help-17

[vale wave]

$3 m^4 n^9 \sqrt {5 m n}$

[split heart]

3 is no longer squared. remember, you divide by 2 when you take them out

[vale wave]

ah

[twin meteor]

Kingboo

[vale wave]

so is that the answer?

or is there more steps

[split heart]

That's it. You can't simplify beyond that

[vale wave]

brb then

oh good it was right

las tone

last one

then i gotta sleep

@split heart

[split heart]

You got this one

[vale wave]

bottom is 9 ans 2

oh wait

latex is frac right

for fraction

[split heart]

\frac

[vale wave]

ok

[split heart]

\frac{numerator stuff}{denominator stuff}

[vale wave]

$\sqrt\frac {16 * 2} {9 * 2}$

[split heart]

Good!

[vale wave]

OH IT WORKEd

[split heart]

Do the numerator too

[twin meteor]

Kingboo

[vale wave]

then

idk

4 and 3 go out

[split heart]

yes, and then?

[vale wave]

then its just 2.2

[split heart]

You got something divided by itself

[vale wave]

2/2*

[split heart]

which is?

[vale wave]

1

so

i have a 4 and a 3

left

[split heart]

4/3 yes

[vale wave]

thats all?

ok

[split heart]

yeah! grats

[vale wave]

hey could i dm you or just ping you whenever cuz i gotta sleep

if you dont mind that is

[split heart]

Just send me a message and I'll try to respond as quickly as I can. If I don't you can always post here

[vale wave]

got it

thanks for everything

.close

[vocal sleet]

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[iron iris]

can someone help me with question 5?

[shrewd zinc]

sure

any guesses before starting?

[iron iris]

[-4,-1]

[-4,1]**

[shrewd zinc]

is it continous at x=-1?

[iron iris]

i dont know what discounitiy that would be

if its open circle its not?

[shrewd zinc]

continuity is just does function have a value at that point

at x=-1 does it have a value?

[iron iris]

i dont think so because the circle is open

[shrewd zinc]

right, so it's not continouus there

[iron iris]

what about at x=1

is that a jump discounitity

[shrewd zinc]

technically it might be called something different because open circles

but yeah discontinous

[grim juniper]

f(-1) is undefined, so how can it be continuous or not continuous there?

[shrewd zinc]

yeah you're right

[iron iris]

for the interval notation

[shrewd zinc]

but not continous at least

[iron iris]

whats the difference between [ and )

[ (

[shrewd zinc]

[ if function is contionus at the ending point

( if not continous at ending point

[iron iris]

ok so would [-4,-1) be correct

[shrewd zinc]

yep

[iron iris]

and then (-1,1)

[shrewd zinc]

right

[iron iris]

thank you @shrewd zinc

are you still in school

[shrewd zinc]

np

yeah but taking a gap sem from uni

[quiet hull]

Yo can I ask a question related to this?

[shrewd zinc]

sure

[quiet hull]

What would this be called?

[shrewd zinc]

no idea xdbut idt anyone would care if you just called it jump discontinuity

not that important imo

[quiet hull]

ah i see

thanks

[vocal sleet]

@iron iris Has your question been resolved?

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[river kettle]

so im trying to solve sqrt(3)*tanx = -1 for 0<=x<=2pi

[river kettle]

so far i have tanx = -1/sqrt(3)

i ran that through my calculator and got -pi/6 as my base angle

should i find where tan is positive

if so its positive in first and third quadrants

[small quarry]

bruh

[river kettle]

what do you mean bruh

[small quarry]

$\sqrt{3}\ tan(x) = -1$, $for\ 0\leq x\leq2\pi$

[river kettle]

what

[small quarry]

Better

[river kettle]

bad tex

[twin meteor]

Fossil

[small quarry]

BETTER

[river kettle]

ok now what

[small quarry]

first I suggest

[paper depot]

$\sqrt{3} \tan(x) = -1$ for $0 \leq x \leq 2\pi$

[twin meteor]

Ann

[small quarry]

to put the sqrt 3 to RHS

[paper depot]

(that will be all my input, do not try to rope me in)

[river kettle]

lol

[small quarry]

to dvide sqrt3 to the whole equation]

[river kettle]

thats what i did

[small quarry]

better than mine, lol

then use the identites

[river kettle]

$\tan\left(x\right)=-\frac{1}{\sqrt{3}}$

[twin meteor]

water beam

[small quarry]

ys

[river kettle]

since when did i need identities here

i was thinking of

[small quarry]

there's sqrt(3) in the denominator

and what u should do

[river kettle]

$\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\frac{\pi}{6}$

[twin meteor]

water beam

[small quarry]

NO

or ys

[river kettle]

what

[small quarry]

somehow

[river kettle]

wot

my calculator is just built different

[small quarry]

nvm u did right lol

[river kettle]

it turns it into -pi/6

[small quarry]

ys

gd

[river kettle]

ok

[small quarry]

u learn ASTC?

[river kettle]

yea

[small quarry]

there's tangent

with neg

as u see, the quadrant I and III leads to tan(x) = +num

and quadrant II and IV leads to tan(x) = -num

u know that right?

[river kettle]

yes tan is positive in quadrant 1 and 3

[small quarry]

so ig u know how to do?

[river kettle]

no lol

[small quarry]

ok

well

u learn trigo identites like tan(theta + pi) = tan(theta) right?

[river kettle]

i think i know how to add the angles to get to positive maybe

so tan is -30 degrees basically

[small quarry]

tan(pi - theta) = -tan(theta)

[river kettle]

and we want to get to positive quadrant 1

so we add 2pi to -pi/6 right

$-\frac{\pi}{6}+2\pi$

[twin meteor]

water beam

[small quarry]

huh

no

or ys somehow

bruh

when u do this kind of question

leave the neg alone somehow

the neg symbol just for quadrant thing

So let me explain how I do

[river kettle]

ok

[small quarry]

first u know there's -pi/6

then the angle must be in quad II or IV

therefore, pi - pi/6

cuz if I did this the angle will leave it in quadrant II (I mean I memroize this for ezier, techniqely is not true, lol)

and there's other angle left on quad IV

[river kettle]

wait hold on pause

[small quarry]

ok

[obtuse sierra]

btw this is true

-pi/6 is just 11pi/6

[small quarry]

ys

[river kettle]

i had a hunch to add 2pi but could you tell me exactly why we do that

its to get back to positive right

[river kettle]

is this for when tan is negative

[small quarry]

ys

[obtuse sierra]

- means we are taking the angle in a clockwise sense

so -pi/6 lies in fourth quadrant

which is just 2pi - pi/6

[small quarry]

I would simply like 3pi/2 + pi/6

just ye

I used to

[river kettle]

wait so like this ?

[obtuse sierra]

correct

[river kettle]

so thats where one solution is

-pi/6 + 2pi

what about the other one

[small quarry]

ye

I alr told

pi - pi/6

[river kettle]

why do we do that?

[small quarry]

Dyssrupt, ur time to explain

[twin meteor]

Dyssrupt

[obtuse sierra]

you know that tan(pi/6) is 1/sqrt(3) right?

[river kettle]

yeah

[obtuse sierra]

sub in pi/6 here

and see what you get

[river kettle]

5pi/6

[obtuse sierra]

and the rhs?

[river kettle]

uh

= tanx?

[obtuse sierra]

the right hand side is -tan(x)

sub x = pi/6 there too

[obtuse sierra]

btw you know this, right?

[river kettle]

not really ive seen it before but i dont really understand it well

[river kettle]

do you think you could show me graphically here

[obtuse sierra]

then trigonometry's gonna be hell

these are the basics

[obtuse sierra]

what do you want me to show?

[river kettle]

like you know how i showed adding 2pi got to quadrant 1

can you show what subtracting pi does

doesnt that get me more negative angles

[obtuse sierra]

Sorry, I dont have a means to draw rn, but you can look up for 'reduction formulas of trigonometry' and maybe learn from there.

[river kettle]

is it like this?

[river kettle]

help lol

[small quarry]

huh

am doing limit thing, lol

wassup

how is it?

I bet he's explain it better than me

[small quarry]

How can I help u, sir

State ur problem

[river kettle]

i donto understand the -pi and why we do it

why we subtract pi from pi/6

[small quarry]

pi - pi/6

u mean this?

[river kettle]

yes

[small quarry]

oh

ok let me explain

Btw

I want to ask u someth'in tho, lol

[river kettle]

ok

[small quarry]

how to get the active role and helpful role, lol

[river kettle]

for active just talk a lot in the server

[small quarry]

I talk a lot, lol

[river kettle]

and helpful idk i think you have to help quite a lot of people before you get it

[small quarry]

ok

[river kettle]

like you need a lot of messages i think

[small quarry]

so umm...

well

u need to know sin cos and tan is simply a triangle

all the things we did is just triangle ratio

[small quarry]

and we present the trigo identies with coor

let me show u someth'in

This is how the thing works

[river kettle]

yeah

whats this part

[small quarry]

the triangle I mentioned

sin cos and tan is trigo ratio

ratio

as u see in the graph shown

that

when an triangle rotate 180 degree (which is pi in radian, lol)

the x and y become neg

when u apply the x' and y' (x' means new x which is -x, y is the same), u get -sin theta

Tha'ts how all the thing works

all about trigo and degree or radian, is about rotating

the change of x and y

the reflection (kinda)

u get it now?

the basic concept

or let me do a conclusion

[river kettle]

hmmmm

ok

[small quarry]

what really makes sin or cos or even tan, become neg or change it state

is x and y

whehter x is neg or pos

y either

Basic concept conclusion

u get it?

(about changing state, I can dm u if u want to learn)

[river kettle]

so how does this tell us to add or subtract pi?

[small quarry]

So, u look at ur answer of tan is neg

which means x or y is neg

[river kettle]

yes

[small quarry]

the main problem: what makes x or y become neg

the rotation

but how we rotate to let x or y become neg

ofc we can't let x and y both neg cuz u know tan = y/x

[small quarry]

cuz as u see this, if x and y become -y, -x, tan still positive

therefore, we found that when the angle rotate to quad II

or I should say reflect

[river kettle]

?

i dont know why i put pi/6 as positive i just pretended it was positive

then subtract pi to get to q3

but wait

that would get me to quadrant 2

[small quarry]

well

we need to use the thing we got before

pi - theta = pi + (- theta)

[small quarry]

This is wrong

[river kettle]

Yes

[small quarry]

I mean not really, since the -30 is correct

[small quarry]

as u see there's a -theta

so u should use -theta to rotate and NOT using the +theta to rotate

understand?

[river kettle]

hmmm

[river kettle]

where u get this property from

[small quarry]

wdym

[small quarry]

Clue 1

[small quarry]

Clue 2

[river kettle]

hm

wait

if i graph

[small quarry]

Why u use this stupid graph

[river kettle]

whatt!!!

[small quarry]

Circle can explain everything

[small quarry]

This graph will make u being stupid

[river kettle]

i dont understand rotating thing

[river kettle]

😦

[small quarry]

HUH

What u can't understand

tell me

[river kettle]

i dont understand how to rotate

the thing

[small quarry]

I alr explain

to u

Let me take u step by step

[river kettle]

Ok

[small quarry]

do u understand the basic concept?

[river kettle]

of ?

[small quarry]

BRUH

[small quarry]

THIS

[river kettle]

of the circle?

[small quarry]

I did the conclusion

read it in every sentence

and don't lie to me

understand or not

it's ok to be not understand

just tell me what u can't understand if u can't understand

[river kettle]

Okay hold on

i dont understand when you say it change state

so if sin or cos is negative then tan is negative yes i get that

[small quarry]

Ys I will explain later if u want, I can dm u

Cuz changing state is another concept, but it still about rotating

[small quarry]

HUH

How u made that conclusion

[river kettle]

because

tan is sin/cos yes

and if sin negative then

-sin/cos = -tan

[small quarry]

oh ys, i read wrong, lol

nvm ur correct

so any thing u can't understand?

[river kettle]

ok i think that part is ok

[small quarry]

So next

about the rotation

Do u understand how rotation affects x and y?

[river kettle]

No

[small quarry]

Look at this

Did u see the triangle

do u see how it changes

[river kettle]

yes

[small quarry]

So now do u understand how it works?

How the rotation affects x and y

[river kettle]

so rotation makes x and y negative

[small quarry]

ys

it can change the pos and neg of x and y

understand?

[river kettle]

yes

[small quarry]

Gd

Then head back to the problem

pi - theta

pi - theta = pi + (-theta)

[small quarry]

Which just like the -30 in this figure

understand?

the theta is 30 degree

?

[river kettle]

I thought it -30

[small quarry]

The degree is -30

then u rotate is with 180

just simple is that

u fully understand now?

[river kettle]

so -30 + 180?

[small quarry]

ys

[river kettle]

so + 180 gets to quadrant 3?

[small quarry]

NO

because it's -30

u can draw the figure

it stays at quad 2

[river kettle]

are we trying to find solution in quadrant 3

[small quarry]

NO

Why in quad III

tan in quad III is pos

turn the line, 180 degree

[river kettle]

so are we trying to find solution where tan is negative

[small quarry]

YES

[small quarry]

This is what u calculate

[river kettle]

like this?

[small quarry]

.

[river kettle]

yes

[small quarry]

Wrong

[small quarry]

U clearly don't know how it works

The rotation must be anticlowise

cuz u have +180 not -180

[small quarry]

YS GD

[river kettle]

Ok but i have question

[small quarry]

so

ys ask

[river kettle]

im confused. So we added 2pi onto -30 which was $-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}$. But this is in quadrant 1 isnt it and tan is positive in quadrant 1

[twin meteor]

water beam

[river kettle]

and we want to find where tan is negative yes

[small quarry]

NO

[river kettle]

WHAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAT

BUTTTTTTTTTTT YOU SAYYYYYYY WE WANT TAN NEGATIVEEEE NOWWW WE WANTTTT POSITIVEEEEEEEEEEEEEEEEEE

[small quarry]

Bruh, that's why I said the formula which dessrupt tought is not gd at all

cuz it will makes a guy like u messy

[river kettle]

so sad

[small quarry]

This formula won't make u to quad 1

[river kettle]

what quadrant it get me to

[small quarry]

4

Let me explain it to u

[river kettle]

ok

[small quarry]

First of all

do u know why we need to have pi - 30?

Do u understand

@river kettle

[river kettle]

to get back to quadrant 2

?

[small quarry]

Do u understand why we need quad 2 to have -tan

Yes or no

[river kettle]

because our solution is $\tan x=-\frac{1}{\sqrt{3}}$ meaning tan has to be negative

[twin meteor]

water beam

[small quarry]

.

am asking u do u understand why quad 2 can make tan become neg

[river kettle]

yes because in quadrant 2 only sine is positive meaning rest is negative

[small quarry]

.

Because the x and y

bruh

[river kettle]

yes thats what i just said

bruh bruh bruh

sine is positive meaning rest is negative

[small quarry]

ok i bet u understand here

and let me introduce how I manage to complete the second answer

about the changing state

[river kettle]

Ok

[small quarry]

This is the changing state

tan -> cot, sin -> cos, cos -> sin

All is x and y

More clear explination

[obtuse sierra]

[small quarry]

Sorry for that Dyssrupt

lol

[obtuse sierra]

lmao

[small quarry]

u appear again

[obtuse sierra]

i sullied for dessrupt

[small quarry]

I gtg

I let u explain

[obtuse sierra]

nah, im busy lol

[river kettle]

Ok wait

i think i figured it out

[river kettle]

yes i figure it out now

.close

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[ruby ocean]

can someone answer this?

this question is basically asking the infinite integer progression sum of n/(2^n)

[vocal sleet]

@ruby ocean Has your question been resolved?

[vocal sleet]

@ruby ocean Has your question been resolved?

[flat whale]

Yes this sum is finite

Usual way is to consider the series of 1/(1-x) and differentiate

,tex .maclaurin

[twin meteor]

riemann

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[vast shale]

Why did they not consider the horizontal component?

[obtuse sierra]

they cancel each other.

[vast shale]

I don’t understand

Ohhhhh

I get it

Thank you so much

.close

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[tired raptor]

⟨√11⟩^5 guys sorry how do I do this

[split heart]

what's with the brackets?

[tired raptor]

idk

(√11)^5

[split heart]

Oh, ok. So what's the issue here?

[tired raptor]

I need to rewrite this using a rational exponent notation

[split heart]

Gotcha

So are you familiar with what the square root sign does?

[tired raptor]

no

i mean

yes

idk what do do with it on this equation t ho

[split heart]

First of all there's no equation there

But more to the point, can you convert the square root into a rational exponent?

[tired raptor]

⟨√11⟩^5

[split heart]

Ignore that for now

[tired raptor]

how do I convert into a rational exponent

[split heart]

I'm asking you if I gave you the expression sqrt(x). How can you rewrite that as an exponent of x

[tired raptor]

X divided by 1 over 2

?

[split heart]

do you mean x/2 or x^(1/2) ?

[tired raptor]

the second one mybad

X to the power of 1/x

?

[split heart]

Okay. So you know that square roots turn to powers of 1/2

[tired raptor]

wait stop

wait

X to the power of 1/2

is that right?

[split heart]

Yes

[tired raptor]

alr

[split heart]

$\sqrt x = x^\frac{1}{2}$

[twin meteor]

TooManyCooks

[tired raptor]

✅

[split heart]

Now. Are you familiar with the properties of exponents?

[tired raptor]

yes

[split heart]

Good. So if I give you x^a * x^b, what do you get?

[tired raptor]

x^ab?

[split heart]

Not quite

[tired raptor]

but if u have the same base dont u just add exponents

[split heart]

Precisely

But you multiplied

[tired raptor]

x^a+b

[split heart]

Yes that's right

Now your question involves powers of exponentials

$(x^a)^b$

[tired raptor]

ohhh

[twin meteor]

TooManyCooks

[tired raptor]

now I get it

11^1/2 times ^5/5

?

11^11/2?

is that the answer

[split heart]

Uh, what

Slow down

[tired raptor]

like

[split heart]

What does (x^a)^b give you?

[tired raptor]

x^ab

ohh so u multiply this time

[split heart]

Cool. Now you have a = 1/2 and b = 5

So what do you get?

[tired raptor]

alr lemme see

11^5/2?

[split heart]

That's right!

Is that the answer you're looking for?

[tired raptor]

i think so ye

thank u bro

.close

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[vocal sleet]

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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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• Be polite and have a nice day!

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[main ferry]

why doesnt this work

[split heart]

properties of log

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-prop/a/properties-of-logarithms

[main ferry]

yes i watched that

how is this not correct

[split heart]

why is x in the exponent

or am i tripping

[river minnow]

Yeah don't put the expressions that high

And who's telling you that 2ln(x) = ln(x) + ln(x) is wrong?

Or any of those steps

[main ferry]

then this should be correct

[split heart]

Who said it was wrong?

[main ferry]

markscheme

also this person

the answer is supposed to be 2ln(y)

[river minnow]

Well ln(y^2) is the same as 2ln(y)

[main ferry]

yes

[outer warren]

who's work in black

[split heart]

They did algebra wrong

[main ferry]

friend

[split heart]

Friend is wrong

[outer warren]

they fked up their algebra

[main ferry]

ahh thank u can u tell me where they went wrong

[outer warren]

$-p-q \redneq -(p-q)$

[split heart]

$-\log x^2 y^2 - \log y^3 = - \log x^2 y^5$

[twin meteor]

ℝam()n()v

TooManyCooks

[split heart]

heh... redneq

[main ferry]

ok thanks guys!

.close

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[snow moth]

Could someone show how they get 21.44 centimeters from 18.20(3pi/8)

they dont really explain anywhere how they got to this number

[livid tapir]

Plug into ur calculator

Units are cus 18.20 is in cm

Oh hold up lol you want more than that

[snow moth]

gonna just assume the answer is what the calculator shows cause my cats decided to hide the calculator

somehow

when i did it on the pc calc i just got something terribly diffgerent

[livid tapir]

Weird

I haven't actually out it in but it's about 9/8*18 so it looked roughly right to me

[snow moth]

found the calculator actually and yah got it

i was replacing pi with 180 cause earlier to find radians it said u could do that

but with this i guess u cannot

[livid tapir]

Oh 💀

[snow moth]

i do not like trig

[livid tapir]

dw there's worse things out there :p

[snow moth]

ANYWAYS thank u for the help lol

im sure there is

[livid tapir]

Np

[snow moth]

i cannot wait

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[karmic glen]

So i got how the solutions did the equations

And I was able to work it out myself

I just dont understand how the units changed from minutes to hours

Because when we plugged in t = 10, units were minutes

Then the final answer became hours

[sly sierra]

yea there's no way the answer is 3.654 hours, if more than half of the radon has decayed after 10 minutes

[karmic glen]

Yeah.. so it should be minutes right?

But it looks like the work is right

[sly sierra]

yea the work looks fine, they just got the units wrong

[karmic glen]

Thank you

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[pine temple]

i dont even know where to start with this one

[vocal sleet]

Please don't occupy multiple help channels.

[flat whale]

closed your other channel

[pine temple]

ok

[pine temple]

<@&286206848099549185>

[tepid moss]

Hm

[pine temple]

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[river kettle]

How can I solve sin(x/3) = 1/sqrt(2)

[river kettle]

without the unit circle

[obtuse sierra]

do you know when sinx = 1/sqrt(2)?

[river kettle]

Not really

Wait

1/sqrt(2) is pi/4?

[obtuse sierra]

x is pi/4

[trail mesa]

remember, there are many more

[river kettle]

So that means sine of 1/sqrt(2) = pi/4 right

[river kettle]

Yeah I’m just finding the base angle first

[obtuse sierra]

no

sin(pi/4) = 1/sqrt(2)

[river kettle]

so sin(x) = 1sqrt(2) = sin(pi/4)

[obtuse sierra]

yes

[river kettle]

okay

And solutions are when sinx>0 so that’s in Q1,Q2

So our first solution is pi/4 and the second will be pi - pi/4 ?

[obtuse sierra]

correct

[river kettle]

so pi/4, 3pi/4

Then x/3 = pi/4, 3pi/4 and I just solve for x

[trail mesa]

now might be a good time time to add on the $+2k\pi$

[twin meteor]

lpieleanu

[obtuse sierra]

I think they need principal values

or no?

[river kettle]

oh yeah well this was actually just part of another question I’m solving

but my original domain for the equation was between -pi<=x<=pi

anyways I got x = 3pi/4, 9pi/4

@obtuse sierra how about solving for sin(-pi/4) between -pi to pi

The solution I’m looking at just has the same one as positive pi/4 but with a negative sign

[obtuse sierra]

pi/4*

[river kettle]

What

That’s what I typed

[obtuse sierra]

typo

[river kettle]

What

I’m correct

[obtuse sierra]

pi/4 and 3pi/4

[river kettle]

How

x/3 = pi/4 , 3pi/4

[obtuse sierra]

oh lmao

it was x/3

sorry mb

[river kettle]

[river kettle]

So what about this

[obtuse sierra]

-1/sqrt(2)

[river kettle]

yeah

[obtuse sierra]

oh you want sin(x) = -1/sqrt(2)

[river kettle]

Well technically

sin(x/3)

But yeah

[obtuse sierra]

just go anticlockwise

as the domain is -pi to pi

you can now include negative angle (not negative really)

[river kettle]

so i just have to add on a minus sign

?

[obtuse sierra]

yeah

as sinx is an odd function, so sin(-x) = -sin(x)

[river kettle]

okay

[river kettle]

Wait does this trick only work if the domain is -pi to pi or if the function is even or odd

[obtuse sierra]

wdym

what trick?

[river kettle]

the negative trick where you just slap on a minus sign

[obtuse sierra]

yeah odd functions

[river kettle]

So even if the restriction wasn’t for -pi to pi

it would still work?

[river kettle]

Am I correct in saying this?

[obtuse sierra]

yeah cuz f(-x) = -f(x), so adding a minus sign, will yield a negative result of the original.

[river kettle]

So like if I have sin(x) = 1/2 which is pi/6, it will have a negative solution -pi/6 as well?

[small quarry]

back

lol

[river kettle]

So let’s say this was cos what would happen

[river kettle]

Lol

[vocal sleet]

@river kettle Has your question been resolved?

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[solid mist]

For ii), the hint was that I needed to take the derivative of the function. But, what does it mean when deals with the "max"

This is what I have with the hints we were given in class

[worthy citrus]

derivatives are intimately tied with maximums and minumums

if a function has a local max or min at a point not on its boundary then the first derivative must be zero at that point

[solid mist]

So do I just need to show that f'(x)=0? Or more specifically do i need to find the x that makes f'(x)=0?

[worthy citrus]

yes precisely

the sign of the second derivative will tell you if its a max or min

[solid mist]

Do I use f(x)=1/sqrt(2pisigma?

or do I need to use the original pdf?

[worthy citrus]

the actual function f(x)

when they write $\operatorname{max}_{x \in \mathbb{R}} f(x) = \frac{1}{\sqrt{2\pi\sigma}}$

[twin meteor]

ΣΑCu

[worthy citrus]

they mean, the maximum value that the function f can achieve across all x values in R, is 1/sqrt(2pisigma)

[solid mist]

Alright, thank you for the helpful hints! I will come back if I get stuck, but I think I need to start with just taking the first derivative and go from there. Thanks again!! 🙂

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[vast shale]

I dont understand specifically after they obtained the individual values of Q's

can anyone explain?

[vocal sleet]

@vast shale Has your question been resolved?

[bleak oak]

specifically there they just plugged in the value of a back in the original calculation for Q

i.e. -15Q(2) = -2a --> Q(2) = -2*315/(-15)

are you asking how they got the original values for Q?

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[reef delta]

i think the answer is 23 i used scale factors im pterry sure theres a typo on the techers side im trnna to help my brother with his homework this was the only quston that was hard for him

[willow hill]

r the diagrams similar?

as in are the ratios the same?

[reef delta]

the quston that the website gave was "Two scale drawings of the same cathedral are shown. Look carefully at the dimensions of the drawings. What is the height of the cathedral on the right?"

[willow hill]

i got 35

can u show me ur working for how u got 23?

[reef delta]

so

what i did was

15 - 6 to get 9

so i did 14 +9

[willow hill]

it doesnt work like that

the ratio of the sides are the same

so, 6:15 is the ratio for the base

therefore, use the same ratio for the height

[reef delta]

ohh i get it now

[vast shale]

no after that, so how they say Qx=Tx... part

seems like this person finished his question anyway, so if u dont mind u can continue explaining

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[worldly bolt]

Bit of a long shot, but does anyone know where I went wrong here?

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@worldly bolt Has your question been resolved?

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@worldly bolt Has your question been resolved?

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@worldly bolt Has your question been resolved?

[worldly bolt]

<@&286206848099549185> Anyone know "basic" cal 2?

[silk kindle]

Hello 👋🏽 . In the expression for r in terms of x, it looks like you wrote 18 instead of 13

[worldly bolt]

Oh my gosh

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@worldly bolt Has your question been resolved?

[crystal sedge]

damn

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[twin citrus]

how to calculate t hi s?

[zealous nova]

thats negative e in the front?

[zealous nova]

and r is any varible?

[twin citrus]

r is positive

and it is not e

it us just a positive constant c

[vast shale]

R also a constant?

[zealous nova]

yea

[twin citrus]

this is origial task

this is wolframalphas answer

which i have no idea what is

[zealous nova]

use exponent property?

[twin citrus]

and thai is desmos graph of similat function

[zealous nova]

log(a^b) is b loga

and then

limit can get into the logs parenthesis

correct

(r/R)^((r/R)^2)

[zealous nova]

u gotta calculate limit of this correct?

[twin citrus]

y

[zealous nova]

.

[zealous nova]

..

[zealous nova]

limit can jump into the inside of the logarithm ryt

e ln( lim x^x^2)

[twin citrus]

ok

what next?

[twin citrus]

its not e but -c

[zealous nova]

yah nvm

so...

x^2 is a lot greater than x

[twin citrus]

yes

[zealous nova]

wait

wait im not rly sure

what to do

someone else will help

[twin citrus]

any chance to use lhopital rule here because this chapter is about that rule ?

[zealous nova]

u gotta differentiate both numerator and denominator in that case ryt

denominator is 1

differentiating that gives u 0

again p/0 issues

[twin citrus]

apply exponent rules i know but apply the limit chain rule i do mot get how to do

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[twin citrus]

how to solve this ?

[paper depot]

did you maybe mean $\int_1^x \sqrt{t^2 - 1} \dd{t}$?

[twin meteor]

Ann

[paper depot]

as-is the notation is incorrect

[twin citrus]

i need to find untegral from 1 to some x under this parabole ( x^2-y^2=1) curve

[paper depot]

that is a hyperbola, not a parabola.

[twin citrus]

yy

[paper depot]

is y = why or is y = yes?

[twin citrus]

yes

[paper depot]

right

so you want to find the shaded area, yes?

which is a triangle minus that integral...

[twin citrus]

yes originally i want that

[paper depot]

ok right

so you do want $\int_1^x \sqrt{t^2 - 1} \dd{t}$.

[twin meteor]

Ann

[paper depot]

probably some kind of trig sub like t = 1/cos(phi) would help.

[twin citrus]

but i do not understand why you untroduced t while the curve equation is y=sqrt(x^2-1)

[paper depot]

you cannot have x appear both in the limits of integration and as the integration variable.

i.e. $\int_1^x (...) \dd{x} = {}$ not kosher.

[twin meteor]

Ann

[twin citrus]

ok then lets change opper x and call a

this is more exapteble for me

[paper depot]

exapteble?

oh you meant acceptable

[paper depot]

so you want $\int_1^a \sqrt{x^2 - 1} \dd{x}$.

[twin meteor]

Ann

[paper depot]

my point still stands:

trigonometric substitution

[twin citrus]

sey for my english witing skills

[paper depot]

x = 1/cos(phi)

[twin citrus]

what is this?

[paper depot]

a substitution

read up on "trigonometric substitutions"

paul's online math notes has it

i think

[twin citrus]

idk who is paul but i found that concept in my book as well

so i need to learn that first yes ?

in order to find that integral

[paper depot]

https://tutorial.math.lamar.edu/

[paper depot]

yes, you need to learn substitutions in general

[twin citrus]

ok then i will leave it as it is for now and will try to solve the problem after i will read that

if something will come back for seeking help

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[burnt totem]

.close

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[rain glacier]

how can i simplify this?

[split heart]

can you factor something in the numerator?

[rain glacier]

for 6 and 24?

umm...

[merry python]

cute

[gritty zenith]

<@&268886789983436800> 😊

[strong grove]

yes

[paper depot]

do not spam.

[split heart]

<@&268886789983436800>

[gritty zenith]

Have you gotten any further?

[rain glacier]

i actually am having difficulties factoring..

[strong grove]

see if there is a number that is factor to both 6 and 24

[heavy flare]

do you know about difference of squares?

yeah first do that

[rain glacier]

isn't the number for 6 and 24, 6 as well?

[strong grove]

yes

take it out and see what u are left

[gritty zenith]

What is common in the numerator

[rain glacier]

remove the 6 in the numerator?

[rain glacier]

6

[gritty zenith]

Great

https://media.discordapp.net/attachments/903463353417072641/1148974421617430650/IMG_2041.jpg

6x^2/6 is what?

[rain glacier]

x^2

[gritty zenith]

-24/6 is what?

[rain glacier]

-4

[gritty zenith]

Great

So you have 6(x^2-4)

Does this ring any bell?

[rain glacier]

ohh, so i have to put the common factor outside the parenthesis, then put the factored numbers or variables inside ?

[rain glacier]

i'm not sure, i'm super slow sorry ;-;

[strong grove]

yes

when u have a common factor u have to place it outside the parenthesis

this is because patenthesis means multiplying and the number outside mulitplied with inside results the same thing so u arent changing anything

[gritty zenith]

%^^

[gritty zenith]

Difference of 2 squares

You might be asking where the second square is

4 = 2^2

So you have 6(x^2-2^2)

Now I’m assuming you don’t know what this is

[rain glacier]

i don't ;-; sorry

[rain glacier]

so i have to split (x^2-2^2) into two parenthesis?

resulting in (x+2)(x-2)?

[gritty zenith]

yes correct ✅

So what will you have as the numerator

[rain glacier]

6(x+2)(x-2)?

what about the denominator?

[gritty zenith]

Notice how you can apply the same principle to the denominator

[elder geyser]

can someone help me

[gritty zenith]

Please find a different channel

[elder geyser]

oh sorry

[rain glacier]

so i'm going to use the difference of squares in x^2-4 as well?

[gritty zenith]

Yes!

[rain glacier]

i have a question, when scenario can i only use the difference of squares?

because in multiplication of functions, i just substituded and distributed

[gritty zenith]

When you have 2 squares that are being subtracted from each other then you can apply it

[rain glacier]

can you give me an example.. i don't seen to understand in words alone.. sorry 🥲

[gritty zenith]

(x^2-1)

Can you apply it here?

[rain glacier]

would that be (x^2+1) and (x^2-1)?

or will it be 1/2?

[gritty zenith]

no

the x^2 becomes just x

Ignore the exponent

[rain glacier]

then (x+1)(x-1)?

[gritty zenith]

Yes!

[rain glacier]

so i have 6(x+2)(x-2) / (x+2)(x-2)

do i cancel out the like terms?

[gritty zenith]

Yes

Correct

[rain glacier]

and all i would have left is 6?

is that the answer?

[gritty zenith]

Yes

Perfect!

[rain glacier]

okay.. thank youuuuu!

.close

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[hard hawk]

how do i determine the value of a b c and d in "(ax^2+bx+c)/(x+d)" using a graph programme like geogebra?

[hard hawk]

knowing that the x asymptote is -1

and that f(x) is more than or equals zero when x [-2,-1) and [2->

[paper depot]

[2->
what does this mean

[hard hawk]

two and more

so from 2 onwards x is positive

d is obviously 1

[paper depot]

so is this like, [2, +∞) but for people who are allergic to the infinity symbol?

anyway idk how to use geogebra to determine your stuff

but it would not be hard to do by hand

[hard hawk]

yeah obviously, but the exercise wants me to use a digital tool

alternatively i could use a calculator programme

but what is it called when you find out values of a function's coefficients like that? extrapolation?

♾️

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@hard hawk Has your question been resolved?

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@hard hawk Has your question been resolved?

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@hard hawk Has your question been resolved?

[hard hawk]

.close

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