help-38

[wraith hinge]

back in

cuz u adjusted the boundaries right?

[tribal fractal]

yeah

[wraith hinge]

so u js plug in 811 and 244 right?

not 1+81x

[tribal fractal]

wait wdym

this was my final calculation

[wraith hinge]

well u orignally subbed in for sqrt(1+81x)

[tribal fractal]

no. just 1+81x

[wraith hinge]

oh ok right right

but u had adjusted it from

3-10 to 244-811 no?

the derivative

[tribal fractal]

yeah

@wraith hinge you there?

[wraith hinge]

yh

im taking a pic

of my work

rq

[tribal fractal]

oh sorry

no worries

[wraith hinge]

all good

since u changed the derivative to follow the u-sub

cant u js do this

[tribal fractal]

yeah

[wraith hinge]

s would that be right

idk tbh, i kinda suck at this

thats js my opinion

[tribal fractal]

does this result in 158.72?

[wraith hinge]

i think so

its the same thing as here

the same equation

[tribal fractal]

hm

[wraith hinge]

what do u think

[tribal fractal]

😮

it works

idk what i did wrong then

[wraith hinge]

its right?

ok

[tribal fractal]

yeah

[wraith hinge]

i can explain

[tribal fractal]

i think i understand what you did

[wraith hinge]

u put 1+81x back into u after doing the derivative

[tribal fractal]

yeah

[wraith hinge]

dont need 1+81x

cuz u adjected the 3-10 to 244-811

so u js put in 244 and 811 in

instead of 1+81(244) and 1+81(811)

[tribal fractal]

but i did that

[wraith hinge]

yeah

im saying

u dont do that

[wraith hinge]

u do this

[tribal fractal]

oh

[wraith hinge]

see, thats what i did

u dont put in 1+81x

u only put in x

[tribal fractal]

OH

[wraith hinge]

cuz u changed the boundaries

u get what im saying?

[tribal fractal]

oh yeah

[wraith hinge]

u sure?

[tribal fractal]

sorry for the trouble

yeah

[wraith hinge]

if u dont understand lemme know

[tribal fractal]

i get it

[wraith hinge]

ok good

[tribal fractal]

so thats why i was told to change the boundaries

so i dont do it later

[wraith hinge]

yepppp

[tribal fractal]

thank you so much!

[wraith hinge]

ofc ofc

[tribal fractal]

late night calculus kinda sucks

.close

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[north prairie]

did i set this up correctly

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@north prairie Has your question been resolved?

[north prairie]

<@&286206848099549185>

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@north prairie Has your question been resolved?

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@north prairie Has your question been resolved?

[vernal wraith]

It should be 2 instead of 0.2

[north prairie]

y\

0.2 = 20% of the isotope remaining

[vernal wraith]

Yes but 20% of 10 is 2

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[orchid wagon]

$f(x)$ is continuous on $[0, 1]$

[solid kiln]

FungusDesu

[orchid wagon]

$\int_0^1 f(x)dx = \int_0^1 xf(x)dx = 1$

[solid kiln]

FungusDesu

[orchid wagon]

$\int_0^1 f^2(x)dx = 4$

[solid kiln]

FungusDesu

[orchid wagon]

evaluate $\int_0^1 f^3(x)dx$

[solid kiln]

FungusDesu

[orchid wagon]

tried IBP on both of the original integrals, but i always end up with a double integral

they dont give derivative of f(x) too

[orchid wagon]

i tried doing IBP on lhs as well

yielding this

$f(1)-\int_0^1 xf'(x)dx =1$

[solid kiln]

FungusDesu

[nova spire]

goodness gracious this is so dumb

[orchid wagon]

but im not sure what to do with this information

[ebon kite]

i assumed a linear function and it worked

but its too handwavy

[nova spire]

from this you have that $\int_0^1(ax+b)f(x)dx= a+b$ for every $a,b$

so with (f(x))^2

this is AGAIN

about 'completing' if you know what i mean

[orchid wagon]

elaborate on this?

[orchid wagon]

yes

[nova spire]

linearity of integral

[solid kiln]

rafilou2003

[orchid wagon]

never heard of

[nova spire]

wut

$\int_0^1(af(x)+bg(x))dx = a\int_0^1f(x)dx + b\int_0^1g(x)dx$

[solid kiln]

rafilou2003

[nova spire]

that linearity

[orchid wagon]

oh that thing

[nova spire]

this is true because of that property

[orchid wagon]

didnt know how its called in english

[nova spire]

and those two

[orchid wagon]

right

[nova spire]

so again

find the correct $a,b$ such that $\int_0^1(f(x)-(ax+b))^2dx = 0$

[solid kiln]

rafilou2003

[orchid wagon]

alright, i can take it from here too

[ebon kite]

this is pretty much guesswork though

[orchid wagon]

i assume this was directed at me

[ebon kite]

is there a nice solution that you can find deductively

[nova spire]

no, the solution is literally 'prove that f(x) = ax+b'

[ebon kite]

lmao

[nova spire]

and it's by doing this

by this point the exercise sheet is like "Do you like this trick yet? Are you not entertained?"

[orchid wagon]

yes, practically most of the exercise sheets i got

chunks of exercises each with their own way of solving

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[devout drift]

If I can define multiple relations for the same ordered pair,the total number of relations for two sets A and B must always be more than 2^mn considering m and n elements in A and B respectively?

[broken pilot]

Can you give an example?

[devout drift]

Let's say we have A × B = {(1,2),(2,4)}

Then n(P(A × B)) = 4

Let's consider {(1,2)} specifically

Then I can define any number of relations for that specific pair?

@broken pilot

[broken pilot]

what are n and P?

[broken pilot]

then it wouldn’t be the cartesian product of anything

[devout drift]

Number of elements in power set of A × B

[devout drift]

Consider A and B

[broken pilot]

so A = {1,2} and B = {2,4}?

[devout drift]

Sure

Sorry my network

Wait a min

I see

I realize my cartesian product can't exist

That was dumb of me

[devout drift]

A × B = {(1,2),(1,4),(2,2),(2,4)}

So now it will have

The power set will have 16 elements

[broken pilot]

Yes

[broken pilot]

The power set of the cartesian product?

[devout drift]

Yes

[broken pilot]

Oh

[broken pilot]

what do you mean by “multiple relations for the same ordered pair”?

[devout drift]

R = {(a,b): a = 2b}
R = {(a,b): a =b - 1}

[broken pilot]

As a subset of A x B?

They specify the same elements, so they are the same sets

[devout drift]

That's what I'm saying

Multiple relations for the same element

[devout drift]

THAT'S EXACTLY MY POINT

[broken pilot]

Ok, the amount of relations is exactly 2^(mn)

[devout drift]

Like we can define infinite relations like this

Yeah

[devout drift]

I'm right

Right

[broken pilot]

Yes, but you know they’re all the same

[devout drift]

The ordered pair is

But can you say the same about the relation?

[broken pilot]

Yes, you can

You noted that they define the same sets, therefore they are equal, and the relations are equal

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@devout drift Has your question been resolved?

[devout drift]

I see

I think I confused the "number of relations" with "number of ways of defining a relation".

[broken pilot]

I think that’s it

[devout drift]

Thanks

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[dusty ledge]

.

Q.no 3

[dusty ledge]

I solved this question i got answer as d) 0

Is it correct

[boreal zodiac]

That seems right. f(0) = 0^2 + 5

[dusty ledge]

so it is ?

[boreal zodiac]

Yes

[dusty ledge]

oh ok one more thing the preimage is the starting set right ?

[boreal zodiac]

The preimage is like the inverse. It's asking what number(s) to input to get this output

[dusty ledge]

yeah like this right ?

f(x)=3+x

[dusty ledge]

..

[boreal zodiac]

Yes. So the preimage of 4 would be 1

[dusty ledge]

yessss

alright thanks

[boreal zodiac]

Sure thing

[dusty ledge]

all clear

cya have a nice one

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[broken pilot]

to clarify, the preimage can be a set of numbers

for example, the preimage of 1 under f(x) = x^2 is {-1,1}

[dusty ledge]

yes , thanks for giving it time

.close

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✅

[dusty ledge]

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[static tendon]

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[tight sage]

Hey I don’t need help with a specific problem but I’m wondering how can you get prepared for a math contest

[tight sage]

I’m doing the grade 11 Fermat contest

The questions don’t seem similar every year so I’m wondering if there even is a way to get prepared

[sullen ore]

The specific questions arent the important thing. The important thing is to just do a bunch of practice problems from previous tests to build up intuition and logical thinking

It’s mainly practice

And a large amount of mathematical intuition

[digital bolt]

do lots of questions anyway

it's practice for thinking outside of the box and being creative at least

and some things will inevitably be useful in lots of contexts

[sullen ore]

I dont know much about the fermat contest but the same idea of constant practice got me good scores in gauss and pascal

Just do a bunch of previous test questions

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[marsh forum]

Ok, so to start

$\frac{k}{n}=q$

[solid kiln]

Why am. I here

[vagrant marsh]

think about where you may have seen that form of sum before

[marsh forum]

I know it's a rienman summation

just trying to figure out the width of the new $\Delta q$

[zinc ginkgo]

Why is this a q

[marsh forum]

just a random variable

I'll change it to x when I integrate

[digital bolt]

the fact that you have 1/n should be a clue

[zinc ginkgo]

Then this also needs to be the same "random variable"

[solid kiln]

Why am. I here

[marsh forum]

so the width of the original intervals is

$\frac{1}{n}$

[solid kiln]

Why am. I here

[zinc ginkgo]

No

[marsh forum]

wait, why not

[zinc ginkgo]

Read this again

Find a, b

[marsh forum]

hmm , so it's from 0 to 1/n, right

[zinc ginkgo]

Nope I was mistaken

[zinc ginkgo]

This is right

[wraith hinge]

@marsh forum Did you study calc from offline coaching ?

[zinc ginkgo]

a is zero

[zinc ginkgo]

!redir

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[zinc ginkgo]

So that implies b=1

[marsh forum]

ah, OK, yeah

[marsh forum]

Depends on what you mean,I learnt most of my calculus at school

[marsh forum]

oh, as k ranges from 0 to n

[zinc ginkgo]

<@&268886789983436800> spam

[wraith hinge]

it is literally relevant

[marsh forum]

$\int_0^1f\left(p\right)f'\left(p\right)dp$ is what I compute, right?

[solid kiln]

Why am. I here

[zinc ginkgo]

Yes

[marsh forum]

hmm, so $\frac{f^2\left(p\right)}{2}$

from 0 to 1

[solid kiln]

Why am. I here

[marsh forum]

which is 35/2

thanks!

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[hot prawn]

Two rockets were launched from the ground. First rocket was launched until it fell back to the ground. Time and height of the first rocket were shown on the table. The height of the first rocket is stated as the function of height (h) over time (t):

$h(t) = -4(t-p)^2 + 64$

Whilst height of the second rocket is stated as the function of height (h) over time (t):

$h(t) = -4t^2 + 32t$

[solid kiln]

sho

[hot prawn]

right table shows height, left shows time

i alrd found it, nvm

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[north prairie]

https://cdn.discordapp.com/attachments/908078008609431674/1210798671721926686/image.png?ex=65ebdf22&is=65d96a22&hm=d8f65e98aa0d871189d8232569a8c4beb812f7f31ec5573e3adb69e7a109959b&

[north prairie]

https://cdn.discordapp.com/attachments/908078008609431674/1210798952786558976/image.png?ex=65ebdf65&is=65d96a65&hm=7aa82598d3d3ef8aba46dc832b24dcb9cceb162dfc65f26cec3d35678c5d668c&

did i set it up correctly

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@north prairie Has your question been resolved?

[north prairie]

<@&286206848099549185>

[vernal raptor]

um wait

a

sec

this is from diffrential equation right?

[north prairie]

uhh

[vernal raptor]

which topic it came from?

phy or diff eqn

;_:

[north prairie]

exponential change

[vernal raptor]

ohhh okay

wait

there are many methods to solve a single question oh my gawd

[north prairie]

yes

oh my gawd

there's an example in the book but there's no initial amount in the formula they used

so i'm just confused

[north prairie]

and when i solved this for t it gave me the wrong answer

so i'm just confused where it went wrong

[vernal raptor]

wait lemme solve then ill see where you got wrong

[north prairie]

way off

[vernal raptor]

wont that be 2 and not 0.2?

[north prairie]

ohhhhhhhhhhhhhhhhhhhhh

[vernal raptor]

..

[north prairie]

i read the formla wrong Lol

[vernal raptor]

and then for simple calc you can give the -ve sigh up and make it ln5

[north prairie]

yea cus y0(y) = y0e^-kt

[vernal raptor]

so you're right just it should be ln1/5

[north prairie]

since the example problem had no y0 they just cancelled it on both sides

ig i didn't catch that

[vernal raptor]

no no

10 g is released

how much we have to find

80%

so

[north prairie]

ya so 0.2 of 10

[vernal raptor]

?

[north prairie]

0.2 * 10 =2

y0 = (10) * y = (0.2)

[vernal raptor]

um not like that

[north prairie]

wut

[vernal raptor]

like 80% of 10

[north prairie]

8

[vernal raptor]

will give you 8

then remaining will be?

2-

so 2=10e^-kt

got it?

if that was 50% then what will you take?

[north prairie]

50%

[vernal raptor]

then what will be your y0?

[north prairie]

10

[vernal raptor]

and y

?

I mean just incase if you have the same question with just given as 50% decay

then what will be your equation to find the time

[north prairie]

0.5

5 = 10e^-kt

[vernal raptor]

yes

good

your question is done

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[halcyon onyx]

may i have a hint for part b

ive been trying for hours and getting nowhere

[granite cove]

We know a_1^2 >= 2. So if we had any a_n with a_n^2 < 2 then we have a_(n-1) such that a_(n-1)^2 >= 2, but a_n^2 < 0

Or in simple words: There exists a smallest n such that a_n^2 < 0

*Excuse the poor notation

[halcyon onyx]

i dont get the n-1 part

how does that mean a_(n-1) >= 2

[granite cove]

a_(n-1)^2 >= 2

[halcyon onyx]

how

[granite cove]

.

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[fallen talon]

i want to know the volume of that object. and i get wrong asnwer the right answer is 25 074 and i get 26 683.
using the 90 degre angle and the 70 degree i calculates the last one on the triangle which is 20 and using cos i got diameter and then half that i got radious then i used V=Ah because i knew A=r^2pi. But i still got wrong answer.

Here is the full question (A cylindrical tube, with a diameter of 10 cm, passes through the floor and ceiling of a room that is 3.0 m high, forming a 70° angle with the floor. What is the volume of the portion of the tube in the room?)

[pearl tendon]

Break the cylinder into 2 parts (technically 3, but the top and bottom parts can be joined into 1) and see if that helps

@fallen talon

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[wraith hinge]

hello had to go earlier sorry

so anyways

you have 10+friction=2gsin45

do you agree?

as we last left off

anyways the friction acts up because the horizontal force of gravity (mgsin(theta)) is greater than the applied force F (10 newtons) so the motion is downwards along the incline, thus the friction opposes that motion

[trim jolt]

@grizzled dock Has your question been resolved?

[grizzled dock]

no

[grizzled dock]

yes

[wraith hinge]

ok so

what is the friction expression you got last time?

[grizzled dock]

2gcos45 * μ

[wraith hinge]

yes!

[grizzled dock]

yay

[wraith hinge]

so putting it in here what do u have for ur expression?

just plug in the thing for thr friction

[grizzled dock]

10+μ2gcos45=2gsin45
(2gsin45-10)/2gcos45 = μ

so because it's acting downwards do I take g as positive ?

[wraith hinge]

yeah

i think thats it evaluate it and it'd be correct

[wraith hinge]

no g is a magnitude and its always equal to 9.81

as a vectorial quantity yeah but you already made the downwards direction as negative

[grizzled dock]

oh thank you I got the right answer!!!

[wraith hinge]

wooo

[grizzled dock]

but if g is gravity it acts down so why is it

Positive

if I'm taking down as negative

[wraith hinge]

this

[grizzled dock]

oh ok so never make it

negative

[wraith hinge]

so like

that's the thing

$\vj g$ is different than what $g$ is

[grizzled dock]

??

what's that

[wraith hinge]

do u know vectors

[grizzled dock]

kinda

[wraith hinge]

ok so ill try to simplify it in the easiest way but like

,,
\vj g = \t{magnitude}\cd\t{direction}\
g = \t{magnitude}

[wraith hinge]

so like

the first one is akin to saying "the gravity acceleration is 9.81 m/s^2 and points downwards" and the second is just "the gravity acceleration is 9.81 m/s^2"

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[wraith hinge]

woops

[grizzled dock]

o_o

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[azure dagger]

a

[azure dagger]

mk so

i got a question

chemistry but its basically math

so

moles are a way to basically talk abt atoms on more of human scale?

nvm

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[umbral dragon]

What are you having problems with

[north prairie]

idk how to start this

ik how to prove they grow at the same rate

what's throwing me off is the second part

[umbral dragon]

Okay so let’s think about how we might approach the task

Firstly, how would we determine the rate at which a given function is growing? Mathematically speaking

[north prairie]

the degree of the function

[umbral dragon]

No not quite

It’s the derivative

[north prairie]

ic

[umbral dragon]

So we can represent the rate of growth in our equation using the derivative. Do you know what we could use to represent the “as x approaches infinity” part?

[north prairie]

lim

with these one divided by the other

[umbral dragon]

lim is correct but notice that you’re supposed to compare the growth to sqrt(x)

[north prairie]

ohh

so

would i just do 1 limit for each

and divide them by sqrt(x)?

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[umbral dragon]

I’m actually unsure if division works - the results I’m getting are throwing me off because I’m not sure how to interpret them. You’re welcome to try for yourself though

[umbral dragon]

Make sure to compare the growth rates, aka derivatives

[north prairie]

mhm

but this is what it's asking me to do right

[umbral dragon]

I believe so yes

[north prairie]

okay

but now i run into a problem

when i do l'hopital's its just an infinite loop

would it be better if i just do (whatever equation) = O(sqrt(x))

[umbral dragon]

Honestly, I’m not sure. Would probably be best if another helper could jump in. Sorry I can’t help further

[north prairie]

no problem

i appreciate your help so far

<@&286206848099549185>

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[zinc ginkgo]

You can write the limit of the square root as the square root of the limit

So find limit of (x+1)/x first as x goes to inf

[north prairie]

gotcha

i got it

thank you for the help

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[mortal sparrow]

I don’t understand how to get the vertex of 4 a using completing the square

[split chasm]

its already in vertex form

vertex can be identified directly from that

[mortal sparrow]

Ok

Thx

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[wraith hinge]

I dont understand one normal vector being enough to describe all normal vectors to a plane simply because they will all be scalar multiples of each other
won't scaling them simply make the magnitudde grow larger/shrink

[zinc ginkgo]

Yes. What do you think "describe all normal vectors" means?

[wraith hinge]

basically one vector that is perpendicular to a plane will represent all vectors that perpendicular to a plane

however, i dont understand how one will represent another because they will be the same if u multiply them by a scalar

[nimble stone]

the only thing that matters is the direction

any vector in the same direction (any scalar multiple of the normal vector) will itself be normal

changing its length doesnt make it no longer normal to the plane

[wraith hinge]

Oh wait

so its initial starting and edning point do not matter?

[nimble stone]

vectors dont have definite start and end points

they can be placed anywhere in space

[wraith hinge]

Ohh okk

Got it, thanks!

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[untold breach]

in a single throw of 3 dice what are the no of sample spaces with the sum of number showing to be 5

[untold breach]

why can't i do it like

Let the nos be x y z

x+y+z=5

Give 111 each

x+y+z=2

Now i go (n+r-1)c(r-1)

Formula

[neon condor]

do you mean the number of outcomes

[untold breach]

ye the outcomes on the dice

[neon condor]

i dont see why you cant do that

[untold breach]

Like (2,2,1)

Would be a sample space

so

4c2 is correct?

[neon condor]

an outcome in the sample space

[untold breach]

Ohh yeeaa

it was damn

I checked the wrong. Ans

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[feral nova]

please help
lets say i have a linear homogeneous recurrence relation and i get 3 characteristic roots: 2,2,2 what will be the form of a_n with these roots?

[lament reef]

if you have repeated roots, you tack on n, n^2, n^3, ... next to your solutions

so $a_n = a2^n + bn2^n + cn^22^n$

[solid kiln]

doaby

[feral nova]

hmm but how come we add a degree to them?

@lament reef

[lament reef]

the simplest reason is bc we want 3 linearly independent solutions since we have a 3rd degree recurrence

and that guarantees it

you could also factor out a 2^n and see for yourself how that works, it's interesting

that's the only reason I'm aware of tho, I'm sure there's a deeper explanation

[feral nova]

ah alright thank you very much

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[shrewd pebble]

I'm not sure what Im doing wrong

or does it mean furthest corners

[wraith hinge]

I believe they are talking about a different "diagonal"

wouldn't the sides be

$$x$$
$$\sqrt{2} x$$
$$C$$

[solid kiln]

vote.gov Register to vote.gov

[shrewd pebble]

https://i.gyazo.com/282f1ec36451a676ce6f3c75194fd8d5.png

this?

oh wait

sqrt x^2+x^2

and c

[wraith hinge]

uuuuh

yeah you're right. i think, i did the math in my head so don't trust me numbers

[shrewd pebble]

tas okay

I end up with this but I can't submit x in the answer

I'll rewrite it, my work is confusing bc i made a lot of mistakes

I'm not exactly sure what to do after this

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[shrewd pebble]

https://i.gyazo.com/14a31ced846809fe72b76ca00c9094f1.png

[shrewd pebble]

I'm not sure how to do this

I've gotten this far

[cobalt cloak]

so

how do you represent the diagonal of a cube in terms of its edges (x)

[shrewd pebble]

(sqrt(x^2+x^2))^2 + x^2 = C^2 right?

so 3x^2

[cobalt cloak]

yea

3x^2=C^2

agree?

[shrewd pebble]

yeah

[cobalt cloak]

then C = xsqrt(3), agree?

by taking square roots of both sides and rejecting the negative answer since how can you have a negative length lol

[shrewd pebble]

OH

ya

[cobalt cloak]

ok great

now we know dx/dt = 13.7, as stated in the question right?

[shrewd pebble]

ya

sqrt(3)*13.7 right?

Got it ty!!

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[cobalt cloak]

yeah

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[nimble wind]

For problem 3

I started the problem like this

Im supposed to take the derivative with respect to P

I used the product rule

but i think im going wrong when i take the derivative of T*P^-1

Let me know if you need anything

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@nimble wind Has your question been resolved?

[nimble wind]

<@&286206848099549185>

[dark roost]

Just simplify the exponents

[nimble wind]

Yeah i get that

[nimble wind]

but it should accept the answers

look at above

it took it in the way i inputed it in problem 1 and 2

[nimble wind]

I have to had messed something up when doing the power rule

[dark roost]

Maybe it is wrong from their end cuz i dont see any other correct solution

[dark roost]

Ya that cud be it

[nimble wind]

Have you done it yourself

[dark roost]

Yup

I dont think as a helper I can give u direct solution of the answer

[nimble wind]

I wanted to ask when i have to take the derivative of this

[nimble wind]

But your answer is the same as mine

[dark roost]

t/p

[dark roost]

yup

[nimble wind]

when p is a variable and T is a constant

[dark roost]

U r majoring in wht?

[nimble wind]

Engineering

im first year tho

[dark roost]

which?

[nimble wind]

Mechanical

[dark roost]

Noice

School?

[nimble wind]

Wake Tech

Community College Lol

[dark roost]

Oh gud

[nimble wind]

Im still in highschool so thats why im at a community college

[dark roost]

Oh

[nimble wind]

Would it be possible for you to show me your work becuase we both got the same solutions which proves that i know what im doing

Dude

you know what it was

i didnt capitalize the T

I just got it right

[nimble wind]

compare the T on the end to this

I swear I hate math sometimes lmaoo

[dark roost]

1/2L * (2p/T)^1/2 * -T/p^2

[nimble wind]

Exactly what i got

[dark roost]

Maybe something is wrong with the question

[nimble wind]

No it was just a variable

i needed to capitalize the T

the simplest thing 🤣

[dark roost]

Lol

[nimble wind]

Thanks for the help man

Are you in college?

[dark roost]

Nope

[nimble wind]

You already got your degree?

[dark roost]

My pleasure

[dark roost]

Nope

10th grade in high school

[nimble wind]

Dude im in 11th

Lmaooo

[dark roost]

lol

[nimble wind]

Crazy were both in highschool

yo ima add your disc

[nimble wind]

you mind if i DM you?

[dark roost]

nope not at all

[nimble wind]

Okay ill close this

.close

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[knotty locust]

in Z_35 there are how many congruence classes

[loud pivot]

i am so lost

what

[knotty locust]

Do you know what modular arithmetic is

[loud pivot]

nvm thanks for the help , got it from my freinf

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[rugged birch]

I’m unsure about the order of transformations in this example, mostly with the 2f in where it fits in the equation

[rugged birch]

basically I researched and like I understand that when discussing order of transformations in for example $y=-2x^2-4$ its written as the reflection over x axis, dilation of scale factor 2, then translation of 4 units down, following pemdas

[solid kiln]

Pheenix

[rugged birch]

although I think it might be due to an understanding of how y=f(x) works in representing equations and therefore the transformations

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[rugged birch]

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[ember escarp]

Hello, can I get some help on how to start the 1st question pls

[fiery crypt]

i assume it's 4?

(the question)

[hallow cliff]

,rccw

[trail ingot]

lol at 5, “use calculus”

[hallow cliff]

Who's finding the integral of √x using geometrical methods

That sounds like hell

Anyways which question are we talking about @ember escarp

[trail ingot]

https://math.stackexchange.com/questions/3913/the-right-weigh-to-do-integrals

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[ember escarp]

Question 4 pls

[ember escarp]

Yea

[hallow cliff]

integrate the left hand side and show us what you get

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[ember escarp]

ok 1 sec

[ember escarp]

i got (4x-A)dx = A^2

[fiery crypt]

that's not right
how'd you integrate it?

[ember escarp]

6x^1/2
add 1 to it becomes: 6x^3/2
divide 6 by 3/2 so it becomes 4

thats what i got

[fiery crypt]

that is true but you got rid of x's exponent
also you did not integrate A

[ember escarp]

so would it be 4x^3/2

then id also integrate A

[fiery crypt]

yes

then you'd have to do the whole replacing x by 4 & 1 thingy

since it's a definite integral

[ember escarp]

yep

ok tyvm

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[crude wigeon]

I’m pretty sure I got something wrong here, that seems like a large near point

[crude wigeon]

And isn’t v supposed to be negative as the near point of someone with long sightedness

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[boreal ingot]

Hi

[boreal ingot]

to check whether above system is linear.
This is a yt video i found on the topic. but it is 1D (Digital Signal Processing)

y(n1, n2) = n * x(n1, n2)

my topic is DSP only but we are also taken 2D, the professor didnt explain much but tomorrow is exam and i cant find any solutions online.
I think it is linear based on the 1D video
needed some clarrification

[vernal wraith]

What's your doubt exactly ?

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[boreal ingot]

Is it correct

is the system linear?

[boreal ingot]

y(n1, n2) = n * x(n1, n2)

[vernal wraith]

X(n1, n2) must be something right ?

What if $x(n1, n2) = sin(n_1^2 + n_2^2)$

[solid kiln]

Solomaniac

[vernal wraith]

Then it's never linear.

The basic concept of linearity is ax1+ bx2 should give ay1+ by2. No matter what dimensions.

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[boreal ingot]

In DSP or (DIP) its the input? Idk the professor skipped trough stuff. I was preparing for exam

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[vernal wraith]

Just remember this.

[boreal ingot]

ok thanks

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[proven flax]

yo solomanic it's me

[proven flax]

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[vernal wraith]

Gotcha 😂

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[static tendon]

n men are to be seated round a circular table. find tha probability that two particular men sit together

[wraith hinge]

hint : total - when they are not together

[wraith hinge]

2/(n-1). Place one person wherever and there are 2 possibilities which work out of the remaining n-1 for the other person

or just tie those two person with rope and merge those two into a person

sure that works as well

[high wave]

hello T_T

i need help

its due later and idk how to do it T_T can someone help me?

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[delicate lance]

I don tget howthey found y prime?

I mean the derivative

Ddi they use the product rule

[compact salmon]

they used the power rule

$$ y = f(x)^2 \rightarrow y'=2f(x)\cdot f'(x) $$

[delicate lance]

whered the 2 at the beginning come from then

if its f(x) times the derivative

[solid kiln]

dp

[compact salmon]

sorry, I forgot about the two

this is the correct formula

[delicate lance]

ohh ok

[compact salmon]

In general

$$ y = f(x)^{n} \rightarrow y'=nf(x)^{n-1}\cdot f'(x) $$

[solid kiln]

dp

[hallow cliff]

Instead of memorising this as a formula, just learn the chain rule

[delicate lance]

OH

Thats the chain rule omg im so blind

[compact salmon]

yeah, of course, this is an application of the chain rule :)

[delicate lance]

Wait so how do i know when to use product rule or chain rule bc i thought i would have to use the product rule

[compact salmon]

In this case technically you could have solved it with the product rule, but it would be very tiring and cumbersome

in general, use the power rule when there are two functions f(x)*g(x) where f(x) is different than g(x)

[hallow cliff]

If it was (x^2+2x+1)(something else), you'd use the product rule. But here its (x^2+2x+1)(x^2+2x+1)=(x^2+2x+1)^2, so you can use the chain rule

[delicate lance]

Ohhhhhhh ok guys i get it

So this is not the answer

[hallow cliff]

It is

[delicate lance]

I meanyes

but how do u simplify it

to be like the other answer

[hallow cliff]

When you have a + a, its the same as 2a

There a=(x^2+2x+1)(2x+2)

The same thing is added with itself, so it's just twice of that thing

[delicate lance]

OHHHHH

[hallow cliff]

So 2(x^2+2x+1)(2x+2)

[delicate lance]

I get it tysm now

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[slow dock]

pls help

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[pale kraken]

does anyone know how to do this?

[slow dock]

anyone knows how to do this ?

[pale kraken]

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[wraith hinge]

Hey, can someone please explain how I'm supposed to do 17?

[wraith hinge]

There's a formula for this I believe, but I don't understand it

can you determine the vector PQ

[wraith hinge]

Yes

ok what is it

[wraith hinge]

<4, 2, -2>

right

so now

I think I'm able to find the parametric equations

The issue is the vector equation

an equation of the line can be described with [
\vj r_0 + t\vj v
]
where $\vj r_0$ is the initial position vector and $\vj v$ is the vector you just find. This parameterises into [
x = x_0 + tv_x \
y= y_0 + tv_y\
z = z_0 + tv_z
]

[wraith hinge]

How's this?

[wraith hinge]

seems good

nice handwriting btw very concise

Thank you

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[autumn shard]

I just started with linear line equations

i have this question but i cant wrap my head around it. i cant understand it.

"A linear line goes through the coordinates (-3,5) and (a, 2). It cuts through the X axis on x = 7. What is a?"

Im lost.

[autumn shard]

Am i supposed to create a new point of (7,0) since i know thats where it cuts through?

[spiral socket]

i dont think that is necessary

remember that linear equations look like $f(x) = ax+b$

[solid kiln]

dqvidutzul

[spiral socket]

and f(x) = y

[autumn shard]

ah that just made it even more confusing haha

[spiral socket]

you have the point (-3, 5) so f(-3) = 5

if you replace x with -3 you have a general rule

wait now im lost too

[autumn shard]

yeah.... im really confused

[spiral socket]

f(-3) = 5 and
f(a) = 2

[low gazelle]

I think this is the best way

[autumn shard]

and then i just do the K formula with these two KNOWN points?

[low gazelle]

yeah

idk what the K formula is but it is probably the same one im thinking of

[autumn shard]

the delta y / delta x formula

[low gazelle]

Oh yeah that works too

I was thinking point slope form but that is fine too

ig theyre technically the same thing if you plug 2 points in

[autumn shard]

hahaha im sooooooooooo confused wow

[low gazelle]

what did u get for this

[autumn shard]

k = -0.5

[low gazelle]

yeah thats the slope so you can plug it into this to get the equation of the line.

We know the point (a,2) is on the line so if we plug it into the equation of the line we can solve for a

[autumn shard]

sorry, i dont understand

[low gazelle]

which part

[autumn shard]

im clueless on what to do after calculating K. I just started with this subject. f(x) is foreign to me

[low gazelle]

You can think of f(x) and y as being the same thing. In your class how do you write the equation of a line?

[autumn shard]

y = kx + m

[low gazelle]

So you have calculated k, which means you know the slope of the line that passes through all the points given in the problem

[autumn shard]

yep!

[low gazelle]

You got -0.5, so we know the equation of the line that goes through all the points has the form y=-0.5x+m

[autumn shard]

im with you

[low gazelle]

Now we want to find out what the value of m is.

We can plug in either (7,0) or (-3,5) into the equation, which will let us solve for m

doesnt matter which

[autumn shard]

how would we do that?

y = -0.5(-3, 5) + m?

[low gazelle]

(7,0) and (-3,5) are points. This means they are of the form (x,y).

The equation of line represents all possible points on a 2d graph that satisfy the condition of y=kx+m. This means that if you plug a point lying on the line into the equation, the equation will be true.

So we would plug in x=7 and y=0 or x=-3 and y=5. If we plug in (7,0) (this just means plug in x=7 and y=0), we know the equation (0)=-0.5(7)+m must be true, which means we can solve for m

That will give m (the y intercept) of the equation of the line

[autumn shard]

ofcourse. so x is just the x point... and not just any var

y = -0.5 * -3 + m?

[low gazelle]

x is a variable in the context of the equation, meaning it can be a range of numbers. One way to think of a line is as a list of points on a 2d graph (satisfying some equation of the form y=kx+b). For any value of x you can think of, there will be a corresponding y value such that the point (x,y) lies on the line.

When we plug in a point like (-3,5), we are choosing the value of x=-3, and we are given by the problem that y=5 is the corresponding y value such that (-3,5) lies on the line.

This is why we can plug in x=-3 and y=5.

x is a variable, that is it is not a known number. When I say (-3,5) has the form (x,y) I mean x=-3 and y=-5 is a possible pair of x and y values, not that we know x=-3 and y=5.

Let me rephrase that last sentence. We will never know what x and y are because they are variables, but we know if it is the case that x=-3 and y=5, then it will be a point on the line

[autumn shard]

okay i understand that now. But how does this give us the value of m?

m is still unknown

[low gazelle]

Yeah so lets plug in a point. The point (7,0) will be easiest since one of the coordinates is 0.

y=-0.5x+b

becomes

0=-0.5(7)+b

Do you see how we can solve for b now?

[autumn shard]

b = -0.5(7) - 0?

[low gazelle]

yeah

So b=-3.5

actually I think there is a slight mistake

It would be b=0.5(7)-0

which gives b=3.5

[autumn shard]

and 3.5 is what i got when drawing in my notebook so the formula is indeed correct

how come we removed the -?

[low gazelle]

In 0=-0.5(7)+b we add 0.5(7) to both sides which gives 0.5(7)=-0.5(7)+b+0.5(7) which gives 0.5(7)=b

[autumn shard]

gotcha!

so now we have

y = -0.5(7) + 3.5

or is it more correct to write
0 = -0.5(7) + 3.5

[low gazelle]

We would write y=-0.5x+3.5. An equation of a line needs the variables y and x in it. We plugged in (7,0) so we could find what the value of m is, it is not needed in the equation of the line.

The 2nd part you wrote is correct, but it is not an equation of the line, its just a factual statement you could say

In general when we want to find the equation of a line we do it in two steps

1. Find the k value (slope/gradient/rate of change) of the line and
2. Find the m value (y intercept) of the line

We find the k value by using the K formula you mentioned

we then plug that into the general equation y=kx+b

finally we plug in any point we know lies on the line and solve for b

We then plug b into the general equation of the line

For the actual equation of the line, we will always have y and x in it

[autumn shard]

but how would we find what "a" is? since its unknown. we cant do -0.5 * a?

[low gazelle]

We now have the equation of the line. If we know a point is on the line (you are given (a,2) is on the line), we can plug it into the equation

What happens when you plug (a,2) into y=-0.5x+3.5

[autumn shard]

2 = -0.5(a) + 3.5

[low gazelle]

Yeah and now since you only have 1 unknown, you can solve for it

[autumn shard]

ohhhhh interesting, i feel so close but i have no idea how haha

[low gazelle]

Yeah when I first did this I basically just memorized the process without understanding why. Since you are just starting this topic it will make more sense why we do this as you learn more. What you want to focus on for this is 1. How to create the equation of a line given some information (the value of k/m or given points lying on the line) and 2. How can you use the equation of the line, which you will probably learn more about if you just started this topic (and why we did everything we did will make more sense)

[autumn shard]

you are being very very helpful btw

can seem to understand how to solve for a 🤔

[low gazelle]

I would start by subtracting 3.5 from both sides

[autumn shard]

2 - 3.5 = -0.5(a) + 3.5 - 3.5?

[low gazelle]

yeah

so -1.5=-0.5a

Now we want to get rid of that -0.5 in front of the 'a' while maintaining the equality.

[autumn shard]

-1.5 + 0.5 = a?

[low gazelle]

When solving for variables we do things on both sides to keep the inequality.

It looks like you tried to add 0.5 to both sides, but the issue with this is that -0.5a+0.5 is not just 'a'

When you have an equation like a=bx, we divide both sides by b (or multiply both sides by 1/b) to eliminate the coefficient infront of x (assume a and b are known numbers and we are solving for x)

When you have something like a=x+b, we eliminate the b by subtracting b from both sides (if b is negative you can add instead to both sides)

So in this case we want to divide by -0.5, since -0.5a/-0.5 = a

[autumn shard]

hmmm okay. i think i understand. i kind of get it but the pattern is not 100% clear yet.

-1.5 / -0.5 = -0.5a/ -0.5

a = 3

[low gazelle]

yeah

[autumn shard]

wow okay

sora, thank you so so so much. i will go ahead and keep studying till this makes perfect sense. now that i have the steps

[low gazelle]

gl

[autumn shard]

thank you!

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[simple jackal]

i am trying to find the cdf of Z = min(X, Y). Is this correct?

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[zinc ginkgo]

Yea looks right

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[shrewd pebble]

I got it wrong and im not sure why

got it

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[crisp gull]

Forgive my writing over it

Idk what to do

Find volume, but how?

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<@&286206848099549185>

I have wrote so much on this page 😭 i still cant find an answer

[tepid hamlet]

the cross section of the pool is a trapezoid right? so you can think of the volume as the area of that trapezoid x the width of the pool (5m)

[crisp gull]

Idk how to find the area of a trapezoid and if i should assume if it is one

Is a trapezoid same as trapezium?

I know that is h(a + b) all divided by 2

[tepid hamlet]

ok maybe easier way to think of it:
the bottom of a pool has a triangular section, you can figure out the dimensions of that and its volume and see how much volume is needed to fill that, then see how much water is left over to fill the box part above it

the volume for that bottom part will be the area of the triangle x the width

(at least if i'm interpreting the diagram you showed correctly)

[crisp gull]

Yeah do u see how i drew like the line connecting that triangle on top but idk its height nor base so i left it

I was thinking that maybe its to do with scale factors but idkkk

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[proven jewel]

why is it = 4 on a),

oh my god

i literally got it as i asked

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