help-49

[fresh sparrow]

@last slate that should more or less be it

[last slate]

well i meant tetration

i think that's the thing

yeah

thank you very much,, wow

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[cerulean arch]

could someone let me know

where im going wrong here in the end

ljke the middle left part

i cant seem to finish up the proof or my algebra is terrible

[misty cairn]

@cerulean arch

[cerulean arch]

yes

[misty cairn]

find the min value where x>f(x)-9, then set a minimum for N so x is always greater

then N<€

does this make sense?

[cerulean arch]

yesss, got it, ty

appreciate it

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[misty cairn]

@cerulean arch looking back, the answer I gave you was wrong

[misty cairn]

you need to express n in terms of epsilon

it should be correct now, I hadnt done epsilon delta in a while, I hope that helps

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[cerulean arch]

ahahahaha no worries, thanks again tho

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[next viper]

Can i make this into simpler fraction by

doing the inner inner outter outter thing

with the fact that there is subtraction?

so 2p / 1-24 ?

[olive matrix]

make $1-\f{25}p$ one fraction first

[grand pond]

hayley is not layla

[next viper]

ž

[olive matrix]

yee

[next viper]

🫡🫡

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[lapis marsh]

 A = { {n \leq275   v  m=2n| n, m \in ℕ}``` what does v mean in the middle?

[hearty rune]

do you have an image?

[lapis marsh]

The cardinality of A = infinity

[hearty rune]

it could mean union?? Im not particularly sure though in all honesty

[lapis marsh]

there is no such thing as V?

[hearty rune]

it can mean 'or'

but again idk if thats what it entails here, just my guess

[tribal temple]

(they likely mean "or" here, if the cardinality of A is infinite)

Strange layout though, usually it'd be stated the other way round, "n and m are natural numbers such that n is at most 275, or m is 2n"

[midnight plank]

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@lapis marsh Has your question been resolved?

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[leaden matrix]

Not sure which one to use, can’t seem to get either one as only a function of x or only a function of y

[feral sedge]

you can assume the integrating factor is x^n * y^m and multiply your equation by that

then suppose it is exact and set up a system of equations

[midnight plank]

@leaden matrix Has your question been resolved?

[leaden matrix]

Tried it out but there’s no n and m that make my equation exact

Can’t find out where I messed up

[feral sedge]

you are tricking yourself

i see at the start of the page you have 8y * x^n * y^m, which you differentiate to obtain 8m * x^n * y^m

this is incorrect and at the start instead of 8y * x^n * y^m it would be more beneficial to write 8 * x^n * y^(m + 1)

[leaden matrix]

Tried it again but i still can’t find an m and n that work

Is there a proper way of doing it rather than just guessing it?

[feral sedge]

there is no guesing here, the problem directly tells you x^n * y^m is an integrating factor

from here, you can algebraically determine the value of n and m

[leaden matrix]

I cant really cancel anything out if I equate the 2 equations tho

[feral sedge]

the idea i you have to compare coefficients

[leaden matrix]

thats what I tried doing here

but it ends up working for 2 of the terms and not working for the other 2 terms

<@&286206848099549185>

[midnight plank]

@leaden matrix Has your question been resolved?

[midnight plank]

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[hidden jacinth]

I found (18!)^2 in base 10 has 32 digits so I think the problem is to find the largest integer n such that 15^n has less than 32 digits and then add 1 to n.

I know you could do this with logarithms, however I am trying to do it without a calculator.

[modern sapphire]

It might help help you to notice that 15^4 is approx 50000

[hidden jacinth]

Ok so 15^8 is approximately 25 * 10^8, 15^16 is approximately 625 * 10^16, 15^24 is approximately 15625 * 10^24. Then 15^25 is about 22000 * 10^25 = 22*10^28, 15^26 is about 33 x 10^29 and 15^26 is about 50 x 10^30, but (18!)^2 is about 40 x 10^30.

So n = 26

and the base 15 number will have 27 digits?

[modern sapphire]

yes

[hidden jacinth]

Ok

for this type of problem should I also estimate the margin of error

to see how far my calculations are off by

because I estimated 15^26 to be about 50*10^30 and (18!)^2 to be about 40 x 10^30. Those seem very close and I could easily have rounded too high or too low

[modern sapphire]

Yes, estimating error margins is the main underlying theme for these kinds of problems

[hidden jacinth]

ok

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[keen seal]

what do i do with the I

[prime hornet]

[prime hornet]

I is the identity matrix

[tribal temple]

(are you meant to find A?)

[keen seal]

ah yes sorry i frogot to mention

i did the inverse but idk what to do with I

i thought of just making it -I in the right

[tribal temple]

The inverse of both sides? If so, then sure

[keen seal]

I end up with 1/13 [MATRIX] -I

and uh

[sharp coral]

,, I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}

[grand pond]

cloud

[keen seal]

yes

but the thing

result is this

and idk how even if i -1 the left up and right down

idk hwo to reach this

[lavish venture]

what did you get for the inverse?

[keen seal]

5 -2

-4 -1

-1/13

[lavish venture]

the determinant of the original matrix was -13

[sharp coral]

double check the bottom right entry as well

[lavish venture]

yea that too

[keen seal]

here i corrected them

which leaves me with 2A + I=

the matrix i mentioned above

[lavish venture]

^

[keen seal]

oh sorry

here its corrected

ok thing is

they did the /2 AND THEN

minus the identity matrix

wait

no

[tribal temple]

Well, if you subtracted the identity matrix first then divided by 2, you will get their answer too

[keen seal]

if they /2 and then minus identity

its a different asnwer

[tribal temple]

Well, if you divide everything by 2, then the identity matrix is scaled by 1/2

Do you have what they did?

[lavish venture]

when was the last time you opened a help channel? i haven’t seen a syrex channel in ages

school started ig

[keen seal]

oh yeah forgot oops\

okay then makes snese

ty!

[keen seal]

yesterdy

for a week now i believe

but yes i was gone for like

3 months

O_O

[lavish venture]

that explains it

[keen seal]

even lost my active role

rip

[midnight plank]

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[keen seal]

what do i do if it isnt p(A)

[valid moat]

whats the given x?

[tribal temple]

What do you mean "it isn't p(A)"?

[keen seal]

basically im not sure what to do here

do i assume x is A?

ah then -2

is it -2 to every entry?

[tribal temple]

Yep, replace all x's with A's (and any constants with the constant multiple of the identity matrix)

[tribal temple]

-2I instead

[keen seal]

ahh okay ty!

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[twilit field]

$\lim_{x\ \to\infty}\frac{2x-1}{x-2}$

[grand pond]

Veni, vidi, perii

[twilit field]

Trying to prove this is 2

so $|\frac{2x-1}{x-2}-2| < \varepsilon$

[grand pond]

Veni, vidi, perii

[twilit field]

or $\frac{5}{|x-2|} < \varepsilon$

[grand pond]

Veni, vidi, perii

[twilit field]

uh, so

$|x-2|> \frac{5}{\varepsilon}$

now what?

so $||x|-2|>\frac{\varepsilon}{5}$

[grand pond]

Veni, vidi, perii

[twilit field]

so $|x|> \frac{\varepsilon}{5}+2$

[grand pond]

Veni, vidi, perii

[twilit field]

is this fine

[hard shard]

watch your step

[twilit field]

wydm

[grand pond]

Veni, vidi, perii

[twilit field]

so |x|> 5/ eps +2

[hard shard]

i dont think this is the most useful route

https://youtu.be/AfrnYS5S8VE?si=e1qhn1WtMR5YUImL try looking at q5

[twilit field]

hmm

Thanks

so this is just 5/(x-2)<eps

which feels sus

[tribal temple]

(Remember the whole “you can choose N” thing here - in this case, such that you can drop the absolute value bars)

[twilit field]

so N= 5+2eps/5

[twilit field]

What am I doing wrong

I'm getting N = (5+2eps)/5

[tribal temple]

Remember you initially had absolute value bars here

[tribal temple]

To have dropped them, you wanna make an assumption on N

[twilit field]

N>0

N>2

[tribal temple]

Yep

[tribal temple]

Max of 2 and this

[twilit field]

okay

so let N= max(2, 5+2eps/5)

then what

[tribal temple]

Then reverse your steps

[twilit field]

yes

so x>5+2eps/5

[runic hamlet]

btw its 3/(x-2) not 5/(x-2)

[twilit field]

oops

yeah

thanks

Yup

done

thanks

would doing an epsilon-N proof for this teach me anything

[runic hamlet]

pain, sure

you would definitely need to be able to bound stuff to simplify it

doing it without that would suck

[twilit field]

Hmm, I'll think about that

[twilit field]

Is that doable in epsilon -N limiits

[runic hamlet]

not only doable, its very essential

for any slightly harder limit

[twilit field]

hmm

okay

Let x^2+1<a

[twilit field]

Can I have a hint

I thought of saying x-1<a and using that to bound everything

[runic hamlet]

x is supposed to go to infty

you cant bound x

[twilit field]

yes

I can bound 1/x though

[runic hamlet]

ok however you got 1/x

[twilit field]

I factored out x^2 from the numerator and denominator

[runic hamlet]

uhm ok

[twilit field]

how else am I supposed to bound this?

[runic hamlet]

show your work

[twilit field]

$\frac{x^2(1+1/x^2)}{x^2(2-3/x-2/x^2)}$

[grand pond]

Veni, vidi, perii

[runic hamlet]

what happened to the 1/2

[twilit field]

oops

yeah

ok, using the 1/2

we get

Ok , I messed up

let me try again

so that simplifies to

$\frac{4+3x}{2(2x^2-3x-2)}$

[grand pond]

Veni, vidi, perii

[twilit field]

ok, no idea

[runic hamlet]

so now you want to bound this by something else which is simpler

and that something else you can then bound by epsilon

so how can you make 4+3x bigger while making it simpler

how can you make 4x^2-6x-4 smaller while making it simpler

and still in a way that the quotient goes to 0

[twilit field]

I could probably re-write this as $\frac{1}{2(x-2)}+ \frac{x-2}{2(2x+1)(x-2)} + \frac{5}{2(2x+1)(x+2)}$

[grand pond]

Veni, vidi, perii

[twilit field]

I could compare the last one to 1/x^2

[runic hamlet]

overthinking

[twilit field]

Make the denominator x^2

[runic hamlet]

yes

[twilit field]

and the numerator 4x?

[runic hamlet]

yes

(each of those for big enough x)

[twilit field]

so (4+3x)/(4x^2-3x-2)<4/x

[runic hamlet]

yes

[twilit field]

wait, so that implies $0<\varpesilon$

right

[grand pond]

Veni, vidi, perii
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[runic hamlet]

what

[twilit field]

$\frac{4+3x}{2(2x^2-3x-2)}< \varepsilon$

[grand pond]

Veni, vidi, perii

[twilit field]

but

$\frac{4+3x}{2(2x^2-3x-2)}< 4/x<\varepsilon$

[grand pond]

Veni, vidi, perii

[twilit field]

I don't see how that helps me

ooh

nvm

x>4/eps

so N= 4/eps

[runic hamlet]

well and N big enough for this

but yes

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

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[blissful drum]

,rotate

[blissful drum]

i dont understand why it's 12 x 3 x 3 not 12 x 3 x 12

[muted panther]

Because you work in the case in which the shirt is pink

And there are 3 possible cases then

For the shirt

[blissful drum]

yeah so the ties cant be red and the waistcoat cant also be red right

[muted panther]

Exactly

[blissful drum]

so what im thinking is so when shirt is pink thats 12 and no red waistcoat so 3 and no red ties so 15-3 = 12 --> 12 x 3 x 12?

[muted panther]

Why the first 12?

How many pink shirts do you have? 3 or 12?

[blissful drum]

3

m

oh

wait hold on let me process this

ok i get it thanks

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[alpine gyro]

Find all the complex solutions to $$iz^3=8$$ in both $re^{i\theta}$ and $a+ib$.

[grand pond]

Good

[hollow oyster]

what have you tired?

[alpine gyro]

I found that $z^3=-8i$, so that $z^3=8e^{i3\pi/4}$ such that $z=2e^{i\pi/4}$, and the one solution is $w_0=(2e^{i\pi/4})^{\frac13}$

[grand pond]

Good

[alpine gyro]

But this didn't give me the right solution.

[alpine gyro]

Is this the right path?

[hollow oyster]

till z^3 its all aright

but $z=\sqrt[3]{8}e^{i\left(\frac{\frac{3\pi}{2}+2\pi k}{3} \right)},k={\ 0,1,2 \ }$

[grand pond]

convergence

[alpine gyro]

I saw my mistake, thought 3pi/4 as 3pi/2.

[alpine gyro]

But isn't $z=2e^{i\pi/2}$ and since it's a 3-root, so $w_k=w_0\cdot( e^{i2\pi/3})^k$, $k\in{0,1,2}$

[grand pond]

Good

[hollow oyster]

what is $w_0$ here?

[grand pond]

convergence

[alpine gyro]

$w_0=\sqrt[3]{z}=\sqrt[3]{2e^{i\pi/2}}$

[grand pond]

Good

[hollow oyster]

and no z is not exactly $2e^{i\pi/2}$ its more of a principal branch but ok..

[grand pond]

convergence

[hollow oyster]

if you let $w_0=2e^{i\frac{\pi}{2}}$ then you can write $w_k=w_0 e^{\frac{2k\pi}{3}} ,k={0,1,2}$

[grand pond]

convergence

[alpine gyro]

But last time I did $z^3=\sqrt{2}(1+i)$ I needed to do this way.

[grand pond]

Good

[hollow oyster]

hm?

[alpine gyro]

Or else the solutions didn’t match.

[hollow oyster]

no like how did you get this?

[alpine gyro]

It was a question I did before

My bad, you're right

[hollow oyster]

np

[alpine gyro]

I didn't need to multiply z with e^i2pi/n

Thanks for the help:)

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[sick lodge]

how did we go from that to this

i tried the b^2-4ac that got me nowhere

[sinful trout]

uh

just

factorize?

[sick lodge]

im trying to know

how he got from first line to second one

[sinful trout]

he

factorized

[sick lodge]

how?

[sinful trout]

what do you mean "how"

[sick lodge]

idk how to factorise

[sinful trout]

the method?

[sick lodge]

from first line to second

[sinful trout]

i'll link a video for that

are you not doing eigenvalues

[sick lodge]

u put the lamda in factor?

[sinful trout]

what?

[sinful trout]

https://www.youtube.com/watch?v=9MUDAr1PnlE&ab_channel=TheOrganicChemistryTutor

watch that

[sick lodge]

ty

[sinful trout]

np

[sick lodge]

ren

can u plz explain me more?

ik what the video is about but it didnt help

ik how to go from second line to first

and from first line to second idk how

[sinful trout]

okay so basically

factoring is a way to turn a complicated polynomial into well its factors

for quadratics such as this in the form x^2 + bx + c, you try to find two numbers such that r+s = b and rs = c

then the quadratic is simply (x-r)(x-s), which can also be written as (r-x)(s-x)

[sick lodge]

how do u find the r and s

its like a system?

[sinful trout]

i guess?

i mean you'll have to solve the very quadratic equation ur doing

[sinful trout]

in situations like these

it's very obvious what r and s are

in others you might use the quadratic formula

since r and s are just the roots of the quadratic equation

[sick lodge]

i tried delta thing

it got very messy

[midnight plank]

@sick lodge Has your question been resolved?

[midnight plank]

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[vocal steeple]

how do i solve this?

[vocal steeple]

can someone to do it visiually for me

[novel lion]

write them in exponents form

then sum

[vocal steeple]

i got 8^13/12

but i feel like it can be shortened more but idk how

[novel lion]

well, thats it

[vocal steeple]

the solution shows this:

[novel lion]

$\sqrt{8} = 2\sqrt{2}$

[grand pond]

Astar777

[novel lion]

cube root of 8 = 2

so thats 4sqrt(2) now

i dont get how you would take out 2 from the last one to make 4*2 = 8

[vocal steeple]

i found this but i dont get how it goes from 8^1+12 to that

[grand pond]

faiyrose

[vocal steeple]

yeah but after that

how does it become 2^3/12

[novel lion]

8 = 2^3

[gentle aspen]

Its correct, just write 8 as 2^3

You will get 2^ 13/4

[slim steeple]

Don't go that way, it will be difficult for them to understand

[slim steeple]

This is better to understand

[gentle aspen]

Gotcha

Just thought we gotta prove LHS =RHS

[slim steeple]

Breaking 8^13/12 as 8 * 8^1/12

[novel lion]

oh bruh

didnt even notice

[slim steeple]

Nah it's fine,that way is even more tricky to understand
They will end up as 2^13/4

Then its
2³ * 2¼

For us it's easy to interpret

[gentle aspen]

Its 2^3 * 2 ^(1/4)

[novel lion]

agreed

[vocal steeple]

i understand now

thanks

.close

[midnight plank]

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[slender zenith]

An archer shoots an arrow at a 72.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow.
(a)
At what angle in degrees must the arrow be released to hit the bull's-eye if its initial speed is 31.0 m/s?

[slender zenith]

answer using vector componenets

[midnight plank]

@slender zenith Has your question been resolved?

[last slate]

still need help ?

[open mason]

do we assume that the effects of air resistance are ignored

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[hoary wind]

can someone help me solve the last one

[midnight plank]

@hoary wind Has your question been resolved?

[hoary wind]

no

[brazen palm]

green one?
Proof that for all natural numbers n,
4•6^n+5n-4 is divisible by 25?

[hoary wind]

yes

[brazen palm]

what have you tried?

[hoary wind]

gave a random value to n

then calculated to see if it is divisible

[brazen palm]

hmmm

[hoary wind]

then we have to demonstrate that is divisible for all numbers

[brazen palm]

have you heard of Mathematic Induction?

[hoary wind]

yes

[brazen palm]

i think MI is very useful for such questions

[hoary wind]

we studied it last week but i kinda got questions since on different exercise we use different techniques

[brazen palm]

oh i see, so you need to use different techniques for this question, am i right?

[hoary wind]

possibly

i just need to get to the point to see if its divisible for all numbers or not

when we swap the number with n

[brazen palm]

then you might try modula arithmetic and binomial expansion to show

seems like doable

[hoary wind]

oh

[brazen palm]

e.g.
4(5+1)^n + 5n - 4
and expand

it looks neat

[hoary wind]

hmm

can u solve it fully so i can see what u did

[brazen palm]

i can try

4(5+1)^n +5n -4
=4(5^n + (nC1)5^(n-1) + (nC2)5^(n-2) + ... + (n(n-1)/2)5² + 5n +1)) + 5n -4
=4(5^n + (nC1)5^(n-1) + (nC2)5^(n-2) + ... + (n(n-1)/2)5²) +20n +4 + 5n -4
=4(5^n + (nC1)5^(n-1) + (nC2)5^(n-2) + ... + (n(n-1)/2)5²)
which is divisible by 25

[hoary wind]

what does the C stands for?

[brazen palm]

ooo, sorry was in another channel,
C is combination

[brazen palm]

it's in binomial theorem
https://www.mathsisfun.com/algebra/binomial-theorem.html

[hoary wind]

oh okay

we didnt use this technique before

[brazen palm]

oh i see, it's not actually that hard 😛

[hoary wind]

yeah but im not sure if the teacher will accept this method

[brazen palm]

hmmm

[brazen palm]

any other methods that your teacher is suggesting?

[hoary wind]

wait imma show how we did other examples

if u don’t understand something lmk

wait hold on i think i mixed with other equations

[brazen palm]

so many 😆

[hoary wind]

yeah, this method is taking 2 pages of writing in order to solve

[brazen palm]

these looks like Mathematical Induction

[hoary wind]

yes they are

[brazen palm]

yea, but i thought you wanna use other methods :3

[hoary wind]

i want this method in order to get accepted from my teacher

[brazen palm]

ahhh i get it now

[brazen palm]

so basically you do the initial case
then you assume case k=n is true,
then

(4•6^(k+1)+5n-4) : 25
( 6(4•6^(k)+5n-4) +6(-5n+4) +5n-4) : 25

can you continue from here?

[hoary wind]

this part is the only part i could actually get done:(, but i cant go further

[brazen palm]

oh ok

lemme try

6(P(k))-30n+24+5n-4
6(P(k))-25n+20
6(P(k))-25(n+1)+5+20
6(P(k))-25(n+1)+25 : 25

[hoary wind]

thanks for helping

[brazen palm]

see if those make sense?

[hoary wind]

imma ask my teacher to be sure

[brazen palm]

sure!

[midnight plank]

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[calm shadow]

I don't know where to start here

[fallow scarab]

,tex .exp rules

[calm shadow]

This is an example I'm given for how to solve the above equation:

[grand pond]

[sinful trout]

i mean

the fourth derivative will just be differentiating the third derivative

and you can use the power rule there

since that is merely x^0.75

[calm shadow]

yes I understand that it's x^(3/4) so far

(or are you saying that's the final answer?)

I was thinking I had more steps

[sinful trout]

no

just

differentiate x^(3/4)

[calm shadow]

I don't know how to

[fallow scarab]

,tex .diff rules

[grand pond]

[fallow scarab]

Power rule

It's valid for all nonzero real numbers, not just integers

[calm shadow]

I gotcha, thank you

Let me see if I can figure it out now

[sinful trout]

except n = 0

for obvious reasons

[fallow scarab]

Oops yes thank you

[calm shadow]

(3/4)x^(-1/4)

easy peasy?

[sinful trout]

okay first off

that notation

is not good

that looks like you've written

[calm shadow]

(3/4)x

[grand pond]

ren

[sinful trout]

use 0.75 instead

but yes

i get what you mean

[calm shadow]

gotcha

[sinful trout]

and your answer is correct

[calm shadow]

thanks

.close

[sinful trout]

gj

[midnight plank]

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[hoary wind]

wait can u explain how this proves that every number is divisible

[brazen palm]

sure, you can open a new channel, we can talk there

[midnight plank]

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[potent forum]

Could someone confirm if my work here is correct...?

for c) "addition" is defined as matrix multiplication. However, matrix multiplication is not commutative, and addition in a vector space must be commutative. Therefore, matrix multiplication cannot serve as the addition operation for a vector space. Thus, it is not a vector space?

[summer terrace]

i agree with this

true, but you didnt check whether it satisifes the remaining 6 axioms of a vector space

remember, theres 10

[potent forum]

c) is correct right?

[summer terrace]

yeah, agreed, nice

[potent forum]

ok, ty!

.close

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[still moss]

anyone here transgnder

happy pride month

@summer terrace

[midnight plank]

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[alpine gyro]

Factorize $$z^3+8=0$$

[grand pond]

Good

[alpine gyro]

I got the roots $(-1+i\sqrt3)$, $-1$ and $\left(-\frac12 + i \frac{\sqrt{3}}{2} \right)$.

[grand pond]

Good

[alpine gyro]

Would it just be $z^3+8=[z-(-1+i\sqrt3)][z-(-1)][z-\left(-\frac12 + i \frac{\sqrt{3}}{2} \right)]$

[grand pond]

Good

[jaunty canopy]

i think you meant -2

[jaunty canopy]

but yeah that's one valid factorisation

(if we're working in $\mathbb{C}$)

[grand pond]

LY

[jaunty canopy]

for reals tho i would prefer keeping it as (z+2)(z^2-2z+4)

[alpine gyro]

I mean, I made -8 to $8e^{i\pi}$, and $w_1$ gives me -1

[grand pond]

Good

[alpine gyro]

If it were -2, why are there no denominator of 2?

[jaunty canopy]

why?

i mean -1 is definitely not a root

cus the cube root of -8 is -2 and not -1

[alpine gyro]

You're right, I made a mistake.

Yeah, I got the answer. Thanks.

[jaunty canopy]

nw

[alpine gyro]

.close

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[crude laurel]

Can someone help we with this problem?

[celest saddle]

just split it up into 2 integrals

bc its a piecewise

because its f(x)=x when x<1 you can integrate x from -2 to 1 and since f(x)=1/x when x>1 you can integrate 1/x from 1 to 7

and add them

[crude laurel]

Oh ok thank you!

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[vocal steeple]

why is (lg x)^2 / lg x = lg x but ln x^2 / ln x is not ln x but 2?

[sharp coral]

because ln x^2 = ln(x^2)

[vocal steeple]

i dont understand

wait i kind of understand

but i still dont get why i get different answers

would it be the same if the second task was (ln x)^2 / ln x?

would it then become ln x

[sharp coral]

it would be ln x in that case

[vocal steeple]

but since its not the whole thing but only x thats ^2 it becomes 2?

[sharp coral]

yes, because ln(x^2) = 2 ln(x)

[vocal steeple]

so (lg x)^2 isnt 2 lg x ?

[sharp coral]

it is not

[vocal steeple]

because its not directly connected to the x?

@sharp coral

[sharp coral]

for (lg x)^2 the square is applied to the output of the function, for lg x^2 = lg(x^2) the square is applied to the input of the function

[vocal steeple]

yeah i get it now

so the parentheses had a lot to say?

@sharp coral

[sharp coral]

yes, parentheses are important

[vocal steeple]

thanks for the help

.close

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[midnight plank]

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[burnt prism]

smth went wrong but idk what and the answers supposed to be whole numbers help

[steep hinge]

https://tenor.com/view/golf-barstool-dance-gif-16821704370590183414

[burnt prism]

https://tenor.com/view/goat-chaeyoung-twice-stare-fake-gif-24766936

<@&286206848099549185>

[median trout]

Isn't y supposed to equal 2z?

Not 2y=z

[burnt prism]

? idk

[midnight plank]

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[umbral bramble]

does this work??

[umbral bramble]

i did impartial fraction decomposition and then integrated it but I think i did something wrong

[timid tangle]

Hmmm looks right to me too

[crystal rivet]

your antiderivatives are incorrect

also is your integrand supposed to be in terms of t while integrating in the domain of x❓ @umbral bramble

[umbral bramble]

I dont know what im doing im just doing it

im trying a different way and doing u sub first

[crystal rivet]

wait

if t is a constant then you dont need to do all of that

can you check what the original question is

[umbral bramble]

what do i do

[crystal rivet]

is it exactly like that from the book or hw

[umbral bramble]

hw

[crystal rivet]

yeah does it say if t is a constant

[umbral bramble]

thats it

[crystal rivet]

hmm

what class is this

[umbral bramble]

calc 2

[crystal rivet]

oh ok then

either there is a mistake or the answer is to treat that whole thing as a constant factor

you can try it assuming its a mistake tho

[umbral bramble]

i think its a typo

[crystal rivet]

ok then

use u-sub

[crystal rivet]

yeah dont forget your 'du' but from there you can use other techniques to solve it

[midnight plank]

@umbral bramble Has your question been resolved?

[midnight plank]

@umbral bramble Has your question been resolved?

[violet ravine]

Denominator is (u+1)(u+2)

1/(u+1) - 1/(u+2)

[umbral bramble]

I figured that one out!

just dont know about this one

how do i even start this

[violet ravine]

Cool

[umbral bramble]

can i U sub thiS?

[violet ravine]

Yep

You could have just substituted 1 + sqrt(x+1) to avoid doing the u-sub twice

[umbral bramble]

Thank you!!

[umbral bramble]

how are they getting a chemical reaction out of this

tf

[sharp coral]

there seems to be a footnote there, they might explain it more there

[midnight plank]

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[drifting root]

help part a pls i have no idea where to start

[strong dome]

oh my gosh what is this

[drifting root]

ive def done this before but completely forgot how to do it

[strong dome]

probably involves some trig stuff that I've forgotten

[midnight plank]

@drifting root Has your question been resolved?

[drifting root]

<@&286206848099549185>

[midnight plank]

@drifting root Has your question been resolved?

[midnight plank]

@drifting root Has your question been resolved?

[midnight plank]

@drifting root Has your question been resolved?

[bright shoal]

Yo

[hollow pier]

@Sho the answer of part "a"

Here is a brief :

We don't have a rule in math to get the area of a triangle with respect to a raduis
So we will work normally and get the area

The common law
Area of any triangle
= 1/2 * s1 * s2 * sin(angle bet. Them)

I have assumed that the length of XP and PY = L

Thus the area
= 1/2 L^2 * sin angle

There is a basic rule in geometry says that if two radii are opened with an angle on a certain curve (which us XY curve in this case) and two cords are opened on the exact same curve then
The measure of the angle bet. The two cords = 1/2 the angle bet. The two radii
Then angle (XPY) = 1/2 (XOY) = theta

Now we have the area
= 1/2 L^2 * sin (theta)

We need a relation between L and raduis

Take a look at the triangle POY, it's an obtuse triangle we can get a relation between them using the cosine rule
Thus
L^2 = r^2 + r^2 - 2r^2 * cos (180 - theta)
= L^2 = 2r^2 + 2r^2 * cos (theta)

(don't forget that cos(180 - theta ) = - cos(theta) )

Now we have to do substitution and every thing is done

[bright shoal]

Aah

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[midnight plank]

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[night gyro]

will the layout be 1/4 x pi x 24 squared?

[night gyro]

nvm i got it

.close

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[full leaf]

you are not getting away with not telling us the answer

the answer is 576 - 144pi

[night gyro]

yep

220.69

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[dawn dagger]

square both sides

[nova yoke]

or, just notice that the sqrt of (x+1)^2 is |x+1|

[dawn dagger]

sure can do that

[gritty hatch]

query 1: How is it possible this was square rooted? I am unaware of any method 2x could be reduced by and they managed to fully eliminate it somehow.```

[dawn dagger]

some people are just uncomfortable with absolute values

[gritty hatch]

Define this

[dawn dagger]

?

[nova yoke]

just ignore my comment if you're not familiar with them

it's just a different way to approach the problem, you'll get the same answer either way

[gritty hatch]

please respond to the prompt```

[fallow scarab]

The second line of this is wrong

Missing a +

[gritty hatch]

where

[dawn dagger]

2nd line implies a polynomial of degree 3

[gritty hatch]

Read this, ignore my prior attempt to expand it.

sadly it appears none of you were willing to answer the original question

I may have to search for help elsewhere or recreate the channel due to this.

Lack of formatting + too many answers + sidetracked.

This simply will not do.

.close

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[grave valve]

I'm presently stuck on a Calculus I problem that involves finding the derrivative of a function:

[grave valve]

I'll show my work soon

My work:

x^2 sqrt(25 - x^2) = x^2 (25 - x^2)^(1/2) <- convert square root into exponent

(x^2/2) [(25 - x^2)^(-1/2)] X (-1/x) <- solve derivatice using identity d/dx (u^n) = n X u^(n-1) X (u prime)

-(x^3/2) [(25-x^2)^(-1/2)] <-final answer

[uncut cloud]

you have to use the product rule

on (x^2 \times \sqrt{25 - x^2})

[grand pond]

Invariance

[uncut cloud]

you can't use the power rule on just one part of a product like that

[grave valve]

I see

[uncut cloud]

e.g.
[\frac{d}{dx} x^6 = \frac{d}{dx} (x^3 \times x^3) \neq x^3 \times 3 x^2 = 3 x^5]

[grand pond]

Invariance

[grave valve]

I'll apply this to my work now, and see if that's what the question is looking for. Though, I am confident now that it is.

Took me a few tries but I got it, thanks! 😄

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[pearl void]

Simplify $( x\sqrt{x} )$ for $( x \in \mathbb{C} )$

[pearl void]

Why is this wrong? $x\sqrt{x} = \sqrt{x^3}$

[grand pond]

Alex

[surreal moon]

Simplify $x\sqrt{x}$ for $x \in \mathbb{C}$ , without removing the square root symbol.

[grand pond]

SWR

[pearl void]

yes that lol thx

[pearl void]

.

[surreal moon]

Consider expressing x in complex polar form and what happens when x³ would exceed 2pi radians

[pearl void]

i dont know what is polar form

[surreal moon]

not familiar with $re^{i\theta}$?

[grand pond]

SWR

[pearl void]

no, i saw that somewhere but not in my book

[surreal moon]

hmm. Not really sure how to help otherwise.

[pearl void]

<@&286206848099549185>

[worthy wing]

Rewrite x as (sqrt(x))^2

[rigid dune]

this is not wrong is it?

[worthy wing]

Yes it is wrong

(-1)sqrt(-1) = -i
sqrt((-1)^3) = i

[worthy wing]

This is the way to simplify correctly

[midnight plank]

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[pearl void]

.close

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[high dock]

not sure what i messed up on

anyone have any input

[summer terrace]

how did you mess this up, you did the previous one well :p

$\lim_{x \to 5} f(x) = \infty , \implies , x=5$ is a vertical asymptote

[grand pond]

Emily

[high dock]

how is 5 vertical?

[summer terrace]

you switched 5 and +oo

[summer terrace]

you wrote as x approaches 5, f(x) +oo

[high dock]

so like this right?

[summer terrace]

yes

you did one like this moments ago, how did you mess it up :p

[high dock]

what math level are u in

[summer terrace]

im not math, im control systems

[high dock]

what in the world does that mean

is that higher than pre calc 😭

[midnight plank]

@high dock Has your question been resolved?

[midnight plank]

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[hasty star]

Just a quick question, so if i got a table, and it is close to linear, mean that there a few point that is slightly away from the line, and i want to get the slope, but i get a different slope every time when i plug the different point, i had a quiz on that, so is that mean i am wrong or what, cuz my friend got like 50, and i got like 51 point something

[hasty star]

Can there be like a range for the slope or it just that I am wrong

Cuz i think i use different number plug in the equation y^2-y^1/x^2-x^1, and i got different result everytime

But it just around that number everytime the result is got

I am a little bit confused

[plush prism]

when you have an almost linear table, then yes, choosing two points and getting the slope, the slope will depend on your choice

there are formulas that take into account every point you have, giving you the most accurate slope

[plush prism]

https://en.wikipedia.org/wiki/Simple_linear_regression

[hasty star]

Ok, i got it , thank you so much

.close

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[primal river]

Question:

Let f (x) = Cax + k

Suppose that the function f (x):

has a horizontal asymptote at y = 70,
crosses the y-axis at -5 and
passes through the point (6, 9).
Calculate the value of x such that f (x)=21.

Note: Your answer must be written in decimal with 3 EXACT decimal places of precision.

[midnight plank]

@primal river Has your question been resolved?

[summer terrace]

i see that f is a straight line

[primal river]

ignore this question can u help me with a different one

[summer terrace]

and a straight line doesnt have a horizontal asymptote

[summer terrace]

may i suggest a new channel

i guess it would be better

[primal river]

i will close this channel and then my life

[summer terrace]

lol

[primal river]

.close

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[wary chasm]

what is the formula for average velocity?

well its change in distance (displacement) over change in time

what is the change in distance?

distance between the bases are all the same: 27.4m

base lines all form right angles

what does this tell you

what shape has all sides the same length and has all right angles

correct

how so

this is a sqaure

yes

you know all distances between every single base and another

displacement is final coordinate minus initial coordinate

final coordinate is 3rd base

initial coordinate is home

displacement, that is

well think about it, what is the only possible direction you can go from home base to reach 3rd (remember this is average velocity)

think of 2 points

you can only connect them with a straight arrow

how many possible combinations of directions is possible between point 1 and point 2

well you probably need an angle, depends on what your teacher wants

plus you dont know which way "north" is

well if you dont have cardinal directions defined then usually you would use what is in the image

also you need to remember to get average velocity, you only have displacement right now

well that is like incredibly incorrect, and also probably a terrible habit to develop

but if your teacher is fine with it then go with something like "west of north" as direction

more specifically since you are given the angles

45 degrees west of north

dont use that lol

just trust me on this

youll need specific angles later on

using generalization like "north west" or "south east" is a terrible habit

cause you just started out

gonna need them later my man

theyre assuming that the person returns to home base

never says that

anywhere

at all

it says average velocity

not change in distance

dont use north west ;-;

that looks like it was pulled from a 1980's version of a physics book

im going to hope hes just trying to simplify it for you right now

that is crazy

gotta get the average velocity

remember that distance is a scalar unit

doesnt have direction

also you should really say 45 degrees west of north

just for your own sanity later on, dont have to write it down but just keep it in the back of your mind

its just angles

straight line means both sides are 180 degrees

wdym purpose

later on youll need trig

to calculate angles

for direction of vector units

yeah

arctangent and stuff

meh

its like

they give you the equation

but you dont actually gotta know how its derived

i dont think

depends on what physics ur in ig

are you in ap or honors?

ah then maybe

but if youre there then i dont see how you havent learned trig

its just basic trig tho so youll get it

if you already did calculus then ur fine

oh

i guess my school was a bit ahead cause algebra 2 is like

8th grade math

riperoni

makes more sense

[midnight plank]

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[twilit field]

could I have a hint

[floral apex]

part a?

[misty gorge]

what’s up phrocks

hope ur doing good

[floral apex]

phrocks?

[misty gorge]

physics rocks

[floral apex]

o

[twilit field]

not really, scored 50% in my first lA exam despite knowing enough to get 100

but other than that yeah

[floral apex]

we must have to rely on elementary row operations not changing rank right

[twilit field]

hmm

[floral apex]

so we get to some kind of thing like

assume wlog m>= n

[twilit field]

Theorm 2.5.3 has something to do with SNF

[floral apex]

$\mqty( I & 0 \ 0 & 0 )$

[grand pond]

jan Niku

[floral apex]

where I is an r x r identity matrix

[twilit field]

which we don't have in our syllabus, so wondering this is doable without the smith normal form

[floral apex]

I think if you were very nervous, just use your elementary matrices

[twilit field]

I was actually thinking of a geometric approach

[floral apex]

write A as some kind of product of a bunch of these

[floral apex]

and then this

actually, yea, this is enough

[twilit field]

When the ranks of two matrices are the same, they span the same space

wait

that's wrong

[twilit field]

hmm

okay

let me try

[floral apex]

do you see what i mean about this.

granted, IDK what im doing

[twilit field]

no, I don't follow

[floral apex]

take rank(A) = rank(B) = r

we do a bunch of rref on A

we end up with an identity matrix, embedded in an otherwise 0 matrix

the identity matrix will have r-many 1s

[twilit field]

how can we be sure of that

am I tripping

[floral apex]

because the rank is the number of linearhttps://en.wikipedia.org/wiki/Rank_(linear_algebra)

i guess in our case itd be

row rank?

in any case

say like $A = \mqty( 1 & 0 & 1 \ 1 & 5 & 2 \ 1 & 1 & 1 \ 0 & 0 & 0)$

[grand pond]

jan Niku

[floral apex]

this is equivalent to $E \cdot \mqty( \imat{3} \ 0 & 0 &0 )$

[grand pond]

jan Niku

[floral apex]

where E is some collection of elementary row operations

[twilit field]

I know that

wait

Maybe I'm messing up my defns

[floral apex]

we know that these dont change the rank, and they're square

it seems they must be invertible, each on their own

[twilit field]

These are my defsn

so can't I have a 2 to the right of the last leading one?

[floral apex]

yea i guess we may have something more screwy

[twilit field]

and have everything else in that row be zero

[floral apex]

something like $\mqty( 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 )$

[grand pond]

jan Niku

[floral apex]

but it wont matter

[floral apex]

no, you can't

[twilit field]

why not

[floral apex]

by definition

[twilit field]

$\mqty( 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 2\ 0 & 0 & 0 & 0 )$

[grand pond]

Veni, vidi, perii

[floral apex]

[twilit field]

this is in RREF

[floral apex]

oh, youre right

hmm

@tribal temple

you may be better suited here lol

[tribal temple]

[prime hornet]

[floral apex]

actually, this isnt a problem, is it

[hard umbra]

the definition of equivalence permits both row and column operations