help-10

[warm shale]

Conflicts

[raw torrent]

How do I find what trigs y can be

Ik there’s prolly 817382971 many but wana find one

[teal turret]

Hmm

If u simplify the right u get

Tan x

Then

$\sin \tan^2 + \sin = y \cdot \tan$

[warm shale]

Stephen

[raw torrent]

O

[teal turret]

Then

[raw torrent]

Do u divide both side by tan

[teal turret]

No

[raw torrent]

[teal turret]

That may lose solutions

[raw torrent]

Ah ok

[teal turret]

If we factor out sin from left what do we get

[raw torrent]

sin(tan^2x+1)

[teal turret]

$\sin (\tan^2 + 1) = y \cdot \tan$

[warm shale]

Stephen

[raw torrent]

There’s a ratio uhh

Sec

[fickle oracle]

whcih is?

[teal turret]

Ye

[fickle oracle]

there is identity

for tan^2 + 1

[raw torrent]

Yes I mean that mb

Sec^2x

[teal turret]

$\sin x (\sec^2 x) = y \cdot \tan x$

[fickle oracle]

u can also do:
(tan + i )(tan-i)

if u know complex nums

[raw torrent]

no idea

[warm shale]

Stephen

[fickle oracle]

just ignore

[raw torrent]

Then do we make sec to 1/cos

[fickle oracle]

ye

so u will have what?

[raw torrent]

so it’s like sinx/cos^2x

[fickle oracle]

tan/cos right?

which is

[raw torrent]

uh

Idk

[teal turret]

$\tan x \cdot \sec x = y \cdot \tan x$

[warm shale]

Stephen

[fickle oracle]

sinx/cosx = tan?

[raw torrent]

Ye

[fickle oracle]

so tan/cos

or what stephen says its the same

since sec = 1 /cos

[teal turret]

So now

Being the tan over

What do we get

And factor it out

We get

[fickle oracle]

what we get @raw torrent

[raw torrent]

One sec

[fickle oracle]

let him answer stephen 😛

[teal turret]

K

[raw torrent]

Do we get 1/cos in the end

[teal turret]

Hmm

Wdym in the end

[raw torrent]

Wat

Ignore that

Is the answer 1/cos

[teal turret]

= y?

[raw torrent]

Yes

[teal turret]

Ye

[raw torrent]

Bet

[fickle oracle]

nice :3

[raw torrent]

Tyty

[fickle oracle]

good luck

[raw torrent]

.close

[obtuse pebble]

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[raw torrent]

[teal turret]

[obtuse pebble]

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[sterile stratus]

Can someone pls tell me how to plug this function into my calculator cause I’m so lost rn when I plug those x values in it doesn’t give the same answer 😭

[zenith raft]

maybe you are in degree mode

[sterile stratus]

No it’s in radians

[zenith raft]

ok what are you doing on the calculator then?

[sterile stratus]

I put pie in the denominator too and it still doesn’t give the same answer

[zenith raft]

why are you adding pi?

[sterile stratus]

cause it’s in the equation

Ik im supposed to add it in the denominator but still 😭 doesn’t give the same answer

[zenith raft]

that should be inside the sin

$\frac{1}{\sin\left(2\cdot \frac{\pi}{4}+\pi\right)}$

[warm shale]

layla

[sterile stratus]

OK ILL TRY THAT IN A BIT

OMG THANK YOU SO SO MUCH IM BRAINDEAD RN @zenith raft HOPE YOULL HAVE A GREAT DAY 🩷🩷🩷🩷🩷🩷🩷🩷THANK YOUUUU

[zenith raft]

hope you have a great day as well

[obtuse pebble]

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[timid silo]

sir?

what is this

[pulsar quarry]

sorry but promotions are not allowed <@&268886789983436800>

[timid silo]

would you like any assistance?

OH shit

[obtuse pebble]

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[hot hazel]

.close

[timid silo]

lol

auto close

[obtuse pebble]

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[regal ivy]

guys can you speak arabic?

[cyan sinew]

I don't, but I could use a translator

as far as I know what the question means and how it could be approached, ill give it a shot

[obtuse pebble]

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[dense prairie]

how do you find all 15 factors of 144 fast

[gaunt walrus]

divide with smaller term?

[obsidian isle]

144 = 2⁴ 3²

[high lily]

the number is pretty small
should be relatively easy to identify factors up to 12, from divisibility rules
then determine the respective factor in the pair

[obsidian isle]

Create a multiplication table with powers of 2 up to 16 and powers of 3 up to 9

Something like this, just filled in

[high lily]

excessive

[obsidian isle]

Systematic

[high lily]

i guess its actually pretty nice here

[obsidian isle]

Creates an interesting geometry

So just list the first 5 powers of 2, multiply each of them by 3 twice

5+5+5=15

[obtuse pebble]

@dense prairie Has your question been resolved?

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[timid silo]

(In a race, the order is drawn among the 58 starters. As the weather report predicts increasing winds, it is an advantage to start early.
Ylva draws her lot first, without showing the result, and then it's Fanny's turn.

"I would have had a better chance of getting starting number 1, if I had been able to draw my number tag first, "thinks Fanny.

Determine if Fanny is right by calculating the probabilities that Ylva and Fan respectively get starting number 1.)

The answer from the book is that the odds are the same(1/58). I also got 1/58.
Is that because since Ylva doesn't show her number,
we cannot simple say that the odds of getting "number 1" are 1/57 for Fanny
(since we dont know if she got number 1 or not)? Because of that, we still keep the 1/58

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

.close

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[lethal magnet]

oops

[lunar raft]

Wrong timing lol

[lethal magnet]

simplified bottom down to (3x-1)^2

not sure where to go from there

[willow crater]

u can have u=3x-1

[rocky trench]

what you might notice, the top is 6 times the derivation of the botoom

[lethal magnet]

mb

.close

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[mossy gyro]

The number of accidents on a certain section of a highway
averages 4 per day. Assuming that this number follows a
Poisson distribution, determine the probability of 1 car
accident in two days?

[drifting wraith]

,rccw

[mossy gyro]

i think my final soln is wrong it cant be 2.68 but is the p(x=1) step right

[drifting wraith]

that step is the solution

it looks right

just it's not 2.68

[mossy gyro]

it says average is 4 perday so mean of 2 days is just 4x2 right

[drifting wraith]

yes

[mossy gyro]

ya cause i copied that part ill fix it

[obtuse pebble]

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[brazen hazel]

Can someone explain this (German)

[alpine bison]

i cant even read the letters

[brazen hazel]

What can’t u read??

[lunar crater]

Fabio and Emma are planing for the city festival a game of chance with dice. the dice are rolled 2x in a row. If the sum (the sum of the numbers on both dice) is greater than 9, the player wins 8€. If the sum is exactly 8, the player gets 2€. The stake is 3€

ich habe es nur übersetzt, um es besser zu verstehen

Weißt du was ein baumdiagramm ist?

@brazen hazel

[brazen hazel]

Ja

Aber ich versteh die Aufgabe irgendwie nd

Like wie kann man da ein baumdiagramm zeichnen

[lunar crater]

warte gib mir ein moment

[brazen hazel]

Ja

[lunar crater]

also du musst 1 kreis W zeichen, dann zu W 6 Pfade zeichnen

Und dann zu jeweils den 6 pfaden nochmal 6 pfaden zeichnen

sorry ich kann es nicht so gut erklären

@brazen hazel

[brazen hazel]

Und wieso 6?

Macht 12 nicht mehr sein weil 2 Würfel

[lunar crater]

nein

Guck mal, du würfelst zuerst den 1. würfeln

würfel*

und dazu gibt es 6 verschiedene ergebniss

dann würfelst du den 2. würfel

dann gibt es wieder 6 verschiedene ergebnisse, aber es gab zuvor auch noch 6 ergebnisse

also musst du 6 mit 6 multiplizieren, d.h. es gibt 36 insgesamte ergebnisse

Die möglichen ergebnisse, um 8€ zu gewinnen sind größer als 9, da geht

4+6
5+6
6+6
6+5
6+4

d.h. die ereignisse davon sind jeweils $\frac {1}{36} , \frac {1}{36} , \frac {1}{36} , \frac{1}{36} , \frac {1}{36}$

[warm shale]

shotgun

[lunar crater]

dann musst du alle ereignisse addieren, nenner bleibt gleich, also 1+1+1+1+1=5

also haben die spieler eine $\frac {5}{36}$ chance, um 8€ zu gewinnen

[warm shale]

shotgun

[lunar crater]

um 2€ zu gewinnen, müssen die beiden augenzahlen zusammen 8 ergeben

die jeweiligen rechnungen sind

2+6
3+5
4+4
5+3
6+2

[lunar crater]

dann geht es wieder zurück nach diesem punkt

warte

es gibt 6 ergebnisse die größer als 9 sind, sorry

also anstatt 5/36 ist es 6/36

[lunar crater]

jedoch hier beträgen die ereignisse zsm 5/36

jetzt die ereignisse 6/36 und 5/36 addieren, die summe ist 11/36

die verlustchance beträgt dann 25/36

[brazen hazel]

Ahhh danke

[lunar crater]

@brazen hazel btw was ist gemeint mit "Erwartungswert"?

[brazen hazel]

Es gibt da so ne Formel um zu rechnen ob das Spiel „fair ist“

[lunar crater]

aja stimmt

gib mir einen moment

pkay sorry

ja es ist nicht fair

und wenn du es fair machen willst dann muss es jeweils 18/36 sein für den verlust & gewinn

[brazen hazel]

Ahhh ja ok sehr danke

[obtuse pebble]

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[rigid lintel]

suppose you want to compute a limit

[rigid lintel]

.close

[obtuse pebble]

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[turbid maple]

what do i do for multiply? as ik how to do the expand one

[pulsar quarry]

constants get mulitply like normally, and for the x part

we use the

law

,, x^a \times x^b =x^{a+b}

[warm shale]

यजतलमाओ

[turbid maple]

so its like the same way as expand?

[pulsar quarry]

yes

because while expanding you multiply them too

[turbid maple]

so you add them always?

[pulsar quarry]

yea

well if b in negative that thing turns into $x^{a}\times x^{-b}=x^{a+(-b)}$

[warm shale]

यजतलमाओ

[turbid maple]

ohh

thanksss

.close

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[lyric totem]

[y'=3x^2y]

[warm shale]

dopediscorduser

[lyric totem]

How’d you go about solving this differential equation? Move the x and ys to separate sides?

[\frac{y'}{y}=3x^2]

[warm shale]

dopediscorduser

[lyric totem]

[lny = x^3 + c]

[warm shale]

dopediscorduser

[lyric totem]

Like that when integrated?

[y =e( x^3 + c)]

[warm shale]

dopediscorduser

[final thunder]

sure

you can also do

\begin{align*}
y &= e^{x^3} \cdot e^{c} \
& = Ae^{x^3}
\end{align*}

[warm shale]

Pure

[lyric totem]

What is A in the equation?

Just some constant?

[final thunder]

yes A = e^c

for some real number c

so A is just a real number (strictly positive tho)

[lyric totem]

Okay thank you

.close

[obtuse pebble]

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[high stag]

we have to calculate the derivative of the function gof with and without the chainrule and i am confused about how to come up with gof

[high stag]

my initial solution was e^(y^2) but i am not sure if this is right

so it was e^y^2 as a solution for gof

[willow crater]

It's correct

[high stag]

okay but how do i derive this function with and without the chainrule how is this meant?

[willow crater]

Maybe u have to find once using chain rule and then not using chain rule

[high stag]

what do u mean by that?

[pseudo swift]

do you know what the chain rule is ?

[high stag]

yes i do

but the question states we should derive it one time with the chainrule and one time without it and im confused about what is meant if we shouldnt use the chainrule on the second try

[pseudo swift]

without chain rule means just differentiate directly, g°f is a function R\{0} x R -> R

and you computed it already

(g°f)(x, y) = e^(y^2)

[high stag]

so i just derive e^y^2 and get 2ye^y^2?

[pseudo swift]

your input is 2 dimensional here

you're looking for the gradient

[high stag]

so i derive it for x and for y ? and put in as a vector?

[pseudo swift]

yea

[high stag]

okay but deriving it without the chainrule just means 2ye^y^2?

as a partial der. in respect to y

[pseudo swift]

yes

I mean you should get the same results with and without chain rule

otherwise the chain rule would be pretty fucked up

[high stag]

yea but it seems a bit to simple to me since usually they want to trick us or something xd

[pseudo swift]

well there's no trick here

[high stag]

if i derive this in respect to x wouldnt the function just stay the same?

[pseudo swift]

no

you'd get 0

[high stag]

since there is no x or would it go to 0

ah okay

thank you very much ^^

.close

[obtuse pebble]

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[hollow river]

What is the matter

[hollow river]

What kind of help do you need. Where are you stuck

!status

[obtuse pebble]

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

[hollow river]

Do you know what a surd is

They are?

Ok

So basically they want you to simplify the number

And write it in surd form

That is why

Simplify it

How can I write √2

[sour sun]

hello

[hollow river]

It's 2^(1/2)

Correct?

[sour sun]

is everything okay in here

[hollow river]

It's the better way dear

And ³√2 can be written as 2^1/3

Hmm

Ok

4^(1/2)

Get it?

Do you know the exponent rule?

[sour sun]

here is the key insight here. When you have $$ \sqrt[n]{\sqrt[m]{a}}}$$ this is equal to $$\sqrt[nm]{a}$$

[warm shale]

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

[hollow river]

So (x^y)^z

[sour sun]

yes

[hollow river]

Yep

[sour sun]

yes

[hollow river]

Yessir

[sour sun]

yep

[hollow river]

Yessir

Yessir

[sour sun]

yes! good job

[hollow river]

Yessir

[sour sun]

yes

you should know both

[hollow river]

Surds aren't popular to use

People use power notation

[sour sun]

nested roots (or surds?) are the same as nested exponents.

[obtuse pebble]

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[timid silo]

How can I prove this sequence is bounded for any given a > 0?

[timid silo]

I know it will be $a_{n+1} = \sqrt{a + a_n}$. I tried seeing if $a_n < 1 + 2a$ but i reached $\sqrt{2 + 2a}$ and i dont know how to show thats less than 1 + 2a

[warm shale]

Master Yoda

[timid silo]

I also proved its increasing, so im looking for an upper bound

[sour jacinth]

If you square them and make some inequality then?

[red ice]

try proving $a_n < a$ for all $a, n$ by induction

[warm shale]

south

[red ice]

wait not a

but induction is the way to go

[sour jacinth]

Since all terms are positive...can we square them and prove that the new sequence is bound so will be the original???

[red ice]

yeah it's along those lines, and induction is just how we formalise it

so assume $a_k < L$

[warm shale]

south

[timid silo]

induction is what im trying to apply, but i need to choose a good L, no?

[red ice]

then $a_{k + 1} = \sqrt{a + a_k} < \sqrt{a + L}$

yeah

[warm shale]

south

[timid silo]

i tried taking L to be 1+2a, but that didnt work out

[red ice]

okok I see where you are now

ah right so we need an L such that sqrt(a + L) < L for all L

[timid silo]

i did this and got $L = \frac{1+ \sqrt{1 + 4a}}{2}$

[warm shale]

Master Yoda

[red ice]

yep me too

[timid silo]

Which is less than 1 + 2a hence my initial choice of L. but i got stuck at trying to show sqrt(2 + 2a) < 1 + 2a

so perhaps i should choose a different L?

[red ice]

yeah it's not true for all a that's why

for 0 < a < 0.309 it doesn't work

[timid silo]

exactly

so is there any other way of approaching this?

[red ice]

I think you just have to use this

as in solve $\frac{1}{2}(1+ \sqrt{1 + 4a}) < 1 + 2a$

[warm shale]

south

[red ice]

you can square both sides since a > 0

or like move the terms around first

as in $\sqrt{1 + 4a} < 1 + 4a$

[warm shale]

south

[red ice]

hey that wasn't too bad

multiply by 2 then minus 1

oh right there are still some values of a that don't work

so it's impossible to show that L < 1 + 2a for all a > 0

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[eager pivot]

shouldnt it be there is a number between -1 and 1 and not 0 and 1?

[red ice]

nope

if x = 0 gives a negative value

and x = 1 gives a positive value

then at some point between x = 0 and x = 1, y must be exactly 0 from the IVT

it must cross from negative to positive

[eager pivot]

ohh ok i was confused becuase i thought that they were saying that those were the f(x) values because they didnt mention the point c itself

tysm

.close

[obtuse pebble]

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[lilac cave]

is this true?

x=-13

y=-x

y=13

[lilac cave]

just checking

[pulsar quarry]

yes

[lilac cave]

ok .close

.close

[obtuse pebble]

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[sweet sparrow]

I need help with this problem

[pallid aurora]

have you drawn out what it may look like?

[sweet sparrow]

No

[pallid aurora]

maybe start with that

[sweet sparrow]

I did

you mean the triangle

[pallid aurora]

draw out the entire diagram

so the triangle and maybe what the rectangle could look like

[pallid aurora]

see anything you could go off of?

if not its all good

[sweet sparrow]

lemme think

I mean the only thing I see is surface area of the rectangle being equal to x*y

[pallid aurora]

could you rewrite y in terms of x?

and then it’s just x(whatever y is in terms of x)

[sweet sparrow]

6-6x/8 ??

[pallid aurora]

yep that works

so the expression x(6 - (6x)/8) is the area of the rectangle

[sweet sparrow]

I see

[pallid aurora]

and now you just need to optimize the output

[obtuse pebble]

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[vestal spoke]

Need help on 22-29

[obtuse pebble]

@vestal spoke Has your question been resolved?

[vestal spoke]

<@&286206848099549185>

[vestal spoke]

…

[karmic palm]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[dapper lark]

Hey

I can help you

[vestal spoke]

Yes please

[dapper lark]

So you're drawing the graph of y=log(x+1)

Basically the +1 gives a translation to the left of 1 unit

Which is why the graph will look like how you drew it

You drew it correctly

The graph of question 23 was also done correctly

Did you just need someone to check your work?

[vestal spoke]

No i was confused on why the translations were how they were. I used to calculator for 22-25 and was confused on why the graphs were how they are

Like in 22 it’s translated 1 unit left because the 1 is positive, but in 23 the 2 is positive but it’s translated to the right?

[dapper lark]

Ah OK I see the co fusion

Because it's f(-(x-2)) and so the - provides a reflection along the y axis and then the -2 gives the translation to the right. You always have to have factored out the coefficient of x, to have f(b(x-c)) rather tha f(bx-c)

[vestal spoke]

Ohhhhh that makes more sense

[dapper lark]

👍

[vestal spoke]

I was also confused as to why 24 is as it is

I thought the +2 was an upwards translation

[dapper lark]

It is

It's drawn wrong

[vestal spoke]

So would it start at (0,2)?

[dapper lark]

It had an asymptote at x=0 because ln(0) is not defined. It normally contains the point (1,0) but in this case has the point (1,1)

[vestal spoke]

I’m a bit confused as to what your trying to say

[dapper lark]

The graph of log(x) has an asymptote at x=0

So you can't say there's the point (0,2)

Right ?

[vestal spoke]

Right

[dapper lark]

Yea and then, the graph of log(x) has the point (1,0)

But then with the transaction of 1 unit upwards it becomes the point (1,1)

Got it ?

[dapper lark]

Like the graph doesn't really start at a point in that sense

[vestal spoke]

But is the translation not 2 units instead of 1?

[dapper lark]

Yes you're right actually

[vestal spoke]

So it would “start” at (1,2)?

[dapper lark]

Yeah

[vestal spoke]

Based on what I’ve learned from 22-24, 25 would be a reflection across the x-axis and a translation 1 unit to the right?

[dapper lark]

Yes exactly

[vestal spoke]

Ok awesome, now for 26-29 I’m basically entirely lost

[dapper lark]

Ok then let's work on it together

[vestal spoke]

Ok let’s do 26

[dapper lark]

First things first, do you know what the domain is

[vestal spoke]

I know what domain is, I don’t know how to find it for logarithms (same for range)

[dapper lark]

So basically, what's inside the log needs to be positive, because 10^(x) can't be 0 or negative

So we need x-3>0 and so x>3

[vestal spoke]

How would I write that

[dapper lark]

D={x€R/x>3}

Or D=(3,♾️) if your teacher uses interval notation

[vestal spoke]

Oh ok yes

The other one looked like I ran my fingers across my keyboard 😂

[dapper lark]

Yeah my grade 11 teacher uses the first one, but this year my teacher accepts both

Anyways,

Back to the question

So for the range,

If 10^x is greater than 10 then x is positive, and if it's 1 then x is 0 and if it's between 1 and 0 then x is negative, meaning that in y=log(x) y can have any real value so I=R or I=(♾️,♾️)

I usually use I for image in French

Do you guys use R for the range?

[vestal spoke]

Yes

[dapper lark]

Ah ok

Well anyways there's your answer

[vestal spoke]

Ok I’m also a bit confused on this one

[dapper lark]

Basically what I'm saying is that, because log(x) is the inverse function of 10^x and so whatever values x can take I the domain of 10^x, becomes the range of log(x)

Which makes the range equal to the set of real numbers

Makes sense?

[vestal spoke]

Ok so the -3 doesn’t really matter in this case for range?

[dapper lark]

Yes

[vestal spoke]

Ok ok, I understand

[dapper lark]

And then for continuity, does your teacher just want you to write whether the graph is continuous or not?

[vestal spoke]

Yes, I forgot to mention I know continuity and increase/decrease

[dapper lark]

Ok good

[vestal spoke]

Also yes for this

[dapper lark]

And for symmetry

[vestal spoke]

And just to make sure this graph is continuous and increasing

[vestal spoke]

This I do not know

[dapper lark]

Yes exactly

[dapper lark]

The graph of log(x) has no symmetry because the function is neither even nor odd

You know what even and odd functions are?

[vestal spoke]

What does that mean?

[vestal spoke]

No

[dapper lark]

Avenue functions have an axe of symmetry along the y axis, and odd functions have a central symmetry along the point (0,0). For example the function y=x² is even and the function y=x is odd

Like just by looking at the graph you can get it

[vestal spoke]

Ok yeah that’s pretty easy

Also makes sense to why the logarithm isn’t even or odd

[dapper lark]

Yeah

And then for boundedness, this is whether the function has a limit, like a specific value of f(x) that it can't pass

But since the range is R then its unbounded

[vestal spoke]

Ok 👍

[dapper lark]

But for example, say you have the graph of 2^x for example, it's range is (0,♾️) and so it has a lower bound of 0

Got it ?

[vestal spoke]

Yes

[dapper lark]

And then Extrema , that's basically if it has local or global maximum and minimum

Do you know what those are?

[vestal spoke]

So none?

[dapper lark]

Yes exactly

[vestal spoke]

Ok cool

[dapper lark]

And then asymptotes, I assume you know that it has the vertical asymptote x=0

That's the only asymptote it has

[vestal spoke]

Yes

[dapper lark]

And then for end behavior, this is the behavior of the graph on each end of its domain, so because the domain is (0,♾️) the end behavior is x->0 y->(?) And x->♾️ y->(?) , and so you find the value that y approaches as x approaches 0 and ♾️ (the two ends of the domain)

Makes sense?

So can you try to tell me what the end behavior would be?

[vestal spoke]

Let me try

As x->3 y-> ♾️

And as x-> ♾️ y->3 ?

[dapper lark]

x->3 y->-♾️ , look at your graph

[vestal spoke]

Ohhhh

Yes

[dapper lark]

Yeah exactly

And then as x approaches ♾️ , y is increasing

So y also approaches ♾️

[vestal spoke]

So I was right?

[dapper lark]

No it's x->♾️ y->♾️

[vestal spoke]

Oh ok

[vestal spoke]

But this one’s correct?

[dapper lark]

No it's like this

[vestal spoke]

Oh it’s just one answer

[dapper lark]

Look:

x->3 y->-♾️
x->♾️ y->♾️

This is it

[vestal spoke]

Oh I’m sorry I read my own answer wrong

[dapper lark]

It's OK:)

[vestal spoke]

Let me try 27

[dapper lark]

Ok

[dapper lark]

Let me read it

The domain is incorrect

The range, continuity, increase/decrease, symmetry, boundedness, Extrema, asymototes are all correct

[vestal spoke]

Is this the domain?

[dapper lark]

Yes exactly

[vestal spoke]

And we’re the end behaviors correct?

[dapper lark]

Yes they were

[vestal spoke]

Ok perfect

[dapper lark]

Are you gonna do 28 and 29?

[vestal spoke]

Let me do the last 2 and I’ll get back to you

[dapper lark]

Ok

[vestal spoke]

Yes I’m goin to try

There they are

[dapper lark]

Let me read it

[vestal spoke]

Oh domain of 28 is (0, ♾️) right?

[dapper lark]

Yes

[vestal spoke]

Asymptote of 29 is negative?

[dapper lark]

Yes

[vestal spoke]

Which changes the bottom end behavior to x-> -3

[dapper lark]

Yes exactly

[vestal spoke]

It’s so nice when I actually understand it

[dapper lark]

Yeah I know

Do you have any other questions?

[vestal spoke]

Let me see

Actually yes

Can you help with some word problems?

[dapper lark]

Sure

[vestal spoke]

30 was an attempt that I got lost on

[dapper lark]

Let me see

So yes, the equation would be 270=93log(d)+65

And so in that case you have to subtract 65 on both sides before dividing by 93

To isolate the value of log(d)

[vestal spoke]

Yes I got that far

[dapper lark]

Oh yeah I saw directly 205/93 so I hadn't realized you had subtracted 65

So from there

10^(205/93)=10^(log(d))

On the left side jt cancels the log, to have just d on the left side

And on the right side you evaluate 10^(205/93) to have the value of d

Makes sense?

[vestal spoke]

Oh that’s because 10^log(d) is just equal to d correct?

Which is how you isolated d?

[dapper lark]

Yes exactly

[vestal spoke]

Ok I understand that

[dapper lark]

My tablet was glitching for a while

Finish the question and show me

[dapper lark]

Yes exactly

[vestal spoke]

Awesome

[dapper lark]

Are we gonna work on 31 and 32?

[vestal spoke]

Yes please but I might understand 31 A

I’m trying it right now

[dapper lark]

Yes exactly

Try to do b)

[vestal spoke]

It was actually pretty easy I think

If that’s right then I can probably do 32?

[dapper lark]

Yes it is right 👍

Try 32

[vestal spoke]

For 32 A

It ends up being a pretty big number yes?

[dapper lark]

Let me see

[vestal spoke]

That’s all of my work for 32 A

[dapper lark]

Yes it gave me 86,948,273,446.949

[vestal spoke]

What’s the short way to write this?

Is it just scientific notation?

[dapper lark]

Yeah, 8.7x10¹⁰

[vestal spoke]

Yes sorry I meant to write it just didn’t

[dapper lark]

Ah ok

[vestal spoke]

Ok let me attempt B

[dapper lark]

Ok let me see

Yes, trusting that you did the calculations properly, this is it

[vestal spoke]

Ok awesome

Do you still have some time?

[dapper lark]

Yeah I do

[vestal spoke]

Ok last one

[dapper lark]

Yes A is correct

Can you try to get the domain and range for B?

And show me

[dapper lark]

Yes exactly 👍

[vestal spoke]

Is it symmetry across y-axis?

For C

[dapper lark]

Let me graph it on desmos and double check

Yes it is

[vestal spoke]

And for D

I need to find a value for x that, when subtracted by 9, would equal 1. Because ln(1)=0 ??

[dapper lark]

Yeah

So x²-9=1

[vestal spoke]

Ok

[dapper lark]

Do it and show me

[dapper lark]

Yes exactly

[vestal spoke]

The last one I’m lost

[dapper lark]

But it's better to start off with writing ln(x²-9)=0 so that it's clear where you're getting the equation x²-9=1, for the purpose of communication

[dapper lark]

Makes sense?

[vestal spoke]

Yes

[dapper lark]

Ok now for D

Sp

It wants H-¹(x) for all x>3

So you know, H-¹(x) means the inverse function of H

Do you know how to get an inverse function?

[vestal spoke]

Yes I’m just a bit lost on how to get the inverse when there is an ln

[dapper lark]

You'd write x=ln(y²-9) and then the next step would be e^(x)=e^(ln(y²-9)) and so e^x=y²-9

[vestal spoke]

So I end up here?

[dapper lark]

Yes exactly

[vestal spoke]

And is that just the answer?

What does it mean by all x > 3

[dapper lark]

It's just that basically this inverse function would only apply to half of the original H(x)

[vestal spoke]

Oh ok

[dapper lark]

Only this half

Let me also graph it to double check

[vestal spoke]

🙌🙌

Thank you so much for all of your help

[dapper lark]

No problem 🙂

If you ever need help with math in the future you can dm me

[vestal spoke]

Really? That’s awesome I’ll keep you in mind

I’ll close this now, I have no further questions

Thanks and goodbye 👋

.close

[obtuse pebble]

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[obtuse pebble]

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[tough hull]

Why did they find out distance like that not with distance formula

[tardy epoch]

which part

[tough hull]

Last part

[tardy epoch]

well you don't know where O is do you?

[tough hull]

Yes

[tardy epoch]

oh

where is it

[tough hull]

Idk

[tardy epoch]

right so how can you use distance formula if you don't know where O is

[tough hull]

Yea

[obtuse pebble]

@tough hull Has your question been resolved?

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[stable urchin]

Hello need help

[stable urchin]

Can someone confirm if this is correct? Because i used other app too but they had different answers so i am not sure if i should use this one

[fierce lagoon]

What's the original angle

2.89°?

Or 2.89 radians?

Like whaddya tryna convert

[obtuse pebble]

@stable urchin Has your question been resolved?

[stable urchin]

This

[stable urchin]

radians to degrees

[fierce lagoon]

Yeah so you just multiply by 180/π

I have no idea what your calculator is doing

[obtuse pebble]

@stable urchin Has your question been resolved?

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[turbid maple]

how do u use the foil method if theres 3 of them for b)

[forest sinew]

you foil then foil

[violet sentinel]

you FOIL the FOIl

[violet sentinel]

yeah that

[worthy flare]

(2x+3)(2x+3)(2x+3) =(2x+3)^3

[forest sinew]

although, youre gonna have 3 terms into 2

so its really like

fmoil

or something

idk im not a coiner

[worthy flare]

foil (2x+3)(2x+3)

then repeat the answer of that

with (2x+3)

[turbid maple]

wdym

ik how to do with 2

[violet sentinel]

Maybe an example will help

[forest sinew]

https://www.youtube.com/watch?v=4QBQtpJbnmE

an example here

[turbid maple]

okk

[forest sinew]

if youre into videos

you will see what happens

you pick two factors to foil

then you have to distribute 3 terms into 2 terms

but its just like a different kind of foiling

[turbid maple]

that was easy

thanks for the video

.close

[obtuse pebble]

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[obtuse pebble]

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[forest fiber]

anyone know why this is wrong?

[subtle sinew]

Check for extraneous solutions

Meaning plug back in your values for x and see if it actually works

[forest fiber]

oh i see

0 wouldnt work since it would become negative

[grim basalt]

||Doesn't log 0 not exist||

[subtle sinew]

Yes

[forest fiber]

thank u

[subtle sinew]

The give away would be log(0 - 1) = log(-1) and log of negative numbers don't exist

[grim basalt]

Yeah but in just log(x)

If x is 0, 0 DNE

[forest fiber]

ye 0 wont work either

[grim basalt]

So when you plug it back in it would be undefined

[forest fiber]

ok ty

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

@pliant tide Has your question been resolved?

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[narrow mesa]

Why is the sum of this N(n+1)/2?

[narrow mesa]

Like where did this even come from?

[dark stirrup]

Start with 1+2+3+4+...+(n-2)+(n-1)+n
But pair them up like (1+n) + (2 + (n-1)) + (3 + (n-2)) + ...
This is equal to (n+1) + (n+1) + (n+1) + ...

So you're basically adding (n+1) a specific number of times.

But there are n terms and you're pairing up two at a time, so you're adding (n+1) a total of n/2 times.

🥇

[narrow mesa]

Ohhhhh I see I see

Wait and does that still work no matter if n's even or odd?

[dark stirrup]

This logic works if n is even. If n is odd, the formula still works, but the logic is a little different.

[narrow mesa]

I see I see
Whats the logic for it when its odd then?

[dark stirrup]

We are still paring up adjacent terms: (1 + n) + (2+(n-1))+...=(n+1)+(n+1)+...

But now that n is odd, we are excluding the middle element.

So we are adding (n+1) a total of (n-1)/2 times.

So this means our sum so far is (n+1)(n-1)/2.

But we now need to add the middle element. If n is odd, then the middle element is (n+1)/2

So the sum of all terms is [(n+1)(n-1)/2]+[(n+1)/2]=n(n+1)/2

[narrow mesa]

Ohhhhh I see I see

And we can do this b/c we're counting the terms with n and using n in the terms themselves at the same time

Aight I think I get it now

Thanks for helping 🙏

.close

[obtuse pebble]

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[dark stirrup]

np

[obtuse pebble]

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[forest fiber]

im kinda lost on this

im trying to solve for x

[high lily]

firstly there are notation issues that need to be fixed

the argument of the log should NOT be in superscript

$\ln^{(\text{stuff shouldn't be here like this)}$

[warm shale]

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

[forest fiber]

wouldnt they be multiplied?

[high lily]

that's not the issue

the issue is that you wrote them up top instead of just normally

you should be writing $\ln(x^2-x-6)$ and NOT $\ln^{(x^2-x-6)}$

[forest fiber]

ok i see

[warm shale]

ℝαμΩℕωⅤ

[high lily]

anyway, you have a quadratic equation. e is just a constant
expand, rearrange to general form
then apply QF

[forest fiber]

whats qf?

[high lily]

quadratic formula

[forest fiber]

$x^2+x-6-e=0$

[warm shale]

dbRev

[forest fiber]

then apply qf here?

[high lily]

just realised i messed up a sign earlier, but yeh

[forest fiber]

i ended up with an imaginary number as well, would this still be correct?

[high lily]

you shoudlnt' be getting imaginary numbers

you messed up your signs

when evaluating the b^2 - 4ac

[forest fiber]

im getting the same thing

idk where im messing up tbh

[high lily]

product of two negatives isn't negative

[forest fiber]

alr i got it finally lol

thank u

.close

[obtuse pebble]

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[obtuse pebble]

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[steady delta]

Hello ?

[lean trail]

you closed the other channel?

[steady delta]

idk what happened

[lean trail]

I couldnt reply

[steady delta]

i did nothing

me to

[lean trail]

but f(x)=2x is clearly wrong considering the given points dont satisfy it

[steady delta]

But that is what i am getting, trying to find f(x)

[lean trail]

can you show me your steps?

[steady delta]

Yeah so Slope is Rise/Run which is 2/1 = 2
so f(x) is also y
so y=2x+b
(1,2) lies on this line so
2= 2+b which means b = 0
so y=2x i.e, f(x) = 2x

[twilit thicket]

slope should be -2

[steady delta]

oh wait

[lean trail]

your points are (0,2) and (1,0)

[twilit thicket]

$\frac{0-2}{1-0}$

[warm shale]

CH3

[steady delta]

Ahhhhhhhh

[steady delta]

Bruh yeahhh

.....

Alright I get it now

[twilit thicket]

so the point (1,2) doesn't lie on the line

[steady delta]

I took the (1,2) as a coordinate

[steady delta]

yeah

now i get it

[twilit thicket]

nice

[steady delta]

Thx @lean trail @twilit thicket 🥹

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[obtuse pebble]

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[gloomy vector]

if x,y,z are 3 real numbers where 3x, 4y, 5z are a geometric series and 1/x, 1/y, 1/z are an aritmethic series, if x/z+z/x=a+(b/c) then find a+b+c

[gloomy vector]

a+(b/c) is like a mixed fraction

[lost tree]

$\frac{x}{z}+\frac{z}{x}=a+\frac{b}{c}$

is that right?

[warm shale]

SilverSoldier

[lost tree]

?

[willow crater]

! status

[obtuse pebble]

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

[willow crater]

Using the first condition u would get y as f(x,z) and then u would substitute that into second condition to get some eqn in x and z

[gloomy vector]

1

[gloomy vector]

yeah

[lost tree]

what equation can u write, knowing that 3x, 4y and 5z are in a geometric progression?

[willow crater]

Ok so do u know what geometric series is

[gloomy vector]

hold up

5z=3x×n^2
4y=3x×n

[willow crater]

Yes

So a=4y/3x

[gloomy vector]

im just gonna guess it will use x/z+z/x=((x+z)^2-2xz)/xz

[lost tree]

well be careful using a coz its already there elsewhere in ur question

[gloomy vector]

ok lemme change it

[lost tree]

okay so n = 4y/3x

[gloomy vector]

yes

and your saying?

uhh does making it 15xz=16y^2 help

[lost tree]

probably

[gloomy vector]

1/y=1+mx/x=1-mz/z
y=x/(1+mx)=z/(1-mz)

[gloomy vector]

hold on lemme explore this idea

y=3nx/4
y=5z/4n

or maybe x=4y/3n, z=4ny/5

3n/4y, m/y, 5m^2/4ny

ok idfk

[willow crater]

If u multiply the 2 eqns , u would get Y^2=15xz

[gloomy vector]

is it not 16y^2=15xz

[willow crater]

It is

It's y^2=15xz/16 which is the same thing as 16y^2=15xz

[gloomy vector]

ok

we use $\frac{x}{z}+\frac{z}{x}=\frac{(x+z)^2-2xz}{xz}$ right

[warm shale]

Skill_Issue

[willow crater]

Do u know what x+z is?

[gloomy vector]

no

[willow crater]

Then why are u going there

[gloomy vector]

ir looked really promising

like im having a hard time understanding how to get the second part of the equation (aritmethic one) into action

[obtuse pebble]

@gloomy vector Has your question been resolved?

[willow crater]

When 3 no.s a,b,c are in AP, a+c=2b

[obtuse pebble]

@gloomy vector Has your question been resolved?

[obtuse pebble]

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[sly tulip]

anyone know how to do this?

[sly tulip]

my answer is r x (q+v)

[kind hawk]

well + makes no sense

[sly tulip]

so?

[kind hawk]

you have the correct idea

but writing q+v makes no sense

you cant add sets in that way

[timid silo]

r* (elements in B - elements common with C)

[kind hawk]

there has to be a different symbol between q and v

[sly tulip]

i see

what should i write then?

[kind hawk]

well which symbols for set operations do you know

you want to "combine" the sets q and v here

[sly tulip]

is it 'U'?

[kind hawk]

you should know a symbol for that

yes

union

[sly tulip]

ohhh

ok nice

thanks

.close

[obtuse pebble]

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@upper shuttle Has your question been resolved?

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[obtuse pebble]

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[willow crater]

They did implicit differentiation with x constant

[inland matrix]

Because it's partial differentiation

So you treat x as constant

Had it been normal differentiation then you would've to consider x as well

[obtuse pebble]

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[scenic fjord]

Hello, may I ask how to do this directly? Normally for me, I’ll do cross product first then only I do dot product

But the answer written in this form and I would like to know how to solve in one step

[weary heart]

that's how you solve it in one step

[scenic fjord]

I can’t und the working, because for me I’ll do the cross product first, then only I do dot product

[weary heart]

this determinant does it for you

[scenic fjord]

Oh, so for this, the determinant is the answer?

[weary heart]

yup

[scenic fjord]

Cause it’s not the same method that I did normally, so I’m a bit confuse.

[scenic fjord]

Okay, thanks

.close

[obtuse pebble]

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[silent furnace]

help

[lost tree]

u might need to send this in a new help channel

[silent furnace]

ok

[obtuse pebble]

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[frigid hawk]

sorry for the fact that it is difficult to understand this task due to the different language

<@&286206848099549185>

😭😭

<@&286206848099549185>

🥀

why doesn't anyone want to help

[robust raven]

если основанием является равносторонний треугольник, используйте теорему Пифагора

[frigid hawk]

MB

The line is perpendicular to the plane ABC, ABC is equilateral, AB = 4√3, O is the center of the circle circumscribed around ABC, MO = 3. Find MВ.

[robust raven]

i told you that in russian lan gto make it more clear

MB^2 = MO^2 + BO^2

AND bo = 2/3 * h, where h is a height in this triangle

$BO=\frac{2}3{}h=\frac{2}{3}\cdot \frac{a\sqrt{3}}{2}$

[warm shale]

Joanna Angel

[robust raven]

where $a=4\sqrt{3}$

[warm shale]

Joanna Angel