help-38

[dusty mist]

subbed values greater than 0 into here

and none of the answers gave me 0 😦

Ik that means they don't intersect, but I dont trust myself

I feel like I messed up somewhere

<@&286206848099549185>

[split chasm]

shouldn't be subbing in integer values like that,
doesn't actually show whether there's a t that actually work

just solve d = 0 and see if there are positive real solutions

[dusty mist]

but the ship is changing directions at 8:30pm

[split chasm]

i mean at the very end

[dusty mist]

while the helicopter changes directions at 9:00

huh

[split chasm]

after you've got their positions at 9

[dusty mist]

yep

did I make mistakes?

or is it fine

[split chasm]

looks fine

[dusty mist]

alr

didnt work

t has to be greater than 0

That means they dont intersect

I def messed up somewhere

oh my god

could this be it

was theta supposed to be 90-16.26

wait no

<@&286206848099549185>

here is the full question

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[little egret]

how i do 10a?

[full dock]

bum chicken

bum chickoen back on vectors again

[little egret]

HELP

we are doing 3d now

and im stuck on 2d

😭

[full dock]

U know i should be able to do this but i dont like geometry

[full dock]

gg

[little egret]

i had a methods test today

it told me to evaluate 16e^9/9

how

[full dock]

evaluate it as in do what

[little egret]

evaluate

[full dock]

o

use ur calculator

ezpz

[little egret]

non calc

[full dock]

let e = 3

ez

thars how u do it as an engineer

[iron pier]

it's probably (a+b)/2 lmao

[simple panther]

Nope, ans would be waaaay off

[iron pier]

I forgot my basic geometry theorems

but it looks like it'd be that 💀

[little egret]

why>?

thats right

[iron pier]

ok see that's the problem 💀 i forgot whatever theorem could let me say such a thing

[little egret]

1/2(a+b)

[simple panther]

🧠 16e^9/9 = 16e^1 = 16e = 16×2.7 = 43.2

[simple panther]

Just use math

😂😂

[little egret]

theres a theorem..

wat...

[iron pier]

ok consider this heuristic argument:

[little egret]

WHATS HEURISTIC

[iron pier]

it'd hold true for a rectangle

[full dock]

math monks guys

[iron pier]

probably

to be honest tho, maybe the question is asking you to prove this result yourself

that'd be a lot more fun

[iron pier]

basically, hand waving not rigorous but lets you see why it's true

[full dock]

i found it the name

[little egret]

there is a way to do it without

[full dock]

the Trapezoid Midsegment Theorem

[little egret]

I DID THIS LAST YEAR

[simple panther]

Try dropping perpendicular from A and D onto BC and using similar triangles

You'll basically prove this theorem this way (hopefully)

[little egret]

i remember finding AD and BC

and BD

i think

[iron pier]

the point of proving the result is to bolster your geometry understanding tho

[little egret]

ohh i meant

theres a way\ to do it wtihout the hteorem

[iron pier]

...by proving the theorem and then using it 💀

[full dock]

we'll call it the bum chicken trapezoid theorem

[iron pier]

beautiful name

let's generalize it to n dimensional euclidean space

[little egret]

.close

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[wraith hinge]

i cant mental math this one

[wraith hinge]

is this correct?

is there a trig identity counterpart for arcsin (u) ?

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@wraith hinge Has your question been resolved?

[fair beacon]

pretty sure no

you got the first term right

[wraith hinge]

2nd?

[fair beacon]

off by a factor

it should be 19 not 29

[wraith hinge]

huh???

how

[fair beacon]

i can't tell where you went wrong because i don't understand anything you've written

[wraith hinge]

my mouse handwriting is truly terrible 😭

(8x-5)/sqrt(7+6x-x^2)

[native shuttle]

Just send original

Is that the original?

[fair beacon]

8x-5 = A+B(-2x+6) = -2Bx + 6B + A

[wraith hinge]

\int\frac{8x-5}{sqrt(7+6x-x^2}

huh

[native shuttle]

Put a $

at the beginning and end

edit it

[wraith hinge]

$\int\frac{8x-5}{sqrt(7+6x-x^2}

[native shuttle]

end

[wraith hinge]

great it cant read mine

$\int\frac{8x-5}{sqrt(7+6x-x^2}$

[native shuttle]

u can edit message

[fair beacon]

it's wrong anyways

[native shuttle]

AT BOTH

beginning and end

[solid kiln]

rivenrivers

[fair beacon]

$\int\frac{8x-5}{\sqrt{7+6x-x^2}},dx$

[native shuttle]

better

[solid kiln]

carburetor

[fair beacon]

B = -4 and A = 19
so you split 8x - 5 as 19 - 4(-2x + 6)

[wraith hinge]

i wonder where i failed my factoring

OH

MY

GOODNESS

[fair beacon]

yeah everything else is right

[wraith hinge]

lksdfgoksdfg

THANK YOU

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[paper jay]

is this diagram correct ?

also i think the answer is a) 110° (corresponding L)

b) 70° (interior angles)

is that correct ?

[obsidian zealot]

the diagram is correct

[paper jay]

what about the answers of a and b

[obsidian zealot]

for a we can analyse quadrilateral KLBC

since KL || BC it's a trapezoid

the interior angles add up to 360

KLC = BKL

if we let KBC = x, 220+2x=360

x=70º

AKL is similar to ABC by AAA

so its also 70 for b

[paper jay]

what is the reason ?

[obsidian zealot]

wdym

for which parte

[paper jay]

for both a and b cus i need to add a reason in my answer

[obsidian zealot]

you can say this

by interior angles of trapezoid
220+2x=360
x=70

[paper jay]

and what about for b?

[obsidian zealot]

umm

by similar triangles (AAA)

although you could write the whole process if there's enough space

[paper jay]

okok thanks

[solid kiln]

Couldn't find an attached image in the last 10 messages.

Mar 19, 2025, 4:55 PM

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[knotty saffron]

find 2 numbers whose sum is a, if the product of the square of one by the cube of the other is to be a maximum

[knotty saffron]

i'm having a hard time with the pattern of problems like these (applications of minima/maxima) can anyone help me out step by step?

[obsidian zealot]

we first need an equation in one variable

cal lthe numbers x and y

[knotty saffron]

i got x+y = a
x^2y^3 = P

[obsidian zealot]

ya

can you make it in terms of x

[knotty saffron]

x = a-y?

[obsidian zealot]

like

define P only using x

sorry 😅

[knotty saffron]

x^2(a-x)^3

like that?

[obsidian zealot]

yea

then u find the maximum

[knotty saffron]

i find the derivative of that, right?

[obsidian zealot]

yeah

then u equate it to 0

[knotty saffron]

okay wait

[obsidian zealot]

im waiting 🛑

[knotty saffron]

i got -3x^2(a-x)^2 + 2x(a-x)^3

sorry i'm quite slow

[obsidian zealot]

thats right

now we factor to find the roots

those will be the critical points

[knotty saffron]

x(a-x)^2[-3x+2(a-x)]

[obsidian zealot]

can u simplify the []s

[knotty saffron]

x(a-x)^2(-5x+2a)

[obsidian zealot]

yaaaaa

so what r the roots

[knotty saffron]

x=0
a-x=0 ; x=a
-5x+2a = 0 ; x=2a/5

[obsidian zealot]

ya

theres only one maximum do u know how to find it?

[knotty saffron]

by finding the second derivative?

[obsidian zealot]

noo

thatll be way harder than necessary

try plugging into the original P

[knotty saffron]

108a^4/3125

ah wait

3a/5

[obsidian zealot]

yess

so whats y

[knotty saffron]

i thought that was y

[obsidian zealot]

oh

yea

i meant like plugging in the x=0/x=a to find that 0 and a are roots of P and thus can't be maxima

the other one, 2a/5, is correct

and you find (2a/5, 3a/5) ❤️

[knotty saffron]

ohh okok

thank you! is it alright to ask another question?

[obsidian zealot]

ok

[knotty saffron]

A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?

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[knotty saffron]

.close

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[jaunty sparrow]

I need to scale this shape down by a factor of 1/24. how should I do that? btw the bottom function says x=

[jaunty sparrow]

hmm I should color code my functions

ight idk if it made it easier to understand at all, but I colored each function the same as it's reflection

as for the dotted line, I just ran out of colors so I made it dotted lol

I guess it would save a lot of hassle if I never had to scale it down at all

I could just scale the rest of the shape up instead (this is only a part of it)

But idk what kinda implications that would have when I want to 3D print this shape

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[jaunty sparrow]

Rip

[jaunty sparrow]

<@&286206848099549185>

[velvet bolt]

🤣 i like this question

[jaunty sparrow]

that makes one of us

[naive crest]

you have an eq y = 8 {-66 <= x <= -56}
wouldn't y = 8/24 {-66/24 <= x <= -56/24} be the scaling down ?

like y' = y/24 and x' = x/24

[jaunty sparrow]

Idk bro I just need to make it smalller

[naive crest]

your points are of the form (x, f(x)) {a <= x <= b}
so I'd say to replace by (x/24, f(x)/24) {a/24 <= x/24 <= b/24}
so (x, f(24x)/24), {a/24 <= x <= b/24}

like your first eq should become y = (0.5714(24x+83)-77.143)/24 {66/24 <= x <= 73/24}

etc

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[jaunty sparrow]

when I have a vertical line, should I make the equation x/24 = 78 ?

[jaunty sparrow]

er

it would need to be x/24 = 78/24

[naive crest]

you replace x = 78 by x = 78/24

and you scale the limits of y

[jaunty sparrow]

ight thanks that worked

now I just need to figure out how to rotate this thing

I'll ask that question when I am ready for it

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[hardy rivet]

tg 30 + tg 60

1 - tg 30 + tg 60

How i simplify this ??

[zinc ginkgo]

use tangent sum formula

or plug in the special values for tan(30 deg) and tan(60 deg)

[hardy rivet]

but how i simplify more than this ?

[zinc ginkgo]

it's not pretty, but you can rationalize the denominator by multiplying the top and bottom by the conjugate

https://www.mathsisfun.com/algebra/conjugate.html

[hardy rivet]

thx man

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[austere matrix]

how does pie actually work

[ionic pendant]

given any circle, if its circumference is $C$ and its diameter is $D$, then $\pi$ (``pi'') is the ratio of its circumference to its diameter: [ \pi = \frac CD ]

[solid kiln]

cloud

[ionic pendant]

it turns out that this ratio is the same for every circle

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[austere matrix]

oh ight

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[solemn sleet]

can someone explain where to start with this

im very lost

i though i could just use trig to find the force on the ramp

[nimble stone]

that is indeed how you do it

[obsidian zealot]

yes

[solemn sleet]

or projections but the answer is 12 and im getting 48

[nimble stone]

how are you doing it?

[solemn sleet]

so i thought sin(30) = 24/ramp

ramp = 48N?

[nimble stone]

got it a bit mixed

it would be sin(30)=ramp/24

[solemn sleet]

but 24 isnt the hypotenuse

[nimble stone]

how did you draw your triangle then?

[solemn sleet]

well i used the triangle here

is that no what ur supposed to do

[nimble stone]

when we are finding components we typically want the force to be the hypotenuse, because then the other sides are our perpendicular components

in this case, perpedicular to the ramp and parallel to it

[solemn sleet]

what do you mean by components

also

the force always need to be the hypotenuse?

[nimble stone]

its because we are looking at the force parallel and perpendicular to the plane, when you draw the triangle as you did, youre taking the 24N to be a component of a larger force, in this case the 48N you calculated

[solemn sleet]

okay so how do i draw the triangle here

[nimble stone]

pretty much this

[solemn sleet]

how did they get this

or oh

is that the end of the ramp?

[nimble stone]

eh

that line is just representing the weight force

[solemn sleet]

ohhhh

how do you know that its 30 degress tho

[nimble stone]

oops

[solemn sleet]

oh!!

okay

thank you, that makes sense

:)

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[rose inlet]

ok so you have that region in the first quadrant under y = x^2 from x = 0 to x = 1

[bleak ridge]

yes

[rose inlet]

for each slice of this solid, perpendicular to the x axis, the cross section is a quarter circle

the radius of that quarter circle is just the distance from the x axis up to y = x^2

which is?

[bleak ridge]

distance from where

[rose inlet]

how far the curve is from the x axis

[bleak ridge]

like at 0 it would be x=0 it would be 0 and at x=1 it would be 1?

[rose inlet]

it would not be one

[bleak ridge]

oh

[rose inlet]

since the curve is y = x^2, the vertical distance from the x axis to that curve is just x^2

[bleak ridge]

okay

[rose inlet]

so the radius of each quarter circle is x^2

[bleak ridge]

okok

so

the equation of a semicircle is

1/4 pi r^2

so i get

1/4 pi x^4

[rose inlet]

exactly

[bleak ridge]

and then i integrate that?

[rose inlet]

!!!

[bleak ridge]

okay awesome thank you

[rose inlet]

and then you need to integrate from x = 0 to 1

np

[bleak ridge]

im just a little confused why

its so different than doing like

a semi circle

like i usually just use the pi / 8 formula for that right

[rose inlet]

yea for semi circle its pi /8

but since we're dealing w quarter

its pi/4

[bleak ridge]

so would it not just be pi / 16 for quarter

how does it become bigger

oh ig i see

bc if the radius is on it then it would be larger vs the whole diameter

[rose inlet]

exactly if ur working with the radius

the numbers r usually bigger cuz the diamater is twice the radius

once u sqare it that factor of 4 comes in

[bleak ridge]

tysm

[rose inlet]

np

[bleak ridge]

so i can just use the pi/4 formula if this comes up again right

quarter circles

[rose inlet]

yes ofc as long as u have quarter circles

[bleak ridge]

cool

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[crimson thunder]

how would one go about this

[west sleet]

Can compare coefficients

Or do polynomial long division

[crimson thunder]

yea im kinda stumped on that ;-;

[west sleet]

Alright then comparing coefficients it is

[crimson thunder]

wait i kinda need to do long division for my test ;-;

[west sleet]

Alright alright

[crimson thunder]

does comparing coefficients work for everything?

[crimson thunder]

no i dont wanna for you into doing anything!

[west sleet]

Uh I mean yea but

Sometimes a bit long

[crimson thunder]

ohh ok

[west sleet]

Actually depends but anyway

[crimson thunder]

yea this is all i got ;-;

[west sleet]

Yes that's correct

Just remember to distribute the negative

[crimson thunder]

oh wdym?

[west sleet]

-(27x^6-216x^3)

So -27x^6+216x^3

[crimson thunder]

ohhh!!

but like how do i subtract 216x^3 from 0x^5

[west sleet]

Uh as seen here we distribute the negative sign

[crimson thunder]

OH SO WE ADD

[west sleet]

Yea

[crimson thunder]

do i pull down the 0x^4?

[west sleet]

And uh we can just add it to the x^3 part of the polynomial

[crimson thunder]

oh so now i bring down the -217x^3?

or do i add it it -217x^3?

[west sleet]

Yea just add it

[crimson thunder]

alr! so -x^3?

and we bring down the positive 8

[west sleet]

Ye

Alright I gtg

But u got it

[crimson thunder]

oh alr!

tyty!!

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[marsh rapids]

I’m confused on turning logarithms into exponential equations and so forth, can someone help me? Specifically this problem!!

[quick tusk]

e^ln(x) = x

[marsh rapids]

Huh

[quick tusk]

ln(x-3) = 4

You can eliminate the ln by exponentiating both sides

since e^ln(x-3) = (x-3)

[marsh rapids]

So I divide by -6?

On both sides

[quick tusk]

yeah, thats the first step

[marsh rapids]

And then do I put it back in exponential form?

[quick tusk]

dividing both sides by -6 gives you ln(x-3) = 4, right?

[marsh rapids]

Yeah

[quick tusk]

Presumably you're trying to solve for x?

[marsh rapids]

Yeah

[quick tusk]

You wan't to eliminate the ln. You can do so by raising both sides to the power of e, euler's number

[marsh rapids]

Do I switch it to e^4 = x-3

[quick tusk]

yeah, because when you take e^ln(something), the e and ln cancel each other out, leaving (something)

[marsh rapids]

And then I solve for x from there?

[quick tusk]

yes

[marsh rapids]

Ok thank you!

Ok I have one more question i need help on (e is tripping me up):

[high patio]

Bring the 5 to the other side?

[marsh rapids]

And then put it into logarithmic form?

[high patio]

Yeah log both sides

[marsh rapids]

Ok is this right?

I’m not very good at math

[quick tusk]

no. don't forget, ln and e cancel each other out

[marsh rapids]

…huh

So is it: n-7 = log 335

[knotty locust]

Why did you write ln on one side and log on the other

[marsh rapids]

I don’t know

[knotty locust]

If by log you mean base e log yes

I.e ln

[marsh rapids]

I’m confused and literally on the verge of tears dude

[knotty locust]

Well that’s rather dramatic

What are your pronouns

[high patio]

Log usually means base 10 tho

[high patio]

Lol

[knotty locust]

Exactly my point

[marsh rapids]

I don’t have control over my emotions because of my autism and I’ve been stuck on this problem since I got the assignment.

[knotty locust]

Let’s see if we can’t clear this doubt then it

e^(n-7) = 335

Is clearly fine right

Okay

Whatever we do now we want to preserve this equality

We want to do the same thing to both sides

Does that general idea make sense it?

[marsh rapids]

So would it be: n-7 = ln 335?

[knotty locust]

What operation occurs to reach this

[marsh rapids]

…I don’t know

[knotty locust]

Okay so why do you think that equation is true

[marsh rapids]

I don’t know

[knotty locust]

Ok

So let’s not randomly ghess

Guess*

e^(n-7) = 335

[marsh rapids]

Sorry

[knotty locust]

This is where we are it

What can we do to this equation

To both sides

At the same time

We can multiply, divide, add, subtract, etc… aslong as we do the same thing to both sides

Obviously the power of e is messing us up yes?

So it’d be nice if we could do something to both sides that would get rid of that

What do you think we should do

[marsh rapids]

ln??

[knotty locust]

Yes we should take log base e (ln) of both sides

Then what is the equation

[marsh rapids]

Uh

n-7 = ln335

???

[knotty locust]

Yes

Great

Now what

It

[marsh rapids]

…Put ln335 in my graphing calculator

[knotty locust]

No

We want to solve for n

[marsh rapids]

…but then I’d find what n-7 is and could just add 7 to both sides and I’d have n

[knotty locust]

Yes

The calculator is also unnecessary here

It’s clearly ~5.8

[marsh rapids]

…

Ok

I’m just dumb

Do I solve for n after that

[knotty locust]

add 7

To get n

Like you said they

[marsh rapids]

How did you get -5.8? Calculator says positive 5.8

[knotty locust]

~

Is not -

[marsh rapids]

Huh

[knotty locust]

It means approximately

[marsh rapids]

Oh I can’t read very well

[knotty locust]

It’s squiggly

[marsh rapids]

Sorry

[knotty locust]

That’s okay it

[marsh rapids]

…do you have to say my pronouns like that…just call me dude or bro…

Or man…

Any work

[knotty locust]

Well I wasn’t sure

What you wanted because you say ask and I asked and you didn’t clarify

Idk how to use

Yours

My apologies dude

[marsh rapids]

It’s fine

So would n roughly equal 12.8

[knotty locust]

Yes

[marsh rapids]

Ok cool

Thanks

[knotty locust]

Np

[marsh rapids]

.close

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[dapper oxide]

is (x, 0, 0) where x is varying, the x-axis? or is it just a point where x can be anything

[orchid wagon]

the locus of (x, 0, 0) when x varies over R is the x-axis

otherwise, (x, 0, 0) is just a point on the x-axis

[dapper oxide]

even though x is varying

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@dapper oxide Has your question been resolved?

[gilded pulsar]

The set of all points (x,0,0) is the x axis

[dapper oxide]

How would u graph (x, 0, 0) when x is varying but (x, 0, 0) is not the set of all points?

[trim jolt]

@dapper oxide Has your question been resolved?

[simple panther]

(x, 0, 0) is just a point. By graphing it, it means that mark all the possible values that that point can be.

For example, at x = 1, it would be (1, 0, 0)
At x = 1.1, it would be (1.1, 0, 0)
At x = 1.2, it would be (1.2, 0, 0) and so on...

Marking all those will give us its graph which would turn out to be the x axis

[dapper oxide]

So is it just a point or the x axis?

How would I graph x=1 OR x=1.1 OR x=1.2

[simple panther]

Depends

If they mention that it's a point or they said x is fixed (or anything similar), then it's a point

If they mention that x is a parameter or ask you the locus of that point, it would be the x axis

Graphing x = 1 or the point at x = 1?

[dapper oxide]

I wanna graph (x,0,0) at x = 1 OR x=1.1.

[simple panther]

Then it's just a point whose x is 1 OR 1.1, y is 0 and z is 0

[dapper oxide]

Not at x=1 AND x=1.1

[simple panther]

Then?

Both together?

[dapper oxide]

Yes not together

[simple panther]

??

[dapper oxide]

In this example when ur graphing all possible points. Are you doing “x=1 AND x=2 AND x=3, etc”

[simple panther]

Yes

From - infinity to + infinity

That would be the x axis

[dapper oxide]

What if I wanna graph x=1 OR x=2 OR x=3, not AND

[simple panther]

-_-

It's just multiple points

[dapper oxide]

Would that also be the x axis

[simple panther]

Yeah, but they won't be together in the same graph

If it's or,

Then it's either 1 or 2

Not both together

So it's either this or that

If it is togethers then we would get the x axis

[dapper oxide]

How would u plot it?

One graph for x=1 and another for x=2?

[simple panther]

Yes

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[signal sundial]

If A is an infinite set, are we allowed to define |P(A)| and treat it as a real number? I have seen some prooofs just say that $|P(A)| = 2^\kappa$ where $|A| = \kappa$

[solid kiln]

rya

[simple panther]

Isn't it supposed to be $\kappa^\kappa$!?

[solid kiln]

Suika

[signal sundial]

I think for powersets its 2^|A|

but usually those are for finite sets

[somber ginkgo]

kappa^kappa would be the set of functions A -> A

[signal sundial]

so im not sure if u can do it for infinite sets

[simple panther]

Oh, ok

[somber ginkgo]

with more formal cardinal arithmetic, you can "sort of" pretend the cardinals are ordinary numbers

[simple panther]

By considering kappa is infinite, I think you can

[marble wharf]

the expression 2^kappa is just defined to be the cardinality of P(A). its not actually exponentiation in the sense of like repeated multiplication or via taylor series or something

[signal sundial]

My lecturer gonna kill me...

[somber ginkgo]

but here you shouldn't be thinking about them like normal numbers

[signal sundial]

Do you think i could find a bijection

[somber ginkgo]

write down relevant bijections/injections/surjections to show that |P(A) x P(A)| = |P(A)|

i gather this is for like an introductory uni course rather than a logic&sets course right?

[signal sundial]

yeah its like all encompassing proofs

sets fxns proofs logic counting etc.

[somber ginkgo]

yeah in which case write down ^

if this were say a set theory course then it's possible to do cardinal arithmetic to show that the sets have the same cardinalities

(also remark - the assumption |A x A| = |A| is technically vacous cus it's always true lol)

[signal sundial]

wait are u sure?

cuz i mean just comparing element in a

u already have |A|

like (1,a0),(1,a1),...(1,an)

then that will already be |A| elements

[somber ginkgo]

A is infinite though

[signal sundial]

ahh right

yes

youre right

i gues they give it as like a property

[somber ginkgo]

ofc it's not true for finite sets, but i.e. N x N has the same cardinality as N, R x R has the same cardinality as R etc.

[signal sundial]

that we should use?

[somber ginkgo]

yh

[signal sundial]

Im a lil behind on lectures, but i think we went into sets/functions a fair amount

I think they would be fine with cardinal arithmetic to prove this

[somber ginkgo]

oh as in don't use what i told u

as in they gave you the assumption that |A x A| = |A| so you should probably use it

but i just wanted to remark that actually if you do more maths in later years @ uni, then turns out this property is always true lol

[signal sundial]

isnt it like u can just find a bijection for A x A and A if they are countably infintie?

[marble wharf]

(isnt |AxA|=|A| assuming aoc or something?)

[somber ginkgo]

(maybe? but we like choice!)

[marble wharf]

(but not all logicians do)

[signal sundial]

Considering Im a first year 5 weeks into this course, do yuo think I should find a bijection or just use cardinal arithmetic without fuilly justifying every step?

[marble wharf]

bijection

[signal sundial]

Just asking because idk if it would be hard to find a bijection

[marble wharf]

cardinal arithmetic is proved by finding bijections anyway

[signal sundial]

alright thankls :)\

[somber ginkgo]

i just brought up cardinal arithmetic to explain why you might have seen some proofs that basically treat |P(A)| like a real number

[signal sundial]

ohhhh okay

but thats like proven through bijection

[somber ginkgo]

yh

[signal sundial]

any ideas on where too start with bijection?

[somber ginkgo]

are you familiar with the proof that N x N is countable? try to mimic that

[signal sundial]

im not familiar with that prof ;-;

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[regal flint]

There are two boxes containing marbles of the same size and mass. The first box contains 5 red marbles and 5 green marbles, the second box contains 6 red marbles and 4 green marbles. Randomly take one marble from the first box and transfer it to the second box, then randomly take one marble from the second box (assuming the marble taken from the second box is red). Calculate the probability that the red marble is from the first box.

[cedar flower]

!status

[trim jolt]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[regal flint]

1

[marble karma]

Alr

so

whats the probability the marble is red

from the first box

lets start here

[regal flint]

sorry im thinking a little bit

[marble karma]

5 red 5 green

[regal flint]

P(get red from box 2) = 5/10 * 7/11 + 5/10 * 6/11 = 13/22
P(get red from box and its origin is box 1) = 5/10 * 1/11 = 1/22
=> P(the marble is from box 1 assuming that it is red) = 1/22 : 13/22 = 1/13

Is it correct?

[marble karma]

no

[regal flint]

why?

@cedar flower help please

@marble karma please explain why it is incorrect

guys

[cedar flower]

sorry I just got home. reading over now

P(a red marble is taken from the second box that originally in box 1 | the marble taken is red) = P(a red marble is taken from the second box that originally in box 1 AND the marble taken is red) / P(the marble taken is red)

yes, I think you are correct

if you can go "oh the actual solution is 1/4" then we can work out where we went wrong - but I don't see a problem w/it

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[sudden seal]

How does one check if a vector is parallell with a plane?

The task is as following:

A plane has a normal vector n = [3,2,-1]. Furthermore, a point P(-2, 1, 0) is inside the plane.

a) Find an equation for plane alfa.

I did this by using Ax+Bx+Cx = 0, and came to the following 3x+2y-z+4 = 0

b) Find the coordinates that cross with the plane and y axis.

(0, a, 0)

Set in values in the equation and solve for a

(0, -2, 0)

c) Research if vector [1,1,5] is parallell with the plane.

How do you start with this?

[cedar flower]

a vector that is parallel with a plane will be perpendicular to the normal vector

[sudden seal]

Alright, so I have to see if v * n = 0?

[cedar flower]

aye

[sudden seal]

where v is the vector and n is the normal vector

Alright great, thanks :)!

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[paper jay]

hi is this correct ?

[undone gorge]

C is wrong

Take triangle OTU

OU=OT (radius)

So it is an isosceles triangle

Now angle TOU=180-angle TOP

Now TOP = 64°

[paper jay]

so 116

[undone gorge]

So TOU = 180-64= 116° (UOP) is a straight line

[undone gorge]

Yeah

Now in isosceles triangle base angles are equal (let's say x)

[paper jay]

so 180-116 divided by 2

?

[undone gorge]

So 2x+116=180

Sorry I was wrong 😭

[obsidian zealot]

you can just use inscribed angle theorem lol

32=64/2

[paper jay]

but is mine correct ?

[obsidian zealot]

yes

[paper jay]

even the reasoning?

[undone gorge]

Yes, I am sorry

[obsidian zealot]

yes

I haven’t checked d tho

[paper jay]

it’s ok thanks

[obsidian zealot]

yea I think it’s all correct?

[paper jay]

thank you

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[rich heart]

I am currenttly on question 2 so sorry for not answering before, if anybody can help it would be hreat

[obsidian zealot]

what do the angle numbers here mean?

[rich heart]

just the angles

I am not sure

[obsidian zealot]

also fyi some of the questions are cut off lol

hmm it doesn’t denote congruency……….

[rich heart]

oh mb

qurstionm 3.2.2

question

I have no idea what to do

[obsidian zealot]

ok let’s just talk in terms of C123 and T2

bc that’s what’s on the diagram

we know C1 + C2 = 90

[rich heart]

yes

c1 tan chord with A

and thats it I think

[obsidian zealot]

have you proven 3.2.3?

[rich heart]

no

[obsidian zealot]

oh bc you need to prove this lmao

[rich heart]

how do I do this

I dont see an resemblance

[obsidian zealot]

is COT isosceles?

[rich heart]

doesnt say

[obsidian zealot]

ok let’s just state all the lines

C1 + C2 + C3 + C4 = 180

T2 + T3 = 180

T1 + T2 = 180

C1 + C2 = 90

so C3 + C4 = 90

and T1 = T3

[rich heart]

yes

[obsidian zealot]

COK and COS are right

is ABC right?

oh it is

[rich heart]

i think so

[obsidian zealot]

C2 + C3 = 90

[rich heart]

im gonna be honest

I think im just gonna give up on this

[obsidian zealot]

C1 = C3 and C2 = C4 then

this is rly fun tbh

[rich heart]

im sorry man

im just no good at math

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[crisp wolf]

(Integral calculus) I have been trying to solve this for a while now, i tried using some substitutions but i always get 2 variables. I want to know what am i doing wrong or what should i consider doing instead

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[crisp wolf]

<@&286206848099549185>

[obsidian zealot]

unless you fix one of the sides, you will always have two variables

[zinc ginkgo]

!original

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@crisp wolf Has your question been resolved?

[crisp wolf]

I'm gonna look into that and confirm with my senior that sent me the question, thanks

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[bronze jetty]

Anyone help to find f(theta) please , i am unable to get a series for cancellations

[bronze jetty]

<@&286206848099549185>

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[sacred jackal]

!showwork

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[bright seal]

5+5-5÷5 ???

[clear cloud]

Idk but <@&268886789983436800> do know

[bright seal]

Why?

Is the answer 1?

[knotty oriole]

Don't post troll problems here.

Use a calculator.

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[bright seal]

Why is it troll?

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[glad patio]

im not sure

how to do this

this is question b)

[dusky thunder]

the surface area is the sum of individual areas of each side

[glad patio]

yes

[dusky thunder]

did you manage to express h in terms of x?

[glad patio]

oh do i just replace h with that

[dusky thunder]

yeah

[glad patio]

this doesnt seem right

[dusky thunder]

here ur second line, u make it 2(2xh) = 4x2h which doesnt make sense. write it as 4xh. even for the first term, write it as 2xh rather than 2x2h

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[hoary idol]

How I finish solving this

[zinc ginkgo]

,tex .cross mult

[solid kiln]

riemann

[gusty bone]

why is there a shortcut for that 💀

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[silver sandal]

can someone tell me where it went wrong?

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[silver sandal]

<@&286206848099549185>

[silver sandal]

ahhh..

ok new approach

rip.

yeah the questions wrong

.close

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[limpid siren]

Find the numbers of factors of n if n = a^x * b^y

[round imp]

Hello

Any other information

[queen burrow]

there's a simple formula for this

assuming a and b are primes

[limpid siren]

Yes, a and b are primes

[queen burrow]

the answer would be (x+1)(y+1)

[round imp]

Ah

[queen burrow]

basic combinatorics

[round imp]

Bruh

[queen burrow]

lol

[round imp]

!nosols

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[queen burrow]

sorry

[round imp]

Aiyo

[queen burrow]

so if you think logically

[round imp]

Anyway try to find an intuitive reason

Why is (x+1)(y+1)

[queen burrow]

you can have 1,a,a^2,.... as the factors

upto a^x

[round imp]

Think of the conditions of p if p|n

[queen burrow]

and similarly 1,b,b^2...b^y

for each factor you need to find a combination

there are x+1 choices for a or it's power appearing in the factor,

and y+1 for b

[round imp]

Let him think ._. but okay

[limpid siren]

All factors of n can be listed as follows:
(a, a^2, a^3, a^4,....................., a^x) + (b, b^2, b^3,............................, b^y) - [x + y factors]

(ab, ab^2, ab^3, ab^4,..................., ab^y) - [y factors]
(a^2b, a^2b^2, a^2b^3,...................., a^2b^y) - [y factors]
........................ - x columns in total
(a^x * b, a^x b^2, a^x b^3,..........................., a^x b^y) - [y facotrs]

These are all the factors. Counting the number of factors we get:
(x + y) + (y * x)

[round imp]

+1 for 1

[limpid siren]

But the number of factors should be (x+1)(y+1)

[round imp]

1 is a factor

[limpid siren]

Of course

[round imp]

Ah yes

[limpid siren]

I was missing the 1.

Thanks

[round imp]

Np

[limpid siren]

.close

[trim jolt]

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[trim jolt]

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[dusty mist]

"a helicopter leaves point (24,12) while travelling at 25km/h in the direction of 343.74 true bearing. What is the position vector and velocity vector of this moving object?"

[trim jolt]

@dusty mist Has your question been resolved?

[dusty mist]

<@&286206848099549185> \

am I mutted 👅

[trim jolt]

@dusty mist Has your question been resolved?

[devout yew]

first you can find the velocity vector. here's a hint: think of a right angled triangle with x and y components equal to that of the x and y components of the velocity, what could the hypotenuse be? and if you knew that, how would you find the x and y components if you new one angle other than 90 inside the triangle?

then you should be able to workout the position vector easiliy

[dusty mist]

would the velocity vector be [-24,7]

[devout yew]

I didn't do the calculation but I'd be surprised if you such a clean answer, what was your method?

[dusty mist]

uh

would it be alright if I could give you the background to the question

its a bit of a read

[devout yew]

yeah, go on

[dusty mist]

the question I asked was a simplified version of question 14

here was my calculations for the velocity vector and position vector for the helicopters motion at 9pm using the 16.26 true bearing

so my first approach was to substitute the theta value with the new 343.74 true bearing

but then I got confused

so I ended up with this

which I hope I got correct

[devout yew]

yeah, this logic works fine for this question

[devout yew]

this is nice because it shows you really thought about the question instead of just plugging in the values to the position vector formula, which would give you the same answer just without the added satisfaction

[dusty mist]

wait I think I got the values messed up

hold on

[devout yew]

I assume you mean the cosine and sine values right?

[dusty mist]

no like the actual x and y components

the blue was supposed to be 24

and the 7 was supposed to be -7 and 7

hence the changed signsd

😭

[devout yew]

ahh the classic sign mixup, nw mate

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[native shuttle]

What do I do after f'(x)

[trim lichen]

post the rest of the question

[native shuttle]

Oh just check for 1-1 and onto

[solid plank]

Why are you derivating it then?

[native shuttle]

To find 1-1

Since if we do f(x1)=f(x2) we can't find

So we need to do f'(x)

And show it's strictly increasing/decreasing

[solid plank]

Oh okay you could do that too yeah

[native shuttle]

But idk what to do after f'(x)

[solid plank]

So equate f'(x) =0

[native shuttle]

hmm ok

I am getting (11x^2-30x-20)/(7x^2+2x+10)^2

For f'(x)

So if f'(x)=0 I need to equate $11x^2-30x-20=0$

[solid kiln]

devthemasked

[solid plank]

Yeah

[native shuttle]

So x=30+-root20/22

[solid plank]

And you're also have to make sure that $(7x^2+2x+10)^2=0$ won't have any variable satisfying it

[solid kiln]

Aria

[native shuttle]

It cannot be 0

Or f'(x) undefined

[solid plank]

Yeah that's why you have to make sure

[trim lichen]

what a crunchy question...

[native shuttle]

wdym by crunchy

[solid plank]

So for some value of x, f'(x) is zero, thus it won't be strictly increasing or decreasing

[native shuttle]

The solution is something else

[trim lichen]

needs a lot of number-crunching

[native shuttle]

Should I send that

[trim lichen]

but i can note the denominator has negative discriminant and so is always strictly positive

and by the looks of it, the same is true of the numerator

[solid plank]

Are you sure that you derivative f'x correctly?

[trim lichen]

so no matter what x is, f(x) will be positive

so just from this alone

no hope for onto-ness

[native shuttle]

Wait why no onto

[solid plank]

yeah you're right

[trim lichen]

cause f(x) = -69 can never happen

[solid plank]

😭😭

[trim lichen]

what

what's wrong

[solid plank]

nothing

[native shuttle]

Did you find discriminant of numerator and denominator

of f(x)

[solid plank]

it would be negative, yeah

[trim lichen]

then why the sobs

[solid plank]

nvm

[trim lichen]

i didn't calculate them but i can tell they are negative for sure

[native shuttle]

Ok so if discriminant is negative the quadratic is positive

[willow urchin]

bc funny number cant happen

[trim lichen]

1 - 4×2×5 and 4 - 4×7×10...

[native shuttle]

yea'

[trim lichen]

oh. sad. many such cases

[trim lichen]

the quadratic has constant sign

whether it's positive or negative, you tell by the sign of a or c (which have to match)

[native shuttle]

ok

ok so codomain is R

and range is +R

[trim lichen]

no

i didn't say what the range is

[willow urchin]

we dont know range = R^+, all we know is range is contained in R^+

[trim lichen]

i didnt claim the range is (0, +infty) -- and it isn't.

what matters is that the range isn't R.

[native shuttle]

Oh

So basically range cannot be -R

[solid plank]

you can say that the range would be a subset of R+

[solid plank]

yeah

[native shuttle]

ok

[trim lichen]

overcomp again

[native shuttle]

so not onto

[willow urchin]

-R is an abuse of notation

[solid plank]

frfr

[willow urchin]

just say the range cant include negative numbers

[native shuttle]

kk

Now how to check 1-1

[solid plank]

Can we adjust the numerator or denominator

I think that might make it simpler

[native shuttle]

how

[solid plank]

Nah we'll have to differentiate

[native shuttle]

yep

I did that

I found roots of f'(x)

Then what

[solid plank]

but I think you did something wrong there

if there are some roots for f'(x), that means it won't be strictly increasing/decreasing

[native shuttle]

There are no roots if D<0 right

[solid plank]

There are no real roots if D<0

So yeah

[native shuttle]

Yea I got roots

[solid plank]

Uhh I got f'(x)= $(11x^2-30x-20)/(7x^2+2x+10)^2$

[solid kiln]

Aria

[native shuttle]

So I saw at those particular roots the value of f'(x)=0 hence it will not be strictly increasing/decreasing

[native shuttle]

yep

[trim lichen]

$f'(x) = \frac{11x^2-30x+20}{(7x^2+2x+10)^2}$

[solid kiln]

Ann

[solid plank]

yeah thanks

[native shuttle]

-20

[solid plank]

Yeahh

[native shuttle]

hmm k

see the answer

[trim lichen]

typo oops

[solid plank]

It's says that function is increasing or decreasing in intervals

[trim lichen]

so they just say the derivative isn't sign-constant

[solid plank]

IT LITERALLY SAYS THAT IT IS MANY ONE FUNCTION😭😭

that means it is not one one

[native shuttle]

Yea I know

But the solution I got is also correct right?