathbb K$, so it is harder to make sense of "angle" there.

[jolly parrot]

qwertytrewq

[frigid lark]

true.

well, in complex its *cis(180°), or *-1

[vital crescent]

Here is a hint:
$$(-1)\cdot x=(-1)\cdot x+0=(-1)\cdot x+(x+(-x))=((-1)\cdot x+x)+(-x)$$
try and go from here, I have extracted a $(-x)$

[late spindle]

guys are u talking about axiom 3

[cloud maple]

alrighty

lemme see

[vital crescent]

yes they are trying to prove the third statement, not an axiom though

[jolly parrot]

qwertytrewq

[vital crescent]

i needed to be a little bit careful, i implicitly used associativity

now it's clearer

[late spindle]

so, -1.x = -x means -1.x + x=0 which means -1.x + 1.x=0 which follows from distributive law

am i missing something

?

[vital crescent]

you showed that -1.x=-x implies 0=0, ur method is "right" but you need to do it in the other direction

go from 0=0 and deduce -1.x=-x

[late spindle]

just reverse the arrows???? they are biconditionals after all

[cloud maple]

this makes a lot more sense

[vital crescent]

you have to prove that they are biconditions (which means that you'd have to invoke the inverse axiom a lot)

but yes you are right. but we are trying to be very precise here

[late spindle]

ok. the supposed "inverse axiom"s you said are just axioms, by the symmetry of equals relation

[cloud maple]

$$0 = 0$$
$$x \cdot 0 = 0$$
$$x \cdot (-1 + 1) = 0$$
$$(-1) \cdot x + 1 \cdot x = 0$$
$$(-1) \cdot x + x = 0$$
$$(-1) \cdot x = -x$$

[jolly parrot]

Klein Bottle

[cloud maple]

I dunno if this would work

[vital crescent]

id put the 0 to the left

[late spindle]

yes

[cloud maple]

in the second line mate? and then all subsequent lines?

[vital crescent]

cuz your multiplication by the field is on the left

[cloud maple]

gotcha

wow you all are so rigorous

this is amazing

[vital crescent]

from the second last line to the last line: it could use a little bit more steps

[late spindle]

multiplication is commutative no?

its one of the axioms

[cloud maple]

i assume by general convention, we keep the field on the left?

[late spindle]

fair enough ig

[vital crescent]

you could define right multiplication, it works. but usually they dont

[vital crescent]

Ok here's how id add to it, one sec

[cloud maple]

should I show that I am subtracting -x from both sides?

woudl that be more precise?

[cloud maple]

cheers thanks so much!

[vital crescent]

\begin{align*}
0\cdot x&=0 &&\text{by part 1}\
(1+(-1))\cdot x&=0 &&\text{ "existence of additive inverse" in a field applied to 1}\
1\cdot x+(-1)\cdot x&=0 &&\text{scalar multiplication distributes over field addition}\
x+(-1)\cdot x&=0 &&\text{scalar multiplication by 1}\
(-x)+(x+(-1)\cdot x)&=-x &&\text{add } -x\text{ to both side}\
(-x+x)+(-1)\cdot x &=-x &&\text{associativity of addition}\
0+(-1)\cdot x &=-x && \text{vector addition with inverse}\
(-1)\cdot x&=-x &&\text{vector addition with 0}
\end{align*}

[jolly parrot]

qwertytrewq

[vital crescent]

phew that was some work

[cloud maple]

You are a legend mmate

thnaks so much

[vital crescent]

np

[cloud maple]

have a great day

.close

[pearl pond]

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[meager iron]

To be fair

!nosols

[pearl pond]

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.

[meager iron]

But its fine i guess

[pearl pond]

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[viral saffron]

Imagine you've got a deck of 48 cards (jokers and 7s removed)

The game consists of having a hand of x amount of cards and, evaluating them, you trying to guess which card will the next card be. You get 1 point by guessing a card that's: same color (red or black), same odd/even, both bigger/smaller than 7, same suit, and 2 points if they have the same exact value.

The idea is to pick which one of your hands you think has the highest chance of giving you more points (so if you have 4 reds and 1 black, you will probably pick the red)

My question is, how many points could I get on average by doing this... I tried to make a simulation but I couldn't make it better than 2 points on average

[pearl pond]

@viral saffron Has your question been resolved?

[storm narwhal]

so you choose a card from x card of your hand and then pick another card from the deck and get point based on these rules?

[pearl pond]

@viral saffron Has your question been resolved?

[viral saffron]

yes

You try to pick the one which will give you the most amount of points

[pearl pond]

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[bitter herald]

Use Heron's formula

[pearl pond]

@buoyant pasture Has your question been resolved?

[buoyant pasture]

Square root is making trouble

@bitter herald

@toxic lichen

[toxic lichen]

maybe it is easier if you rewrite heron's formula as $$S=\frac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$

[jolly parrot]

Ann

[timber tartan]

nah man i think u can just keep it like that

just define s a b c

[buoyant pasture]

Still I can't apply addition?

[toxic lichen]

"apply addition" means?

[buoyant pasture]

Ohh you meant i have to multiply

Hmm interesting

Never saw such formula from high school

Thanks

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[solid crescent]

Hello, is my answer to this Q Correct and do I get full marks?

Does it definitely correct?

[meager iron]

Irrelevant but why'd you write π like that?

[fair creek]

it would help if you could show us the question properly

[meager iron]

But yeah its correct

[solid crescent]

Ok

But

I wrote it like that is that okay?

[meager iron]

It was asked to find the volume

[meager iron]

Hmm

It may be confusing to others

[lost peak]

π

this is pi

[meager iron]

Π or like this for the capital letter

[lost peak]

yes

either or are fine you don't have to be perfect

[solid crescent]

My version is fine too?

[lost peak]

I mean

it's subjective

[meager iron]

Looks more like an Ohm to me

Aka omega

[lost peak]

Ω

[meager iron]

Ohm with a tail

[sonic hound]

It's basically scribbling.

[lost peak]

like I can't say it's wrong

but in my eyes it's wrong

[meager iron]

I get that theyre tryying to write it in 1 stroke

Like how I write x but it looks like a u

[sonic hound]

I also write in a single stroke.

[lost peak]

you can do the capital in a single stroke

square without the bottom part

[lost peak]

sometimes it looks like a κ

[solid crescent]

@sonic hound do I get 3/3 marks?

[lost peak]

bigger problem for me since κ is a letter in my alphabet but still

[sonic hound]

What's the question?

[meager iron]

Can we agree on this

[sonic hound]

Yes.

I write it like the second one.

[lost peak]

that 4th is perfection

[solid crescent]

Are these correct!

[solid crescent]

And this one?

For these Q

/12

How many marks do I get ?

[meager iron]

Note that 3h44m is 128% of last week's, which we can denote as x

[solid crescent]

<@&286206848099549185>

[meager iron]

@solid crescent

For no 7

[solid crescent]

Do these Q Look correct to you?

They seem correct?

What marks have I got for these Q in total

From the ones I've sent?

Will you tell me?

@median plume

@rough forge

[median plume]

Hi

[solid crescent]

Can u tell me

How many marks

From these Q?

[median plume]

No you have to times it by 1.28

Not 0.28

[solid crescent]

Huh?

[median plume]

If the screen time is 38% better how do you get a lower sclre

[solid crescent]

/0.38

Dude.. Do you even know about GCSE maths?

@junior pewter

@∆•×

@outer echo

[cobalt hinge]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[solid crescent]

I'm saying

[median plume]

Yeah

[solid crescent]

Is it correct?

[solid crescent]

This isn't foundation tier

you should try foundation tier

[median plume]

You timed by 0.72*

My mistakes

Times by 0.72 not 0.28

[solid crescent]

Do you know how to do foundation tier?

[median plume]

And you find last week

[median plume]

Yes

[solid crescent]

Gcse maths

How?

I will be doing AQA

GCSE Maths

Foundation Tier

[median plume]

I do edexcel

[median plume]

It's fine I understand it

[solid crescent]

The highest grade you can get is a 5

[median plume]

I know haha

[timber tartan]

ts mf fantasizes abt gcse math

[radiant terrace]

This is a user that is ban evading.

They are banned on this acct now too.

.close

[pearl pond]

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[timber tartan]

damn

[pearl pond]

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[balmy igloo]

b). More screenshots of my work are coming:

[balmy igloo]

is the solution that i am supposed to obtain

but i am not going anywhere with that last result

it looks similar at least but the numerator is definitely not the same

error found, its 2cos(2θ) not 1/2..

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[pearl pond]

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[buoyant pasture]

Sum n=1 to infinity 1/n! Z^{}n

[buoyant pasture]

Radius of convergence

[blissful cloak]

you meant (\sum_{n=1}^{\infty}\frac{z^n}{n!})?

[jolly parrot]

PajamaMamaLlama

[buoyant pasture]

Yeah

Thanks

[toxic lichen]

this is just $e^z - 1$ no?

[jolly parrot]

Ann

[versed mica]

think they want radius of convergence

use ratio test

[buoyant pasture]

Ohh i didn't know it is same as real

I thought complex doesn't have same sum

[blissful cloak]

or the radius of convergence is the distance to the nearest singularity in the complex plane (my wording might be off there)

but how many sigularities does e^z-1 have?

[buoyant pasture]

1/R===limit n tends to infinity|an+1/an|

Here an is 1/n!

so it is 1/R==limit n tends to infinity |n!/n+1(n!)|

Which is just 0

So R is infinity

Means entire functions

[buoyant pasture]

0 singularity

Thanks

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[pearl pond]

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[earnest finch]

$$\begin{aligned}
&f(x) = \tan(x) \
&f: \qty(-\frac{\pi}{2}, \frac\pi 2) \mapsto \bR \end{aligned}$$Can I use this to say that $\bR$ has the same cardinality as $\qty(-\frac\pi2, \frac\pi2)$?

[jolly parrot]

@earnest finch

[random ermine]

yes

f is a bijection

[earnest finch]

Thanks!

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[pearl pond]

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[earnest finch]

-# you're hired?

[slate tide]

!status

[pearl pond]

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

[finite jolt]

yes im hired

[earnest finch]

Welcome to the job

[finite jolt]

so i have these proofs for each of them

but like

they seem wrong

Basically the logic is that its not closed under addition because u+v must be in neither u or v

and same logic for u,v,w

But like

How does that not work when F has 2 elements first of all

[slate tide]

if possible can you latex your solutions

[finite jolt]

uh idk how to use texit

my proof for 12 is basically:

If one subspace contains the other the union is trivially a subspace. Otherwise, take u+v=w where u is the member of one subspace called U and v is the member of the second subspace called V. If w was in V, then w-v=u must also be in V, so for any u in U it must also be in V, which implies U is a subset of V which we assumed was false. Apply it again for implying any v in V must be in U to get a contradiction proving that the union is not closed under addition

And my proof for 13 is basically:

If one subspace contains the others the union is trivially a subspace. Otherwise, let u+v+w=x. Assume that this is in U. Then v+w must also be in U. As we assumed v and w were any elements of V and W, then V+W is a subset of U. Since V+W contains both V and W, V and W are both subsets of U, which we assumed was false. Repeat for other 2 to conclude that u+v+w is not in the union which is a contradiction

[grim fractal]

I think is fine

I would add that because $w \in U \cup V$, $w$ has to either be in $U$ or $V$.

[jolly parrot]

Ari

[grim fractal]

Are you assuming that u + v + w is always in U for all u in U, v in V, and w in W? You should probably clarify that

[finite jolt]

oh yea i forgot to mention that

but also

how does this proof break down

[grim fractal]

Because generally, a single v+w being in U does'nt mean that V and W are contained in U

[finite jolt]

if F is {0,1}

[grim fractal]

hm I'm not sure

let's try to think of a counterexample

[finite jolt]

lets take F^3

the subspaces of this are

[grim fractal]

sure

[finite jolt]

{0}, {0,{0,1}}, {0,{1,0}}, and F^3

so

hm

no this seems to follow the rule

[pearl pond]

@finite jolt Has your question been resolved?

[humble basin]

https://youtube.com/shorts/fTfAW8jIQDE?si=EYOChmyV347TPoE1

[pearl pond]

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[pearl pond]

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[unborn cave]

truly not sure what i'm doing wrong here

[copper epoch]

So what was your answer?

[unborn cave]

-4, 0, 4

[copper epoch]

What's your reasoning?

[unborn cave]

it's asking where the points of inflection are on the graph, that's where i see them

[copper epoch]

Yes, but do you know what an inflection point is?

[unborn cave]

where the graph changes right?

[copper epoch]

You need to be more precise

[unborn cave]

like it changes from increasing to decreasing etc

[copper epoch]

No, that's not an inflection, those are the extrema of the functions

Minimum/maximum

[unborn cave]

oh

i must have it mixed up then

[copper epoch]

In fact, around 0 is always increasing

Inflection points are where the derivative changes from increasing to decreasing or viceversa (in case it is differentiable)

On the original graph, it's the points where a function goes from convex to concave or viceversa

[unborn cave]

so like where the concavity changes?

okok

[copper epoch]

yes

[unborn cave]

but if i don't have the function how can i get those precise points?

like it looks like it is changing somewhere around -2, 0, and 2

but i don't know exactly where

[copper epoch]

We don't have the analytic expression, so the best we can do is take an educated guess

-2 and 2 seems fair to me

[unborn cave]

i tried -2, 0, 2 and it still says only some parts are correct, so i'm not sure how to be more specific

[copper epoch]

Does it give 0 only as a correct solution maybe?

[unborn cave]

it says "you have submitted only part of the correct answer" when i try only 0 😕

[copper epoch]

Seems weird to me

[unborn cave]

yeah this software sucks i've had this issue a ton

[copper epoch]

-1 0 1? -3 0 3?

[unborn cave]

still no, might have to take the L on this one

i'll try a couple decimals

but thank you for your help!

[copper epoch]

That shouldn't be necessary normally though

Very weird

No way to check the solution?

[unborn cave]

i agree, if i get it wrong a couple more times sometimes it'll pop up lemme see

it wanted -2.5, 0, 2.5

🙄

[safe siren]

I was literally just going to say that 💀

that's a dumb question tho

[copper epoch]

2.5 is not even marked on the graph

[safe siren]

I had to zoom so far in to see it

[copper epoch]

That's just a bad question

[unborn cave]

its the software tbh, like this one

i referenced the solution for part d after getting it wrong a few times and it says 1, 4, 5, 6

but when i type that in exactly like that it wont even submit bc it says it needs to be a constant

[safe siren]

that's sucks, my homework software was so good you could just not input the whole answer in and it wouldn't mark you down but it would still tell you if you got it right or not, so I used to basically get full marks all the time for no effort 💀

[unborn cave]

that must be so nice

ill never takea class that uses this again

[safe siren]

does this count for your final grade?

[unborn cave]

yeah it does

and our midterm and final use the same software

[safe siren]

💀 that's so fucked

maybe you could use the report a problem button at the top and they won't count the question if they fine problems with it or something?

actually tbh I don't think they care enough to do that 💀

[unborn cave]

i deadass dont either

oh well

i just hope it doesnt come up on my final

[pearl pond]

@unborn cave Has your question been resolved?

[safe siren]

it probably will but it'll probably get drowned out by the other questions and only be a couple of marks anyways

[pearl pond]

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[midnight haven]

Is the way I solved this inequality right?

[exotic scaffold]

the steps are right but your answer is slightly off

4/(1/2) isnt 2

remember when you divide by a fraction you multiply by its reciprocal

[midnight haven]

Like keep change flip?

[exotic scaffold]

yeah

4 / (1/2) = 4 * 2

[midnight haven]

Is the way I did it like this right?

[exotic scaffold]

yes thats correct

[midnight haven]

Ohhhhh aha

Thank you!

[exotic scaffold]

so your final answer would be x >= 8

yup no problem

[midnight haven]

.close

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[winter thistle]

simone is facing north at the entrance ofa tunnel throuhg a mountain. she notices that a 1515 m high mountain in the distance has a bearing of 270 degrees and its peak appears at an angle of elevation of 35 degreres. after she exits the tunnel, the same moun tain has a bearing of 258 degrees. asuming that the tunnel is perfectly level and straight, how long is it to the nearest metre?

[winter thistle]

im having trouble figuring out how to draw the diagram

for this question

[pearl pond]

@winter thistle Has your question been resolved?

[pearl pond]

@winter thistle Has your question been resolved?

[pearl pond]

@winter thistle Has your question been resolved?

[pearl pond]

@winter thistle Has your question been resolved?

[hazy bear]

The mountain was at the bearing of 270° which means it was directly in his West

After coming out of tunnel, the mountain is now at bearing of 258° which means 12° clockwise from his West

[hazy bear]

Upper one is in xy-plane.
Lowe one is in xz-plane.

M=mountain base
A=your initial position
A'=your final position
H=mountain top

You have to find AA'

@winter thistle

[pearl pond]

@winter thistle Has your question been resolved?

[alpine belfry]

.

[pearl pond]

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[carmine peak]

yea for question 5 statement 2 im getting it as false??

[toxic lichen]

how so

[carmine peak]

oh ok

so basically

they’re disjoint sets

so it’s basically the same question as S1

so the union must also be the same

gang sm1 help me out here

[cosmic charm]

what should A - B = empty set mean

[rose yarrow]

What’s happening

[carmine peak]

disjoint

sets

[cosmic charm]

no?

[carmine peak]

what

[cosmic charm]

if A and B are disjoint then A - B = A

since A and B have no common elements

then taking out B (which is literally nothing) leaves you still with A

[carmine peak]

ohhh right i confused it with intersection

ohhhhhhhhhh thanks thanks

[cosmic charm]

if taking out B from A leaves you with nothing

that means A is the same as B

[rose yarrow]

Is the correct answer D ?

For 5

[carmine peak]

yerp

[rose yarrow]

Yea

It’s pretty easy

[carmine peak]

also have a doubt in 4 if i can ask?

[rose yarrow]

I can help

[rose yarrow]

Lemme check

[carmine peak]

i’m sending

my working

[rose yarrow]

So is the answer incorrect ?

[rose yarrow]

For you

[carmine peak]

ye

[rose yarrow]

I see

[carmine peak]

my ans is like 110-44

but it isn’t even in the set

[rose yarrow]

Alright lemme do it myself in paper

Give me a moment

[carmine peak]

bet

[rose yarrow]

@carmine peak I hope this helps.

Had some AI help but it’s pretty clear

[carmine peak]

yeye

thanks

[pearl pond]

@carmine peak Has your question been resolved?

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[flint drum]

Hi

[flint drum]

Need help understanding all these two questions

Mostly just the first question

[smoky musk]

@flint drum What is the average rate of change?

[flint drum]

I forgot how to do that again

Something like

Y2 -y1 over

X2-x1

So I pick to points on the graph

Of the original function

And then solve it and I get my answer?

I don't quite understand the question

[smoky musk]

Yes

[flint drum]

Is it like

The X values

Are already give to me he's?

Yes*

?

But then how would I do b?

[smoky musk]

Probably the question asks for "instanteous rate of change"

[flint drum]

How do I do that

[smoky musk]

use the derivative function.

[flint drum]

Not sure I get it

I wanna know how to do this btw

Since my exam is tomorrow

And it was worth 3 points on the last year's exam

[smoky musk]

The derivative function is given in the graph.

[flint drum]

1,-2

?

[smoky musk]

go on

[flint drum]

Yeah

That's what I don't get

What is there else to do

What formula do I use

To proceed

Cause it isn't he other one

[smoky musk]

f'(1)

[flint drum]

Since that's 2 points

I only have 1 point

[flint drum]

I don't get it

How am I supposed to derive a equation

That I don't know how to get

[smoky musk]

The instanteous rate of change is simply the value of the derivative function at a certain point.

[limpid lily]

There's a graph of f', the derivative of f.

The derivative of f tells the instantaneous rate of change of f.

So, if you're looking for the instantaneous rate of change of f when x = 0, you look at what f' is when x = 0.

[flint drum]

Yeah don't understand it so the 5 marks can go in the bin whatever

We move on

I'll ask someone in real life before the exam to help

I have another thing that i want to understand

How do I solve this

[smoky musk]

The question already shows you how to solve it step-by-step.

I assume that the function is odd

[flint drum]

I don't understand how to interpret the integral graphically and the determine? Part

[limpid lily]

Yeah, it's odd because it's symmetric with respect to the origin.

[flint drum]

Also I know it's symmetrical

So -4 to 0 is 7

So the other side is -7?

I don't know how to do it

Ok yk what it's fine

.close

[pearl pond]

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[midnight haven]

How would i prove that?

[limpid lily]

Which one?

I assume that fg means f(g(x)) in their notation.

[lost peak]

synthesis or multiplication

[limpid lily]

For 1.29, a hint is to consider T = {1}.

Not necessarily for the proof, but to get an idea.

[midnight haven]

Yes

[midnight haven]

well there must be an s in S st B(s) = 1

[limpid lily]

For 1.30, think of cables going from elements in R to distinct elements in T through alpha and beta. Is it possible for alpha to send two cables to the same S value?

[midnight haven]

must a and b be onto?

[limpid lily]

Why beta?

[midnight haven]

Because of every T there must be an S no?

[limpid lily]

Right.

If there's a left-out T, the composition isn't onto.

Why alpha?

[midnight haven]

Because there must be an r for every S so we get every T?

[limpid lily]

Why does there have to be every S?

[midnight haven]

Because if theres an S missing we might not get all T?

[limpid lily]

Well, consider T = {1}.

Let's say that S has 1,000 elements.

Wait a minute.

OK.

So, S has 1,000 elements.

[midnight haven]

Oh yea maybe you dont need all of S only the ones that are necessary for T

[limpid lily]

Right.

In the example case, you can have all R going to the same S, since all the Ss go to 1.

[midnight haven]

but then is a Onto no?

[limpid lily]

No, onto means all possible output values are produced from R.

So, you'd need alpha(R) = S.

But we don't need alpha(R) to be S. It can be {5} or something.

Like alpha(x) = 5. beta(x) = 1.

Since T = {1} and beta(alpha(x)) = 1, you have beta alpha as onto.

But S can be the integers from 1 to 1,000.

If alpha(x) = 5, it doesn't cover all of those, so alpha isn't onto.

[midnight haven]

i drew a diagram

i get it

[limpid lily]

OK.

[midnight haven]

Thank you!

[limpid lily]

You're welcome.

[midnight haven]

for 1.30, does One to one require that all of R has a designated value?

Or can it be a subset or R

[limpid lily]

Yeah, each element in R has a unique value in T.

No element in R shares its T value with another element in R.

[midnight haven]

Yes but do all elements R have to have an element in T?

[limpid lily]

Yes, unless we're dealing with partial functions. They usually don't when they're asking these questions.

If beta alpha is one-to-one, then it's a total function.

!occupied

[pearl pond]

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

[midnight haven]

they dont say that Every S has a T

idk

[limpid lily]

Ahh.

Well, when you're at this stage, they're dealing solely with total functions.

[midnight haven]

what does that mean

[limpid lily]

That means that every element of the domain produces a value from the function.

Like if the domain is the real numbers, any real number input will produce an output.

A partial function can have some real number inputs having no output.

Like when a function is undefined.

Like log(0) or something like that.

[midnight haven]

hmm

but theyre not saying that theyre total

Theyre just saying that functions are rule assignments from S To T

[limpid lily]

You can prove it for partial functions as well.

[limpid lily]

From this ^

[midnight haven]

So one to one mean that for every evelemt R theres a different T

[limpid lily]

Yes.

[midnight haven]

so b and a are one to one?

[limpid lily]

Why?

[midnight haven]

Because b has to be one

and so a must be aswell because we cant have a(s1) = a(s2)

[limpid lily]

Why does b have to be one-to-one?

[midnight haven]

Because if we get duplicate values from B it wouldnt be

[limpid lily]

Why would duplicate values from beta cause beta alpha to not be one to one?

[midnight haven]

B(a(s1)) =/ B(a(s2))

[limpid lily]

Hint: What if R = T = {1}, and S = {1, 2}?

[midnight haven]

well if a is one to one then there can only be one s reached so B doesnt need to be one to one?

[limpid lily]

Why doesn't beta need to be one-to-one?

[midnight haven]

Because we only get one value as imput

[limpid lily]

Well, yes, but I mean in a more general manner, with any sets.

[midnight haven]

we need at least 2 to contradict One to one ness?

[midnight haven]

well it already showed that B must not be

[limpid lily]

Oh, OK.

What about alpha?

To clarify, beta can be one-to-one but it doesn't have to be.

[midnight haven]

Yesss

[midnight haven]

i does because if its not we could get the same S and so we would get the same T for different R

[limpid lily]

Why would that make beta alpha not one-to-one?

[midnight haven]

cause then we have a(r1) = a(r2) for different r, then b(a(r1)) = b(a(r2)) for r1 =/ r2

[limpid lily]

Right.

[midnight haven]

Thank you!!

[limpid lily]

You're welcome.

[midnight haven]

.close

[pearl pond]

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[pearl pond]

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[buoyant pasture]

radius of convergence

[pearl pond]

@buoyant pasture Has your question been resolved?

[grim fractal]

Have you tried Lagrange Remainder Theorem

[pearl pond]

@buoyant pasture Has your question been resolved?

[buoyant pasture]

What's that?

[pearl pond]

@buoyant pasture Has your question been resolved?

[pearl pond]

@buoyant pasture Has your question been resolved?

[pearl pond]

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[pearl pond]

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[low skiff]

I have a and b, how do I find c?

[dusty flame]

(1, -3)

${f(1) = -3 = a + b + c}$

[jolly parrot]

k

[low skiff]

oh right

Ok got it thx

.close

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[mortal spade]

I need help with part b

My idea was that we can have 2023^3 +1 as n

Then 2023³ = 1 mod (n)

Oh wait

That is incorrect

[toxic lichen]

the exponent is n not 3 though

[mortal spade]

That was incorrect

I mixed up a few things

Does the property a^n = b^n mod k work for fractional n

I cannot see the images

Why is it deleted

????

n= 2023³-1 works

My idea

If n= 2023^m -1 is divisible by m then it workse

Is that correct

<@&286206848099549185>

Oh wait

That is like the initial thing

What am I doing

So any number which has previously satisfied the eqn can be used to find the next n ???

<@&286206848099549185>

[near mauve]

well prove it, what makes u say that

[mortal spade]

I have proved that I think

Wait I am not sure

Yes I have proved that

But I am still not sure if this is right or not

As I was able to verify for only 1 number

[near mauve]

then that is not a proof

you probably observed one and concluded it works for all other?

[mortal spade]

I have a proof

I am not sure if it is correct

[near mauve]

well then you can show the proof

[mortal spade]

I made an error took +1 instead of -1 and the entire thing is wrong

I will give this another try

.close

[pearl pond]

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[pearl pond]

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[sonic mist]

Yo is there any other way to solve this other than drawing a circle around B, A E corners?

[sonic mist]

Because I'm not bright enough to think about drawing a circle during the exam probably

Just a little hint would be very nice

[timber tartan]

Can you translate it

[sonic mist]

ABC 90° triangle
According to the information above, m(BED)=a ?

a in degrees

[timber tartan]

Okay so

Assign angle BAC the value k

[sonic mist]

Oke

[timber tartan]

What would be angle ECD then

[sonic mist]

90-k ?

Tthen edc is also k

[timber tartan]

So what's angle CDE

Yes

Now what's angle EDB

[sonic mist]

180-k

[timber tartan]

Good, what about EBD

[sonic mist]

Hmm

180 - ( a + 90 + 90 - k)

= k - a

?

[timber tartan]

Wait

It is 180 - (a + 180 - k)

Okay so k - a

Yeah

Finally angles EBA and AEB

[sonic mist]

Woh I think i see it

Give me a second

Nope lets continue

Uh
Eba = 90 - k + a
Aeb = 90 - a

[timber tartan]

Okay