precalculus

[pale pond]

thank you!

[carmine elbow]

Does anyone here know how to do half life equations?

[past meadow]

what about them?

[carmine elbow]

I have to find the decay constant k

Hold on

[unique hill]

which question specifically

[carmine elbow]

Question 14

[past meadow]

theyre all the same, just diferent numbers, so it doesnt really matter

[unique hill]

sneaky u help cora since u answered first

[past meadow]

sure

well

we can substitute into the formula first

[carmine elbow]

Like this?

[past meadow]

hmm ive kinda forgotten about this to be honest, but im pretty sure just looking at the formula you can substitute 25 in for a(t) as well, since its time we know at time 5.27, theres half of 50 remaining, right?

nighty could you check if im right if you're still there? im not perfectly sure if im right

[unique hill]

all is substitution

i havent tried it yet but i'll look at it

[past meadow]

it sounds right, from there you can divide both sides by 25, then take ln of both sides, and solve from there

[carmine elbow]

?

[past meadow]

are you familiar with logarithms?

[carmine elbow]

I am

I’m a visual learner. That’s why it looks confusing to me

[viscid thistle]

theres a 3blue1brown video explaining logarithms

its pretty cool

for visual people like you

[carmine elbow]

Can you show me?

[viscid thistle]

https://www.youtube.com/watch?v=sULa9Lc4pck

[past meadow]

o 3blue1brown is great

[tardy ridge]

Yo I don't understand how to not brute force to solve this

if you are on a cube

6 vertices

you start at a vertex

after moving to another vertex 8 times, what is the probability that you get back to the original vertex

[carmine elbow]

@viscid thistle is this specific to my problem?

[unique hill]

@past meadow i havent done these in a while, but did u get 25=50e^(5.27)k

cuz half life then mass is halved

[past meadow]

yeah i did nighty

[tardy ridge]

whoops looks like this is occupied

[unique hill]

ok noice

[viscid thistle]

@carmine elbow no but its interesting

[carmine elbow]

You’re right. Since it’s half, it should be 25 = 50e^k(5.27) @unique hill

[past meadow]

yeah thats right, do you know how to work from there cora?

start by dividing by 25

[carmine elbow]

I think so. I’ll work it out and show you how I do it

I want to make sure I know what I’m doing

This is what I have so far

[unique hill]

holup

[past meadow]

where did your e go

[unique hill]

the 5.27 should be outside

wait

[past meadow]

in the second to secoond last line you take ln of both sides, and your e just dissapears

[unique hill]

yea

nono that's correct

ln e = 1

[carmine elbow]

the e is naturla log so I just took the natural log of both sides

[past meadow]

ok so you have ln(a)=ln(e^b)

that means ln(a)=b

but you have ln(a)=ln(b)

[unique hill]

your last step is incorrect

[carmine elbow]

lne = 1

[unique hill]

from second last

[carmine elbow]

right?

[unique hill]

yes

@past meadow that's not what he's having trouble with btw

cora, your 5.27k should be outside of the ln

[past meadow]

yes that was what i was trying to say

[viscid thistle]

can somebody help with vectors

asap

[unique hill]

channel's taken btw

[past meadow]

kordal try a #❓how-to-get-help channel

[unique hill]

you shouldnt have an extra k in the denominator

ok forget the three last steps

from 1/2 = e^(5.27k), you can bring the 5.27k to the front of the ln using log laws

[carmine elbow]

There’s an extra k because I’m bringing that whole section over to the left and dividing it by the ln(1/2)

[unique hill]

from 1/2 = e^(5.27k), you can bring the 5.27k to the front of the ln using log laws
@carmine elbow

ignore the last 3 steps you have

[past meadow]

then you have ln(1/2)=5.27k

[carmine elbow]

Ok

[past meadow]

do you understand how we got that?

[unique hill]

and the rest is same concept rly

[carmine elbow]

Yes. And I saw the extra k that you guys were talking about

[unique hill]

ok so you know how to get the k out right?

[carmine elbow]

The k is erased

[unique hill]

yup that's correct

then calculator does the work

[carmine elbow]

Ok!

I got -0.1315

[past meadow]

thats right

[carmine elbow]

Thank you for your help!

[south forum]

is the answer to both of these no?

i'm trying to study and the bottom one with the graph is confusing me

doesn't it have to pass the horrizontal line test?

[unique hill]

are u asking for both of these questions

[south forum]

yes

7 and 8

i think the answer to both is No

because i know for sure 7 doesnt pass the horizntal line test

[unique hill]

nah only the first one is not

[past meadow]

7 is definitely no

[south forum]

i cant tell by looking at the graph if 8 does or not

[past meadow]

8 is probably yes

[south forum]

alright

[unique hill]

8 is yes

correct

[harsh condor]

I have the question:

eep

I solved it tho

Find x, 2ln(x) = sqrt(ln(x))

And so I found x by doing:
(2*ln(x))^2 = ln(x) [squaring both sides]
4*ln(x)^2 = ln(x)
4 = ln(x)/ln(x)^2 [divide both sides by ln(x)^2]
4 = 1/ln(x)
1/4 = ln(x)
e^(1/4) = x [Bring the old e into here]

But I was wondering

[willow bear]

yes you did miss a solution

[pale bison]

you gotta be careful when you divide

[willow bear]

be careful*

[pale bison]

ty

discord font is too small not my fault

[harsh condor]

Ye, so there would be the negative solution too

[unique hill]

pub u can change the font

[pale bison]

why would that be true?

i'm just making lame excuses for my shit english

[unique hill]

ok gotcha

[pale bison]

are you typing an essay bRo

[unique hill]

:\

[harsh condor]

Yes.

Instead squaring both sides, if I decided to, bring e into it immediately.

e^(2ln(x)) = e^(sqrt(ln(x)))
x^2 = e^(sqrt(ln(x)))

What could I do?

Is e^(sqrt(ln(x))), like death or something? can I not break it down?

[pale bison]

yes, it's death

[harsh condor]

rip

[pale bison]

4*ln(x)^2 = ln(x)
minus ln(x) on both side then factor would be the best idea to avoid division

the fact that it looks like a quadratic would help you to identity when stuff like this happens

[harsh condor]

4*ln(x)^2 - ln(x) = ln(x) - ln(x)
4*ln(x)^2 - ln(x) = 0

[pale bison]

keep going

[unique hill]

another essay incoming

[pale bison]

it better be a good one

i mean it's superb that they can use parenthesis unlike many many other users in this discord

[unique hill]

true

[harsh condor]

I've done some sinful things in this essay.

"imagining" ln(x) to be z temporarily to solve the quadraitc
4z^2 - z = 0
8z - 1 = 0 [cheat a bit]
8z = 1
z = 1/8
4(1/8)^2 - (1/8) = -1/16

Vertex = (1/8, -1/16)

4(z - 1/8)^2 = 1/16
(z - 1/8)^2 = 1/64
z = sqrt(1/64) + 1/8 [replacing z with ln(x) again]

ln(x) = sqrt(1/64) + 1/8

[unique hill]

aight take ur time

you dont need the vertex

[harsh condor]

I use vertex to find quadratic, since I never remembered the quadratic formula.

Also a bit quicker imo

[unique hill]

you dont need the vertex

[pale bison]

4z^2 - z = 0
then,
z(4z - 1) = 0

...

[unique hill]

yea

[harsh condor]

Shiet

[unique hill]

it's not a quadratic

[pale bison]

sorry that i said quadratic, that probably mislead you

[harsh condor]

ripparoni

So:

[unique hill]

so you're good now?

[harsh condor]

4*ln(x)^2 - ln(x) = ln(x) - ln(x)
4*ln(x)^2 - ln(x) = 0

[unique hill]

nvm

[harsh condor]

Is that part correct?

[unique hill]

then,
z(4z - 1) = 0

follow from here

yes indeed

[harsh condor]

Idk how it got there tho....

[pale bison]

factor out the z

[harsh condor]

Oh

[pale bison]

ripparoni

[harsh condor]

ripparoni indeed

[unique hill]

you're good now?

[harsh condor]

nope, idk how z(4z - 1) helps?

[unique hill]

= 0

[harsh condor]

I can only go to ln(x)(4ln(x) - 1) = 0

[unique hill]

that equals 0

nono factor out the z

which is already done

[harsh condor]

I can like divide 0/z, but thats illegal

[unique hill]

$z(4z-1)=0$

[harsh condor]

z(4z - 1) = 0

[pale bison]

latex gore

[obsidian monolith]

nighty:

[unique hill]

i swear sqaure brackets worked earlier

i mean curly

[pale bison]

z is ln(x) btw nighty, so i guess it's fine if they decided to change it back

[unique hill]

yes but isnt it faster if u get the values of z and then sub into ln(x)?

[harsh condor]

Idk, what to do after that?
(4z - 1) = 0/z
4z = 0/z + 1
z = 1/4

[unique hill]

nani

[harsh condor]

feels illegal

[unique hill]

it is

what's 0/z

???

[harsh condor]

0/z is 0

[unique hill]

you're cancelling out a value of z

[harsh condor]

Though that is technically the correct answer

[unique hill]

cuz u divided it by z

so therefore u got one less solution

[harsh condor]

Yeah

[unique hill]

dont divide

factor it out

[harsh condor]

z(4z - 1) = 0
But the z is already factored out isnt it?

[unique hill]

z(4z-1)=0, so it's obvious that z=0 or 4z-1=0
i think you're having one of those mind blanks

[pale bison]

i think you meant or

[harsh condor]

OOoh, yeah there is that way of thinking lol.

[unique hill]

ty

corrected @pale bison

[harsh condor]

4z - 1 = 0
4z = 1
z = 1/4

[unique hill]

for one of the sln yes

[harsh condor]

I have compeltely forgotten that z(4z - 1) = 0 is solveable.

I was completely relying on algebra

And not thinking rip

[unique hill]

alrighty so you can continue from here now?

now that z = ln(x)

[harsh condor]

Yeah so:
ln(x) = 1/4
or
ln(x) = 0

And so
x = e^1/4
or
x = e^0

[unique hill]

x=1 for last, yea

[harsh condor]

Yea, damn i feel smooth brained today

[unique hill]

glad u got it

gj

[harsh condor]

Yeah thanks for sticking along

[unique hill]

no problem

[harsh condor]

Helps heeps

[unique hill]

:D

i need to go so cya

[harsh condor]

Aight cya!

[random steppe]

Hi, I'm having some trouble determining which function is leading and lagging for:

y1 = 5cos(2x + pi/3) and y2 = 4sin(2x + pi/3)

Is it true that if the phase difference (a2 - a1) is negative, then then y2 leads y1 ??

[harsh condor]

Idk, physics so I can't be sure but from what I know about basic trig:

cos(x) = sin(x + pi/2)

So, cos(2x + pi/3) = sin(2x + pi/3 + pi/2) = sin(2x + 5pi/6)

So looking at:
y1 = 5 * sin(2x + 5pi/6)
y2 = 4 * sin(2x + pi/3)

I'd imagine, y1, to look "slightly" like y2, but behind by pi/2. Since adding more to the x, makes it go more to the left.

To find the exact offset,
Given sin(ax + b), the offset from 0 will be b/a.

[random steppe]

hi, thanks for replying! My working is fairly similar as I also converted to cosine to sine

I found the offset to be pi/4 radians

[harsh condor]

That sounds about right

[random steppe]

But i wasn't sure whether if the phase difference (a2 - a1) is negative, then then y2 leads y1

as in, here the phase difference is: pi/3 - 5pi/6 = -pi/2

so do you know if having a negative (or positive) phase difference implies y2 is leading y1 (or vice versa for positive)?

[harsh condor]

Well, I would assume that just knowing the offset wouldn't tell you which wave is shifted more than the other.

Since there is also the period of the wave.

One could be faster than the other, though in this case, both waves have the same period

But what I can tell from this is that, y1, probably is slightly behind y2.

I would assume (a2 - a1) would tell you which one leads the other if, both have the same period.

[random steppe]

Yeah i agree they would definitely have to have the same period to be comparable

but also in the same sense, we would just divide by k: (a2-a1)/k so that gives the 'offset'

[harsh condor]

Whats K?

[random steppe]

k is the coefficient of x

[harsh condor]

The period?

Well not the period

But 2pi/k = period?

[random steppe]

wait no so (a2-a1)/k gives us the translation along the x axis to get from y1 to y2 (or vice versa)

[harsh condor]

I dont think (a2 - a1)/k would tell you the difference of offsets between the two. Since there is the other x's coefficient too.

[random steppe]

(a2 - a1)/k would only give the offset of the 2 functions when k (the coefficient of x) is the same for both functions

that was actually the formula we were given

[harsh condor]

Not entirely sure, but I assume there is some proper formula for the phase difference.

[random steppe]

phase difference = a2 - a1, where a is in (kx + a)

i'm just confused over how we would determine which function is leading or lagging analytically from the functions only

(like without sketching a graph)

thank you for your help so far by the way

[harsh condor]

From the functions only, I would convert both to sine waves.
And check whichever one has the larger offset.

Which ever one has the larger coefficient is probably lagging behind.

So e.g
y1 = sin(ax + b) and
y2 = sin(ax + c)

(b - c) > 0, it probably means y1 is behind y2
(b - c) < 0, it probably means y1 is ahead y2

Of course we have this is only true when both have the same period.

So yeah I think your formula checks out in these general situations. Tho idk the specifics about phase shifts so there might be specifics I'm missing.

[random steppe]

That's okay, i understand it better now

Thank you for your help Lang!!

😃

[pale pond]

i know i asked the same question yesterday, but i just got back to doing it and im still stumped

can anyone help with 7?

[lost mesa]

@pale pond express tan and sec in terms of sin and cos

[uncut mulch]

combine fractions

apply trig identity relating sec and tan

factorise, simplify

simplify further

[quartz garnet]

there are different approaches

i wouldve multiplied (sec(x) - 1)/(sec(x) - 1) to the first fraction

[lost mesa]

that works too

then simplify the bottom and whatnot, right?

[quartz garnet]

uhuh and the 1/tan x will cancel out and then sec/tan = csc

[lost mesa]

mhm

[vague minnow]

hey I need some help on how to begin to graph this equation system

[muted granite]

Having difficulties building a polynomial with 2,-2, -1+3i, and p(1)=54. It can only have real coefficients and zeros. The -1+3i is throwing me off. Thoughts?

[patent beacon]

I'm guessing you mean "roots"?
If -1 + 3i is a root of a poly with real coefficients, then -1 - 3i is as well @muted granite

Wait, you can only have real zeros? Then what are 2,-2,-1+3i?

[muted granite]

@patent beacon Yeah only real zeros. I think it wants me to take 2,-2,-1+3i and turn it into a polynomial that equals 54.

[patent beacon]

@muted granite How would you take 1 and make a polynomial out of it?

p(1) = 54 is not "a polynomial that equals 54"

[vague crystal]

@vague minnow oh boy those are fairly ugly

I'm pretty sure you can factor the second fraction

although....

$x^2 + y^2 - 625 = 0 -> x^2 + y^2 = 625$

[obsidian monolith]

bela:

[vague crystal]

that turns into a circle with radius 25

[stuck lark]

$x^2 + y^2 - 625 = 0\implies x^2 + y^2 = 625$

[vague crystal]

not sure if that helps at all though just spitballing obvious relationships

[obsidian monolith]

RokettoJanpu:

[vague crystal]

oh thank you pog

forgot what that operator was

i'm 95% sure you can factor that fraction into something cleaner but not sure

[muted granite]

@patent beacon I am not sure exactly what to do. But this is the question.----Build a polynomial function p(x) of least degree having only real coefficients and zeros: 2,-2, -1+3i, and p(1)=54. ( Give your final answer completely multiplied out in normal polynomial form )

[patent beacon]

The polynomial has real coefficients.
The polynomial has these zeroes: 2,-2, -1+3i
p(1) = 54

Worth mentioning, the polynomial does not have real zeroes. The zero -1 + 3i has a conjugate zero, so -1 - 3i is also a zero

[vague crystal]

hmmm

i'm not really sure how you can eliminate the i

also oop

made an oopsy

OH MY GOD

nevermind absolute moron i am

make mathematical errors, i must

This is actually a very cool and clever problem

I found a polynomial but p(1) = -39 for mine

[hollow oar]

OOOOOOOOOOOO

[viscid thistle]

Write an arithmetic sequence that gives the nth positive x-intercept of the graph of f(x)=cos18x. Leave your answer in terms of π.

how do I do ^^^

[twilit rock]

Wait a bit

[viscid thistle]

@viscid thistle u said /8 for the other equation?

in geo-trig chat

ya

i didnt wanna interupte u

dw bout iot

[orchid pier]

hello i have a homework problem i dont know how to solve. arccos(cos(pi+1))

[viscid thistle]

pi+1

[harsh condor]

use a calculator

jk

[viscid thistle]

arc cos and cos just cancel out

[harsh condor]

It's not

[viscid thistle]

yah.....

[harsh condor]

PI + 1 gets shrunken into the range of cos.

[twilit rock]

Yaaa

[viscid thistle]

no its still within the range?

[twilit rock]

arccos range is 0 to pi

[viscid thistle]

its from 0-2pi

and 2pi can rotate x amount of times

[harsh condor]

Oh yea

[viscid thistle]

pi+1<2pi

[harsh condor]

Should be then.

[orchid pier]

so the answer is pi+1?

[viscid thistle]

yah

[harsh condor]

Wait no!

That's not right.

[viscid thistle]

why not?

[harsh condor]

PI + 1 = 4.14...

[viscid thistle]

they didnt specify for a domain?

[harsh condor]

The actual answer arccos(cos(PI + 1)) = 2.14159...

[orchid pier]

no it says find the exact blue

value

[viscid thistle]

how does that work

@harsh condor

[harsh condor]

cosine basically takes a dip at pi/2

[viscid thistle]

yah

[harsh condor]

So at pi/2 it's gone to zero

[twilit rock]

arccos range is 0 to pi

[viscid thistle]

x=p1/2-x

2pi*

[twilit rock]

No pi

[viscid thistle]

yah

30=330

[twilit rock]

arccos not cos

[viscid thistle]

or pi/6=11pi/6

[harsh condor]

Just use unit circle.

[viscid thistle]

yah

but is there a range for arccos?

thats only on calculators

[harsh condor]

There should be a range for arccos

-1 to 1

[viscid thistle]

if its 0->pi then its pi-1

[twilit rock]

Cool

[viscid thistle]

@orchid pier was there a range for the question

[orchid pier]

no all it says is arccos(cos(pi+1))

find the exact value

[twilit rock]

Ans is pi - 1

[viscid thistle]

k it probs ask for the range then

[twilit rock]

2.142

[harsh condor]

Yeah answer is pi -1

Whats your reasoning.

[viscid thistle]

bc there are 2 answers if its just 0->2pi

but only one for 0->pi

[twilit rock]

arccos has only 1 primary range 0 to pi

[viscid thistle]

oh yah ur right

[orchid pier]

ok so to show my work i should write arccos(cos(pi+1))=arccos(cos(2pi-(pi+1)))=arccos(cos(pi-1))?

im trying to read your picture

@twilit rock

[harsh condor]

That seems about right.

[twilit rock]

Yaaa

Correct

[orchid pier]

ok thanks everyone i appreciate it

[twilit rock]

Cool

[muted granite]

@patent beacon (x-2)(x+2)(x+1-3i)=54 would that be the answer?

[twilit rock]

What's the question???

[muted granite]

@twilit rock Build a polynomial function p(x) of least degree having only real coefficients and zeros: 2,-2, -1+3i, and p(1)=54. ( Give your final answer completely multiplied out in normal polynomial form )

the (1-3i)^2 doesn't work. But I suspect there is a multiplicity in there.

I know it wants the final answer to look like a polynomial that you could do synthetic division on but the -1+3i and =54 is throwing me off.

[patent beacon]

I agree that polynomial has the zeroes you need.

The question asks you to multiply out the polynomial, so why not go ahead and check if it has real coefficients?

[vague crystal]

i tried multiplying that out earlier

i got a polynomial but the p(1) was wrong

[twilit rock]

Here u go

?

Cause immaginry roots occure in pairs

[vague crystal]

doesn't work

[willow bear]

bazinga

please check what you send people

[vague crystal]

i'm probably just really bad at arithmetic

i'll check later

[willow bear]

because it'd be really disappointing if you made a mistake you yourself didn't see

@twilit rock

[twilit rock]

Here u go

Sorry for before

Fine??

[willow bear]

no

[twilit rock]

Why?

[willow bear]

a factor of (-1+3i) doesn't give -1+3i as a root any more than a factor of 10 gives 10 as a root

[twilit rock]

???? Whatttm
?

[vague crystal]

ah fuck i just realized

i'm absolutely braindead

[willow bear]

the polynomial you wrote down has neither -1+3i nor -1-3i as roots, bazinga.

[vague crystal]

i forgot you can just....take a polynomial and multiply it by a factor if you want it to equal a certain value

[willow bear]

(-1+3i) on its own is a constant

it does not give any roots

[vague crystal]

I literally spent so long trying to figure out why I couldn't find a p(1) = 54 polynomial

and i'm just now realizing that you can just multiply the polynomial by a constant

[willow bear]

if you want -1+3i as a root, then you need (x+1-3i) as a factor.

[vague crystal]

[twilit rock]

Ooohh .. ya ya .. I meant the above one sorry LOL

[willow bear]

the blind leading the blind...

[twilit rock]

OOO come on .. it was just a small mistake .. I just missed an x for god's sake ...

People here are not that stupid to not understand what I ment

[willow bear]

very bold assumption right there

[vague crystal]

i mean

i spent half an hour multiplying expressions

[twilit rock]

Its the answer right?

[willow bear]

no

that now doesn't have the right roots

[vague crystal]

I think it does

[willow bear]

no it doesn't.

(x - 1 + 3i) gives 1 - 3i, and not -1 + 3i, as a root

[twilit rock]

Oh ya ya..

Let me do it once more

[willow bear]

you can't pull a "they'll understand what i mean" here

in math you've really got to take care to not say incorrect things

[twilit rock]

Ya sorry

Can you check this one??

@willow bear

@vague crystal

[willow bear]

ok this looks fine now

[twilit rock]

Ya.. sorry for before .. I was eating and solving lol

🤣

[muted granite]

@twilit rock Where did the -18/16 come from? so confused

[twilit rock]

-18/13 bro

You can multiply any constant to a polynomial without changing d gree or roots

To make p(1) = 54

[muted granite]

ooh ok. So the question is asking "create a polynomial with these zeros that equals 54?"

[copper matrix]

help, i need help with vectors

https://gyazo.com/b35b1c9be4ea341d72e89cb3fcfdb1c4

[twilit rock]

Sure

[copper matrix]

khan academy is making me go insane

Magnitude is root of X squared + y squared
Direction is inverse tangent of y divided x

[twilit rock]

@muted granite yes

[muted granite]

thank you.

[copper matrix]

but i cant get it

[twilit rock]

Basically p vector + wind vector = a vector

Then do

[copper matrix]

how do i find it tho

[twilit rock]

Let me show you bro

[copper matrix]

to find both X and Y for A should be easy since all i have to do is [5COS(30), 5SIN(30)]

but idk how to find x and y for P

it only gives me 6 m/s

[twilit rock]

Is it 3m/sec

First one?

?

And 56 degree??

[copper matrix]

uh

@twilit rock so by 56 degree ur sure that its in the first quadrant?

doesnt https://gyazo.com/d33a4b2f6e00ccc27fc3da8125f4a0c9

mean its in the fourth quadrant?

also can u show me what u did to get the first answer o3o, (thanks for helping me btw)

[twilit rock]

It's in 4th

See if you can understand

[copper matrix]

so if its in 4th

[twilit rock]

Yaa

[copper matrix]

wouldnt it be 360 - 56

[twilit rock]

360-56 from 1st quadrant answer then

I gave the absolute angle

360-56 anticlock

[copper matrix]

304

kk ill tyr

[twilit rock]

Yaa

Correct?

[copper matrix]

ill try it

ok so i did

[twilit rock]

And?

[copper matrix]

5cos30) = 4.33
5sin(30) = 2.5
root 4.33^2 + 2.5^2 in my calculator

and i didnt get 3

4.99

for the wind spped

[twilit rock]

What's the given answer??

[copper matrix]

i havent answered it yet

[twilit rock]

Then answer 3 and check

It's correct

Go on

[copper matrix]

if i get this wrong i gotta do the whole 10 question test again but heres go nothing

[twilit rock]

🤣🤣

[copper matrix]

ph

[twilit rock]

Practice is good u know

[copper matrix]

it was third quaddrant?

ive done this test like 17 times my man

cause i cant get 100%

[twilit rock]

🤣🤣

It's correct check 3

I double checked it

[copper matrix]

https://gyazo.com/88d654918e1a6dd52b20cd5d06d05905

ms was correct

secpnd wanst cause im stupid

[twilit rock]

🤣🤣🤣

Have patience my man

[copper matrix]

sigh this is not how life is supposed to work

im asian im supposed to have an A in everything, but i got a B ._.

sigh asian life hard

[twilit rock]

I am asian too bro

[copper matrix]

u got the good genes

lucky o3o

[twilit rock]

Bro take a break chill a little ..

U will be fine

[copper matrix]

my mom's belt says otherwise

[unique hill]

relatable

[copper matrix]

https://gyazo.com/87418c18fbadbb15cff2fc546c32311d

my answers are right, right?

magnitude is -1,1
direction is 45
https://gyazo.com/c54e5cb32d930d5f86f5d434b8024f6f

since its -x,y its on the second quadrant
180 - 45 = 135 degrees

what was the formula to change degrees to radians? @twilit rock

[twilit rock]

180 degree = pi radian

So x degree is pi / 180 times x radian

[copper matrix]

so

(pi / 180) x (135 x 2pi)

?

sigh im so dumb

[twilit rock]

pi/180 * 135

135 degree to radian

Cool?

[copper matrix]

that gives 2.4

[twilit rock]

Nooooo

Yes yes sorry

[copper matrix]

https://gyazo.com/ec88e75e0e71ad616228ad5c2918ee9d

2.4 is over 2

HMM

"In what direction is the ring getting pulled? ** Assume 0 is the rightward direction.** "
does this mean i just should use 45 degrees instead of 135?

since if i change 135 degrees to radians it gives 2.4
if i do 45 degrees to radians it gives 0.8

[twilit rock]

Fine.. anticlock wise is positive

Remember

[copper matrix]

meaning that?

btw im really sorry im bothering you with this, this late at night o3o

[twilit rock]

What's the question??

[copper matrix]

nvm

i was right but im retarded

and i doubted myself

and then i got the question wrong cause i changed it

[twilit rock]

Lol.. it's ok...

Bro sleep now .. you may be tired

I am also feeling sleepy lol

😅😅

[copper matrix]

thanks for your help

you should sleep, ill stay up trying to do tihs by myself

since the deadline is in 1 hour

sorry for being a burden

and thanks for your help

[twilit rock]

Omg deadline shit ..lol

I will help then

[proud gate]

I have just barely gotten into conic sections and wondering.
Is the famous y=x^2 where the 2d plane intersects the cone at a prevose exact angle?

[copper matrix]

https://gyazo.com/1a00bcb47f74fc6f1a72442e17503b7b

@twilit rock thank you so much, I did it by myself, but there was one question which I didnt know how to do, but then I remembered what you showed me with the earlier question and i was able to figure it out

[unique hill]

now you can avoid that belt

[copper matrix]

MHM

YAY

[twilit rock]

@som awsmm broo

You really work man..

[copper matrix]

now i have to wash the dishes

or else i get belt'd

[twilit rock]

🤣🤣🤣🤣🤣🤣🤣🤣🤣

Lol omg ...

I wish I could help u do that lol

[unique hill]

:(

[willow bear]

...

doesn't sound like a healthy situation to me

[heady jewel]

prolly kidding

[copper matrix]

Yeah I’m kidding

I only used to get “disciplined” when I was younger

but it’s normal in asian families

[twilit rock]

I feel you Bro...

I am asian too lol

😅😅

[proud gate]

Is that the vector section on precalculus on khanacademy i see?

[orchid pier]

hello i have a home work problem.

Given cos alpha = sin beta = 1/3 with -pi/2 < alpha < 0 and 0 < beta < pi/2 what is tan(alpha - beta)

[willow bear]

uh huh. and what's giving you trouble here?

[orchid pier]

i am coming up with no solution, is that right? i think if i used the co tangent it would be zero

but i dont know

my end result is -9(sqroot2)/4 divided by zero

[willow bear]

ok yeah you're right

tan(α - β) is undefined

[orchid pier]

if i used the cotangent sum and different formula, would it be zero? or still undefined

[willow bear]

cot(α - β) would be 0

[orchid pier]

the question asks for tan(a-b), would the right answer be undefined or using the cot function to get zero

[willow bear]

the right answer would be undefined because you are not asked for cot

you are asked for tan

[orchid pier]

ok thank you very much ive been doing this problem over and over again

[proud gate]

How does the equation x^2 + y^2 = r^2 plot a circle?

Idk why it does it

Like for a visual representation

[willow bear]

are you familiar with the distance formula

[odd dagger]

a circle is the locus of points that are equally far away from some point

[willow bear]

@proud gate

[proud gate]

Yes

Using Pythagorean theorm right?

[willow bear]

i mean

yes

okay so now

the expression $\sqrt{(x-0)^2 + (y-0)^2}$ represents the distance between which two points?

[obsidian monolith]

Ann:

[proud gate]

(x,y) and the orgin(0,0)?

[willow bear]

yes

sorry for the late reply. i was afk

so

don't you agree that the equation $\sqrt{(x-0)^2 + (y-0)^2} = r$ describes all the points at distance $r$ from the origin?

[obsidian monolith]

Ann:

[proud gate]

Yes

[willow bear]

and so this equation describes the circle of radius r centered at the origin

does that make sense to you

[proud gate]

I believe so

Im doing soome checking

I dont have much experience with both x and y values in same side of a equation

[willow bear]

you're already overthinking it

think about what this equation is actually saying

"the distance from a point (x,y) to the origin is equal to r"

isn't it obvious that a point fits this description if and only if it lies on the circle of radius r centered at the origin?

[proud gate]

I believe so

I guess now im just trying to figure out it graphs it.
But i guess it does the distance formula a few times to find the points right?

With bounds of the radius

[willow bear]

??

you're trying to figure out what?

[proud gate]

Well you know, a function is defined by having 1 output for some input.
And since a circle isnt a function i just figure its graphed with a different method.

Sry idk how to explain xD

[willow bear]

a circle is graphed by drawing a circle

[proud gate]

So a graphing calculator just draws a circle at the h,k point with a radius of r?

[willow bear]

maybe? or maybe it uses some general algorithm that it uses for every other equation

[indigo silo]

a circle of radius r is given by the equation x^2+y^2=r^2

[willow bear]

congrats, you added precisely nothing new to the conversation

[indigo silo]

oh

i didnt read it sorry

wait did i not

Well you know, a function is defined by having 1 output for some input.
And since a circle isnt a function i just figure its graphed with a different method.
@proud gate

but i just said how to graph it with a function

[willow bear]

How does the equation x^2 + y^2 = r^2 plot a circle?
this is the original q

[indigo silo]

oh

i didnt read that far lol

i would have to draw to explain

just use pythagoras

[odd dagger]

the conversation ended lol

ann answered like 3 hours ago

[indigo silo]

oh

[muted granite]

are 1,1,(2+√2),(2-√2) the zeros for x^4-6x^3+11x^2-8x+2? thanks.

[uncut mulch]

are you unsure about the methods used to reach those values?

[muted granite]

im iffy because when i did synthetic division I had to finish with the quadratic formula.

I know for sure the 1,1 are zeros.

[uncut mulch]

what's the issue with using the quad formula?

and/or are you also unsure whether you did synth division correctly

,w factor x^4-6x^3+11x^2-8x+2

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[versed basalt]

How to calculate 2^8/5 and similar things with calculus ?

[uncut mulch]

your zeroes look ok

[muted granite]

Cool thanks. Looks like i took it a step further than the calculator.

[uncut mulch]

,w solve x^4-6x^3+11x^2-8x+2

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[uncut mulch]

@versed basalt be more specific

[versed basalt]

@uncut mulch how to calculate 2^8/5 on pen and paper exam?

[drowsy quail]

You're kidding right?

[versed basalt]

i wish

[drowsy quail]

😅

dw

[uncut mulch]

is that: $\frac{2^8}{5}$ or $2^{8/5}$

[obsidian monolith]

ramonov:

[versed basalt]

ofc the latter lol

[uncut mulch]

you don't 'calculate' it

just leave it like that

[versed basalt]

if you don't believe me

answer is permitted 1816 to 1820

[uncut mulch]

and no calculators are allowed?

[versed basalt]

nope

[visual wasp]

does $(x^2-1)^3 = ((x-1)(x+1))^3=(x-1)^3(x+1)^3?$

[obsidian monolith]

Admin:

[drowsy quail]

Yes it does

[visual wasp]

Thanks

[drowsy quail]

Np!

[viscid thistle]

for this hyperbola equation: https://gyazo.com/1b0c09ce4a961aadde8d1a8844e086fc

would the foci be (0, 2+ sqrt29) and (0, 2- sqrt29)

and the asyntomptes would be y = 5/2x +2 and y= -5/2x +2

[smoky needle]

hello, can someone explain to me how to solve this problem? I don't really get what it's asking..

[patent beacon]

Can you draw the graph of |x| + |y| = 1?

If you were to flip the graph over the x-axis, or the y-axis, would it look the same?

[smoky needle]

Hi yes. It would be symmetric!

Would that be the answer to it?

[viscid thistle]

Hello
I had to prove that two consecutive Fibonacci numbers are coprime

I made that but I don't know if it's right

Can you tell me if it's wrong

[muted granite]

if -1+3i is a give zero for a polynomial, would it be expressed (x+(-1+3i)) or (x-(-1+3i))? What about the conjugate?

[patent beacon]

If a is a zero
Then (x - a) is one of the factors

If the polynomial has real coefficients and has a complex root,
Then the conjugate of that root is also a root

[muted granite]

create a polynomial function p(x) of least degree having only real coefficients and zeros: 2,-2, -1+3i, and p(1)=78.

Can someone double check to see if I did correctly? Thanks. I think I am having troubles translating the complex number into zeros.

[fleet yew]

,w expand (x+1-3i)(x+1+3i)

[obsidian monolith]

Display results online and refine query

Requested by AMD#4434

[fleet yew]

@muted granite ???

you expanded wrong

,w expand (x2-4)(x2+2x+10)

[obsidian monolith]

Display results online and refine query

Requested by AMD#4434

[fleet yew]

your process looks alright

but be careful about those algebraic errors

[obsidian monolith]

PhysicsMonster:

[fleet yew]

Lim dont exist lol

[willow bear]

that's where you're wrong amd

the limit does exist

$\cos(\sqrt{x+1}) - \cos(\sqrt{x}) = -2\sin(\frac{\sqrt{x+1}+\sqrt{x}}{2}) \sin(\frac{\sqrt{x+1}-\sqrt{x}}{2})$

[obsidian monolith]

Ann:

[willow bear]

$\sin(\frac{\sqrt{x+1}-\sqrt{x}}{2}) = \sin(\frac{1}{2(\sqrt{x+1}+\sqrt{x})}) \to \sin(0) = 0$

[obsidian monolith]

Ann:

[willow bear]

so by squeeze theorem the entire thing also goes to 0

@fleet yew @stable pasture

[stable pasture]

ah ic

thanks

[obsidian monolith]

PhysicsMonster:

[willow bear]

divide num and denom by x

[stable pasture]

whoops gotcha

[willow bear]

arcsin(x)/x and arctan(x)/x both approach 1

[stable pasture]

i have a few more, i'll post

[obsidian monolith]

PhysicsMonster:

[stable pasture]

^this one

[willow bear]

yeah ok so what's giving you trouble with this one

[heady jewel]

@viscid thistle can you repost again

i dont get the first line which states

$F_n=k[F_{n+i}]$

[obsidian monolith]

Lionel:

[heady jewel]

rather use induction

or you could just do repeated subtraction

like gcd(a+b,b)=gcd(a,b) and you will reach f_2=1 which would directly give your result

[stable pasture]

yeah ok so what's giving you trouble with this one
idk I cant simplify it

also $$\lim_{x\rightarrow \frac{\pi}{4}}{\left(2x\tan{x}-\frac{\pi}{\cos{x}\right)}$$

[obsidian monolith]

PhysicsMonster:
Compile Error! Click the reaction for details. (You may edit your message)

[stable pasture]

got ^this one by substitution

need help in the first one tho

[willow bear]

first suggestion: turn the limit into a limit at 0, those tend to be easier to analyze

perhaps $\varphi := \theta - \frac\pi4$

[obsidian monolith]

Ann:

[willow bear]

this substitution

[viscid thistle]

Ok so I have to prove that Fn and Fn+1 are comprime so I supposed that they are and I used induction

In France the first line means Fn = k mod Fn+1

Because if they are comprime their integer division has a remainder

[willow bear]

reste
remainder

[viscid thistle]

Yes thanks

[indigo silo]

About the limit above approaching pi/4 just use lhopital bruh

[willow bear]

no

[muted steeple]

Hi guys, can someone help me clear up an ambiguity that i have about $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

So i'm solving $\sqrt[3]{y\sqrt[]{y}}$ and i get $y^{\frac{1}{2}}$ using the rational exponent manipulations...

but using that rule i can't seem to get the same, answer.

with radicals

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

i get something like $\sqrt[3 \cdot 2]{y \cdot y}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[serene heath]

how are you getting that

[muted steeple]

https://cdn.discordapp.com/attachments/363224154469826562/698850048418644058/169117441631322114.png using that rule

i can see clearly what's going on with rational exponents and get correct answer, i just can't see what's the thing is with radicals.

[serene heath]

that should give you 1/2 as well

what is y* sqrt(y)

[muted steeple]

i know

y

[serene heath]

y^3/2 right?

[muted steeple]

oh

yes

[serene heath]

so you have $\sqrt[3]{y^{3/2}}$

[obsidian monolith]

the one n only:

[muted steeple]

oh lmao

such a clutz

thank you

[serene heath]

u see your mistake?

[muted steeple]

yep

[serene heath]

np

[stable pasture]

no LHopital plz

I wouldnt have asked the problem here if I wanted a lhopital solution

[willow bear]

did my substitution suggestion help

[stable pasture]

well that was the first thing that had come into my mind, and I did that... but I can't reduce it to a solvable form...

[novel dirge]

thought I sent the wrong picture

but I didn't

anyway, can someone explain this to me?

my first answer was 14/21

because it says he wants to buy a bday card and the company has 21 cards of which 14 are bday cards

but in the solution, it says that the desirable result are the ones that aren't bday cards

[lost cove]

Yes, it is asking for the probability that one selected card out of the 21 is not a birthday card

[serene heath]

read the question

[willow bear]

you want the probability of a NON-birthday card.

[novel dirge]

oh

damn

I focused on the description

wow, feel like an idiot now

makes sense now that it is asking for non-bday

[astral mountain]

at the start of this video the guy says that 1/x isnt continuous

https://www.youtube.com/watch?v=dHwrzLDmdT8

but from what I've learned it is

[muted granite]

@fleet yew thanks

[frozen needle]

you shouldn't take this channel too seriously x')

[willow bear]

bprp lmfao

[astral mountain]

because 0 isn't in the domain

im I correct?

[willow bear]

imagine thinking bprp would ever be rigorous

[astral mountain]

[rigid beacon]

yea it isn't not continuous at x=0 @astral mountain

[astral mountain]

[frozen needle]

0 isn't even in the domain, the continuity at 0 isn't a subject of discussion to have

what you can say though is that however you extend the function at 0, the extension won't be continuous at 0

[patent beacon]

You can integrate functions that have discontinuities. sin(x)/x comes to mind

But 1/x isn't integrable, asking about an integral over x = 0 makes no sense

[fallen crane]

Can anyone help me figure this out?

I know that it is just trial and error, but I am trying to look for a quicker method than just going through all my plausible options, because I want to keep my work neat

Also, I need to know how to do it using the matrix/box method, so if anyone know hows to use the box method, that would be great 🙂

[patent beacon]

Far from trial and error, though it may seem like it. Try the form:
ax + by

[fallen crane]

Wait, what do you mean?

[patent beacon]

(3x + 4y)(ax + by)

= 3ax² + (4a + 3b)xy + 4by²

[fallen crane]

The questions says that we have to figure out what A and B are though

[patent beacon]

This gives you this system:
3a = A
4a + 3b = -6
4b = B

[fallen crane]

oh, alright

ya makes sense now

thank you so much 🙂

[willow bear]

4 vars 3 eqs

underdetermined ass

[fallen crane]

hol up

ive been trying to do it now with the system

but im still a bit confused

Which numbers would I input and how would I show where I got those numbers from?

[willow bear]

your system is underdetermined. it does not have a unique solution.

[fallen crane]

How would I figure out the equation?

[willow bear]

you can't

[fallen crane]

I asked my teacher, and he said that there is an answer and that it is not undetermined

[willow bear]

there are some things left unstated here

a lot actually

is it implicitly assumed that the coefficients have to be integers or what

[fallen crane]

yes

[willow bear]

even then there will be infinitely many solutions
like (a,b) = (3, -6) or (a,b) = (6, -10) or (a,b) = (9, -14) etc...

,w expand (3x+4y)(3x-6y)

[obsidian monolith]

Display results online and refine query

Requested by Ann#0413

[willow bear]

,w expand (3x+4y)(6x-10y)

[obsidian monolith]

Display results online and refine query

Requested by Ann#0413

[willow bear]

,w expand (3x+4y)(9x-14y)

[obsidian monolith]

Display results online and refine query

Requested by Ann#0413

[willow bear]

so there are at least 3 possible solutions

how's one meant to know which one your teacher had in mind?

[fallen crane]

He said that any is fine, but preferably the one that has the most simple numbers

like this one

[harsh heart]

hmmm

[willow bear]

so...

[harsh heart]

i think i know

[willow bear]

okay so you were given a problem without a unique solution.

that's fine i guess.

the wording made me think a unique sol was expected tho.

[fallen crane]

ya I see how it can be confusing, my teacher has kind of a repuation for poorly wording his questions

anyways, how would I show my work in figuring out this equation?

[willow bear]

dunno

you could write what kaynex wrote

[novel dirge]

I need some help with probability again

watched the Khan academy lessons, but got stuck trying to do more difficult problems

[lost mesa]

ask

[novel dirge]

if we randomly pick 5 numbers from a lottery with 50 numbers, what are the chances of getting numbers 7,13 and 33?

I thinkI can start by getting 5C50

[willow bear]

"5C50"?

that's 0

did you mean 50C5

[novel dirge]

or is it marked with K?

I think it is supposed to be combinatorics

but Idk if I should even do that

[willow bear]

nCk is for combinations
"combinatorics" is the name of the entire field of mathematics

anyway

[novel dirge]

combinations

that

[willow bear]

does the order of numbers matter

[novel dirge]

it is a lottery

[willow bear]

is 1, 2, 3, 4, 5 a different outcome from 5, 4, 3, 2, 1

[novel dirge]

so I believe it does

but it is nowhere written

[willow bear]

[novel dirge]

just my guess because it is a lottery and lotteries usually take order into account

we didn't really go over this with our teacher, just got the basic formulas for combinations, permutations and variations

but now Idk if I even have to use that for this problem

if it was just to get chances of taking those 3 numbers from 50, I could just write 3/50 and get it

but here, I am not sure what to do

[willow bear]

it really depends on how the lottery works

[novel dirge]

lol

@willow bear what I wrote is the entire problem/question

it is nowhere stated how the lottery works

[lost mesa]

could you possibly email your teacher to confirm

because it's kinda vague

[novel dirge]

I can't

she doesn't want to help us

[lost mesa]

[novel dirge]

and her explanations just confuse me even more

I already ranted about her here a few times

but to shorten the story, she is doing everything she can to mess with us

[lost mesa]

well

then you could find the answer for both combination and permutation

[novel dirge]

gives us just some definitions and tells us to do homework

[lost mesa]

that way, there's no room for a wrong answer if you get both of them right

[novel dirge]

now after several homeworks, she told us she is going to grade it

[lost mesa]

talk about shitty teaching damn, sorry man

[novel dirge]

and even worse is that she is our class teacher( Idk if that is the right translation)

but that aside, how do I approach the problem?

even if I use permutation and combination to get 5 out of 50

how do I get the other 3 numbers out of that?

[lost mesa]

i can't really help u rn, im a bit busy so i can only pop in here and there

[novel dirge]

np

[lost mesa]

sorry

[novel dirge]

don't worry

I'll just wait here until somebody answers

honestly, I am kinda ashamed that I have to bother strangers to explain something to me that should be basic and easy to understand

[tardy ridge]

wait where is the question

can you copy the exact question word for word

[novel dirge]

yes

[tardy ridge]

whoops sorry

[novel dirge]

What is the probability that by drawing 5 numbers on a lottery which has 50 numbers in total, we will get numbers 7,13 and 33?

[tardy ridge]

ok assuming by 50 numbers, she or he means from numbers 1-50

so lets say you have drawed your desired numbers, 7 13 and 33

there are 47 choose 2 ways to draw the other two random rumbers

and there are 50 choose 5 ways to draw in total

so you would do 47 choose 2 / 50 choose 5

[novel dirge]

I lost you at the second line

sorry

[tardy ridge]

So this is actually hypogeometric distribution, you can imagine it as 3 red balls and 47 blue balls

do you know hypogeometric distribution

[novel dirge]

never heard of it

[tardy ridge]

ok i will first explain how to do the question, if you don't get it search up hypergeometric distribution and come back

[novel dirge]

ok

just one question

[tardy ridge]

so you want to choose the 3 red balls and 2 of the 47 blue balls

[novel dirge]

why are we first looking for the 3 numbers?

instead of 5

just seems a bit counter intuitive to me

not saying it isn't right, just that I want to understand the logic behind it

[tardy ridge]

ok one sec I think I'll finish it

maybe it'll make more sense

[novel dirge]

ok

[tardy ridge]

so you want 3 red balls (or 33, 7 and 13) and 2 blue balls

since you are choosing 5 balls in total, even if you get all 3 red balls, you will still have to fill in the two spots with non - red balls (or blue balls)

so 3 choose 3 ways to pick the three red balls

and 47 choose 2 ways to pick the 2 blue balls

(there are 3 red balls in total and 47 blue balls in total)

over the total amount of ways to choose any 5 ballso

which would be 50 choose 5

so what it would be is

3 choose 3 * 47 choose 2 / (50 choose 5)

if you don't understand this part you can look at hypergeomtric distribution

[novel dirge]

so 3 choose 3 ways to pick the three red balls
@tardy ridge why 3 out of 3?

[tardy ridge]

There are only 3 red balls, and you need to choose all of them, and 3 choose 3 =1

[novel dirge]

oh

so is there even any need to write that?

[tardy ridge]

well for the sake of explanation

if you the question was worded differently

like choose 2 red balls in 5 picks and there are 3 red balls in total

it would be

3 choose 2 * 47 choose 3 / (50 choose 5)

[novel dirge]

so we basically divide the number of balls we want to get (the 5 ) with the total number of possibilities?

[tardy ridge]

https://www.youtube.com/watch?v=BCeFgnh6A1U

[novel dirge]

thanks

in the end, we want to get the number of options that fit our conditions/ total number of options ?

[tardy ridge]

yeahos

[novel dirge]

I think I kinda get it now

thanks

[tardy ridge]

nice

[carmine elbow]

Can someone help me with this problem?

[tardy ridge]

,rotate

[tardy ridge]

well find the circumference

that thing is spinning 7200 circumferences every minute

@lucid spindleXskull

@carmine elbow

[carmine elbow]

Ok

[sharp marsh]

How do I find tan θ = 1

[lost mesa]

sovling for theta?

[carmine elbow]

so for the circumference, do you use 2pi r^2?

[sharp marsh]

Degrees and radians ya

[tardy ridge]

Bro

no

[carmine elbow]

Just asking

sheesh

[lost mesa]

@sharp marsh tan(x) = C, x = arctan(C)