help-0

[raw shard]

y’ = dy/dx = f’(x)

[peak coral]

So if I was to put my question in it would be ?

[raw shard]

do you want to see the answer?

i assume the less work the better for you lol

but just in case i asked

[peak coral]

Yes if you could please I’ve looked at my working again and am struggling to see the part I’ve gone wrong

I don’t mind the working at all, infact I need it so I can understand it just no point working towards the wrong answer lol

[raw shard]

,w solution to y’’+4y’+8y = sin(3t)

[ocean seal]

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[peak coral]

Ah, so it would appear the first bit was correct but not my particular solution?

I just can’t seem to figure out a way in which I’d have my denominator as 145 ?

[raw shard]

at the very least your second derivative of the particular solution guess is wrong

it had coefficients of 6

well, -6

the negative part is correct

it should be Y’’_p(t) = -9Acos(3t)-9Bsin(3t)

[peak coral]

Oh of course

Sin3t > 3cos3t > -9sin3x

[raw shard]

yeah

i actually messed up my derivative at first as well lol

[peak coral]

So it would then be -9Acos3t-9Bsin3t

[raw shard]

yes

[peak coral]

So then I get A=-12B

[raw shard]

it’s A = 12B

[peak coral]

Oh yeah my mistake lol

[raw shard]

my system of equations was\ $-12A-B = 1 \ 12B-A = 0$

[ocean seal]

quantum

[raw shard]

too much editing

of course i could have divided the negative out but i didn’t for some reason

[peak coral]

So it is A=12B? No ?

The bottom one

[raw shard]

yes

[peak coral]

Ohhh

Then -144-B=1

Giving -1/145

Then A=12B so A = 12/145

-12/145 that is

[raw shard]

A = -12/145

,w 12A+B = -1, 12B-A = 0

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[peak coral]

Ah I see thank you very much

[raw shard]

no problem

[peak coral]

To be honest I’m a bit too confused to even attempt the bottom bit

[raw shard]

i can help

[peak coral]

Where it asks about showing A=B however this is different A and B as the A and B we were discussing is the questions p and q if that makes sense

[raw shard]

by the bottom bit, you mean the part at the very bottom right

just making sure

[peak coral]

Yeah

C)

[raw shard]

could you take a better picture

[peak coral]

Sure

[raw shard]

that part is a bit too blurry

[peak coral]

Hopefully that’s readable

[raw shard]

it is

wait

what do they even mean

by A = B

[peak coral]

Exactly lol

[raw shard]

yeah uh

[peak coral]

A ≠ B if I’m not mistaken no?

[raw shard]

i can’t help because they made this solution sheet terribly

[peak coral]

They have indeed lol

[raw shard]

well, do you need help on anything else?

[peak coral]

Erm just this one question actually it’s sort of related

[raw shard]

ok

[peak coral]

(-1+2j)^10

Evaluate the following

[raw shard]

are you in some sort of electrical engineering program or something?

[peak coral]

Mechanical engineering lol

[raw shard]

because that’s the only place i’m aware they call i j

oh

[peak coral]

Yeah it is ahahahha

[raw shard]

well anyways

[peak coral]

My lecturer is an electrical engineer

And says he’ll write j however i is also acceptable

So I’ve always just learnt by j

[raw shard]

well it doesn’t matter for me because i know what you mean

i would say either use the binomial theorem or suffer and multiply it out manually

lol

of course, just do ^5 first

then square it afterwards

[peak coral]

Problem is using binomial theorem is also giving me a “wrong” answer

As online it says I should be getting around 3116i however mine is in the 9000’s

[raw shard]

what are you getting

oh

i can try using the binomial theorem then

wait a second i forgot

[peak coral]

And for some reason when I did the (^5)(^2) it’s giving me 400i

414 to be exact

[raw shard]

you can just square (-1+2i), square it again, then multiply it by (-1+2i) to get (-1+2i)^5

then square it again after that

not sure how i forgot that lol

[peak coral]

Square it twice and multiply the answer by itself?

[raw shard]

yeah

that’s (-1+2i)^5

wait uh

[raw shard]

do ((-1+2i)^2)^2 * (-1+2i)

[peak coral]

Yeah that’s working somehow

[raw shard]

oh nice

[peak coral]

How did u know that would work?

[raw shard]

because the exponents add

[peak coral]

I can’t seem to comprehend how that would =^10

Ohhhhh

[raw shard]

((-1+2i)^2)^2 * (-1+2i) = (-1+2i)^5

square it again and it’s (-1+2i)^10

[peak coral]

I’ve never ever thought about doing it that way before lol thanks very much

[raw shard]

no problem

you can ask more questions, but whenever you’re done, do .close

[peak coral]

I think that’s all from me you’ll be glad to hear, thanks very much you’ve been a great help

[raw shard]

you’re welcome

@peak coral wanna make sure you don’t forget

[peak coral]

Yeah no worries

.close

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[peak coral]

Thanks again

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[versed mica]

this is all the information i have, how to solve for the question mark area?

[versed mica]

i’ve got an answer, but i’m not sure if i solved for it correctly

first, i found the intercept (3.2,1.6) by graphing

then, i found the area between x=4 and the edges of the circles with two integrals.

integral 1:
upper bound: 4
lower bound: 3.2
integrand: 1.6-sqrt(4 - (x-2)^2)
dx

integral 2:
upper bound: 4
lower bound: 2
integrand: 4-sqrt(16 - x^2)
dx

answer to integral 2 minus answer to integral 1

then, 16-4pi-answer above = a

then, 2pi-a= the answer

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@versed mica Has your question been resolved?

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@versed mica Has your question been resolved?

[cursive badger]

let's see, 16=(y-4)^2+x^2 and (x-2)^2+y^2=4

I'll take your word for it, that they intercept at (3.2,1.6)

the integral from 0 to 3.2 of the first semicircle minus the second one so

$\int_0^{3.2}{\sqrt{4-(x-2)^2}-(4-\sqrt{16-x^2})}dx$

@versed mica

yeah, that should be it

I'm not sure you'll be able to integrate that algebraically though

no need for two integrals

forgive me again, I am an idiot

you only need the ? area

$\int_0^{3.2}{\sqrt{4-(x-2)^2}-(4-\sqrt{16-x^2})}dx$

so this

[ocean seal]

Scythe

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[strong fractal]

Can someone who is good with logic gates let me know if I'm understanding the question correctly? Thanks 🙂

[bleak ridge]

That looks right to me @strong fractal

[strong fractal]

Ok, thanks bud

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[alpine sable]

hi

[lapis ingot]

well the scale is 5cm/1000km

so just multiply the distance by the scale

[alpine sable]

nvm its 4

[lapis ingot]

do .close

[alpine sable]

.close

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[weary hemlock]

Hello im stuck at question c

Can anyone give me some hints? Thanks in advance

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[weary hemlock]

.reopen

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✅

[lost ledge]

what the length of KL

[lone heart]

@weary hemlock Has your question been resolved?

[weary hemlock]

KL is not given. Is there anyway to find KL?

.close

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[ember vigil]

Hello why it isn't answer D?.
It is a sum of 3's.
And I guess it 2022/2 times

[glad delta]

It isnt 2022/2 times, that would be the case if the last one was 2-(-1), instead the last one is 4-1 so thats one time less

[ember vigil]

Makes sense tnx👍

.close

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[humble wren]

hello, if i have 2 arrays, a=[0,1,2,3,4,5,6,7,8,9,10] and b=[5,4.5,4,3.5,3,2.5,2,1.5,1,0.5,0] is there a formula to calculate the value from b that represents, let's say the value 3 from a? knowing 0 from a is represented by 5 from b and 10 from a is represented by 0 in b?

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[humble wren]

.close

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[hollow pelican]

hi guys how to solve this question?

[hollow pelican]

my ans is wrong

this is the correct ans

[finite linden]

It’s 3x^2 instead of x^2 in the numerator

So there’s a multiple of 3 which cancels the 1/3 from ur answer

[lone heart]

@hollow pelican Has your question been resolved?

[hollow pelican]

oh so

[hollow pelican]

it should be du/3 instead of du/6?

[finite linden]

du/2

X squared is du/6 so 3x squared is du/2

[hollow pelican]

ohh ok

[lone heart]

@hollow pelican Has your question been resolved?

[hollow pelican]

i still cant get the ans tho

[vague coral]

but you are basically done

u = 2x^3 + 1

[hollow pelican]

but is still the wrong ans

the ans is this

[raw shard]

the numerator is the exact derivative of what’s in the square root

there’s no need to multiply by a constant

oh my gosh

i’m dumb

[hollow pelican]

huh

[raw shard]

$u = 2x^3+1 \ \dd{u} = 6x^2 \implies\frac{\dd{u}}{2} = 3x^2 \ \int \frac{3x^2}{\sqrt{2x^3+1}} \dd{x} = \frac{1}{2} \int \frac{1}{\sqrt{u}} \dd{u}$

wow that’s a lot of latex

and i messed up

amazing

[ocean seal]

quantum
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[raw shard]

whatever this is good enough

[hollow pelican]

aight thanks

[raw shard]

wait a second

that is the answer

their answer is correct

thought it was wrong for a second lol

$u = 2x^3+1 \ \dd{u} = 6x^2 \implies\frac{\dd{u}}{2} = 3x^2 \ \int \frac{3x^2}{\sqrt{2x^3+1}} \dd{x} = \frac{1}{2} \int \frac{1}{\sqrt{u}} \dd{u} = \sqrt{u}+C = \sqrt{2x^3+1}+C$

[ocean seal]

quantum

[raw shard]

@hollow pelican if you’re done do .close

[hollow pelican]

ok thank u

.close

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[snow portal]

Does anyone wanna help me attack this fun problem😅

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[timber geyser]

help withthis question plz

[timber geyser]

i differentiated dP/dt to get 0.02+0.03cos(0.6t)

then input t=0

which gives 0.05

which is wrong

[velvet pelican]

Hmm, I think that looks about right

Is it something to do with units?

[timber geyser]

i dont think so, one sec ill paste the answer

i dont know where i went wrong, my method seems right to me

[velvet pelican]

So you should get 0.015

[timber geyser]

ye

[velvet pelican]

Idk your thing looks right to me

[bleak ridge]

,w d/dt 1+0.02t+0.05 sin(0.6t)

[ocean seal]

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[timber geyser]

its probably an error on the ms then i guess

[bleak ridge]

,calc 0.03 * cos(0.06 * 0) + 0.02

[ocean seal]

Result:

0.05

[bleak ridge]

,calc 0.05 * 1000000000

[ocean seal]

Result:

5e+7

[bleak ridge]

Hmm

,calc 0.03 * cos(0.06 * 1) + 0.02

[ocean seal]

Result:

0.049946016198056

[bleak ridge]

,calc (1 + 0.02(1) * sin(0.6 * 1))-( 1 + 0.02(0) * sin(0.6 * 0))

[ocean seal]

Result:

0.011292849467901

[bleak ridge]

I dont see how you could get 0.015

[timber geyser]

ye, its probably an error on the marking then

[bleak ridge]

Yeah prob

[timber geyser]

thankyou!

.close

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[fast mantle]

very dumb q but can someone help me count number of triangles

[fast mantle]

it's between 22-25

[raw shard]

i counted 25

[fast mantle]

thanksss

[true sonnet]

i counted 23 somehow

[ancient saddle]

I counted 25 too

[true sonnet]

i feel dumb now xd

[fast mantle]

dont worry bc same lol

thank you guyss

.close

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[hollow pelican]

hi guys for this question can i times in the (x^2-x) to the (x^2-3/2x^2)^8

[hollow pelican]

then solve from there

like this

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@hollow pelican Has your question been resolved?

[hollow pelican]

or issit like this

<@&286206848099549185>

[wanton turtle]

This is incorrect

[wanton turtle]

On the right track here, there's an extra term in your last step

(x^2 - x) dx is du/3, so the last term won't have (x^2 -x) since it has already been taken by du/3

[hollow pelican]

so i should do like this?

[wanton turtle]

No

[wanton turtle]

The substitution here is perfectly correct

The last step on the left side of the page shouldn't have (x^2 -x)

It should be (1/3) integration (u)^8 du

[hollow pelican]

oh i am abit confuse, is it like this?

[wanton turtle]

Yup

du missing in the last step there

[hollow pelican]

ok

but like

im confuse

why the whole term just dissapears

[wanton turtle]

Alright I'll try to give an extremely simple example which you can relate with this question

Suppose y = a . b . c and d = b . c

Can I say y = a . d?

[hollow pelican]

yea

[wanton turtle]

Why is it not y = a . b . d?

[hollow pelican]

cuz d is b.c

[wanton turtle]

Because then it becomes y = a . b^2 . c, an extra b is multiplied

[hollow pelican]

i see

[wanton turtle]

So the du/3 term here already accounts for the x^2 - x term

[hollow pelican]

Ohh

[wanton turtle]

If you reverse the substitution here you'll see that the question changes, it becomes (x^2 - x)^2

[wanton turtle]

Just like this

[hollow pelican]

I see

[hollow pelican]

understood thanks!

[wanton turtle]

Be sure to .close the channel

[hollow pelican]

okay

.close

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[hollow pelican]

for this question can i do the du like this?

[wanton turtle]

You're simplifying the substitution too much

The original question already has a 2x dx term, which you also got in term of du

so take -du = 2x dx and you're good to go

[hollow pelican]

okay

[hollow pelican]

for this du, is du/2 correct?

[wanton turtle]

2 du

Multiply both sides by 2 and you get 2 du = (8x - 2)dx, which is already present in the numerator of the question

[hollow pelican]

ok

[hollow pelican]

the 2 can put outside of the integral sign?

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@hollow pelican Has your question been resolved?

[wanton turtle]

Yup, constants can be taken out

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[sterile vale]

Hello, I am asked to work with change of basis matrixes

[sterile vale]

And I am asked that for the coordinates (2,1) in a basis B1, the first coordinate in another basis B2 is 0

How do i do this?

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[sterile vale]

<@&286206848099549185>

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[crisp iris]

How come this limit is not equal to 1?

[crisp iris]

Here is my demonstration, what's wrong with it?

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[royal meadow]

you can't just split the limit up like that unless both of the bits are finite?

the base goes to 1 from below and the exponent goes to infinity

so you can get some weirdness

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[tiny quartz]

Wrong discord so sorry

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[zinc gate]

Is there any common notation for the product of all elements of a vector? As in $\prod_{i=1}^n x_i$ for an $n$-dimensional vector $\vb{x}$

[ocean seal]

mummiedanser

[glass lichen]

yes, what you just wrote

[zinc gate]

Yeah i meant like something less big

I need to write it a lot in my document and it looks cluttered

[glass lichen]

$P(\vec{x}):=\prod_{i=1}^nx_i$

[ocean seal]

Mosh

[glass lichen]

then just refer to P(x).

[zinc gate]

Ah i suppose that would be best

Thanks!

.close

[lone heart]

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[echo peak]

i'm not fully sure how to factor the problem into the form that it wants, when i factor it i get h(t)=16(-t-3)(t-6)

[wary stream]

You don't need to factor like that, you can just complete the square

[echo peak]

oh ok thank you

[lone heart]

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[pallid grove]

can someone explain to me how to solve this cubic equation

[glass lichen]

factor theorem should be trivial

then it's just factoring a quadratic after the long division

[pallid grove]

well i dont know what that is

can u explain

[glass lichen]

"that" is so specific.

[pallid grove]

reference to your initial statement

[glass lichen]

well how did you show q=-5?

[pallid grove]

subbed in x-4

x for -4

[glass lichen]

why?

[pallid grove]

cause x = -4 is given by x+4 is a factor

[glass lichen]

yes, that's factor theorem

x-a is a factor of p(x) iff p(a)=0

[pallid grove]

ok but why long division is that the same as synthetic division

i looked up how to solve and i got synthetic division which is convoluted

[glass lichen]

synthetic is a shortcut for long division.

the same way as any other polynomial long division..

[bright hedge]

do you know how to do polynomial long division when its just numbers as the coefficients

[raw shard]

https://www.mathsisfun.com/algebra/polynomials-division-long.html

[glass lichen]

Google how to do polynomial long division, or refer to your notes

So make notes

Not my problem if you choose to be a bad student and not take notes. Notes really do help when you're practicing problems.. as well as allow you to retain taught material

[lone heart]

@pallid grove Has your question been resolved?

[mystic mulch]

You're being very aggressive towards people trying to help for free. Just giving you the answer is not a good way to teach so he's trying to point you in the direction you need to go.

[lone heart]

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[hollow pelican]

HI guys how to do 3.7b

[raw shard]

e^(x+1) = e*e^x

sqrt(ab) = sqrt(a)*sqrt(b)

[hollow pelican]

oj

ok*

[hollow pelican]

do i need to do u sub for this question?

[raw shard]

no

@hollow pelican close the channel if you’re done

[hollow pelican]

ohh

i am still confuse

havent solve it yet

[raw shard]

oh ok

[hollow pelican]

they used u sub tho on online calculator

[glass lichen]

yeah, cause Symbolab is stupid

[hollow pelican]

ohh

[raw shard]

$\int \frac{3}{\sqrt{e^{x+1}}} \dd{x} = \int \frac{3}{\sqrt{e \cdot e^{x}}} \dd{x}$

oh gosh

[ocean seal]

quantum

[raw shard]

$\int \frac{3}{\sqrt{e \cdot e^{x}}} \dd{x} = \frac{1}{\sqrt{e}} \cdot \int \frac{3}{\sqrt{e^x}} \dd{x} = \frac{1}{\sqrt{e}} \cdot \int 3e^{-\frac{x}{2}} \dd{x}$

[ocean seal]

quantum

[hollow pelican]

ohh

[sand pawn]

$$ \frac{3}{\sqrt{e}} \int{e^{-\frac{x}{2}}} dx$$

[ocean seal]

👑Overlord Prince Khan😈

[hollow pelican]

okayy

[sand pawn]

then follow it up:

https://youtu.be/RqRQYX1emOA

here, a = -1/2

[raw shard]

oh darn i just remembered that a u sub is required lol

[sand pawn]

lmao

happens man

no worries

[hollow pelican]

ok

[sand pawn]

Close this if you're done

.close

[lone heart]

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[lone heart]

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[alpine sable]

hi

[alpine sable]

<@&286206848099549185>

[steep briar]

10 minutes ain't equal to 15, lol. Anyways, you're trying to find x?

[alpine sable]

ya

[steep briar]

Okay, so the goal is to figure out the scale factor from the triangle ABE, to the triangle ADC.

[alpine sable]

o

[steep briar]

So we know the "hypotenuse" side of the small triangle is 40, but what about the big triangle?

[alpine sable]

isnt the hypotenuse 25

[steep briar]

The side that the 95° angle isn't touching

[alpine sable]

o

uh

[alpine sable]

dont know the big

[steep briar]

(hypotenuse is usually the side opposite of the 90°, but here 95° is the closest to that)

[alpine sable]

o ok

[steep briar]

Well we see that 95° for the small triangle is the angle ABE, but for the big triangle it's the angle ADC

[alpine sable]

ya

[steep briar]

So the side AB corresponds to the side AD, while BE corresponds to DC

[alpine sable]

oh

i got 32 for x and it was wrong and now i see why

i didnt think much of the angles-

[steep briar]

Well it's just to get a bearing of what side corresponds to what

[alpine sable]

o

[steep briar]

From this logic, what side of the big triangle would correspond to the side AE?

[alpine sable]

bc?

[steep briar]

(To draw the connect again from what I said above: AB__E AD__C and A__BE A__DC)

[alpine sable]

is it bc?

[steep briar]

Well we know the side has to have A

[alpine sable]

oh

ac

[steep briar]

Since A corresponds to the same point on both triangles, and yes, it'd be AC (so AE~AC)

So we know the length of AE is 40, what about AC?

[alpine sable]

50?

[steep briar]

Yep, and so the scale factor?

[alpine sable]

40/50

o'

[steep briar]

Well to go from the smaller to bigger it'd be 5/4

[alpine sable]

5/4

[steep briar]

Yea

Now we're looking for x, which is the bottom side

[alpine sable]

5/4 times 16

[steep briar]

Yep

[alpine sable]

can i ask 1 more

[steep briar]

I'll have to see it first

[alpine sable]

ans is 20

okay

[steep briar]

Agreed

[steep briar]

Oh same deal,

[alpine sable]

what happend here

[steep briar]

So we see AB~AD, so what'd the scale factor be?

[alpine sable]

the other side isnt given

[alpine sable]

7/5?

oh wait

7/12?

or 12/7

[steep briar]

Ye 12/7

[alpine sable]

is theans 12

the ans

[steep briar]

Yea

[alpine sable]

alrighty ty!

[steep briar]

Np, gl!

[alpine sable]

<:

.close

[lone heart]

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[ionic thistle]

1 APPLE TREE produces 1 APPLE SEED a day.
10 APPLE SEEDS make 1 APPLE TREE.
Johnny has 2 APPLE TREES. If he keeps compounding his APPLE SEEDS into APPLE TREES, how many days before he has 100 APPLE TREES? Hint: on the 5th day, Johnny would have 10 APPLE SEEDS and could thus make 1 APPLE TREE on that day. Attached is a picture of how I tried to calculate this

[ionic thistle]

I figured he’d have 10 APPLE TREES by day 20, but after that it gets fuzzy

Each X represents an apple tree being created. And each tally represents a seed

[lone heart]

@ionic thistle Has your question been resolved?

[ionic thistle]

<@&286206848099549185> I’m in need.. Is it true that by day 35 Johnny would have 100
APPLE TREES

[lone heart]

@ionic thistle Has your question been resolved?

[alpine sable]

The hint makes our calc alot easier

Every 5 days 1 apple tree is added

let d be days by which 100 trees be produced

d / 5 = 100

like -
1 apple tree per 5 days = 1 apple tree / 5 days = 1 apple tree * 98 / 5 * 98 = 98 apple trees / 490 days

oh i forgot about the 2 apple trees he already has

sorry

let me edit the one above

[ionic thistle]

It’s okay!

Thank you!

.close

[lone heart]

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[alpine sable]

can someone please confirm if i have this question correct please this is for i)

[alpine sable]

yeah you're good

[alpine sable]

wait i got it right?

Oh wait no

but 2(-3)^2 makes 18 not 12 👀

The form is x - a

So a would be 3

[alpine sable]

what about this

[alpine sable]

yeah but that would make it - 3 right

Nope

the - is infront of the 3

If you subtract a negative it's a positive

x - (-3) = x + 3

so it's just a 3

x - 3

what about this but 2(-3)^2 makes 18 not 12 :eyes:

It does make 18

so i have used the wrong number for the function then?

hmm?

If you do (x - 3)^2 you get x^2 - 6x + 9

2(x^2 - 6x + 9) = 2x^2 - 12x + 18

[lone heart]

@alpine sable Has your question been resolved?

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[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[alpine sable]

this is the graph and the minimum spanning tree generated based from above.

the minimum cost of connecting the computers is 50 and if computer a must be connected to computer b, the difference it makes to the answer is 2
are they all correct based on the instructions given above?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[vale wigeon]

what tree do you get if edge AB is required to be included?

[alpine sable]

included in the minimum spanning tree?

102

[vale wigeon]

??

if you just add AB to the tree you've constructed you won't have a spanning tree anymore.

no, you should rerun kruskal's algorithm and get a new tree that starts out with AB.

[alpine sable]

how do i do that + i don't know what you mean

the site that i use already says that the minimum spanning tree is 50

[vale wigeon]

oh, so you're using a website instead of applying kruskal's algorithm yourself??

[alpine sable]

ye

[vale wigeon]

do you even know how kruskal's algorithm works?

if you did, i'm sure you would have no trouble modifying it to force an edge to be included.

[alpine sable]

we were only taught about finding the minimum spanning tree. i do not know how kruskal's algorithm works

do you have to code it or something?

[vale wigeon]

okay, so you were not taught kruskal's algorithm at all?

no, you do not have to do any coding.

[alpine sable]

nope

[vale wigeon]

so you were not taught kruskal's algorithm, but are expected to apply it.

do i understand you correctly?

i mean, the algorithm itself is not very complicated. it's one of those algorithms that you can (and probably should!) execute by hand

it's just weird that you are expected to do something that you haven't been taught to do.

[alpine sable]

the definition of kruskal's algorithm was taught yes but the actual application of it was not

[vale wigeon]

what do you mean

can you show what you're talking about with "the definition of kruskal's algorithm"?

from my pov, knowing the description of an algorithm implies knowing how to apply it - unless there are some steps in the description you don't know how to execute.

[alpine sable]

i really don't know how to apply it

[vale wigeon]

show me the description you were taught

we will go through it step by step and you will tell me which step you do not know how to execute

[alpine sable]

i cannot recall it (low comprehension). the whole session was recorded but is not published yet

[vale wigeon]

ok so what you're saying is not that you don't know how to apply the algorithm, it's that you forgot the algorithm itself.

[alpine sable]

yes. plus i have no idea of it in the first place

[vale wigeon]

in fact the algorithm is pretty simple:

write out all the edges in order from smallest to largest weight. considering the edges in that order, attempt to include them one by one into your spanning tree. if including an edge would not create any cycles with what's been included before, then include the edge; otherwise ignore the edge and move on to the next.

do you understand this? (y/n)

[alpine sable]

n

[vale wigeon]

okay, which part do you not understand?

[alpine sable]

"considering the edges in that order, attempt to include them one by one into your spanning tree." this is where i draw the vertices, connect the edges, and draw the whole tree right?

[vale wigeon]

no, the "whole tree" is what you will have at the end of the algorithm.

if you want, you can make a physical drawing of your graph and then include edges by tracing them over with a different color (like red)

[alpine sable]

how about this?

[vale wigeon]

this is your graph, with no edges marked as of yet

[alpine sable]

"if including an edge would not create any cycles with what's been included before, then include the edge; otherwise ignore the edge and move on to the next." This is the part where i mark the edges right?

[vale wigeon]

yes

[alpine sable]

is this right?

[vale wigeon]

no

but since you're not sharing any of your steps, i can't tell you where you went wrong.

[alpine sable]

this step "if including an edge would not create any cycles with what's been included before, then include the edge; otherwise ignore the edge and move on to the next"

[vale wigeon]

well ok sure but like

there's 8 edges in the graph

so you should've done that 8 times

once for every edge

in the right order too

[alpine sable]

is this one right?

[vale wigeon]

no...

[alpine sable]

i don't get what you're saying

[vale wigeon]

ok look

can you write out all the edges in increasing order?

[alpine sable]

CE - 10
DE - 12
BD - 12
BE - 15
AD - 16
BC - 23
AC - 37
AB - 52

[vale wigeon]

okay, great

so now

you start with nothing

the first edge you will be trying to add is CE

does adding CE to your set of edges make any cycles?

[alpine sable]

yes?

[vale wigeon]

really?

[alpine sable]

no

[vale wigeon]

so which is it?

yes? no? don't know?

[alpine sable]

don't know

[vale wigeon]

do you know what a cycle is?

[alpine sable]

no tbh

[vale wigeon]

and why in blazes is this only coming up now????

[alpine sable]

srry if this is making you mad

[vale wigeon]

i'm not mad i'm just confused

if you didn't know what a cycle was then why didn't you say so the moment i mentioned cycles

[alpine sable]

cause what i know of cycles is when from vertex a to e's edges are all connected together

[vale wigeon]

...no, that's very far from the truth

[alpine sable]

then what is it?

[vale wigeon]

a cycle is a path that starts and ends at the same vertex (and doesn't visit any edge more than once)

for example, in your graph, ACBA is a cycle

[alpine sable]

ahh

so CE is not a cycle

so far

[vale wigeon]

of course it's not a cycle yes

okay, so we include CE. our set of edges is now {CE}

the next edge is DE

does adding DE create any cycles?

[alpine sable]

yes

[vale wigeon]

are you sure?

if yes, name the cycle that is created.

[alpine sable]

CEDE

[vale wigeon]

that is not a cycle

your path visits the edge DE twice

nor does it start and end at the same point

[alpine sable]

i'm having trouble understanding

what path visits DE?

does it make a difference if it's ECDE?

[vale wigeon]

CEDE, the path you just wrote...

CD isn't even an edge in your set

this is roughly what a cycle looks like, in case you were confused by my notation.

the way to describe a cycle is to name all the points it visits, in order

kind of like you would describe an itinerary

[alpine sable]

i'm still lost tbh

so DE is not a cycle

[vale wigeon]

a single edge is never a cycle

so yes

[alpine sable]

so basically all the edges in the graph is not a cycle, right?

i'm too dumb for this

[vale wigeon]

i am having trouble getting through to you

i will not make any comments on your intelligence

[alpine sable]

so CE - 10 = not a cycle
DE - 12 = not a cycle
BD - 12 = not a cycle
BE - 15 = not a cycle
AD - 16 = not a cycle
BC - 23 = not a cycle
AC - 37 = not a cycle
AB - 52 = not a cycle

no?

[vale wigeon]

no edge creates a cycle ON ITS OWN, no...

but that doesn't mean we can just add all edges

ok

let me reproduce your graph and show you what im talking about

or at least try to

here is your graph again

i will be drawing edges included in the spanning tree with thicker lines

the first edge under consideration is CE. since we have no edges added yet, adding CE will not make any new cycles.

the next edge is DE, which also does not create any new cycles as you should be able to see

and neither does BD

HOWEVER, if we were to add BE, we would get this picture.

do you see what is wrong with it?

[alpine sable]

DEBE has the same verfex

[vale wigeon]

when you say "DEBE" do you mean the path that goes from D to E, then from E to B, then from B back to E?

or are you trying to talk about the edges DE and BE?

[alpine sable]

This one

[vale wigeon]

okay so again

DEBE isn't a cycle

a cycle cannot visit the same edge more than once

and a cycle has to end at the same place it starts

your path, once again, doesn't do either of those things

yknow what, if you're struggling with going through kruskal's algorithm this much, i'm not even going to try to ask you to modify it.

because i am almost certain that'll go right over your head.

[gentle flame]

I don't know if this will help, but imagine you're standing on some vertex

Then a cycle is simply you going around the tree and coming back to where you started

Subject to the condition that you visit no vertex more than once

[alpine sable]

okay so CE - 10 = will not result in a cycle
DE - 12 = will not result in a cycle
BD - 12 = will not result in a cycle
BE - 15 = will result in a cycle
AD - 16 = will not result in a cycle
BC - 23 = will result in a cycle
AC - 37 = will result in a cycle
AB - 52 = will result in a cycle

right?

Here are my answers based on the instructions/questions above. the minimum spanning tree in the graph is 50 and if computer a must be connected to computer b, the difference it makes to the answer is 86 since the instructions require that the tree starts out at AB. correct? (y/n)

[lone heart]

@alpine sable Has your question been resolved?

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[midnight sand]

How would you do this

[glad sluice]

test?

[midnight sand]

no

[glad sluice]

ok so

for Terraced houses, what's the mean number of total bedrooms?

[midnight sand]

1.9?

[glad sluice]

or rather, number of total bedrooms

[midnight sand]

a.. idk

[glad sluice]

emphasis on total

ok how is mean defined?

[midnight sand]

do u do 70x1.9

[glad sluice]

yup!

[midnight sand]

no

[glad sluice]

what

[midnight sand]

ok

[midnight sand]

idk the definition of mean

im shiet at defining words

[midnight sand]

133

[glad sluice]

[midnight sand]

[glad sluice]

mean is basically (sum of all values/number of values)

[midnight sand]

my englsih is very poor sorry

[midnight sand]

Okay

[glad sluice]

you should know that before attempting the question

anyhow

[midnight sand]

wait, i know what mean is

but idk how u word it

[glad sluice]

[midnight sand]

ok wat do u do next then

[glad sluice]

find the sum of values for the total dataset

and the number of total values is already given

[midnight sand]

im in secondary, i should kno wat mean is i think

[midnight sand]

wdym

so

multiply

the thingy

[glad sluice]

what

[midnight sand]

20x3.1

[glad sluice]

yes

[midnight sand]

10x3.5

then add it up?

[glad sluice]

precisely

[midnight sand]

okai

[midnight sand]

230

[glad sluice]

now what

[midnight sand]

idk

[glad sluice]

that is the sum of values

[midnight sand]

ok

[glad sluice]

what is the number of values

[midnight sand]

100

[glad sluice]

and mean is?

[midnight sand]

OooooHHhh

230/100

[glad sluice]

yep

[midnight sand]

why didnt i think

[glad sluice]

that is a good question

[midnight sand]

i should bac to primary

[midnight sand]

anyways ty

.close

[lone heart]

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[rose plover]

In a Triangle: The Side a is a quarter of the Side b and the side c is one cm longer than the side b. The circumference is 19cm. Find a, b, c.

[hexed viper]

What is scope?

[rose plover]

Scope is the circumference

[hexed viper]

Can you relate the variables with equations?

[rose plover]

sure i need an equation and an answer of a, b and c.

[hexed viper]

“ The Side a is a quarter of the Side b “

What is this algebraically

[rose plover]

1/4

[novel dune]

Can I help?

[rose plover]

sure everyone can help

[hexed viper]

It’s an equation, it should have an equals sign

[rose plover]

b = 1/4 a

Thats what i know

[hexed viper]

a is a quarter of b

[rose plover]

Its a textexample

I need to understand the text to form an equation and then find a, b, c.

[hexed viper]

Yes

[hexed viper]

You incorrectly formed the equation

[rose plover]

tf

[novel dune]

Euler2 plz check my solution

[rose plover]

im not that advanced

[hexed viper]

Daniel it’s okay

My slice of cake is a quarter of the entire cake

slice = (1/4)cake

Can you check what you wrote and see that it doesn’t match

[hexed viper]

Here

[rose plover]

b is the cake and a is one fourth, right ?

[hexed viper]

Yes

[rose plover]

i can't understand this

[hexed viper]

You can form your own equations now

Use your own brain

[rose plover]

i tried but i cant

[hexed viper]

How does b and c relate

[rose plover]

b = 1 + c and c = b - 1

[hexed viper]

You did the same mistake again

You’ve got it backwards

c is bigger than b right?

[rose plover]

yes

but by one

[hexed viper]

Your equation is wrong then

Cause image c=4 and b=5

It fits your equation but c is one smaller than b here

c is one bigger than b

c=b+1

[rose plover]

it cant be

in the question it says

[rose plover]

c is one cm longer than b

shouldn't it be b = c + 1

[hexed viper]

Yes and I’m saying you have a fundamental misconception about how that translates when you try to turn it into an equation

No

c=b+1

If b is 7 what is c?

According to my equation

[rose plover]

6

oh wait

8

[hexed viper]

And that’s right

It should be 8, it’s one bigger than 7

[rose plover]

oh yeah right

[hexed viper]

You did the same thing for the other one too

a=b/4 is correct

Because if b is 8, a is 2 which is what you want

[rose plover]

????? where did this equation come from

[hexed viper]

The first sentence

We already did this

“b = 1/4 a”

You said this

This is wrong

Same mistake again

[rose plover]

shouldn't it be

a = 1/4 b

[hexed viper]

Yes

(1/4)b is the same as b/4

[rose plover]

oh yeah

cus u can put b in one of the fractions

[hexed viper]

Yeah

[rose plover]

it's like (1/4) b = (b/4) = (1/4b)

[hexed viper]

You can’t put it in the denominator

[rose plover]

oh ?

ok

[hexed viper]

Just the numerator

[rose plover]

so a = b/4

[hexed viper]

Yes

Now last equation

What about the scope

[rose plover]

the scope is 19cm

[hexed viper]

Yeah but that’s English

What’s that as an equation

[rose plover]

a + b + c

[hexed viper]

Yeah but that’s not an equation

You should have an equals

[rose plover]

a + b + c = 19cm

[hexed viper]

Okay now you are basically done

Because you have expressions for everything

[rose plover]

oh you use the a and c

and b = 4a

ooooohhhhhhh

[hexed viper]

Yeah exactly

[rose plover]

ok so

b/4 + 4a + b+1 = 19 ?

[hexed viper]

You want it all in one variable though

[rose plover]

how

[hexed viper]

You turned the b into a 4a

But you could have just left it as a b

[rose plover]

oh 4a + 1

[hexed viper]

Yeah that works too

So a+4a+4a+1

[rose plover]

yeah

then its

9a = 18

a = 2

[hexed viper]

Yep

[rose plover]

then if a = 2

then i use expressions again

lets see

ok i got it

a = 2, b = 8, c = 9

[hexed viper]

Yeah and you can see that works

Well done, you can work things out easily so have more self assurance in the future

[rose plover]

thank you Euler2

.close

[lone heart]

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[burnt wren]

how can i solve this

[lone heart]

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[burnt wren]

i am kinda stuck solving this i thought rationalizing the denominator would works but i dont really know

[glossy briar]

@burnt wren I'll try and help

[burnt wren]

okay thanks

[glossy briar]

Being honest I'm at a museum with no paper

[burnt wren]

oh wait i think just rationalizing helps

[glossy briar]

You know what you could do

Root 5=5^1/2

[burnt wren]

ok

[glossy briar]

And (a+b) ^2=a^2+b^2+2ab

Then try

(A+b) ^1/2=a^1/2+b^1/2+(ab) /2

Where a=5 and b=3

To simplify the denimonator

Sorry @burnt wren I will try at home

[lone heart]

@burnt wren Has your question been resolved?

[lone heart]

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[rich fractal]

I started learning these trig functions a few days ago and I'm happy with my progression, I tried solving this problem myself but I'm getting a different answer than the lecturer I'm following.
So this is the question: https://snipboard.io/QCeA3g.jpg
And this is my solution: https://snipboard.io/KSB4tA.jpg
What wrong did I do there?
The lecturer used this cancelling technique which I'm not really aware of: https://snipboard.io/dcXQrW.jpg
And got this answer: sin x - cos x
However I'm getting cos x - sin x which is completely changing my final answer.
So whats wrong there?

[gray isle]

is nonsense

by "cancelling" those sin(x) and cos(x) like that,

you just get back to what you had in the line before which is
1/cos(x) - 1/sin(x)

you're also completely ignoring the denominator of the original big fraction

the teacher multiplied multiplied the numerator and denominator of the big fraction by the lcd (of the smaller fractions), which makes simplification a lot more efficient

you could instead first combine the numerators and denominator into single fractions and continue simplifying from there to achieve the same result

[rich fractal]

Ohh thank you

[lone heart]

@rich fractal Has your question been resolved?

[burnt wren]

how do you do this

<@&286206848099549185>

[lone heart]

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[quasi ruin]

Discrete Mathematics: If an integer greater than 1 is a perfect square,
then its cube root is irrational.

[novel dune]

This statement is false

[quasi ruin]

My counter-example: 8^2 = 4^3?

[novel dune]

Same thinking 👍

[quasi ruin]

Suppose I already have a counter-example, do I need to make a formal Proof by Contradiction? I am trying to do that and I am stuck right now :(.

Proof by contradiction: Suppose there is an integer n, where n > 1 such that n is a perfect square and n's cube root is rational.

(1) By definition of perfect square, let x = n^2 where n is an int.

(2) By definition of rational numbers, let cube_root(x) = p/q where p and q are integers and q != 0.

(3) By substitution: cube_root(n^2) = p/q

n^2 = p^3 / q^3

n = sqrt(p^3 / q^3)

...and I am stuck here 😦 😦 😦

[lunar sail]

You need to disprove this statement, right?

Then, if you have a counter-example, you're already done

You can just say:
The statement is false. Counterexample: $64 = 8^2$, and $64^{1/3} = 4$. 4 is not irrational, therefore the statement is false.

[ocean seal]

Roberto Tan

[lone heart]

@quasi ruin Has your question been resolved?

[quasi ruin]

Ok. Thank you very much. That solves the problem.

[lone heart]

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[dull onyx]

its not pinning but whtever

[half epoch]

@dull onyx use the quadratic formula

[dull onyx]

to find the sign?

[half epoch]

You get two candidates for f, only one satisfies f(0) = 1

[dull onyx]

basically

there were two questions

one to find the sign and the second to find what f(x) actually is

i used the quadratic for the second and the teacher said first of all i cant use the quadratic 💀 for the second question nor can i find what f(x) is to answer the first one

[half epoch]

Hm, there might be some trick to find the sign of f without actually finding f first, but I'm not aware of it. I'd find f then find the sign

[dull onyx]

yea thats what i thought of as well

he said i cant use the quadratic to find what f(x) is for some reason but i dont remember why

so i just completed the square but like isnt it the same shit

[half epoch]

Um, did they elaborate on what the problem with using the quadratic equation is?

It works just fine here btw, the only reason I can see for them to tell you not to use it is because they don't want you to use it

[dull onyx]

he said that we’re assuming f(x) is of the form y=mx+b but i literally did not understand what he meant

[dull onyx]

ah okay thank u!! i used it on my exam and he gave me my marks but said to not use it😭

[lone heart]

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[polar sequoia]

abt logarithms

[polar sequoia]

i already have the answer for it since im remaking exercises but i forgot how i got to it

<@&286206848099549185>

[lone heart]

@polar sequoia Has your question been resolved?

[lone heart]

@polar sequoia Has your question been resolved?

[polar sequoia]

im so stupid the height was in KM and i calculated it w meters

.close

[lone heart]

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[lone heart]

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[neat wharf]

I don't understand where 3.450 came from, can someone help?

[neat wharf]

Wait im an idiot

Nvm

13.8 / 4 = 3.45

.close

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[harsh ocean]

Is this method correct?

If not using the secx+tanx method?

[half epoch]

Well the answer looks wrong, and I spotted that when doing DI you integrated sinx to get cosx, you should have gotten -cosx