help-38

[maiden hare]

We also know it decreases constantly at a rate of 1 per second

[covert oxide]

ok

[maiden hare]

That means the velocity at t = 3 is 2 less than at t = 1

[covert oxide]

ok u lost me

WAIT NO I GET IT

ok keep going

[maiden hare]

And since the average is 0, then those must be opposite

[covert oxide]

yes

[maiden hare]

So let's take v1 and v3 again

v1 is the velocity at t = 1, and v3 is the velocity at t = 3

We know v1 = -v3

And also v3 = v1 - 2

[covert oxide]

yea

[maiden hare]

Can you solve that?

[covert oxide]

solve for wha

[maiden hare]

Find the values of v1 and v3

[covert oxide]

how

[maiden hare]

v3 = -v1 and v3 = v1 - 2

You have the same thing that is equal to two different things

[covert oxide]

how

[maiden hare]

Hm I guess you haven't seen many equations

You can just replace the v3 in one of them by the v3 in the other:
-v1 = v1 - 2

[covert oxide]

oh

like a composition?

[maiden hare]

Not sure what you mean by that

If a = b and a = c then b = c

[covert oxide]

-v1=v1-2

liike that?

[maiden hare]

Yes that's what I wrote

[covert oxide]

ok

then

can i say+2

then -v1+2

=v1

then +v1

so 2=v1+v1

v1=1?

[maiden hare]

Yes

[covert oxide]

ok

[maiden hare]

So you have the velocity at t = 1

Now the initial velocity is at t = 0

[covert oxide]

how

wouldnt it be 2

cause if acceloration decreases each sec

wouldnt t=0

be2

[maiden hare]

I said it's at t = 0, not that it is 0

But yes you're correct

[covert oxide]

huhhhhhhhhhhhhhhhhhhhh

[maiden hare]

So it's 2, then 1, then 0, -1, and -2

[covert oxide]

ok

how do i expain that

[maiden hare]

Idk

[covert oxide]

😭

im doing quiz corrections

so i have to explain it

[maiden hare]

Wait you're a teacher?

[covert oxide]

no

a student

[maiden hare]

Wdym quiz corrections

[covert oxide]

like i mesed up on a quiz

[maiden hare]

Oh

[covert oxide]

so my teacher toldme to explain every questiion

to getsome points back

[maiden hare]

Honestly it's hard to say without knowing what the course is supposed to teach

[covert oxide]

its kinematics

[maiden hare]

Did you understand the average velocity thing?

[covert oxide]

kinda

[maiden hare]

If you have two points in time where the position is the same, then the average velocity over that duration is 0

[covert oxide]

ok

so how do i explain the whole thing

[maiden hare]

Additionally, if the velocity's rate of change is constant, in other words if the acceleration is constant, then the velocity at the midpoint between the two times is 0

Because both sides need to cancel out

[covert oxide]

yea

[maiden hare]

It's more general than that, for example if the position at t = 1 is 5 and at t = 7 is 17 then the average velocity is (17-5)/(7-1) = 2, and that would be the velocity at t = 4, the midpoint between 1 and 7

Assuming constant acceleration of course

[covert oxide]

But what would I say for my problem

I'm just lost

[maiden hare]

Well, maybe you can say that the positions are symmetric about t = 2

since the positions at t = 1 and t = 3 are the same, and also at t = 0 and t = 4

That means the velocity at t = 2 is 0

[covert oxide]

but we need the initial

[maiden hare]

You have the acceleration

[covert oxide]

how do we know velocity at t=2 is 0

[maiden hare]

It's the point where the position is the greatest

It's where the car turns back

[covert oxide]

bro this akes no sense

when time is 2

the position is 2

and to find velocity dont we divide

so why wouldnt it be 1

[maiden hare]

You're just finding the average velocity between t = 0 and t = 2

[covert oxide]

why

arent we trying to find initla velocity

[maiden hare]

I mean when you divide 2 by 2 to find 1

That's the average velocity over the first 2 seconds

Nothing more

[covert oxide]

why are we trying to find average velocity

what does it have to do with initial velocity

[maiden hare]

That's not what I'm saying

I'm saying that's what you are doing

[covert oxide]

ok

so how do i find initialvelocity

[maiden hare]

right here

[covert oxide]

so what do i need to do

to find initial velocity

[maiden hare]

Well you find the velocity at one point in time and use the acceleration to find it at all points in time, including t = 0

[covert oxide]

ok

how do i find it at 1 point

[maiden hare]

By using the average between 2 points

[covert oxide]

why are we finding average

if we want to find the velocity at one point

[maiden hare]

Because acceleration is constant

[covert oxide]

but u said 1 point in time

[maiden hare]

Maybe this will clear things up:
we know the velocity over the first second is 1.5,
we know the velocity over the next second is 0.5,
what's the velocity at exactly t = 1?

[covert oxide]

1.5

[maiden hare]

No

[covert oxide]

HOW

u said ove the first second is 1.5

[maiden hare]

Hold on

[covert oxide]

so how is the velocity at 1 second not 1.5

<@&286206848099549185>

[covert oxide]

so then the velocity over the first second is or isnt 1.5

[maiden hare]

When I say "velocity over" some duration, it's average velocity

It is 1.5

But that's not the same as the velocity at t = 1

[covert oxide]

so then whats on t=1

[maiden hare]

Do you agree that the velocity over the first second is the average between the velocity at t = 1 and the velocity at t = 0?

[covert oxide]

yes

[maiden hare]

So then if v=1.5 at t=1, since you know a=-1 that means v=2.5 at t=0

That would mean the velocity over the first second is 2, which is incorrect

[covert oxide]

ok

then how do i find at v=1

[maiden hare]

You mean v at t=1?

[covert oxide]

oh yea

[maiden hare]

Be precise with wording in this kind of problem

You know the velocity over the first second is 1.5 because of the positions, so you need to find v0 (v at t=0) and v1 (v at t=1) such that the average of v0 and v1 equals 1.5

[covert oxide]

and how do i do that

[maiden hare]

Keep in mind you have the acceleration still, so you also know v0 is one more than v1

In equations that's
(v0 + v1) / 2 = 1.5
v0 = v1 + 1

[covert oxide]

ok

so

(2v1+1)/2=1.5?

[maiden hare]

Yes

[covert oxide]

ok

then

do *2

both sides

2v1+1=3

-1 both sides

2v=2

divided by 2

v=1?

[maiden hare]

v1=1, yes

[covert oxide]

ok

so initial velocity is = 1

[maiden hare]

No, initial velocity is v0

[covert oxide]

oh

so

v0=2

cause accel decreases by 1

[maiden hare]

Yes

[covert oxide]

ok

now how do i explain that all at once

[maiden hare]

I mean... you've solved it twice now

[covert oxide]

PLEASE

PLEASEEE

[maiden hare]

I literally don't know how you could explain it in your own words

[covert oxide]

😭

[maiden hare]

I would say something like:

• we know p0 = 0 and p1 = 1.5 and a = -1
• (v0+v1)/2 = p1-p0 = 1.5
• v1 = v0 - 1
• so (2v0 - 1) / 2 = 1.5
• solve: v0 = 2

[covert oxide]

whats p = to?

[maiden hare]

p is position

p0 at t=0, p1 at t=1

[covert oxide]

ok

tysm

[maiden hare]

This is a graph of the position and velocity

You can see at t=0 that v=2

Velocity in blue

[covert oxide]

yea

[maiden hare]

Hopefully that gives you a better understanding of it?

[covert oxide]

yea i think im starting to understand

[covert oxide]

this helps

[maiden hare]

In particular there is a point where v = 1.5 but it's not at t=0 or at t=1

It's between the two, and because v decreases linearly (it's literally just a line in the graph), it's exactly in the middle

[covert oxide]

ok

[maiden hare]

Can't believe I spent 2h30 on this lol

[covert oxide]

lol

[maiden hare]

If you learn derivatives in the future, you'll learn that velocity is the derivative of position and acceleration is the derivative of velocity

aka rate of change

[covert oxide]

ok

[covert oxide]

wait quizk question

p1=1.5?

position 1 is equal to 1.5

[maiden hare]

Ye

[covert oxide]

ok

[maiden hare]

Position at t=1

That's in the table

[covert oxide]

so then should i say t=1=1.5

[maiden hare]

No that's not a good notation

It pretty much reads 1=1.5 which is nonsense

[covert oxide]

so then should i say

t1=1.5

[maiden hare]

Why t? I chose p because it's the first letter of position

[covert oxide]

ok

how is position 0 =0

[maiden hare]

That's in the table too

[covert oxide]

position 0

woudlnt position 1 be = 0

and then position 2 = 1.5

[maiden hare]

If you want to start at 1, sure, but it says in the table that time starts at 0 so why mix?

[covert oxide]

idk if my teacher wil get what i mean tho

[maiden hare]

Just write it in plain words: p0 is the position at time t = 0

[covert oxide]

ok

then write the thing u told me to above?

[maiden hare]

If you understand it, sure

[covert oxide]

what should i say v0 is

and v1

[maiden hare]

If p0 is the position at time t = 0, then v0 is the velocity at time t = 0, as simple as that

[covert oxide]

ok

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[trim jolt]

Please don't occupy multiple help channels.

[kindred lark]

How do I find out radius

[wraith hinge]

use the fact that they are right triangles

[trim jolt]

@kindred lark Has your question been resolved?

[kindred lark]

So 8^2 = R^2 + 6^2

[wraith hinge]

no

draw the triangle first

then use pythagorean theorem

they will always give you questions where you know everything except the one you're trying to find

they gave you 2 out of 3 side lengths of the triangle, the 3rd one you try to find

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[brazen gazelle]

how can i get my logarithmic to look curved on the y axis

it’s just going straight

[trim jolt]

@brazen gazelle Has your question been resolved?

[brazen gazelle]

<@&286206848099549185>

can anyone help

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185> bruh hello

can someone help

<@&286206848099549185>

[narrow tide]

hm

your problem resolved?

hello?

Sorry but I'm going to have to close this channel due to inactivity.

.close

[brazen gazelle]

no bruh

my question hasnt been solved

ive been looking for a helper but noone is helping me

<@&286206848099549185>

<@&286206848099549185>

[brazen gazelle]

where the fuck are the helpers

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[keen void]

-2 + (3/2)^2

What is an easy way to solve this?

[keen void]

.close

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[inland fox]

Can someone help B i dont get it

[delicate bobcat]

do you know general formula for the coordinates of vertex

for quadratics

in general

[inland fox]

Ya

[inland fox]

Like this??

[delicate bobcat]

yes

-b/(2a) is for x-coord

[inland fox]

But how about the other coordinate

Ya how about Y?

[delicate bobcat]

just find value of y when x = -b/(2a)

or

y coord of vertex is given by

-Delta/4a

[inland fox]

Wait what?

Oh yea also i remembered my teacher said that if i find the x coordinate i should just input it into my equation

[delicate bobcat]

Delta = discriminant

[inland fox]

To get the Y coordinate

[delicate bobcat]

yep

[inland fox]

Oh so that is equal to this?

[delicate bobcat]

yes

try both ways and see

[inland fox]

Oh Okay thanks let me try

[inland fox]

Wait so -a+b/2/4a

[delicate bobcat]

lemme use a1x^2 + b1x + c1 to distinguish coefficients

a1 = 1
b1 = (a+b)
c1 = ab

D = b1^2 - 4 * a1 * c1 = (a+b)^2 - 4 * 1 * ab = (a+b)^2 - 4ab

= a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a-b)^2

y of vertex = -D/(4a1) = -(a-b)^2/(4 * 1) = -(a-b)^2/4

[inland fox]

Ohhh wait omg i get it now

Tysm!

Just adking why u cant cross this out

Why do i have to make it a fraction first

And is my B correct,

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[inland fox]

.close

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[main hare]

is this equation for an angle bisecotr of two line correct?
instead of r1 and r2 put in R1 and R2, i just wrote r accidentally

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[main hare]

<@&286206848099549185>

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[royal orbit]

.close

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[smoky kite]

Based on this

[smoky kite]

When they swap two of the rows to find the determinant

Can that only be done when there are zeros like in this example

Or can you swap rows in any situation as long as you include the -

[inland mural]

yes

you can always swap rows or columns and include -

determinant is a volume form

which means switching a row or column u get a - sign

[smoky kite]

Ah okay great thanks

Would you say what they’ve written is sufficient working, only because if someone’s marking it they’d have to follow it somehow right

[inland mural]

maybe first line to second line is more unclear than exchanging the rows? I'm not sure how mark schemes work

[smoky kite]

That’s what I thought

Thanks anyway

.close

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[true coral]

does drichlets theorem imply that there is an infinte amount of primes in the form ab+1?

[true coral]

Im pretty sure this is true, just want to be sure

[inland mural]

yes

actually proving infinite number of primes of form pn+1 where p is a prime can be done in a very elementary way

[true coral]

Oh i think i Know how

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[wraith hinge]

I'm not sure what im doing wrong here tbh

I got $3,977.04 for my Present Value

which is apparently incorrect

These are my inputs in the financial calculator

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[mystic veldt]

[mystic veldt]

i want someone to check my solution please

that would be R in the question

okay youre back

[wicked drum]

I don't think it's right

But it does give you the correct answer

[mystic veldt]

how is that wrong then

[wicked drum]

It's the same problem again

The r and dr and not the same r

[mystic veldt]

what?

i taken them different this time

[mystic veldt]

.

[wicked drum]

The r in the 2πr the radius of the ring you made correct?

[mystic veldt]

huh?

ive not taken R as radius of my element this time

[wicked drum]

This ring

[mystic veldt]

ive taken anything else which is 'r'

[wicked drum]

The dr should be RdTheta

Where theta is the angle between the axis of the sphere and R

Remember what we did last time?

[mystic veldt]

yes but that was for volume and i revised it and understood it

like we used Pythagoras and all

[wicked drum]

We had to change the dr to dz because dr wasn't correct?

[wicked drum]

Yes

The problem is

dr means differential change in r but in your diagram dr is not a change in r

[mystic veldt]

yes

also the problem we did last time

[wicked drum]

So the thing you're writing as dr is not actually dr

It's something else

[mystic veldt]

okay then what should we do next

[wicked drum]

Integration of 2πr dr gives you area of a circle

Cuz small change in r would be in the direction of r

[mystic veldt]

yes

[wicked drum]

Like slowly increase the r

[wicked drum]

The same thing we did last time

[mystic veldt]

okay

so listen

the problem we did last time

i tried to find an alternative for that

and this is what i did

this method is also correct right?

you just tell me if this is right then i will try to find the answer of the ongoing problem with this method also like youre saying

[wicked drum]

What is z

Like which part

[mystic veldt]

the distance from the bottom axis

to the cental point

just like you took

but instead of pythagoras i used trig

[wicked drum]

How did you get cos theta = z

[mystic veldt]

like i had sin theta

from that you can get cos theya

theta

(by making a triangle)

and then cos theta was also equal to z/r

so i equated them

there are 2 triangles look closely

[wicked drum]

This should be z

Other than that everything is correct

Can you like draw the triangle from which you got z?

[mystic veldt]

okay i just realised that was not a right angled

😭😭😭😭 i messed up again

[wicked drum]

Well luckily 2 mistakes cancelled out and you got the correct answer

[mystic veldt]

wait are there any problems like these in 12th ncert?

[mystic veldt]

😂

[wicked drum]

Idk

[mystic veldt]

okay what grade you in?

[wicked drum]

12th

[mystic veldt]

yea so havent you started applications of integrals?

[wicked drum]

I'm done with it

[mystic veldt]

is it easy?

[wicked drum]

They didn't ask anything like this in school

[mystic veldt]

yea

[wicked drum]

So I think there's not

But also they changed syllabus and stuff

So maybe they removed jt

I know all this volume and area integrations because I've been doing this in physics a lot

Centre of mass specially

[mystic veldt]

yea thats why im practicing this

cuz of physics

[wicked drum]

Then there's stuff they make you derive in magnetism and electrostatics

[mystic veldt]

oh ok

ok i will go through my work again with the hint you've provided above

🙏

.close

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[white lantern]

Hi, I have a question. If proof of 𝛼 ⊢ 𝛽 have 20 lines and proof of 𝛽 ⊢ 𝛾 30 lines, how many lines does a proof of 𝛼 ⊢ 𝛾 will have?

[vapid lynx]

is this a "real" question of like a one-off question

could just be going for 51 since 𝛼 ⊢ 𝛽 and 𝛽 ⊢ 𝛾 is just 𝛼 ⊢ 𝛾

[trim lichen]

at most 50, but there might be a proof that doesn't go through beta...

[white lantern]

What about ¬𝛼⊢¬𝛽 or ¬𝛽⊢¬𝛼?

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[white lantern]

.close

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[sly steppe]

how do I simplify this?
I'm not 100% sure which formula/ rule to use this

[gray rapids]

2sinacosb is sin(a+b) isnt it? so middle term becomes sin^2(a+b)?

that would be double angle identity

leaves you cos^2b

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[wraith hinge]

Hello ,if the system called f y(n)=2x(2n) and if df/dn=0 then why is it time variant ?

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[winged hinge]

is n supposed to be time?

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[wraith hinge]

Yes

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[wraith hinge]

<@&286206848099549185>

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[rapid kestrel]

Can someone explain this to me?

[rapid kestrel]

<@&286206848099549185>

[torpid depot]

!15mins

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[torpid depot]

!status

[trim jolt]

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

[rapid kestrel]

I don’t know where to begin

[rose grotto]

if the pace is 7 cans in 3 days how much would you expect the dog to eat in a single day??

[rapid kestrel]

7/3 ≈ 2.33

[rose grotto]

yeah

so how much would you think he would eat in d days

[rapid kestrel]

Wouldn’t it be 7/3 + 7/3 x D

Since 7/3 cans in a day multiplied by the days

[rose grotto]

yeah there u go

exactly

[rapid kestrel]

I can’t see that in the choices or am I blind lmao

[rose grotto]

hm. that indeed isnt in the choices

[trim jolt]

@rapid kestrel Has your question been resolved?

[rapid kestrel]

So what’s the answer

[fair patio]

so he eats 7/3 per day, so in 3 + d he will have eaten the regular 7 cans + some 7/3 * d

[rapid kestrel]

Which one in the choices is it?

[fair patio]

k

[rapid kestrel]

Does it make a different if the * d is on the numerator or the denominator?

[fair patio]

$7 + \frac{7d}{3}$

[solid kiln]

Charles of Palestine

[fair patio]

yes because if it is the numerator then it is multiplied by 7/3, if it is the denominator it is in the division

also because the number of eaten cans is directly proportional to the number of days

the more days pass, the more cans are eaten, no?

so it is a direct relation, so it is in the numerator

aye @rapid kestrel ?

[rapid kestrel]

Yeah I got it

[rapid kestrel]

Jazak allah khayran man

How do I close this

[fair patio]

type .close, also thank you brother, i am muslim too : D

[rapid kestrel]

Alhamdullah me too, inshallah what’s happening in Palestine can be finished soon

Also thanks for the explanation

.close l

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[tame notch]

This correct ?

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[nova spire]

do we agree that {j,j+1,j+2,j+3,j+4,...} means "all integers bigger or equal to j" right?

[tame notch]

Yes

But until they reach 3*i

Right ?

i is 2

So 3*2 is 6

Sorry, I just started college haha

When does j+1 and so on stop then ? Never ?

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[daring brook]

i got (4, 4.016)

either i got it wrong or idk how to format the answer

anyone

wait i got it

[tidal forge]

If u want to check the answer, use desmos

[daring brook]

.close

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[wraith hinge]

hey

[wraith hinge]

for part b im just a bit confused

im gonna send my steps so far in a second

so for Qi = i
deg Q1 = 1 meaning that its in the form of ax+b
deg Q2 = 2 means that its in the form ax^2 + bx + c (used different variables to differentiate the numbers tho)

now im kinda stuck, ik that aw = 1 (comparing coeffecients from left side to right side) but i dont know how to find the exact value of say a for example

if anyone can help pls lmk and @ me so i get the notification

[marble wharf]

wlog you can just assume a=1

if you have a solution with for example a=17, then you also have a solution where you divide Q1 by 17 (which makes the new leading coefficient equal to 1) and multiply Q2 by 17 (so in total the equality is still true)

[wraith hinge]

so both a and w are 1?

assuming that a*w = 1 meaning both values are 1

but it could also be that a = 5 and w = 1/5 and a*w = 1 so we cant really make that assumption?

[marble wharf]

it will end up with an equivalent solution

[wraith hinge]

right

[marble wharf]

for fun, try solving both of them

first assume a=w=1 and solve for the rest. then assume a=5,w=1/5 and solve for the rest again

compare the solutions you get

[wraith hinge]

so its fine to assume?

also for bz

its a constant

so the only real values it can be are

b = 1 z = 6
b = 2 z = 3
b = 3 z = 2
b = 6 z = 1

does that help in any way

[marble wharf]

well thats under the assumption that everything is integers

why should that be true

[wraith hinge]

imean were already assuming a=w=1 so why not assume some more lol

[marble wharf]

well for starters you make only one assumption and see where it gets you

if at some point you cant go further then you can try assuming something more

that said, the assumption I made is completely different from having everything be integers

also you forgot stuff like (-1)*(-6) btw

[wraith hinge]

true

ok gimme like

5 min

lemme try a=w=1 and a= 5 w = 1/5

i got 2 different values

for az + by = 3
when a=w=1 its
z + by = 3
when a=5 w=1/5 its
z + by = 3/5

im really stuck lol idk how im meant to find the values of these

[marble wharf]

you have three equations and three variables left

(I mean the faster way to do this whole exercise would be to guess a root of P and then set Q1 from there and do polynomial division but lets ignore that)

[wraith hinge]

yeah

dont really wanna get into that cuz prof didnt mention it

so im just tryna use like basic rules of polynomials

[wraith hinge]

yeah but none of the others get affected by a and w

@marble wharf sorry for ping lmk if u dont want me to ping but is this even really solve-able?

[marble wharf]

it should be. but I dont want to work through the details tbh

[wraith hinge]

what should i do then lol cuz im completely stuck

[marble wharf]

what three equations do you have

[marble wharf]

(and like I said, this would be faster from the start)

[wraith hinge]

ill do that then

how do i do it tho

[wraith hinge]

P(x) = x^3 - 2x^2 + 3x + 6
from that we know that Q1(x) has to be in the form of ax + b
and Q2(x) has to be in the form of wx^2 + yx + z

thats as far as i got

i tried expanding then solving for the values but got stuck on a roadblock

[marble wharf]

sry I have to go

[wraith hinge]

damn

<@&286206848099549185> can someone help since hes gone? i can resend the question and everything

[trim jolt]

@wraith hinge Has your question been resolved?

[wraith hinge]

<@&286206848099549185> sorry for double ping but i really need help

[umbral dragon]

Send the equations

Try a=1 and w=1

[wraith hinge]

i did but that doesnt really help

[umbral dragon]

Then just resend the equations

So i can see them

[wraith hinge]

ok one sec

3 part b is my question

this is where i reached

aw = 1
ay + bw = -2
az + by = 3
bz = 6

[umbral dragon]

Okay, so we have 4 equations for 5 variables

What do you get if you insert them into each other?

[wraith hinge]

more equations that can isolate 1 variable

[umbral dragon]

We can find the relationship of two variables

[wraith hinge]

right

[umbral dragon]

From that we can choose some value i suppose

[wraith hinge]

so for example aw = 1 we can do a = 1/w

?

and then waht from there

cuz thats where i got stuck

[umbral dragon]

Hmm yeah

[wraith hinge]

u get stuck too?

[umbral dragon]

Probably, but it’s all in my head rn, and that makes it hard to tell

[wraith hinge]

the guy that was helping before u mentioned polynomial division

[umbral dragon]

Yep

Basically choose some value for x such that the thing is 0

[wraith hinge]

what thing

[umbral dragon]

The polynomial

This is just trial and error

But try x=-1

[wraith hinge]

so substitue it into x^3 - 2x^2... ?

[umbral dragon]

Yes

[wraith hinge]

-12

then what

[umbral dragon]

+6

Not -6

The result should be 0

[wraith hinge]

oh right

yeah

[umbral dragon]

This means that you can factor out (x+1)

[wraith hinge]

how

oh

cuz x = -1

[umbral dragon]

Yes

[wraith hinge]

wait x - 1

is

in the form of ax + b

[umbral dragon]

Correct

[wraith hinge]

so that could be one of the products

Q1 = x + 1

[umbral dragon]

Yes

And now divide by x+1

To get Q2

[wraith hinge]

divide the polynomial by x + 1 right

ok wait lemme just watch a quick yt video about polynomial divisions cuz been a while sincei u sed them lol

another thing

so this is for a coursework

how do i say i got the value x = -1

do i just say something like through trial and error find a root of x^3 - 2x... and x = -1 fits the conditions or something like that

[umbral dragon]

Yeah probably, i don’t really know myself

[wraith hinge]

thats fine

i messed up my polynomial division

imma do it again one sec

x^2 - 3x + 6

is Q2

[umbral dragon]

It is indeed

[wraith hinge]

does this look good?

or is there anythingu can point out

[umbral dragon]

I think that’s good. But I’m no undergraduate, so I really have no way of knowing

[wraith hinge]

no problem lol as long as theres no missing info

[umbral dragon]

Yeah I think it’s fine

[wraith hinge]

really appreciate ur help bro i was stuck on this question for like an hour and a half

[umbral dragon]

No worries

[wraith hinge]

thanks again

cya

.close

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[umbral dragon]

Cya

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[solid kiln]

nick

[prisma drift]

sum from j=1 to i (i) is i²

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[wraith hinge]

Hi, I'm new to the topic. What do I have to do if both results give me 0 when I substitute them?

[stoic iron]

both results of what

[wraith hinge]

if i substitute 3 in the t it give me

t - 3 = 3 - 3 = 0
3 - t = 3 - 3 = 0

[stoic iron]

thats fine? im not sure what you're asking

[wraith hinge]

im asking how to solve that

the problem itself

[stoic iron]

you havent described the problem

[wraith hinge]

State whether the given functions are continuous to 3, if not continuous, say why:

[stoic iron]

and what does it mean to be continuous at 3

( i know im just asking you)

[wraith hinge]

the main topic is limits

i dont know if that say you somethign

[stoic iron]

i mean, can you describe what it means for f to be continuous at 3

[wraith hinge]

that doesnt exits?

[stoic iron]

what is the definition of being continuous at a point

[wraith hinge]

okay

thanks

.close

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[wraith hinge]

btw, if you wont help someone, just dont answer them

[stoic iron]

im literally just asking you to recall the definition of continuity

[wraith hinge]

fuck you man

[cunning cliff]

[wraith hinge]

what?

[cunning cliff]

<@&268886789983436800> dont think this is appropriate

[west nest]

😂

[wraith hinge]

are you that kind of people that never answer?

but just know how to send memes

[stoic iron]

genuinely do not understand why you're mad at me for trying to make sure you know what the definition of continuity is

[west nest]

What's the point of knowing the answer if you don't know why the answer is the answer

[dire charm]

.reopen while we figure out what's going on

.reopen

hmm?

Anyway, there doesn't seem to be a real mod problem other than the asker is rude and won't accept help, but that really brings its own punishment ...

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[unborn glen]

hello I have a really quick question in probabilities

[patent crow]

[unborn glen]

When he says probability function in general he means P(x)?

the question is if a function is f(x)=x^2/c x=0,1,2,3 find c so it can be called probability function

[delicate jasper]

its discrete?

$\sum_{x=0}^{3} \frac{x^2}{c} = 1$ i guess

[solid kiln]

Fucktalogist

[unborn glen]

so I have to compine the probabilities so that all together get me 1

the exercise is this

[unborn glen]

translated says this

I found c=26 but I think thats wrong

[delicate jasper]

,w sum of (x²+3)/c from x=0 to x=3

[solid kiln]

Results provided by WolframAlpha

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Requested by fractalogist

[delicate jasper]

,w (26/c)=1 solve for c

[solid kiln]

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[unborn glen]

oh okay

thank you very much

.close

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[proven narwhal]

Is my answer for the 1st question correct?

[proven narwhal]

better version

[ripe sparrow]

Looks correct and well written to me

[proven narwhal]

thx

.close

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[steady pine]

How to solve it

[steady pine]

I get that the modulus is or imaginary or zero which doesn't make any sense

Or that the argument is 90 degrees which can't be because it's given that z1 is in the 1st quadrant

[tepid rock]

Modulus imaginary?

How'd you get that?

Also whatever else is given, you should post that too.

[steady pine]

That's the only equation

[steady pine]

Because I converted it into polar

Which means -2r=r^3

And the answers to that are r=0, plus minus isqrt2

[tepid rock]

There's no cis in rhs anymore.\
$|e^{\iota \theta}| = 1$

[steady pine]

rhs?

I mean fine I can write it as an exponent

[solid kiln]

! What the hell am I doing here?

[tepid rock]

You do realise there's a modulus?

There's Not?

Fuck.

This is the other dude.

With modulus my bad!

[steady pine]

You got confused with the guys you are helping to? lmao

[tepid rock]

I'm sorry...

[steady pine]

That's totally ok

I don't understand what you mean by rhs

[tepid rock]

Right hand side

Anyways

How'd you get imaginary modulus again?

[steady pine]

.

[tepid rock]

Still doesn't give imaginary modulus does it?

You didn't even consider modulus.
You just rewrote that equation.

No modulus involved in your work whatsoever.

You'd need to do\
$|-2re^{i \theta}| = |r^3 e^{-i 3\theta}|$

[solid kiln]

! What the hell am I doing here?

[tepid rock]

And from here, you don't get " imaginary modulus "

[steady pine]

What

Wait wait

re^itheta

isn't r the modulus?

isn't modulus the distance of the complex number from the origin?

which is denoted by r?

[tepid rock]

Yes, r is the distance.

But -2r cis(theta) is the complex number. (this is what you've written.)

Also note

[steady pine]

Yes

[tepid rock]

r is the modulus for z

Not 2z or -2z

[steady pine]

I know

But if I have two complex numbers equal, it means their moduleses and arguments are equal, no?

[tepid rock]

Precisely.

But you barely took the modulus.

You didn't take any modulus.

Did you?

[steady pine]

What

I literally wrote it

[tepid rock]

For a complex number 2z, the modulus is merely 2r (if it's r for z)

[steady pine]

Here

[tepid rock]

No -

The modulus for z is the same as modulus for -z

$|z| = |-z|$

[steady pine]

Rightttttttttttttt -2r is negativeeeeeee

[solid kiln]

! What the hell am I doing here?

[steady pine]

I am so fucking stupid

Ok fine but

What about the argument?

[tepid rock]

I didn't know this was the issue you were facing lmao.

[steady pine]

My brain is not braining sometimes

And I guess r^3-2r=0 has real roots?

Yes sqrt2

Ok but what about the argument?

[tepid rock]

Argument of z?

[steady pine]

Yes

[tepid rock]

Theta = -3 theta? Hm?

[steady pine]

The arguments

[tepid rock]

Where'd that equation come from?

[steady pine]

Shouldn't I put them equal like the moduleses?

[steady pine]

Stupid logic of mine?...

[tepid rock]

Well you only set the left and right hand sides of the equality sign equal.

Nothing else unless you know what you're doing.

Two complex numbers being equal definitely mean their arguments are also same.

But in this case the two numbers aren't e^(i theta) and e^(-3 i theta)

[steady pine]

Ho?

[solid kiln]

! What the hell am I doing here?

[tepid rock]

There's clearly something you missed here...

[steady pine]

Oh wait

The - sign changes the direction

So it also changes the argument

Maybe I got it

Gimme a sec

Oh gosh is it supposed to be long?

Me stupid

I got theta equals to 45 degrees

Which seems legit

[tepid rock]

,w -e^{i pi/4} = e^{-3ipi/4}

[solid kiln]

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[tepid rock]

Checks out.

[steady pine]

Let's go

Thank you so much

.close

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[onyx plover]

Can someone help me ro check if this true? , and then if it's convergent

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@onyx plover Has your question been resolved?

[zinc ginkgo]

!original

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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.

[zinc ginkgo]

Why does your summation not have a limit on n but the term being subtracted does?

[onyx plover]

Well the question is proof that Un = summation (k= 1, n) of ((1/sqrt(k) -2sqrt(n+1))
Is convergent serie

@rie.mann

[onyx plover]

Hmm because it's go to infinity and i wanna calculate the limit, and also the Serie start from 1

@zinc ginkgo

[zinc ginkgo]

Yea so you don't put infinity on the upper limit

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[onyx plover]

Ops it's closed

[nimble stone]

huh

[onyx plover]

So i should put n upper the summation?

Instead of infinity

[onyx plover]

Is that true?

[zinc ginkgo]

Use .reopen next time

[onyx plover]

Oh well , i already reopen it

So , tbh idk where you exactly mean that i shouldn't put infinity

[zinc ginkgo]

Do this

And keep the limit

[onyx plover]

Alr let me edit it and show you

[onyx plover]

Well it's like that (I just I forgot to write limit in first line near summation, i add it)

So according to that Un is not convergent?

Because we get limit is = -infinity

👀

[zinc ginkgo]

You seem to have lost the 2sqrt(n+1) term

[onyx plover]

Yea , because it's const , it's like , for example limit(x->infinity) x+5 and 5 Will not effect on the results (I think)

[zinc ginkgo]

No it's not a constant

2sqrt(n+1) depends on n

The summation also depends on n

So you have to take the limit of the two terms together

[onyx plover]

Hmm so N->+Infinity and also K->infinity?

[onyx plover]

Un = summation of 1/sqrt(k) -2sqrt(n+1)

= Summation of 1/sqrt(k) + summation of -2sqrt(n+1)

And now we add limit of both of them

And now
Limit (k->infinity) Summation (k==1,n) 1/sqrt(k) = -infinity

+limit (n->infinity ) summation of (k==1, n) 2sqrt(n+1) = -infinity

And now -infinity - infinity = -infinity?

[zinc ginkgo]

k is just a summation index, not being taken a limit of

Write out u_n for n=1,2,3 and you'll find a pattern

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[worldly temple]

Need help with this question 3 and 4

[worldly temple]

I mainly need an explanation on how to solve this please 🙏

[low gazelle]

Do you know what 27^(1/3) is

[worldly temple]

Am a bigger I have avoided maths for a long time so am just starting form the basic and building up

Beginner

[low gazelle]

Seems like you started on the right track. 27^(-1) is 1/27. Since we are taking it to the -1/3 power you have to also take the cube root

[worldly temple]

Ahh thanks you

[low gazelle]

$x^{\frac{a}{b}} = \sqrt[b]{x^{a}}$

[solid kiln]

Cure Miracle

[low gazelle]

You can use this to solve the rest of the problems

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[smoky pebble]

a

[smoky pebble]

hey

i worked out ag its 16.18 (i think) im not sure how to do the rest

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[fast summit]

$\int :\frac{e^{-5x}}{cos2x}dx

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[pure drum]

How would I do this

[wraith hinge]

full pic

whats the cropped out part

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[brazen idol]

i found the integral but now im missing calculating it

[wispy sparrow]

,w integrate 1/sqrt(x+4) from 1 to infty

[solid kiln]

Results provided by WolframAlpha

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

Requested by chlamydia3844

[wispy sparrow]

you mean this?

[brazen idol]

yes but the value

im at this point

[wispy sparrow]

it... doesn't converge

[brazen idol]

yes

but its sitll asking for a value

and i cant put does not exist

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[sudden stream]

T

[sudden stream]

Okay so basically I have to make a thing where I check if the value output (lets call it output) is within 1% of a value (lets call it final value)

now this is not a math class but this is a math problem

so I came up with 100-(output/final value)*100

And I was wondering if that's adequate enough or if there's a better way

So like the idea is

1/100 is 0.01, or 1%

so 50/79 is 63% (after *100)

but I wanted to know if its withing

so 100-63 so 50 is within 37% of 79

So hopefully that explains what I'm talking about fully

[shrewd ridge]

it's wrong or right, it's not like you can't make something that always works

it makes sense

i mean, you misunderstand the task or something, but that formula is not far from right

[sudden stream]

Okay I see

Can you direct me to the formula

[shrewd ridge]

what if output = 504 and final = 500

[sudden stream]

mmmm

yeah that's a good point

[shrewd ridge]

100 − 100.8 = −0.8

[sudden stream]

So maybe the entire thing should be

|100-(output/final value)*100|

[shrewd ridge]

i agree

[sudden stream]

So

Now it shows from both greater and less than the final value how close the output number is

is there any other flaws in this formula that you see?

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[wary flax]

,tex $\theta = \frac{\pi}{3}$

[solid kiln]

Vignesh Saravanakumar

[wary flax]

Should the equation above, graphed in 3D, be a half or a full plane? When I convert this equation to rectangular coordinates it is a full plane. Additional points like (-1,pi/3) still satisfy the equation. But my textbook states that it is simply a half plane.

[zinc ginkgo]

-1 is not a valid polar coordinate value

for the ray value that is

[wary flax]

I didn't know that, ill have to revist old topics before doing this more complex version. Thank you for your help.

.close

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[rugged latch]

need help, not sure how to start. this section is on poisson distributions, but not sure if it relates to this problem

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[shrewd ridge]

i would pretend it's poisson yeah

like each game takes the same time

@rugged latch

[rugged latch]

Yea this is what I ended up getting

[shrewd ridge]

sure

same