help-27

[strange nimbus]

And let's convert the root to a power.

How do you do that?

[quaint rampart]

5^1/3

Before multiplying

Oh

After would be 5^5,5/3

[strange nimbus]

[5^x = 5 \sqrt[3]{5} \cdot 5^{4,!5}]
[5^x = 5^{5,! 5} \sqrt[3]{5}]
[5^x = 5^{5,! 5} \cdot 5^{\frac13}]

[quaint rampart]

Isn’t thst just 5^2,5

[woven radish]

Chai T. Rex

[strange nimbus]

No, it won't be that.

Remember that you have to add the exponents.

[quaint rampart]

Oh I was doing it the other way

I was adding it to the one in the root

[strange nimbus]

So, how would you add 5,5 to 1/3?

[quaint rampart]

Multiply 5,5*3 then add it to the top

N7merstor

Numerator

[strange nimbus]

No, you can't do that.

Remember, the rule is (a^{b + c} = a^b \cdot a^c).

[woven radish]

Chai T. Rex

[strange nimbus]

You need to add the exponents, not multiply them.

[quaint rampart]

5,5/1 + 1/3

[strange nimbus]

OK, so what fraction does that simplify to?

[quaint rampart]

Lemme see

6.5?

[strange nimbus]

Nope, 5,5 + 1 = 6,5, but you have 5,5 + 1/3.

First, convert 5,5 to a fraction.

[quaint rampart]

5,5/1

I did that

[strange nimbus]

OK, now make it an integer on top and bottom.

[quaint rampart]

Then I multiplied the bottom ones

[quaint rampart]

What

[strange nimbus]

Well, 5,5/1 has a decimal on top.

[quaint rampart]

Yeah

[strange nimbus]

So, we can convert it into an integer by multiplying the top and bottom by 10.

55/10.

[quaint rampart]

Ohhh integer means full number?

[strange nimbus]

Right, a number without anything after the decimal point.

[quaint rampart]

Right

[strange nimbus]

So, what does 55/10 simplify to?

[quaint rampart]

5.5/1

😭

Oh

[chilly quest]

Sorry to interrupt , do we need log to solve this question or no

[strange nimbus]

No, log won't be needed.

[chilly quest]

Oh okay thank you

[strange nimbus]

No problem.

[quaint rampart]

11/2

[chilly quest]

Cuz I wanna learn too

[strange nimbus]

Good.

So, what's 11/2 + 1/3?

[quaint rampart]

35/6

[strange nimbus]

OK, good.

So, what's x?

[5^x = 5^{5,! 5} \cdot 5^{\frac13}]
[5^x = 5^{\frac{35}6}]

[woven radish]

Chai T. Rex

[quaint rampart]

5,83 approximately

Idk if I need to keep on going

[strange nimbus]

Unless they say to use a decimal or to round, I'd leave it as a fraction.

[quaint rampart]

Nvm it repeats itself

[quaint rampart]

Right

[strange nimbus]

Yeah, to get a decimal that stops, the denominator has to be just 2s and 5s multiplied together.

6 has a 3 in it.

[quaint rampart]

,w 5^(35/6)

[strange nimbus]

So, like if you have 40, that's 2^3 * 5, so it has just 2s and 5s, so 1/40 has a decimal that stops.

[quaint rampart]

Whoops

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by refinebss

[strange nimbus]

Yes, but x is just 35/6.

,calc 35/6

[woven radish]

Result:

5.8333333333333

[quaint rampart]

Dude I don’t really understand is that ok

[strange nimbus]

But leave it as 35/6 unless they say to make it a decimal or they say to round it.

That's OK.

[quaint rampart]

Yeah

[strange nimbus]

Now let's check it.

[quaint rampart]

Ok

[strange nimbus]

[\frac{0,2^{x + 0,5}}{5} = \sqrt[3]{5} \cdot 0,04^{x - 2}]

[woven radish]

Chai T. Rex

[strange nimbus]

So, let's type that in with (35/6) for x and see if the sides are equal.

,calc (0.2^((35/6) + 0.5))/5

[woven radish]

Result:

7.4854854098249e-6

[quaint rampart]

This is my teachers explanation

[strange nimbus]

Oh, OK.

[quaint rampart]

I didn’t understand it tho but

The answers right

[strange nimbus]

Well, a few things needed to be understood to do it the way I showed.

[quaint rampart]

Idk wtf is going on here

[strange nimbus]

You needed to see that 0,2 and 0,04 are squares and square roots of each other.

[quaint rampart]

Yeah

Way easier than whatever he did

[strange nimbus]

And you needed to change things to fractions.

Like 0,2 = 1/5 and 0,04 = 1/25.

With that, it's easier to see them as squares and square roots of each other.

[quaint rampart]

0,04 is not 1/25

Is it?

[strange nimbus]

It is.

4/100.

[quaint rampart]

Wow

Ur right

[strange nimbus]

You can then also notice that 5 keeps showing up.

And that's kind of what guided me along the way.

That and getting the xs on one side and the other stuff on the other side.

[quaint rampart]

Right

Yeah if I find out the order and sequence then I’m good

I know how to do most of the like calculations

[strange nimbus]

Oh, that's what your teacher did on the second line.

[quaint rampart]

?

[strange nimbus]

They must have noticed the 5s.

Because they rewrote everything in terms of powers of 5.

But it's OK if you don't notice that at first.

[quaint rampart]

How’d he get 5^-1

[strange nimbus]

From the denominator on the left.

[quaint rampart]

Where’d the 0,2 go

On the numerator of left

[strange nimbus]

It's the 5^-1 in parentheses.

They noticed that 0,2 is 1/5.

And 1/5 is 5^-1.

So, that's how they got (5^-1)^(x + 0,5).

Does that make sense?

[quaint rampart]

Really?

Yeah ig

[strange nimbus]

[(0,! 2)^{x + 0,! 5}]
[\qty(\frac15)^{x + 0,! 5}]
[\qty(5^{-1})^{x + 0,! 5}]

[woven radish]

Chai T. Rex

[strange nimbus]

Do you see how each step follows?

[quaint rampart]

Yeah now I do

[strange nimbus]

Anyway, their method requires you to see that everything is a power of 5.

[quaint rampart]

Imma try to do the next question

Brb

[strange nimbus]

OK.

[chilly quest]

Can u also send me the other questions please

I wanna learn

[supple knot]

like uhhhh https://www.homeschoolmath.net/worksheets/examples/exponents_worksheet_8.gif

or uhh https://www.functionworksheets.com/wp-content/uploads/2022/07/exponential-word-problems-with-solutions-pdf-1.png

[strange nimbus]

There's also this unit about solving basic exponential equations (close the donation request and stuff like that): https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:exp.

The section right before quiz 2 is like the earlier problem.

[chilly quest]

Thanks

[chilly quest]

Thanks

[chilly quest]

What about I wanna do sth like harder

like the question that guy asked (maybe not hard for u)(but I wanna see like uhh ye)

[devout snow]

@quaint rampart Has your question been resolved?

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[arctic pulsar]

can someone help on any of these since i ahve no idea what im doin

[strong tiger]

for the first 2 questions you need to know the formula of theta

and the third one you need to know the property that the diameter subtends an angle of 90 degree on the circumference

[arctic pulsar]

whats the formua tho

[strong tiger]

theta = (lenght of arc)/radius

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[thorn lagoon]

I’m struggling to understand why my teacher used the formula she did for number 13, could someone explain it to me?

[chilly quest]

U need to find the area of the QRA sector

And remove the triangle QRA

[thorn lagoon]

Wait qra do u mean wat

Qta*

[chilly quest]

Oh yes sorry

[thorn lagoon]

Why do I have to remove the triangle?

[chilly quest]

To find the shaded area

[thorn lagoon]

I need to find the area of the triangle sector?

[chilly quest]

So first we find the area of that purple region

Then we remove the area of the triangle inside

[chilly quest]

And we can get the area of this shaded region

[thorn lagoon]

OHH

Thank you that helped a lot

Have a good day

[chilly quest]

no problem

Have fun

[thorn lagoon]

.close

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[prime whale]

is there a particular reason in this proof the summation went to N -1 instead of N?

[prime whale]

is it becuse we only want first N terms and we start at n = 0?

or is it deeper than that

[vast drift]

there are N terms if u start from 0 and end at N-1

[prime whale]

k figured thats what it was

thanks

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[winter meadow]

What does it mean by horizontal line representing the sum? Just like the sum of terms 1-10? If so did I answer c correctly?

[winter meadow]

This is uhh alternating series test

I’m mostly just super confused on series and stuff so maybe I’m just overlooking an easy answer

[acoustic leaf]

you plot it on a graph where the x-axis is n and the y axis is S_n

[winter meadow]

Ohhhhhh

[acoustic leaf]

and the actual sum is on a horizontal line y = S

[winter meadow]

So I need to enter it as a stat plot?

[acoustic leaf]

that would work

[winter meadow]

Shiiii

So what exactly would the horizontal line be?

[acoustic leaf]

you would enter it as a function y = S (where S is the value of the sum)

[winter meadow]

So where n=.7604 the sum of the first ten terms?

I’m sorry if these are dumb questions I’ve just missed a lot of class recently and idk what’s really going on

[acoustic leaf]

it should be the sum of the full infinite series

which seems to be given to you in the exercise

[winter meadow]

Guh

Oh wait

Rha

OH

WAIT

OK YEAH OK

Me when I’m stupid

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[winter dragon]

Im on the mclaurin series and i got everything but the -2 at the end chatgpt didn't help

[supple knot]

!show

[devout snow]

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

@winter dragon Has your question been resolved?

[winter dragon]

mb was doing otheerr problems

In the box is the problem sorry for the bad work idk how to use latex

Tell me if you need clarification

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[dark shoal]

hey guys, if I have $\int_{a}^{b} |x^2| ,dx$ does that evaluate to just$ |1/3b^2-1/3a^2|$?

[woven radish]

Michele

[dark shoal]

like for any integral is it just the absolute value of the normal integral

[pseudo basin]

no to both

$\int_a^b |f(x)| \dd{x}$ in general need not be equal to $\absv{\int_a^b f(x) \dd{x}}$ (though you \textit{can} write down a version of the triangle inequality that relates these)

[woven radish]

Ann

[sonic kelp]

the |x²| case is that x² is always positive or 0

[pseudo basin]

in your specific example, $|x^2| = x^2$ anyway, so your integral equals $\frac{1}{3} b^3 - \frac{1}{3} a^3$ (not $b^2$ and $a^2$ there)

[sonic kelp]

so you can remove the modulus in this case

[woven radish]

Ann

[sonic kelp]

yeah

[dark shoal]

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[pseudo basin]

bruh doorslam

[midnight echo]

lmao

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[unkempt heath]

??? What's the shaded area

There isn't anything else we know

[strange nimbus]

So, the numbers are areas?

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@unkempt heath Has your question been resolved?

[soft umbra]

Is that a rectangle?

[devout snow]

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[median arrow]

Guys, can you help me?? Find a when a equation has two roots (x)

[median arrow]

had this on a mock exam

[grand edge]

We have secretly a quadratic equation

Just express sqrt(4^x - a) as some variable

[median arrow]

yeah i ve got it

[grand edge]

Nice

[median arrow]

but i cant go next

[grand edge]

How do we know whether a quadratic equation has two roots

@restive river can you stop

[median arrow]

D > 0

[grand edge]

Yep

[median arrow]

and sqrt var is terrble

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Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[median arrow]

found t

and i think is not good to return t to its initial value

please get out here, you neglect ethic rules and sense

but it it is true its blow my top

[grand edge]

This is incorrect

[median arrow]

i ve found this limit but i thoughyt its not enough

[grand edge]

We defined t as a square root, right?

[median arrow]

yeah

[grand edge]

And we need two distinct values of t

Which correspond to two real roots

So we have to have t > 0

If t = 0, then the two roots are equal and then there's only one real root

Idk if I explained it well

[median arrow]

oh

[devout snow]

@median arrow Has your question been resolved?

[median arrow]

a>3 is a result?

or it is a part of it

ohhh

now

no

i forgot the task

a = (2.75;3)

??

[grand edge]

yeah

[median arrow]

i cry

i coulg got 4 balls if i thought over a last step

one step carl 😭

[median arrow]

thank you 🐘

[grand edge]

no worries

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[woven radish]

lin.fei_. ₊˚❀₊

[warm lantern]

just 2 cars

[glossy dew]

do you think people having exactly 2 cars can have more than 2 cars

[warm lantern]

^

[glossy dew]

i have 10 dollars

does that imply i have atleast 10 dollars

waht

i think they mean exactly by default then

hmmh

but they explicitly mentioned at least one car

[warm lantern]

if the given information about myself is that "I have at least one car", and "I have two cars"

then that means I have two cars

not three

not four

not one

not zero

etc

have you clarified this exact situation with your teacher?

same question, etc

get a 4th opinion

me and blurple being the 2nd and 3rd (assumedly, and in no particular order)

[raven leaf]

just two cars I think

assuming that you have correctly quoted the problem, the problem also uses "has two cars" to mean "has exactly two cars"

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[north roost]

what have u done so far?

[rugged spoke]

found until the volume

like the integration part

i dont know the next step

[north roost]

assuming ur integral is set up correct

then just integrate

[rugged spoke]

idek if its correct hehe

[north roost]

u are using disk method, right?

[rugged spoke]

shell method

[north roost]

dy?

[foggy ermine]

Did you use parametric integration?

[rugged spoke]

urm....

whats parametric integration

[north roost]

the rotation is about x = 5

why are u using dy?

[rugged spoke]

because its respect to y no?

[north roost]

u're using washer then, right?

[wispy oyster]

they intended to use shell but they did it wrongly

[north roost]

oh

[wispy oyster]

you have to do it wrt x since it is rotated about x=5

your radius becomes (x-5) and height becomes √x - x/2

with thickness dx

[foggy ermine]

Its pi * the integral of f(x))^2

[north roost]

this, if u want to do shell

bound of integration is 0 to 4

[foggy ermine]

Subtract upper function from lower function

[foggy ermine]

0 to 5 ?

[wispy oyster]

0 to 4 for shell

0 to 2 for washer

one is wrt x and one is wrt y

[foggy ermine]

How come?

[north roost]

cuz its dy

[wispy oyster]

look at the point of intersections when expressed in terms of f(y)=x

[foggy ermine]

It's about axis

x

[north roost]

?

the question asks for rotation about x =5, parrlel to the y-axis, no?

[wispy oyster]

Newton, maybe you need to revise your calculus \j

anyways @rugged spoke do you get the idea?

[rugged spoke]

i think so?....

mostly

[foggy ermine]

x=2y, could you just make y the subject and use washer's?

y=x/2

[rugged spoke]

i can try

[wispy oyster]

when revolving around the y axis or any axis that is a translation of the y axis (namely x=k where k in R), your thickness is dx so everything has to be in terms of x

for shell method

[rugged spoke]

ohhhh

thank youu

[foggy ermine]

It's rotation about x = 5 so why use shell (dy) instead of washer's (dx)?

[north roost]

shell is dx, washer's is dy

[devout snow]

@rugged spoke Has your question been resolved?

[foggy ermine]

Dang it, I read the question wrong.

[devout snow]

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[zinc flint]

hi

[zinc flint]

is this a true statement

[north roost]

yes

[zinc flint]

this is from deepseek cause gpt kept getting thre answer wrong

[zinc flint]

why is it true?

[zinc flint]

all my lectures say nothing about this so im not sure the proof

[north roost]

epsilon-delta proof?

[zinc flint]

any kind of proof showing that the statement is true

[zinc flint]

is this the proof to show it?

[north roost]

this but

${0 < x-a< 0}$ instead of ${0 < |x-a| < 0}$

[woven radish]

k

[zinc flint]

sorry but how does this prove it

i havent gone that far into limits although i kinda know how the epsilon delta definition works

[north roost]

oh wait

its infinity

just do something similar

and show that i can always pick a bigger number that the one choose by making x within delta from the positive side

[foggy ermine]

Could you prove divergence?

[zinc flint]

no clue how to do that

[foggy ermine]

Do you know the definition of a convergent sequence/function?

[zinc flint]

would i do this for the same case on the other side

[north roost]

yes

[zinc flint]

nope

[zinc flint]

ah okay

[north roost]

do 0 < x-a < delta or -delta < x -a < 0 depending on left or right limit

[zinc flint]

hm interesting

ill give it a go

thanks

.close

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[plush furnace]

why could he pull out a negative sign????

[north roost]

cuz its like a constant

-1

[plush furnace]

two different answer

[plush furnace]

i did it the top way, he did it the bottom way

bottom was is the answer

[north roost]

distribute

[plush furnace]

oh is it like this

[north roost]

correctly

😭

[north roost]

-(4-x^2)^2 + (4-x)^2

[graceful cosmos]

∫ f(t) dt = F(b) - F(a)

-∫ f(t) dt = -F(b) + F(a)

[plush furnace]

oh i had to distribute it

[north roost]

yes

[plush furnace]

alr thanks

.close

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[dim dust]

why does the line "normal" always bisect the angle between the 2 foci?

[dim dust]

its normal to the line tangent to the point chosen to enclose the triangle between the 2 foci

<@&286206848099549185>

[rare kernel]

you can think of it from an optics perspective if u want

think of one of the lines passing through the focus as an incident ray

and the line passing through the other focus as the reflected ray

so angle of incidence = angle of reflection

and both these angles are measured with respect to the normal

[devout snow]

@dim dust Has your question been resolved?

[dim dust]

it makes sense visually but how would that be proved using geometry

[rare kernel]

not really aware of the proof

you can search it up tho

proof of reflective property of ellipse

[dim dust]

aight

tysm

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[lofty monolith]

How do I even start with making C0 the subject

[lofty monolith]

It looks like they’ve skipped so many steps to me

[summer summit]

remember lne = 1

[lofty monolith]

My first attempt lmaoo and it looks completely different to the answer

[wind mason]

Try to factor out a e^{C_0} term.

Or rather, just set each side equal. Then multiply by e^{C_0}.

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[stark shale]

i'm doing triple integrals, spherical coords.
in this problem i'm struggling to find the angle phi.

my reasoning is, i have to find phi that doesnt depends on rho, and it doesnt depend on theta.
what i did:
z= rho cos(phi)
so
rho cos(phi) = 2r
from the cone.
so i have
phi = atan(1/2)

i don't know if this is correct.

[raven leaf]

sorry what are you trying to do here

[stark shale]

as i wrote in my second line, i'm trying to find phi

the bounds of integration for phi

[raven leaf]

how are your coordinates defined?

[stark shale]

i am using standard spherical coords

i think it's standard?

i'm just beginning studying these things.

rho, phi, theta

[raven leaf]

but if you draw a section of the cone by x=y=0, then to the edge of the cone, you could have a triangle

should*

[stark shale]

i dont understand what you want to tell me...

[raven leaf]

I'll draw a diagram

[stark shale]

yes

[stark shale]

mmm no... what you wrote is z^2 i think

[devout snow]

@stark shale Has your question been resolved?

[azure flint]

@stark shale Hey, I totally understand how you're feeling, and I’m really sorry if I made you uncomfortable in any way — that was never my intention. This is actually my first time using the app, and I guess I got a little too curious and excited. I didn’t really know the best way to start a conversation, so I just kind of dove in without thinking it through. I realize now that friendship takes time and can’t be rushed, and I respect that completely. Thank you for being honest with me — I appreciate it. I’ll take your advice and try joining some public chats, get to know the space better, and just enjoy talking about shared interests naturally. Again, I’m really sorry for coming off too strong. I genuinely hope you have a great experience on here, and maybe we’ll cross paths again sometime under better circumstances 🌷

[graceful cosmos]

Let's relabel x² + y² = r² for radius reasons

Considering everything to be positive, I'll freely take the √ and we have
z = 2r

It's bounded above by the plane z = 2, so it's also bounded radially by r = 1

[raven leaf]

my bad, I didn't see the z^2

the idea still holds though

[devout snow]

@stark shale Has your question been resolved?

[devout snow]

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[stark shale]

.reopen

[devout snow]

✅

[stark shale]

is my reasoning correct then to find phi?

[graceful cosmos]

Oh I didn't see you typed out exactly what I did. Yes your reasoning is good.

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[candid drift]

guys ik this isnt a science server but i have a simple question 😔 i cant figure it our, im dumb :((

[candid drift]

during anaphase in mitosis, ik the chromatids get seperated from the spindel fibers, but after telephase carrys out and the stuff after that does it end up with 46 or 23 chromosomes im confused?

[candid drift]

pls someone answerr 🤒

[supple knot]

dafuq

[wispy oyster]

46 chromosomes

[supple knot]

#old-network has a link for bio server

#old-network message

[wispy oyster]

you end up with 23 chromosomes in meiosis

[candid drift]

oh tyty

UR SO SMART MIA

TYSMMM

!!

[candid drift]

oh yeah haploid cells are only produced in egg cells and the other one

,close

.

ERM

how do u close

!

[wispy oyster]

.close

do that

[candid drift]

ohh

im dumb

.close

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[candid drift]

tysm anyways

!!

[devout snow]

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[zenith whale]

how do i start with this problem

[livid sapphire]

try finding the angle first in relation to the given arc length

s = 2pi(r)((angle)/360))

r will cancel out in the end, and you should get a constant answer

[zenith whale]

i dont understand the number its given me

is 0.3 the arc legnth?

[livid sapphire]

0.3r is just the arc length, where r is the radius of the circle

[livid sapphire]

yep!

[zenith whale]

is it 30?

i dont understand what to do with the arc legnth

am i supposed to assume the radius is 1?

[livid sapphire]

oops sorry, wrong one, the formula should be the simpler one
angle = (arc length)/(radius)

[livid sapphire]

nopee, unless mentioned, leave the radius as is for now

try using the formula above to get a constant angle

[zenith whale]

angle=0.3/r?

[livid sapphire]

you're quite there!! don't forget that 0.3 is being multiplied by r in the given as well

[zenith whale]

i dont understand what the next step is

[livid sapphire]

angle = (0.3r)/r

r cancels 🎆

angle = 0.3

but right now, it's in radians so convert it to degrees

0.3 to degrees

[zenith whale]

170?

or 17?

whats the angle

or how am i supposed to get it

[supple knot]

Angle is n degrees

Pretend it's some arbitrary value like 100

Then once you figure the answer for arc length, do the problem again with n instead of 100

[zenith whale]

i have n/360 * 2pi * 6.5 so far

[supple knot]

Looks right, just simplify

[zenith whale]

x*4680?

i dont understand

like how am i supposed to find the arc legnth without a angle

how am i supposed to do this without trig functions

[elder nova]

prolly just means dont write ur final answers in terms of trig functions

[zenith whale]

how do i find it

im really confused

like its 8cos(30) but how do i get it exact

[elder nova]

unit circle, cos(30 deg) is sqrt(3)/2

ur supposed to mem them but if u draw a right triangle u can figure it out

[zenith whale]

is it just sqrt(3)/2 *8?

[errant zephyr]

yup basically

memorising the unit circle, even just the first quadrant will really help u

[zenith whale]

i still dont understand this

.close

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[devout snow]

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[slender birch]

can i get some help with this one?

[slender birch]

???

<@&286206848099549185>

[soft umbra]

!15m

[devout snow]

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.

[hybrid snow]

So you know how when we have some random line

y = mx + b

[slender birch]

yea

[hybrid snow]

we can break that up

[slender birch]

how so?

[hybrid snow]

So consider m as a vector <Δx, Δy>

So like

For example if m is like 5/2

Your vector is <2,5>

[slender birch]

in our case isnt it 3/11?

im not fully sure what that means though. whats a vector?

[hybrid snow]

A vector, in layman's terms, is something that indicates a direction

[slender birch]

oh

ok, so in this case t

[hybrid snow]

Well t is a parameter

[slender birch]

yea

[hybrid snow]

Technically there are position vectors that indicates a location

But consider this

Your slope is -3/11, so your direction vector is <11,-3>

[slender birch]

that makes sense

[hybrid snow]

Whats nice about these lines is that you can take any point on that line and add t times the direction vector

[slender birch]

im still a little bit confused though

like if the vector is -3,11, then how do i go and get two equations that are equal to that?

[hybrid snow]

You're getting ahead

[hybrid snow]

I would start by choosing a point where t = 0

In your case it's (-6,-5)

[slender birch]

yea

[hybrid snow]

Im gonna rewrite that as a position vector

<-6,-5>

[slender birch]

so x = -6, and y = -5?

[hybrid snow]

Actyslly hold on I realized the other point is at t = 3

[slender birch]

its the other one 5,-8

i think

thats t=3

[hybrid snow]

So you need to take the direction vector <-3,11> and divide it by the difference in parameter (which is 3, because 3 - 0 = 3)

[slender birch]

so is it -3/11?

[hybrid snow]

Well here

Think about it

You need to take 3 "steps" to get from <-6,-5> to <5, -8>

[slender birch]

what do yo umean 3 steps?

[hybrid snow]

3 steps in a direction

In the direction of the direction vector

[slender birch]

oh, so there kinda needs to be a mid point?

[hybrid snow]

Well no like

Okay let me restart

[slender birch]

ok, sorry

[hybrid snow]

When you have a line, there are two "characteristics": a point, and a direction

[slender birch]

ok

yes

[hybrid snow]

So when we parameterize a line, we can say it's <a,b> + <Δx, Δy>t where (a,b) is a point on the line and m = Δy/Δx, and t is the parameter

[hybrid snow]

So now look at when t = 0

<a,b> = <-6,-5>

[slender birch]

oh, so it would be (-6,-5) , -3/11

[hybrid snow]

Well

Slow down a bit

[slender birch]

oh, sorry

[hybrid snow]

We can say any other point <x,y> can be expressed like this

Since we know that <a,b> is just the point when t = 0

[hybrid snow]

And what I mean by this is:

<x,y> = <a,b> + <Δx, Δy>t

When t = 0, we get a point <x,y> = <-6,-5>

Si we can fill in that equation

<x,y> = <-6, -5> + <Δx, Δy>t

You know that when t = 3, <x,y> = <5,-8>

So you can find the direction vector

Whats nice about characterizing a line this way is that things get paired in the order they are listed

So like

<x,y> = <-6, -5> + <Δx, Δy>t

You can break it up into two equations:

x = -6 + Δx • t
y = -5 + Δy • t

Which is why thinking about vectors is really nice

[slender birch]

ok, thank you

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[inner cargo]

Trying to learn calculus early how did he get the 6(x+2)

[heavy current]

by factoring 6x + 12

[pseudo basin]

6x + 12 = 6(x+2)

this is not really a calculus thing in and of itself

[inner cargo]

Last question hehe how did he get 2.8

[primal ferry]

that's not a decimal point

if that's what you're asking

[inner cargo]

Oh

[primal ferry]

that is being used as a multiplication sign

2*8

[inner cargo]

Mb brah

[primal ferry]

npnp

[inner cargo]

.close

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[primal ferry]

happens more often than youd think 😭

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[tepid fulcrum]

how would i go about solving a differential equation in the form\ $y'+y^2p(x)=q(x)$
i have tried doing substitutions with $y=\tan(x)$ and function derived from it but i'm not getting anywhere

[woven radish]

pixel

[tepid fulcrum]

Let $P(x)$ be the antiderivative of $p(x)$\
$y=-a\tan(aP(x))\
-a^2p(x)(1+\tan^2(aP(x)))+a^2\tan^2(aP(x))p(x)=q(x)\
-a^2p(x)-a^2p(x)\tan^2(aP(x))+a^2p(x)\tan^2(aP(x))=q(x)\
-a^2p(x)=q(x)\
a=\sqrt{-\frac{q(x)}{p(x)}}$
this will only work if $a$ is a constant, so $p(x)$ and $q(x)$ need to be the same (scaled) line

here is one attempt ^^

[woven radish]

pixel

[tepid fulcrum]

on the third line i just sub it into the differential equation

[devout snow]

@tepid fulcrum Has your question been resolved?

[devout snow]

@tepid fulcrum Has your question been resolved?

[proper tartan]

search up bernoulli's equation

its a form of that

[devout snow]

@tepid fulcrum Has your question been resolved?

[tepid fulcrum]

okay, thanks

[devout snow]

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[vale iris]

Hello.

I have following practice problem that i need to solve in matlab:
2nd order linear DE with Robin boundary conditions.

• Equation: u′′(x)+7u′(x)−u(x)=sin x
• Domain: x∈(0,π)
• Boundary Conditions: u′(0)−3u(0)=0 and u′(π)+3u(π)=0
• Method: Grid method/Finite difference method
• Discretization: n=100 subintervals

I wrote some code and got solution (picture).

It seems right to me but i would like some confirmation from someone else

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[terse bolt]

Im having trouble with this question

[terse bolt]

here is what i have done so far

Using Fermats little theorem we can distribute $x^p$ into $xx^{p-1}=x$

[woven radish]

oklmaosad

[terse bolt]

then you have x-a, which means it will always be reduceable as there is always going to be an a-a situation as $a \in \mathbb{Z}_p$ so you will end up with an $a-a$

[woven radish]

oklmaosad

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[lusty dragon]

Hey! This might be a bit of a dumb question, but how do I find critical value on my calculator? I have a Casio fx-115es plus.

[upper ermine]

Assuming you mean f'(x) = 0 or undef and not critical values for statistical tests

im not sure specifically about this calculator, but you could always just graph the derivative and find points where its undefined or 0

[lusty dragon]

Sorry I meant for statistical tests, I'll send a pic of the question I need it for

[upper ermine]

ah

[lusty dragon]

I've been looking through youtube videos to try and figure it out for about 40 min now lol, it's ok if you don't know how but it doesn't hurt to ask haha

[upper ermine]

sorry i cant help with your calculator then

shouldnt be too hard to navigate though uhh

try to go into statistical mode and find some distribution menu?

[upper ermine]

for a z-test, just use a function analagous to invNorm

[lusty dragon]

I've been looking for an invNorm function or something like that but I can't seem to find it

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[bleak arrow]

If I have the surface x+y+z=1, x>0, y>0, z>0, how would I find the boundary curve C without graphing it?

[solar orchid]

For x=0, you get y+z=1, and y=0,get x+z=1, z=0, x+y=1
And then you can have equations for boundaries

Also, confine x,y,z to belarger than zero

[bleak arrow]

https://www.desmos.com/3d/upbsrcsaze

im trying to input x>0, y>0, z>0 but it gives me an error

[solar orchid]

Do you want to derive equation and plot it at demos?

[bleak arrow]

yes

i already have equations but it doesn't give me the first octant

check the link i sent

[devout snow]

@bleak arrow Has your question been resolved?

[bleak arrow]

I am kinda confused. You said that we put x = 0, y = 0, z = 0 to find the boundary lines. But I thought x>0, y>0, and z>0? How can x be = 0 and > 0 at the same time?

[solar orchid]

In your case, since you want boundary , but x,y,z >0.
I thought that question doesn't have exactly answer, since you can always find another curve C that is more close to boundary.

[bleak arrow]

What if we have a problem like this:

"z = 9-x^2-y^2 above the xy plane", the curve of intersection is x^2+y^2=9 at z = 0, here, we have z>0, but we don't use that

[devout snow]

@bleak arrow Has your question been resolved?

[devout snow]

@bleak arrow Has your question been resolved?

[solar orchid]

I want to make sure what the problem is . Do you mean solving what curve of equation z = 9-x^2-y^2 ?

[devout snow]

@bleak arrow Has your question been resolved?

[bleak arrow]

getting the boundary curve of the surface: z=9-x^2-y^2, where z>0

[devout snow]

@bleak arrow Has your question been resolved?

[devout snow]

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[heavy current]

Let $\lambda = \int_{0}^{1} f(t) dt$ and define $g: [0, 1] \to \R$ by $g(x) = \int_{0}^{x} f(t) dt - \lambda x$. Then $g(0) = \int_{0}^{0} f(t) dt - \lambda \cdot 0 = 0 + 0 = 0$ and $g(1) = \int_{0}^{1} f(t) dt - \lambda \cdot 1 = \lambda - \lambda = 0$, so we obtain $g(0) = g(1)$. Furthermore, we have

\begin{align*}
\omega - \lambda \ dx = f \ dx - \left(\int_{0}^{1} f(t) dt \right) dx = f \ dx - \lambda \ dx = dg
\end{align*}

which is what we wanted to show. Lastly, if $\lambda' \in \R$ is another number with the same property as $\lambda$, then

\begin{align*}
\omega - \lambda \ dx = dg = \omega - \lambda' \ dx \implies (\lambda' - \lambda) \ dx = 0 \implies \lambda' = \lambda
\end{align*}

so $\lambda$ is unique.

[heavy current]

does this argument seem fine?

[woven radish]

higher!

[heavy current]

my primary concern is that I haven't used the fact that f(0) = f(1) anywhere

[faint gorge]

dumb question but if you differentiate g(x) couldnt you find lambda using f(0) = f(1)

[heavy current]

I think I'm the dumb one here, because I don't see what you mean

[faint gorge]

well i am not that familiar with differential forms, but the last sentence tells me that you want to actually find lambda not just show it's unique, so my very first and vague thought was to use g(x) for example since it contains lambda and then by using f(0) = f(1) somehow figure out what lambda is

[heavy current]

well like, we don't know what g is going to be a priori, right?

I defined g out of thin air

because it works

[heavy current]

(kinda, the hint gives you an idea of how to define g, but the point remains that we don't have it to begin with)

[faint gorge]

ok fair

[devout snow]

@heavy current Has your question been resolved?

[finite obsidian]

How’s dg defined here

Once you’re given g

[heavy current]

dg is the differential of g

[finite obsidian]

So it should be basically like, g’dx right?

[heavy current]

yeah, cause g is only a function of x

dg = (f - lambda) dx, yeah?

the way I've set this up

[finite obsidian]

Yep

I dunno where exactly the 0 on ends condition of f would come in here fr

[heavy current]

I feel like what I wrote works

the f(0) = f(1) thing feels useless

[autumn fjord]

sorry, is g given here?

[heavy current]

no

I defined it

[finite obsidian]

So for the latter part, you can’t necessarily assume they’re both equal to the same g I think?

[heavy current]

ah

I see

[finite obsidian]

But if (f-lambda’)dx = dg, and it’s zero on 0, 1, then \int_0^1 f dx - lambda’ = 0, right?

[heavy current]

do you mean g or g'?

[finite obsidian]

Whatever the g is, it says lambda’ has to be that integral

[autumn fjord]

why is lambda just not 0 for all f

[heavy current]

you're asking me?

[finite obsidian]

Since it’s fdx not df

[autumn fjord]

oh treu

[finite obsidian]

Ye

[autumn fjord]

okay yeah. sorry. i'm tired.

[finite obsidian]

So you didn’t need to show g=this function

[heavy current]

I don't understand

[finite obsidian]

You could’ve just done something like this to show lambda = integral of f?

[autumn fjord]

lambda = int_0^1 f(x)dx

yippee

[heavy current]

but is that not how I defined lambda

oh, you're saying that this shows uniqueness?

if you find it that way

instead of pulling it out of a hat

[autumn fjord]

what happens when you integrate dg @heavy current

[finite obsidian]

g(0) = g(1) = 0, so \int_0^1 dg is 0 yeah?

[heavy current]

yes

I agree

[autumn fjord]

so when you integrate the other thing you should also get 0

[finite obsidian]

So the integral of (f - lambda)dx is 0

Which is: (integral of f) - lambda

So that says lambda = integral of f, without any sort of like, what g is

[heavy current]

wait, am I dumb or is this supposed to be lambda*x?

[finite obsidian]

But saying what g works explicitly also says there’s always such a lambda, g for any such f

[finite obsidian]

It’s outside the integral mf

[autumn fjord]

x evals to 1 and 0

int (f-lambda) = int f + lambdax | ^1_0

[heavy current]

ic

[heavy current]

then what I did should be fine for existence, is it not?

[finite obsidian]

Ye, just saying this is a touch overcomplicating it idk

[heavy current]

ic

[finite obsidian]

I don’t see why the zero condition on f is present at all

Very curious

[heavy current]

but you said that I failed to show uniqueness, yeah?

cause it need not be the same g that works for lambda'

[autumn fjord]

uniqueness comes from showing that lambda has to be int f

[heavy current]

I see

so the way you and Sharp did it kills uniqueness at the same time

oh, I guess I can just do that for my uniqueness part though

[autumn fjord]

i am too tired to read sharp's messages. you should still provide an explicit g to show existence.

[heavy current]

I have one dw

if you're up for another problem, I'm trying to solve the very next exercise but find myself kinda clueless

[autumn fjord]

but i'm sure everything that sharp said is good

[heavy current]

it's basically this one but ++

[autumn fjord]

yeah i'm up for it

[heavy current]

I have no idea how to find lambda or g this time

[finite obsidian]

Ye

[finite obsidian]

Well, d omega = 0 is an important part

What’s c_R,1*

[autumn fjord]

radius r

circle

maybe

just a guess

[finite obsidian]

Btw, in light of last night discussion, proposition says closed 1 form is a scaling of d theta, modulo an exact form

[heavy current]

I don't know what the heck the subscripts are

[finite obsidian]

So it says we get that it’s 1 dimensional in that quotient

Curious how that works

[heavy current]

c is usually a singular n-cube in Spivak's book

oh, hold on

4-23 defines it like this

[autumn fjord]

if you have c be the circle of radius R then it will pull back to a 1-form and use the previous part

[finite obsidian]

The circle

So, this makes it a map from [0, 1] and f(0)=f(1) since it’s a circular path

[autumn fjord]

heck yeah

i was right

[finite obsidian]

For the one form pullback

[heavy current]

I see

[finite obsidian]

So, the prior lemma applies

And you know exactly what lambda is

[heavy current]

hm

I guess I still need g

[finite obsidian]

You get g by the same procedure too

So, start by showing that integral doesn’t depend on R, perhaps?

Since, if that’s the case, surely that helps out a bit in making g what it “should” be?

[heavy current]

the details seem kind of murky to me, but I guess I gotta write it down first

lemme see where I go with this

[devout snow]

@heavy current Has your question been resolved?

[heavy current]

.close

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[median patio]

idk what to do

[upper schooner]

What happens as n gets larger and larger?

[median patio]

um

[upper schooner]

(f being continuous means you can Riemann integrate it, so what does that imply?)

[median patio]

the approximation is less accurate?

[upper schooner]

(And not as n grows, it wouldn’t be!)

[median patio]

wdym by larger

[lethal pollen]

As you add more subintervals basically

Because n is the number of subintervals

[median patio]

so if there is more subintervals would the approximation be more accurate..??

[lethal pollen]

Yea

And if n approached infinity

[median patio]

exact approximation??

well almost ig

[heavy current]

https://upload.wikimedia.org/wikipedia/commons/e/e3/Riemann_Integration_and_Darboux_Lower_Sums.gif

more subintervals, better approx c:

[median patio]

hm how does that apply to the question tho

[lethal pollen]

Ok nvm

It's a limit question

the limit of the expression as n->inf is the value of your integral

[median patio]

how do you know its n-->infinity

[lethal pollen]

Because as you add more subintervals you get closer to the actual value

So as n approaches inf you're going to approach the actual value

[median patio]

hmm

how does it imply that you need to add more subintervals tho

cause it just says n subintervals

[lethal pollen]

Because they don't want an approximation

They want the actual value of the definite integral

[median patio]

ohh

ty

.close

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[solid osprey]

let $ABC$ be a triangle where $AB<AC$, let $H$ be the orthocenter and let the bisector of angle $BAC$ intersect the circumcircle with diameter $AA'$ at point $M\neq A$, let there be point $X$ on $AM$ such that $HX\perp AM$, let $P$ and $Q$ be the intersections of $HX$ and $MA'$ to $BC$ respectively, prove that $PMQ$ is an isoceles triangle

[woven radish]

skissue.in.a.teacup

[solid osprey]

i posted this yesterday, i thought that you could prove that MA'=MY and since A'Y//PQ then MQ=MP, but i forgot that you needed to prove Y (extension of altitude at A intersecting the circumcenter) is on line MP

is there any way you can salvage this or is there another way

[devout snow]

@solid osprey Has your question been resolved?

[devout snow]

@solid osprey Has your question been resolved?

[devout snow]

@solid osprey Has your question been resolved?

[lament kraken]

@solid osprey

you there

I have an idea but it probably wont work

If we prove BPM should be congruent to CQM were done

And CBM is CMB as AM is bisector of BAC

So we obtain MB is MC as well

We might be able to use orthocenter and the fact that AA' is diameter to give is similar triangles or ratios to prove the rest

I am not sure tho probably will fail

[solid osprey]

hmm

[lament kraken]

@solid osprey i gtg soon good luck

[jaunty shuttle]

Hello, is it okay if i ask some questions here?

[junior flax]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lament kraken]

!occupied

[devout snow]

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

[somber dawn]

You can ask questions in the math help (avaliable) category

Should be one category above

[lament kraken]

Also @solid osprey where was the qn from

[jaunty shuttle]

I see, thank you

[solid osprey]

uhh it was from a p small oly

[lament kraken]

Oh ok

[devout snow]

@solid osprey Has your question been resolved?

[lament kraken]

@solid osprey any progress

[solid osprey]

no

i have no idea how to use congruence

[lament kraken]

hmm ok let me attempt it

[solid osprey]

cause like it feels like proving congruence needs to prove the first statement first

[lament kraken]

Mb youre right

[solid osprey]

.close gtg sleep

[devout snow]

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[devout snow]

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[empty nexus]

question 5

[spring salmon]

Can you show your work so far

Where are you stuck

[pastel pasture]

FTC

[empty nexus]

im confused about the 2x part

[mystic scarab]

Well, FTC answers your question, doesn't it?

[empty nexus]

is this wrong

[mystic scarab]

Yes, it's wrong

[empty nexus]

why

[mystic scarab]

You have to use chain rule, because the upper bound is not x, rather a function of x

[real belfry]

there's nothing wrong with subbing in a variable as a limit, have you evaluated the integral?

[empty nexus]

so you have to also multiply by 2?

[real belfry]

and it makes sense right, it would be a pretty boring f(x) without any x in there, t is just a placeholder for your function where you need to place in the limits for the area

[empty nexus]

or divide?

[mystic scarab]

Exactly

[mystic scarab]

No no, chain rule says to multiply by the derivative of the inside function

[devout snow]

@empty nexus Has your question been resolved?

[devout snow]

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[floral lodge]

"In a resturant, there are 12 sandwhiches of different meats. Each sandwhich can be served in half or full, with light or dark bread. There are also 18 different veggie-add ons, where each can be added or removed. Finally, one may choose one of 3 dressings or none. In how many different ways can a sandwhich be ordered?"

[floral lodge]

My attemt:

12 meats, i.e 12 options, 2 options for the size of the sandwhich and 2 more for the type of bread

so far, $12\cdot2\cdot 2$

[woven radish]

Pen

[floral lodge]

For the veggies, there are 18! ways to choose a veggie combo

Lastly, there are 3 dressings but one also has the option to not order any dressing, meaning that there are 4 options for the dressings

This leads me to conclude that there is $12\cdot2\cdot 2\cdot 18! \cdot 4$ways to order a sandwhich

[woven radish]

Pen

[floral lodge]

Which is... 1229255751499776000 ways

the correct answer is 50 331 648 possible ways

I do not understand where I went wrong

[versed juniper]

isnt it 2^18 not 18!

[floral lodge]

That does give the correct answer

[versed juniper]

[floral lodge]

2^18 would be whether to have the veggie or not?

repeated 18 times for all veggies?

[versed juniper]

yes, 18! is how many permutations of veggies lol

[floral lodge]

gosh, you are right

in my head I was thinking about the veggie combinations, at first you can order 18

then 17

then 16...

[versed juniper]

well even if you did it with combinations you have to remove dupes

veggie A and then veggie B is the same as veggie B then A

*in this problem

[floral lodge]

yup

how many dupes would there be?

[versed juniper]

18! - 2^18 👍

[floral lodge]

so big number, gotcha

combinatorics is going to make me go bald before life gets a chance

thank you for the quick and helpful help, ten

the help was a ten out of ten, ten : )

.close

[devout snow]

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[devout snow]

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[restive elk]

BTW the inflection point is (2,-47).

[restive elk]

And the cubic is x^3 - 6x^2 - 15x - 1

[versed juniper]

plug in sin x first?

and what does one cycle mean

0 to 2pi?