help-27

[dry leaf]

#help-27 message

[restive river]

Ohoh sorry mb

But bro this equation

3g-t=a3

How is it correct?

[glossy dew]

he admits that he did many mistakes

[restive river]

Well the thing is not everything can be understood if u brainstorm it right from scratch

When I solved HC Verma it was very rigorous

But after I completed a lot of it following the correct steps and methods did i actually connect the dots to understand why and how

My suggestion is to correct your mistakes first and make the general approach a second nature and then try to understand the rest

As for how to solve this question i think the power= constant approach will be best

You find the v for block 1 and then v=d/t will do it

And a1=0??? Wtf is going on 😭😭

[dry leaf]

It's not correct, I just said that

#help-27 message

[dry leaf]

But you couldn't find the one that I'm looking for

[dry leaf]

Yes I get it but I'm only trying to understand what's the mistake in my solution

[glossy dew]

a lot of things

and im not sure if the isolated system method would work

[devout snow]

@dry leaf Has your question been resolved?

[devout snow]

@dry leaf Has your question been resolved?

[devout snow]

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[normal sigil]

hi, can you explain how to prove whether a function is partially differentiable.
I am asked to prove it for this function, and so far i think i have proven that it is partially differentiable for IR without (0,0), but i dont understand how to do it for (0,0).
I only need to show that it is partially differentiable twice.

[supple knot]

Use limit definition of partial derivative

[normal sigil]

meaning this?

[runic prawn]

yes

[normal sigil]

so if i show that all partiall derivatives of f(x,y) exist by calculating them i have proven that the function is partially differentiable?

[devout snow]

@normal sigil Has your question been resolved?

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[devout snow]

Please don't occupy multiple help channels.

[somber garnet]

oops

is my duiagram wrong

im so confused

.close

[devout snow]

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[devout snow]

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[waxen steeple]

i know it's principle by inclusion exclusion but im a little bit confused on how to find the probabilities of eg selecting at least a red and a yellow, why would it not just be (8C2) (1/4)^2 (3/4)^6 ?

[devout snow]

@waxen steeple Has your question been resolved?

[devout snow]

@waxen steeple Has your question been resolved?

[devout snow]

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

[lunar harbor]

(Why would it be that?)

[devout snow]

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[fiery mist]

help pls. i don't know how to get the domain and range

[autumn girder]

!status

[devout snow]

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

[fiery mist]

1

[autumn girder]

Ok, do you know what domain and range are, just in general

?

[fiery mist]

yeah

both of the graphs domains are limited

or has a restriction

right?

[autumn girder]

Yes for both f and g the domains are restricted

[fiery mist]

yeah for f(x) range is [-8, oo), and g(x) range is [-7,oo)

[autumn girder]

Can you tell me what are the domains and ranges of f and g respectively?

Oh I didn't even need to ask lol

[fiery mist]

😭

[autumn girder]

Check again for g

[fiery mist]

the domain and range of g?

[autumn girder]

The range is correct

I thought you wrote domain so asked you to recheck it

Mb

[fiery mist]

all g

[autumn girder]

Can you also state the domains though

[fiery mist]

sure

domain of f(x) is [1,oo), and g(x) is [-9,oo)

[autumn girder]

Correct

Now, (f + g)(x) = f(x) + g(x), you know that from the note you have in the image

[fiery mist]

yeah

[autumn girder]

Good, But f(x) can't be defined for any x < 1

[fiery mist]

yeah..

[autumn girder]

So by definition, (f + g)(x) should also be undefined

Does that make sense?

[fiery mist]

no😭

[autumn girder]

Ok, why?

[autumn girder]

for x < 1

[fiery mist]

uhm so the domain and range of (f+g)(x) should be undefined?

[autumn girder]

?

No

I'm saying that the Domain of (f + g)(x) is going to be $x \in [1, \infty)$

[woven radish]

@autumn girder

[fiery mist]

ohhh

[autumn girder]

Yea because it can't be below that, but it's defined for every number above that

[fiery mist]

yeah bc the domain of f(x) starts from 1

?

[autumn girder]

Exactly

[fiery mist]

okok

i got it

how about for the range

[autumn girder]

Wanna take a shot at it yourself?

[fiery mist]

i tried.. but i got the wrong answer

[autumn girder]

Ok well, so you see that they're both increasing functions right?

[fiery mist]

yeah

[autumn girder]

So we're pretty sure that the right bound for the range is gonna be infinity

because these functions will keep increasing infinitely

there's no denying that

[fiery mist]

uhuh

[autumn girder]

Ok cool

Also from the arrow on the top right

of both of the functions

we can tell that they never decrease

[fiery mist]

yeah

[autumn girder]

i.e. their lowest values are the ones we see on the graphs in front of us

aka their minima

[fiery mist]

-8 and -7??

[autumn girder]

Yes, those are each of their respective minimas

but those aren't the minima of (f + g)

recall that, (f + g)(x) is only defined in the range $[1, \infty)$

[woven radish]

@autumn girder

[fiery mist]

so i should add the values of f(1) and g(1)?

[autumn girder]

Good

Yes

[fiery mist]

omg i get it now

[autumn girder]

Cool

[fiery mist]

i have a problem with the composite ones

though

[autumn girder]

Shoot

[fiery mist]

do u still have time? or

[autumn girder]

Yea I do have time

[fiery mist]

itd be great if u help me out with those 🥹

[autumn girder]

What are you confused about with them

[fiery mist]

for f(g(x))

do i only solve the domain and range for g(x)?? bc it's the input for f(x)??

idk im kinda confused tbh

i might not be understanding it well enough

[autumn girder]

$$f \colon [1, \infty) \to [-8, \infty)$$
$$g \colon [-9, \infty) \to [-7, \infty)$$

[woven radish]

@autumn girder

[autumn girder]

yea?

[fiery mist]

yeah

[autumn girder]

So, when we try to compute f(g(x))

we're pluggin in the output of g(x) into f(x)

Correct?

[fiery mist]

yeah..

[autumn girder]

So what we're doing is we're trying to map over the range of g using f

does this make sense?

[fiery mist]

uhhh

sure

[autumn girder]

Don't say sure, unless you are sure of it

[fiery mist]

no im not sure

[autumn girder]

Ok

So, output of g(x) can only be in $[-7, \infty)$, that is what range means, right?

[woven radish]

@autumn girder

[autumn girder]

But notice that there is a part of this range

which isn't a valid input for f

[fiery mist]

yes

[autumn girder]

More precisely

$g(x)x \notin [-7, 1)$ if we want $f \circ g$ to be defined

[woven radish]

@autumn girder

[fiery mist]

so do we start at [-2,oo)?

for the range

???

[autumn girder]

We'll get to the range

but do you understand the domain rn

[fiery mist]

oh so u were talking about the domain

okok

yeah i do

[autumn girder]

we're talking about domain of f, and range of g

[fiery mist]

okok

[tepid rover]

Hi

[autumn girder]

From this, can you find the sub-domain of g where $f \circ g$ will be defined?

[tepid rover]

I also need help

[woven radish]

@autumn girder

[autumn girder]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fiery mist]

sub domain?

never learned about that

[autumn girder]

subset of domain

[fiery mist]

uhhh

[1,oo)?

[autumn girder]

What is the value of $g(1)$?

[woven radish]

@autumn girder

[fiery mist]

-2

[autumn girder]

Does -2 belong to the domain of f?

[fiery mist]

no

it does not

[autumn girder]

So

for which smallest value of x

Does g(x) belong to the domain of f(x)

[fiery mist]

uh

5???

wait is it 7

g(7) is equal to 1

which belongs to the domain of f(x)

[autumn girder]

Correct

$f \circ g$ will be defined for all $x \ge 7$ because g is an increasing function and f has a continuous domain

[woven radish]

@autumn girder

[fiery mist]

right

[autumn girder]

And the range remains that of f

because, remember, that $f(g(7)) = f(1) = -8$

[woven radish]

@autumn girder

[autumn girder]

And then f(x) is increasing

and continuous

so

you get the range

$[-8, \infty)$

[woven radish]

@autumn girder

[autumn girder]

making $f \circ g \colon [7, \infty) \to [-8, \infty)$

[woven radish]

@autumn girder

[fiery mist]

oh so we can find the range from the domain of f(g(x))?

[autumn girder]

Yes

[fiery mist]

ahh

[autumn girder]

Well you also require the procedure or the graph to get the minimum value

and the maximum value

and the values where it isn't defined

[fiery mist]

okok

[autumn girder]

but yea, if you have Domain, and either procedure to get to range or a graph showing the relation between Domain and range

you can get to it

[fiery mist]

thanks man

you helped alot

i thought i was cooked for my finals

[autumn girder]

Glad to be of help

[autumn girder]

Best of luck for your finals

[fiery mist]

have a great rest of ur night/day!

[autumn girder]

Thanks

You too

[fiery mist]

tyty

[autumn girder]

!done

[devout snow]

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

[fiery mist]

.close

[devout snow]

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[devout snow]

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[jagged lily]

help.
“As more rectangles are added, the estimate increases, but is still an underestimate of the actual value. What does this say about f’(x) and f’’(x) over the interval?”

I can’t visualize this 😭 help

[jagged lily]

Where do you even start

[drifting sinew]

This is very hard to do without some specialized knowledge, so I'll try my best to help you work through it.

First, assume that you're taking a lower RAM.

Then, what happens as you add more rectangles?

[jagged lily]

RAM?

[drifting sinew]

RAM stands for Riemann Approximation Method. It's synonymous to using the different rectangles (and other shapes) to approximate integrals.

[jagged lily]

Ok

[jagged lily]

The estimate goes up

[drifting sinew]

And how will the estimate compare to the true value of the definite integral?

[jagged lily]

The estimate is always gonna be lower than the actual value

[drifting sinew]

Why?

[jagged lily]

Hmm

That’s a good question

Idk why

[drifting sinew]

You are correct, I just want you to formalize the reason into words so we can use the same explanation later.

[jagged lily]

???????????

[drifting sinew]

First, understand that the lower RAM simply uses the lower y-value of the rectangle's endpoints as the height of the rectangle.

[jagged lily]

Yes

[drifting sinew]

Now, can you try explaining why the lower RAM is always lower than the true value of the integral.

[jagged lily]

Because the RAM uses the lower y-value at the rectangle endpoints

[drifting sinew]

Cause:

• The lower RAM uses the lower y-value at the rectangle endpoints. #help-27 message
• We are adding more rectangles within an interval to estimate a definite integral. #help-27 message
  Effect:
• The estimate increases. #help-27 message
• The estimate stays below the true value of the definite integral. #help-27 message

It's good to understand that the effect is primarily caused because we use the lower y-value of the rectangle's endpoints.

[drifting sinew]

!original
Some information seems to be is missing.

[devout snow]

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.

@jagged lily Has your question been resolved?

[devout snow]

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[devout snow]

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[frigid torrent]

how do i solve (-1-i)^10 for complex numbers

[frigid torrent]

i already got the modulus which is route(2)

[soft umbra]

Wdym, To simplify?

[frigid torrent]

sorry i didnt write the ^10 bit

just changed it

and I got to route(2)cis(x)

then i expanded which is just cos(x)+isin(x)

which is -1/route(2)

which is -45 degrees

now what?

im up to

[soft umbra]

looks fine till now

[frigid torrent]

now what

[soft umbra]

are you struggling with how to proceed?

[frigid torrent]

how do i remove the exponetial

[soft umbra]

I see

[frigid torrent]

route 2 becomes 32

[soft umbra]

it's just -45* 10

because you rotate 10 times

[frigid torrent]

that becomes -90

same as -90

[soft umbra]

and don't forget the length, which you have found already

[frigid torrent]

sorry i dont get you

what length?

[soft umbra]

This

[frigid torrent]

yep

so 32cis(-90?

[soft umbra]

correct

[frigid torrent]

is that thr answer

how do i rmeove the -90\

the negative

[soft umbra]

yes, now convert it into polar form

[frigid torrent]

isnt it already polar form?

[soft umbra]

Imagine you rotate clockwise about 90 degree

what would the coordinate be?

If the original coordinate is (32, 0)

[frigid torrent]

0,-32

[soft umbra]

correct

convert it, and that's all

[frigid torrent]

convert it to?

[soft umbra]

a+bi form

[frigid torrent]

oh alr

thank you so much

why does chatgpt say 32cis(90)

thats wrong right

[soft umbra]

don't use gpt for math tho

They're currently not equal to doing math.

[frigid torrent]

ok

have a good day or night

.close

[devout snow]

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[calm python]

Fluid Mechanics Question: may I ask whats the difference between these 2? are they not both Velocity head?

[thick ledge]

In what context is the second formula written?

[calm python]

i just saw it here, im still trying to understand

[acoustic leaf]

the second one is the change in velocity head

[calm python]

correct me if im wrong, so this formula is used when the water that is being pumped is along the Pump Centerline??

[thick ledge]

Yeah

It's true for any flowing liquid

That is irrotational

[calm python]

ok thx both of you, ill give it a try

[thick ledge]

Also of course it should be incompressible and non viscous fluid

[thick ledge]

No worries

[devout snow]

@calm python Has your question been resolved?

[calm python]

may I ask why is Zd only measured at the centerline?

Isnt it suppose to be like this? since is Zd - Zs?

[devout snow]

@calm python Has your question been resolved?

[devout snow]

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[devout snow]

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[cyan hearth]

im unable to find g(x), anyone has a hint?

[cyan hearth]

this is how i found p(x)

-p(x)=w'/w

so i had to find w'

but the left one im unsure

[meager sequoia]

you know a solution to the ODE (y_1 for instance) and you know p(x). Therefore, you can obtain an equation for g(x)

[cyan hearth]

yea yea i realised i was being a dum dum

its just rearranging

and plugging y=exp(6x)

[meager sequoia]

nice! glad it worked out

[cyan hearth]

thank you :>

.close

[devout snow]

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[devout snow]

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[sick bolt]

Can someone walk me through how my teacher got these answers

[fossil locust]

similar triangles

[sick bolt]

Elaborate

[fossil locust]

https://www.youtube.com/watch?v=m9p37TE53SI - try watching this

[sick bolt]

Bruh I ain't got time I have a quiz in 5 min give me a summary

[fossil locust]

so for all of these triangles, the largest triangle is twice as big as the small one

because the large triangle has side length 2y, and the small triangle has side length y

so x is twice of 10 in the first question

[sick bolt]

How do you know it's 2 times larger

[fossil locust]

I suspect you don't understand what the marks mean

[sick bolt]

Equal

[fossil locust]

ok so you do know

[fossil locust]

try drawing the small triangle on the top and the largest triangle separately

draw them as two triangles

[fossil locust]

then one has side length y

the other has side length y + y = 2y

[sick bolt]

Oh I think I see it

[devout snow]

@sick bolt Has your question been resolved?

[sick bolt]

.close

[devout snow]

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[devout snow]

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[thick ledge]

@spring oasis translation please

[spring oasis]

is being done, beep beep boop boop

4. Let ( B = {(0,1,1); (1,1,1); v} ) be a basis of ( \mathbb{R}^3 ) such that the coordinates of the vector ((4,6,4)) in the basis ( B ) are ((3,2,1)). Find ( v ) and the coordinates of the vector ((2,2,0)) in the basis ( B ).

[woven radish]

938c2cc0dcc05f2b68c4287040cfcf71

[spring oasis]

(4,6,4) = 3(0,1,1) + 2(1,1,1) + v

[thick ledge]

Yes

[spring oasis]

(4,6,4) - (0,3,3) - (2,2,2) = v

[thick ledge]

Yep

[spring oasis]

,w {4,6,4}-{0,3,3}-{2,2,2}

[woven radish]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[thick ledge]

[spring oasis]

v = (2,1,-1)

[thick ledge]

Yeah

[spring oasis]

(2,2,0) = a(0,1,1) + b(1,1,1) + c(2,1,-1)

[thick ledge]

Yeah

[spring oasis]

i) 2 = b + 2c
ii) 2 = a + b + c
iii) 0 = a + b - c

[thick ledge]

Yes

[spring oasis]

a + b = 1

[thick ledge]

True

[spring oasis]

b = 2 - 2c

[thick ledge]

Yes

[spring oasis]

a + (2-2c) = 1

a + 2 - 2c = 1

a - 2c = -1

[thick ledge]

Yes

[spring oasis]

a = 2c - 1

[thick ledge]

Yeah

Why take such a long route

[spring oasis]

yeah this is nonsense

what do you recommend

[lilac stone]

and you reply him every line

[thick ledge]

c is 1 from when you got a + b = 1

[spring oasis]

right, I didnt noticed

because a+b = 1

[thick ledge]

For moral support

Yes

[spring oasis]

2= b + 2c
2 = b + 2

b = 0

[thick ledge]

Yes

So a is?

[spring oasis]

2 = a + b + c

2 = a + 0 + 1

a = 1

(a,b,c) = (1,0,1)

[thick ledge]

Noice

Can you tell me why we won't get infinite solutions for a, b, c?

[spring oasis]

we got three unknowns and 3 equations

[thick ledge]

That doesn't always gurantee a unique solution

[lilac stone]

(determinant is not 0)

[spring oasis]

because the system is consistent

[thick ledge]

Again that isn't concrete

You can have 0 solutions as well

[spring oasis]

the matrix is non square

[thick ledge]

No it is

[lilac stone]

isn't it 3 by 3?

[spring oasis]

i thought it was a 3x3 augmented with 1 column

[thick ledge]

That's A augment B

Where A is 3 by 3

And B is column matrix

He's talking about B

Also btw my bad det not equal to 0 satisfies uniqueness

[spring oasis]

even if det(A) = non zero, we can still get no solutions because the system can be inconsistent

[lilac stone]

I mean, 0 determinant implies the whole vector space going down several dimensions, which could only have 0 or inf solutions

[thick ledge]

Yes mb

[thick ledge]

Not really

[spring oasis]

even if A is invertible, we can still have no solutions for A|B if the system is inconsistent mate

[spring oasis]

elaborate

[thick ledge]

https://math.stackexchange.com/questions/791750/how-do-we-know-that-if-ax-b-has-a-unique-solution-a-is-invertible

[spring oasis]

very informative mate

im upvoting the emily answer jjajaj

[thick ledge]

[spring oasis]

I was reading your message @lilac stone why did u deleted it mate

whatever

[spring oasis]

why do you ask me this?

we were already done with the exercise, it was like extra question?

(2,2,0) = a(0,1,1) + b(1,1,1) + c(2,1,-1)
(a,b,c) = (1,0,1)

[thick ledge]

Yeah just so that you understood everything nicely

[spring oasis]

(2,2,0) = (0,1,1) + (2,1,-1)
(2,2,0) = (2,2,0)

[spring oasis]

i appreciate the help @lilac stone @thick ledge

[thick ledge]

[lilac stone]

When we consider a system of linear equations, we're basically finding the solution vector x in the equation Ax=b, where A is a square matrix with a bunch of constants, b is another vector.

[spring oasis]

what happens if det(A) = 0

[lilac stone]

Ax is the vector x after the space had transformed according to A

[spring oasis]

we have infinitely many solutions?

[thick ledge]

Then there's 2 cases it can either have zero solutions or infinite solutions according to rank of A augment B

[spring oasis]

ok

I think I get it tho

.close

[devout snow]

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[lilac stone]

if det(A)=0, this means that everything in the original space got volume multiplied by 0

and he left

[devout snow]

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[noble acorn]

Hey quick question, how would you interpret this task; there are 3 million «personal cars» and half a million «business vans», 400k of the personal cars are hybrids, and 800k are electrics pick a car at random, given that you pick a personal car, what are the chances of that car being neither electric nor hybrid? Would you include the chances of pulling a personal car in the first place? Rounded the numbers for ease of handling

[thick ledge]

No

Because they've given to you that you're choosing out of the personal cars

This is what is called conditional probability

[noble acorn]

Ok, task b phrases it slightly differently, within the business vans there are 40k electrics and 1800 hybrids, given that you pull an electric vehicle(either van or car) what are the chances its also a van

[thick ledge]

What's your doubt here

[noble acorn]

The math isnt the hard part honestly, its moreso the interpretation, since this is a question from uni level statistics course for engineers, i might be overthinking it

[thick ledge]

Drawing a Venn diagram or a tree diagram substantially helps

[limber steppe]

how about my answer?

[thick ledge]

That's correct

[noble acorn]

I know, i ended up making a tree, but the problem is that im not sure whether to use the cumulative probability or just the chances of conventional or alternative propulsion

[thick ledge]

What don't you get

[limber steppe]

thanks, where are you from?

[noble acorn]

Btw task c is a bit more involved, if you pull 12 random vehicles, what is the probability that you get less than 4 electric cars

[thick ledge]

I'd like to not disclose that thanks

[noble acorn]

I tried binomial distribution, but im not 100% sure

[thick ledge]

First of all is your previous doubt cleared

[limber steppe]

cool if you have a lot of problems, I can help you, and how about?

[noble acorn]

Yeah ✅ already handed in my answer so i cant really do anything about it now

[thick ledge]

Ah I see

[thick ledge]

No thanks

[noble acorn]

Just came home, im really beat, this was task one so i ended up overthinking it a lot, but regarding task C, how would you solve it?

[thick ledge]

Is this using the same data?

[noble acorn]

Yeah

[thick ledge]

Cuz 3 million cars, 12 vehicles seems strange

Anyways

Yes it is binomial distribution

P( X < 4)

[noble acorn]

Yeah, i forgot to specify that its directly hinted that the changes in probability over 12 cars wouldnt change much so i guess they accept fixed chance

[thick ledge]

[noble acorn]

1 sec i have a harder one i got fully stuck on

[lament kraken]

bro what you on

[noble acorn]

Its clearly a neural network

[lament kraken]

can ya respect others privacy or smth

[noble acorn]

It has AI in the name

[lament kraken]

indeed

[limber steppe]

I am human. tell me your problem..

I like football and I like Vini very much.

sorry , in Eglish, give me problem.

[lament kraken]

shouldn't have disturbed you

[lament kraken]

skill issue

[noble acorn]

Lets call the largest distance(measured in cm) between two points on a potato as the length of the potato. As the table under shows, you have for given lengths of potatoes(x) picked five potatoes with weight given as y(in grammes)
Length x | 6.0 | 6.4 | 6.8 | 7.2 | 7.6|
Weight y | 82 | 77 | 88 | 100 | 100 |

Find the empirical correlation-coefficient to the data in the table what does it tell us about the length and weight?

[lament kraken]

🥔 yum

[noble acorn]

Sorry if it doesnt translate well, its given in a foreign language lol

[thick ledge]

I'm not well-versed in statistics enough to solve this sorry
I don't know what is an empirical correlation coefficient

[noble acorn]

My guess is that its a pearson correlation coefficient

[thick ledge]

Are you just copy pasting chat gpt answers

[thick ledge]

!nogpt

[devout snow]

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

[limber steppe]

Compute 𝑟 accurately to conclude how length and weight relate for potatoes. If you need assistance with any part of the calculation, let me know!

[noble acorn]

[limber steppe]

ok, you understand gpt answer is not correct? so I can help you by myselt.

myself.

[thick ledge]

I'm sorry I think you should ping helpers for this

[limber steppe]

and if I solve, can you pay about my answer?

[noble acorn]

Yeah no worries

<@&286206848099549185>

[oblique dune]

Why is chatgpt selling answers in chat?

[noble acorn]

I have no clue

[limber steppe]

I am human.

[oblique dune]

What is the question?

[noble acorn]

Here

[devout snow]

@noble acorn Has your question been resolved?

[devout snow]

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[devout snow]

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[wispy geyser]

am i right with A here?

[wispy geyser]

am i right with C here?

its 24/25 so its 0.96 right

[lunar harbor]

Yes

[thick ledge]

Yes (nobody saw that)

[wispy geyser]

tysm

.close

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[smoky locust]

Can I have help and advice for maths gcse

[autumn girder]

!da2a

[devout snow]

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

[smoky locust]

I have gcses over the next few weeks what topics should I prepare for that are the hardest/more likely to come up

Don't know what country you are from so do you know gcses

[south temple]

past papers are essential

focus on what you struggle with

[smoky locust]

I don't have any weaknesses strengths

I'm about a 7 right now aiming for 8

I'm about to do past paper shall i send when I'm done?

[south temple]

sure

[smoky locust]

For an 8 I need about 180 marks over 3 papers

For a 7 I need about 145 over 3 papers

[south temple]

focus on the back questions for the higher marks

[smoky locust]

I'll let you know when I start and stuff

[devout snow]

@smoky locust Has your question been resolved?

[smoky locust]

@south temple I've done the paper how can I send it

[south temple]

its best to mark yourself as it helps you to understand what the examiner is looking for

but send what you get wrong

[smoky locust]

Ok I will

@south temple if I got it right but with a different method do I get the marks

[south temple]

yes

[smoky locust]

Ok nice

[south temple]

for ervey wuetsion except one where the answer is given in the question

[smoky locust]

@south temple

Sorry most of them are sideways

[south temple]

For this one you can find the lengths of all the sides from the dimensions given as the triangles are isosceles. Then you can just add them all up

[smoky locust]

For which one

[south temple]

The first one

Part b

11b the 7-x multiplies by the k both parts so you get 7k-xk on the right

[smoky locust]

Oh right it's a bracket

Sorry I'm quite stupid

[south temple]

Nah you’re fine

[smoky locust]

I'm stupid I make so many mistakes

[south temple]

The best way to combat that is to just check over your answers once you are done

Try working out the other way if you can like expanding brackets of a factorise question

For 8 you got it very close. You just have to multiply the mean weight by the number of counters, add the three up and divide by 100 to get the mean

For 14 the only recurring digit is the one with the symbol above it so that would be 0.233333333 and so on

[smoky locust]

I'm so stupid I didn't even notice that

[south temple]

It’s a mistake you won’t make in the exam

All practice is good practice

For 16 I tend to say let f(x) = y then solve for that to reduce complexity. If you do your method you also need to reverse order of operations so do the subtraction first

[smoky locust]

Hopefully but I'm quite stupid so maybe

[south temple]

Practice makes perfect

And you’ll be able to focus better in the exam

[smoky locust]

I'm bad at focus in exam

I'm just bad at exams

[south temple]

In the end results aren’t everything

If you get what you want then that’s all that matters

[smoky locust]

How

[south temple]

There are always opportunities that open up in the future

In a few years no one will care what you get really

Just that you passed

[smoky locust]

My parents will be so mad at me for my exams

It is so annoying

[south temple]

If you do the best you can then nothing more can be expected

They probably did worse

Remember a 7 is an a

[smoky locust]

No they did very well they expect me to be really good and stuff

I'm just stupid overall

[south temple]

Half the country fail maths and English so I wouldn’t say your doing too bad

A 7 would be a great result

[smoky locust]

I hope so

I'll look over what you said properly after I eat

[smoky locust]

@south temple I'm going to do another paper thanks for the advice

[south temple]

Np

Glad to help

[smoky locust]

@south temple instead realised I don't know how to do surface area to volume

[south temple]

Find lengths then multiply

Gtg for today tho

Gl for exam tmr

[devout snow]

@smoky locust Has your question been resolved?

[devout snow]

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[limber ravine]

please help on this!! gcse vectors and i can’t figure out what to do next

[small quartz]

Hey

[limber ravine]

hey

[small quartz]

Familiar with the intercept theorem?

[limber ravine]

no i don’t think so😭

this is gcse maths so like US grade 10? i think

[small quartz]

I'm not sure, it's basic though

[limber ravine]

what is it i might know it

[small quartz]

Can you figure out the ratio MB:AC

[limber ravine]

5/6 b : x b?

that’s the only thing i can think of

is that right or wrong?

hellooo

[small quartz]

Familiar with this?

[limber ravine]

i’m really sorry no😭

[small quartz]

and this https://en.wikipedia.org/wiki/Intercept_theorem?

[limber ravine]

i don’t think that’s taught in the gcse course

[small quartz]

So I'm not sure what you should use here

I don't really know what gcse is

[limber ravine]

uk exams my maths paper 1 is tomorrow 9am

i need to use other vectors to find the proof for AC and prob will be an algebraic simultaneous equation in there🙏

if that’s any help

i’ve got a bit further

[worn flax]

find MN as a vector

then use similar triangles

NBM is similar to NAC (make sure you understand why)

oh tbh you don't even need to find OM as a vector i think

[limber ravine]

thank you!!

[limber ravine]

would i need to find NC vector?

[worn flax]

dont think so

[limber ravine]

alright thank you!!!

[limber ravine]

actually thank you so much i could cry this is the first time ive understood a vector and my exams in literally 9 hours ❤️❤️

[worn flax]

good luck!

you mostly did the right thing trying to figure out the vectors of the individual line segments, that's always a good thing to do

[restive tide]

ive been waiting for like 35 mins for help😭

[worn flax]

you just needed to look at the big picture here, and spot the similarity

[restive tide]

[limber ravine]

okk i’ll remember that for tomorrow🙏

[limber ravine]

thank you!!

[worn flax]

my friend says good luck (we're typing this from a uni in London)

[limber ravine]

haha thanks

[limber ravine]

that’s the goal 🙏🔥

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[wraith pumice]

can someone explain the steps for this

[wraith pumice]

all the bad handwriting stuff is me so it's not part of the problem I just didn't know how to do it

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[restive river]

Am i solving this right

[drifting sinew]

Does lg refer to the logarithm function?

[restive river]

Yes

base 10

[rare kernel]

log x^2 = 2 log x

[drifting sinew]

This doesn’t work, sadly.

[pseudo basin]

when you raise 10^(both sides) you have to write 10^LHS = 10^RHS without breaking either one up into individual terms.

[drifting sinew]

Do you want a hint as to what you can do instead?

[restive river]

Wait what do u mean?

[restive river]

Yes please

[drifting sinew]

If two logarithms are being added, you can commit them like this:

[pseudo basin]

$a+b=c$ becomes $10^{a+b} = 10^c$ and \textbf{not} $10^a + 10^b = 10^c$.

[woven radish]

Ann

[drifting sinew]

$$\log(a) + \log(b) = \log(ab)$$

[woven radish]

@drifting sinew

[restive river]

Like this?

[pseudo basin]

that works too

i do feel like i need to re-emphasize my point though because it is something more general that you need to keep in mind when doing algebra

[restive river]

Yes please do

[pseudo basin]

when applying an operation to both sides, you have to apply it to the entire LHS and the entire RHS -- and in general you cannot do it term-by-term, though in some cases you can.

i remember explaining this to another helpee in the context of quadratic equations and rooting both sides -- give me a moment to find it again

[restive river]

Ohhhh

Alr

I got it

[pseudo basin]

and 10^x is also a nonlinear function and does not distribute over addition,

in the sense that 10^(a+b) is not the same thing as 10^a + 10^b

it does simplify to 10^a * 10^b, but that is a different matter.

[restive river]

Ahhhhh okay thank you

I think i got it now

.close

[devout snow]

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[stable swift]

Okay so I don't undertnad where the fractions are coming in from?

factors of 5: 1,-1,5,-5

I get that

factors 2: 2,-2,1,-1

okay

[brave plaza]

you might wanna read https://en.wikipedia.org/wiki/Rational_root_theorem

[stable swift]

Okay I read it

still not getting where the fractions are coming from

[fossil locust]

plusminus (factors of 2)/(factors of 5)

so plusminus (1, 2)/(1, 5)

[stable swift]

yeah I got that

[stable swift]

right here

but I guess I pick to do fraction with these two

by random?

[fossil locust]

so you can choose any of 1, 2 for the numerator

[lilac crescent]

Is this the entire polynomial?

[stable swift]

whats the reasoning here I don't get it

[fossil locust]

and any of 1, 5 for the denominator

1/1, 1/5, 2/1, 2/5

and of course all of those but negative

[stable swift]

yes

yeah like I don't understand where 2/1 is coming from at all

or 1/5

[lilac crescent]

How 2 is a zero of that polynomial?

[stable swift]

I have the two lines and now I am just making random fractions?

[fossil locust]

you're making all possible fractions out of the factors

[fossil locust]

it's not

but it's a candidate

[lilac crescent]

I see

[winter patrol]

its an unideally worded question

[fossil locust]

these are all the possible candidates, i.e no other rational number can be a zero of f(X)

[stable swift]

Yeah I just don't get where the fractions are coming from or how to get them. Everything else is explained except that

[lilac crescent]

Why

[winter patrol]

for what south just said.

[fossil locust]

to list all possible zeros

it's worded just fine

the context is clearly the rational root theorem

[winter patrol]

idelly the question should say
list all candidates for rational zeroes

[fossil locust]

OH

yeah subtle difference but important

[lilac crescent]

Make sens

[fossil locust]

how to get the fractions is the easy part

where the fractions comes from involves explaining the proof

[stable swift]

I have no idea what you mean

show me how to get the fractions

[fossil locust]

I just told you?

[fossil locust]

start from here

[stable swift]

Okay lets start over because I don't know what exactly you're responding to at any time

How do I get the fractions? Where are they coming from? Show me

[fossil locust]

you said you understand that the numerator can be the factors of 2, so 1 and 2 right?

and the denominator can be the factors of 5, so 1 and 5 right

[stable swift]

Okay but you said not to use the factors for the fractions

[fossil locust]

I don't remember saying that

[stable swift]

too many cooks in the kitchen

[lilac crescent]

As south said , you get the multiples of the constant and the greatest term

[fossil locust]

you mean the factors

[lilac crescent]

Yep , sorry

@stable swift

[stable swift]

Okay so I am making factors of first and last numbers

then making fractions out of those factors?

[lilac crescent]

Yep

[stable swift]

okay lol

thx

[lilac crescent]

All possible fractions

With +-

[devout snow]

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[loud cave]

need help with this

[wind mason]

Do you know how to break it down into cases?

[loud cave]

yes

[wind mason]

So do it.

[loud cave]

wait

i was a bit lost here

[wind mason]

Aha, there's also case(s) involving x=1

[loud cave]

wait could you elaborate?

[wind mason]

Its using the interval method

You take all of the roots (or asymptotes) of the function and make a sign chart

Usually that is for things > or < 0

But you're on the right track

[loud cave]

where does the x=1 come from?

[wind mason]

^

[loud cave]

ohh

i got until here

<@&286206848099549185>

[devout snow]

@loud cave Has your question been resolved?

[devout snow]

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[unique kelp]

Is there any 1-1 mapping between R and R^2? Just out of curiosity

[sand dove]

yeah there must exist some

[acoustic leaf]

we know it definitely exists, it's a bit tricky to explicitly write it down

[unique kelp]

Huh

Tricky or impossible?

Or unsolved?

[supple knot]

https://en.wikipedia.org/wiki/Space-filling_curve

[acoustic leaf]

not impossible, but all of the answers are inherently not very simple

[supple knot]

that?

[unique kelp]

Technically thats only a part of R^2, no? Or is that analogous in some way

[unique kelp]

Makes sense

[lilac moat]

you can just compose with a map of R2 to a unit square though

[sand dove]

You can try to create a bijection between [0,1] and R

then if you have a bijection between [0,1] and [0,1] x [0,1]

it'll be easy to make one between R and R^2

[unique kelp]

That seems difficult in its own right, though maybe that's just cause I have studied whatever field of math this is

Haven't*

[sand dove]

making a bijection between (0,1) and R is easy, take f(t) = tan(pi (t-1/2))

[lilac moat]

proving it exists is much easier, from the Cantor Bernstein Schroeder theorem

do you know it

[unique kelp]

Oh valid didn't think of asymptotic functions

[sand dove]

then for going between (0,1) and [0,1], you just need to store those extra end points inside

so make for example f(0) = 1/3, and f(1) = 2/3

[unique kelp]

No lol

[sand dove]

then f(1/3) = 1/9, f(1/9) = 1/27, etc...

and f(2/3) = 8/9, f(8/9) = 26/27, etc...

and f(x) = x for the others

[lilac moat]

for two sets A and B
If there is an injection from A to B
and an injection from B to A
then there is a bijection between A and B

[unique kelp]

Remind me what an injection is?

[lilac moat]

f(a)=f(b) implies a=b

[sand dove]

no two different inputs have the same output

maybe better formulated that way

anyways

[unique kelp]

Cool

[lilac moat]

this is the contrapositive, if a!=b, then f(a)!=f(b), which can be more intuitive sure

[sand dove]

so anyways with those

[lilac moat]

So, with that in mind, let's find an injection from R to R2, and from R2 to R
I'll do the hard part, you do the easy part

hard part:
R to R^2, f(x)=(x,0)

okay you do the easy part

sorry thats backwards

[sand dove]

that's not very injective

[wicked turtle]

another way, enumerate the rationals in (0,1), then right-shift that sequence by two slots in order to make room for 0 and 1 (and map each irrational to itself)

[lilac moat]

lol I did the exact opposite of what I wanted

fixed

[unique kelp]

Ok r to r2 would just be f(x) = (x,x)?

Oh

[sand dove]

sure why not

[unique kelp]

You changed it

[lilac moat]

I was joking about it being hard, that direction is much easier than the other direction

[unique kelp]

So then we need r2 to r

Hmmmm

[lilac moat]

you willing to accept that all real numbers have exactly 1 or 2 decimal expansions?

that's not a trivial fact, but it makes life easier for this

[unique kelp]

Wait lemme think

[sand dove]

they have exactly one decimal expansion that doesn't end with repeating 9s

[unique kelp]

I wanna see if I can get this

[sand dove]

try to go for that

[lilac moat]

yeah. that's not OBVIOUSLY true, but it is true, and if you're willing to accept it, this problem becomes significantly more manageable

[unique kelp]

Something like s_i is the ith digit of f(x,y)

Then s_i = x_(i/2) if i is even, y_(i/2-0.5) if i is odd?

Oop

Like just zipper in the digits sort of

[lilac moat]

yeah!

just have to figure out how to write that precisely

but you got it

[unique kelp]

Sure

Ok cool I should go now but thanks everyone

This has been enlightening

[lilac moat]

if you want to learn more about this kind of stuff

the first chapter of Munkres's Topology is not about topology, it's about sets and logic

and deals with cardinality in a way that doesn't require cardinal arithmetic (arithmetic with infinite numbers)

[unique kelp]

Screenshot taken :)

Ok bai

.close

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[storm kiln]

<@&268886789983436800>

[devout snow]

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[storm kiln]

#prealg-and-algebra #geometry-and-trigonometry #precalculus

[eternal iris]

Can mods not just ban them and auto remove all messages they sent?

[wicked turtle]

sometimes the bot fails to clean up all the messages

[devout snow]

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[hasty ether]

idk if this counts as math in this server but it is propabilities, and im very confused on what this says and what is N and n... i tried searching up online but i cant even find my asnwer as idk what it is it about... sorry of my poor english

[thick ledge]

N is the total number of outcomes for an event

n is the number of favourable outcomes for an event

[hasty ether]

what does that mean? favouruable outcomes? like what the question wants ?

[thick ledge]

Yep

[polar chasm]

Event: even number is rolled on a 6 sided dice
Favorable outcomes: 2 is rolled, 4 is rolled, 6 is rolled
Total outcomes: 1 is rolled, 2 is rolled, ..., 6 is rolled

n = 3
N = 6

P(A) = n/N = 3/6 = 1/2

[thick ledge]

[polar chasm]

just an example

[hasty ether]

so the probability of an event is equal to OH ooooo okay

so

the multiplication rule

is refering to basically how we view the ratios

like the wanted value on top?

[polar chasm]

The one which says P(A and B) = P(A) * P(B)?

[hasty ether]

ah no its this one, i was super confused in linking both paragraphs but okok so its just how we do normal ratios

[thick ledge]

That's only true for independent events no?

[hasty ether]

aaa yes

im still trying to understand all of this still

[polar chasm]

Using the rolling dice analogy again, roll dice twice, so there are 2 components. Now you can create a table like this:

this table will have 6 rows for the first roll and 6 cols for the second roll

total of 6 x 6 = 36 outcomes

and it works similarly for higher dimensions

[hasty ether]

ooooo okay so this is just proper name for that process

okay yah!!! i got it, makes sense now, ty <3333

.close

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[pearl grotto]

yo

[pearl grotto]

idk where to start on B

i got
s = h

u = ?

v = 0

a = 4/9g

t = ?

[devout snow]

@pearl grotto Has your question been resolved?

[thick ledge]

For which block

[pearl grotto]

B

[thick ledge]

Find the time of flight for B until it hits the ground

Then use the same time to find the velocity of A at that instant

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@pearl grotto Has your question been resolved?

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[maiden bridge]

https://projecteuler.net/problem=13
Is there any clever way to solve this problem?

[maiden bridge]

print(str(sum(int(i) for i in nums.split()))[:10])
This is my approach

just took the sum of numbers and printed the first 10 digits

[kind tiger]

i imagine if youre trying to be a mega sweat and doing it in C then you could speed it up in some ways

but otherwise i wouldnt say so, i could be wrong of course

[maiden bridge]

I was reading the problem thread on project euler

[kind tiger]

yeah thats the most obvious thing you can do, but then theres some level of computation in taking only those last 10 digitis of each number

[maiden bridge]

makes sense

would just increase the computation time?

[kind tiger]

so i wasnt sure if its acc worth it (at least in python)

idk might make it faster

[maiden bridge]

also this method works just for this set of numbers. might not work for some other set of 50 digit numbers

[stone stump]

you could easily have carries which change the result

[kind tiger]

huh?

[kind tiger]

no it would work for any numbers

[maiden bridge]

no i am talking about this method

[kind tiger]

yeah...

that would work for any set of numbers

[maiden bridge]

this would cause problems if we use this method

[kind tiger]

if you write the numbers each as
a1 = b1 * 10^10 + c1
a2 = b2 * 10^10 + c2
....
then the sum of ai = (sum of bi) * 10^10 + (sum of ci)
so the last ten digits of sum of ai will be the same as the last ten of sum of ci, and ci are precisely the last 10 digits of each of the ai

and obv this can be generalised to the last n digits if you wanted, could even work in a different base etc etc

[maiden bridge]

wait

don't we have to find the first 10 digits?

[mortal vale]

yeah thats what im thinking

[maiden bridge]

yup

[kind tiger]

OH the front 10

mbmb

[maiden bridge]

yes the front ones

[kind tiger]

whoops 😅

[maiden bridge]

i submitted the 10 in the front and it worked

i did think of working with mod 10^n

but that would only work if we were talking about last n digits right

[kind tiger]

yeah

[mortal vale]

i think one approach is to add the the first digits of each number together, then the second, and so on. after the 12th digit of each number, you can then apply a check to see if the sum of the 13th digits of all the numbers can carry over

if it cant carry over, u can simply return the result

[maiden bridge]

do you mean i should sum in reverse?

like from the front

[stone stump]

given that you are adding 50 numbers you can at most have a carry of 4. so just some playing around with that will give you the result

[mortal vale]

yes, but starting at the 13th digit, u can add a check that stops the calculation if it cant carry over, which can save on computation

but at this point its also relevant talking about how the numbers are stored

in memory

[stone stump]

no wait the fuck am I saying

much bigger carries are possible

[kind tiger]

this can be made more general, i.e. not using the specific knowledge of the problem

[maiden bridge]

highest would be 450 right

if all the digits are 9

no wait

there are 100 of these

so 900

[maiden bridge]

makes sense

[stone stump]

oh whoops 100 numbers

[kind tiger]

||if you have k numbers where the longest number has m digits and you are interested in the front n digits
the maximum value that the last l digits can add up to is k(10^l - 1) and we want that to be < 10^(m-n)
k(10^l - 1) < 10^(m-n)
10^l < 10^(m-n) / k + 1
l < log10 (1 + 10^(m-n) / k)
l = floor( log10 (1 + 10^(m-n) / k ) )
and you only need to worry about the m - l front digits||

[stone stump]

yeah 900 sum, so up to 90 carry

and of course that carry could apply to a 900

to give you 990, so a 99 carry to the next

[kind tiger]

applying that to this particular problem
||k = 100, m = 50, n = 10
so l = floor( log10 (1 + 10^40 / 100) ) = floor( log10 (1 + 10^38) ) = 38
so we only need to sum the front m - l = 12 digits, anything after the 12th (from the front) digit is not possible of affecting the front 10||
which pretty much lines up with what the post says (they only use 11 because my calculations assume worst case, i.e. the last digits of all numbers are 9s

[maiden bridge]

thats crazy

[maiden bridge]

yupp

[maiden bridge]

i kind of got overwhelmed on seeing this, but looks easy when i read it twice

[mortal vale]

keep in mind that the sum of the 13th digits can affect the 11th digit, which could cause it to carry over to the 10th digit

but this is not guranteed, so we check if it happens

[kind tiger]

oh right right

true

hm

you could just send n->n+1 and then itd affect even fewer cases

[maiden bridge]

oh yea in the case of 900 for example