help-23

[obtuse plover]

not anymore

[lean otter]

lol

[obtuse plover]

unfortunately i only have this

thanks for the help tho, i appreciate it

.close

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[rare vale]

How did number 3 happen?

[rare vale]

<@&286206848099549185>

[upper rivet]

could you be a bit more specific? which part of 3 exactly

it's just using $\tan^2{x} + 1 = \sec^2{x}$

[flat frigate]

nebula40

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@rare vale Has your question been resolved?

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[devout steeple]

how would I take the derivative of 7(6 + x)^(3/2)

[thin bridge]

power chain rule

[devout steeple]

power rule and chain rule combined?

[thin bridge]

yes

[devout steeple]

would first step be

$\frac{3\sqrt{7(6 + x)}}{2}$

[flat frigate]

Price

[thin bridge]

no

only the (6+x) is being raised to the power

7 doesn't belong under the radical

[gentle herald]

Both the functions r different

Oh mb
I get it

[thin bridge]

for the chain rule component you should also be multiplying by the derivative of the inner function,
but the derivative of the (x+6) here is 1 so not really important if you don't include that here
(it would be depending on what you have)

[devout steeple]

so if it was $7(6 + x^2) ^{(3/2)}$ would we have

$\frac{21\sqrt{(6 + x)}}{2} * (2x)$?

[flat frigate]

Price

[thin bridge]

not quite,

you'd have 6+x^2 under the radical
feels like you did a copy/paste job but forgot to change that

[devout steeple]

oh yeah lol

[devout steeple]

thank you

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[solar needle]

i am confused

[solar needle]

how did we show this has fewer than 3 pivots

and why is that the critera for not being invertable

[safe radish]

@solar needle Has your question been resolved?

[safe radish]

@solar needle Has your question been resolved?

[split ether]

When you row reduce, you can subtract rows 1 and 2 from row 3, making every entry in row 3 equal to 0 since row1 + row2 = row3

So if there are any pivots, there will be in rows 1 and 2 but not in row 3

Hence the amount of pivots is less than 3

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[brave wolf]

How to prove ∃y∀xFxy ∴ ∀x∃yFxy using natural deduction? My problem is that I cant figure out how to deal with multiple quantifiers. I can try doing existential elimination, but I then cannot figure out what B should I conclude from ∀xFxc to get something useful

[brave wolf]

the existantial elimination rule is this, the problem is that B cannot include c

oh nvm I think I see it

.close

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[brave wolf]

After pondering for more than a hour, I always see the solution at the moment I post the problem here. Strange

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[idle raven]

I have a physics question

[idle raven]

What is the dielectric constant of a perfect conductor???

[plain bobcat]

its undef

[idle raven]

undefined? or infinite?

[plain bobcat]

whats the problem your doing

send

[idle raven]

IT's a problem I came up with myself

[plucky elk]

#old-network for physics server

[idle raven]

k sorry

.close()

.close

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[warm wren]

I have to calculate the angle ASD:

If I were to draw a coordinate system and label the sides of the bigger squares as one, I would be able to calculate the angle by assuming that the smaller square's sides are 2/3rd of the bigger ones.

This way I can calculate the slope of the two straights by using tan, then subtracting them from 180° as its a triangle.

But how do I get to the conclusion, that the smaller square's sides are 2/3rd smaller than the bigger ones? Would it be appropiate to draw the diagonal, draw it out of the square, stack the bigger squares until it reaches the corner of the bigger square and then use tan again to calculate the side, as I know that the bigger ones sides are 1 and the angle is 45? However wouldn't I need to have an scaled drawing which would be wrong as I dont have one?

Or do I need to use the given straights?

Another way would be like I depictured, by Saying the bigger squares sides are one and the smaller ones are x

[safe radish]

@warm wren Has your question been resolved?

[warm wren]

<@&286206848099549185>

[narrow bronze]

@warm wren Could you send me the words in questions also, as if length of AB is not fixed, angle ASD can be arbitrary.

(Tag me if you want me to reply so I won't miss your message)

[warm wren]

I have nine points, A B C D E F G H and I. They make out four squares ABHI, BCFG, CDEF. The points ABCD are in this order on a straight g and G, H are on a beam s, with starting point B. The straights AF and DI intersect in point B

@narrow bronze

[narrow bronze]

@warm wren AF and DI intersect in B? (I think that you mean S)
Anyways, the length of AB is still arbitrary...

[warm wren]

ah yeah

I meant s

my bad

since the smaller squares side arent given, I could simply define them as x or one, couldn't I? @narrow bronze

[narrow bronze]

I mean the ratio between AB and BC

[warm wren]

its not given

[narrow bronze]

AB can be infinitely large s.t. angle ASD is in 0 degree.
and as shown in the figure, angle ASD is more than 0 degree.
so you can see that angle ASD varies with length of AB (or AB:BC)

@warm wren

[warm wren]

In the exercise it is written, that you have to solve the exercise by using coordinate geometrie or a circle around ABHI and a thales circle around the rectange

What I came to notice is, that AH is the same length as AS and that CS is the same length as CB

@narrow bronze

And as they ask me to use circles, the positions seem fixed

Like in the picture above

[warm wren]

still there?

[warm wren]

Ah and the angle stays the same, as long as AB is smaller than BG

abd as long its a normal number

[narrow bronze]

ok, let me try

angle IHA = angle ISA

[warm wren]

How @narrow bronze

[narrow bronze]

angle in same segment

[warm wren]

what?

[narrow bronze]

CH = CF = CD -> HFD is a circle with center C -> angle DCF = 2 angle DHF

[narrow bronze]

angle in circumference x 2 = angle at center

problem solved

[warm wren]

@narrow bronze could you elaborate a bit further? I know that DCF is 90 degrees and when DHF is 2 times DCF, then ASD is 135. But how do you know that DCF is twice the angle of DHF? I mean like in the drawing

DCF wouldnt be an triangle in the thales circle

[narrow bronze]

https://thirdspacelearning.com/gcse-maths/geometry-and-measure/angle-at-the-centre-is-twice-the-angle-at-the-circumference/
check the 2nd diagram.

[warm wren]

ah thx

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[noble glacier]

Is this permutation

[noble glacier]

Number 2

[sinful grove]

Yes

[noble glacier]

Thanks

[noble glacier]

Is number 1 also permutation, number 3 fundamental and 4 is factorial?

[sinful grove]

Number 1 is combination (n choose k)

[noble glacier]

Oh

[sinful grove]

(The order doesn't matter, we are choosing a president and vice president from a group of 18 people)

[noble glacier]

It cant be combination since its say it can be only of the 3

[sinful grove]

What do you mean by "its say it can be only of the 3"?

Ohh you're talking about 3

nvm

[noble glacier]

Yes

[sinful grove]

Yea 3rd is addition principle

[noble glacier]

Ok

Ok

So it would 7x6x5x4x3x2x1 =

[sinful grove]

Ahh yes

[noble glacier]

Which would equal 7!

[sinful grove]

Yes mb

[noble glacier]

U gud

Thanks bro

I will close now

.close

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[manic heart]

I'm confused on what to do

[worthy hemlock]

Set them equal to each other, solve for x

Then plug that x value you got back into one of the equations to get the y value

[manic heart]

how would I fo that

do *

[worthy hemlock]

Can you set the equations equal to each other?

[manic heart]

waitt

[plain bobcat]

sub one of the equations for y

[manic heart]

so u add 3x - 2x?

and do 1 + 6

[worthy hemlock]

That's not an equation

[worthy hemlock]

.

That's the first step I'm asking

[manic heart]

wdym

I'm not understand on what ur tryna say

[plain bobcat]

ok imagine this y=5 y=x what is x equal to

[manic heart]

understanding *

[worthy hemlock]

It means what it means, can you set the equations equal to each other

[manic heart]

5

[worthy hemlock]

You are given two equations, both are equal to y

You can set the equations equal to each other

[plain bobcat]

ok good what did u do there you knew that y=5 then you subed y=x with 5 so then 5=x right?

same thing here

[manic heart]

but it doesn't say here y's value

so how would I figure out x if I don't know y

[plain bobcat]

it deos

does

y=3x+1 y=-2x+6

y is equal to those things

thats a value

[worthy hemlock]

You don't need to figure out the actual y value. The equations are already y =

You need to set both equations equal to each other

[manic heart]

so like x on one side and normal numbers on one side?

[plain bobcat]

no just equal to each other

let me give u another example

y=x+2 and y=2x+2

[worthy hemlock]

https://www.youtube.com/watch?v=Vk6hEP1KwGo

[plain bobcat]

then u make

2x+2=x+2 and solve it

since y is equal to both things

[worthy hemlock]

Simple video that does the same concept you are applying

[plain bobcat]

u can sub y and solve for 1 variable

[worthy hemlock]

https://www.youtube.com/watch?v=UlexMTWrkxI&ab_channel=CharlieBurrus

[manic heart]

ok I'll watch the video

[safe radish]

@manic heart Has your question been resolved?

[safe radish]

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[flat perch]

​For the polynomial function g, the rate of change of g is increasing for x<2 and decreasing for x>2. Which of the following must be true?

​The graph of g has a minimum at x equals 2 .

​​The graph of g has a maximum at x equals 2 .

​The graph of g has a point of inflection at x=2, is concave down for x<2 , and is concave up for x>2

​The graph of g has a point of inflection at x equals 2 , is concave up for x less than 2 , and is concave down for x greater than 2 .

[flat perch]

!status

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[flat perch]

4

1. The graph of g has a minimum at x equals 2: This statement cannot be determined from the given information. The fact that the rate of change of g is increasing for x < 2 and decreasing for x > 2 does not provide information about the presence of a minimum at x = 2.

2. The graph of g has a maximum at x equals 2: This statement cannot be determined from the given information. The rate of change of g being increasing for x < 2 and decreasing for x > 2 does not provide information about the presence of a maximum at x = 2.

3. The graph of g has a point of inflection at x equals 2, is concave down for x < 2, and is concave up for x > 2: This statement cannot be determined from the given information. The rate of change of g does not directly indicate information about the concavity or the presence of a point of inflection at x = 2.

4. The graph of g has a point of inflection at x equals 2, is concave up for x less than 2, and is concave down for x greater than 2: This statement must be true. Since the rate of change of g is increasing for x < 2 and decreasing for x > 2, it indicates a change in concavity at x = 2, which corresponds to a point of inflection. The concavity being up for x < 2 and down for x > 2 follows from the given information.

got this as reasons

Answer is number 4

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[junior wagon]

I want to show that finite sets always have a maximum. This is my idea:

[empty gyro]

ordered sets I assume?

(totally ordered)

[fair dagger]

I think he means subset of reals.

[junior wagon]

yes

[empty gyro]

As a mathematician, it is my job to nitpick

[fair dagger]

true

[junior wagon]

my idea is to use that sup(A) = max(A), which Ive already shown

but I need to show that max(A) exist

[fair dagger]

I just knew because yesterday we had this issue a bunch and it basically boiled down to this "fact".

[empty gyro]

Do you know the definition for finite? The actual mathematical definition?

[junior wagon]

no

no

[fair dagger]

I mean there can be different definitions.

You should have at least one. 🙂

[empty gyro]

A set A is finite if it is

1. equal to empty set, or
2. there exists a bijection f: A->{1, 2, 3, ..., n} for some natural n

[junior wagon]

riight

Sooo

[empty gyro]

There's two ways to approach this problem, based on your technical depth

First, does empty set have a max?

[junior wagon]

yes

i mean, max of a set means there is an element greater than all other elements in this set

so...

there is nothign to be greater than

idk

[empty gyro]

That is definition of upper bound

[junior wagon]

This is what I can find from the book

Oh.

Maybe greater than or equal to ?

[empty gyro]

hm?

[junior wagon]

So a0 >= a for all a in A

[empty gyro]

also, a0 in A

[junior wagon]

Right

[empty gyro]

so, again, does empty set have a maximum, and if so, what is it?

[junior wagon]

Oh it doesn’t

Since it is empty

Sorry I think I need a break lol. Been studying all day. I don’t think I will be very productive

.close

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[eternal flint]

<@&286206848099549185>

[eternal flint]

i need help

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[worthy hemlock]

Do not open a channel pinging helpers

Post your math question you are stuck on, and wait patiently for help to come

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[stray birch]

can u help me

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[untold basin]

any elp

[untold basin]

ik this is squeeze theorem

but my trig and unit circle stuff is rusty

[fringe viper]

Are you allowed to use a calculator for this?

[untold basin]

yea

graphic calc

the only rule my prof has for any lecture/hw/exam is no l hopitals rule lol

[fringe viper]

all you really need to do here is to evaluate sin and -cos at x=3pi/4

[untold basin]

so plug in 3pi/4 for x for sin and -cos and thats it?

[fringe viper]

yes

you will see that they will have the same value

and then you can use the squueze theorem

[untold basin]

ohh okay

thank you

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[junior wagon]

Im back

[junior wagon]

The empty set has no maximum, because it is empty.

Anyway(since Im not sure where this is going), my approach is to use induction.

The base case is the set with n=1 element. S = {x1}. The max of this set is just x1.

Assume that a set with n=k elements satisfies having a maximum. We need to show a set with n = k+1 elements has a maximum.

If we consider the set S' = S - {x_k+1}, where S = {x1, x2, ..., x_k}. Then by induction hypothesis, since S' has n = k elements, S' has a maximum.

So now we consider this extra element x_k+1.

There are two cases:

If x_k+1 > the maximum of S', then we update the maximum of S' to be x_k+1.

If x_k+1 <= maximum of S', then we keep the maximum of S' as the maximum.

Either ways, we produce a new max of the set.

That is my idea.

If this idea works, then I should just add the empty set as another base case.

[fringe viper]

What if A only has one element?

[junior wagon]

That is covered in the base case

[fringe viper]

a finite, bounded set contains its supremum no?

like you said earlier that max(A) = sup(A)

[junior wagon]

I am trying to prove that a finite set always contains maximum

[fringe viper]

my bad, thought you were trying to prove Question 4

[junior wagon]

Yeah, I did already show max = sup

[junior wagon]

But I need to show this

In order to say a max exist

[fringe viper]

theres a problem with this though

Q4 says to show that there exists an x different from s=sup(A)

if A only has only element

you cant have two different elements in it

or ?

[junior wagon]

Uh ok

So

It is just false

[fringe viper]

i think it holds when A has more than one element

but fails in this one case

[junior wagon]

Yeah

Ok

[fringe viper]

also fails

for empty set

[junior wagon]

I will ask for clarification

I think the prof means to say A has more than one element

[fringe viper]

ok

otherwise the induction proof seemed solid to me

[junior wagon]

Yeah, but I think my main proof broke down

Because I didn’t see that x is different from s

[fringe viper]

for sets with more than one element

do you still want to prove it

[junior wagon]

Yeah

[fringe viper]

how were you going to do it?

[junior wagon]

I’m not sure. I need to come up with something new

[fringe viper]

do you want a hint?

[junior wagon]

Yes

[fringe viper]

if s is the max of S, consider S\ {s}

[junior wagon]

Oh so we can consider the max of this new set

[fringe viper]

yessir

[junior wagon]

That will be the x

Ok thanks.. I’ll write up later

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[idle tiger]

can someone explain to me how i got this wrong

it asks "if one student is chosen at random out of ALL SENIOR students from school 1-4, what is chance they're enrolled in calculus II"

162/388...

[cold aurora]

ALL SENIOR STUDENTS FROM SCHOOL 1-4

That's your sample space.

[idle tiger]

okay yes i know that. and out of that sample space, it asks for the probability that a student is enrolled in calc ii..

the sample space total is 388 seniors and the seniors in calc ii is 162

am i misunderstanding something

[cold aurora]

How is it 388?

[idle tiger]

?

[cold aurora]

You only have to consider students from school 1 and 4.

[idle tiger]

because it says the total of each seniors

[cold aurora]

Not all schools.

[idle tiger]

wait why

[cold aurora]

FROM SCHOOL 1 AND 4

[idle tiger]

OH..

I JUST SAW THAT..

oh my god

thank you so much

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[steep sail]

I just want good resources for studying image compression from DCT to the end (JPEG compression)
I have watched this video but I need more detailed explanation as I'm making an advanced video about the topic

https://www.youtube.com/watch?v=0me3guauqOU

[safe radish]

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[steep sail]

<@&286206848099549185>

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[proud haven]

huh

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[lament kite]

i genuinely have no idea how to do this

[lean thorn]

can you show your work for how you got that answer?

[lean thorn]

,rccw

[lament kite]

i wasn’t in class for these notes so i’m struggling 💀

[lean thorn]

so implicit differentiation is pretty straight forward. You're taking a derivative with respect to x in your problem, so then if you see any x values you take the derivative normally, and if you see a y you'll have to use the chain rule (find the derivative of y as if you were taking the derivative with respect to y then multiply it by dy/dx

looks like you were on the riht track but I think you overdid some stuff

oh wait derp you need to use the product rule too lol

[lament kite]

yeah that’s what i did idk what im doing wrong

[lean thorn]

$9x^2y + 6y^2x = -6$
$\\18xy + 9x^2y' + 12yxy' + 6y^2 = 0$
$\\y'(9x^2 + 12xy) + 6y^2 + 18xy = 0$
$\\y'(9x^2 + 12xy) = -6y^2 + 18xy$
$\\y' = -\frac{6y^2 + 18xy}{9x^2 + 12xy}$
$\\y' = -\frac{6y(y + 3x)}{3x(3x + 4y)}$
$\\y' = -\frac{2y(y + 6x)}{x(3x+4y)}$

[flat frigate]

MellowDramaLlama

[lean thorn]

This was a mistake right here. It should've been 12xyy' + 6y^2

[safe radish]

@lament kite Has your question been resolved?

[lament kite]

wait how come in the 4th step，when you move 18xy to the other side it’s positive?

ok i think that was just an accident

idk what i did. wrong

[lean thorn]

yeah that was a mistake on my part lol

my b

oh this might be a small syntax error? You see how the negative is outside of the fraction? That distributes. Try answering the same but with e^y - 19

[lament kite]

e^y - 19 didn’t work

i’m a little confused though because shouldn’t it be -e^y + 19 bc you moved it to the other side?

[lean thorn]

yeah but the negative distributes

$y' = \frac{-e^y + 19}{xe^y + 5} = -\frac{e^y - 19}{xe^y + 5}$

[flat frigate]

MellowDramaLlama

[lean thorn]

that should be the answer. Can I see how you're inputting it?

[lament kite]

it’s ok i got it right

[safe radish]

@lament kite Has your question been resolved?

[lean otter]

uh i need help

@safe radish i need help'

[lament kite]

can i ask another question 😭

i’m kinda lost , i knew i didn’t get the answer but i just tried submitting it anyway

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[lament kite]

.reopen

[safe radish]

✅

[lament kite]

<@&286206848099549185>

[lilac patio]

simplify it

assuming you differentiated right

y^5=x^6?

[lament kite]

well i didnt

it takes equivalent answers i’m saying i knew i didnt do it correctly

[untold sky]

dy/dx looks ok

but not only did you change (5y^5)^2 to (5x^5)^2 at the bottom

when did the question give you d^2y/dx^2=0?

[lament kite]

?

i thought it was quotient rule

[untold sky]

it is the quotient rule

but why did you equate the second derivative to 0

the second derivative of some function isn’t always 0

[lament kite]

what’s it equal to then

[untold sky]

differentiate wrt x both sides of dy/dx=6x^5/5y^4

where do you get a 0 on one side

because you somehow managed to get a 0 on one side then solved for dy/dx instead of d^2y/dx^2

[lament kite]

well i didn’t get a 0 i just assumed it was 0 i don’t rlly know how to solve for d^2ydx2

[untold sky]

well the right side is the blob of quotient rule

what’s the left side going to become

what’s d/dx (dy/dx)

[lament kite]

oh

d^2y / dx^2

[untold sky]

ye so that’s the thing you’re looking for

it’s not 0

and the rest is just wrestling through the blob on the right

[lament kite]

so right is just regular quotient rule? do i not multiply y by dy / dx

[untold sky]

nah, still have to multiply by dy/dx where needed

[lament kite]

well then what do i do w the dy/dx

[untold sky]

you calculated it didn’t you

just sub it in

[lament kite]

oh

i’m lost again

[untold sky]

it’s all just algebra now

keep going

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[maiden sphinx]

lim_(x->2) x^2/(x - 2) how do I calculate the limit formally

[maiden sphinx]

I dont get an answer with L hospital and I wonder I cant use it

[restive plover]

because that requires the limit to exist in the first place

[quasi bison]

no

well ok it does actually

but also your original limit MUST be of the form 0/0 or infty/infty !!!

[restive plover]

notice that the denominator approaches 0

but the numerator does not

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[maiden sphinx]

Ah yes of course thanks

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[formal shell]

How does -3b + 2a / 3 become + in second equation? I dont understand.

[ionic blaze]

-*-=+
-(x-y)=y-x

[upbeat ridge]

They just only switched places

[formal shell]

Ah i see. Thank toy

.close

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[formal shell]

.reopen

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[formal shell]

@ionic blaze but how did -3b + 2a turn to 3b - 2a?

[safe radish]

@formal shell Has your question been resolved?

[grizzled pasture]

By multiplying with (-1)

(-1) (-3b+2a)
= (-1)(-3b) + (-1) (2a)
= 3b + 2a

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[tidal lance]

Need help with these.

[tidal lance]

<@&286206848099549185>

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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.

[tidal lance]

👍

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[tidal lance]

.solved

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[molten hawk]

I'm struggling to understand the vector equation

[molten hawk]

did they miss an equal sign?

[lean otter]

Me thinks they are two equations

x-1 = 2y + 2

z = 4

So your z coordinate is always 4, x and y change relatively

[molten hawk]

hmm ok thanks

[lean otter]

Welcome

[molten hawk]

.close

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[molten hawk]

.reopen

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[molten hawk]

I'm thinking the vector equation would be <2y+3, y , 4 >?

[lean otter]

Hmm??

[molten hawk]

or would it be <2y+3,y,0> since theres no change in z

[lean otter]

It would be (2y+3 , y , 4)

If its the coordinates youre asking?

[molten hawk]

not really sure, I'm trying to find distance between two lines

oh wait

it would just be <1,2,0>

thank you

[lean otter]

Okay welcome

[molten hawk]

.close

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[latent ore]

Hey could I have a hand with green section letters I,j,k,l - a link to videos or an explanation on this part of the topic would help me out a lot

[latent ore]

I says (root 2)^3 if you can’t see it

[lean otter]

I dont have a video but i can help with the questions?

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@latent ore Has your question been resolved?

[lean otter]

@latent ore Ping if reply

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[lean otter]

Anyone know how to integrate this

[lean otter]

Or what strategy u use

[calm bridge]

try to write the numerator as A(derivative of denominator) + B

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[fair dagger]

!status

[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[fair dagger]

What do you mean?

[safe radish]

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[lean otter]

Having trouble with integral of 5/(x^2+5)

[lean otter]

Keep getting 5root5 arc tan (x+1/ root5)

[queen ermine]

$\int{5/(x^2+5)}$

[flat frigate]

LoloitsJo

[lean otter]

Yea that

Yes that also

[unique bison]

dx

[queen ermine]

How do you write with a numberator and denominator?

[unique bison]

what

[queen ermine]

$\int \frac{5}(x^2+5)}dx /$

[lean otter]

Uh

Can I get some help

[flat frigate]

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

[sinful grove]

$\int\frac{5}{x^2+5}dx$

[flat frigate]

Normed

[queen ermine]

Do you know what the antiderivative of this is

[lean otter]

No, I’m confused about the numerator

[queen ermine]

$\int\frac{1}{a^2+x^2}dx$

[flat frigate]

LoloitsJo

[queen ermine]

this is something you have to memorize

this is equal to

$(\frac{1}{a})arctan(\frac{x}{a})$

[lean otter]

Yes but I’m getting the wrong answer also that doesn’t account for when the numerator is not 1

[queen ermine]

wait

[lean otter]

It’s wrong

[flat frigate]

LoloitsJo

[lean otter]

Don’t know how I’m messing up

[queen ermine]

no no no

you are over complicating it

it's simpler than this

Write your original integral

Then Factor 5 out of the integral and show me your work okay?

[queen ermine]

this is wrong and too complicated

[lean otter]

How do I do that

I thought I did factor out the 5

[queen ermine]

Just write the orignal integral and then take five out of the integral

DO that then send a picture

[lean otter]

The trig substitution WORKS, it’s literally how u do the proof

[queen ermine]

thereee

[lean otter]

I don’t wanna just be a monkey who memorizes the shite without knowing how to derive it somewhat

[queen ermine]

now you can rewrite the denominator as X^2 + (sqrt(5))^2 write

[queen ermine]

I did it with arcsin but it's the same process with arctan

[lean otter]

Ok then what, trig sub?

Sorry I’m really ducking stupid

[queen ermine]

i haven't learned that yet

[queen ermine]

this will become

[lean otter]

Why are u helping if u don’t know the answer

[queen ermine]

I do but you are doing it a different method

I am using the antiderivative of arctan

[lean otter]

Ok well what next anyway

[lean otter]

?? I don’t get it

What did u get for this

[queen ermine]

okay now you have the right form

$\int\frac{1}{a^2+x^2}dx$

[flat frigate]

LoloitsJo

[queen ermine]

equals

$(\frac{1}{a})arctan(\frac{x}{a})$

[flat frigate]

LoloitsJo

[queen ermine]

try that

[lean otter]

It’s wrong

[lean otter]

What did u get, I can already see the right answer

I just don’t get how to get there

[queen ermine]

that should be it

[lean otter]

Well 3 sources, my professor, integral calculator and my textbook disagree with you

[queen ermine]

idk

[lean otter]

This is correct answer

[lean otter]

U came here saying I’m over complicating it and u didn’t even get right answer

[queen ermine]

actually

i got that answer

[queen ermine]

this is the same thing

once you rationalize the square root of 5

There

[lean otter]

Oh my got I’m gonna go die now

Ok thx

[queen ermine]

LOL

Np

[lean otter]

.close

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[crude star]

I dont understand this

if x = {1, 2, 3}

then 2^x = {2, 4, 8} right

[final halo]

you are not doing 2^x for each element in X

it is notation for all subsets of the set X

[crude star]

I dont understand

[crude star]

but the example

it says let x = {1, 2, 3}

[final halo]

and then they listed all 8 subsets

[crude star]

wait a min

are they j referrin to

power set

of the set

[final halo]

this is exactly the power set yeah

[crude star]

oh brah

I get it

ty

.close

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[gilded shell]

I can’t figure out the answer

[lean otter]

Isnt the function increasing on x=-1 and x=0?

[stoic dune]

f'(-1) is definitely not a negative number

[lean otter]

So how you got a negative derivative

[stoic dune]

Recall that f'(x) represents the slope of the tangent at x

Do those words make sense?

[gilded shell]

no

[stoic dune]

https://youtu.be/9vKqVkMQHKk?si=mZyMrV8Qd_yf4wuh

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[quartz sleet]

so in question 1/a

i calculed P(1) its ez

i got P(1)=2-3+1=0

so x=1 its racine of this polynome becaujse P(1)=0 so x-1 its a factor of this polynom

now i need to do the 2x^3-3x^2+1/x-1

so i have problem on this one

i dont know how to calcule it right

[safe radish]

@quartz sleet Has your question been resolved?

[gentle oak]

On va parler fr, tu connais delta ?

Ou le discriminant ?

[quartz sleet]

Yes

[quartz sleet]

Oui

[gentle oak]

Eh bien vu que 1 est une racine évidente du polynôme la deuxième racine est 1/a

[quartz sleet]

Oui x=1

Est une racine

[gentle oak]

Nan je dis de la merde

[quartz sleet]

💀

[gentle oak]

Dsl

[quartz sleet]

Oui

Non ecoute

[gentle oak]

,w expand (x-1)^3

[quartz sleet]

So maintenant j ai besoin de fair cette division esst je ne sais pas comment et est ce que vrai ou faux

[gentle oak]

If you don’t speak French very well we can switch to English

[quartz sleet]

Yea its better

[gentle oak]

Ok

Sorry but I don’t think I can help you

I’m very sorry

[quartz sleet]

Ok

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@quartz sleet Has your question been resolved?

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[carmine saffron]

hello

[stoic dune]

What info did you start with?

[carmine saffron]

the one on top of the page

[stoic dune]

Just A, B, Cinverse?

[carmine saffron]

blue and red is mine

yes

C inverse times 1/2

[stoic dune]

Right. So you need a matrix X that, when multiplied with A, gives that matrix

What size does X need to be?

[carmine saffron]

i think 2x2 but it doesnt specify

[stoic dune]

You're correct that, because A doesn't have an inverse, you can't use the method in red.

2×2 wouldn't work. Remember, X has to multiply A

And give a 2×2 output

[carmine saffron]

ok

yes exactly

because B-C matrix is 2x2

but I don’t know a method to find X without using A inverse

[stoic dune]

Pretend X =
(a b c)
(d e f)

Carry out the multiplication

[carmine saffron]

ok

the resulting is 3x3

<@&286206848099549185>

[safe radish]

@carmine saffron Has your question been resolved?

[safe radish]

@carmine saffron Has your question been resolved?

[safe radish]

@carmine saffron Has your question been resolved?

[flat frigate]

User cancelled member selection.

Sep 13, 2023, 2:49 AM

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@carmine saffron Has your question been resolved?

[carmine saffron]

closed

.close

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[velvet apex]

4. Determine whether each relation is a function. Would appreciate any help just got stuck on this

[formal sphinx]

a consistent way to do this would be with the vertical line test

but this is a linear function and it doesn’t have anything “weird” that would make it not a function

[velvet apex]

The linear and nonlinear lecture is for tomorrow and for some reason this assignment is too, i would appreciate if you could explain to me now

[restive plover]

you can use the vertical line test here

do you know that?

[velvet apex]

Yes i know the vertical line test

How do I graph it?

[restive plover]

The normal way you graph a function

Do you know how to graph a function?

[velvet apex]

no not yet, lesson ended early today, all this should’ve been today, and I have to submit this by midnight to the portal soo

[formal sphinx]

uh do you know the Cartesian plane?

The coordinate system?

[velvet apex]

Yes

[formal sphinx]

alright, so plug in an x into the function and you’ll get a y

For example plugging into 2 gives you -5, so you plot the point (2,-5)

[safe radish]

@velvet apex Has your question been resolved?

[safe radish]

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[crystal zodiac]

i need to simplify this expression: $(tan^2a)/1+tan^2a$

[flat frigate]

Shade

[crystal zodiac]

can anybody help?

[gritty glacier]

$2\tan^2 a$

[flat frigate]

Bettim

[gritty glacier]

i dont think your question is right?

[crystal zodiac]

why not

[karmic hedge]

What is 1+tan^2a

[gritty glacier]

okay i get it you messed up with paranthesis

[crystal zodiac]

the denominator

wait a sec

[karmic hedge]

Simplify it

[gritty glacier]

$\frac{\tan^2 a}{\tan^2 a + 1}$?

[flat frigate]

Bettim

[karmic hedge]

obviously

[crystal zodiac]

exactly

thanks

[gritty glacier]

right

[hard wedge]

hi guys need help (rearrange equation) dm me 🙂

[gritty glacier]

#❓how-to-get-help

[crystal zodiac]

bruh it's what i just asked

[karmic hedge]

Use your identities

[gritty glacier]

what is the relation between sec^2 theta and tan^2 theta

[crystal zodiac]

i only have $sin^2+cos^2 = 1$ and $\frac{\sin}{\cos}=tan$

[karmic hedge]

Divide both sides by cos^2

[flat frigate]

Shade

[crystal zodiac]

whats theta

and sec

[gritty glacier]

a variable

you never learnt abt theta in trig?

[gritty glacier]

secant

what are the trig ratios you have learnt about?

[crystal zodiac]

i'm in italy we dont call it theta

[crystal zodiac]

i only have these to use $sin^2+cos^2 = 1$ and $\frac{\sin}{\cos}=tan$

[flat frigate]

Shade

[gritty glacier]

what is $\frac{1}{\cos a}$

[flat frigate]

Bettim

[crystal zodiac]

arcos

[gritty glacier]

what

[crystal zodiac]

cosecant sorry

[gritty glacier]

isnt arccos inverse trig?

no its secant

[crystal zodiac]

yeah

[crystal zodiac]

oh ok right

[gritty glacier]

so now

learn this trig identity

$\tan^2 a + 1 = \sec^2 a$

[flat frigate]

Bettim

[crystal zodiac]

dude

[gritty glacier]

do you want me to teach you how we get this?

[crystal zodiac]

the exercise says i can only use the 2 fundamental identities

[gritty glacier]

okay ten

[flat frigate]

Dyssrupt

[gritty glacier]

ten

then

[crystal zodiac]

bruhhhhhhhhhhh

u cant do that

[calm bridge]

why?

[gritty glacier]

$\frac{\tan^2 a}{\tan^2 a +1}$

[calm bridge]

you said you had the first one

[flat frigate]

Bettim

[gritty glacier]

we have this right

[crystal zodiac]

yes

[gritty glacier]

can i write this as $\frac{\frac{sin^2 a}{cos^2 a}}{\frac{sin^2 a}{cos^2 a} +1}$

?

[flat frigate]

Bettim

[crystal zodiac]

yes and thats what i did

[gritty glacier]

now take LCM in the denominator

[crystal zodiac]

LCM?

[gritty glacier]

wait

[crystal zodiac]

do you mean to factor it?

and divide it by the fraction?

[gritty glacier]

can i write this as $\frac{\frac{sin^2 a}{cos^2 a}}{\frac{sin^2 a + \cos^2 a}{cos^2 a}}$

[flat frigate]

Bettim

[gritty glacier]

?

this?

what you call this step?

[crystal zodiac]

denominatore comune

[gritty glacier]

okay

that

[calm bridge]

huh?

mcm?

[gritty glacier]

hes italian so different education system

[calm bridge]

oh ok

fair

[gritty glacier]

most common multiple ig?

[crystal zodiac]

denominatore comune sorry

[gritty glacier]

nevermind, then name doesnt matter

[crystal zodiac]

ok so

[gritty glacier]

now from that step

we can write that as

[crystal zodiac]

we can simplify cos

[gritty glacier]

can i write this as $\frac{sin^2 a}{cos^2 a} \times \frac{\cos^2 a}{\sin^2 a + \cos^2 a}$

[crystal zodiac]

no

you have to invert denominator and numerator

[flat frigate]

Bettim

[crystal zodiac]

ok now it's correct

[gritty glacier]

right?

[crystal zodiac]

yes now i just simplify

[gritty glacier]

now what is sin^2 a + cos^2 a

and you can cancel the cos^2 a

[crystal zodiac]

1

from the fundamentale trig statement

[gritty glacier]

so you get sin^2 a

[crystal zodiac]

the denom becomes 1

[gritty glacier]

thats your simplest answer

[crystal zodiac]

perfect, thank you very much

it was part of a bigger expression and at the end sin^2 simplified

.close

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[lean otter]

If I have two pants, one red and one blue, and two shirts, one grey and one black, can we use the combination formula to find pairs? If I use the combination formula, the answer is six, but according to the Cartesian product of sets, the answer should be the number of set A multiplied by the number of set B, which is 2*2 = 4

[hard crest]

how do you get six with the combination formula?

clearly it should be 4 right

[lean otter]

Its 6

n!/(r!(n-r)!)
2 shirt + 2 pants so n = 4
4!/2!(4-2)!
4!/2!2!
(4*3*2!)/(2!2!)
(43)/2
23 = 5

[hard crest]

yeah so what extra things are you counting here?

[lean otter]

it maybe combination of 2 shirts and 2 pants

[hard crest]

you've calculated the number of ways to pick 2 items from a set of 4

yeah unless you want to wear pants on your head that's not the appropriate thing to calculate

[lean otter]

dude
see what i was wasking

I just asked can we use the combination formula or not
If yes why is the answer wrong

[hard crest]

the way you're using it you'll count the pair of two shirts, and the pair of two pants

[lean otter]

.close

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[lean otter]

thanks

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[silver raptor]

Hello, just a question on proving this trig identity. When I go from the last part on the left to the first part on the right how did I get the negative sign from the denominator to the numerator?

[silver raptor]

Did the guy in the video just multiply both parts by -1?

[flat frigate]

Dyssrupt

[silver raptor]

I understand that, that seems to be from the 2nd from bottom on the left to the bottom on the left for the denominator

Ok well I see what its saying now, but I am just wondering did we get that by multiplying the num and den by -1

[calm bridge]

yeah

[silver raptor]

Thank you @calm bridge

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[stray axle]

so i've completed the square for the first line to get (x+2)^2+(y-1)^2+(z+2)^2=4 so i have a center of (-2,1,-2) and radius of 2 while for the 2nd line has a radius of 2 also and a center at the origin but im not sure whats the next step to find volume? how do i find the overlap these two have and then presumably create an integral for it

[desert juniper]

the obvious way would be the integral of the volume. You would first need the intersection of the spheres.
Is this enough information to continue?

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@stray axle Has your question been resolved?

[stray axle]

.close

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[astral lodge]

Hello, can someone help me with square root laws

[astral lodge]

I have -3sqrt(3) + 4sqrt(6) : 2sqrt(2)

I don't really remember the laws

[frigid spruce]

um is that divide part with the whole or just with the 4sqrt 6

[astral lodge]

I know how addition works

The dividing part is the problem

[frigid spruce]

$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt\left(\frac{a}{b}\right)$

[flat frigate]

yajatk07
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[frigid spruce]

you see this?

[astral lodge]

um

yes

[frigid spruce]

4sqrt(6) : 2sqrt(2)

now you can this with

this part

[astral lodge]

umm, isn't just only the roots

what about the 4 and the 2

[frigid spruce]

they will get divided normally

just like numbers get divided

4/2

=2

and then do the sqrt part seperately

if youre having problem

[astral lodge]

so that would be

-3sqrt(3)

4 : 2 = 2

and the divison is sqrt(3)

but I'm lost on the signs

[frigid spruce]

it becomes -3sqrt(3)+2sqrt(3)

[astral lodge]

oooh, and that is -sqrt(3)

Thank you very much!

[frigid spruce]

np

[astral lodge]

Have a nice day!

.close

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[frigid spruce]

you too

!

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[safe radish]

@lean otter Has your question been resolved?

[bronze river]

i guess i can help in it just give me a min

[calm bridge]

you see its the big root outside

so we will differentiate it

then apply chain rule for inside

first its the product x*sqrt(....)

so apply product rule for that

then inside that root is another function, so chain rule

then its product of x^4*sqrt(..)

so product rule

then for inside the root chain rule

[bronze river]

You have this function: f(x) = ∛(x√(x^4 + 2)) / (4√(8 / (x + 2)))
Imagine you have a number x. First, you take the fourth power of x and add 2. Then, you take the square root of that result. After that, you multiply it by x. This gives you the top part of your fraction.
On the bottom, you have 4 times the square root of 8 divided by the quantity (x + 2).
So, f(x) is essentially the top part divided by the bottom part.
It's a bit complicated to simplify further without specific values of x, but this is what the function does for any value of x you plug in.

[mystic lion]

gpt 🫡

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@lean otter Has your question been resolved?

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[lean otter]

How do i fully understand / remember this

[lean otter]

any tips?

[thin swan]

you are trying to memorise this for a test?

[lean otter]

sure

[thin swan]

are you allowed to have calculators?

[lean otter]

No idea

But i wont use them anyways

[thin swan]

i mean you could always like

write it down in the calculator

[obsidian oracle]

Like you could find a mnemotechnic sentence

Don’t Answer If Rudolph Comes

The stupider, the less likely to forget

[fair dagger]

Just make sure you understand why these conditions are true.

This looks like it is post school, so you probably saw all the proofs.

Also just work through some examples for yourself to make sure you spot the right approach quick enough.

Generally there is not one test that will do everything. Like when integerating you just need to be a bit creative. Sometimes very creative.

I recommend strongly against simply memorizing exact sentances or formulations. I think this will not be helpful when it comes to perfoming a practical computation.

[safe radish]

@lean otter Has your question been resolved?

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[crude star]

I dont rlly understand this counting rule

[fair dagger]

I have never heard of the nm rule let alone an extended nm rule.

I can try explain the final equation though

[crude star]

sure

isnt this all related to the final equation anyways?

[crude star]

nm rule is just n * m

[fair dagger]

it should be.

Do you understand what you are trying to count?

[crude star]

not rlly

I rlly dont understand anything from this whole paragraph

[fair dagger]

You have a set A with cardinality n and you take m elements in some order.

Example

A = {a,b,c}

so n = 3

For k = 2 we can construct

(a,b), (a,c), (b, a), (b,c), (c,a), (c, b)

you take any k elemenet from a and order them.

There are 6 options.

[crude star]

thats permutation right

[fair dagger]

This is what we want to figure out.

[crude star]

wait thats combination nvm

because order doesnt matter for combinations

[fair dagger]

I think the text calls it permutation.

[crude star]

but for permutation I thought

[fair dagger]

I am from europe, we didn't really use any of these names for that.

[crude star]

look for combination the order doesnt matter

[fair dagger]

oh

order definitely does matter here

[fair dagger]

^

[crude star]

so wont the permutation be (a, b), (a, c), (b, c) das all

[fair dagger]

(a,b) and (b, a) not same

[crude star]

so what would b an example of when the order doesnt matter

oh so

combination would be

[fair dagger]

I think just (a,b), (a,c), (b,c) if you do nto care about order.

[crude star]

(a, b), (a, c), (b , c)

right got it my bad

[fair dagger]

Does this make sense?