help-33

[flat raft]

measurable function from Omega to R, yeah?

then what is $1_\emptyset$?

[elfin berry]

Saccharine

[flat raft]

what is $1_\Omega$?

[elfin berry]

Saccharine

[flat raft]

substitute into the definition of Q

this is supposed to be the easy part

wtf

do you not understand what the definition of Q is

what did you get

I'm not sure what you're doing

showing that Q(omega) = 1 using the definition should NOT be the hard part

what is this

have you taken any math classes before this

any proof-based ones

linear algebra?

abstract algebra?

real analysis?

an intro to discrete math?

do you understand everything that the question has stated

0 < E[X] is just saying the expectation is positive

E[X] = int X dP is saying that the expectation is defined as the Lebesgue integral of X with respect to the probability measure P

E[X] < infinity just means that X is integrable

I'm literally so confused

I don't even know what to tell you

I'll probably just give you the answer for why Q(emptyset) = 0 and Q(omega) = 1 but honestly I don't know what to say or do to help you understand

$1_\emptyset(\omega) = 0$ literally everywhere

[elfin berry]

Saccharine

[flat raft]

$Q(\emptyset) = \frac{\mathbf E[X 1_\emptyset]}{\mathbf E[X]} = \frac{\mathbf E[X \times 0]}{\mathbf E[X]} = 0$

[elfin berry]

Saccharine

[flat raft]

no like I'm just baffled why it's not immediately obvious that E[0] = 0

even if you go through the machinery of the construction of the Lebesgue integral, I fail to see why you need to use a random inequality to conclude that

like bruh

no see you're just following what I wrote without understanding

why it's like

did you like skip everything or something

I'm so confused

[noble python]

yea seems good to me bos

[marsh citrus]

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[still temple]

I got this as my answer rip

[brave spire]

....

3V/2, no?

[turbid badger]

i am going to high school ... 😦

[brave spire]

Wait oh i am not supposed to do calculus here am I

[turbid badger]

welp i'm 13

[brave spire]

Am in the same boat buddy

[turbid badger]

I HATE Math...

[brave spire]

Okay no I love math

[turbid badger]

bruh

i cant even do 2 pie r

[still temple]

this is calculus

[brave spire]

Oh i am doing the right thing then 🙂

[turbid badger]

2TTr i cant even solve it dude

[brave spire]

Hmm..

[still temple]

,w implicit derivative Psqrt(V^3)=500 with respect to T

[elfin berry]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by _____#6266

[still temple]

calculator doesn't even work

.close

[marsh citrus]

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[still temple]

Given that the raduis is 2mm. Find the area

[glass silo]

The area of?

[still temple]

square

[glass silo]

Well if they give that it's a square, you can find the side length easily...

[still temple]

yes

but the trick is you can't find the l

because the two circles touch

[cunning jackal]

The circles are touching each other and also the sides

[still temple]

That's why

[cunning jackal]

That's why?

[still temple]

Can you just help me find the area

lol

[cunning jackal]

Well this was supposed to be the hint lmao

[still temple]

Triangle

?

Equal triangle

each sides is 2mm

[cunning jackal]

Try drawing all radii you can

Which you think can help you find the length

[glass silo]

...
are you sure about this comment(?)

[still temple]

Can you just give me the answer

Please

I'm doing this for fun

Not for school

[cunning jackal]

What's the fun if you're just told the answer???

[glass silo]

Well, you can find it for the fun of it

[still temple]

No I want to understand the solution

That's what I mean

write the solution

Also half my screen is broken, my mouse doesn't work (I'm using keys), and I'm writing an essay

Give me another hint

[glass silo]

What's a diameter?

[still temple]

2r

A=4xl

[cunning jackal]

Look at the upper two circles

[still temple]

Yes

[cunning jackal]

And how they 'fit' in

[still temple]

The side length is 4

[glass silo]

Using this, can you find the side length of the square?

[cunning jackal]

Not quite

[still temple]

4 x WHAT

I mean 8

I'm dumb

8

[cunning jackal]

Yes, you got it

[still temple]

Okay but how do I find l

A=8xL

[cunning jackal]

Bro u just found the side

And now you want the area

[still temple]

I want both sides

ofc

[cunning jackal]

It's a square

[still temple]

No it's not

The question says it's a rectangle

[glass silo]

[still temple]

Excuse me you need to leave

[cunning jackal]

You're the one who said that

It's a square

[still temple]

Ok you know what I mean

Can you help me please

[cunning jackal]

Alright if it's a rectangle then let me think

[still temple]

okay 🙂

[cunning jackal]

Form the triangle by joining 3 centres and then use trigonometry

To figure out the length from the centre of the bottom circle to the upper base of the triangle

if that makes sense @still temple

The height would be 2 times the radius + part of the green line inside the triangle

Which is nothing but the altitude of the equilateral triangle

@still temple

[still temple]

Omg

Yass

Law of sines

I think

Idk pythag

[cunning jackal]

No need to use that, recall what the angles of a equilateral triangle are equal to

[still temple]

60

[cunning jackal]

Yes

So look at the right small triangle

[still temple]

60 30

[cunning jackal]

Yes

Nice

[still temple]

Idgaf about special triangles

[cunning jackal]

You can apply cos(30°)

To find the altitude

[still temple]

altitude?

[cunning jackal]

You know the hypotenuse is 4

[still temple]

what's that

[cunning jackal]

And want the adjacent side

[still temple]

okay

[cunning jackal]

The green part of the line inside the triangle

[still temple]

yup

I get it

thanks. Is there any way I can redeem like a thank you point or something?

[cunning jackal]

Idk man

But yeah btw sorry for my earlier solution

Because

This figure

Can never be a square

[still temple]

Ik

[cunning jackal]

I should have realised that

[still temple]

It's okay

You're still the one helping me

[cunning jackal]

👍😄

[still temple]

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[marsh citrus]

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[marsh citrus]

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[light frost]

You know that one pair of sides and one pair of angles are the same

What more information do you need to prove that the two triangles are congruent?

[vapid wyvern]

I got it

[light frost]

yup

[marsh citrus]

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[sleek lake]

100³

1 million

[high pine]

1 m = 100 cm ---> 1 cm = 0.01 m
cube both sides:
1 cm^3 = 10^(-6) m^3

[marsh citrus]

@analog pier Has your question been resolved?

[marsh citrus]

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[long cape]

(cf)'=cf'
what is this identity called?

[high pine]

it's rather property

that tells you can take out constant

when dealing with derivatives

[glacial hedge]

Linearity of the differential operator/derivative

[long cape]

that's a mouth full

i was hoping for "constant rule" or something nice, like Power Rule, Product Rule, etc.

[glacial hedge]

Maybe you can. It just never occurred to me because you'd probably never have to detail it

[high pine]

constant rule rarther refers to the property that derivative of a constant is zero

[broken forum]

maybe constant multiple rule

ayeee

common sense ftw

[long cape]

ooooo

ty

[glacial hedge]

"trivial application of Leibniz's formula/theorem for derivatives" is that better ?

[broken forum]

lmaoo

[long cape]

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[normal ginkgo]

Hello

Um

I guess the answer is No Solution right???

hehe

Because it could be solved by Elimination method if it had a second equation with it

also

we cant take any common between 2a and 5b

sooooo

There is no solution

right?

[nova totem]

If there are more variables than equations, then there are an infinite number of solutions

[normal ginkgo]

??

how

[nova totem]

Try to isolate a here

a = 1/2 (7 + 5b)

You can choose any number for b

And there will be a corresponding a

[normal ginkgo]

aaa

what

??

[marsh peak]

E.g. a = 3.5 and b = 0 is a solution

[nova totem]

But you can choose an infinite number if b's

[knotty trellis]

all the points on this line are solutions

[nova totem]

In fact if you were to graph the equality, it's a line. Lines have an infinite number of points

[normal ginkgo]

sooo

example

2a + 5x + 9x = 2

there are also infinite solutions for this one?

[marsh peak]

Yup

[nova totem]

Yes

[sharp cypress]

how can i prepare for my math olympiades?

[marsh peak]

If a and x are the unknowns

[normal ginkgo]

hmm

[marsh peak]

I think asking that in #competition-math will be more helpful

[normal ginkgo]

thank you

:)

.close

[marsh citrus]

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[normal ginkgo]

wait

.reopen

[marsh citrus]

✅

[normal ginkgo]

what if my teacher wrongs it

what do i say

in that case

what do i tell her

that its infinite

[marsh peak]

Yes

[proper ridge]

because it is?

hmm

i guess

you can express b in terms of a

then you say since it works for any values of a

therefore there are infinite solutions

[marsh peak]

Yup

[normal ginkgo]

ok

i have another question

this

[still temple]

sup baby

what's the benefits from learning math and be a pro at it ?
will i buy a BMW by solving math issues 🤓 ?

[low light]

Do you know the formula of x³+y³?

It's (x+y).(x²-xy+y²)

Now, we know that x+y is 3

That means x²-xy+y²=4

[normal ginkgo]

yes

[cunning jackal]

use (x+y)³, it's better

For this problem

[normal ginkgo]

how

it was 3 rn

[cunning jackal]

(x+y)³=x³+y³+3xy(x+y)

[normal ginkgo]

ik both formulas

but idk how to derive the answer

[cunning jackal]

All you have to do is put now ..

Put the values in

[low light]

No, x+y is 3, when we wrote it down as 3.(x²-xy+y²)=12, x²-xy+y²=4

[cunning jackal]

We know x+y, we know x³+y³

[cunning jackal]

That just leaves xy as the unknown in the equation

[normal ginkgo]

???

[normal ginkgo]

how is 12 - 3 = 4

me no understand

rip

[cunning jackal]

@normal ginkgo have u tried this

[normal ginkgo]

um

leme try

12+3xy(3)
12+9xy
9xy= (-12)
xy = (-12/9)
xy = -1.33?

@cunning jackal

reee

its not the answer

!!!

[cunning jackal]

Incorrect

Why did u magically equate 12+9xy to 0?

Its equal to (x+y)³, not 0

[normal ginkgo]

ohhh

but

wait

ok

(3) ^3 = 12+3xy(3)
27 = 12 + 9xy
27-12 = 9xy
15/9 = xy
1.66??

bruh

@cunning jackal

[cunning jackal]

Yes

[normal ginkgo]

its not correct

its not one of the options

[cunning jackal]

What is 5/3?

[normal ginkgo]

wtf

hwow

how it is same

wait

wait a min

[cunning jackal]

15/9=5/3

[normal ginkgo]

sowry

ur right

boss

[cunning jackal]

Divide numerator and denominator by 3

[normal ginkgo]

its actually

i have not slept for 16 hours

doing just math

so my brain perfomance

[cunning jackal]

You should sleep

[normal ginkgo]

is a lil impacted

[cunning jackal]

You can't do math like this

And it's no fun or no benefit

[normal ginkgo]

but

[cunning jackal]

Doing math like this

[normal ginkgo]

its exam

after tomrow

[cunning jackal]

You'll perform worse if you're lacking sleep

[normal ginkgo]

:/

[marsh citrus]

@normal ginkgo Has your question been resolved?

[normal ginkgo]

Is middle term to be broken here?

[still temple]

No

I think what u can do is let a variable equal to x^2

[normal ginkgo]

??

[still temple]

So like let a=x^2

Then the equation becomes a^2 + 11a - 180

[normal ginkgo]

ohh

[still temple]

And then you can factor it like a quadratic

[normal ginkgo]

do we put back values later on

for example if answer is

11 a

we do it

11x^2

?

[still temple]

Is factorable?

[normal ginkgo]

??

[still temple]

Is it factorable?

The equation is it factorable?

[normal ginkgo]

what you mean by factorable?

[still temple]

Like for ex. x^2 - 4x + 4 becomes (x-2)^2

Do you understand how i got from the original equation into factored form?

[still temple]

Once you get your roots/solutions then yes

[normal ginkgo]

ok

[still temple]

Does that help?

[normal ginkgo]

yes

Need help with this

too

Wanna ask

how

is

LMN + MNO

= 180

[marsh citrus]

@normal ginkgo Has your question been resolved?

[marsh citrus]

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[dawn onyx]

Hey, we have 3 independent situations A B and C

[dawn onyx]

We know that P(A) = 1/2

And

I simplified it to

And the question is

What is P(C)

Any tips or help on what to do next bcus im a lil stuck

<@&286206848099549185>

[past maple]

Use distributive law to get C outside of the parentheses

[dawn onyx]

And then apply the P(A)+P(B)-P(A/\B) rule?

[past maple]

No, I don't think that is useful

Instead use $P((A\cap B^c)\cup (A^c\cap B)) = P(A\cap B^c) + P(A^c\cap B)$

[elfin berry]

d឵

[dawn onyx]

I got this btw

[past maple]

C is independent of the rest

[dawn onyx]

How do you get that out of the thing I got?

[dawn onyx]

How do you mean?

I mean I know what it means but how does it help us?

OOOOOOW like

Nvm

[past maple]

If X and Y are independent then P(X /\ Y) = P(X)·P(Y)

[dawn onyx]

If you have independent A B and C, Is that for every sort of combination of for eg A and B in relation to C, like in my exercise? Like P(C/(A/Bcomp))=P(C)*P(A/Bcomp). ?

[past maple]

Yes

That is not hard to prove

[dawn onyx]

Does that change if C is included in the combination?

[past maple]

Yes

[dawn onyx]

Can you show one please?

[past maple]

If $X$ and $Y$ are independent, $X$ and $Y^c$ are also independent. Write $X = (X\cap Y) \cup (X\cap Y^c)$, then $P(X) = P(X)\cdot P(Y) + P(X\cap Y^c)$. You also have $P(Y) = 1 - P(Y^c)$, and mixing both equations you end the proof

[elfin berry]

d឵

[dawn onyx]

Ooook thank you 🙏

.close

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[dawn onyx]

.reopen

[marsh citrus]

✅

[dawn onyx]

Yeah still stuck lmao

My fault

[dawn onyx]

So I implemented this but I dont know how to get rid of B

[past maple]

Apply this again

[dawn onyx]

Is this right?

[past maple]

Yes

[dawn onyx]

Thank you 🙏

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[rancid saffron]

How do I solve no.5?

[rancid saffron]

(Work copied, not mine)

[wind light]

Use the properties of log

[rancid saffron]

I’m gonna need a better explanation that that sorry

<@&286206848099549185>

[coarse vale]

@rancid saffron what the difference between the compounded and compounded continuously interest?

Do you have any idea of it?

[rancid saffron]

I know how to do the regular continuous one! Pe^(rt)

The non continuous… I can’t remember but I can pull a sheet out if you want

[little sequoia]

t = ln(1971.71/1551) / 0.04

For the continuous case right?

[rancid saffron]

Is that the equation I need for no 5?

[little sequoia]

No that's the solution t.
You look like you already have the correct equation.

[coarse vale]

Yeah! He is already using the right formula

[rancid saffron]

hold on I don’t get it.. to find t is it this equation?

[little sequoia]

@coarse vale I thought so as well. Compounded continuously derives from the fact that you are evaluating a limit as n goes to infinity of
(1 + 1/n)^n

[little sequoia]

Yes.

[rancid saffron]

Ah ok! That’s what I needed!

[little sequoia]

What's different now is that instead of solving for either the final amount A or something like that there's a little twist.

You have all the amounts and interest. You want to how to find how much TIME it will take to go from your current value of $1551 to a new total $1971.71 after getting compounded interest.

[rancid saffron]

Oh wait in the equation for t should ln be e?

[little sequoia]

Thus you have to "undo" this exponential equation, meaning you're going to need to use logarithms.

[rancid saffron]

Ahhh

I see

[little sequoia]

Not quite e. You're going to need to use log to get that t "down" from the exponential.

[rancid saffron]

Makes sense

[little sequoia]

The classic x = e^ log(x) except going in the opposite direction according to the equation.

[little sequoia]

Awesome. Good luck!

[rancid saffron]

Thank you so much!

Have a nice day! Thank you for helping, God bless and Jesus loves you very much!

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[marsh citrus]

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[still temple]

I need to find the angle between (ACD) and (BCD) ,
so I need to build 2 perpendiculars from DC to those 2 plans, but I dont know where to start building them

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

.close

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[stray helm]

For b), how do I find it with so limited information

[stray helm]

Can anyone give me a clue

[marsh citrus]

@stray helm Has your question been resolved?

[stray helm]

<@&286206848099549185> 🥺

[little sequoia]

This is a problem that sometimes appears on the physics GREs but in a less complicated way.

For a since they give you the solution for free, just find the derivative and plug both that and the original solution to v into the equation given in the question.

You should find both sides match.

The second one involves plugging in t = 5 min and t=10 min and comparing formulas

for c) they're asking you to take the limit as t goes to infinity.

[stray helm]

I've got the answer for a) but i just dont know how to do b

Do i just substitute 5 and 10 into the formula?

[little sequoia]

Because it's an exponential with a negative sign it'll just die off and cause the whole term to go to zero. So the only thing remaining is v = g/alpha

[little sequoia]

Yes

[stray helm]

But I dont know the value of constant a

[little sequoia]

Good point. You'll have to look at both cases when a is greater than 0 and a less than 0.

If a is greater then it's the first answer is sent. If a is negative it'll cause v to go to infinity.

[stray helm]

Oh am I supposed to use limit here?

[little sequoia]

The constant is probably greater than zero as it seems this question would then make no sense if a is negativen

[little sequoia]

Yes

Limit as t goes to infinityn

[stray helm]

Oh oke I'll give it a try

Thanks for the help!

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[little sequoia]

No problem. I believe alpha here is the drag coefficient.

So it should be positive anyway. The question just awkwardly avoids telling you upfront @stray helm

Good luck

[marsh citrus]

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[grim hemlock]

I am confused with finding the truth value of these statements:

𝑝 is an even number.
The product of 𝑎 and 𝑏 is a hundred.
X > 3

Are these statements true/false or would one state that there is not enough information?

[proud ice]

not enough info

[grim hemlock]

.close

[marsh citrus]

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[marsh citrus]

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[soft wigeon]

I need help with 51

[soft wigeon]

I think I messed something up cuz x becomes very weird number

[desert dirge]

cant really read that easily

[soft wigeon]

If a geometric sequence has t5 = 3x + 2 and t7 = 7x-22 with common ratio r = -3, determine t6

I'm tryna solve x

[desert dirge]

when you multiplied them how did you get that quadratic?

[soft wigeon]

Uh t6 cancels out

[desert dirge]

yeah, but how are the numerator and denominator being multiplied suddenly?

[soft wigeon]

O whops

[desert dirge]

you should just have (7x-22)/(3x+2)=9

then multiply and solve

[soft wigeon]

Ok I got it lol

It's -2

.close

[marsh citrus]

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[still temple]

$f(x)=\sqrt{3x^2+4}$

[elfin berry]

okokok

[still temple]

find domain

3x^2+4 greater than equal to 0

subtract 4

divide 3

x^2 greater than equal -4/3

idk what to do now

welp, the square of all real numbers are positive right...

[still temple]

wdym

[tight island]

what are you even doing here

why can't you say the domain is the entire real line? all numbers work..

[spark dawn]

basically, what you want here is to find all values of x for which the thing inside the square root is not negative

however, $x^{2} \geq 0$ for all real values of $x$

[elfin berry]

noh

[spark dawn]

so, it follows that $3x^2 + 4 \geq 0$ for all real values of $x$

[elfin berry]

noh

[still temple]

can you show this on a number line for the domain

[spark dawn]

the part you need to pay attention to is "for all real values of x"

can you express all real values of x on a number line?

[still temple]

so i have x^2 = -4/3

now what

[spark dawn]

that's not useful

[spark dawn]

this is the key to it

[still temple]

this this is the work you show

[spark dawn]

hm?

[spark dawn]

we're using that to show that it's impossible for $3x^{2} + 4 < 0$

[elfin berry]

noh

[spark dawn]

*for all real values of x

[still temple]

ok so what is the work written out in order

[spark dawn]

if it's impossible for $3x^{2} + 4 < 0$, then it should follow that $\sqrt{3x^2 + 4}$ is "possible" for all real values of x

[elfin berry]

noh

[spark dawn]

and i'll leave that for you to figure out

[marsh citrus]

@still temple Has your question been resolved?

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[ornate silo]

Explanation please ?

[sleek lake]

half of a square is a rectangle that specifically has one side twice as long as the other

so algebraically the longer side must be 4, for the perimieter to be 12

[ornate silo]

got it. thanks

[sleek lake]

they said they folded it but they could as well cut it

[ornate silo]

${2}x.{2} + x.{2} = {12}$ so yeah got it

[ornate silo]

yup one square turned into 2 rectangle, makes sense.

[sleek lake]

it doesn't really become 2x+x, it's x + 2x + 2x +x = 12,

[elfin berry]

Bunnings

[sleek lake]

well done

[ornate silo]

i know mistyped that

Thanks.

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[cloud dawn]

−11π/6 = Theta

I'm struggling on finding the csc of theta.

[cloud dawn]

This is how I have my triangle graphed for the unit circle

[stark trail]

can you tell me what csc is in terms of either sin or cos?

[cloud dawn]

csc is hyp/opp

sin

[stark trail]

csc=1/sin

yup

[cloud dawn]

I tried 1/-1/2

one sec

[stark trail]

simplifies to?

[cloud dawn]

I got -2, but that’s incorrect

[stark trail]

its not

it is correct

perhaps you are doing the wrong angle?

could you send the original question?

[stark trail]

theta = -11pi/6

not 11pi/6

[cloud dawn]

Ohhhhh yeahhh

So its a pos pos triangle

Lemme just get the rest of these answered and I’ll be good to go

Yep that seemed to fix my issue

Thanks!

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[cloud dawn]

Alright I am back at it again because I am lost and there is no help video

[cloud dawn]

So from what I understand from my course materials, wouldn’t this be a quadrangle?

Like (0,1)?

[distant peak]

I have no idea what a quadrangle is but it's asking the value of sec(0) and csc(0)

[cloud dawn]

csc in a calculator works as sin^-1?

[stark trail]

csc is 1/sin

yes

[distant peak]

sec(x) = 1/cos(x)

csc(x) = 1/(sin(x))

its not sin^-1

:/

[stark trail]

sin^-1 is 1/sin

poor notation though

[distant peak]

everyone use sin^-1 for inverse of sin in terms of operation of function, not multiplication

[stark trail]

on your calculator sin^-1 probably functions as sine inverse

[distant peak]

again sin^-1 isnt the same as 1/sin(x)

on a calculator tho

[cloud dawn]

So 1/sin(0) would be 1/0, which is DNE or undefined?

[distant peak]

there is an other formula for csc I think

[stark trail]

yeah I realize that, I pointed out that it was poor notation. normally the power going in between the trig function and the angle is what indicates the special behavior, so sin^-1 (not sin^-1(x)) is kinda ambigous

csc(0) is undefined

[cloud dawn]

And then 1/cos(0) is 1/1 so 1?

Hey that worked

I’m still curious though

What is the sin^-1, cos^-1, etc for?

[stark trail]

inverse sin and inverse cos

arcsin and arccos

[cloud dawn]

Okay I see

For some reason I was under the impression is was for csc sec and cot

[distant peak]

inverse of sin and inverse of cos for the law o

which means :
$g \circ f = f \circ g = Id \implies g = f^{-1}$

[elfin berry]

Ｈｅｒｅｌｓ

[cloud dawn]

Yeah I’ve got no clue what any of that means

Maybe later in my math journey

[distant peak]

you will know later, but keep in mind that sin^-1 is the inverse of sin, not for the multiplication we know

but for a special sign for function which is called o

thats a bit abstract yea i know

[cloud dawn]

Gotcha gotcha. Thanks for the help though, seems pretty straightforward to me!

👍

.close

[marsh citrus]

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[distant peak]

np

[marsh citrus]

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[stark pecan]

$4^6\sqrt{3} + 2^4\sqrt{32} - 3^6\sqrt{192} - 2^6\sqrt{192}$

[native niche]

Same logic

[stark pecan]

💀

Bro wdym?

Why does sqrt not work

[native niche]

Oh myb

[glass silo]

\sqrt{stuff}

backslash

[stark pecan]

Oh wrong slash

[elfin berry]

KenWasDum

[native niche]

This works, if you can use calculate use this

[glass silo]

Also did you want like
$4^{6}\sqrt{3}$
or
$4^{6\sqrt{3}}$
?

[elfin berry]

@glass silo

[stark pecan]

Finally

[native niche]

No!

[stark pecan]

This

[glass silo]

Just making sure

[stark pecan]

Wait no

Bruh

Lemme just take a picture

No.6

[glass silo]

,rccw

[glass silo]

$4 \sqrt[6]{3} + 2 \sqrt[4]{32} - 3 \sqrt[6]{192} - 2 \sqrt[6]{192}$

[elfin berry]

@glass silo

[stark pecan]

Yees

Thats right

[glass silo]

For reference to do n'th roots, it's \sqrt[n]{x}

Anyways, you've learned some TeX

[stark pecan]

Professor pickles liked your help

[glass silo]

Anyways, erm, note that 32 = 2^5

And also try to factor 192 similarly

[stark pecan]

Its 2

⁶√192 is 2

[glass silo]

...are you sure?

,calc (192)^(1/6)

[elfin berry]

Result:

2.401873910352

[stark pecan]

Wait thats 64

[glass silo]

There is a factor of 64 there though

[stark pecan]

64×3

=192

⁶√64 =

2

[glass silo]

And as you noted, $\sqrt[6]{64} = 2$

[elfin berry]

@glass silo

[stark pecan]

Yep

Aren't you gonna help me with my question?

[glass silo]

Am I not already

[stark pecan]

What is the first step?

[glass silo]

You're trying to simplify, I'd assume is the question (that wasn't explicitly asked)

[stark pecan]

Arent i gonna do somethin like 1/6 + 1/4 - 1/6 -1/6

And like im gonna change their underwear

I forgot if its denominator or nominator

[glass silo]

And well now you've simplified $\sqrt[4]{32} = 2\sqrt[4]{2}$ and $\sqrt[6]{192} = 2\sqrt[6]{3}$, you can use those to make your life a bit easier

[elfin berry]

@glass silo

[glass silo]

Put all that into your original and also combine the terms and you should end up with
[
4 \sqrt[6]{3} + 2 ( 2 \sqrt[4]{2} ) - 5 ( 2 \sqrt[6]{3} )
]
and, well, there's only a tiny bit of work there

[elfin berry]

@glass silo

[stark pecan]

Oh i just simplify and not actually solve em

So i dont do the 1/6 + 1/4 - 1/6 - 1/6?

[glass silo]

I don't think you do(?)

I mean, there's not much you can do really

[stark pecan]

Coolswagaliciousauce

.close

[marsh citrus]

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[spare ether]

how would you solve c and d

[glass silo]

Parts b, c and d of the question?

https://tenor.com/view/joker-very-poor-choice-words-gif-22861860

(Considering that there’s also an angle BCD )

[spare ether]

mb bro

wait i just solved b

[glass silo]

Cool cool, you’ve got an idea of how to do c then?

[spare ether]

what would be the first step ?

[glass silo]

Well, given a and b, I’d just sin^2(t) + cos^2(t) = 1 it

[spare ether]

ooohhh

ok

thx

i dont know the rule works but it works in the question lmao

[spare ether]

why do you do this rule to find x ?

[glass silo]

Well, you use it to get the equation they've stated there - from there, you solve that equation to find what x is

(there are values that you'd end up ignoring as solutions for logical reasons)

[spare ether]

I’ve gotten to this part but how do u make it 56 x squared

[spare ether]

how does sin^2(a) + cos^2(a) = 1 allow you to find x

OH WAIT

i got it

it =1

[glass silo]

Saves me from writing it out

[spare ether]

im still confused on how x would be found from squaring cos and sin though

ok i got 4 and 2root10 for x

[glass silo]

Well, you know that $\cos(\alpha) = \frac{x^{2} - 32}{4x}, \sin(\alpha) = \frac{x^{2} - 16}{4x}$ and that $\cos^{2}(\alpha) + \sin^{2}(\alpha) = 1$, which gives you, as you said
[
\pqty{\frac{x^{2} - 32}{4x} }^{2} + \pqty{ \frac{x^{2} - 16}{4x} }^{2} = 1
]
which gets you the equation they said after rearranging, $x^{4} - 56x^{2} + 640 = 0$

[elfin berry]

@glass silo

[spare ether]

oh so u just plug that rule in just because it like works

[glass silo]

Pretty much, you have the angles in terms of x and you know that the rule holds, so you're forced to have that equation

Solve for x^2, and then for x

[spare ether]

yea

i got the two possible answers for x

but how do you know both are correct

[spare ether]

^^

[glass silo]

That's what I'm wondering - I think a property of triangles

They say give reasons for your answer - we already know the negative answers of x don't work as lengths can't be negative, but then we need to be able to rule one of those out

[spare ether]

yeah

[glass silo]

Well, if we consider like the x = 4 one

[spare ether]

hmm

[glass silo]

2 + 4 = 6

https://tenor.com/view/big-head-shaq-manly-knows-algebra-gif-12056072

[spare ether]

fr

whered u get the 2

[glass silo]

(but memes aside, notice this triangle)

[spare ether]

yea

?

i notice it

[glass silo]

You have that in a triangle, that the length of one side is less than or equal to adding the other two sides together

[spare ether]

oh what

rlly

[glass silo]

And if you have equality, you're in a straight line

Yep, triangle inequality

[spare ether]

what abt an equilatreal triangle

[glass silo]

With respect to here, that's saying that, taking 6 and then the other two sides, that 6 is less than or equal to 2 + x, and equality only if B lies between P and C

[spare ether]

but B doesnt lie between P and C

[glass silo]

if all sides of an equilateral triangle are x, you basically have the statement x <= 2x, which is true as x > 0, and strict

[glass silo]

Exactly

x = 4 would imply it does, but it doesn't

So rule that one out

[spare ether]

o

hey i have a questoin

referring back to that sin squared cos squared rule

could i just substituet htme back in

would that work

[glass silo]

Which do you mean by "them"? the values of x?

[spare ether]

yeA

[glass silo]

Well for any x values we get, they would work in the sin^2 + cos^2 equation we have

That's how we found them to begin with

[spare ether]

oh yea

mb

[glass silo]

But it isn't enough to rule any of them out

Which is why we're trying to reason

[spare ether]

using this triangle inequality ?

[glass silo]

This here btw is effectively the statement "the shortest distance between any two points is a straight line", if that helps to see why that's true

"triangle inequality" is just the more fancy way of saying it

[spare ether]

hmm

my brain is kinda confused

[glass silo]

Alright, let's say I want to get from point P to point C

[spare ether]

uh huh

lemme repsaste

[glass silo]

Which would be shorter - going directly to C, or going from P to B first and then from B to C?

[spare ether]

going directy to c

[glass silo]

That's effectively what I was saying before - and that if they're the same distance, then really you have that B must be in between P and C on that line, if you get me?

[spare ether]

what do u mean by B must be in between P and C on that line, wouldnt it just be on P

[glass silo]

Well B would be 2 units away from B, so they can't be the same point if that makes sense

[spare ether]

um

could u like draw where B must be in between the line if they r the same idstance

[glass silo]

Erm, I'm gonna steal this from Brilliant and hope it helps me

But if you had that x = 4

You'd have a straight line

[spare ether]

wtf brilliant has math ?!?!

i thought it was like ted ed stuff

[glass silo]

They do

https://brilliant.org/wiki/triangle-inequality/

[spare ether]

what is strict inequatliy

[glass silo]

They also have some examples (though, as before, it might be that you're already familiar with this but it was stated differently, rather than the "fancy" name of triangle inequality )

[spare ether]

ooh

i kinda get it

[glass silo]

stuff like 1 < 2, you're less than something (or greater than something) but can't be equal to it

[spare ether]

u cant do any wiggly wobbly way to get to another point faster than just using the line on the triangle

[spare ether]

oh ok

[glass silo]

Yea that's it

[spare ether]

how would u use that to figure out which answer is valid tho ?

because dont they technically both fit

[glass silo]

Well, to take your question again

[spare ether]

2+4 or 2root10 >= 6

[glass silo]

"...with a point P within the square..."

[spare ether]

uh huh

[glass silo]

You'd have problems with that if you had x = 4

Actually, on that note, maybe as well

,calc (2 * sqrt(5) + 2) - 4

[elfin berry]

Result:

2.4721359549996

[glass silo]

Oh no that isn't a contradiction, nvm

[spare ether]

huh

why -4

[glass silo]

Don't worry about that - I was testing triangle PBA and the sides of length 2sqrt{5} and 2 against the side of x

[spare ether]

o ok

repasting

so are both solutions valid ?

is there any other way to prpove it

[glass silo]

No - only the 2sqrt{10} one is valid

[spare ether]

o

but how can prove ?!?!

[glass silo]

There may be, but I can't think of it right now

[spare ether]

oh ok

[glass silo]

Other than saying "point P being inside the square makes triangle PBC non-degenerate (i.e. a triangle, not a straight line), but choosing x = 4 would get us that PBC is a straight line, which is a contradiction"

You could probably do similar with triangle PBA as well I think actually no you can't

[spare ether]

ooog

[spare ether]

how would u know it would be straight line from x=4

?

[glass silo]

Because you'd have 2 + 4 = 6, and remember the statement here that the inequality is strict if the triangle is non-degenerate (not actually a straight line)

[spare ether]

o m g

i get it!!

yippee!

thank u

[glass silo]

Anyways, to compare the picture here...

[spare ether]

uh huh

i have a question

how would u write this

as working out

/proof

[glass silo]

...with here, if it's of any help

[glass silo]

Oh you already got it nice nice

[spare ether]

yes thx

[glass silo]

Erm, I would say something along the lines of "considering triangle PBC, we can't have x=4 else we would have the triangle degenerate/it would violate the triangle inequality/[something like that]"

[spare ether]

oooh ok

tysm

thank u big much!

[glass silo]

Was a pleasure, hope that helped for ya

[spare ether]

.close

[marsh citrus]

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[marsh citrus]

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[left night]

I need some help With HALF ANGLE AND POWER REDUCING IDENTITIES, I have no idea where to start, I do have the sheet that has the identities, this is more for study for a test i am taking tomorrow for Precal

[rotund walrus]

.close

[marsh citrus]

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[left night]

.reopen

.close

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[serene inlet]

how would i simplify this?

[serene inlet]

16^3/4

[novel fjord]

convert it to a root

you could alternatively use some exponent rules to split it into two different roots and multiply

[marsh citrus]

@serene inlet Has your question been resolved?

[still temple]

@serene inlet 16^(3/4) would be equivalent to having the 4th square root of 16 to the 3rd power. the 4th square root of 16 is 2, and 2 to the 3rd power is 8

[serene inlet]

this is what i have now

[still temple]

are you able to do the rest?

[serene inlet]

i can basically input ths into my calculator right?

ahh shi

wait im not alowed to use calc

[serene inlet]

how would i solve it?

[still temple]

you dont need a calculator

[serene inlet]

do i just do 16^3?

[still temple]

You could also write it as (4throot16)^3

[still temple]

No

though its the same, its much easier to simplify if you put the cube outside of a paranthesis the 4th root of 16 instead of just 16

[serene inlet]

im confused

[still temple]

one sec i had a stroke typing that

@serene inlet

[serene inlet]

would this be the answer?

the simplified version?

[still temple]

no

[still temple]

you can still go further @serene inlet do the rest try it by yourself

[serene inlet]

did i do this problem right?

the 2 kinda confused me

[still temple]

well 1/5 is correct, but the way you got there is not

[serene inlet]

ya…

did i rewrite it correctly?

[still temple]

where did the square next to the square root of 25 come from

[serene inlet]

from the exponent

<@&286206848099549185> why would the answer be 2? im confuzzled

[soft wigeon]

Wdym

Can you show entire question?

[marsh citrus]

@serene inlet Has your question been resolved?

[serene inlet]

.close

[marsh citrus]

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[mystic heath]

It's been a while since I've done a problem like this can you verify that I'm correct?

[elfin berry]

keto11

[mystic heath]

also I forget how to use wolfram alpha to check, could ya'll demonstrate this?

[still temple]

,w int_{pi/6}^{pi/2} int_{0}^{sin(theta)} 6rcos(theta) dr dtheta

[elfin berry]

Results provided by WolframAlpha

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Requested by Toby#2275

[mystic heath]

https://tenor.com/view/bing-gif-25601964

thanks!

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[tame cedar]

Does anyone know how to get M? And what is M?

[marsh citrus]

@tame cedar Has your question been resolved?

[main idol]

you're supposed to find a reasonable value for M

[tame cedar]

ok

So f'(x) <= M

[main idol]

simplify this

[tame cedar]

i thought it was already as simple as it could get

[main idol]

2(1) = 2

[tame cedar]

How did you get 2

[main idol]

2(1) means 2 times 1

,calc 2 * 1

[elfin berry]

Result:

2

[tame cedar]

No I mean like

Where did you get the value

[main idol]

i'm just telling you how to simplify

[main idol]

yes

[tame cedar]

What comes next?

[main idol]

pick something convenient that fits this

there's no single "right" answer

[tame cedar]

1

[main idol]

you should be computing a derivative

[tame cedar]

f(x) = sqrt(x)

[main idol]

right

use that too to find M

i should have included a and b in the previous screenshot

[tame cedar]

cool so i choose any x in between a and b

and put that into the derivative

[main idol]

no

it really should read "for all $a \le x \le b$. "

[elfin berry]

riemann

[main idol]

right. so find a convenient M that satisfies this inequality

for that range of x

[tame cedar]

36

Yay I got it

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[oblique musk]

Yo

[oblique musk]

Can anyone help me with this integration

[broken forum]

split into two integrals

[oblique musk]

You mean like this

But am stuck in the 2nd integral part

Anyone

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[glass dust]

Matrices and distance from 2 planes.
I have a point in 3D space that is not on either of the planes.
I know the "normal" way to calculate the distance of point from a plane using projection onto the normal of a plane.

I'm wondering if I can somehow better use matrices to find out the distances from both planes at the same time?

[marsh citrus]

@glass dust Has your question been resolved?

[marsh citrus]

@glass dust Has your question been resolved?

[glass dust]

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[marsh citrus]

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