help-38

[wet ocean]

that could work, but there's a way easier route

pay attention to the big triangle

it's equilateral

[wraith hinge]

Yep

[wet ocean]

since it's equilateral, all sides are equal

and all angles are equal as well

[wraith hinge]

Wait so they’re all 90??

[wet ocean]

they couldn't be 90, they sould add up to 180, remember?

[wraith hinge]

Oh sorry

[wet ocean]

no problem

3 times an angle is equal to 180

3x = 180

[wraith hinge]

X=60

[wet ocean]

right! thats it

with that, we already know angle A, B and C

they're all 60 ;)

[wraith hinge]

Yep so now we find the sin cos and tan

[wraith hinge]

For number 1 and 2 they’re both 60

[wet ocean]

we need the cos of BAD

[wet ocean]

did you do cos(60)?

[wraith hinge]

Wait so 0.86?

Or 0.87?

Wait no

[wet ocean]

yes, thats cos(30)

[wraith hinge]

0.5 right?

Sin is 0.87 according to the calculator

[wet ocean]

let's pay attention to something real quick

line CB is cut in half by point D, right?

[wraith hinge]

Yep

[wet ocean]

so, since the side is cut in half, we can confirm that the angle above it is also cut in half

since there's a straight line running from D to A

*keep in mind that this only works with equilateral triangles

[wraith hinge]

So do I divid C to B by 2?

This is confusing

[wet ocean]

well, wee need to find the angle BAD

and that's half of angle BAC

do you understand?

[wraith hinge]

Yep

[wet ocean]

whats angle BAC?

[wraith hinge]

180?

And the D makes it 90

[wet ocean]

we already figured it out that, it's angle A

[wet ocean]

180 would be a straight line

[wraith hinge]

But each angles are 60

[wet ocean]

right

A = 60

BAC = 60

[wraith hinge]

Ah yes

[wet ocean]

ok, so if BAC is 60, and BAD is half of BAC

what's angle BAD?

[wraith hinge]

BA is 60 and D is 90 so 150?

[wet ocean]

here, see the angles

point B, A and C form the angle of 60, right?

[wraith hinge]

Yeah

[wet ocean]

the points B, A and D cut the angle of 60 in half

[wraith hinge]

So 30

[wet ocean]

so what would be angle BAD?

correct!

BAD = 30

ok?

[wraith hinge]

So cos(30)

[wet ocean]

yes

[wraith hinge]

0.86

[wet ocean]

yes

ok, let's go for 2

[wraith hinge]

So sin(30)

[wet ocean]

yes!

[wraith hinge]

0.5

[wet ocean]

nice

number 3

sin(ABC) - cos(CAD)

[wraith hinge]

Sin(60)-cos(30)

[wet ocean]

yes!

[wraith hinge]

1.11*10^-16

[wet ocean]

yeah, it might be something like that

my calculator resulted in error :/

1e-34

[wraith hinge]

Aight thank you so much!

[wet ocean]

no problem!

[wraith hinge]

Sorry if I took so much of your time

[wet ocean]

i'm not occupied with anything, don't worry about that ;)

[wraith hinge]

Bye!!

[wet ocean]

bye :)

[wraith hinge]

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[velvet fox]

I know this limit is usually -inf. Is this a special case?

[wraith hinge]

what is the notation $\log_{0,3}$ supposed to be saying

[velvet fox]

Log of base 0,3

[wraith hinge]

ah wait is that decimal?

ok

[wraith hinge]

for bases between 0 < a < 1 the logarithm effectively gets flipped across the x axis

so the limit diverges to infinity indeed

[velvet fox]

thank you

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[keen void]

Let OACB be the parallelogram to the right. The sides OB and BC are bisected
three equally long stretches where the dividing point P ∈ OB is closest to O, while
the dividing point Q ∈ BC is closest to C. Show that the points of intersection P′,
M and Q′ of the diagonal OC with the lengths AP, AB, and AQ respectively divide
this diagonal into four equally long parts.

Tip: Choose a = OA and b = OB as base. Due to the symmetry, it suffices to show that p′ = OP′=1/4OC=(a + b)/4

Express OP′ in two ways and take advantage of the fact that a and b are linearly independent.

[maiden hare]

Well, have you tried expressing OP′ in two ways?

[keen void]

Yes

In the hint it says OP' can be written as OC/4 and (a+b)/4

Based on this

[maiden hare]

Yes that's what you need to show

[keen void]

I honestly have no idea how i am supposed to prove this

I can clearly see that it is based on the picture :(

[maiden hare]

Well we know P' is on the diagonal

How can you express that in terms of vectors?

[keen void]

I'm not sure what you mean? P' can be expressed as OC/4

since i can see that the vector a + b = OC

however P' is 1/4 the way on OC

[maiden hare]

You can't use the result to prove itself

"P' can be expressed as OC/4" is the result you need to show

You can't use that

[keen void]

Hmm alright

[maiden hare]

How do you express a point on a line?

[keen void]

in x and y?

If it is R^2 that is

[maiden hare]

Ideally using vectors

[keen void]

P' = OP + PP'

[maiden hare]

Ok but that's not really what I asked

Say I have a line that goes through the origin and a point (3,5)

How would you express another point on that line?

[keen void]

Uhm, i think i'd either multiply or divide the 3 and 5 by lambda, which iirc is the leaning coefficient

[maiden hare]

Right, can you write that using vectors?

[keen void]

3x + 5y

[maiden hare]

That's not a point

[keen void]

Then i don't know :(

[maiden hare]

You just said "multiply or divide the 3 and 5 by lambda", so why did you write x and y and no lambda?

[keen void]

I'm not sure, i was guessing

I just know that if i want another point on that vector, it needs to have the same incline/recline

so i take each x and y and multiply them by lambda

[maiden hare]

Ok, but that's not what you did with "3x + 5y"

a point (3,5)

multiply or divide the 3 and 5 by lambda

Come on, what does that give you?

[keen void]

lambda(3,5)

[maiden hare]

Yes

Or (lambda * 3, lambda * 5)

Let's use q instead of lambda

[keen void]

OK

[maiden hare]

So a point on the line that goes through the origin and (3,5) can be expressed as (3q,5q)

[keen void]

yes

[maiden hare]

Now in your problem, you are told to choose (OA, OB) as a base

[keen void]

Yes

[maiden hare]

Or rather {(OA, 0), (0, OB)} {OA, OB}, since it's a set of two vectors

What can you say about the diagonal OC

[keen void]

Well OC is (OA + OB)

[maiden hare]

As a vector, that's right

What about as a line?

[keen void]

That it can be writte as, (qOA +qOB) ?

Where q <= 1 ?

[maiden hare]

What is the + doing here?

[keen void]

A summary of two vectors? I’m not sure.

I just know that the line starts at O and ends in C

[maiden hare]

Ok you need to use the base

[keen void]

the line splits the base in half

[maiden hare]

If {OA, OB} is your base, then A = (1,0) and B = (0,1), by definition

[keen void]

ah

yes

So the line starts in (0,0) and ends in (1,1)

[maiden hare]

Right

How can you express a point on that line then?

[keen void]

A point in the line would be q(x,y) or (qx, qy)

[maiden hare]

Again, why are you writing down x and y? We never mentioned them

[keen void]

Feels natural because it's 2D the same as graphs

q(OA, OB)

[maiden hare]

Are you using OA and OB as vectors here?

[keen void]

yes but i might as well write

q(a, b)

since a = OA and b = OB

[maiden hare]

Ok, as usual, I'm not sure what you're confused about

Let me backtrack

[keen void]

I simply don't understand the question

[maiden hare]

That it can be writte as, (qOA +qOB) ?

That's actually correct, but I'm not sure you understand exactly why

[keen void]

That's because if i were for example multiply by q=0.5 i would get half the length of the total vector

[maiden hare]

Ok yeah but then why not just say qOC?

[keen void]

oh

Yea that is also right

since (qOA+qOB) = qOC

[maiden hare]

Yes

Ok let's do this in two ways, one using vectors only and the other using coordinates with the base {OA, OB}

So a point on OC can be qOC

In the base, OC = (1,1)

So a point on the line would just be q(1,1) = (q,q)

Does that make sense?

[keen void]

yes

[maiden hare]

Alright, so that's one way to express OP'

Now you need a second way

Do you understand why I asked how you can express a point on OC?

[keen void]

So we can find OP' ?

[maiden hare]

Right but precisely why express a point on OC?

[keen void]

Because P' can be expressed as a coordinate?

[maiden hare]

The mind works in mysterious ways...

How is P' defined exactly?

[keen void]

as OC/4

[maiden hare]

No

You went back to the result you need to prove

Again, you can't use that

It would make no sense to give an answer to the problem that way

[maiden hare]

Any point can be expressed with coordinates, you're just saying that P' is a point

Read the problem statement again and find how P' is defined

[keen void]

P' = qOC

[maiden hare]

The problem statement doesn't mention that

[keen void]

Then i have literally no clue

I don't even know what i'm trying to prove anymore. I'm trying to prove something that has already been proven?

[maiden hare]

This is a hint, it's not a definition, it has not been proven in any way

That's for you to do

That's the whole point of this problem

[keen void]

OP' = OC / 4
OQ' = 3(OC) / 4

OP' = OA + OB / 4
OQ' = 3(OA + OB) / 4

OP' = (OA/4) + (OB/4)
OQ' = (OA/4) + (OB/4)

OM = OC/4 = 1/4 * (OA+OB)

[maiden hare]

I don't know what you're trying to say

[keen void]

Me neither

I don't have a clue what i'm doing

I'm supposed to prove that AP, AB and AQ divide the thing into 4 equal parts

And then it says that because of the symmetry i only need to prove the formula

And give two ways on how OP' can be written

[maiden hare]

Yeah

[keen void]

OP = qOC is one way? And now i'm looking for another?

[maiden hare]

Yes

[maiden hare]

You understand that this is what you need to prove, yet you said P' is defined using this

That just doesn't work

P' has an actual definition in the problem statement

[keen void]

Yeah but i have absolutely zero clue what i'm doing

[maiden hare]

Give me the definition of P'

[maiden hare]

You can't just repeat yourself and expect me to say that's correct...

[keen void]

But i don't know what the def is?

[maiden hare]

By "problem statement" I mean everything above the hint

Where is P' mentioned in the problem statement?

[keen void]

It isn't`?

[maiden hare]

It is...

I can't teach you to read

Even your translation mentions P'

[keen void]

Let OACB be the parallelogram to the right. The sides OB and BC are bisected
three equally long stretches where the dividing point P ∈ OB is closest to O, while
the dividing point Q ∈ BC is closest to C. Show that the points of intersection P′,
M and Q′ of the diagonal OC with the lengths AP, AB, and AQ respectively divide
this diagonal into four equally long parts.

I don't see P'

Do you?

Only time it is mentioned is

Show that the points of intersection P′,
M and Q′ of the diagonal OC with the lengths AP, AB, and AQ respectively divide
this diagonal into four equally long parts.

How does that transle into a definition ?

[maiden hare]

It literally says "points of intersection", names P' among other points, as well as the relevant intersecting lines

Is that not a definition for you?

When two lines intersect, you can define a point as the intersection

I don't know how else to put it

You can rewrite the problem statement as a clear list of definitions and a "show something" at the end if you want

[keen void]

I just don't understand how someone can solve something they don't understand to begin with

How am i even supposed to elaborate to an answer when the question is so unclear

[maiden hare]

I mean... that's just reading comprehension

Just because it has maths in it doesn't make it a maths issue

[keen void]

So i just can't read?

[maiden hare]

• OACB is a parallelogram
• P is on OB such that 2OP = PB
• Q is on BC such that 2CQ = QB
• P' is the intersection of OC and AP
• Q' is the intersection of OC and AQ
• M is the intersection of OC and AB
• Show that OP' = P'M = MQ' = Q'C

This is what you should have understood from the problem

[keen void]

Yes i understand that

[maiden hare]

You do now or did you already earlier?

[keen void]

I did earlier

[maiden hare]

Then how did you not see the definition of P'

[keen void]

I can see that just based of the picture alone

P' = OC/4

[maiden hare]

...

[maiden hare]

I'm not sure you do

[keen void]

I don't think guessing myway to an answer is how i solve something like this

The directive of the question is to unclear

"Show that OP' = P'M = MQ' = Q'C" like how? We can clearly see on the picture that it is

it even states that

OA, BC and OB and AC

are parallell to each other

So how am i supposed to prove something that clearly already has been proven in the statement in the quiestion

It's like someone asking; Prove that 2+2 = 4

[maiden hare]

A picture is not a proof? The equalities could be very close but not exact

[keen void]

They have to be the same

because it states

that they are divided into 3 exactly the same long

"The sides OB and BC are bisected
three equally long stretches"

With that info they have to be exactly 1/4 away from each other

[maiden hare]

Wdym these are not the same line segments

You're not making any sense

P is on OB

The question asks you to show something about P', which is on OC

[keen void]

If someone gave you a cube with each side are all the same length, And then they drew two vectors diagonally from each corner which resulted with a point in the middle. And then they ask, prove that the point in the middle is in fact in the middle

^ this is exactly how i feel about this assignment

The proof is already in the question to begin with since they say that p' = OP' = OC/4 = a+b/4

[maiden hare]

The parallelogram's sides are not all the same length, and P is not the intersection of the diagonals... You could say something like that for M, but not for P

[maiden hare]

You really need to differentiate between a hint and the actual problem statement

They never state that "p' = OP' = OC/4 = a+b/4" is true

They give you a hint that it should be true

[keen void]

Ok

[maiden hare]

The actual proof is left to you

Again, that's the whole point of this problem

[keen void]

I don't have a single clue on how to prove that then

It's completely incomprehensible to me

How am i supposed to understand this after >1 week into linear algebra

[maiden hare]

Well, the hint also says that you should express OP' in two ways, which is what we started to do earlier

This has little to do with linear algebra, it's a geometry problem, probably early high school level...

[keen void]

Great, guess i'm under high school level then

[maiden hare]

The only thing about linear algebra here is that they talk about a base, but you can easily solve this without using a base

I'm no teacher, but to me it feels like a tricky problem for an introduction to vectors

Assuming you learn about vectors early on in high school, yeah, that's high school level

[maiden hare]

P' is the intersection of OC and AP

This is what you need to use

If a point is the intersection of two lines, then it is on both lines at the same time

We showed earlier that, because P' is on OC, it can be expressed as qOC

Now, P' is also on AP, so... how else can it be expressed?

[maiden hare]

Clarification: the vector OP' can be expressed as q*OC

P' would be expressed with coordinates of some sort

[keen void]

(qOC, qAp)

where q is the point on the line P'

P' = q(OC, AP)

[maiden hare]

You keep mixing up different concepts... since when is q a point and P' a line?

[keen void]

q is lambda

idk

[maiden hare]

q is a number

P' is a point

[keen void]

Yes

[maiden hare]

OC and AP are both vectors

not coordinates

Earlier you found how to express a point on OC

Now you just need to find how to express a point on AP

[keen void]

OP' = qOC

isn't this a point on a vector?

[maiden hare]

What does that mean

[keen void]

qOC is a point on OP'

[maiden hare]

Ok don't mix up lines and vectors

Technically, OC is a distance

(OC) would be a line

OC with an arrow above would be a vector

Throughout this I wrote OC for both the line and the vector

P' is a point on the line (OC)

The vector OP' is equal to some number q multiplied by the vector OC

[keen void]

well it was defined to us such that a bold letter such as a, b, c ... z or for example AB with a arrowover it or a letter with an arrow over represents Vectors

[maiden hare]

Right that's why they use p', a, and b, in bold, to represent the vectors OP', OA, and OB

[keen void]

points are represented the same way graphs are made, for example A = (1, 2) which is 1 step x axis and 2 steps y axis

[maiden hare]

With curly braces?

[keen void]

No, ()

[maiden hare]

Ok yeah that's sensible

[keen void]

Something like that

where if i wanted A i'd just summarise the point on O + M

[maiden hare]

I don't get this figure

Why are there 5 coordinates

[keen void]

Idk? my teachers example

[maiden hare]

...

In the plane, points only have 2 coordinates, you just write two numbers in the parentheses

[keen void]

I know

But then again, my teacher doesn't speak eng or swedish properly so I can't tell what he says

[maiden hare]

Idk what this is but it's not helpful

[keen void]

Either way, i don't know how this helps me with my current question

How else P' could be written as

[maiden hare]

I guess it could be a representation of two 5D vectors and the vector that averages them

[maiden hare]

as a point on AP

To get from the origin to a point on AP, what can you do?

[keen void]

P' = qAP

[maiden hare]

No that doesn't work, AP doesn't go through the origin

[keen void]

How does that equal going through the origin?

You just multiply AP with q which is between 1 and 0

to get the length from A to P'

[maiden hare]

I guess you have a common misconception about vectors

Vectors are only a magnitude and a direction

They don't "start" anywhere

If you scale the vector AP, you just get the same direction and a different magnitude, but then if you use that to get a point, you have to start somewhere

By default you start at the origin

So the point P' corresponds to the vector OP'

The vector AP has no corresponding point in the picture

It would be somewhere above and to the left of the origin

[keen void]

OP + PP'

[maiden hare]

Yeah that gives you OP'

[keen void]

Like this question is too hard for me

He expects us to understand this after >1 week = 3 hours of lectures

This doesn't reflect the lectures at ALL

[maiden hare]

Welcome to university I guess

[keen void]

How should i solve a problem i haven't been taught

It makes no sense

[maiden hare]

you're supposed to be familiar with vectors

[keen void]

i know what they are, i know how to add them

that doesn't help me in this question

[maiden hare]

Clearly you don't quite understand what they are

[keen void]

clearly

[maiden hare]

You are on the right track, the problem is that I asked for "a point on AP", not specifically P'

[keen void]

A point on AP is q(OA + OP)

[maiden hare]

No

OA + OP would give you a point on AC

Scaling that would give you a line through the origin and through that point

That's not the line AP

[keen void]

qAP

[maiden hare]

.

[keen void]

q(OP + PA)

Like i don't know?

I can't keep guessing

[maiden hare]

OP+PA is just OA

[keen void]

how the fuck is OP+PA OA?

[maiden hare]

The vectors?

[keen void]

How is the blue line

equal to

OA

[maiden hare]

Clearly we aren't talking about lengths, right?

The vector OP plus the vector PA equals the vector OA

I thought you knew how to add vectors

[keen void]

Why do i have to guess myself to the answer?

Like how am i supposed to know this

i haven't been taught

i can't be guessing

I still don't understand the question or how to solve it

[maiden hare]

Imagine starting from the origin O and going somewhere on PA

What path do you take

[keen void]

I got too P and then follow PA

or i can go to P' and go left or right

or i can go to A

and follow AP

[maiden hare]

Right

In vector form, what does it mean to "get to P", and then to "follow PA"?

[keen void]

It means OP + PA

[maiden hare]

Surely there is a difference between going from P to A and following PA

[keen void]

Well you have to go the entire distance to P so OP is OP

qPA

[maiden hare]

Yes

[keen void]

where q dictates the pointn on PA

[maiden hare]

Exactly

So OP' = ?

[keen void]

OP' = OP + qPA

[maiden hare]

Yes, and that's a second way to express OP'

Now, it's not the same q in both ways, so let's use r instead

OP' = qOC and OP' = OP + rPA

There can be only one value for q and for r

[keen void]

So we get a matrix equation?

[maiden hare]

You get a system of two equations with two unknowns

Two equations because it's in 2D, so you have two coordinates for everything

[keen void]

So i have to convert OP', OC, OP and PA to their respective coordinates ?

and then find q and r?

$OP' = qOC \
OP' = OP + rPA$

[solid kiln]

Merineth

[keen void]

Or sorry

i mean their length

ig they don't ahve a length since they are vectors

[maiden hare]

Either that or deal with vectors directly but that might be harder

[keen void]

so quick question

OC would that be the cordinates of

C?

(1, 1)

[maiden hare]

Yes, using the base {OA, OB}

[keen void]

am i thinking right when

$1/4 = q(1.1) \
1/4 = (0,1/4) + r(1.0)$

[solid kiln]

Merineth

[keen void]

,rotate

[maiden hare]

That's all kinds of wrong

You are using 1/4, I assume because of the hint, but I told you multiple times that it is never explicitly stated that OP' = 1/4 OC

Then, you are equating 1/4, a number, to something that results in a vector

Then, the coordinates of the vector OP are not (0, 1/4)

[keen void]

$OP' = q(1,1) \
OP' = (0.\frac{1}{3}) + r(1,0)$

[solid kiln]

Merineth

[keen void]

This would be my best guess

[maiden hare]

Oh and the coordinates of PA are not (1,0)

[keen void]

Is it not? it's one step to the right 0 up?

[maiden hare]

No, that would be OA

[keen void]

So the coordinates for OA and PA are different?

[maiden hare]

They are different vectors, so yeah of course...

[keen void]

oh PA is

OP + PA

Nah i don't know

I can't tell the difference

[maiden hare]

Well then you need to take an introductory course on vectors

OA = OP + PA

PA = OA - OP

[keen void]

fine

[trim jolt]

@keen void Has your question been resolved?

[trim jolt]

@keen void Has your question been resolved?

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[azure zenith]

I don't know what to apply here, i tried integration by parts but got stuck there

[azure zenith]

Got this

[pearl plaza]

maybe now let u=xf(x)

you will get another integral(x^2f'(x))

[azure zenith]

I figured it, apparently it needed the outcome in a different form

Gucci

.close

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[open kiln]

im having some toruble with part c

so im trying to find the y component of F_31

heres my work so far

this is what the answer should be

im solving for F_31y

[open kiln]

maybe im using the wrong r here actually

but im only using the direction of the y component, right?

<@&286206848099549185>

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[wraith hinge]

10.The area of a circle is 38.5 cm2
. Find the diameter of circle

[lethal anvil]

what's the formula for the area of a circle?

[umbral maple]

i wonder

[wraith hinge]

pie r^2

[lethal anvil]

yep good. It's just pi not pie 🙂

but yeah it's A = (pi)r^2.

So you have the area, so you know what A equals

so you just need to solve for r

[wraith hinge]

ya

i did it b y doing

38.5/22/7 =r^2

but the answer is coming wrong

I mean I cant solve any further

[lethal anvil]

well you can, you can take the square root of both sides

to isolate r

$A = \pi r^2$
$\\frac{A}{\pi} = r^2$
$\ \pm \sqrt{\frac{A}{\pi}} = r$

[wraith hinge]

i didn;t get it

[solid kiln]

MellowDramaLlama

[wraith hinge]

what is this +_

[lethal anvil]

plus/minus

when you square root a value you get both a postiive and a negative value

BUT

[wraith hinge]

I like buts

[lethal anvil]

we can throw away the negative value since we can't have a negative radius

[wraith hinge]

ladys buts

[wraith hinge]

ook

[lethal anvil]

uhhh, congrats?

[wraith hinge]

lol

just joking

[lethal anvil]

so we're left with $r = \sqrt{\frac{A}{\pi}}$. Plug in your A and you'll find your radius.

[solid kiln]

MellowDramaLlama

[wraith hinge]

$\sqrt{\frac{5.5}{7}} = r$

[solid kiln]

JAAT

[wraith hinge]

this

I am left with

[lethal anvil]

check your math again. I'm getting 12.25

then you need to square root that with a calculator

[wraith hinge]

$\sqrt{\frac{38.5}/{22/7}} = r$

[solid kiln]

JAAT

[wraith hinge]

with this? ur getting 12.25

[lethal anvil]

$\frac{38.5}{\frac{22}{7}} = \frac{38.5 \times 7}{22} = \frac{269.5}{7} = 12.25$

[solid kiln]

MellowDramaLlama

[lethal anvil]

but regardless a calculator wil help you tremendously here. Are you not allowed to use one?

[wraith hinge]

hmh m

correct

but

then u gotta square it

and the answer is 7

[lethal anvil]

,calc 2 * sqrt(12.25)

[solid kiln]

Result:

7

[lethal anvil]

yep! 🙂

[wraith hinge]

okk

wait

how?

7*7

is not 12.25

[lethal anvil]

so once you square root it you get 3.5, and the diameter is 2 times the radius, so 3.5 * 2 = 7

[wraith hinge]

ok

ok

ok

ok

got it

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[wraith hinge]

byee

.reopen

wait

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✅

[wraith hinge]

wait

@lethal anvil

[lethal anvil]

what's up?

[wraith hinge]

.Find the value of:
(3/8) ÷ (3/4)

Find the following multiplication:
(6/13) × (36/ ̶5)

Calculate –2
1
9
– 7.

Find the sum of ( ̶9/10) and (21/15).

help me with these once too

[lethal anvil]

nah I could only do your problem, but someone else can step in. I gotta go get some dinner

best of luck!

[wraith hinge]

dinner?

at 6 in morning??

where do you live?

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[frosty glacier]

Stuck on the first hw question lol

[frosty glacier]

How do I know whether it is negative half or -3?

[burnt mulch]

Note that if x = -1/2 is a factor, then (2x+1)^3 must be a factor

now consider the leading coefficient

[frosty glacier]

Aaaaahhhh

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[devout garden]

Oh i closed the other one sry

So this one

[blazing geode]

Emm

It is xy^2

[devout garden]

oh yeah

just re write like

Ohhh

Let me see

(x^3y^6)^1/3

and then xy^2 ok perfect

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[hallow crane]

hi, how would i do this without calc?

[hallow crane]

i thought of assigning f(x) = 2^(-x/2) and then every other option is just

f(x/2), f(x/4), f(x/8), ...

not sure if that's the way to go

and i guess thus i can say something about horizontal stretches

[lusty delta]

yeah pretty much

[hallow crane]

but not sure if i'm being dumb

what's the final conclusion

after identifying that?

[lusty delta]

2^(-x) is positive and decreasing from 0 to 10 so it makes sense to say that a horizontal stretch would increase the area

[hallow crane]

oh yeah fair enough

and for the converse it would decrease the area

[lusty delta]

yep

[hallow crane]

thanks, nice observation

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[wraith hinge]

3 lines, y = 0, y = x^2, and y = -1/5x + 6/5, whats the area over the interval [0,1]

[wraith hinge]

<@&286206848099549185>

[wraith hinge]

idk

[torpid pasture]

find which function is greater over the interval

then integrate its difference

bc the area between curves will be the integral of top function minus integral of bottom function

[wraith hinge]

alr

[drowsy portal]

@wraith hinge

U in calc2?

[wraith hinge]

ye

@drowsy portal

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[leaden nexus]

did u try that

show ur work

[wraith hinge]

wait mb i did do it

[leaden nexus]

show

[wraith hinge]

close this chat

i finished it

[leaden nexus]

.close

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[frosty glacier]

I'm trying to factorise h(x)

[leaden nexus]

what is h(x)

[frosty glacier]

Image still loading

My math is not mathing

[leaden nexus]

alr

damn, nvm, I don't remember this stuff, sorry

I saw factorise and got excited

[frosty glacier]

Lol

[sly pewter]

wait I remember root of mulitplicity

hold on let me see if I

[leaden nexus]

I miss HS, rip

gotta go back and study this eventually

[frosty glacier]

I'm kinda confused on 2 equal roots

[sly pewter]

oh

[frosty glacier]

That means multiplicity two right?

[sly pewter]

well derivatives are involved 😭 my math class teaches that next semester I only know some stuff

[sly pewter]

x^2 has root of multiplicity 2

at x =0

[frosty glacier]

Bro that isn't helpful

Sorry if I'm rude

[sly pewter]

ok sorry i didn't know what you meant 🤩 i will leave

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[astral tangle]

A=(x,√+3), (x,√(-3)

B= (√+2,y)(√-2,y)

I guess option D is correct

But given answer is B

[leaden nexus]

is this y mod 0 cong to 3?????????

[astral tangle]

no no

Y^2=3

[leaden nexus]

ohh okay

okay in that case u can lowkey just graph both and just see how many times they intersect

[astral tangle]

4

[leaden nexus]

yup

but u said the answer is 0?

huh.

weird..

[astral tangle]

I don't know it is given in the answer sheet

[leaden nexus]

maybe somebody else can find where we are both mistaken......

[astral tangle]

But who?

[knotty locust]

elements in A are of the form (x, +/- sqrt(3))

Agree?

Similarly, elements in B are of the form (+/- sqrt(2), y)

if we want these to have a common intersection

we can freely choose x, and y

and count all the possibilities

[astral tangle]

Of course

So my answer is 4

But given answer is b

Thanks

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[full dock]

For part B, I’m assuming I’ll need the length of AD first to find out the area, but how to I rearrange the formula c^2 = a^2 + b^2 - 2abcosC so I get b^2 as the subject and get 2abcosB instead of cosC?

[sly pewter]

$b^2 = a^2 + c^2 -2ac*cos(C)$

[solid kiln]

MoonCaik

[whole coral]

You don't need the length of AD btw

[full dock]

Hm but in this case I don’t have cosC or technically cosD in this context so how can I use it?

[full dock]

I don’t?

[sly pewter]

yeah

[median torrent]

You're looking for area?

And You're given all three sides right?

This is Heron's formula

[full dock]

I don’t have AD

[sly pewter]

heron's formula?

[whole coral]

You could find the area of triangle ABD and use that it's part of ABC

[sly pewter]

what about 1/2absin(C)

for me, i would use law of cosines then 1/2absin(C)

[full dock]

I’ve never heard of it

[sly pewter]

wait no

[median torrent]

You do need AD to find the area im p sure

[sly pewter]

no yo udon't wait

[jade aspen]

does the angle ADC is equal to 90 ° ?

[sly pewter]

area of full triangle - triangle ADB

[median torrent]

You can find AD by using law of cosines

[sly pewter]

1/2absinC

[median torrent]

Then use heron's formula to calculate area

[haughty oasis]

why not 1/2 ab sin C on the big triangle?

[sly pewter]

no yo udon'ttt

[sly pewter]

yeah

[full dock]

But how do I find the area of ADB if I don’t have AD?

[sly pewter]

1/2absinC

[median torrent]

You can get AD with law of cosines and then use heron's formula!

Heron's formula is the one that uses semi perimeter

[full dock]

i don’t have sinC tho, how do I find that?

[median torrent]

You dont know the angles in the triangle except for the 60 degree

[sly pewter]

sinC is sin(60)

[median torrent]

You dont have a right angle where AD meets the base

[full dock]

Isn’t that sinB?

[sly pewter]

1/2*8*3*sin(60)

what?

[whole coral]

[it's worth noting that in "(1/2)ab * sin(C)", the "a", "b" and "C" are really placeholders]

[haughty oasis]

1/2 ac sin(B) is an equivalent formula

[full dock]

ohh

so I can swap them around?

[sly pewter]

nono

not really

wait acutally

yes?

that is so real i just realized that

this emoji is so rage

[full dock]

like if I wanted sinA and I had 1/2absinC I can get it down to 1/2bcsinA?

the discord italic thing ruined the asterisk

grrr

[knotty locust]

do \*

[whole coral]

Put spaces or backslash the asterisks

[full dock]

does this swappy thing work for the law of cosines too?

[sly pewter]

I think kind of????

[haughty oasis]

yep

[sly pewter]

there's 3 versions

[whole coral]

[which are interchanging the roles of a, b and c (and their respective angles A, B and C)]

[full dock]

ooh okay

That’s good to know that I can change the formula when I need to

so for the area of ADB I got 12sin60

[whole coral]

You can simplify that more

[full dock]

12*sqrt(3)

wait

lmao

hold on

i don’t have my calculator with me I’ll get it

it usually does most of the work

Is it 6*sqrt(3)

[whole coral]

You probably should know the "special angle" values

[full dock]

hehe why memorize when my calculator does it

[whole coral]

Grrrr

Anyways, you're happy with finding the area of triangle ADC now?

[full dock]

okey i got it yaa

thank u charr and everyone else

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[supple copper]

What is det A

What does that tell you about the column space of A

The column space is a set

Ok we might get sidetracked down this road

But what I wanted to say is that a (square) matrix with a determinant of 0 means it is not full rank

So it is possible for some b’s to not be in the column space

In the case that A was non-singular (ie det A ≠ 0) then any b you picked would be in the column space

Yeah anyway, with part a you’ve shown that Ax=b has infinitely many solutions for x

Then yes it must be in the column space of A

And hence it is in span{u,v,w} since that is the column space of A

Why does part a have the last line saying a vector = a set

Wait why are there curly brackets

[supple copper]

No it should say $\vec x\in\left{\begin{pmatrix}x\y\z\end{pmatrix}= \begin{pmatrix}2\-1\0\end{pmatrix}+\alpha \begin{pmatrix}-1\1\1\end{pmatrix}, \alpha \in\mathbb{R}\right}$

[solid kiln]

Frosst

[supple copper]

Use copy text

Not highlight and copy

[solid kiln]

Mycobacterium

[supple copper]

I think so, the first part just gives the reader an idea of what the structure of the elements look like

Sometimes you may even see

$\vec x\in\left{\begin{pmatrix}x\y\z\end{pmatrix}\in\mathbb R^3\middle|\begin{pmatrix}x\y\z\end{pmatrix}= \begin{pmatrix}2\-1\0\end{pmatrix}+\alpha \begin{pmatrix}-1\1\1\end{pmatrix}, \alpha \in\mathbb{R}\right}$

[solid kiln]

Frosst

[supple copper]

Not necessarily

You can put a lot of stuff on the right

$\vec x\in\left{\begin{pmatrix}2\-1\0\end{pmatrix}\cdot \alpha \begin{pmatrix}-1\1\1\end{pmatrix}, \alpha \in\mathbb{R}\right}$

[solid kiln]

Frosst

[supple copper]

This becomes a scalar

But yoy might not catch it

You need double \\ for next line in a matrix

You’re missing it for the first x y z

And the last one

[solid kiln]

Mycobacterium

[supple copper]

Has at least 1 solution yes

But has a solution => has at least 1 solution

So you aren’t wrong

Yes

The idea is that the span of the column vectors of a matrix is the column space of the matrix

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[void wasp]

i need help to make a sign line

[void wasp]

i think its called that in english at least, not sure though

heres an example

on all my other tasks, when its f’(x) ive always found top and bottom points and used - - - if the growth is negative, and —— if its positive, but in this one its not doing it

i thought you were only supposed to take when it hits the x axis when it was f(x) and then have —— if its above y = 0 and - - - if its below y = 0

can someone help me explain where i got this wrong?

[delicate belfry]

Fortegnskjema er snakk om den deriverte, endringen av funksjonen, ikke hvilke interval funksjonen har sine positive/negative verdier.

Men måten du gjorde på, å finne positive/negative verdier, er på riktig spor. Du skal bare bruke det til den deriverte, ikke selve funksjonen.

[void wasp]

men er ikke det den deriverte?

oppgave b

[void wasp]

men kan du ikke lage et fortegnsskjema til en normal funksjon?

[delicate belfry]

Svaret ditt ser fint ut. Stemmer det ikke med løsningsforslaget?

[void wasp]

nei, det stemmer ikke, bilde nr 2 viser fasiten

var derfor jeg var helt forvirret

[delicate belfry]

Var ikke det ditt svar?

[void wasp]

nei

[delicate belfry]

Har dårlig tid, men, du skal se når grafen går oppover (positiv, solid strek), og når den går nedover (negativ, stripete line).

[void wasp]

ja

det var det jeg trodde

[delicate belfry]

Du får bare prøve deg med andre, må gå nå.

[void wasp]

åja vent

bare glem d

jeg skjønte hva keg gjorde feil

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[dark turtle]

i have these 3 matrices and i'm trying to determine if theyre linearly independent. my process is as normal, but i'm not sure if my logic is correct. when equating coefficients i find a contradiction and that's how i determine they're linearly independent. is this legit?

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[viral oasis]

yes

that’s correct

@dark turtle

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[rigid tangle]

So there's a block with
circumference length of the base is 52 cm.
The length and the height difference is 8 cm.
And the widsth minus the height is -2cm, what is the volume of the block?

idk how to solve this

[rigid tangle]

im so bad

that makes half the circumference 26 right

So L + W = 26

[wraith hinge]

Is the circumference longer than the height or is the opposite true?

[rigid tangle]

No circumference is not longer than height

[wraith hinge]

so the circumference of the base is shorter than the height by 8cm
is it a cylindrical block

and the width is the diameter right

if there is a drawing in the question it would help

My calcs dont match up

[rigid tangle]

no its not cylinder

[rigid tangle]

Noo its not

i would draw wait

its like this

[wraith hinge]

So it is a rectangular block

And by circumference you mean perimeter

ok i solved it using simultaneous equation

r u there

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[rigid tangle]

sorry i was doing other tasks

[rigid tangle]

oh yeah its called rectangle

sorry

[rigid tangle]

would you help me how

wait for a bit, im almost done

Hello @wraith hinge

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[rigid tangle]

.reopen

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✅

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[hallow crane]

hi, dumb question but isn't this wrong?

[hallow crane]

if anything shouldn't it be +2 * all three

[trail ingot]

yea that sounds wrong

$|A\cup B \cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$

[solid kiln]

chmonkey #1 simp

[shell barn]

This is PIE

[hallow crane]

oh okay so i'm not the only one

thanks-

honestly i got that formula i think

[trail ingot]

[hallow crane]

if i think only about A, B, and C

not sure if that's what they meant

at least i got close to it

[hallow crane]

the difference was this^

thanks btw- @trail ingot

.close

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Use .reopen if this was a mistake.

[trail ingot]

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[hallow crane]

@trail ingot (sorry for the ping)

[hallow crane]

but wait this works right?

so i guess definitionally they just consider different regions

in actual PIE, you'd have that the green represents green + orange

[trail ingot]

am i dumb or is this not right...

o maybe it is

[hallow crane]

😭

am i dumb too 😭

i'm pretty dumb lately

it doesn't make sense to me either

i use the actual thing too

[trail ingot]

we should have 175 = 64 + 72 + 88 - exactlytwo + 8 by inclusion exclusion amirite

[hallow crane]

yeah i think

that's what i would've done too

[trail ingot]

[proper kernel]

if you think of adding/subtracting these areas as adding/subtracting the circular areas,
you can intuitively consider 64 + 72 + 88 - 175 - 8 - 8 as getting you the green areas only

first, 64 + 72 + 88 ends up = outer areas + 2 * green areas + 3 * red areas
then - 175 removes one whole area to be = green area + 2 * red area
then - 8 - 8 removes the red area twice to be = green area

[hallow crane]

hi matt

wait okay yeah that makes sense

but what's wrong with the actual inclusion exclusion

that's because it counts the three way intersection too?

in the "exactly two"

[proper kernel]

whats wrong with the formula is that it absolutely sucks to use rather than this simple approach

I havent checked the formula yet but it should be correct

[hallow crane]

no i mean the actual PIE formula

which u generalize for any n-set

[proper kernel]

which one

you have two to three formulas mentioned right now

[trail ingot]

i see why it's wrong now

[hallow crane]

|A U B| = sum of first intersections - two way intersection

|A U B U C| = sum of one way intersections - two way intersections + three way intersects

and then so on

https://en.wikipedia.org/wiki/Inclusion–exclusion_principle

[proper kernel]

thats vague

you literally have a wikipedia article about it so that formula is correct, what are you asking for

[hallow crane]

i know it's correct

i'm curious why it didn't work

or rather what went wrong

in using that

[proper kernel]

it didnt work?

[hallow crane]

yeah

probably used it wrong

but was curious where so

[trail ingot]

A intersect B etc includes the intersection of all 3

[hallow crane]

oh yeah

wait we're dumb

[trail ingot]

yes

[proper kernel]

bruh

[hallow crane]

wait

[proper kernel]

also since theyre only giving you the green areas and the red areas as separate, that means you need more than just the formula to solve this

its a system of equations

[trail ingot]

so this is fine actually, it's a little different than inclusion exclusion

[hallow crane]

wait but actually using the formula is more involved right

it's not just |A U B U C| - |A n B n C|

[proper kernel]

what would that accomplish

[hallow crane]

cuz u gotta get rid of the only A, only B, and only C

[proper kernel]

|A U B U C| - |A n B n C| = outer areas + green areas

[hallow crane]

that would leave everything there while removing the three way intersection

[proper kernel]

you then need to remove the outer areas

[hallow crane]

yeah that's what i meant

probably more complicated

[proper kernel]

which is an approach that the color areas thing I mentioned does anyway

typing AB for A and B, A + B for A or B,
PIE: |A| + |B| + |C| - |AB| - |AC| - |BC| + |ABC| = |A + B + C|
youre told the value of |A|, |B|, |C|, |ABC|, |A + B + C|
you need to find the value of the green areas which ends up being: (|AB| - |ABC|) + (|AC| - |ABC|) + (|BC| - |ABC|)

|A| + |B| + |C| - |AB| - |AC| - |BC| + |ABC| = |A + B + C|
|A| + |B| + |C| + |ABC| = |A + B + C| + |AB| + |AC| + |BC|
|A| + |B| + |C| + |ABC| - |A + B + C| = |AB| + |AC| + |BC|
|A| + |B| + |C| - 2|ABC| - |A + B + C| = |AB| - |ABC| + |AC| - |ABC| + |BC| - |ABC|
|A| + |B| + |C| - |A + B + C| - 2|ABC| = (|AB| - |ABC|) + (|AC| - |ABC|) + (|BC| - |ABC|)
64 + 72 + 88 - 175 - 2(8) = green area

so its not a system of equations but its still involved

[hallow crane]

right right, i get it now

thanks matt-

[proper kernel]

np

[hallow crane]

also thanks @trail ingot

.close

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• Show your work, and if possible, explain where you are stuck.
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[loud mica]

Calculus 1, I’m very confused on how to set this up

[loud mica]

after I can get the equations figured out I’m sure I could solve it, but idk how

[hallow crane]

not sure if i'm being dumb

but why do you need calculus here?

[loud mica]

I’m just saying the class I’m in for convenience of the tutors here

You need to find where the slope is equal to 0 as calculators are not allowed.

[loud mica]

So, by using derivatives. but the other part is definitely solvable just by being not slow (unlike me)