help-27

[untold crow]

nevermind I feel so lost

<@&286206848099549185> Taylor Polynomial. Question: \ Given function: \ $f(x) = \sin(-10x^{2})$ \ Find the Taylor polynomial of f(x) of degree 8 and centered at x0 = 0.

[woven radish]

HqppyFeet

[thin inlet]

What is the taylor series for $\sin(x)$?

[woven radish]

Moosey

[untold crow]

,tex .maclaurin

[woven radish]

HqppyFeet

[untold crow]

godammn it

this

[thin inlet]

Yes, and so you know $\sin(x)=\sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$ and since you're asking about a Maclaurin series for $\sin(-10x^{2})$, what would we plug in for $x$ into the Maclaurin series for $\sin(x)$, and what is the maximum degree of that polynomial we want?

[woven radish]

Moosey

[untold crow]

the maximum degree of polynomial that we want is 8.

and we would plug in -10x^2 into x into the maclaruin series for sin(x)

[thin inlet]

Mhm. I suppose you could write out just a few of the terms, because you're only looking for an 8th degree polynomial. To be clear, I'd write out the first few terms of $\sin(x)$ Maclaurin series, plug in $(-10x^{2})$ in for them, and get rid of all the terms strictly greater than degree $8$.

[woven radish]

Moosey

[thin inlet]

I mean it lists it on the Maclaurin table too, so just write out those!

:)

then plug in (-10x^2) in for each x

then get rid of all degree terms strictly greater than 8 (if there are any)

[untold crow]

so only when k = 0 and 1?
I did spoil myself with looking at the answer, but when plugging into the maclaurin series, the first two terms was the answer...

and I don't fully understand why 😅

[thin inlet]

well what happens to the Taylor polynomial when we try to add $\frac{(-10x^{2})^{5}}{5!}$ (the third term)

[woven radish]

Moosey

[untold crow]

the polynomial is greater than 8

[thin inlet]

mhm

[untold crow]

but what we found is a polynomial less than 8

[thin inlet]

we stop at the first term where it's greater (we do not include it in the Taylor series)

[untold crow]

this method does feel weird at first, but I guess I just have to accept it lol

[devout snow]

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[grave hill]

please help i dont know how to even start

[near trout]

do you know what $\binom{40}{4}$ means

[woven radish]

Kaisheng21

[grave hill]

its a vector

[near trout]

no

not in this context

[grave hill]

or coefficent

nCr

[near trout]

yes

so do you know how to calculate that

[grave hill]

yes thanks so much

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[frosty oasis]

can someone help me out with this

not sure where to begin

[spiral oracle]

The y int would be when x is zero so if you plug that in...

[frosty oasis]

-57

o

i began by solving for rational roots

[spiral oracle]

so the coord would be...

[frosty oasis]

(0,-57)

[spiral oracle]

right

[frosty oasis]

and th en

i plug that in to find b

[spiral oracle]

I think you have to solve for roots to find the x

[frosty oasis]

?

oh

[spiral oracle]

wdym plug it in

[frosty oasis]

wait nvm

i would just get the same thing

ignore that

[frosty oasis]

ok

so the potential rational roots would be plus minus

1, 3, 19, 57

[spiral oracle]

assuming thats correct you can rule out 19 and 57

[frosty oasis]

alright

yea im lost

what do

[spiral oracle]

how did you find the potential rational roots?

is there more info im missing?

[frosty oasis]

took 57

and 1

did the theorem thing

[spiral oracle]

which theorem thing?

[frosty oasis]

uh

[jade oak]

Rational root theorem

[frosty oasis]

rational root theorem

took the factors of 57

1,3,19,57

over factors of 1

ok im just gonna say it's 3

[spiral oracle]

yeah its prob 3

[frosty oasis]

look at the graph it doenst look like 1

ing*

[spiral oracle]

it looks like 3

thats prob my best guess

[frosty oasis]

could you also help out with this

[spiral oracle]

find where both are positive or both are negative so when you divide it both a positive or >0\

does that make sense?

then do the opposite to find when <0

if you know how, a sign chart might be good

[frosty oasis]

not sure

if x was -8, then f(x) would be positive

[spiral oracle]

so you see the intercepts of f(x)?

[frosty oasis]

but would would g(x) be

yea

-8* i mean

[spiral oracle]

basically from -infinity to -7 everything would be positive right? so you would kinda mark a plus there

lemme see if I can find an example

basically you would write out your roots or int

then between and beside them you would write out if it's positive or negative

[frosty oasis]

ah

[frosty oasis]

could you explain that

[spiral oracle]

its basically how you start it

[frosty oasis]

i understand that from neg infinitity to -7 f(x) would be positive

[spiral oracle]

right

[frosty oasis]

why would g(x) be as well tho

like wher eis it

the grpah itself

cuz either neg/neg and pos/pos

for x>0

maybe idk

[spiral oracle]

it wouldn't its just you find the parts of f(x) that are, and then you can kinda visualize where everything is what

for this prob ig you dont really have to do a sign chart

but bascially you see how from neg inf to -7 f(x) is positive and g(x) is negative so k(x) would be negative

[frosty oasis]

yea

[spiral oracle]

just keep doing that with each part to find out the intervals

[frosty oasis]

so that wouldnt satisfy k(x)>0 then right?

(-oo,-7)

[spiral oracle]

right so thats not in the interval

[frosty oasis]

oh i thought u said it was

got confused

(-7,-1)U(4,8)

would tha twork, both neg i assume

[spiral oracle]

close

so what about from -1 to 4?

[frosty oasis]

idk

[spiral oracle]

wait no sorry your right

[frosty oasis]

f(x) pos g(x0 neg

?

[spiral oracle]

my bad

I was thinking about negative

[frosty oasis]

so would my answer for pos be correct

or is there more

[spiral oracle]

I believe it is

it should be good

[frosty oasis]

and for neg (-oo,7)U(-1,4)?

oh and

-8 pos infinity

[spiral oracle]

good

[frosty oasis]

i mean positive 8

[spiral oracle]

yeah thats good

[frosty oasis]

so for neg theres 2 intervals

pos 3

k thx

[spiral oracle]

yes 3 int

np

[frosty oasis]

preciate it

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[mild basin]

can anyone explain this to me pls?

[mild basin]

the answer is as follows but I'm still confused

[pseudo basin]

do you want the answer explained in detail

or do you want something independent of it (which may or may not resemble it in part or in full)

@mild basin

[mild basin]

sorry, was afk

back now

I am open to any suggestions here

explained in detail, or independent if it's easier than the above method

ty @pseudo basin

[pseudo basin]

ok, i've numbered all the steps

can you tell me the earliest one that confuses you

@mild basin

[devout snow]

@mild basin Has your question been resolved?

[mild basin]

Tyvm, checking it over now

[pseudo basin]

why the half hour delay?

[mild basin]

Sorry I got called to eat

What do the || double pipes mean?

||AB||^2

[pseudo basin]

length

aka magnitude

aka norm

whichever of these three words is familiar to you

[mild basin]

OK

Is that with or without the ^2?

They all seem to include ^2

It would not be correct if you remove the 4 pipes from AB?

[pseudo basin]

it would become nonsensical

[mild basin]

OK

It can’t be (AB)^2

[pseudo basin]

no such thing as the square of a vector

anyway $\nrm{v}$ denotes the length of $v$.

[mild basin]

Oh

[woven radish]

AnnGhost

[pseudo basin]

the ^2 is independent of that, notation-wise.

common as the squared length may be.

[mild basin]

Why no such thing as the square of a vector?

Vector is starting point and tail iirc

Like (3,4)

x,y

Just a point on 2d plane

[pseudo basin]

yeah sure and?

[mild basin]

Or (3,4,5) point on 3D

[pseudo basin]

dot product exists

[mild basin]

We cannnot square that point?

Doesn’t make sense

[pseudo basin]

if you're talking about squaring each of its coordinates

you can do that

but it doesn't have much meaning geometrically

[mild basin]

OK

[pseudo basin]

dot product exists, and one may take the dot product of a vector with itself, but for the love of god don't call it the "square of a vector".

[mild basin]

For parts 3 and 4 why was AC chosen?

Oh nvm

That was the question.

[pseudo basin]

are you asking why they worked out |AC|^2 before |BD|^2?

[mild basin]

Well, that too

Would it change the answer?

I would guess yes

Since the question asks for AC + BD and not BD + AC

Are these lengths commutative?

[pseudo basin]

they're NUMBERS, avid.

is addition of real numbers commutative, or is it something that you just now decided might somehow fail to be so?

[mild basin]

So it wouldn’t matter the order?

Well.. Multiplication of matrices are non commutative

But that’s different

[pseudo basin]

multiplication of matrices is non-commutative.

yes.

but even for them, addition is commutative still.

[mild basin]

Technically is AB a matrix? Or still just a vector

Yeah

[pseudo basin]

AB itself is a vector.

|AB|^2 is a number.

it is a number.

it is a number.

[mild basin]

Why capital letters?

[pseudo basin]

the capital letters denote points here.

[mild basin]

I thought notation is capital for matrices generally

[pseudo basin]

two capital letters concatenated, properly with a little arrow on top, means a vector.

[mild basin]

Lowercase for scalars and variables etc

[pseudo basin]

no, capital letters do not always mean matrices.

[mild basin]

OK

[pseudo basin]

you have it the wrong way around.

matrix => uppercase letter

uppercase letter ≠> matrix

[mild basin]

OK

[devout snow]

@mild basin Has your question been resolved?

[devout snow]

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[acoustic pebble]

This is my last question of the night (about to attach work) but i cannot figure out how to approach any kind of shadow problem

[acoustic pebble]

This is what I was able to figure out myself and I think its correct

I just cannot figure out how to relate everyhting together

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[gray isle]

uh I have a question about 3d trig, its a proof question

[gray isle]

its just q7...

i cant rly understand why its times instead of 300 over cot blah blah and i dont get why the cots are minused from each other...

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[somber kernel]

How would you continue the proof from here for this question?

[somber kernel]

theres 3 images

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[eager hemlock]

Don’t know where to begin and can’t even understand

[tacit wedge]

Do you know what a factorial is?

(n!)

[eager hemlock]

No

[eager hemlock]

Ah yeah I kinda remember it

[tacit wedge]

SO what would 8! be?

in terms of the number

not the exact value

[eager hemlock]

So basically 8! Is 8x7x6x5x4x3x2x1

[tacit wedge]

yes

and what is 4!?

[restive river]

the fraction be shortened due to multiplication

[eager hemlock]

Over 4x3x2x1 power 2

[tacit wedge]

yes

yes albert

so you cna cancel 4x3x2x1 with 8!'s 4x3x2x1

as 4x3x2x1 appears both in the numerator and denominator

then you have another 4x3x2x1 left over

because its (4x3x2x1)^2

and you can see that 8 = 4x2

so you can cancel the 8

[gleaming socket]

Better approach is to square it first.

[tacit wedge]

Similarily with 6

3 is a factor of 6

and you get 2 left over on the top

[eager hemlock]

Wait I’m sorry, In the denominator, do I power the 4 first?

[tacit wedge]

I wouldn't

I would just write it out

as 4x3x2x1 * 4x3x2x1

and cancel out terms with the top

[eager hemlock]

So basically it’s 8! / (4!) x (4!)

[tacit wedge]

yes

[eager hemlock]

So simplified would be 8x7x6x5 / 4!

[tacit wedge]

yes

and you can further simplify that

do you know how?

[eager hemlock]

Alright thanks

Yep

[tacit wedge]

ok good luck

[eager hemlock]

Would you be free for another question?

[tacit wedge]

surei ll try my best

[eager hemlock]

I have 2 more questions one is so simple but I got the answer wrong somehow

9 and 10

9 sin should be opp / hyp

Which would be b/c (D) but the correct answer is c somehow

[tacit wedge]

I think its an error because yes it should be b/c

maybe inform your teacher

[eager hemlock]

Alright then 10

[tacit wedge]

what did you try for 10?

I think the most common error for a questionl ike 10 is that when expanding, they don't apply the minus sign to all terms in the second polynomial

it should be 6a^3 - 5ac^2 + 12c - 4c - (-3a^3) - (-2ac^2)

and - (-) turns into +

[eager hemlock]

So beginning would be 6a^3 - (-3a^3) which would be 9a^3

[tacit wedge]

correct

and a good observation is that H) is the only option that has 9a^3 so it's the answer but, it's best to solve the whole problem and check

[eager hemlock]

So final answer, 9a^3 - 3ac^2 + 8c

[tacit wedge]

yes

[eager hemlock]

Will that was quite simple

Problem with there ACT questions is it gets my mind twisted, I manage to get the harder ones but get fucked on the simpler one

Thanks man

[tacit wedge]

Np

good luck

[eager hemlock]

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[frigid brook]

I need help setting these up

[devout snow]

@frigid brook Has your question been resolved?

[frigid brook]

<@&286206848099549185>

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[frigid brook]

.reopen

[devout snow]

✅

[frigid brook]

<@&286206848099549185>

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[frigid brook]

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[heavy yarrow]

Question: Roll a fair standard 6−sided die until a 6 appears. Given that the first 6 occurs before the first 5, find the expected number of times the die was rolled.

Not sure howw to start it as I thought the die rolls are independent so would just be 6?

[topaz axle]

it means you can't roll 5

so it becomes 5 instead of 6

just guessing

[heavy yarrow]

oh ok so then is it 5 rolls instead?

[topaz axle]

woah

simulation says 3

[heavy yarrow]

woah

interesting

[topaz axle]

it makes sense, like you actually stop if 5 or 6 appears

so 1/3

and then there's no further bias from only caring about half of those

well sorta, i don;t fully get it

[heavy yarrow]

hmm yeah ok I think that makes sense

[patent marsh]

I need to restudy prob and stats.... I have trouble understanding it lol

[heavy yarrow]

like you basically are rolling a die and stop when you get a 5 or 6 and then you know you get a 6

yeah so then 1/3 probability then 3 rolls

So I think if the numbbers switched like youre rolling a die untill a 5 appears, given that a 5 appears before a 6 what is the expected number of tosses

thaat should be the same

any 2 numbers on a die then I think

thanks!

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[coral dome]

why is this wrong?

[coral dome]

.close

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[grand thicket]

Find the domain, all increasing and decreasing intervals, all local extrema, the intervals of concavity, all points of inflection, end behavior, and sketch a graph.
f(x) = e^2x-4e^x

[devout snow]

@grand thicket Has your question been resolved?

[devout snow]

@grand thicket Has your question been resolved?

[supple knot]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[grand thicket]

I am stuck on sketching a graph with the information I have

[timber pebble]

i think sketching a graph would be the last step

have you found the other pieces?

[devout snow]

@grand thicket Has your question been resolved?

[devout snow]

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[jade oak]

!status

[devout snow]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[granite vortex]

1

Hi

[jade oak]

Employ the chain rule

[granite vortex]

I can see the chain rule with the dy/dt part but i dont see what the correlation is between those g(2) g(3) stuff

[jade oak]

Some of them wont be needed but some of them you can plug into whatever you get for y’

[granite vortex]

is the chain rule for the question df/dg * dg/dt?

[jade oak]

You should have a composition in there

[hybrid elk]

write it out using lagrange notation

[granite vortex]

wait o

im dumb lmao

10 got it thank you @jade oak @hybrid elk

.stop

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[restive river]

will i come across a situation where both of the results ( highlighted ) becoming true?

[wooden zodiac]

Wdym

[lavish nimbus]

That will be contradictory, a can't be < 0 and > 0 simultaneously.

[wooden zodiac]

That's what I thought

[restive river]

yeah

thx for answering my dumb qn

💀

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[marble quail]

How do i draw graphs without having a calculator?

[marble quail]

For example, x^(1/2)

or x^(3/2)

or something as complicated as x^(-1/2)

[twin wedge]

find y'

and check when y increases/decreases

[marble quail]

oh yeah good idea

thanks!

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[faint igloo]

Not sure what to do here

[restive river]

how to find the derivative of this?

[faint igloo]

I've tried doing $\frac{1}{x^2+2y^2}\le \frac{1}{x^2+y^2}$ but I can't seem to figure out what to do about the top

[woven radish]

clacii

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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[faint igloo]

this channel is already taken but i'll help you anyways

so you want to find $f'(x)$

[woven radish]

clacii

[faint igloo]

derive the RHS with respect to x so:

$\frac{d}{dx} (x^2 lnx^2 -3)$

[woven radish]

clacii

[faint igloo]

you can then expand and remove the -3 because derivative of a constant = 0

so now we have $\frac{d}{dx}x^2lnx^2$

[woven radish]

clacii

[faint igloo]

using the product rule of differentiation

$2x\cdot lnx^2 + x^2 \cdot \frac{2}{x}$

[woven radish]

clacii

[faint igloo]

simplifying to give: $2x(lnx^2+1)$

[woven radish]

clacii

[faint igloo]

@restive river all good?

.close

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[marble quail]

Does someone have a good explanation to the rational root theorem ? Having a hard time understanding it

[marble quail]

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[glass skiff]

how do you answer the second part

[dark talon]

Can someone explain how he got r=3?
It's Q 10 btw

My bad

[glass skiff]

I got linearly dependent by doing row ops

x3 is free

but idk how to find the scalars it's asking for

[devout snow]

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[mental edge]

this tells me z=7/2i

no?

[supple knot]

yes

[mental edge]

and the solution to it is a dot no?

[supple knot]

what does dot mean

[mental edge]

1 solution

with pi/2 angke

[supple knot]

no

try any nonzero real part for z as well

[mental edge]

why

[supple knot]

try it

[mental edge]

im not sure wym

[autumn geode]

conjugate property

[supple knot]

z = (anything) + (7/2) i

but anything nonzero

[mental edge]

what property is it

[mental edge]

yes and?

[supple knot]

plug that z into here

[autumn geode]

it flips sign of the imaginary part, by definition, so the real part stays the same.

misleading word use by me ngl

not a property

mb

[mental edge]

thats why we have x-x

so real part of z is 0

[mental edge]

yeah it obliterates

[autumn geode]

obliterates lmao

[supple knot]

so there's not just one solution

[supple knot]

did you do this?

[mental edge]

if i do something like

2+yi-2+yi

it still will be 2yi

[autumn geode]

👍

so its doesnt just work for 2

[mental edge]

yeah its that for any real value

so its always just 7/2pi

[autumn geode]

well yeah any real

[mental edge]

yes

[autumn geode]

so whats z

[mental edge]

so reals are 0

[autumn geode]

what

bruh

[mental edge]

and z=7/2pi

[autumn geode]

z is a complex

with x being real

and iy im

[supple knot]

z = x + iy

[autumn geode]

same thing

ur real can be wtv, itll cancel

u j said that

right?

[mental edge]

yes

so?

[autumn geode]

ok so

whats z

[autumn geode]

follow this form

[mental edge]

a line

[autumn geode]

what

[mental edge]

the result is infintie

[autumn geode]

ok so

what is z

[mental edge]

x+7/2pi

[autumn geode]

whats x

[mental edge]

a vòaue

gvalue

[autumn geode]

well

define x € R

idk how to latex

[mental edge]

yes R

[autumn geode]

deal w it

so there u have it

[mental edge]

is R

[autumn geode]

yep

[mental edge]

thx

[autumn geode]

so z= x + 7i/2 , x€ R

yeah?

[mental edge]

ys

[autumn geode]

there u go then

[mental edge]

thcx

[autumn geode]

👍

[mental edge]

.close

[devout snow]

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[mental edge]

.close

[devout snow]

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[heavy terrace]

I get 0 here

[heavy terrace]

am I right?

I just take out n^3 from each term under the cubic root

and take it infont

and I get 0-0

[restive river]

Yes

The anwser is 0

[heavy terrace]

oh ok ty!

.close

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[gloomy valve]

Could someone explain how I can figure out of T is one to one

[devout snow]

@gloomy valve Has your question been resolved?

[gloomy valve]

<@&286206848099549185>

[devout snow]

@gloomy valve Has your question been resolved?

[graceful cosmos]

@gloomy valve
On the contrary, T is one-to-one if dim(image(T)) = 2

Or in other words, if the output can produce two vectors that are linearly independent

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[nova whale]

A conservative force field in n dimensions is a function F: R^n → R^n for which there exists a function V: R^n → R such that F = -∇V. In this case, V is called a potential function.

(a) Consider the force fields F1 and F2: R^2 → R^2 defined as follows:
F1(x, y) = (y, x), F2(x, y) = (-y, x).
Determine whether F1 or F2 are conservative forces and, if so, find their respective potential functions.

[nova whale]

can anyone help me with this

[abstract lynx]

I’m too young to know the answer srry

[nova whale]

appreciate it

[sullen island]

well you want to find a V : R^2 -> R with - grad V = F1

first off, what would grad V look like ?

@nova whale

[sullen island]

yeah -grad V is that

[nova whale]

yh

[sullen island]

so we have $$\begin{bmatrix} -\pdv{V}{x} \ -\pdv{V}{y} \end{bmatrix} = \begin{bmatrix} y \ x \end{bmatrix}$$

[nova whale]

yh

[woven radish]

aPlatypus

[sullen island]

darn that's ugly

anyway

[nova whale]

this is what i've done so far

[sullen island]

wait so you found it ?

[nova whale]

idk if i did

i've never faced such a question before

[sullen island]

seems to work at least

[nova whale]

idk quite how this is t rue

[sullen island]

I mean if you found a suitable function you're done

[nova whale]

but can you just swap them like that

how does (-x,y) turn into (y,x)

[sullen island]

it's not (-x, y)

it's a number

V(x, y) returns a single number

[sullen island]

but anyway, the general method from there is to solve those two very little partial differential equations

$$\begin{cases} \pdv{V}{x} = -y \ \pdv{V}{y} = -x \end{cases}$$

[woven radish]

aPlatypus

[sullen island]

those two things have to be true at the same time

[nova whale]

in this case

not normally

[sullen island]

that's what you do in general also

ofc I plugged in the functions of your exercise

[nova whale]

could you perhpas do a very dillegent way of the F_2

i am a bit at a loss

[sullen island]

?

where does F_2 come from ?

wdym

[nova whale]

, F2(x, y) = (-y, x).

the second one

it is stated above

[sullen island]

ok sure

$-\nabla V = F_2$ means that $\pdv{V}{x} = y$ and $\pdv{V}{y} = -x$.

If we integrate these two equations, we get that $V(x, y) = xy + F(y)$, and $V(x, y) = -xy + G(x)$ for arbitrary functions $F$ and $G$

therefore, we get that $xy + F(y) = -xy + G(x)$, which is impossible

$F_2$ is not conservative

[woven radish]

aPlatypus

[sullen island]

yeah sorry I thought of F_2 the finite field for some reason

there

[nova whale]

it has to equal y

because of negative negative?

[sullen island]

yeah

[nova whale]

if you intergrate the nabla disappears

[sullen island]

yeah, the first equation integrated wrt x

the second wrt y

[nova whale]

mmh ok

this one makes a little sense

why is V(x,y) two things @sullen island

@sullen island i have anothing question as well, just ping me when u r on

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[devout snow]

@nova whale Has your question been resolved?

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[restive river]

how do i do this

[dense jay]

could express 1/16 as 2^a

know how to go about that?

[restive river]

well im pretty sure it was sum like find the prime number or sum

i didnt really understand what they were on about

[dense jay]

what is 16 as a power of 2

[restive river]

4

[dense jay]

yeah, 2^4, so whats 1/16 as a power of 2

[restive river]

fuck if i know that

.00390625

i think

[dense jay]

$\frac{1}{x^a}=x^{-a}$

[restive river]

if i did tht right

[woven radish]

AℤØ

[dense jay]

say what now

[restive river]

oh

LMAO

i plugged that into calc

ok nvm i did that way wrong

oh i get what ur asking

yeah it flips the exponent

to neg

[dense jay]

so what is 1/16 as a power of 2

[restive river]

Bro i dont get this

what am i suppsoed to be solving for like

2^-4

[dense jay]

well, if 2^t=2^(-4) theres only one thing t could be

[restive river]

if this was a bigger number tho like howd i go about it

cuz like tis a little number so its just sum yk already

[dense jay]

logs usually

[restive river]

so the t is -4

[dense jay]

it is

[restive river]

thx

.close

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[restive river]

How do i solve this

i tried doing i but got like a fraction rthen having to devide by that fraction or sum and was a whole mess

[unreal thunder]

log to the what base do you think will work here?

[graceful cosmos]

Can we see your work?

[restive river]

i was doing 7x⋅ln(3) = ln(4)

so far so good yeah?

then divided by ln(3) on both sides

sure

so i got 7x = ln(4)/ln(3)

.

hm?

where did u get ln(1)?

divide both sides by 7 instead

so

7x⋅ln(3) = ln(4)

then divided by 7 on both sides

(x⋅ln(3))/7= ln(4)/7

?

[graceful cosmos]

You're done. Divide by 7,
x = ln(4) / 7ln(3)

[restive river]

i suppose change of base would be faster

$\f{\log_c(a)}{\log_c(b)} = \log_b(a)$

[woven radish]

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

[restive river]

wtf

wut

supposed be a divide on LHS

[lament cradle]

\frac

[restive river]

its going to be log base 3 and argument 4

once you do change of base

but yeh, kaynex is right

,w ln(4)/(7ln(3))

[woven radish]

Results provided by WolframAlpha

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Requested by sherby11999

[restive river]

u already have the answer

or change of base

,w log_3(4)/7

[woven radish]

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[lament cradle]

$$\log_3(3^{7x}) = \log_3(4)$$
$$7x = \log_3(4)$$
$$x = \frac{\log_3(4)}{7} = \frac{\ln4}{7\ln3}$$

[woven radish]

b0ngl0rd

[lament cradle]

good illustration of change of base actually

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[fiery remnant]

Pretty sure how to do this just making sure

[restive river]

I assume you have an answer in mind then. Could you show us your work?

[fiery remnant]

I am thinking it’s the 8i

[restive river]

8isqrt(7)? Show your steps just to be sure

Actually nah not necessary

8isqrt(7) is correct

Since you're apparently not here I'll just close this channel

.close

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[devout snow]

@restive river Has your question been resolved?

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@restive river Has your question been resolved?

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[brazen star]

!status

[devout snow]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@burnt tide Has your question been resolved?

[smoky nimbus]

<@&268886789983436800>

[wicked rover]

.close

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[umbral forge]

Consider the following function:

f(x) = x^2 +2kx + 1 where k is a real number

For what values of k does f-1(x) have two assymptotes?

Express your answer in the form a<k<b or k<a,b<k as appropriate

[umbral forge]

ive found the inverse function: +/- √(x-1+k^2) -k

but first the function has to be one-to-one, so idk

[restive river]

The function above will never be one-to-one

Which is why I am confused

[umbral forge]

yeh same. also dont quadratics have no asymptote anyways

[restive river]

For a second there I started to doubt whether I knew the definition of "asymptote"

[brittle burrow]

i wonder if it means 1/f(x)

[restive river]

Hmm, that's worth wondering but very unlikely

[umbral forge]

ohh that could be very possible

[restive river]

Though in this scenario it's the only feasible explanation

[umbral forge]

yeh, ill try it out. wish me luck and thanks for the help

.close

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[jaunty mantle]

im struggling to get the minimum part

[jaunty mantle]

im very confused why the GS orthogonalisation has v'_jw_j on the bottom

shouldn't the bottom be w'_jw_j?

[supple knot]

You're probably right

[jaunty mantle]

Ok I figured it out

But somehow it works?

[devout snow]

@jaunty mantle Has your question been resolved?

[supple knot]

Huh interesting

https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling_Nachtergaele_and_Lankham)/09%3A_Inner_product_spaces/9.05%3A_The_Gram-Schmidt_Orthogonalization_procedure
Looks like this has a similar proof too

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[jaunty mantle]

.close

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[untold crow]

Anyone good at Taylorpolynomials?
I wanted to double check with you guys and ask if the only mistake I made was the "+" sign in front of the second term (when n=1)?
Don't get demotivated by language barrier. They are asking for the Taylor polynomial of degree 8, with centre at x0 = 0.

[untold crow]

I used this by the way.

[sullen island]

your 2nd term (with x^4) should have a minus

[untold crow]

yup, realized that. Anything else?

[sullen island]

rest sounds good

[untold crow]

Like.. it's correct for me to include the third term yes?

[sullen island]

yes

the 8th degree term is included

[untold crow]

yeah, good to double check. Thank you :3

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[bronze island]

Hi guys,

Which would be quicker, to row a boat 5 miles then turn back and row 5 miles in the opposite direction when there is no river current, or to do the same thing when there was a current (meaning you'd be rowing downstream for 5 miles, then upstream for 5 miles the other way)?

[rugged sparrow]

well, it would depend on the speed of stream and the boat

[bronze island]

hmm, I wasn't given those in the question. I was just told that you can row at velocity v and the river has velocity x

[topaz axle]

i think it would universally be slower with current

any speed of current and boat

[rugged sparrow]

$T_{rc}=10\cdot\frac{x}{x^{2}-y^{2}}$\$T_{nrc}=\frac{10}{x}$

[woven radish]

B-eard

[rugged sparrow]

These are the times assuming x to be speed of boat and y being speed of current

rc- river current
nrc- no river current

[bronze island]

ah ok interesting, I just assumed that any speed gains caused by rowing downstream would be cancelled out when you turn back and row upstream

so I assumed they'd average out and both scenarios would take the same time

[topaz axle]

i was thinking about set destinations, this a different scenario

[bronze island]

So here is the exact question

You are an excellent rower and you can row at velocity v on a lake in your hometown. Today you decide to row on a straight section of the river Thames. The Thames is flowing out to sea at a velocity of x. You always row a distance of y meters from the boathouse then turn around and row back.

ia) Does the round-trip take you more or less time on the Thames or the lake?

[rugged sparrow]

well

we can figure out what is faster

[topaz axle]

right, that's not different, i'm just confused af

[rugged sparrow]

$\frac{1}{x}>\frac{x}{x^{2}-y^{2}}$\$-y^{2}>0$

[woven radish]

B-eard

[rugged sparrow]

If we assumed T_nrc>T_rc then we would arrive to a contradiction

But yeah this is true if and only if x^2>y^2

[bronze island]

wish I knew how to put that into a calculator 😅

ok, I must admit, that is beyond me. So, I guess I was wrong to assume that the differences in speed caused by rowing down/upstream would average out?

[rugged sparrow]

wait

Let me explain you

assume speed of boat to be x

and the speed of river to be y

So, the effective speed downstream would be (x+y) while that upstream would be (x-y)

you follow?

[bronze island]

yeah

[rugged sparrow]

So, we have time=distance/speed

Just apply it

[bronze island]

that's what I initially tried, so I somehow ended up with the formula: 2y/((x - current) + (x + current) / 2, with y being the distance to the boathouse and x being the speed of the boat

[rugged sparrow]

$T_{rc}=\frac{5}{x+y}+\frac{5}{x-y}=5\left(\frac{1}{x+y}+\frac{1}{x-y}\right)=5\left(\frac{2x}{x^{2}-y^{2}}\right)=10\cdot\frac{x}{x^{2}-y^{2}}$

[woven radish]

B-eard

[rugged sparrow]

that is really unreadable without latex

[bronze island]

sorry yeah, I haven't used latex before

ok so the formula at the very right of your post is what we use

[rugged sparrow]

yes

[bronze island]

10 x ( x / x^2 - y^2 )

[rugged sparrow]

if you don't use latex better use * to show multiplication

Okay, so similarly you can calculate time whitout river current

speed upstream as well as downstream is same

[topaz axle]

When you row downstream the current saves you time proportionally to time it takes, and speed of the current
When you row upstream, "the time it takes" is increased

so it doesn't cancel, it's universally worse if there's any current

[rugged sparrow]

no

[topaz axle]

ok why

[rugged sparrow]

if the speed of boat is less than the speed of current, then T_rc is less than T_nrc

$x^{2}<y^{2}$\$T_{rc}<T_{nrc}$\$\frac{x}{x^{2}-y^{2}}<\frac{1}{x}$\$x^{2}>x^{2}-y^{2}$\$0>-y^{2}$

[woven radish]

B-eard

[topaz axle]

it's safe to say this never happens

[rugged sparrow]

Oh wait yes

I get it

The boat won't go upstream

[topaz axle]

if the current is faster, the trip you take upstream takes infinite time

yeah

[bronze island]

so the tldr is that if there is a current, the additional speed you gain from rowing downstream isn't as much as the speed you lose from rowing upstream?

[topaz axle]

yes

[bronze island]

ah ok interesting

[topaz axle]

the time increase is more than time decrease more accurately

because the current has more time to to affect you

[bronze island]

oh that makes sense!

so kinda like those things at airports, the flat escalator things that you walk down

if you walk down them, you exit quicker and thus the speed increase from the escalator had less time to affect you

but if you walk against them, they're slowing you down and thus is takes you longer to reach the end, meaning they had more opportunity to slow you

I think I get it

[topaz axle]

exactly that

[bronze island]

that's interesting. Thanks both, you guys helped a lot!

guess we're done here, but idk how to unoccupy this channel 🤔

[topaz axle]

type .close

[bronze island]

thanks

.close

[devout snow]

Channel closed

Closed by @bronze island

Use .reopen if this was a mistake.

[devout snow]

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[brazen star]

n^3 + 121 is divisible n + 11, what is the greatest value of n?

[somber shore]

hi
you can use this
ax+b | P(x) then ax+b | P(-b/a)
(just so you know if P(-b/a) is a fraction just take the common denominator and write it as 1 fraction and then ignore the denominator)

[placid rover]

You should post what uve tried

!status

[devout snow]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[placid rover]

etc etc.

[devout snow]

@brazen star Has your question been resolved?

[brazen star]

wait i've done it, i forgot to close

[devout snow]

Channel closed

Closed by @brazen star

Use .reopen if this was a mistake.

[brazen star]

mb

[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[tender lotus]

How is this equal to e??

[deep vortex]

What would you have thought it is equal to?

[tender lotus]

1^infinity which is an I.F

[deep vortex]

Many would say 1

[deep vortex]

By saying this is 1^infinity, you are taking the limit of the bracket first

[tender lotus]

How did we get rid of the I.F though and got e

[pseudo basin]

"indeterminate form" is not a value.

it is

[tender lotus]

ik

I mean i don't know what the limit is equal to (i got I.F) I couldn't get rid of it

[wooden veldt]

What's your definition of e?

[tender lotus]

A constant approx equal to 2.7 😅

[wooden veldt]

Well then you can take this limit as the definition of e

It converges to some number, we call that number e

[tender lotus]

So there's no proof?

[wooden veldt]

If you have no other definition of e then no

[tender lotus]

Hmm understandable

Thank you

.close

[devout snow]

Channel closed

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[restive river]

a right triangle with one acute angle of 45∘ and the hypotenuse of length 10 units, find the lengths of the other two sides (the legs).

[restive river]

using trigonometry

[smoky nimbus]

Did you draw a diagram?

[restive river]

No

only this is given

[smoky nimbus]

That should be your first step

[restive river]

okay

[smoky nimbus]

Draw a diagram with that info

[restive river]

i'll draw

i dont have camera on my pc to show you...

[smoky nimbus]

You can use paint to draw it and send a screenshot of

[restive river]

i'm drawing on my notebook

ok

after this

?

[serene pulsar]

sin(a) = opp/hyp, and cos(a) = adj/hyp, so you can sub in the values for the angle and hypotenuse, and then solve for the unknowns. On 45° right triangles, the sides should be the same length, so if you get different answers for each, you know something went wrong.

[smoky nimbus]

Don't forget to label the hypotenuse

Because you know that too

[restive river]

i don't know what is opp and adj... i know only base and perpendicular

[restive river]

ok

[smoky nimbus]

If you know SOH CAH TOA, you can find the ratios based on the angle

[restive river]

I dont know what does that mean

i want to find only the legs

not ratios

[smoky nimbus]

You would use that to find the values of the legs

If you are using trig

[restive river]

i'll try

[smoky nimbus]

If you don't know that, then you would use properties of a 45 - 45- 90 special right triangle

[restive river]

what should i name the opp and adj

[smoky nimbus]

It's in relation with the angle

[restive river]

i should name opp and adj as x and y?

[smoky nimbus]

Yeah you can do that

[restive river]

ok

can i solve the problem using pythogoras

?

so both sides will be x

right>

?

[smoky nimbus]

You can do that too

[restive river]

wait i'll do that and send the pic

[smoky nimbus]

There are multiple different ways to solve, but you said solve using trigonometry

[restive river]

But i want it using pythogoras now... pls?

like whatever method is easy we can do...

[smoky nimbus]

Like I said, there are multiple methods to solving it. You specified trig, which was why I started off with giving you SOH CAH TOA

[restive river]

now i want pythogoras

i finished it in pythogoras... can u check

?

???

[smoky nimbus]

Yes that's right

[restive river]

I'm sorry for that...

tyvm now can i .close the chat

?

[smoky nimbus]

If you're done with it

[restive river]

i'm new i don't know that's why

will the bot ask me whether i got my solution or not?

[smoky nimbus]

No

[restive river]

okay

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[restive river]

tyvm

bye

.close

[devout snow]

Channel closed

Closed by @unique cloud

Use .reopen if this was a mistake.

[restive river]

.close

[smoky nimbus]

You just need to do it once

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[smoky nimbus]

.When you see this, you are good

[restive river]

ok

[restive river]

.close

[devout snow]

Channel closed

Closed by @unique cloud

Use .reopen if this was a mistake.

[restive river]

you r so good at teaching

thank you

.close

[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[gilded coyote]

I've completed all of my exercises in Basic Mathematics so far, but I'm struggling with this proof.

If n is even, then (-1)^n = 1

Working with what I know so far, I can't seem to figure it out.

If n is an even number, then by definition, n = 2m where m is any positive integer.

Then,

(-1)^n = (-1)^2m
So, (-1)(-1)... (-1)(-1) for a total of 2m times.
Then, (-1) will always be multiplied by itself a multiple of 2.
Basically, in the lowest case, if m = 1; then, 2m = 2

So, we'd have (-1)^2m = (-1)(-1), which is positive 1 by definition.

So, whether it’s 4, 6, 8, etc., we will always be able to split it into pairs of:

(-1)(-1).

For example, 4 would be (-1)(-1)(-1)(-1)
Associating, = [(-1)(-1)] * [(-1)(-1)]
Which will be 1 * 1, which is always positive.

This would obviously hold for all even numbers, since it will always be possible to put it into pairs of 2.

My issue is that I can’t figure out how to formulate it with the tools I have.

All we have so far is the basic properties of addition and multiplication (that they’re associative and commutative). We have the distributive property and the usual proofs with them. We have the laws of exponents. We have proved the sums and products of negatives and positives.

We have proved, for example, that a^2 is always even if a is even. We have proved that a^2 is always odd if a is odd.

In a sense, I can “see” what’s happening, but I just can’t figure out how to write it down

If it’s possible, could you guide me toward the right answer rather than just explicitly giving it?

[pseudo basin]

We have the laws of exponents

@gilded coyote you can use one of those

particularly $a^{pq} = (a^p)^q$

[woven radish]

Ann

[gilded coyote]

Okay, I'll try that, thank you.

[jaunty mantle]

Another way would be splitting $(-1)^{2n}$ into $(-1)^n \times (-1)^n$

[woven radish]

Frosst