help-10

[nocturne minnow]

Do you know polynomial long division?

[versed eagle]

yes

oh shit

thats all i do

[nocturne minnow]

Yep

[versed eagle]

oh lmao

ok thanks

<3

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[obtuse pebble]

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[rigid fog]

How are vectors perpendicular when it comes to 3 dimensions? So for 2 dimensions, v=<3,5>, one perpendicular to it would be <-3,5>. How would it work for <a,b,c>?

[brisk grove]

Take <1, 0, 0> and <0, 1, 0>

Can you try to draw those vectors

[rigid fog]

yea

itd look like a right angle

[brisk grove]

Try imagining a cube

and take 3 sides that meet in a corner

Those 3 sides are perpendicular

[rigid fog]

Mm, don't understand

[brisk grove]

Don't understand what?

Here are 3 perpendicular vectors

In 3 dimensions (this is a drawing on a flat surface, but you can imagine it)

[rigid fog]

Okay

[brisk grove]

So you get it?

[rigid fog]

No

[brisk grove]

Or are you having trouble visualizing it

[rigid fog]

vectorz r perpendicular if their dot product = 0

[brisk grove]

That's right

So? where are you having trouble

[rigid fog]

Im not sure, I did it fine with vectors with 2 dimensions

[brisk grove]

Again, you are not telling me what's the problem.

What are you confused about? What do you not understand? I can't help if you don't tell me

[rigid fog]

I guess solving the equations

so

I need a vector that has length x, and is perpendicular to another vector

so for 2d, I was able to set up the equations for both and solve them

[brisk grove]

Show me the original question

[rigid fog]

the 2d one?

or 3d?

[brisk grove]

3D, if that's what you are asking about

[rigid fog]

yea

so vector x = <18, 3, 9>, find vector a that is perpendicular and has length 4

[brisk grove]

The difference here is that in 3D you have an infinite number of solutions

There are an infinite amount of vectors that are perpendicular to any given vector.

[rigid fog]

okay

well to get the length

its 4/|x| (vector a)

[brisk grove]

Here is an example

[brisk grove]

What?

Let's just take a general vector first

v = <a, b, c>

and we want v to have length 4, and be perpendicular to x, right?

[rigid fog]

yee

[brisk grove]

So what are your equations?

[rigid fog]

18a + 3b + 9c =0

and

sqrt(a^2+b^2+c^2) = 4

[brisk grove]

Great

Now notice, you have 3 variables but only 2 equations

[rigid fog]

yee

[brisk grove]

So there are an infinite number of solutions. All you need is to find one of them

Can you try and do that?

[rigid fog]

yea

[brisk grove]

Also there is probably an easier way, which I could tell you

The problem is that you can't just randomly pick c = 6 for example, because that might result in no solutions

So first just try finding a vector, doesn't matter of what length, with 18a + 3b + 9c = 0

[rigid fog]

a = -1, b = 3, c =1

[brisk grove]

Okay,

now you want to take that vector

And WITHOUT changing it's direction, make it have length 4

[rigid fog]

oh

yea

so just 4/sqrt(11) <-1,3,1>

[brisk grove]

I think you got the length wrong

It's not sqrt(9)

Yes

Great!

[rigid fog]

My b

[brisk grove]

And you're done

[rigid fog]

ah, alright

[brisk grove]

You can write it as

<-4/sqrt(11), 12/sqrt(11), 4/sqrt(11)>

But that doesn't really matter

[rigid fog]

alright, well tysm

I appreciate the help ❤️

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[obtuse pebble]

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[dawn thunder]

are there any shortcuts I can take besides just evaluating the entire thing before taking anti dervatives?

[glossy basalt]

i guess you can factorize the inside on the sqrt?

opps, seems not

[amber obsidian]

hmmm

arc length question

[obtuse pebble]

@dawn thunder Has your question been resolved?

[tardy epoch]

Just do the algebra

[obtuse pebble]

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[vocal marsh]

help

[vocal marsh]

how does this happen?

can I get that result by using Bhaskara?

[nocturne minnow]

You can but it's not needed

Just factor by GCF

Then apply zero product property

[vocal marsh]

what is GCF

[nocturne minnow]

Greatest common factor

[vocal marsh]

oh

lol

okay, thannks

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[gritty saffron]

not sure what to do for the entire problem; for a i dont know the equation to use to find the position

[gritty saffron]

how do i find acceleration?

[obtuse pebble]

@gritty saffron Has your question been resolved?

[dawn thunder]

the equation for avg V is v= delta x / delta t

which means there of delta x = v * delta t

displacement is just the area on a V x T graph

[gritty saffron]

is displacement = position?

[dawn thunder]

acceleration is given as delta V / delta T

i think

[gritty saffron]

how do i find delta v?

[dawn thunder]

delta v = V(final) - V(initial)

[gritty saffron]

so should i find the area of the graph to find the positions of the particles?

sorry im a little confused here

[dawn thunder]

yes exactly

[gritty saffron]

if im finding the area, how do i find the area of something like t = 6 since its not marked on the graph?

[dawn thunder]

in that case look at t = 4 and t =8

notice how velocity remains constant

in which case you can assume velocity does not change at t=6

and the same can be applied to all other times t not labeled on the graph

i hope this clears it up for you

[gritty saffron]

ah so if i wanted to find 6 it would be something like position = (1)(5) ?

then it would be something like 5 meters?

[dawn thunder]

no, i think in this case time would be the entire interval from where you are counting from

the 5 is correct but 1 should be 6 to represent change in time

since change in time = T(final) - T(initial)

final in this case is 6s

initial in this case is 0s

[gritty saffron]

ohhh

[dawn thunder]

keep in mind you have a triangle and area under x-axis

[gritty saffron]

so just checking, the answer would be found something along the lines of like position of t = 1 is (1)(-5)
t = 2 is (1/2)(1)(-5) and etc?

[dawn thunder]

which means its not just rectangles and positive area

[dawn thunder]

yea exactly

[gritty saffron]

oh wait would t = 2 be (1/2)(2)(-5) instead?

[dawn thunder]

wait

[gritty saffron]

to represent this

[dawn thunder]

t = 2 should just be (-5)(2-0)

since that area before t=2 is just a rectangle

[gritty saffron]

thought it was a triangle

[dawn thunder]

draw a line going from t=2 straight downward

[gritty saffron]

so is the little upside down triangle for 3?

[dawn thunder]

the area should just be everything to the left of it

[dawn thunder]

yes exactly

[gritty saffron]

so t = 3 would be 1/2 (1) (-5) or 1/2 (3) (-5)?

[dawn thunder]

it would be the area of the rectagle you found from t=2 + this triangle t=3 which is 1/2 (1)(-5)

[gritty saffron]

oh can i add the areas to find t = 3 after i use 1/2 (1) (-5)?
like (1)(-5) + (-5)(2) + 1/2 (1) (-5)

area of t = 1 + area of t = 2 + area of t = 3

would that work?

[dawn thunder]

you wont have to add area of t1 in that case since t2 already has t1 included

consider each time point its own

its just you can combine previous areas to make the algebra a bit simplier is all

its still a better idea to calculate each area for each time point on its own

[gritty saffron]

so if i did 1/2 (3) (-5) i wouldnt have to do all the addition thing?

oh wait

nvm

i think i got the general idea of it now

[dawn thunder]

yup

because to t=3 there is a rectangle and a triangle

[gritty saffron]

i think combining previous areas would make things easier for me

[dawn thunder]

just be sure you dont add the areas that are already found like u did here

[gritty saffron]

right

[dawn thunder]

you should be all set good luck!

[gritty saffron]

thank you so much for the help i appreciate it

:)

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[obtuse pebble]

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[static valve]

Stuck on D

Idn how to make it = 58kg

Is 90 1 over 2 not 90.5?

I’m getting like 99.9 lol I messed something up

[violet sentinel]

can you show your work ?

[static valve]

I would if i knew what to do lol

i have 0 idea how this = 56 but i understood the ones before

[solemn epoch]

I think the answer is 58

[static valve]

ya that's what i meant typo

im just not getting that answer

doing the above formula

[solemn epoch]

Use 'Percent Decrease' formula given in your book. Well its same as Increase

Percent Change = (New-Old)/Old * 100 percent

We know the 'new value' and the 'percent change'. Take Old price as any variable

[static valve]

so they want me to guess? lol

[solemn epoch]

?

[static valve]

when i mean i dont understand what im doing wrong i have 0 idea lol

to get 58

[solemn epoch]

Percent Change = 90.5 percent
New value = 110.49 kg
Old value = x kg

Now put all that in the formula

And solve it

[static valve]

idn why its not clicking

@ruby fulcrum someone just show me cause i stg it makes no sense

like i need to see it

[obtuse pebble]

@static valve Has your question been resolved?

[solemn epoch]

I have an alternative method. Do you wanna see that?

[static valve]

please

[solemn epoch]

Way easier. (I usually ignore all formulas while doing these types of problems)

And use variable instead

[static valve]

ill wait cause thats confusing

i was doing 110.45x0.905 which is 99.95

[solemn epoch]

You were doing 90.5 percent of 110.49 kg.(which they didn't ask for)

[static valve]

is there any way to do that answer simpler?

[solemn epoch]

I don't think so

I used variable method but its still coming to that hard fraction part

So yea

[static valve]

ya i have 0 clue what u did

tbh

ive never seen that

even in class the prof didnt do that

[solemn epoch]

I used the formula given in ur book

I have written that above

90 1/2 percent in fraction is 181/200

And 100 percent is 1

Percentage can be confusing sometimes so I converted them into fraction

[static valve]

considering i been on this for 1 hour and sitll dont understand i need help lol

so lets say i dont make it a fraction

and leave it as 90.5

can i do the same math>

[solemn epoch]

Yea

But

90.5x/100 = 110.49-x

At one point you have to convert it into fraction

90.5 of x means (90.5/100)*x but also we cannot write decimals in fractions. So, 181x/200

Just multiplied both numerator and denominator by 2

[static valve]

is that 381? at the bottom left?

ok i see u added them together

[solemn epoch]

181x = 22098 - 200x
From RHS, -200x goes LHS and becomes 200x
181x + 200x = 22098
381x = 22098
x = 22098/381

[static valve]

ok let me try e and see if i have finally learned

OH SHIT

so ur right i make it into a fraction its so much easier

56 over 200 than 200 x 231.75 = 46350/256 = 181.05

ty so much @solemn epoch

i know it took awhile but the light bulb is working now

[solemn epoch]

No worries

Haha well. Practice a lot

[static valve]

trust im trying my best I make sure i understand everything to a T have not done math serious in my life lol

[solemn epoch]

Yeah. Not doing is better without understanding

[static valve]

if my percent is over 100% do i still do 600 over 100 or am i doing 600 over 1000?

and than turning it into a fraction

[solemn epoch]

f right?

[static valve]

ya

[solemn epoch]

Oh finally did it without picking up pen

[solemn epoch]

You know that 100 percent is 1 right?

100/100 = 1

So 600 percent will be 6*100/100 = 600/100 = 6

Now do what i did

The previous percentage was a bit hard but this one is easy

600 percent is basically 6

[static valve]

dw thats why i got 23.5 and its wrong

go g and h

so just have to figure out F than im done

[solemn epoch]

Gl

[static valve]

isnt 600 just 6 over 100 now?

fraction wise?

[solemn epoch]

Previously we found out, 90.5 percent i.e. 90.5/100 = 181/200. Now 600 percent in fraction is 600/100, basically 6

[static valve]

the 6 throws me off

i got it doing 600 over 100 than i multiply 100 by 24.92

than divide by 700

= 3.56

[solemn epoch]

Yep that's correct

[static valve]

so when its over 100% i dont have to x2 each side

[solemn epoch]

600/100 = 600/100

[static valve]

ahh

[solemn epoch]

Left with 6/1 = 6

Now 6 = (24.92 - x)/x

/x from RHS goes to lhs and gets multiplied with 6

6x = 24.92 - x

-x goes from RHS to LHS and becomes+x

6x+x = 24.92

7x = 24.92

X = 3.56

100 makes it complicated

You can make it way simpler, then why not do that

[static valve]

ngl for my brain it makes more sense lol

[solemn epoch]

No worries

Need help with 'h' or you are done?

You can close the channel by using the command .close if you are done

[static valve]

done had some worded questions but flew threw them now that i understand

just 2 more questions ima seen if i need help if not ill close

[solemn epoch]

Aight

[static valve]

ok im all good

was something easy

[solemn epoch]

Nicee

[static valve]

ill be back on the weekend lol

ty

[solemn epoch]

[static valve]

.close

[obtuse pebble]

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[obtuse pebble]

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[hasty viper]

alright so

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[hasty viper]

which one are u at

[icy timber]

2

[hasty viper]

ok send work

what have you done so fa

far

[icy timber]

1*

im on 1

[hasty viper]

ok

okok

[urban patrol]

do you know chain rule?

[hasty viper]

do you know what g(f(x)) is

[icy timber]

yes composition

of function

[hasty viper]

ok so you start with that

[icy timber]

square root of x/4

do you compse g into f

[hasty viper]

sqrt(x)/4 or sqrt(x/4)

just making sure what ur trying to say

which 1

[icy timber]

yes

second one

[hasty viper]

its the first one

[icy timber]

oh okay

[hasty viper]

since u replace x with the function

[icy timber]

ohhh

[hasty viper]

ye

now you do that again since now u have g(f(x)) and u need h(x) of that

[icy timber]

so plug h into that

[hasty viper]

plug that into h

[icy timber]

ok

[hasty viper]

the outside function is waht ur plugging the inside into

[icy timber]

oh

[hasty viper]

ye

[icy timber]

4(sqrt(x)/4 +16

so answer is

[hasty viper]

simplify that

[icy timber]

first one

sqrt x +16

since the 4s cancel right

[hasty viper]

yea

[icy timber]

ok thank you

if i have more questions should i make a new help thing

or should i just leave this one open

[hasty viper]

either or but I suggest making a new one

this one will be like all the way down

[obtuse pebble]

@icy timber Has your question been resolved?

[obtuse pebble]

@icy timber Has your question been resolved?

[obtuse pebble]

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[modern palm]

heres my work

what do i do next?

could a = 0i+ 2j + 1k
and b = -1i + 0j + 0k?

that would satisfy my equations(i believe)

but is that correct?

[polar fossil]

a + b would be <-1, 2, 1> then which isn't v

[modern palm]

ah you are correct

[modern palm]

any tips?

am i correct in saying that $b_i = -1$

[warm shale]

hamr

[polar fossil]

yeah that's forced

since a isn't going to help you get any i

[modern palm]

yes

[polar fossil]

i think if you just think about like

hmm

for a and w to be parallel

then aj and ak need to be in a 2:1 ratio

that's all

you don't need to worry about the square root of 3 stuff

[modern palm]

i see

[polar fossil]

your bj • 2 + bk = 0 is good

those two eqns should be enough to get you the solution

[modern palm]

ok thank you

[polar fossil]

oh yeah along with the ones you wrote down earlier

[modern palm]

.close

[obtuse pebble]

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[old ermine]

does limit exist for c

I’m so confused on this topic

it is approaching but there’s a whole

[agile oriole]

Yes, it does

[fathom flicker]

here's something

limit existing sometimes has nothing to do with continuity

[agile oriole]

the limit doesn’t care about what the value is at the point, or even if it’s defined at that point

[fathom flicker]

a function you can say is continuous, if at every point, the limit approaching that point is the same as the function evaluated at that point. At x=-8 this is not the case, but that just means it is not continuous, not that the limit DNE

[old ermine]

thankyou so much

what about f(-8)

[fathom flicker]

is the function defined at x=-8

[old ermine]

no

there’s a whole

[fathom flicker]

look above the hole

there is a colored in dot

meaing..?

[old ermine]

it exists okkkkkk

so like is there any part where it doesn’t exist

[fathom flicker]

where what doesn't exist

[old ermine]

I’m so sorry but like any part of the question

where limit doesn’t exist

[fathom flicker]

surely yes

in order for the limit to exist at a point

you must approach the same value from the left as from the right

is there any points you can see where this does not happen on your graph

[old ermine]

part d

[fathom flicker]

part d isn't a point

[old ermine]

x=-2

[fathom flicker]

sure that works

the limit does not exist at x=-2

( the two-sided limit )

[old ermine]

for part d the limit exists

???

[fathom flicker]

idk

you tell me

[old ermine]

yes it does

[fathom flicker]

go back and reread what I said here

[old ermine]

but it’s approaching isn’t it

like the black point

from the right

so like if the graph has a breaking point

it means the limit doesn’t exist

[fathom flicker]

what about from the left

[old ermine]

from left it does exist

[fathom flicker]

what exists

I want you to start talking in full sentences please, it might clear up some of your confusion here

you are wondering if the limit as x approaches 6 exists. I ask of you, is the limit approaching x=6 from the left, the same as the limit approaching x=6 from the right?

[old ermine]

no it’ not the same

like the limit from the right isn’t the same as the limit approaching from left

I’m so sorry for that

my problem is that for lim(x-> 6)

there is a limit approaching from right

but

is this gap considered as approaching or not

because there is a gap but the whole on 6 is being covered

[fathom flicker]

yeah that gap means it isn't continuous

right

it is a jump discontinuity

also though the limit doesn't exist because coming from the left we approach the empty hole

[polar fossil]

not even layla would argue with that one

[fathom flicker]

and coming from the right we approach the filled in hole

[old ermine]

ok thankyou so much

.close

[obtuse pebble]

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[lost hawk]

Hello! can i have help about my mathematics please?

[royal basin]

http://dontasktoask.com/

[lost hawk]

hello! my problem is about GCF and factoring

i cant keep up

[royal basin]

the way things work here is that you begin by posting your problem

[lost hawk]

how sorry

[royal basin]

do you have a problem/question/exercise that you want us to look at?

[lost hawk]

yes

[royal basin]

ok then post it here...

[lost hawk]

about the factoring one

[royal basin]

... this is way too low-resolution sorry

i cannot read this at all

[lost hawk]

oh

[agile oriole]

can you copy it or take a picture from closer to it

[lost hawk]

its 6x + 10

[timid silo]

do you have to calculate x?

[lost hawk]

about gcf / factoring

[agile oriole]

What is the full original question

[lost hawk]

my teacher just showed it only numbers

sorry the original picture is low resolution

my question is now how to get the gcf of the following:

.close

[obtuse pebble]

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[lost hawk]

.reopen

[obtuse pebble]

✅

[dire plinth]

between 6 and 10 just take the greatest common factor

if you don't know how to do that just list all the factors of 6 and 10

and find the biggest one

[lost hawk]

factoring

it really struggles me

[lost hawk]

2

[dire plinth]

then you can take it out

6x + 10

you can factor out the 2

[lost hawk]

like that

what to do with that

gcf polynomial

<@&286206848099549185>

.close

[obtuse pebble]

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[vast magnet]

i have the complex exponential but i dont know what do to to get to the correct notation

[vast magnet]

currently i have -1/32(sin(7theta) +21sin(3theta)+35sin(theta))

[kind hawk]

think about e^(7it) and (e^(it))^7

you are missing a sin(5 theta)

and the coefficients also aren't correct

[vast magnet]

ok i redid it i just dont know how to write into the sum and how to get rid of the even values of n

[agile oriole]

what did you get after you redid it

[vast magnet]

well the first fraction was just wrong

2^7 is 128 not 64 my bad

and i just put in the 7sin(5theta)

[obtuse pebble]

@vast magnet Has your question been resolved?

[obtuse pebble]

@vast magnet Has your question been resolved?

[obtuse pebble]

@vast magnet Has your question been resolved?

[agile oriole]

What did you get in the end

[vast magnet]

idk i don’t know how to make the summing part remove the sin of even thetas

just gave up

[agile oriole]

you can just manually define a_n

[vast magnet]

oh well

.close

[obtuse pebble]

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[river pivot]

how can i write:

[river pivot]

$\frac{\sqrt{2}}{32}$

[warm shale]

marty

[river pivot]

with a base of 2, with some exponent b

how can i think what b will have to be

i understand if b is -5, we get 1/32

i thought maybe -4sqrt(2), but that didn't work the way i wanted

[royal basin]

$\frac{\sqrt{2}}{32} \neq -4 \sqrt{2}$

[warm shale]

Ann

[royal basin]

you want to find $b$ such that $\frac{\sqrt{2}}{32} = 2^b$, yes?

[warm shale]

Ann

[river pivot]

yep

[royal basin]

@river pivot

oops

sorry didnt see that

okay so

how about just sqrt(2) itself?

what's sqrt(2) as a power of two?

[river pivot]

i don't know how that is represented as a fraction

[royal basin]

that... is not what i asked, at all.

[river pivot]

well, if you're asking sqrt(2) to the power of 2, then that is 2

[royal basin]

no.

i am asking $\sqrt{2} = 2^?$

[river pivot]

okay, so 2 to the power of sqrt(2)

[warm shale]

Ann

[river pivot]

idk what that is

1/2?

[royal basin]

do you know the relationship between roots and fractional exponents?

[river pivot]

i think i don't then. I know sqrt(x) = x^1/2

[royal basin]

yes that's enough for this.

sqrt(x) = x^(1/2)

[river pivot]

and 3sqrt(x^2) = x^2/3

[royal basin]

... do not write the cube root as 3sqrt.

nobody will understand you.

everybody will read that as if you meant 3 times the square root.

[river pivot]

just wanted a quick way to express that without having an alternative good way

[royal basin]

generally, $\sqrt[n]{x} = x^{1/n}$.

[warm shale]

Ann

[royal basin]

anyway

$\frac{\sqrt{2}}{32} = \frac{2^{1/2}}{2^5} = 2^?$

[warm shale]

Ann

[river pivot]

negative 5 gives 1/2^5

so i assume we have to put something under 5 ?

[royal basin]

no you are overthinking again...

[royal basin]

do you know exponent laws?

[solemn epoch]

Use the Quotient Rule

[royal basin]

^ here's a reminder of exponent laws

[obtuse pebble]

Channel closed

The owner is missing!

[river pivot]

what

[solemn epoch]

ehh

[river pivot]

my response to this not ur answers

[solemn epoch]

Same

[obtuse pebble]

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[somber quail]

Hi
I have an issue with a 3D coordinate/ 2D projection.
Basically I have a line for which I have the 2D coordinates on screen and also know the real length between them that is 7 cm. I know there is an extension of that line by 11.7 cm by a third point which is not visible on screen. All 3 points are collinear in real life. I know the roll and pitch of the stick on which all three points lie. I need to correct the 2d projection pitch to make it completely flat out so i can make 2d assumptions of the entire stick. Basically calculating depth and other stuff by using pixel distances.

[somber quail]

So this is an abstract image of the points known and the marker point x’, y’

the real life distance between x1,y1 and x2,y2 is 7 cm

the same for x2,y2 and x',y' is 11.7 cm

The line is basically a pointer which can have any roll or pitch

I cannot directly keep the pointer there as I don't know the third coordinate. I need to be able to find that third point and then also correct the pitch so the stick is facing the camera in proper 2d plane.

All help is much appreciated.

<@&286206848099549185>

[timid silo]

What the hell the question book want from us?

I mean can you please share the abstract question?

[somber quail]

its not an abstract question, it is an issue that I am dealing with in another system that i need to engineer and it requires this insane 3d coordinate play. and i needed some help.

[timid silo]

bloke I mean when I said "abstract" it means what is the original question.

[somber quail]

Well this is the original question. I do not have a question book that has a question that I am solving. It is a real life system which needs me to do math. And the question I raised is the issue I am having. I am sorry for the confusion. 🥲

[timid silo]

Hey I am getting very sleepy so I think to accompish this we should work together dm me.

[somber quail]

Sure.

Okay SUMER is AFK it seems.
Anyone else can help me here?

<@&286206848099549185>

[obtuse pebble]

@somber quail Has your question been resolved?

[obtuse pebble]

@somber quail Has your question been resolved?

[obtuse pebble]

@somber quail Has your question been resolved?

[mint hull]

mid point

[obtuse pebble]

@somber quail Has your question been resolved?

[somber quail]

Mid point as in?

@mint hull

[dim root]

@somber quail try to structure your question a bit better so we can understand it. the length of a curve is this

[obtuse pebble]

@somber quail Has your question been resolved?

[obtuse pebble]

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[silk patrol]

Is anyone familiar with proofs?

[rancid yacht]

send the problem

[silk patrol]

Using this definition

It's legit not possible correct?

Am I just dumb?

[rancid yacht]

i'm not familiar with number theory proofs but you could just calculate the mod 19 of powers of 2 until it equals 18

and i did get a solution

[silk patrol]

We have never learned mod

😦

[rancid yacht]

basically the remainder when you divide it by some number

it would go 2, 4, 8, 16, 13 (since it is 16*2-19), etc

[silk patrol]

Can you maybe elaborate xD

I'm still lost

[rancid yacht]

basically what i was saying is you can calculate the remainder of the numbers when you divide them by 19

[silk patrol]

But a remainder means that there is a fraction?

Or sorry like

You have a whole number multiplied by a fraction

[rancid yacht]

well like 16 mod 5 is the remainder when you divide 16 by 5, which is 1

and 32 mod 7 would be 4

etc

[silk patrol]

16/5 = 3 * 1/5 correct? So is mod is the one?

[rancid yacht]

yes

[silk patrol]

But going back to the question, you would need to use mod and there is no other possible way?

To get a solution/integer of k

[rancid yacht]

well as i said i'm not too experienced with number theory

but the solution comes pretty quick

[silk patrol]

Wait a second...

I got the answer

I'm so dissapointed in myself

Lmao sorry to waste your time and appreciate it

[rancid yacht]

you didnt waste any time lol

youre good

and np

[silk patrol]

.close

[obtuse pebble]

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[vagrant obsidian]

Hi

[visual gulch]

Hey, got any math question?
(If not, type .close )

[terse quartz]

A business firm is owned by five partners, P1, P2, P3, P4, and P5. When making
group decisions, each partner has one vote and the majority rules, except that P1 has
veto power and therefore must vote yes for the motion to pass.
Find the Banzhaf power distribution of the weighted voting system.

can someone pleaseee help me

[vagrant obsidian]

.close

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[round sable]

Pls how do I do this

[round sable]

I know generally what I’m supposed to do but idk if I’m right

please

<@&286206848099549185>

⁉️

<@&286206848099549185>

[obtuse pebble]

@round sable Has your question been resolved?

[obtuse pebble]

@round sable Has your question been resolved?

[obtuse pebble]

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[honest flicker]

I forgor to work on my math hw plz help

[honest flicker]

Idk what counts as a quadratic function

[violet sentinel]

quadratic function is one of hte form ax^2 + bx + c, where x is your independent variable

in your was your quadratic equation is n^2 + n

[obtuse pebble]

@honest flicker Has your question been resolved?

[honest flicker]

Umm idk wat ur saying

[obtuse pebble]

@honest flicker Has your question been resolved?

[obtuse pebble]

@honest flicker Has your question been resolved?

[stiff cape]

my question

[obtuse pebble]

Please read #❓how-to-get-help

[obtuse pebble]

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[obtuse pebble]

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[slender sundial]

can someone help make a diagram?

[slender sundial]

i tried but im finding difficulty since they gave the resultant vecctor and not the individual vectors

[obtuse pebble]

@slender sundial Has your question been resolved?

[obtuse pebble]

@slender sundial Has your question been resolved?

[wide rose]

Isn't this a sum of vectors?

[somber marten]

I think so

[obtuse pebble]

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[fair canyon]

guys I know how to solve this question, can you make sure this is the best way, is there an alternative way to solve it quicker

$\textbf{calculators not allowed in exams}$
So if they have given a problem like this for practice, I am pretty sure there is a way to easily solve it without wasting 15 mins on this

I was gonna solve the cubic equation and do $Tr (M) = \lambda_1 + \lambda_2 + \lambda_3$ and then |A| = product of 3 eigenvalues

I am having problems with solving this manually as the roots of this cubic equation are not whole numbers
@dapper prism

[fair canyon]

Vieta's formulas give the sum and product of roots given the coefficients

.close

[obtuse pebble]

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[fair canyon]

.reopen

[obtuse pebble]

✅

[warm shale]

Schrodinger

[fair canyon]

.close

[obtuse pebble]

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[lethal blaze]

Is the answer in the number 3 is there is no real solution??

[cinder lintel]

which part

[lethal blaze]

It is system of non linear equations

[cinder lintel]

i mean which q

[lethal blaze]

Number 3

[cinder lintel]

there is

a solution

[lethal blaze]

Using quadratic formula??

[cinder lintel]

yes

replace either x or y

wait i mis judged signs

[lethal blaze]

With x=-y-3 or y=-x-3??

[cinder lintel]

x=-y-2

[wide rose]

heyyo

did you try factorizing first equation?

[lethal blaze]

what do you mean?

[cinder lintel]

oh lol

imsginary roots

[lethal blaze]

so no real solution right?

[cinder lintel]

ye

[wide rose]

(3x+6y)(x+y-1)-11y=12

[lethal blaze]

in number 2 there is no real solution too right?

[wide rose]

6-8y=12

why no real solution?

(i just replaced x+y's with 2)

oh wait

[cinder lintel]

from there u get y 0

x doesnt correspond

[rich silo]

There are, double check!

[cinder lintel]

in number 2 there is solution

[lethal blaze]

wait there is??

[wide rose]

i got 6-8y=12 in 3

[lethal blaze]

my answer in number 2 is y=1-i?

[lethal blaze]

how?

[wide rose]

simplified this: (3x+6y)(x+y-1)-11y=12

x+y=2

3(x+2y)-11y=12

3(2+y)-11y=12

[lethal blaze]

is this for no.3??

[cinder lintel]

huh

[wide rose]

no 3

[cinder lintel]

ah

x+y is -2

[lethal blaze]

wait let me send my solution

[cinder lintel]

not 2

[wide rose]

sorry

1 moment

3(x+2y)(-3)-11y=12

-9(-2+y)-11y=12

18-20y=12?

[cinder lintel]

y=3/10?

x=-23/10

[wide rose]

yeah

[cinder lintel]

does that even satisfy

[wide rose]

should

[cinder lintel]

it does not

[lethal blaze]

yes this is correct

[cinder lintel]

theuh

i misplaced x and y

smh

[lethal blaze]

I already do the checking

so 3 has a solution

for number 2

wait im going to send my solution

This is what I got so far in number 2

[cinder lintel]

its =9y^2

you write as -9

[lethal blaze]

Ohh yeah I forget the squre

There

Then what do I do next

[agile oriole]

Solve for y

You can start by factoring y out

There are 2 real solutions

[lethal blaze]

-9(y+2)??

[agile oriole]

There’s also a factor of y

[agile oriole]

I meant it should be -9y(y-2)

[lethal blaze]

ahh yeah

then my y=2?

[agile oriole]

or

[lethal blaze]

hmm

y=9?

no?

[agile oriole]

you can always substitute back into the equation to check your answer

[obtuse pebble]

@lethal blaze Has your question been resolved?

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[woeful tree]

Hello guys, can anyone help me with this question. I've plotted the curves but I'm finding it hard to find the volume when it's rotated w.r.t x=2.

[obtuse pebble]

@woeful tree Has your question been resolved?

[shrewd wasp]

Since there are parts on the right and left part of the y axis, you can use two inner radius separately for the washer method. But I prefer you to use the shell method.

[obtuse pebble]

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[woeful tree]

.reopen

[obtuse pebble]

✅

[woeful tree]

Can you elaborate on how to choose two inner radii for washer method?

[shrewd wasp]

If you choose any x from the right-hand side division, the radius becomes 2-x. If you choose from the left, 2+x.

[woeful tree]

okay thanks

.close

[obtuse pebble]

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[shut lagoon]

The topology contains all finite subsets, N and {}. Then they give you an example of a union of finite subsets (which are open) give an infinite subset (which is not open). This cannot be because we require that a valid topology produce an open set from a union of arbitrarily many open sets.

In T or not in T

Usually in topology you would call elements of the topology open

It's a loose interpretation of open

Yes

The difference with the definition after this is you consider ANY subset

So infinite as well

[obtuse pebble]

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[cloud sparrow]

i need help in proving this

[cloud sparrow]

not sure where to start

[obtuse pebble]

@cloud sparrow Has your question been resolved?

[ornate oar]

Well, first try moving that 2 into the summation by changing the bounds

[cloud sparrow]

sorry, got no clue how to do that ;-;

[ornate oar]

What is $\binom{p}{0}$ and $\binom{p}{p}$

[warm shale]

DAILI

[cloud sparrow]

both equal one

[ornate oar]

So if you take $r=1$ and

[warm shale]

DAILI

[ornate oar]

r=p

what does the summand evaluate to

can you see then that they basically ask you to prove $\binom{2p}{p} = \sum_{r=0}^p \binom{p}{r}\binom{p}{p-r}$

[warm shale]

DAILI

[cloud sparrow]

ahh i get it!

[ornate oar]

This is a pretty common binomial identity

[cloud sparrow]

yeha

[ornate oar]

There's an easy combinatorial proof

[cloud sparrow]

think i get it

[ornate oar]

Also a systematic one using generating functions if you know that

[cloud sparrow]

$\binom{p}{r} = \binom{p}{p-r}$

[ornate oar]

See here, I guess https://math.stackexchange.com/questions/148583/combinatorial-proof-of-summation-of-sum-limits-k-0n-n-choose-k2-2n

[cloud sparrow]

we use this?

[ornate oar]

You can use it

There are a lot of proofs

[warm shale]

SirGareth

[cloud sparrow]

ayye got it!

thanks man

.close

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[inland thicket]

guys

number 2 is true

but is number 1 true? :3

no number 2 not true?

[sage geode]

Not necessarily

The first equation is not true in the general case but the second equation is

[inland thicket]

oh

but how first?

2x^1/2?

[sage geode]

Yes, it should be $2x^{\frac12}$

[warm shale]

A Lonely Bean

[inland thicket]

🇭🇷🇭🇷🇭🇷🇭🇷

[sage geode]

Otherwise I could pick a counterexample like x = 2

[inland thicket]

Math War

https://tenor.com/view/kim-jong-un-kim-jong-un-icbm-gif-25209750

[frank monolith]

⚔️ ⚔️

[thin star]

not this shit again

[inland thicket]

https://tenor.com/view/飛彈-missile-launched-attack-weapon-gif-17042418

Ait 🤙

.close

[obtuse pebble]

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[graceful marten]

Hiaaa

[graceful marten]

How would you draw such a curve

Ik it has x=1 x=-1 x= Imaginary x = imaginary

How do you know where to plot the 1's?

Do you assume since 2 there are 2 imaginary,
(x-1) (x+1) is only counted so u only draw thoes 2 so make it an x^2 graph?

[eternal thistle]

$x^4-1<0 \iff x^4 < 1 \implies |x| < 1$

[warm shale]

tales

[obtuse pebble]

@graceful marten Has your question been resolved?

[wary bobcat]

find the real roots (1 and -1 are roots so you factorise it to find it has no other real roots)

then find turning points (triple turning point at x=0)
hence it has either a u shape or n shape (because the power is even)

then notice the x^4 has a positive coefficient so it must be a u shape (you see this by letting x get very big)

[obtuse pebble]

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[crimson zodiac]

2x/|x-1| < 1

[crimson zodiac]

why is it only x < 1/3

And not also x < -1

[fossil crag]

First, writing it as 2x < |x-1|, distinguish the case when x > 1 and x < 1

[crimson zodiac]

Wdym

[fossil crag]

Ok you can do it in another way as well, but you must have gotten "x < 1/3 OR x < -1"

Not an AND

[crimson zodiac]

Yes i think its or

[timid silo]

\dm
The absolute value function is defined such that [
\abs x = \env{dcases*}{\hp - x &if $x \ge 0$ \ - x &if $x < 0$}
]

[crimson zodiac]

I squared both sides

[fossil crag]

Ah this step is probably very unneeded

2x < max(x-1,1-x)

So 2x < x-1 OR 2x < 1-x

[crimson zodiac]

Yeah so that still gives u

1/3

and -1

[fossil crag]

yep, as an OR

[crimson zodiac]

Yeh

So why is it only 1

How do u decide which one it is

[fossil crag]

We don’t decide, we check the union of those statements

As it so happens, x < -1 IMPLIES that x < 1/3

[crimson zodiac]

Oh ok

Ic

[fossil crag]

So x < -1 is superficial while dealing with 'OR'

It's like if the requirements for going to a party were "(Be there on time) OR (Be there on time AND bring 1000$)"

[crimson zodiac]

That makes sense

[fossil crag]

Well being there on time is enough

[crimson zodiac]

Thanks

[fossil crag]

No problem 👍

[crimson zodiac]

.close

[obtuse pebble]

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[mossy sandal]

I have tried plugging in time in the force formula and then dividing by mass. then after getting the x and y accelerations, i try to integrate them for velocity at time 2 but haven't been successful? I am so confused, please help

[obtuse pebble]

@mossy sandal Has your question been resolved?

[mossy sandal]

<@&286206848099549185>

[mossy sandal]

<@&286206848099549185> May someone please help me, it has been 40 minutes

[timid silo]

this is pretty messy

[mossy sandal]

Yeah sorry

[timid silo]

so tell me what u did

in bried

brief**

[mossy sandal]

I did force over mass to get acceleration

plugged in 2 seconds for t

then I got the acceleration for x and y axis

so i got the magnitude, by sqr rooting and squaring the ax + ay under the sqroot

which got me 90.55 m/s^2

then I assumed that I could just integrate it for velocity, and got 90.55t, and plug in 2 seconds and got around 181 m/s which sounds pretty off and isn't even close to any of the options.

[timid silo]

owww

[mossy sandal]

I must have some weird misconception

[timid silo]

thats not how u gotta do that bro

[mossy sandal]

Please teach me the ways lol, I thought I was doing sum but ig not

[timid silo]

u did fine till finding the acceleration expression

[mossy sandal]

mmm

[timid silo]

but after that u gotta integrate to find velocity

u know that right?

[mossy sandal]

I'm a little confused, sorry.

By acceleration expression you mean the 90.55 i got?

[timid silo]

take it slow

nope

u dont have to put t=2 after finding acceleration expression

[mossy sandal]

I see

So should I integrate the force over mass expression(acceleration) and then plug in t = 2?

[timid silo]

integrate the acceleration expression with respect to t

(time) take upper limit as 2 and lower as 0 and then calculate

simple

[mossy sandal]

I see, I was just doing some weird order then

[timid silo]

its easy after integration

good luck bro

[mossy sandal]

Ok, yeah I'm getting different numbers now

Appreciate the help, thank you so much honestly

[timid silo]

i used to get confused a lot before with these things

just ask for help when u cant proceed

not being able to do is far better than making mistakes

[mossy sandal]

Ok, so I integrated the functions independently using a calculatorand got 60 for the x acceleration, and -20 for the y acceleration

[mossy sandal]

Yeah, I can see that I was messing up halfway through lol

[timid silo]

acceleration?

u got velocity bro

[mossy sandal]

oh yeah yeah nvm

[timid silo]

after integration

[mossy sandal]

so with these velocities, I did the square root over vx^2 + vy^2 and got 63

which is an actual answer choice lets go

Once again, thank you very much. I was getting worried my question would go unnoticed. Thank you

.close

[obtuse pebble]

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[spark stump]

No idea where to start

[timid silo]

use work energy theorem

[thin star]

why not just conservation of energy?

[timid silo]

the eqn will be=> 4mg=1/2mv^2+3mg

solve for v ull get it

[spark stump]

Hm?

[timid silo]

yeah yeah

[spark stump]

If this was an exam how would I know which formula to use?

Whats conservation of energy

:D

[thin star]

Energy before = Energy after

[spark stump]

My exam is in 3hrs and I don't know much

Oh

[timid silo]

u know ke+pe = constant in a conservative system

[timid silo]

wow thats tough

i can help tho

[thin star]

do you know formula for kinetic energy?

[spark stump]

50*0=50v?

Uh ke=1/2mv^2

[thin star]

yes

[timid silo]

god please help this poor soul