help-26

[arctic moon]

Couldn't you factor that thought?

though*

(x+1)(x-1)

oh

wait nah

because multiplied that would equal -1

So basically

Try factoring first, if you don't have the ability to factor, check if the denominator equals 0 when set to the limit, if it can't do either of those, use limit theorems

So for example with the earlier question

the factors of the top are (x-19)(x+16)

the factors of the bottom are (x+16)(x-16)

cancel out x+16

plug in 16

the answer is 0, so limit does not exist?

[radiant tapir]

double check your factoring for the numerator

your signs are flipped

[arctic moon]

Its a -16 and a 19, so it would be (x-19)(x+16) no?

or do you not flip the signs when putting them in factorized forms?

[radiant tapir]

it can also be 16 and - 19

which one adds up to 3?

(x-16)(x + 19) or (x + 16)(x - 19)?

[arctic moon]

the former

[radiant tapir]

yep

[arctic moon]

So you don't flip the signs

Idk where I learned that

[radiant tapir]

not sure lol

when in doubt just FOIL out your factorization and see if it matches

if it was t^2 - 3t - 304 then you'd be correct

[arctic moon]

So

35/32

[radiant tapir]

yep exactly

nice work

[arctic moon]

So wait, is this how you'd go about it?

Because assuming both factorization and limit theorems work for the problem, they should give the same answer?

[radiant tapir]

specifically what do you mean limit theorems? Can you provide how that works for this problem?

without factorizing?

[arctic moon]

Like, you could use VI, and then use III and IV

then fill in the approached limit for the x values

I would do it in latex but it takes an eternity, but I can if that doesn't make sense

[radiant tapir]

so the issue with doing limit theorems without factorizing first is that if you were to use rule VI you'd get 0 in the denominator which is undefined

[arctic moon]

Ahh

So because you get 0 in the denominator

you can't use limit theorems

[radiant tapir]

yet

[arctic moon]

Or you'd have to change the form right

[radiant tapir]

if you can find a way to eliminate a factor, then you can use the limit theorems as desired

[arctic moon]

but at that point

its just easier to factorize

which is why it does it?

[radiant tapir]

you factorize so you can use the limit theorems

notice the exception in rule VI: limit as x approaches a of g(x) cannot equal 0

[arctic moon]

Ah

Yeah

[radiant tapir]

so in order to use that rule, you have to do some algebraic manipulation

[arctic moon]

Which would be factorizing?

But doesn't factorizing give you the answer? So why would you use a limit theorem there?

Or I guess in what instance would you use the factorization, and then the limit theorem?

The reason I ask this kind of stuff is because it seems like in calculus if you don't understand all the "what ifs" or how it can be done, you don't really actually understand the theorems or rules, and then you can have them pull up a question you didn't see the method to before and get absolutely wrecked.

So I'm always scared of that lol

[radiant tapir]

I can respect that. A word of advice, I found that sometimes in calculus the "whys" and "what ifs" can be better understood later. Some concepts won't fully click in your head until you venture over to a real analysis course

but if it helps consider this

[arctic moon]

I'm going for a business degree, its just my Bachelor's degree path requires this calculus course, and potentially one more. I'm not planning to become a full on mathematician per se, but I just always get scared of not fully grasping the concepts and doing horribly on exams and stuff.

It seems in calculus classes, professors love throwing those "what ifs" questions at you and don't really follow the guidelines of what they cover directly within lectures.

[radiant tapir]

You have 6 limit theorems there to play with.

Number 6 might work because we have a rational function. Let's plug in the value of the limit.

Oh no we get a 0 in the denominator, so we can't use rule 6 yet.

Is there any algebraic manipulation we can use? Looks like factoring will work. We've eliminated a factor in both the numerator and denominator.

Let's try rule 6 again. Hey this time we don't get 0 in the denominator. Plugging in the value we get an actual answer!

[arctic moon]

Where would the plugging in occur after the factorization, though?

[radiant tapir]

Math, and I mean this loosely, is all about what ifs and edge cases. A lot of times when you get into more rigorous proof courses the edge cases matter a ton because just one example disproving a proposed theorem can make it false

[arctic moon]

because once you cancel out the x-16s you are left with a x+19 and x+16

OH!

[radiant tapir]

so the "what ifs" are important, since many concepts in math have a lot of factors at play and a lot of rules to work around

[arctic moon]

It just clicked, they just skipped a step

In the examples

They factorized it down to

$\lim_{x\to 16}\frac{x+19}{x+16} = \frac{\lim_{x\to 16}x+19}{\lim_{x\to 16}x+16}$

[thorny flame]

Huntifer

[arctic moon]

Then plugged in the 16 for x

They just skipped over the distribution of step VI

[radiant tapir]

yep pretty much

you'll start doing that too lmao

[arctic moon]

🤦‍♂️

[radiant tapir]

eventually

[arctic moon]

This has happened twice today lol

where the homework would just like.. Skip a logical step in showing it

and then I would be like "What, how did they just magically do that"

[arctic moon]

Yeah, which makes sense. Theorems are theorems because they are known to work in all examples listed within the parameters, if they don't work within the parameters listed, they're false theorems.

That's why there are theorems for solutions if certain theorems don't apply

It just can become incredibly confusing, and its something that is very hard to learn sometimes within the confines of worrying about time and or grades or other classes

Because it seems like a lot of calculus is basically just a billion and one theoretical rules applied that you have to remember to get an outcome, and as it gets more and more complex sometimes there are a lot of ways to a single answer.

[cloud rapids]

barely learning calculus?

[arctic moon]

Yeah, early calculus

[cloud rapids]

good luck

lol

[arctic moon]

I mean I did calculus 1 in high school

I struggled with that class

[cloud rapids]

i mean

you just need a understading of the fundamentals

that been

algebra

trig

algebra 2

[arctic moon]

Of which I'm not super great at

[cloud rapids]

and you good

[cloud rapids]

get baldor for algebra

1

[arctic moon]

I didn't slack off learning those, its just been a long time

[cloud rapids]

good ass book

[arctic moon]

See, I don't have enough time outside of school to just

also re-study old stuff fully

[cloud rapids]

you do any sports

?

[arctic moon]

Yes

[cloud rapids]

oh nice

[arctic moon]

I work too

Which is mainly why

I cut back on extracurricular activities so I had more time to study

Like I've been studying for this exam for 7 hours today so far

[cloud rapids]

study smart not hard

g

@arctic moon

is the exam about>

math

I assume

[arctic moon]

Yes, its covering the stuff that I'm doing in the practice exams

and have been practicing for the past 2 or so days

[cloud rapids]

ok ok

good luck

dont stress about it

and yeah

have fun with the exam

if any questions ping the helpers

[arctic moon]

I would have fun, if it wasn't for a grade lol

[cloud rapids]

[arctic moon]

Any graded exams take out basically all the fun I have out of learning

because I hyperfixate on doing well with grades

[cloud rapids]

damn

[arctic moon]

Oh speaking of which for this question

part d. it becomes (1/7)/(3/7)

Is there some simplier way to reduce this than doing the whole ad/bc thing?

Thats easier to remember? Or is it just remembering that rule

[sweet shard]

No

[arctic moon]

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[old garden]

can someone help. I can not seem to get the correct B and C values at all

[keen matrix]

flashbacks

[old garden]

btw

i have been on this

since we last talked

haha I am crying on the inside

[keen matrix]

lmao aight let's see if I embarrass myself this time

so you have a and d values right?

[old garden]

let me check what this one would be

i havent looked at it

i just know i dont know how to get b and c correct

a=5 and d=0

[keen matrix]

ok now it's a sin, so the "5-point pattern" for positive sin is mid-max-mid-min-mid

do you see that pattern occuring on your graph?

[old garden]

yes

I do

[keen matrix]

Ok so then a is positive

now, where does the period start?

[old garden]

at 0

[keen matrix]

ok now what is the period?

[old garden]

the period is pi

[keen matrix]

so pi=2pi/b

[old garden]

oops sorry

[keen matrix]

why sorry 😅

[old garden]

wait so is it not pi?

[keen matrix]

P=pi and P=2pi/b

so pi=2pi/b

now solve for b just like before :))

[old garden]

okay let me try!

b is 2

i think

bare with me and my fried brain

[keen matrix]

same honestly

should be going to bed

[old garden]

me too man

[keen matrix]

so a=5
b=2
c=0
d=0

see if $y=5\sin(2\theta)$

is correct

[thorny flame]

XxMrFancyu2xX

[keen matrix]

if not ill just go die

[old garden]

LMAO

yes it is

i think it gave me an easier one

[keen matrix]

YES

[old garden]

after getting the same question wrong 5 times

[keen matrix]

IM BACK IS BUSINESS BABY WOOO

[old garden]

wooO!O!!

[keen matrix]

ok sorry got a little excited there

[old garden]

no honestly

this is a W

[keen matrix]

ong

[old garden]

ight can you help with one more

the next one

[keen matrix]

im hella tired so i cant garuantee correct

but i can try

[old garden]

thank you so much

so it's sine

and yeah. that's all i got

[keen matrix]

ok what is a and d?

[old garden]

how do I get A again

[keen matrix]

1/2(max-min)

[old garden]

it's 1/2(max-min)?

yeah

thank you

[keen matrix]

yep :))

[old garden]

so 4

A:4

[keen matrix]

and d?

[old garden]

and what's d again 😂

[keen matrix]

1/2(max+min)

[old garden]

thank you

so 1

D: 1

[keen matrix]

yep (sorry if I sound dead—its midnight)

[old garden]

don't worry bro. me too.

[keen matrix]

next what's our 5-point pattern?

[old garden]

i have like no sleep

in the middle of moving to a new home

[keen matrix]

starting from mid

[old garden]

yeye

let me see

[keen matrix]

damn 💀

[old garden]

it seems to go positive

[keen matrix]

mid-max-mid-min-mid

this means a is positive

ok all good to proceed :))

[old garden]

great

so c and b

are left

[keen matrix]

now where does a period begin?

[old garden]

period begins at 3pi/4?

[keen matrix]

now find where the next period begins

the distance between these two points is the period

[old garden]

wait is it 3pi/8?

[keen matrix]

no... 7pi/8

for sin period begins are (c,d)

cosine is (c,a)

[old garden]

no i meant for period beginning

sorry, didn't clarify

[keen matrix]

oh no did I mess this up?

I think a is negative

[old garden]

oh really

[keen matrix]

fuck man I need go to bed

[old garden]

wait yes!

it goes down

[keen matrix]

I think it goes mid-min-mid-max-mid

[old garden]

i guess we should find period first

to see if it goes up or down

yeah you're right

[keen matrix]

ok so c=3pi/8 :))

now find period

set equal to 2pi/b

and Bob's your Uncle :))

now

i need sleep

[old garden]

UhhHhh sorry I am lost when it comes to finding the period

what do i set equal to 2pi/b

c?

[keen matrix]

the period is for what duration does the function not repeat

the distance between two endpoints of an oscillation

[old garden]

oh right

[keen matrix]

ok gn now man :))

[old garden]

gn

i got it from here

thank you

sleep well

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[neon iron]

Oops, kinda a bad photo

But how would 1e and g work?

Here is a much better photo

And I needed some help with 12 :p

[quick gorge]

for 1e, try plugging in some positive number for b

[neon iron]

Would we get square root?

Sqrt (b)?

[quick gorge]

why would we get square root?

[neon iron]

Oh, I misread my notes for fractions

Undefined

[quick gorge]

yep

if you think abt it, you can’t do a to the power of anything to get a negative number

[neon iron]

Ahh, ic

[quick gorge]

for g, you’ve basically solved it

$ln(x) = log_e(x)$

[thorny flame]

Kihei

[neon iron]

Ahh

lnb/lna

[quick gorge]

mhm

[neon iron]

Yay

How would I approach Q12?

Last question :3

[quick gorge]

hmm is it asking to simplify?

[neon iron]

Express each of these in terms of log a and log b

[quick gorge]

I think you should use these

first it’s the fraction one

[neon iron]

log1-logab^4

[civic monolith]

hey

the first thing you should know is:

[neon iron]

Asuna

[civic monolith]

(1/(ab⁴)) = (ab⁴)⁻1

so it is easy to do

[civic monolith]

log(a.b⁴)^(⁻1)

now

the expoent you can put in front of the log

(-1).log(a.b⁴)

[neon iron]

Would this be correct

[civic monolith]

now you can apply log(a.b) = loga+ logb

[neon iron]

Oh, +4log(b)

[civic monolith]

-1. (loga + logb⁴)

-1*(loga +4.logb)

-1*loga -4logb

-loga - 4 logb

[neon iron]

OHHHH

I see why it’s -4

Tysm

[civic monolith]

anytime

[neon iron]

New SAO coming

[civic monolith]

Really? I didn't know about it

[neon iron]

It’s the movie I think

Asuna perspective

[civic monolith]

Nice

[neon iron]

Yayay

Tysm again

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[autumn plaza]

Hi can someone help

[autumn plaza]

I keep getting 240

Oh its volume

Not area

Oops

Can i have some help

This is all ive done

[granite apex]

Use Pythagoras theorem

Connect the centre of base with vertex of cone

[mellow arrow]

Try to use similarity

[topaz sinew]

@autumn plaza Has your question been resolved?

[mellow arrow]

@autumn plaza You there?

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[distant ingot]

real quick

[distant ingot]

How do I fully factorise and simplify “i”

[ionic oar]

Difference of squares

[mellow arrow]

write 3x^2 as a perfect square

[distant ingot]

I have to use surds in this question

[mellow arrow]

yes

[ionic oar]

Yes

[distant ingot]

what 3x(x) or something?

[autumn plaza]

Omg is that cambridge

[distant ingot]

omg yes

[ionic oar]

$(\sqrt 3 x)^2 - 2^2$

[mellow arrow]

no

[weary valley]

i need help with sets

[distant ingot]

ahhh

[thorny flame]

NEONPerseus

[autumn plaza]

Ayy we love cambridge

[thorny flame]

B-eard

[weary valley]

where can i get help with a set theory question

[distant ingot]

dude

are you in year 10 too or?

[topaz sinew]

Please read #❓how-to-get-help

[distant ingot]

is the answer

[autumn plaza]

yes 😭😭

[mellow arrow]

noo

[distant ingot]

(sqrt 3 + 2)(sqrt 3 - 2)

[distant ingot]

Omg

That’s so rare

wth

[ionic oar]

yes but with xs

[distant ingot]

oh yeah

3x

mb

[jade thunder]

Hey I studied from that textbook too

[thorny flame]

B-eard

[distant ingot]

British? Aussie?

[jade thunder]

Hong Kong doing IB

[distant ingot]

Ahh

I see

[autumn plaza]

u got exams coming up?

[distant ingot]

mhm

[autumn plaza]

im struggling too bad rn

[distant ingot]

week 8

im on week 6 rn

[autumn plaza]

tomorrow for me

[distant ingot]

you in advanced?

[autumn plaza]

yeah

[distant ingot]

I got semesters and i need to study 38 exercises

[distant ingot]

damn me too

I literally just joined

2 weeks ago

[autumn plaza]

really u moved up?

mines not that bad

[distant ingot]

we’re doing chapters 1 to 5

not as hard as you think tho cause they’re pretty much linked together

[autumn plaza]

im struggling with the past exams im too dead

[distant ingot]

gahd

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[maiden gale]

hi can anyone send me questions of linear equation in one and two variables?

[topaz sinew]

Please don't occupy multiple help channels.

@maiden gale Has your question been resolved?

[valid marsh]

please close this channel

[neon iron]

.close

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[bitter mortar]

Gram-Schmidt:

[bitter mortar]

Solution

It's in danish, however the math is in math, so hopefully its understandable.
What happens to the \frac{-3}{7} in lin 3 on the 2nd picture?

Seems to disappear in the next line, the 3 is neglected. However symbolab agrees with the solution

[pastel oracle]

I think they end up canceling out when dividing the vector by its norm

these problems are always so tedious

[bitter mortar]

Problem is, I get the norm of the vector to 7. Then we have -3/7 * 7, which somehow turns to a 1/7 in line 4.

[pastel oracle]

the norm is 3, not 7

which you wrote

[bitter mortar]

I did get 3 on the right side, after the 3/7*7 😄

[pastel oracle]

but I would still think q2 should be multiplied by -1/7, not 1/7 as they have it...

[bitter mortar]

I guess its just fucked

Skipping it :/

Thanks anyways ^^

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[shell orbit]

https://i.imgur.com/rMdIJBo.png

[shell orbit]

I dont really understand what theyre asking me to do here

its a 2 part question

I dont understand what theyre asking with 'To do this, we will first give an equation or inequality for which the solution precisely corresponds to all the values of x for which the function f(x) is defined. Give this equation or inequality.'

specifically the part 'for which the solution precisely corresponds to all the values of x for which the function f(x) is defined'

[pastel oracle]

well when is the function not defined?

[shell orbit]

When the denominator is zero or inside the square root is a negative

[shut obsidian]

When is the denominator zero?

[shell orbit]

I cant think of any value for x to make it zero...

Its probably really obvious

[shut obsidian]

Well, x^2 has to be 7, right?

[shell orbit]

Yes

Ah so like sqrt(7)?

Cause then sqrt(7) squared is 7 and sqrt(7-7) = 0 so the denominator is zero, right?

[shut obsidian]

plus/minus sqrt(7), yes

[shell orbit]

Ah true

But im still not really sure what to put as the answer

Like I still kinda dont understand the phrasing of the question

[shut obsidian]

Now, for which values of x is the square root negative?

[shut obsidian]

This isn't the final answer yet

[shell orbit]

x = √(7) ∧ x = -√(7)

[shut obsidian]

At those values, the square root is 0, yes, now, at which values is the square root negative?

[shell orbit]

x<7

wait no

[shut obsidian]

We are looking for $x$ that satisfy $x^2 < 7$

Then, the square root is negative

Can you express that in terms of x?

[shell orbit]

x < √7

[shut obsidian]

That's one side of the inequality, yes, can you give a left bound on it too, now?
... < x

[shell orbit]

x > -√7 ∧ x < √7

[shut obsidian]

Yes, $-\sqrt 7 < x < \sqrt 7$

So we don't want x to be between -sqrt(7) and sqrt(7) and we don't want x to be sqrt(7) or -sqrt(7)

[shell orbit]

Ah yea theyre the same

Just noticed

[shut obsidian]

Thus, we want $x > \sqrt 7$ or $x < -\sqrt 7$

(Since we don't want $-\sqrt 7 \leq x \leq \sqrt 7$)

[shell orbit]

Okkk I think I get it

Let me try to see if I get the correct answer 1 moment

[shut obsidian]

Intuitively, you want x to be large enough so the -7 doesn't make its square negative, or negative enough so the -7 doesn't make its square negative (the square will always be atleast 0)

[shell orbit]

Oh whoops it didnt accept my answer

I think we solved part 2 aswell haha i think this is the answer for that

So what do I put the answer for part 1

[shut obsidian]

Well, it wants all x for which f(x) is defined, so the domain

That's what we calculated. Perhaps it wants it in a different format?
I'm pretty sure the solution, that $x > \sqrt 7$ or $x < -\sqrt 7$ is correct, since WolframAlpha also agrees

[shell orbit]

https://i.imgur.com/pXKiYEb.png

This is the error message it gave when I put it in

[shut obsidian]

Maybe it wants it in some not-simplified form?
[x^2 - 7 > 0] or [x^2 > 7]

[shell orbit]

https://i.imgur.com/aSefzlC.png

Yeye you were right

This was what it wanted

I think I get it now though!!

Thank you so much for your help, you are very patient and kind

❤️

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[shut obsidian]

np

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[flint fossil]

I set up a base case using n = 1

then I tried to set n = to a variable K, and am now trying to move from k to k + 1

[spare yacht]

ethy just do it tf

[brisk brook]

https://tenor.com/view/mattdog-dog-nose-big-nose-gif-24274385

https://tenor.com/view/dog-crying-gif-23859252

https://tenor.com/view/dog-walter-walter-dog-bull-terrier-long-nosed-dog-gif-26541605

get good kid

[neon venture]

what have u tried ?

[flint fossil]

But I don't think I'm allowed to take the intersections of both sides

[neon venture]

taking the intersections of both sides ????

The only intersection you should do is $\left(\bigcap_{i=1}^{k}A_i \right)\cap A_{k+1}$

[flint fossil]

oh

[thorny flame]

Herels

[flint fossil]

But how can I show the implication holds?

Because I'm just arbitrarily doing that

[neon venture]

$a \in A_{k+1}$

[thorny flame]

Herels

[neon venture]

so from the intersection here, $a \in \left(\bigcap_{i=1}^k A_i \right) \cap A_{k+1}$

[thorny flame]

Herels

[neon venture]

and its done

[flint fossil]

But my initial assumption is that the statement is true for some arbitrary positive integer $k$, i.e., if $a\in A_n$ for all $n\in \mathbb{N}$, then $a\in \cap_{i=1}^n A_i$ for all $n\in \mathbb{N}$ such that $1\leq n \leq k$.

[thorny flame]

Ethy

[flint fossil]

So how do you go from this to $a \in A_{k+1}$

[thorny flame]

Ethy

[neon venture]

I don't go from this to that, they said a € A_n for all n

[flint fossil]

do you know why "The two ns are not the same (note the 2 quantifiers) so your statement for P(n) is incorrect." because doesn't this assume that my statement for p(n) is correct here?

[neon venture]

brb, im in class rn

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[flint fossil]

.reopen

[topaz sinew]

✅

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[fossil widget]

can someone explain this notation to me?

[long stirrup]

2 values in X

ordered

(6,2) → 3, (2,6) → 600

or (R,X)

[odd pagoda]

do you know general function notation f:A->B?

[fossil widget]

I think it means a function which maps a value from A to B?

[odd pagoda]

yes

so + is a function which takes in pairs (x,y) in XxX and gives another element z in X

which in usual function notation would be written as +(x,y)=z

but that sucks as notation so we write it as x+y=z

[fossil widget]

but what is XxX?

[odd pagoda]

the cartesian product of X with itself

[fossil widget]

ahh

I see now, thank you both

.close

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[haughty wren]

Hello! I need help finding the particular solution to:
y‘‘-2y‘+y=e^t

[haughty wren]

I found the complementary solution, but since the particular solution has to be linearly independant, i don‘t know what to do

[topaz sinew]

@haughty wren Has your question been resolved?

[topaz sinew]

@haughty wren Has your question been resolved?

[hidden violet]

Hi there @haughty wren, so you found the homogeneous solution and you want to find the particular solution correct?

[haughty wren]

Exactly

[hidden violet]

Alright, what is the homogeneous solution

Because that's important

[thorny flame]

imTyp0

[hidden violet]

yup, alright so normally we would make a guess of

$y_p = Ae^t$

[thorny flame]

jpsz

[haughty wren]

I tried Ae^Bx and that too

Didn‘t really work

[hidden violet]

yea, but we can't make that guess since both are in the homogeneous solution

we have to guess $y_p = Ax^2e^t$

[thorny flame]

jpsz

[hidden violet]

Does that make sense?

[haughty wren]

How come?

[haughty wren]

Yeah cause linearly independant, but why ^2?

[hidden violet]

we step each guess up by an x just as we did Ae^t, Axe^t

[haughty wren]

Oh.

Is that valid for any guess type? Like for sinx+cosx and polynomials?

And would you multiply both sine and cosine by an additional x if it‘s the case?

[hidden violet]

Well, it is similar. When you guess the particular for sin or cos. You have to do xcos(x), (Ax+b)cos(x), (Ax^2 + bx + c)cos(x) and so forth

A, b and c are just constants that you would guess. Thats a generalization

does this make sense?

[haughty wren]

Kinda yeah thanks. Do you happen to know a website where I could read up on this more?

[hidden violet]

Hmm, I don't. Here are some notes tho. You probably can look up Undetermined Coefficients to see more on it tho

This might clarify what we were talking about

[haughty wren]

It did. Thanks :D I‘ll keep on practicing

[hidden violet]

Awesome!

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[polar blaze]

While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. What are the two smallest positive values of x such that the sine of x degrees equals the sine of x radians?

source: 2002 AIME II P10

[rigid ivy]

!status

[topaz sinew]

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

[polar blaze]

i don't know where to start

[rigid ivy]

The professor tried to calculate sin(x) in degrees, but it was accidentally in radians. However, the answer was still correct

[polar blaze]

so i set an equation and i got sin(x) = sin(x/180*pi)

if i inverse sin both sides i get x = x/180*pi

now what

that gets me x=0

and it has to be positive

the problem states the answer is in the form of $\frac{m\pi}{n-\pi}$

[thorny flame]

starlight

[odd pagoda]

dont just blindly use the inverse sin

remember that sin is periodic

[polar blaze]

but what else am i supposed to do?

the period of sin x = 2pi and the period of sin(x*pi/180) = 360

so the first 2 solutions are in the first 2 periods

im technically not supposed to use graphing caculators but for reference here

yea ik this question is exceptionally hard it's aime p10 after all

<@&286206848099549185>

ok ig insanely hard problem

ping me i have to go soon

bump to prevent inactivity

[topaz sinew]

@polar blaze Has your question been resolved?

[pseudo needle]

@polar blaze

[polar blaze]

yes

perfect timing lmao i just finished dinner

[pseudo needle]

note that sin(theta) = sin(pi-theta) = sin(2pi+theta)

these are the least positive angles such that they're equal to each other

[polar blaze]

but idk how to solve the first part which is what is the first sol

[pseudo needle]

what?

[polar blaze]

what is the first solution

(without using graphing calculator)

[pseudo needle]

do you mean the lesser one?

[polar blaze]

yes

and any solution

cuz like

[pseudo needle]

it's 180pi/(180+pi) I believe

[pseudo needle]

do you know how to solve it given this is what matters

[polar blaze]

hm that?

uh

[pseudo needle]

what is that in response to?

[polar blaze]

how does that help

[polar blaze]

this

cuz im dealing with pi/180*x

[pseudo needle]

so let's let like, I don't know, a be the least angle that we're looking for

ok

x

[polar blaze]

no way i can convert a simple x to that

yep ok

[pseudo needle]

x is equal to pi-pix/180

[polar blaze]

?

where did you get that

[pseudo needle]

x is equal to xpi/180 radians right

[polar blaze]

yes

[pseudo needle]

so we want an x such that sin(x) = sin(xpi/180)

[polar blaze]

yes

[pseudo needle]

and obviously that gives the solution of zero

[polar blaze]

thats where i got to

[pseudo needle]

but zero isn't positive

[polar blaze]

exactly

[pseudo needle]

and since sin(x) = sin(pi-x)

and since sin(x) = sin(2pi+x)

[polar blaze]

yep

[pseudo needle]

we instead want x such that sin(x) = sin(pi-xpi/180)

and sin(x) = sin(2pi+xpi/180)

[polar blaze]

oh

[pseudo needle]

now do you see it?

[polar blaze]

i see

[pseudo needle]

because any value like sin(x) = sin(4pi+xpi/180) is obviously greater than those two

[polar blaze]

yeah

[pseudo needle]

so can you solve it now?

[polar blaze]

i'll see

[polar blaze]

no can't solve it still

cant tell why it helps

[pseudo needle]

just take the inverse sine of both sides at that point

and solve it like a linear equation

[polar blaze]

owh

OH

[pseudo needle]

can you solve x = pi-xpi/180

[polar blaze]

wow you're so smart

[pseudo needle]

this wasn't too hard

[polar blaze]

this is why im bad at math

lmao

[pseudo needle]

there's many things i can't do

[polar blaze]

im unfamilliar with trig identities so

[pseudo needle]

it's alright because you're challenging yourself!

[polar blaze]

im new to trig

[pseudo needle]

I see

[polar blaze]

im assuming the 2nd sol would be x+2pi

right

[pseudo needle]

what

sin(x) = sin(2pi+xpi/180) this is the second smallest one

[pseudo needle]

that one would be the third solution

[polar blaze]

oh ok i see

thanks!

.close

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[rain mango]

what is the formula to solve this ?

[haughty wren]

graph g(x) on the same plane that you have f(x) on

it should help you see some things

[rain mango]

is related to "Monotonic Sequences" but it is not the introduction so no idea of what to look for

[haughty wren]

i'm telling you, graph g(x). You'll definitely notice what you need

[topaz sinew]

@rain mango Has your question been resolved?

[rain mango]

i pass from that, to this. i don't remember how to graph, I only want to know how to complete that guide before taking one exam

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[icy pilot]

$4/30mi *

[icy pilot]

How do I convert 4 dollars per 30 miles

into how much 1 mile costs

dimensional analysis

[keen venture]

1 mile × |4 dollars / 30 miles|

Miles cancels out, leaving you with dollars

[icy pilot]

Honestly I don't get the math

.close

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[neon iron]

What is special about $$\sqrt{x^3}$$

[thorny flame]

Brandon H

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[orchid sonnet]

i need help with this question

(im banned from the physics server)

.close

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[inner crater]

I don't get that part where did they get 2^17 from

<@&286206848099549185>

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@inner crater Has your question been resolved?

[sweet shard]

"Twice the value of the previous prize"

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[stark oyster]

asks me to do it with substitution, I can t really see any substitution maybe some trigonometric one?

[stark oyster]

but that x^2 seems to make it worse

[noble laurel]

Where is this question from?

Can you post the original question

[stark oyster]

thats the thing this is the whole thing somebody sent me and my first thought was that It can't be done in the was he asked me

so I am asking is it even possible by the seems of it

[silk delta]

I think you can try integration by parts?

[noble laurel]

you can't integrate this using elementary functions

[stark oyster]

got it!

.close

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[snow token]

When finding the two straight lines of f(x)=2x-1+|3x-6|. I get y=5x−7 for x > 2 and y=−x+5 for x ≤ 2. Is there an actual difference when saying y=5x−7 for x ≥ 2 and y=−x+5 for x < 2? Why can't I place the x ≥ 2 instead of > for the line y=5x−7, and x < 2 for the line y=−x+5?

[gray ridge]

from what i learn, if f is continuous at x=2, then it doesn't matter

[snow token]

Ok thanks

.close

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[inner crater]

.reopen

[inner crater]

How about the 17 exponent

[topaz sinew]

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[modern epoch]

6 years will remain .75, 3 years will be 1.061, 1 year will be 1.336

What I got for Part A wanted to double check

But idk what B wants me to do I’m confused

[topaz sinew]

@modern epoch Has your question been resolved?

[modern epoch]

<@&286206848099549185>

[grim rain]

after 6 years, 0 mg

[halcyon spade]

it says that it has a half life of approximately 6 years

that means after 6 years it halfs its mass

[modern epoch]

Yeah

[halcyon spade]

so 1.5/2 = 0.75mg

[modern epoch]

This is what I got to double check

[halcyon spade]

1.5 / 2 ^ 1 for 6 years

1.5 / 2 ^(0.5) for 3 years

1.5/2 ^ (0.166)

for 1 year

@vast bay your results are correct

[modern epoch]

Ty

Do you know part B

Cause that’s where I’m confused at

[halcyon spade]

make a function that models the amount of radium remaining after t years

so basically

f(t) = something right

if we have t = 6

f(6) = 0.75

f(3) = 1.061

f(1) = 1.336

we need to make a function

that works like this

[modern epoch]

That solved for all?

[halcyon spade]

no xD

we need to make a function right

@vast bay what does the graph of half life look like?

[modern epoch]

Half every 6 hrs

Yrs*

[halcyon spade]

it might look something like this yeah?

[modern epoch]

Yeah like a extreme decay

But function is asking for a graph?

[halcyon spade]

kinda yeah

do you remember how to calculate the thing

so if i said find me the mass after 3 years

what would you do to calculate?

starting mass is 1.5

and half life = 6

[modern epoch]

1.5(1/2)^3/6

[halcyon spade]

hm

[halcyon spade]

i did it like this

so your starting mass

[main iron]

im in 9th

[halcyon spade]

9th grade?

[main iron]

yeah

[halcyon spade]

i guess we are the same then

xD

[main iron]

k

[halcyon spade]

@vast bay so it would be for after 3 years
1.5/ 2 ^ (3/6) right?

[modern epoch]

Yeah

[halcyon spade]

notice that 3

if i said "after t years"

what would it be now?

[modern epoch]

T/6?

[halcyon spade]

1.5 / 2 ^ (t/6) ?

[modern epoch]

Yes

[halcyon spade]

exactly!

[modern epoch]

Oh so that’s the function?

[halcyon spade]

it should be yeah

f(t) = 1.5 / 2^(t/6)

[modern epoch]

f(T)=1.5(2)^ (t/6)

Could it be written as that as well

[halcyon spade]

that divide sign

is important

[modern epoch]

Alr gotchu

Oh yh cuz mine is

Multiplying right

[halcyon spade]

wait

now lets make it look "prettier" @vast bay xD

[modern epoch]

Alr fs fs

[halcyon spade]

you could write it as

1.5 x 2 ^ (-t/6)

[modern epoch]

Alr I’ll do that

[halcyon spade]

that just makes it look fancier

[modern epoch]

But it’ll just mean the same thing right?

[halcyon spade]

yeah

try it with t = 6

1.5 x 2^(-6/6)

1.5 x 2 ^ (-1)

1.5 x 1/2

0.75

same answer

[modern epoch]

Alr ty man

Thx for the help

.close

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[halcyon spade]

np

[halcyon spade]

.close

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[neon iron]

i got the row reduced form

now i dont understand what c is tryna get me to do

write the solution of the system in vector form

the x_0 is the constant

then the span part are the directions

https://math.libretexts.org/@go/page/63386

is x0 (4 0 3 2 0) ?

@neon iron

[pastel oracle]

assuming your work is correct, yes

[neon iron]

ok ty

row reduced matrix was checked dw

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[neon iron]

Help plz

[muted spire]

!status

[topaz sinew]

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

[neon iron]

1

[muted spire]

okay

what do you know about a tangent line to a circle ?

[neon iron]

Angles from tangent to radius are 90°

[muted spire]

ok its good

now name the tangent point B

[neon iron]

Oh I get it now

90-29

[muted spire]

yep just an isosceles triangle

[neon iron]

Ok thank you

👍

[muted spire]

🤝

[neon iron]

.close

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[rugged idol]

just asked this in a channel that I already closed but how is it possible to say the sum goes to k instead of n ??
In my book the sum goes over every k in X(Omega) and that should be from 0 to n or not ?

[rugged idol]

damn in another example it even uses k too , Im so confused rn

[drifting swift]

$\sum_{i \in \bR} P(X = i, Y = k - i)$

[thorny flame]

Ann (glomed)

[drifting swift]

if you want to include everything

like EVERYTHING everything

maybe $\sum_{i \in \bZ} P(X=i, Y=k-i)$

[thorny flame]

Ann (glomed)

[drifting swift]

X=i happens with positive probability iff 0 ≤ i ≤ n

and Y=k-i happens with positive probability iff 0 ≤ k-i ≤ n

do you follow y/n

[rugged idol]

Im sorry missed your response

[rugged idol]

yes

ah sorry for answering late. I think I got it, but still its weird. I'll ask a tutor about that. Thanks a lot anyways @drifting swift !

.close

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[woeful depot]

An idea to show that f is C1
I thought about doing a series expansion but it seems complicated

[odd pagoda]

series expansion (which needs infinitely often differentiable) to show C1

[woeful depot]

ah ye alpha between 1/2 and 1

[woeful depot]

i know that
f is Cn => can perform series expansion at order n

[odd pagoda]

so you know that this implication is only one way and want to use it in the other somehow?

[woeful depot]

but for n=1 it's an equivalent, right?

[odd pagoda]

is it? maybe. not sure

I would just diff and show derivative is cont

diff is obvious

[woeful depot]

,w derivate sin(pi*sqrt(x))/x^alpha

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[woeful depot]

hum there's 1/(2x^a-1)

the rest is obviously continue

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[fiery gyro]

evaluate.

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What step are you on?
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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

[sweet shard]

,w int 0 to pi/2 arccos(cos(x)/(1+2cos(x))

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@fiery gyro Has your question been resolved?

[sweet shard]

Try using some of the identities for arccos
https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions

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[oak root]

What is limit of the sequence (a_n) where a_n is the solution to f_n(x)= x^n - x - 1 = 0 (here f defined when x>=1)

[oak root]

We know that (a_n) is decreasing

And a_n > 1

[sweet shard]

,w roots x^2 - x - 1

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[sweet shard]

is a_2 = (1/2) ( 1 + sqrt(5) )?

[oak root]

Yes

[sweet shard]

usual thing to is to assume a limit exists and solve an eqn

but i don't know if that'll work here

[oak root]

is the limit 1? Because a_n >1 and decreasing but I don’t think it’s enough ??

[sweet shard]

No it's not enough

[oak root]

So what can I use

Squeeze theorem ?

[sweet shard]

Maybe try factoring x^n-1

Then taking that sum to the limit as n goes to inf

And you know that'll equal limit of a_n

[oak root]

x^n -1?

[sweet shard]

Yes

[oak root]

I dont Know how to do that, are we trying to find an expression for a_n?

[sweet shard]

Don't think that's necessary

[oak root]

ok thanks

so the sum is

sum of x^k when k goes from 0 to m-1

@sweet shard so what are we doing here, once we get that geometric sum; we can see that its limit is +infinity when m approaches infinity?

uh

wait

no i dont know

[sweet shard]

Just giving you ideas to try

[oak root]

idk

[oak root]

bruh idk

<@&286206848099549185>

[sweet shard]

?

[oak root]

i don't know what to do with that sum

[sweet shard]

If a_n converges, the left side isn't infinity

[oak root]

the problem is that i don't understang why is it related to a_n

like why do we calculate this

[sweet shard]

a_n^n - 1 = a_n

the limit of the right side is what you're looking for

[sweet shard]

Yea this approach proves that limit is 1

[oak root]

so it's limit of (a_n)^n-1 when n approaches infinity

and (a_n)^n - 1=(a_n - 1)*(a_n^m-1 + ....+1)

since a_n > 1, (a_n)^n semms to be infinite when n is bigger

How do i find the limit of (a_n)^n - 1 when i don't know how a_n is expressed

@sweet shard

<@&286206848099549185>