help-23

[rustic hound]

alright, let me try

not sure how that helps though

am I supposed to notice something?

[rancid wagon]

adding 0x^2 makes long division easier

and convenient

[rustic hound]

well, I don't know what to divide this one by

well, I wanna find the solutions

does dividing the polynomial by something that doesn't give the remainder 0 help?

[rancid wagon]

nvm

[rustic hound]

the problem asks the rational solutions to this equation, but it seems like there are 3 irrational solutions

[dim herald]

maybe Cardano lol

at last

Where did you get this problem from?

[rustic hound]

some algebra exercises book (which is not in english)

this one asks what is this set equal to

[dim herald]

have you already tried splitting the 5x term?

[rustic hound]

well

I don't see how I should do it in a way that it helps

the answer at the back of the book says that there are no rational solutions, I checked wolframalpha and this is indeed true, but I'd have to conclude that

somehow

[dim herald]

There is a formula for 3rd degree polynomials but I don't know if you can justify the solution using it

[rustic hound]

yeah, I'm not using that abomination

[shy temple]

The rational roots must be (factors of constant term)/(factors of leading term).
So just show that 1 and -1 are not roots

[dim herald]

oh yeah

exactly

[rustic hound]

got it, thanks a lot!

.close

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[whole acorn]

how does ax^2 + bx + c = 0 turn into that

[whole acorn]

wait nvm i got it from a website

why do i add (b/2a)^2 tho

where do i get that

[thin bridge]

it's completing the square

[whole acorn]

what square

[thin bridge]

adding that term turns the left side into a perfect square

I suggest you look up completing square

[whole acorn]

ok

ohhh

i get it now

so it's simply

completing squares like this

ok now i know where it got the name quadratic equation

or something

and this +- symbol was used because a root of a number can be positif and negative right?

welp

t

y

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[stray socket]

Draw 3 line: AB, BC, and DC

And AD

So 4

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[peak python]

I don't know how to solve this, i get DV/DT = -kV^3 but that's all

[sharp crane]

(a) or (b)?

its separable. you just need to rewrite it to
$$\frac{1}{V^3} \dd{V} = -k \dd{t}$$

[flat frigate]

nvx

[sharp crane]

and then integrate both sides

[safe radish]

@peak python Has your question been resolved?

[peak python]

ah i dont think we've learned integration for connected rates yet, also thanks again c:

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[toxic isle]

What would be the No and Na be? and what is the significance level?

[shy temple]

What do you mean No and Na? Na is the chemical symbol for sodium by the way. It helps if you're clear.
Anyway, the model assumption is that the carcinogenic insecticide comes from N(mu,sigma^2). The hypotheses are H0: mu = 0.08 ppm, H1: mu > 0.08 ppm.

[toxic isle]

like null hyp and test hypothesis

and thankyou

.close

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[hasty pasture]

going through notes and wanted to ask: https://cdn.discordapp.com/attachments/860810557429776434/967438742753259540/unknown.png

[hasty pasture]

https://cdn.discordapp.com/attachments/860810557429776434/967438826802929704/unknown.png

how did they get 2.064?

like what's the formula

[shy temple]

Use R command qt(0.975,24)

[hasty pasture]

where does 0.975 and 24 come from?

also i don't know how to use qt(0.975,24) in a calculator

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[coral mantle]

Can someone help me to do this question

[coral mantle]

So I have done a) and b)

But what should I do in c)

[final halo]

you now need to solve for N

and see what happens as t goes off to infinity

[coral mantle]

But How do I do that

I tried

[woeful plume]

Solve the differential equation, find the const. of integration, and then take the limit

[coral mantle]

But get 500 00

And that cant be right

[tall sphinx]

What do you get after integrating

[coral mantle]

Wait is this right

[tall sphinx]

about a million you should get if my subconscious is right

Sorry, i cant read that. what do you get after integrating

[coral mantle]

Which part

[tall sphinx]

...

you need to integrate it

this is a first-order nonlinear ordinary differential equation

[woeful plume]

Find N as a function of t

[coral mantle]

Have not Done right

[tall sphinx]

Not to be rude but you do know integration and all that right?

[coral mantle]

Yeahh, kinda. It was just long time

Ago i did

So I feel a bit lost

Sorry :/

[tall sphinx]

right, dont worry we have all been there

do you know what you need to do to get the answer

[coral mantle]

Not Really

You think you can show me a few steps

[tall sphinx]

sure

[coral mantle]

Appreciate

It

[tall sphinx]

so, what is a diff equation. a diff equation shows the rate of change over time you could say

here we see how the population changes

[coral mantle]

Okay I will just erase my answer on C and follow ur steps

[tall sphinx]

if my f(x) = x^2 the rate of change is 2x

same stoory here, so to get the f(x) we need to integrate 2x

this is a first-order nonlinear ordinary differential equation so there is not a lot to it

[coral mantle]

Okay

[tall sphinx]

i can write it out one sec

[coral mantle]

Thank you so much, means a lot

[tall sphinx]

shortly done, it was a bit tricky integrating actually

[coral mantle]

Yeahh

[coral mantle]

Came up with anything yet?

[tall sphinx]

@coral mantle Yeah nearly there, its the large numbers that are tripping me up

i have to use a calculator to make sure im getting things right

would be much nicer if they didnt do millions

[coral mantle]

Yeahh true

[tall sphinx]

i have to take back my words on this being simple, often they are but not in this case lol

[safe radish]

@coral mantle Has your question been resolved?

[coral mantle]

Hahahah yeah its quite hard 😦

[tall sphinx]

i have to disappoint here, im unable to solve it

running into some stuff i dont know how to do

The last few steps when its integrated and im solving for t

the last part is wrong btw, but thats how far i got

,rotate

[tall sphinx]

if we assumed the last part was right

you would take n(0)=250 000

so calibrate for C

and then you have your expression

you can take the limit as t goes to infinity

sorry i couldnt solve it, but thats how it is sometimes in math

[coral mantle]

Its okay, I appreciate

[coral mantle]

Have I done this one right?

[safe radish]

@coral mantle Has your question been resolved?

[safe radish]

@coral mantle Has your question been resolved?

[coral mantle]

.close

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[lean otter]

Hi everyone, I’m having a hard time understanding variations inference. Particularly why we can choose likelihood p(x given z) but p(z given x) is intractable. Why is that?

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@lean otter Has your question been resolved?

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[tough heath]

Hello, i'm trying to solve the following problem from Sheldon Ross' book on probability: "suppose that 5 members are to be selected from a group of 20 individuals consisting of 10 subgroups of 2, what is the probability $P(N)$ that all 5 chosen are from differing subgroups?"
I understand the approach taken in the book, but i am trying to approach it from the angle that the probability of $P(N)$ is equivalent to $1 - P(N^c)$. Here's what i did:

$P(N^c)$ is equivalent to $\bigcup_{i=0}^{10} A_i$, where $A$ is defined as the event that the $i$th subgroup is selected. I calculate the amount of outcomes for event $A$ as being equivalent to a multi-stage experiment in which we choose 5 members to form a group by selecting 1 of the 10 subgroups, then 3 members from the remaining 18 individuals to complete the group, as such: $10\times\binom{18}{3}=8160$.
However, the solution given in the book tells me that this is overcounting, because in the book $\lvert{N}\rvert$ is $8064$, and the sample space is $\binom{20}{5}=15504$, so there should be $15504-8064=7440$ outcomes for $N^c$. I'm not sure where the overcounting comes from.

[flat frigate]

Doctor Blythe

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[tough heath]

<@&286206848099549185>

[tall bough]

don't ping helpers

[tough heath]

well, the message says "after 15 minutes, feel free to ping helpers", i apologize if that's not what i was supposed to do but it was all i had to go by.

[grizzled fossil]

You are supposed to ask your question and show/explain what you have tried/where you are stuck

Like it says

[tough heath]

sorry, i explained it. It is in the post above. Should i repost?

[grizzled fossil]

Seeing how the channel is closed, yes

[tough heath]

i see, i apologize for the empty ping.

Hello, i'm trying to solve the following problem from Sheldon Ross' book on probability: "suppose that 5 members are to be selected from a group of 20 individuals consisting of 10 subgroups of 2, what is the probability $P(N)$ that all 5 chosen are from differing subgroups?"
I understand the approach taken in the book, but i am trying to approach it from the angle that the probability of $P(N)$ is equivalent to $1 - P(N^c)$. Here's what i did:

$P(N^c)$ is equivalent to probability of $\bigcup_{i=0}^{10} A_i$, where $A$ is defined as the event that the $i$th subgroup is selected. I calculate the amount of outcomes for that union as being equivalent to a multi-stage experiment in which we choose 5 members to form a group by selecting 1 of the 10 subgroups, then 3 members from the remaining 18 individuals to complete the group, as such: $10\times\binom{18}{3}=8160$.
However, the solution given in the book tells me that this is overcounting, because in the book $\lvert{N}\rvert$ is $8064$, and the sample space is $\binom{20}{5}=15504$, so there should be $15504-8064=7440$ outcomes for $N^c$. I'm not sure where the overcounting comes from.

[flat frigate]

Doctor Blythe

[safe radish]

@tough heath Has your question been resolved?

[tough heath]

i'm starting to think it might be that the intersection between each individual union for any A_i and A_j is being counted multiple times. If so, and if i have to use the generalized axiom of probability of a union to calculate this, then i guess it's not worth doing since there's 9 unions. But is that the case?

.close

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[broken nimbus]

Im having trouble trying to solve a

[safe radish]

@broken nimbus Has your question been resolved?

[marsh walrus]

try splitting each into x and y components

[broken nimbus]

is there an easier way? this is related to calculus btw

[marsh walrus]

this is the way i know how to do it

its just trig

you may choose them to be like uhh

idk 20 degrees and 160 degrees

so you know the direction of one of the resulting components

(off set equal about 90)

that saves you a bit of work

or maybe not since its not asking direction

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[quiet tapir]

How do I use algebra to prove that this doesn't exist

Cosine may not be zero I suppose

[marsh walrus]

use definition

with left and right and sides

well, you dont even really need to do that

do one or the other

show it diverges for left or right hand

[quiet tapir]

How do I "show"?

[marsh walrus]

what class?

just calc i?

you could probably get away with just arguing if so

idk what exactly the proof looks like here

well i mean youd just show tanx is unbounded around pi/2

[quiet tapir]

I can't do that without using a graph tho can I?

[marsh walrus]

in calculus its usually assumed you know how trig functions behave

[quiet tapir]

True

[marsh walrus]

vertical asymptotes are just classic places where functions arent bounded

[quiet tapir]

Yeah that's true

[marsh walrus]

if you have to be really formal about it

id do like

$\lim _{x \to \sfrac \pi 2 ^+} \tan (x) = - \infty$, but $\lim _{x \to \sfrac \pi 2 ^-} \tan (x) = \infty$. As the left and right hand limits disagree, the limit does not exist

this is really uhh

[flat frigate]

jan Niku (join us for @pomo)

[marsh walrus]

i mean each individual limit doesnt exist

if were being pedantic

(limits dont equal infinities)

[quiet tapir]

Yh I thought so too

Arguing using definitions should probs be enough during a questioning

Thx bud

[marsh walrus]

np

if you really really want to like

show it

show its unbounded

its just a more like

its not a calc i argument

usually

im sure you could figure it out, they just dont usually teach it

[quiet tapir]

Yh I haven't been taught that yet

.close

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[novel kettle]

I am stuck on this problem even though it seems super simple. I believe there is an issue with my process rather than the calculations themselves.
This is a triple integral problem. First I convert the equations to cylindrical coordinates.
Then I find the intersection between the sphere and cone (a circle).
This becomes my domain of integration and is in polar coordinates.
My triple integral thus becomes

V = (0, 2π)(0, 8)(r, sqrt(128 - r^2))∫∫∫ r dz dr dθ

for this problem.

[novel kettle]

My answer is wrong though. I know I do the integral correctly so really I mess up before then.

z = sqrt(x^2 + y^2) --> z = sqrt(r^2) --> z = r
x^2 + y^2 + z^2 = 128 --> z = sqrt(128 - r^2)

sqrt(128 - r^2) = r
128 - r^2 = r^2
2r^2 = 128
r = sqrt(128/2) = 8

[feral linden]

I will use x=rcos(t) ,y=sin(t) and dxdydz=rdrdtdz . You also did this? If so let’s see whether we get the same answer

[novel kettle]

Yes I used cylindrical coordinates

I am really sorry

I did make an error with solving the integral. The integral is set up properly.

I found the answer, sorry haha

Thank you for the assistance

.close

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[shrewd cairn]

hi V is vector space and M,N are subspaces of V determine if these properties are true for orthogonal complement i Think that A) is not true from the definition of it i found in the scripts but im not sure if the b) is true as well because of the backwards implication

[feral linden]

$M=(M \cap N) \oplus (M \cap N^{\perp}) \subset (M \cap N) \oplus (M \cap M^{\perp}) \subset (M \cap N) \oplus 0= M \cap N$
So $M \subset N$

[flat frigate]

Cogwheels of the mind

[feral linden]

So the field, this inner product need to be standard enough, any subspace M, the intersection of M and M^orthogonal should be 0. I am not sure that it holds for any field. Probably we can find a counterexample in a finite field for some quadratic form

[safe radish]

@shrewd cairn Has your question been resolved?

[shrewd cairn]

wait im kinda confused which of these is this 😅

[feral linden]

?

The whole space $V=N \oplus N^{\perp}$

[flat frigate]

Cogwheels of the mind

[feral linden]

Intersection M both sides it gives you my first line

[shrewd cairn]

wait like is this a help or this is answer to one of the a,b,c

oh okey

[feral linden]

I was answering b

a is false like you said

[shrewd cairn]

and b is true then right ?

[feral linden]

It’s true if it’s inner product of vector spaces over R or C

If it’s an arbitrary quadratic form over an arbitrary field we might be able to find a counterexample

c is true

[shrewd cairn]

okay

[shrewd cairn]

and some proof or something

cuz i dont know why it is

[feral linden]

By definition

[shrewd cairn]

oh okay thanks ill look it up

[safe radish]

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[cunning hornet]

how would i find the angle between the two planes?

[cunning hornet]

more specificly, how would i set the right side of the equations to zero?

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@cunning hornet Has your question been resolved?

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@cunning hornet Has your question been resolved?

[lean otter]

3x-4y+z-6=0

Move the 6 to the other side

Is that what youre asking

$\cos\theta =\left|\frac{\vec n_1 \vec n_2}{|\vec n_1| |\vec n_2|}\right|$

[flat frigate]

Em

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[hollow patrol]

https://media.discordapp.net/attachments/387449305885048852/967927079884714064/Screen_Shot_2022-04-24_at_4.16.24_PM.png

[hollow patrol]

Hello! This is my math homework and I am unsure how to do it, I don't understand what the constraint/function value means.

https://media.discordapp.net/attachments/387449305885048852/967927079884714064/Screen_Shot_2022-04-24_at_4.16.24_PM.png

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[flat frigate]

Rœmer

[hollow patrol]

.reopen

[safe radish]

✅

[hollow patrol]

i don't get it at all

[gleaming anvil]

do you know the domain of the trig functions v

?*

[safe radish]

@hollow patrol Has your question been resolved?

[hollow patrol]

like the quadrants? then yes

[worthy hemlock]

Why did you ping helpers?

#❓how-to-get-help

Not your channel

Also you need to wait at least 15 minutes before pinging helpers

Read #❓how-to-get-help

Then read them

Read it then you'll know where to go

No

#❓how-to-get-help

I am tagging you to the channel

Read #❓how-to-get-help

Alright then

Bye

[safe radish]

@hollow patrol Has your question been resolved?

[plucky elk]

.close

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[median eagle]

s

[median eagle]

In abstract algebra

Can anyone please help me with this question

[safe radish]

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[tame jasper]

CanI get some help on where to start I'm a bit confused?

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[fathom parcel]

If i have instead of x x_1+x_2 can someone tell me what this would seem like?

[fathom parcel]

I thought you would just replace x by x_1+x_2 but this does not seem to be correct.

[iron flare]

if arctanhx = 1/2 ln(x) there are no solutions to x

they never intersect

[fathom parcel]

i mean if i have artanh(x_1+x_2) is it then just 1/2ln(1+x_1+x_2/1-x_1+x_2)?

[iron flare]

there should be a bracket at the bottom but yeah

(so that its -(x_1+x_2)

[fathom parcel]

like i have x=x_1+x_2 and now i want to use artanh on both sides. so i get this

so instead of x i have beta obviously

but i do not get to this equation by doing so, even tho i should.

You see a way to get to this equation by using artanh on x=x_1+x_2?

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[safe radish]

Please don't occupy multiple help channels.

[simple delta]

.close

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[queen pier]

How would I solve this? Would I have to add a "y" variable to the left side to solve the ratio?

[trim trench]

no need

notice that the angles are the same

[queen pier]

?

i know the angle bisector theorem

but

what ratio would i put

to solve for the x

[trim trench]

therefore x+x=12 then

[queen pier]

can we just assume 2x = 12??

[trim trench]

wait no

sorry

use pythagoras theorem

to find the remaining side

@queen pier

then you can use angle bisector theorem

[queen pier]

(9x^2)+(_)^2=c^2

[queen pier]

what would i put in the blank?

OH WAIT

IM DUMB

MB

ITS 12

[brisk ember]

no 15

hypotenuse 15

now you can use the pythagoras theorem to solve for x

[trim trench]

noope

[brisk ember]

am i wrong?

[queen pier]

hyp is 15

[brisk ember]

yea

nvm about the part with pyth to solve x

[trim trench]

I think we should use angle bisector

[trim trench]

sure

now you got 15

you can use angle bisector

which the other side is 12-x

[queen pier]

9/x=15/12-x

ty

[trim trench]

here you go

please close it by .close

[brisk ember]

kasper

can you help me with mine?

[trim trench]

make a new help channel

but first @queen pier close this

@brisk ember Make a new help channel

[queen pier]

.close

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[royal iris]

Length of a major axis of 20 and the length of the minor axis is 12, assume the center point is (0,0) and the major length is horizontal. what is the length from the center of the room to the focal points

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[royal iris]

<@&286206848099549185>

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[royal iris]

.closed

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[bold ferry]

An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side.

[bold ferry]

This unit is about trig so should I assume it is a right triangle?

[tawdry fox]

no its not a right triangle

[crude nest]

you can make it into two right triangles though

[bold ferry]

ohh

I solved it

[bold ferry]

thanks

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[crimson heron]

I dont really know how to approach this question..

[sharp sky]

its basically asking you the velocity

[crimson heron]

umm why

<@&286206848099549185>

[marsh walrus]

just write the direction

they dont turn or anything

what are you confused about?

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[somber kiln]

Can someone help me understand how the last statement is justified?

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[safe radish]

@lean otter Has your question been resolved?

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@lean otter Has your question been resolved?

[grizzled fossil]

Don’t understand the question

You are basically asking what is this data - how are we supposed to know?

[lean otter]

How do I label this thing?

[grizzled fossil]

Its fine if that is what the data actually is

Which you were just like “I presume” it is

[lean otter]

Chur. I am a bit tired doing something I have no idea about lol

How do I close this thing

[grizzled fossil]

Type the command .close

[lean otter]

.close

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[bright notch]

What are the steps for figuring this out?

[tall bough]

how old are you

[bright notch]

21 and failed math

Trying to relearn

[tall bough]

let me find a YouTube vid

[bright notch]

I found one

I did other problems

But I cant do this one

[tall bough]

https://youtu.be/HLMGPoiXeNI

[bright notch]

Oh

How does that work

Okay

Why wouldnt I take from the 2 at the end

To make 3 13

If it was 2013 - 1026 could I take the 1 and put on 3 to make 13 minus 6 to get 7?

But since next number is a 0 I have to go into negative numbers?

Okay ill take a look now

.close

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[restive condor]

do you just sub in 3 for t?

does anyone know what this means?

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[gray plaza]

Would someone please explain me this statement:

∼ (P ⇒ Q) = P ∧ ∼ Q.

[gray plaza]

How this statement conveys that P is true and Q is false?

[grizzled fossil]

it is explained in the text

only way P implies Q can be false is if P is true and Q is false.

[gray plaza]

Yes, I understood this.

But how this statement conveys that P ∧ ∼ Q. P is true and Q is false?

[pallid nymph]

P and not Q

thats what its saying

[gray plaza]

Yes, But why P = True?

[pallid nymph]

ok it's saying that P does not imply Q if P can be true while Q is false

it's like a situation

[gray plaza]

Gotcha

Thank you @pallid nymph @grizzled fossil 😊

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[proud igloo]

i would like help with this question

[thin bridge]

consider vietas formulas

(sum and products of roots)

[proud igloo]

im not really sure how to apply it to this question

[final halo]

Let k be one of the roots

Then we can let k+1 be the other root

Now you should be able to write the equation in a factored form involving k, can you do that?

[proud igloo]

Sorry i dont know how to do that either lol

[final halo]

If $a$ and $b$ are roots of $f(x) = x^2 + px + q$, then $f(x) = (x-a)(x-b)$

[flat frigate]

iCaird

[final halo]

Do you remember seeing that?

[proud igloo]

yes

[final halo]

So now we can do that for k and k+1

Give it a go

[proud igloo]

i really dont have a clue on what to do but would you make x^2-mz+n= k(k+1) ?

[final halo]

Not quite

[final halo]

Look at this carefully

[proud igloo]

sorry im struggling, i really dont know what to do

.close

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[gusty halo]

Hey I need help with Geogebra, specifically parameter functions

[gusty halo]

I have this parameter with these different variables, and I am wondering if it is possible to write a command that would find what value of velocity would make the curve a intersect with the point G

Is that possible, or is it possible with scripting (not sure what that means but I would assume it is programming related), or is it not possible at all?

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@gusty halo Has your question been resolved?

[gusty halo]

<@&286206848099549185>

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@gusty halo Has your question been resolved?

[obtuse jackal]

I don't know about Geogebra, but you can do that with desmos. Just see where (20 - f(170) = 0) where f depends on v instead of x

[gusty halo]

I can try that, but this is a curve, so I would have to turn it into a function first

[obtuse jackal]

I just put f because I didn't want to write the formula, but you could just write t = x / (v cos 20.3) and then graph y(t) with your formula

since the curve is implicit there is a bit of extra work, but it's doable

also technically with a bit of math you can rewrite this as a proper y(x)

[gusty halo]

okay, thanks a lot!

[obtuse jackal]

if you struggle with this feel free to ask here

[gusty halo]

I will, thanks!

.close

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[brisk comet]

Hello, I'd like to ask for a bit of clarification regarding this application of FODEs for the problem given to us here in our reference book.

I'm trying to figure out how to answer the SCENARIO portion. I already got eqn 4 (the one that's u(t)) through SODE methods (wasn't sure if there was a way to get an answer with the same form with FODE methods given to us).

However, I'm stuck here with getting c_1, and c_2, which are the unknown coefficients in equation 4. I figured c_0 can be obtained using by doing having t=0 in eqn 4. This would result in u(0) = u_0 = eqn4 right hand side. This would make it to where the term with sin would be cancelled to 0. We would eventually arrive with c_0 = a_0 - u_0 - c_1.
From there though, I'm stuck. I don't know how to get c_1 and c_2. I assume you'd do something in eqn 3, but I can't see what I'm supposed to do. So, I've just been stuck here trying to derive and figure out how to get c_1 and c_2.

[safe radish]

@brisk comet Has your question been resolved?

[patent fulcrum]

from what i think

differentiate eqn 4
you will get a d(u)/dt eqn

substitute that in eqn 3

you will have (something)sin + (something)cos = (something)sin + (something)cos

then you can equate those somethings and solve for c1 and d1 (or as u call it c2)

[patent fulcrum]

@brisk comet

[brisk comet]

oh right it was d_1, woops

sorry for that

aight ill try that, thanks!

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[fierce dust]

<@&286206848099549185> this is what i tried but its not correct

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@fierce dust Has your question been resolved?

[potent bay]

120?

why?

@fierce dust

[fierce dust]

yeeeeeee i just realized my mistake ,-,

its 108 not 120

how do i solve this

isnt it -tan(48)?

[potent bay]

the 48 is the top angle

its tan(-66)

[fierce dust]

we always use the one on the x-axis?

[potent bay]

no

[fierce dust]

the one on the side not in the middle?

[potent bay]

we need this slope

[fierce dust]

if we want AC then Tan(66) right

[potent bay]

yeah

[fierce dust]

i get it tysm

[potent bay]

cuz you imagine the unit circle there

so if you put the circle around C

[fierce dust]

i see

[potent bay]

thats the 66

and its negative cuz its downwards

[fierce dust]

damn turns out its so simple

alr tysm!

.close

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[narrow kraken]

Hi

[narrow kraken]

So we have to use l hopital’s rule

But I’m really confused by the + sign next to the 0, I know it means that we’re approaching 0 from the right but how would I plug that in?

[feral linden]

Just In(x) over 1/x

To see that the limit is 0

Using lhopital like you said

[narrow kraken]

Wouldn’t that result in undefined? Cause if we plug in 0 for x we get 1/0

[queen parcel]

You're not plugging in 0

You're seeing what happens as x gets closer to 0

[narrow kraken]

So can we plug in 0.00000001

[queen parcel]

yes, to get a feel for how the function behaves

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[hollow moon]

Help me

[hollow moon]

Does integral from 0 to 1 of 1/ cube root x

converge of riverge

and which test do i use

[lean otter]

$\int_{0}^{1} \frac{1}{\sqrt[3] x}$

[hollow moon]

Can I use comparison test

I dont understand if cube root grews faster than x

So I am confused

which will be bigger

when in denominator

[flat frigate]

Frustrated Cat

[hollow moon]

x^1/3 correct

I am a bit confused

So is that greater than 1/x

can you help me

I THINK I / X IS GREATER

srry caps

[zinc berry]

what are you even trying to do

[hollow moon]

converge or diverge

[lean otter]

He is trying to day wether the integral converges or diverges

[hollow moon]

Does the function 1 / x^1/3 convery or diverge

from integral 0 to 1

[lean otter]

I am pretty sure it diverges because it's literally discontinious at 0 . But I am not very sure how to say it more systematically

[hollow moon]

i am trying to use comparison test

[lean otter]

Oh

[zinc berry]

It does not diverge. I don't understand what you are actually trying to accomplish. Finding that antiderivative is not difficult

You rewrite it as x^(-1/3) and then do the thing.

[hollow moon]

is 1/x greater or less than 1/ cube radical of x

tha's all I want to know

you'tr typing so long

u good

[lean otter]

Oh right it does converge

[hollow moon]

how

[lean otter]

Calculate the anti derivative and do the normal business u will get a finite value tho

[hollow moon]

ok

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[safe radish]

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[tough basin]

Im having trouble figuring out whats going on here

[woven anchor]

what about it do you not understand?

[tough basin]

I'm confused on why d = 4, why we are doing n=4q, n=4q+1, n=4q+2, and n=4q+3

[woven anchor]

there are 4 consecutive integers

so there will always be every single possible remainders when divided by 4 within those 4

[tough basin]

So if it were, for example, 5 consecutive integers, then d would = 5? Then respectively, n=5q, n=5q+1, n=5q+2, n=5q+3, n=5q+4?

[woven anchor]

depends on what it's asking for you to prove

in this case you're proving that 8 always divides the product

and 4 is like a close relative to 8

because they're 1 power of 2 apart

if you're proving that 3 always divides the product of 4 consecutive integers, then choosing 4 would not be a good idea

[tough basin]

So how do you go about choosing what number to use for d?

[tough basin]

Using this as an example

unless that kinda question is unlikely to be asked..

[woven anchor]

it's not tight

because 3 always divides 3 consecutive integers

and if it divides 3 consecutive integers, clearly, it will divide 4

so it's unlikely

but you would look at the remainders when divided by 3

since you care about 3, and not 4 in this case

[tough basin]

ahh

okay!

Thanks for you help

understand it a bit better now

.close

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[slender pagoda]

i need help lol

[slender pagoda]

dont know where else to continue

all thats given is the angle of elevation from 2 different points and the distance between those 2 points

theres no other variable its all just angles i worked out the formula already if only a was given it would be already solved

but yeah

[fair beacon]

You have successfully written x in terms of a in two different ways already
Just using that is enough to solve the problem

[slender pagoda]

u mean the x for the scenario a and b? yeah i can combine add those and make the equation into 2x = blabla

ended with ths

[fair beacon]

Yes but these are just simultaneous equations in a and x; you don’t need to do anything fancy

Just combine the two equations and solve for a first, then solve for x

Or whatever approach you like

[slender pagoda]

i kinda dont know how to solve for a

its just all angles i dunno what to do

[fair beacon]

x = tan(51) a - 36
x = tan(35) a
Combining gives
tan(51)a-36=tan(31)a

Which is just a linear equation in a

[slender pagoda]

OHH wait ill try that real quick

YEAHH it worked thank you so much, dont know how i didnt think of it

.close

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[slender pagoda]

thanks!!

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[thin token]

complete the figure to make a right-angled triangle and then remove the rectangle of sides 3 X 15

[viral gate]

i dont understand

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[viral igloo]

area of the big triangle is (9+3)×(9+10)/2 = 540ft^2
now remove the area of the rectangle:
540 - 9×3 = 513ft^2

@viral gate

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[stiff storm]

Is there a quicker way of finding the year than calculating the singular increases every year?

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@stiff storm Has your question been resolved?

[stiff storm]

<@&286206848099549185>

[buoyant shadow]

sure

you're looking for n such that 1.03^n × start = 30000

which is just log

[stiff storm]

Oh ok

Thank you!

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[undone osprey]

Can someone help me?

[undone osprey]

show that the integral converges for p>1 and diverges for 0<p<=1

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[shy temple]

For p =1, you can integrate it to get ln ln x and that will diverge

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[plucky elk]

did you try integration by parts

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[lean otter]

Im very confused on this question as a whole

[agile fable]

Which part do you find confusing?

What "equivalent expression" means?

[lean otter]

basically it's asking me to simplify the top one right?

[agile fable]

yep

[lean otter]

issue is idk where to even start

[agile fable]

To begin, you see how everything is multiplying or dividing, right?

No additions, subtractions...

[lean otter]

oh yeah

[agile fable]

This means that we can just move things around "safely"

So first start grouping similar things, q's, w's and known numbers

[lean otter]

wait wait

im so dumb

first i gotta do the 4(q6)^2

16^64q?

[agile fable]

That's not exactly right. First, note that the square only applies to q^6

[lean otter]

cause the ()?

[agile fable]

yep

Second, do you know how to compute (a^b)^c ?

[lean otter]

no

[agile fable]

$(a^b)^c=a^{bc}$

[flat frigate]

MateoKz

[agile fable]

so you can use this

[lean otter]

wait how do i use it?

in which part?

[agile fable]

for (q^6)^2

[lean otter]

oh

so first i would do 2^6

64

then q^2

[agile fable]

you should not do 2^6

the ^6 doesn't affect the 2

look again at the problem

[lean otter]

an exponent can't multiply times another one?

[agile fable]

what do you mean?

[lean otter]

for the (q^6)^2, it wouldn't be 2^6 because the 2 doesn't 2x2x2x2x2x2 right?

[agile fable]

(q^6)^2 = (q^6) * (q^6) = q^(6*2) = q^12

[lean otter]

ahh so just 6x2

because the ^2

[agile fable]

yeah, (a^b)^c = a^(b*c)

[lean otter]

okay okay

so now that we got q^12

do i add the 4 to that?

[agile fable]

why would you add it, it's multiplying, right-?

[lean otter]

so i do 4*q^12

[agile fable]

well, that's correct

but there are other things that you still need to work on

[lean otter]

should I leave it like 4*q^12/w-5 for now?

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[orchid cargo]

I don’t understand 8c)

Isn’t the degree just the highest power

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@orchid cargo Has your question been resolved?

[dapper forum]

yes

[orchid cargo]

but i cant play the power of 1 and 3

can I

@dapper forum

[dapper forum]

It is asking for the degree of polynomial. It is 3 and you can add 2x + 9x = 11x

[orchid cargo]

oh ok

thank u

[dapper forum]

You're welcome. The degree of a polynomial is the highest power of the variable in a polynomial expression. fyi: https://byjus.com/maths/degree-of-a-polynomial/

[orchid cargo]

👍

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[stoic chasm]

real quick: Is t(2) = f'(2)?

[stoic chasm]

and how do I solve this

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[upper badge]

hint please

[plucky elk]

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

[upper badge]

x'(sigma(fi) - n) where x' is the mean and n is the number of classes ¯_(ツ)_/¯

not really sure where to start

[plucky elk]

you can start by explaining your variabels

[upper badge]

i did... fi is the same as in the ques

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[upper badge]

sorry... still didnt get it... just a hint on how to start

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[warped heart]

I think that option 5 is true, but i need help with reasoning about the other options.

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[blazing dew]

why is this true?

[final halo]

if $ax^2 + bx+c = dx^2+fx+g$ then we must have that $a=d$,$b=f$,$c=g$

[flat frigate]

iCaird

[final halo]

in your case we have $(4a-1)x^2+(4a-4)x = 0x^2 + bx$

[flat frigate]

iCaird

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[pine berry]

can someone help

[ruby pelican]

first find f(6,4)
then you can set f(x,y) equal to that value and rearrange into whatever form you need

[pine berry]

f(x,y)=4 ?

[ruby pelican]

yep

so sqrt(x+y^2 -6) = 4

[pine berry]

oh I see

thanks

[ruby pelican]

np

[pine berry]

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[worn kettle]

How would you find the apothem and perimeter of this?

[final halo]

Try using Pythagoras

[worn kettle]

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[vagrant sand]

for a question like P(Y=2|X=1) cant i just use bayes theorem?

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[vagrant sand]

ok

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[vagrant sand]

<@&286206848099549185>

[grizzled yacht]

Yes

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[arctic bloom]

can anyone help me see this application of MVT to this composition of functions, in the context of chain rule for curves. i cannot see how they got x(t+h) - x(t) on one side.

[arctic bloom]

im thinking of seeing it as a composition as g_0 (s) = f(s,y(t+h)) and g(s) = g_0(x(s)), and seeing that g is differentiable on (t,t+h). but when i apply the single variable MVT to g, i dont get the result that i want

[plucky elk]

just apply it for the x-variable

y(t+h) just goes along for the ride

[arctic bloom]

oh.. you know what i see it now ..

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[simple oyster]

.close

[simple oyster]

this is my question

i was wondering if this was right

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[old ibex]

need help moving forward on this problem

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@old ibex Has your question been resolved?

[old ibex]

<@&286206848099549185>

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@old ibex Has your question been resolved?

[light shoal]

well, 1 and c^2 are not functions of v

the only nontrivial step here really is going from the 3rd to the 4th line

[old ibex]

oh i might be stupid

[light shoal]

you have to justify why you can bring the limit inside the square root

it is because square root is a _______ function

btw c is definitely not zero, it's the speed of light, a constant

[old ibex]

oh so i dont need to add that

so what step do i need to go back to

[light shoal]

well first, you should explain why it's important that v is approaching c from below and not from above

[old ibex]

i have that seperately

[light shoal]

ok

[old ibex]

but essentially juust cause you cant go faster than light

[light shoal]

well that's the physics reason

the math reason is that if v were bigger than c, then what would happen when you take the square root?

[old ibex]

it would be a negative number

so you couldnt

[light shoal]

right

so next question, why can you bring the limit inside the square root?

[old ibex]

idk i think its a limit rule

let me pull it up rlly quick

i might have just been stupid i don't see it anywhere

[light shoal]

the general property is $\lim_{x \to a}f(x) = f(a)$

[flat frigate]

OurBelovedBungo

[light shoal]

that's not true in general, but if f is ____________ then it's guaranteed to be true (fill in the blank)

[old ibex]

if f(x) is = to a?

[light shoal]

a property of the function f

f has to be what kind of function in order to do this?

[old ibex]

umm i'm honestly not sure

[light shoal]

continuous

[old ibex]

oh okay

[light shoal]

you can move the limit inside if the function is continuous

ok so lastly, as i mentioned earlier, both 1 and c are constants

[old ibex]

yes

we need to limit as x approaches v i think

[light shoal]

so what does that tell you about the limits involving 1 and c^2 ?

[old ibex]

we can set them equal to a?

so c?

[light shoal]

or putting it another way, they don't vary at all with v

so taking the limit as v approaches some value has no effect on them

[old ibex]

okay

[light shoal]

1 is just 1

c^2 is just c^2

regardless of what v is doing

[old ibex]

okay

so we just disregard the limits on them?

[light shoal]

yep

so you're left with just the one limit involving v^2

[old ibex]

so the limit of L would be equal to root of (1-((limv^2)/c^2))

[light shoal]

yes

now what is the limit of v^2

as v approaches a?

[old ibex]

would it just be c?

[light shoal]

how would it be c

there is no c involved in "limit of v^2 as v approaches a"

[light shoal]

oh wait sorry

i forgot we're letting v approach c, not a

limit of v^2 as v approaches c

keeping in mind that squaring is also a continuous function

[old ibex]

ohhh so we can do (lim v)^2

[light shoal]

therefore $\lim_{v \to c} v^2 = (\lim_{v \to c} v)^2$

[flat frigate]

OurBelovedBungo

[light shoal]

yep

and that last limit is really easy

[old ibex]

wait that would just be c right?

[light shoal]

yes

[old ibex]

Awesoem

[light shoal]

inside the brackets is c

so overall it's c^2

[old ibex]

oh cause then its 1-(c^2/C^2)

[light shoal]

so overall your limit is equal to $\sqrt{1 - \frac{c^2}{c^2}}$

[flat frigate]

OurBelovedBungo

[light shoal]

which simplifies to what?

[old ibex]

and thats why the limit is there

root of zero

which isnt possible

[light shoal]

yep

sure it is

[old ibex]

that makes a lot of sense

[light shoal]

square root of zero is just zero

so your overall limit L is zero

[old ibex]

so we just say lim L as v approches C is zero?

[light shoal]

yes that's what you have shown

[old ibex]

awesome thank you so much

[light shoal]

so as an object approaches the speed of light its length approaches zero

[old ibex]

this makes a lot more sense than it did when my teacher talked about it

thank you so much for the help

[light shoal]

the math isn't too bad but physically it seems crazy 😁

[old ibex]

it's just a lot of steps for such a small equation

[light shoal]

but that's because we're never traveled near the speed of light so it's beyond anything that is familiar

[light shoal]

true but after you have some experience you'll see that some of the steps are obvious and skip right past them

but for now it's good to go through all of them