help-27

[long pasture]

multiply by x/x

oh wait

infinity

hmmm

[solemn carbon]

Yeah

[long pasture]

ah, sandwich theorem

[livid geyser]

do u know squeeze theorem

ye

[winter patrol]

find a way to apply the limit property for
sin(ax)/x (or similar) as x→0

[solemn carbon]

Aight ty

.close

[devout snow]

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[livid geyser]

X goes to infinity

[devout snow]

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[wooden axle]

how many words will be there with (a,b,c,d,e,f) and each word will have least one vowel. 3 letter in a word

[wooden axle]

Answer is 96

[livid geyser]

what counts as a word

If it's just any sequence of letters where order matters then the number of words whose only vowel is a alone is greater than 96

[stone stump]

given that you didnt set a maximum length of the word, clearly infinitely many

[livid geyser]

Oops

Where you can also only pick each letter one time* (still greater than 96)

[wooden axle]

3 letter in a word

[livid geyser]

ok ye that works out

u can do it by cases

1. only a in word
2. only e in word
3. both a and e in word

For each case u pick out the other letters, then permute all the 3 letters

[devout snow]

@wooden axle Has your question been resolved?

[wooden axle]

I tried but didn't get 96

[livid geyser]

What'd u get

[wooden axle]

By letter A/E we have 72

72+24

Ohh i got it

[livid geyser]

Noice

[wooden axle]

AE4
A4E
4AE
Same
EA4
E4A
4EA

Total 96

[wooden axle]

.close

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[mellow glade]

this is what i tride

and i got theta = pi/2

which means its not iscocoles

[devout snow]

@mellow glade Has your question been resolved?

[mellow glade]

<@&286206848099549185>

[tiny oar]

maxima minima the topic i hate
i can't help u with that sry

[mellow glade]

.....

thanks

[eager lodge]

a and theta are dependent

[real pewter]

mot rlly, a is a fixed constant

right ?

we just need to prove b = a for that given theya

b = asin(0/2) *2

you get b = sqrt(2) a

yeah nvm i think a is dependant

wait nvm

idk

wait i think u can do it

witht eh other perimwtre for@ula

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[glossy drift]

could anyone explaim this to me?

[sand dove]

If k^2 - 3 is bigger or equal to 1/n for some natural integer n

then it means k^2 - 3 is bigger than ?

[glossy drift]

n-1?

didnt understand bruh

[devout snow]

@glossy drift Has your question been resolved?

[glossy drift]

<@&286206848099549185>

[twin wedge]

0<1/n<1

0<k²-3<=1

[glossy drift]

where did that come from?

[wheat pawn]

you know that k^2-3 = (r+1)/n
Since you are calculating nC(r+1), you know that n>=(r+1)

since you also are calculating (n-1)Cr, you know that n-1>=r (which is equivalent as before)

you also know that n-1>=0, so n>=1

and that r>=0

that's what you got on the right of the line

now, since you're dealing with combinatorials, both are positive. Since you're also dealing with combinatorials, they both are non-zero

From the very first equality, this means that k^2-3 > 0

which means that k^2 > 3, so k>sqrt(3). You got your lower limit for the interval

since r+1<=n, (r+1)/n <=1

since k^2-3<=1, k^2 <=2, thus k<=2. You got your upper limit for the interval

[devout snow]

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[glossy drift]

ty

[devout snow]

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[pearl void]

$\frac{-u_{2}x_{0}y_{1}+u_{1}u_{2}y_{1}-u_{1}u_{2}y_{0}+u_{2}v_{1}x_{0}-u_{1}v_{2}x_{1}+u_{1}x_{1}y_{0}-v_{1}x_{0}x_{1}+v_{2}x_{0}x_{1}}{\left(u_{2}-x_{1}\right)\left(v_{1}-y_{0}\right)-\left(u_{1}-x_{0}\right)\left(v_{2}-y_{1}\right)}$

[woven radish]

Finky

[pearl void]

how can I simplify the top and bottom expressions? most likely in terms of dot/cross products?

(x = <x0,x1> u = <u1,u2> etc.)

[devout snow]

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@pearl void Has your question been resolved?

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[restive river]

can you use hopsital rule for the limit x->0

[restive river]

if it is x->0^+

?

Can i use the rule here l'hospital

[cold gust]

you can only use hospital if the limit is indefinite

like 0/0 etc

in this case you can

yes

[restive river]

but what will even happen

i cant differentiate the x away here

It will always be divided by x^n

or wait

i can differentiate 1 and x irrespective of each other

[devout snow]

@restive river Has your question been resolved?

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[proud perch]

the square root is multiplicative

[pale cedar]

Yea

[proud perch]

that means that:
sqrt(xy) = sqrt(x)sqrt(y)

so let's look at sqrt(8)

[pale cedar]

Ok

[proud perch]

in order to simplify this, we want to set 8 = xy, such that x is a perfect square

is 8 a multiple of a perfect square

[pale cedar]

4 and 2

[proud perch]

great!

so then you wish to write:
sqrt(8) = sqrt(4)sqrt(2)

[pale cedar]

Ok

[proud perch]

but what is sqrt(4)

[pale cedar]

2

[proud perch]

so then sqrt(8) = 2sqrt(2)

does that make sense?

[pale cedar]

Yes

But how do I minus the 6sqrt50

And att the 10sqrt18

[proud perch]

now let's look at sqrt(50)

can you write 50 = xy

[pale cedar]

K

[proud perch]

with x being a perfect square?

[pale cedar]

Hmm

I don't know

It's easy when it's a small number

[proud perch]

can you find two numbers that make 50

[pale cedar]

No

I mean yes

10 and 5

[proud perch]

sure that's a good start

but 10 is just 5x2

[pale cedar]

Yes

[proud perch]

so 50 is just 5x5x2

(5x5)x2

[pale cedar]

Ok

[proud perch]

do you see the square

[pale cedar]

So then the 25

[proud perch]

yes

so then sqrt(50) = sqrt(25)sqrt(2)

but what is sqrt(25)?

[pale cedar]

5

[proud perch]

so sqrt(50) = 5sqrt(2)

[pale cedar]

Ohh

So for 18 can I do 6 and 3

Then 6 goes to 2 and 3

[proud perch]

not quite

[pale cedar]

Then 3 and 3

So it escapes

[proud perch]

it needs to have a SQUARE factor

[pale cedar]

Ok I'm confused now

[proud perch]

does 18 have a SQUARE factor

[pale cedar]

Uhm

Yes

9

[proud perch]

9 times what?

[pale cedar]

2

[proud perch]

so then sqrt(18) = ?

[pale cedar]

3sqrt2

[proud perch]

yes

[pale cedar]

Ok

How about the ones on the outside though

[proud perch]

let's put it all together now:
sqrt(8) = 2sqrt(2)
sqrt(50) = 5 sqrt(2)
sqrt(18) = 3 sqrt(2)

[pale cedar]

Ok

[proud perch]

we need to simplify:
5sqrt(8) - 6sqrt(50) + 10sqrt(18)

substitute

[pale cedar]

Yea

Ok

Wait

For example for the sqrt 8

So I multiply the 2 by 5

[proud perch]

i see what you're saying. let me write it out

[pale cedar]

Ok

[proud perch]

5 x sqrt(8) =
5 x (2 sqrt(2)) =
5 x 2 x sqrt(2) =
10 x sqrt(2)

[pale cedar]

Where did the 2 5 and 3 go

[proud perch]

just focusing on the first term

now we can move on to the rest of them

[pale cedar]

2 sqrt2

5sqrt2

3sqrt2

[proud perch]

right. let's go over the concept of substitution

[pale cedar]

Ok

[proud perch]

let me use a really simple example:
x = 3.
what is 2+x?

[pale cedar]

5

[proud perch]

ok. and that's the concept you're missing

wherever you see an x, you REPLACE it with 3

so 2+x becomes 2+3

[pale cedar]

Yea

But if I substituted the sqrt8

Won't it just be

5(2sqrt2)

[proud perch]

yes

[pale cedar]

ok

[proud perch]

and you simplify that to 10 sqrt(2)

[pale cedar]

Ok

Is the final answer 10sqrt2

[proud perch]

no

because we still have the other terms

[pale cedar]

No like

The other ones

Are -30sqrt 2

And then 30sqrt2

[proud perch]

oh i see. yeah nice work!

[pale cedar]

Ok

There's another one

[proud perch]

mb, didn't realize you already worked ahead

[pale cedar]

I'ma try it on my owner

Own

Then come back

[pale cedar]

Hey

I just did one

Can u check if I'm right

What I got was

0

<@&286206848099549185>

[daring nebula]

i got 12√2 -18

what did you do?

[pale cedar]

So

I first simplified all the sqrts

Sqrt32 becomes 4sqrt 2

Sqrt81 becomes 6sqrt2

Sqrt12 becomes 2sqrt3

Sqrt27 becomes 3sqrt3

I then plugged them in

So it went from

The original problem

To

[daring nebula]

what about the numbers in front of the square roots though?

[pale cedar]

3(4sqrt2) - 2(6sqrt2) + 3(2sqrt3) - 2(3sqrt3)

[pale cedar]

Then this is equal to 0

[daring nebula]

aa

[pale cedar]

Where did I go wrong?

[daring nebula]

i simplified the square roots with the base

in front of the aquare root

[pale cedar]

I did what the helper above showed me

[pale cedar]

But where did I mess up in my method

[daring nebula]

i think the part where you only simplifed the sqrt but you have to simplify with the base in front of it

[pale cedar]

Wdym

[daring nebula]

because its written as

3√32 right?

but it could also be written as 3 multiplied by √32, so for example like 4x you would need to simplify the 3 with the square root

sorry im really bad at explaining

[pale cedar]

nono you're good

[pale cedar]

But isn't that what I did

[pale cedar]

Here

[daring nebula]

i mean here i have something different

[pale cedar]

Wait

OHHHHH

Sqrt81 is 9

OH

Ohhhhhhhhhhhhhh

[daring nebula]

YOU GOT IT

[pale cedar]

YEAAA

[daring nebula]

[pale cedar]

SO THEN ITS -18+12SQRT2

[daring nebula]

but im also not happy with the last 2

[pale cedar]

Wdym

[daring nebula]

YWA

[daring nebula]

thisnone

[pale cedar]

OKOK

[daring nebula]

and the one below

[pale cedar]

How come

Isn't sqrt12 2sqrt3

[daring nebula]

it is but

then you have 3*2√3 and i dont know if u have it but then that would be

6√3

the other one is also 6√3

3 I MEMAN

[pale cedar]

So it cancels out right

[daring nebula]

yes

so -18 + 12√2

or 12√2 - 18

[pale cedar]

yeaaaa

[daring nebula]

[pale cedar]

Thank you

[daring nebula]

youre welcome!!

do youneed help with anything else or are we done here?

[pale cedar]

Uhm

Can you help me on one more

[daring nebula]

sure!!

[daring nebula]

how did you approach the problem?

[pale cedar]

I multiplied 3 and 2

Got 6

That's it

[daring nebula]

o

i did

$32+3(-2 \sqrt{5})- \sqrt{5}2- \sqrt{5}(-2\sqrt{5}$

[woven radish]

shotgun

[daring nebula]

and got 6-6√5-2√5+10

[pale cedar]

Ok

[daring nebula]

then i did

(6+10) - (6√5+2√5)

[pale cedar]

So like foil?

[daring nebula]

yes

FOIL method

[pale cedar]

Ok

[daring nebula]

then i got 16-8√5

yyayayayy

[pale cedar]

Ok

[daring nebula]

[pale cedar]

You fr are very good at explaining

[daring nebula]

not really but thank you!!!

[devout snow]

@pale cedar Has your question been resolved?

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[tacit wedge]

for interval B, I always get -10m/s but it should be positive?
Here's what I did:

I found interval A final velocity is 20m/s so
interval B final velocity is 2(∆x/∆t )(both for interval B) + interval B initial velocity which is (150m-100m)/(20s-10s)*2 - 20m/s which turns out to be -10m/s.

[unborn relic]

need help

[tacit wedge]

why do i always get that final velocity of B is negative

when it sould be positive

in this graph

[unborn relic]

yo u wanna share

channels

[tacit wedge]

I'm down

[unborn relic]

alr

[tacit wedge]

I don

I dont mind

[unborn relic]

do the inital velocity plus acceleration and * by time

inital+(acc*time)

[tacit wedge]

I cant use that formula

it's split into parts and i cant do anything with acceleration

at the part that im in

also jay help 21 is open

if u want to take it

@unborn relic

[restive river]

Hello, I'm currently in Honors Algebra 2 and Honors Geometry; would you guys recommend I take Honors Precalculus over the summer before 10th Grade?

[tacit wedge]

💀

.close

[devout snow]

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[unborn relic]

.reopen

[unborn relic]

oh alr thanks

[devout snow]

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[lament cradle]

do you know how to find the mean

[dreamy palm]

yes

[lament cradle]

so we have a set of data points, and we have the actual mean. we can create an equation representing the mean here.

once you write that out, you can solve for x.

[dreamy palm]

oh ok

so i equate it to (x+1) and solve for x

[lament cradle]

yes, x + 1 is the given mean.

[dreamy palm]

thanks

.close

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[stark coral]

On the x intercept, how did the 2nd step happen?

Where'd the denominator go?

[frosty cradle]

0 divided by anything is 0

[stark coral]

so he carried the denominator to the left side, making it be multiplied by 0?

[frosty cradle]

I guess that's one way to look at it

[devout snow]

@stark coral Has your question been resolved?

[devout snow]

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[short gorge]

😧

what do i solve for here & how

[timber pebble]

say i gave you the function f(x)=4

and i tell you its defined for all x except x=2, where its equal to a

what should a be to make f continuous?

[short gorge]

2?

[timber pebble]

why 2?

[short gorge]

because its definied for everything except for x=2

[timber pebble]

we want the function to have no holes right

[short gorge]

yes

[timber pebble]

but i plucked a point out

i made a hole

whats the coordinate of the hole

right now theres a point at (2, a)

[short gorge]

ohhh

so would it be 2,4?

[timber pebble]

right, thats where the hole is

but if a were 4…

[short gorge]

Oh okay okay i get it

[timber pebble]

wed have poked a hole

[short gorge]

kinda

[timber pebble]

then filled it back in

[short gorge]

so what would i do here?

[timber pebble]

well you want the point you add back in to be equal to what the function would have been at that point

what would f have been? if we didnt poke a hole at x=-1

aka whats the coordinate of the hole

[short gorge]

2???

[timber pebble]

so here

$f(x) = \frac{x^2+5x+4}{b(x+1)}$

[woven radish]

jan Niku

[timber pebble]

oh i see the problem

[short gorge]

hm

[timber pebble]

well, no

here

lemme rewrite it for you

this should make it more clear

[short gorge]

Okay okay ty

[timber pebble]

$f(x) = \frac{(x+4)(x+1)}{b(x+1)}$

so start here

[woven radish]

jan Niku

[timber pebble]

here's what i want you to do

tell me where the discontinuity is, what value of x

notice its removable

and tell me what the value of the function is at that x value after removing the removable discontinuity

[short gorge]

OH

x+4

right?

x+4/b

?

[timber pebble]

okay so you noticed you can cancel that factor

where was the removable discontinuity?

what x value

[short gorge]

OH

1

well -1

right

[timber pebble]

okay, so you have a hole in the function

at x=-1

f(x=-1) is not defined

but thats okay, its just a hole

you can fill it by defining this value specifically

so leave f(x) as it is for all x except x=-1

[timber pebble]

after you removed the removable discontinuity, whats the value of the function at the (old) problem x value?

you said $f(x) = \frac{x+4}{b}$ after removing the discontinuity

[woven radish]

jan Niku

[timber pebble]

this is just defined and OK for x=-1, right?

[short gorge]

yeah yes

But idk where to go from there

[timber pebble]

well, whats f(-1)...

lemme make a graph

@short gorge https://www.desmos.com/calculator/hzupspt5fr

the red line is the function

if you mouse over, youll see that f(-1) is Undefined

but thats fine, we can add the purple point manually

we just need to get b right ...

[short gorge]

OH

I GET IT

so it would be

sqrt -3

and sqrt 3

right?

[timber pebble]

hmm i dont think so

is it

f(-1) would be 3/b

oh, maybe youre right lol

[short gorge]

AYYY

ty for the help!!

.close

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[lament schooner]

Answer for 2- is -12 and answer for 2+ is 12 why is there such a big gap

even graph calculators show this as false

[arctic temple]

@lament schooner

[arctic temple]

it simply means function discontinuous at x=2

[unreal eagle]

try graphing it for a better understanding

[lament schooner]

so would this be correct

[arctic temple]

yes

[arctic temple]

^

[lament schooner]

but graph shows me this

which is not correspondant of my results

[unreal eagle]

thats correct, look at the graph at negative y too

approach 2 from both sides

[arctic temple]

it is

[lament schooner]

oh

[lament schooner]

mb i didnt see

i didnt know a function like that was possible wtf

to just have a huge Y gap

tyty

xx

[lament schooner]

yeah haha just realiozed

tyty

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[lament schooner]

is it possible algebraically

[unreal eagle]

just solve f(x) = g(x)

[lament schooner]

that gives x^3 = 1-sin(x) right

[winter torrent]

yeah you won't be able to find an explicit intersection point

you'll have to use some reasoning skills to prove that there is one

and some theorems

[lament schooner]

okay ty

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[stiff sphinx]

I’m confused on who should get the last spot but I’m guessing Joe

Since he’s more farther right on the distribution

<@&286206848099549185>

[real pewter]

if last spot = better, then joe

[stiff sphinx]

Ok thanks

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[brisk grotto]

how do i approach solving this system of equations

a^2+bc=1
ab+bd=2
ac+cd=0
bc+d^2=4

[devout snow]

@brisk grotto Has your question been resolved?

[dapper tiger]

c(a+d) = 0
so either c = 0 or a+d = 0 ie a = -d
if a = -d
then a²+bc = 1
but bc+d² = bc+a² = 4, which contradicts
so it's c = 0
so a² = 1 and d² = 4

b(a+d) = 2 is our last equation to use

you have either a = 1 or a = -1, and d = 2 or d = -2

and you find b in each case

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[heady eagle]

Hey guys! Could someone tell me where I went wrong in my working?

[heady eagle]

It seems that the answer was

[main gull]

Why did you multiply by 2?

When you were isolating for y

[heady eagle]

The way i'm taught is to get rid of the coefficient of x and then add it back

so I can factor it

[main gull]

The 2y -> y part

[heady eagle]

Oh shit yeah. Its supposed to be 1/2

[main gull]

Yes

[heady eagle]

because I need to divide both sides

Thank you!

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[untold oak]

How do I solve this question?

[untold oak]

Without using L’Hopital because we haven’t reached that point

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[untold oak]

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[restive river]

Hey I am looking how do I get a quadratic formula in order to solve for x

[pseudo basin]

what are the width and height of the big rectangle in terms of x?

[restive river]

Oh no I just need to solve x

[pseudo basin]

...

this is a step along the way though...

[restive river]

In the question I had there was no need to solve that

YEAAHH

[pseudo basin]

how can there be no need to know the big rectangle's dimensions

[restive river]

I have no idea??!

I can send the original paper with the answer, I just need someone to help me explain the logic

[pseudo basin]

ok sure whatever

[restive river]

How to get to that

Oop

Its horizontal

[pseudo basin]

,rccw

[pseudo basin]

is this work yours?

[restive river]

Yes

[pseudo basin]

it's wrong

[restive river]

Which one?

The purple is the correction

[pseudo basin]

the length of the rug is not 5+x but 5+2x

and the width is not 3+x but 3+2x

[restive river]

How??

[pseudo basin]

look at it

you've drawn it so the length is vertical

so going from the top down you have the top margin (x), the mosaic length (5) and the bottom margin (x)

x + 5 + x

that's 5+2x

[restive river]

OH

Oh my gosh

Is that what I have been stuck on

[pseudo basin]

might have been that

the purple equation is cut off but it appears incorrect too...

[restive river]

Oof yeah! I think so too

I think I rushed to copy it from the board from another version of the same paper

With different units

But thanks! (Off topic: Also I noticed you know two languages I am studying lol (french and russian) very cool)

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[weak cove]

Could anyone double check my proof here please?

[terse shale]

yeah looks fine I think

neat little algebra trick

[weak cove]

Are you sure that I stated my strong induction statement correct?

that is the part that I am wondering most about

if the inequality should be

1<=n<=k

[terse shale]

oh sure so if you phrase it that way then yes you should actually be computing x^(k+1) + 1/x^(k+1), not with n

because you assume it is true for 1 <= n <= k and then want to show it for k+1, yeah?

[weak cove]

yes for k+1

[terse shale]

so you'd just want to change "We show this implies that ..." to use k+1 and not n+1, and likewise change the rest of that chunk of the proof to use k+1

[weak cove]

so but

oh okay

since it is less than or equal to

we still can compute that product claiming it is an integer

x+1/x still given

[terse shale]

yeah exactly

[weak cove]

and x^k+1/x^k assumed

[terse shale]

yep!

[weak cove]

okay good point ty

[terse shale]

after that I think it's ok

at the end maybe could be slightly better worded perhaps but that's like a little pedantic maybe

[weak cove]

how so?

[terse shale]

like "Therefore by strong induction we have that if x + 1/x is an integer, then x^n + 1/x^n is also an integer for all n in N" or something

as is it says "thus we have shown our inductive step and .." which I perceive as a little more clunky but still fine

so this isn't a problem in the proof or anything as someone reading it should understand what's going on perfectly well but I think it is good to be really explicit

[weak cove]

okay that is fair

ty

I will change that aswell

[terse shale]

in the end the way proofs are written are totally to personal taste

[weak cove]

ture

true

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[weak cove]

ty again btw

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[thick inlet]

hi

[thick inlet]

I understand that it cannot be less than 0

x+2-1 > 0 and -x+2-1> 0 are the equations I chose

[brittle burrow]

*-x-2-1

[thick inlet]

yeah ahaha

I just noticed that

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@thick inlet Has your question been resolved?

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[cerulean crater]

Number 2… my answer doesn’t make sense it can’t be zero

[cerulean crater]

Oh fudge… I forgot pi equals 180 degrees and 2pi is 360 degrees which makes it the same as start— I’m so stupid sorry guys

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[fair gazelle]

I want roots of this equation

[finite briar]

find possible roots using rational root theorem, then do factor theorem

then polynomial long division

[fair gazelle]

I tried hit and trial
For x= 1,-1 and -1/2

Bt none worked

[finite briar]

do you know any root theorems?

[fair gazelle]

I forgot ig

[brittle burrow]

loll

[brave vapor]

The root are not integers

[finite briar]

....

[rich summit]

,w 8x^5 - 8x^3 + x^2 + x + 1 = 0

[brave vapor]

1real and 4 imaginary roots 💀

[fair gazelle]

Does this hav any values of x which satisfy the solid angle for cos theta ?

[woven radish]

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[brittle burrow]

[rich summit]

there isn't even a closed form

[finite briar]

huh?

[finite briar]

why do u need them btw? this full question?

[brave vapor]

Apply bisection method and wish you don't get stuck in a loop

[finite briar]

what is that

[brave vapor]

Cheat code to solve any polynomial equation given you have a lot of time

[finite briar]

hmm am gonna goggle it

[fair gazelle]

I actually wanted to solve this particular thing

2(cos2theta + cos3theta + cos5theta)= -3

[finite briar]

BRUH

[brave vapor]

Oh

[finite briar]

https://xyproblem.info/

[brave vapor]

Combine the cos

[fair gazelle]

I did all the expansion nd got an equation in cos

[finite briar]

cos3theta is $\cos^{3} \theta$?

[woven radish]

ItzKraken

[fair gazelle]

No

Cos(3theta)

[brave vapor]

Yeah combine the cos

Use cosC+cosD formulae

[fair gazelle]

@brave vapor
Lemme show u wht i did

[brave vapor]

I bet you expanded the cos(3theta) and cos(5theta)

, w simplify (cos(2x) + cos(3x) + cos(5x))

[restive river]

If am not wrong, try to find one root by hit and trial or some sort of then

Convert that root into factor like (x - alpha) alpha is root

And write it five times, then try to multiply stuff with it to make the original polynomial, this will get you a new polynomial with degree 4 or so

[fair gazelle]

Yupp

[brave vapor]

He tried that, the root is not an integer

[woven radish]

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[brave vapor]

You weren't supposed to do that

[fair gazelle]

Then wht do i use ?

[restive river]

[brave vapor]

https://cdn.discordapp.com/emojis/930589662945366096.png?v=1&size=64&quality=lossless

[fair gazelle]

Bruh
This is wht i used

[brave vapor]

Mb

Wait, you combined which two cos? (3x) and (5x)

[fair gazelle]

Yea

Coz it gave integer

Otherwise its odd

[finite briar]

how do u get a quintic equation

[fair gazelle]

My soln is pretty messed up
Bt i'll send

[brave vapor]

, w solve 2(cos(2x)+cos(3x)+cos(5x))=-3

[woven radish]

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[brave vapor]

https://cdn.discordapp.com/emojis/888244820378792006.png?v=1&size=64&quality=lossless

[finite briar]

[fair gazelle]

Wht hppnd lol?

[brave vapor]

You sure the question is correct?

You need to solve this equation right? 2(cos(2x)+cos(3x)+cos(5x))=-3

[finite briar]

have they specified anythin about theta being complex for some stupid reason?

[fair gazelle]

I've got the main question from where i got this all
I'll send
M sry
It might b a mistake from my side

[brave vapor]

https://cdn.discordapp.com/emojis/1017709558065532948.png?v=1&size=64&quality=lossless

[fair gazelle]

So here basically wht i did was ...

[finite briar]

[fair gazelle]

The value of the det is supposed to be
-(a+b+c)(a^2-b^2-c^2 nd smth smth)

So since this is equal to zero,
I equated the thing with iota in it

The first line

[brave vapor]

, w simplify (-(b-c)^2-(a-c)(a-b))

[woven radish]

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[brave vapor]

Is that the det value

I mean

(a+b+c) (-a^2+ab+ac-b^2+bc-c^2)

Is that the det value?

[fair gazelle]

Yess

[brave vapor]

Cool

Go on

[fair gazelle]

This is also equal to 3abc -(a^3+b^3+c^3)

[brave vapor]

, w expand (a+b+c)*(-a^2+ab+ac-b^2+bc-c^2)

[fair gazelle]

So i jst did a+ b+c = 0

Coz the other part cannot be zero

[woven radish]

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[brave vapor]

Okay

[fair gazelle]

It will become way too complicated as well nd the question might not b tat complex

[brave vapor]

Since 3abc=a^3+b^3+c^3, a+b+c=0

[fair gazelle]

Yep

U can say

[brave vapor]

So you need to solution of cosx+cos2x+cos3x=0 with sinx-sin2x+sin3x=0

[fair gazelle]

But there's iota as well

So on squaring both sides, we get the whole squares nd with a negative sign

[brave vapor]

The coefficient of iota is sinx-sin2x+sin3x right? That needs to be equal to zero

Since a+b+c=0

[brave vapor]

Don't square

[fair gazelle]

U missed the + ig

How else to simplify then ?

[brave vapor]

a+bi=0 equate to a=0 and b=0, you compare the coefficients

[brave vapor]

When you are square say (a+bi) you will end up with a^2-b^2+2iab, you get another iota in the equation

, w cosx+cos2x+cos3x=0

[woven radish]

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[brave vapor]

, w sinx-sin(2x)+sin(3x)=0

[woven radish]

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[fair gazelle]

Bt in this particular case
Since a+b+c =0
I can send all the iota terms i.e the sin terms on the lhs
Nd square both sides
The iota jst gets eliminated without any additional iota

[fair gazelle]

Here,
The first step

[brave vapor]

You will then need to deal with squares of sin and cos

[fair gazelle]

Hmm yea

[brave vapor]

That makes the question a pain

[fair gazelle]

So ok
Lets do it separately

[brave vapor]

Equating will work

You may not need to even solve the trigonometric equation, you can just plug in the options

[brave vapor]

Oh wait

We skipped a solution in the expansion of determinant

[fair gazelle]

Which one ?

The other bracket or wht ?

[brave vapor]

This is the expansion of determinant right? What if a=b=c

That's another solution

So if you go that route

[fair gazelle]

I don't think they can be ever equal
Or can they?

[brave vapor]

They can be

a^3-3abc+b^3+c^3 => a^3-3a^3+a^3+a^3=0

It holds perfectly

[fair gazelle]

But look at the values of a b c in the question

[brave vapor]

Are you well versed with euler formula for complex number?

[fair gazelle]

The complex numbers...

[fair gazelle]

Yea

[brave vapor]

So a is e^(i theta), b is e^(-2i theta) and e^(3i theta)

[fair gazelle]

Yea

[brave vapor]

They all will be equal when e^(i theta) is equal to one

And that's possible when costheta is equal to 1

So the answer is theta=2kpi

[brave vapor]

Option A

[fair gazelle]

Yeah
Ok
Got it

Thanks a lot

[brave vapor]

Anytime

[fair gazelle]

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[neat bridge]

How di I calculate each compnent and the resultant?

<@&286206848099549185>

[desert yew]

Use the angles (and slope) to decompose F1 and F2 into components, then sum in each direction

[neat bridge]

so for example the x component of F2

how would I find that

[desert yew]

https://www.youtube.com/watch?v=VZm6mxu2xlk

[neat bridge]

7 cos (36.9) sin 30

?

[desert yew]

Sorry, don't have time to check exact numbers. Maybe someone else can

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@neat bridge Has your question been resolved?

[neat bridge]

<@&286206848099549185>

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[short plaza]

Hello,
lim x- infinity ((x^2 +1)/(x^2 -1))^x^2
The reponse is e^2 but how can i find that?

[devout snow]

@short plaza Has your question been resolved?

[cinder mesa]

replace x^2 with some t

so $\lim_{t\to\infty}({\frac{t+1}{t-1}})^t$

[woven radish]

TimK

[cinder mesa]

ig you get it

[short plaza]

Okk i will try

I dont know what i have to do after

[cinder mesa]

so it's $\lim_{t\to\infty}({1+\frac{1}{t}})^t$ but $\lim_{t\to\infty}({1+\frac{2}{t}})^{t+1}$

[woven radish]

TimK

[short plaza]

Is it not that?

[restive river]

No

It's

e

And ans is e^2

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[restive river]

The radiator in a car holds 8 L. The coolant consists of 3/10 glycol and the rest is water. To increase the glycol content to 3/5, you drain a little coolant and top up with glycol. How much coolant must be drained?

I went this far:

3/10*8 = amount of glycol

Rest amount (water) = 7/10*8

Then I got stuck. How do I figure out how much to spill out when the glycol and water is blended??

According to the solutions, the answer is 24/7 L, which is approximately 3,4L.

[restive river]

Oh sorry I meant 3/108 and 7/108

What

How do u write multiply

3/10 * 8

And

7/10 * 8

Yeah

[devout snow]

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[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[karmic hound]

Ello

[karmic hound]

I have many, I don’t know what to do…

Anyone wanna help me one by one please

[wheat pawn]

without the instructions, we dont know what you have to do either

[karmic hound]

Oh

Multiply and sumplify

For the second one I got

6+9c

For my answer

[wheat pawn]

that's not correct

do the multiplication again paying closer attention

[karmic hound]

What??

Oh

-9c

6-9c

[devout snow]

@karmic hound Has your question been resolved?

[karmic hound]

<@&286206848099549185>

[mighty pine]

yes

if you need help

it is the same form as ( a+ b ) ( a - b )

( a + b ) ( a - b ) = a ² - b ²

a = sqrt ( 6 ) and b = 3c

[open granite]

who ping

[mighty pine]

so yes ! , ( sqrt ( 6 ) + 3c ) ( sqrt ( 6 ) - 3c ) = sqrt(6) ² - ( 3c ) ² = 6 - 9 c ²

and the same thing for sqrt ( 7 ) and sqrt ( 2 )

you'l find 2 - 7 = - 5

[karmic hound]

Wait what

I got 9 for that

[karmic hound]

I simplified on the bottom they are all divisible by 3.

🫃🏻

How does it look? Legit lr nah??

[devout snow]

@karmic hound Has your question been resolved?

[karmic hound]

<@&286206848099549185>

I need help 🫢

[mighty pine]

no

close

not + 7

but - 7

cauz u have sqrt ( 49 ) but there is - before

[karmic hound]

OH

Yes I se

The rest,,

[mighty pine]

you can't say that

[karmic hound]

What do I do haha

I was confused there

[raw knoll]

U can't combine them

[devout snow]

@karmic hound Has your question been resolved?

[karmic hound]

Cant you multiple them?

Multiply the outside and inside numbers to each other

[raw knoll]

They said + not x

Why would u multiply a +

[devout snow]

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[dreamy birch]

What type of discontinuity does f have at x=0?

[cold gust]

jump i think

plug -0.01 and 0.01

[dreamy birch]

how do i determine a discontinuity?

like i just learned this today

[cold gust]

if you know all the discontinuities and understand basic limits its easy to determine it

jump, removable, asym

[dreamy birch]

plugin -0.01 and 0.01 into what?

.close

[devout snow]

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[devout snow]

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[mellow lintel]

need help with this one:

[devout snow]

Please don't occupy multiple help channels.

[midnight bane]

.close

[devout snow]

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[unborn vector]

3 is wrong though right?

[brittle basin]

yeah n could be 2, 5, 8 etc.

[unborn vector]

Ok cool

[brittle basin]

I would email your professor

[unborn vector]

That’s what I thought

It’s possible he did it on purpose

[brittle basin]

my discrete math book would include incorrect proofs and then tell you why on the next page. It was really annoying.

[unborn vector]

Hah that’s awesome

I’m trying to formulate how I would say n congruent 1 (mod 3) is incorrect.

I feel like saying (2+3k) not congruent 1(mod 3) forall integers k

I think it sounds right but I’m not sure @brittle basin

[brittle basin]

I mean that is right

because (2+3k) is congruent to 2

(mod 3)

[unborn vector]

Is there a better way to write that?

I feel like I’m forgetting something that would formulate it better

Wait I think I got it

[wooden wraith]

if it's not true you can just provide a counterexample

[unborn vector]

Yeah true

I’d like to point out a way to show how all counterexamples would be constructed though. Feels more rigorous to do so

That feels more complete to me

Thx for the help tho

[brittle basin]

,rccw

[unborn vector]

Hmm wait maybe the if is bad to include

Is it going to cause some problems because of it being a conditional statement?

I think I’m further confusing myself

[brittle basin]

I mean you could write as you had it as (2+3k) and show that is not congruent to 1 nor is it divisible by 3, I guess

but like tatpoj said you just need one counterexample

[devout snow]

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[modest inlet]

if i get two answers for x, can there be two answers for the measure of an angle?

[wise hemlock]

hard to know exactly what you mean without seeing the question, but yes you can often have multiple angles that are solutions to the same problem

[wise hemlock]

your set of solutions looks good to me, both fit the question

[modest inlet]

alright thanks

.close

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[restive river]

how can i solve this?

[restive river]

i have to determinate (f o g)

[mighty knoll]

do you know how the circle symbol works?

(idek the name in english <_<)

[restive river]

composition

[mighty knoll]

Ahh right

so yeah, do you know how that works?

[restive river]

$f(x) &= \sqrt{x-1}
g(x) &= \frac{1}{x}
(f \circ g)(x) &= \sqrt{\frac{1}{x} - 1}$

[woven radish]

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

[restive river]

i did this but not sure

[mighty knoll]

that is correct

[restive river]

aight, but the question is, that's it?

or would you do something else?

the problem ends in there?

[mighty knoll]

if they asked you to just do composition between the two then yeah

it's really not that complex :)

[restive river]

ohh okay okay in that case, thanks haha

[mighty knoll]

you put one function in another. Sometimes things cancel out, in this case no

[restive river]

.close

[devout snow]

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[devout snow]

@half cape Has your question been resolved?

[devout snow]

@half cape Has your question been resolved?

[half cape]

<@&286206848099549185>

[devout snow]

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[vague garden]

Need help with c, not sure how to solve.

[wooden zodiac]

is z a complex number?

[vague garden]

no?

this is calculus 1

[wooden zodiac]

You should put numerator in common denominator first