help-33

[rocky lava]

a > 1

right

or smth else

@past frigate

<@&286206848099549185>

[marsh citrus]

@rocky lava Has your question been resolved?

[marsh citrus]

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[rocky lava]

.reopen

[marsh citrus]

✅

[rocky lava]

<@&286206848099549185>

[rocky lava]

my bad

but pls help

i’ve been waiting for so long

i will actually forever be indebted to you

[rocky lava]

sir knief, please

[still temple]

how can we help

[rocky lava]

thank you so much

is this right

[wary kite]

yes mr president

a>1 true

[rocky lava]

aight

[wary kite]

and yes

a is negative

and a<-1

[rocky lava]

what

oh

so it’s not |a|<1

[wary kite]

well yes

but

a is also negative

because it’s concave down

for part b

[rocky lava]

what

does that mean

[wary kite]

like it’s going down

[rocky lava]

oh

[wary kite]

like a frown

[rocky lava]

so then it’s a<-1

and not a<1

1 should be negative

[wary kite]

yea

[rocky lava]

alright

and then also

[wary kite]

well wait

no

[rocky lava]

what

[wary kite]

it’s stretched outwards

so it should be between 0 and -1

[rocky lava]

so

what

[wary kite]

my fault

[rocky lava]

oh so it’s

a<1

[wary kite]

but it has to be <0

[rocky lava]

oh

so it’s -1<a<0

alr

and then

for 28

is that always true

[wary kite]

which

[rocky lava]

28

because it’s a vertical compression and that makes it wider

[wary kite]

well stretch

not compression

but it will be wider yes

[rocky lava]

oh aight

aight thanks

.close

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[zinc sparrow]

I am incredibly stuck on uncertainties

Im guessing there’s a rule that means if you square an uncertainty it’s the same as multiplying it by 2, kinda like indices but I don’t understand the +2 whatsoever

[whole sleet]

Are you supposed to get this through differentiation?

[zinc sparrow]

No I was taught differentiation after uncertainties so I don’t think they’re related

but then I forgot all about uncertainties

[marsh citrus]

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[harsh iron]

hey!
I'm having trouble with quadratic formula. I got 2.5 and 2.69, but that's incorrect, and I'm now confused - actually I was always confused, but that's ok

[left wadi]

can you show your work?

2.5 and 2.69 doesnt feel like right answer at all

[harsh iron]

it's in my book, but I can try type it out

[harsh iron]

it really doesn't

[left wadi]

you can send a pic

[harsh iron]

I don't have my phone on me atm sadly

[left wadi]

ok wait

what did you get under the sqrt?

[harsh iron]

-5^2-4x1x-1

[left wadi]

so its equal to?

[harsh iron]

-25+4

[left wadi]

no

because its (-5)^2

so its 25

not -25

[harsh iron]

ohh that makes sense

so 25+4, which is 29

[left wadi]

anyway if youd get -21 there would be no solutions

[left wadi]

yeah

maybe try solving now

[harsh iron]

so, -(-5)/2 is 5/2, and the other one would be sqrt 29/2

[left wadi]

no 😭

[harsh iron]

I'm so lost here 😭

[left wadi]

first is -(-5)+sqrt(29)/2

[torpid cairn]

NO NO NO

[left wadi]

and second -(-5)-sqrt(29)/2

[torpid cairn]

solve D first

[left wadi]

he solved D first

[left wadi]

here

[torpid cairn]

what did he get

[left wadi]

29

[torpid cairn]

show me your work

[left wadi]

bruh read the chat

we just went over this

[harsh iron]

oh thats what happened, that's where I got lost

[torpid cairn]

well what did you get?

[harsh iron]

so the answers would be 5.19 and -0.19?

[harsh iron]

so I just missed this fact, good to know

thank you all! :)

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@still temple Has your question been resolved?

[still temple]

what

oh

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[remote halo]

anyone here can with me with my data management homeowrk?

[marsh citrus]

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[thin cove]

how do I solve for

[thin cove]

continuity and discontinuity here

no test value given

<@&286206848099549185>

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[lyric kelp]

arcsec(7)

gives error

[desert dirge]

it shouldnt

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[dull yew]

reference angle

[lyric bay]

?

[desert dirge]

dont think so

[dull yew]

thats not my answer

but

you would add 360

ti -461 right

and then do again so its positive

and then to the x axis

wait no

i tihnk its right

[dull yew]

is it right

cuz -461 + 360 = -101

then +360 again

which is 259

that to the x axis

is 259-180=79

because 259 still in quadrant 3

[desert dirge]

-461+360=-101

this is in quadrant 3
the angle from the negative x axis to there is 79

from the positive x axis is 180+79=259

unless i misunderstand reference angle (I was)

im not that familiar with it

ah okay

yeah 79

[dull yew]

bet

thanks

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[muted solar]

hi

[marsh citrus]

Please don't occupy multiple help channels.

[muted solar]

we want to see if this series converges or diverges

we want to use direct comparison test

an is the series, what would bn be?

[lyric bay]

Talyor Swift?

Just a guess lol

[agile idol]

you can maybe split the sum into 1/e^n and cos(n)/e^n

then determine if each one converges

if at least one of them diverges then the entire thing diverges

[muted solar]

cos(n)/e^n???

[muted solar]

how do i find out is that diverges or converges

[agile idol]

all I did was split the numerator

[agile idol]

the first term should be a geometric series and the second one can use direct comparison I think

let me write it out

[muted solar]

oh yeah

cuz 1/e^n

then r = 1/e

and thats less than 1 to its converges

[agile idol]

|r|<1 yes

$\sum_{n=1}^{\infty} \frac{1}{e^n} + \sum_{n=1}^{\infty} \frac{\cos(n)}{e^n}$

[elfin berry]

Triaxyz

[agile idol]

I believe for the second term you can compare it to $\sum_{n=1}^{\infty} \frac{(-1)^n}{e^n}$ because it's an alternating function from -1 to 1

[crystal lintel]

[elfin berry]

Triaxyz

[crystal lintel]

i would like to suggest $$\sum_{n=1}^\infty \left|\frac{\cos(n)}{e^n}\right| \leq \sum_{n=1}^\infty \frac{1}{e^n}$$ instead

[elfin berry]

sl²a²y

[crystal lintel]

but i would also like to suggest not splitting it up in the first place

just use the fact that $0 \leq1+\cos(n)\leq 2$ for any $n$

[elfin berry]

sl²a²y

[crystal lintel]

also no

[agile idol]

?

you mean adding for example a convergent value to a divergent value doesn't yield a divergent value?

[crystal lintel]

no

i think what you wanted to say was "if exactly one of them diverges then the entire thing diverges"

we can have $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ divergent but $\sum_{n=1}^\infty a_n+b_n$ convergent

[elfin berry]

sl²a²y

[agile idol]

that is true but you didn't answer my question

[crystal lintel]

this one?

[agile idol]

yes

[crystal lintel]

i said no

that is not what i meant

[agile idol]

didnt see the answer mb

[crystal lintel]

@glass silo

[marsh citrus]

@muted solar Has your question been resolved?

[marsh citrus]

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[fading cosmos]

how do i know which formula to use definitional or computational..

[slim spire]

the top one makes more sense to think about but you usually use the bottom when doing problems

[fading cosmos]

so i can use any will that make a difference in my question or no??

also in some questions its not very clear if either they have given frequency or X so how am i supposed to figure if its grouped data or ungrouped

[marsh citrus]

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[lethal galleon]

I need help with part B

[lethal galleon]

My work is mostly on matlab, so ill just post a snippet of my code

[marsh citrus]

@lethal galleon Has your question been resolved?

[lethal galleon]

<@&286206848099549185>

[amber birch]

Use properties of expectations: E[Y] = E[40X - 0.4X^2] = E[40X] - E[0.4 X^2] = 40 E[X] - 0.4 E[X^2]

Then given the expected value and variance of a binomial distribution

You can figure it out

[lethal galleon]

i did that but my answer is still wrong

[amber birch]

What did you do next then?

[lethal galleon]

i understand that the E(X) = nxp which is just 80x0.92

i plug that into Y

[amber birch]

Yeah

[lethal galleon]

and i should get the answer

but apparently its wrong

[amber birch]

Ahhhh $E[X^2] \ne (E[X])^2$

[elfin berry]

south

[amber birch]

Those two are very, very different

[lethal galleon]

right i see

[amber birch]

Do you know the formula Var X = E[X^2] - (E[X])^2

[lethal galleon]

i think i do, i rmb doing something similar for a different question

let me check my notes

[amber birch]

Yeah it definitely should be in your notes

Anyway that's how to find the value of E[X^2]

You know Var X is np(1 - p)

And you know E[X] = np as you told me

[lethal galleon]

ah i see where i went wrong

ok thank you!

[amber birch]

No worries

[lethal galleon]

yep i got the correct answer now

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[orchid wigeon]

hello! I meed help understanding the geometric demonstration of the next problem:

[orchid wigeon]

so

i have find for which r (the radius created by the modulus of z-3i) the system |z|=1 and |z-3i|=r has only one solution

and my teacher said that this is happening only when the circles formed by both the modulus are tangent

and i will get r=2 or r=4

and my question is, what really is the intersection between these circles?

and why do we get the answer by intersecting them?

some materials that explain this would really help me

[cunning jackal]

Essentially all points that satisfy the above given modulus equalities lie on the respective circles. We want there to be a number which satisfies both equalities so it should lie on both circles

If it is lying on both circles, it is nothing but their intersection point

Simple observation will tell you that the circle can either cut at two points, not cut at all, or cut at exactly one point

If they cut each other at exactly one point then there will be exactly point which lies on both circles, hence exactly one z which satisfies both equalities

Hence only one solution of the problem

So we want the circles to 'touch'

[orchid wigeon]

yes … but i still have a question

lets say the circles are not tangent

why is there no z that satisfies the equations?

i mean… two circles are still formed

[cunning jackal]

As I said, the point which satisfies the equation should always lie on the circle

And for it to simultaneously lie on two circles

It HAS to be the intersection point

[orchid wigeon]

do you have any materials reffering this?

i mean, i understand that this right

but something seems off in my mind

considering the circles are tangent in that point

how do i know the same z satisfies the equations?

and the intersection between the circles is only the peak of the modulus

that shouldnt really represent something

but it does

[cunning jackal]

When we say it lies on the circle |z|=1 we mean it's distance from the origin is 1

So it contains all points being at a distance of 1 from origin

i.e. a circle

[cunning jackal]

Why shouldn't it

The circle represents all points at a certain distance from a certain point

And when we say a point is at that very distance from the centre

We mean it's ON the circle

The point where they 'touch' is at a distance of T( whatever that symbol is) from 3i

[orchid wigeon]

bruh i think i lack information somewhere

[cunning jackal]

And at the same time

At a distance 1 from 0

So it satisfies both the equalities

[orchid wigeon]

i mean i understand the equality between two functions represents the points where they intersect

however when it comes to complex numbers , the problems are brain rotting me

[cunning jackal]

Forget about complex numbers for a moment

Think of it like coordinate geometry

[orchid wigeon]

okey

[cunning jackal]

They've given you a point (0,3) and want a point at a certain distance away from it

That should represent a circle

[orchid wigeon]

alright…and my radius is variable

[orchid wigeon]

yes

[cunning jackal]

Yes

And then we have another circle centred at origin

Which represents the first equation

Now say the circles don't touch, that means AMONG ALL the points on the one centered at (0,3), none is lying on the other circle, that means among them none is at a distance of 1 from the origin, that means among them none is satisfying the first equation

It is only when they intersect

That we get a point

Which simultaneously lies on both

[orchid wigeon]

so the intersection would satisfy both circles equations ? (x^2+y^2=1 and x^2 + (y-3)^2=r^2

[cunning jackal]

Yes

[cunning jackal]

You can go ahead and set the D=0 there of the quadratic you get

You'll arrive at the same answer/conclusion

[orchid wigeon]

hm…

something starts to sort in my mind

but it still comes back to the question “ it’s the same complex number that creates both moduluses?

[cunning jackal]

Wdym

[orchid wigeon]

i mean , as you can see , the complex number can be represented as a line in the complex plane. If we take r=2 the circles are “exterior tangent”. and the intersection between the two circles is given by a conplex number that is pointing up and one that is plinting down

pointing*

[cunning jackal]

Not quite, there's only one position vector for one complex number

[orchid wigeon]

now i know im lacking information somewhere

[cunning jackal]

We start from origin

Yeah I recommend clearing up the basics of the argand plane

[orchid wigeon]

ok!

i will look through my 10th year book 💀

thank you for helping me out!

.close

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[lyric kelp]

1/arccos(7) gives

Error on my calculator is thta normal

[sinful thistle]

!show

[marsh citrus]

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

[lyric kelp]

idk how i can show my work

if its just1/arccos(7) gives error

[amber birch]

It's not that

It's -arccos(1/7)

If x = arcsec 7, sec x = 7
cos x = 1/7

[lyric kelp]

ahhh

okay tytyt

[amber birch]

npnp

[lyric kelp]

.close

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[stiff swallow]

I have a little question about variables, suppose I’m solving an equation and I get 2+5x, there the equation can still be solved?, as far as I understand 2 and 5x are different things, so those values can’t be added, subtract, multiply or divide among themselves, please could they help me better understand that part?

[quaint elm]

a couple of things:
2+5x is an expression not an equation because it doesn't have an equal sign
you can simplify expressions but for this one you have the right idea that this cannot be further simplified.

[stiff swallow]

in which cases an expression could be simplified?

[quaint elm]

if you have "like terms" for example 4x + 7x could be simplified to 11x

and no matter what value x is, those two things will be the same

[stiff swallow]

Another example, in the case of 2+5x=7, I think that could be solved by passing the terms on the other side of the same, right?

[quaint elm]

yes, in this case you have an equation and so you're allowed to manipulate each side however you want, as long as you do the same to both sides

so you could subtract 2 from both sides, as you're suggesting

[stiff swallow]

Ok thanks!

Thank you for helping me understand that point.

.close

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[dreamy swift]

Hey, what's the use of injection in set theory ?

[vernal forge]

as in injective functions?

or what injection are tou talking about

[dreamy swift]

in set theory

[vernal forge]

this thing?

[dreamy swift]

yes

[dreamy swift]

I mean...

what is both an example of application and an analogy of injective functions in human interactions?

For example let's consider a table of relations (human number and names) as a function,
10 folks have a name , then it's not injective on an imaginary set I don't know how to define it but that is defind by possible names (some king of regex for the names) : because two different people can have the same name, it can't be injective

But the relation between names and people is a surjective function kinda because one name can correspond to more than one person.

What do you think? How right is it? Do you have practical examples closer to the truth maybe?
In every potential example there should be the mention of a range, to be compared to function types it seems

[marsh citrus]

@dreamy swift Has your question been resolved?

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[drifting ingot]

where am i going wrong in this intergral

[marsh citrus]

@drifting ingot Has your question been resolved?

[drifting ingot]

<@&286206848099549185>

[marsh citrus]

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[drifting ingot]

.reopen

[marsh citrus]

✅

[drifting ingot]

help

pls

[desert dirge]

youre missing a 1/y

this one disappeared in the next line

[drifting ingot]

damn it should be 1/y^2?

[desert dirge]

yeah

[drifting ingot]

tysm

.close

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[visual wolf]

Find the sides of a parallelogram if their lengths are 8:19, and the diagonals of the parallelogram are 15 cm and 25 cm.

[visual wolf]

mistake

[ruby cape]

actually before i give any hints

what exactly are you stuck on

do you need somewhere to begin or are you partway

[visual wolf]

I'm stuck at the beginning

[ruby cape]

ok so

[visual wolf]

I can only say that the sides are 8x and 19x

[ruby cape]

lets say that we have two unknowns, the sides of the parallelogram

yeah perfect

so now its in terms of one variable, yes?

[visual wolf]

yes

[ruby cape]

now you just have to use whatever is in your toolset to attack the problem

hint: Pythagorean theorem

draw altitudes

[visual wolf]

so

from triangle ABD need to find AD?

oh no

but from the triangle I only have 8x

[ruby cape]

ok now

is there some information here that youd like that would help you?

what information would you really like to have

[visual wolf]

I need to somehow get x

Perhaps need to make some kind of proportion?

[ruby cape]

yeah you want x, but if you tried to use Pythagorean theorem to solve for x

what happens?

try to apply Pythagorean here

[visual wolf]

but I only know the hypotenuse

[craggy veldt]

so (BF) is perpendicular to (AD) right?

[ruby cape]

you dont only have the 8x

[visual wolf]

yes

[ruby cape]

you also have the 19x

[craggy veldt]

ok thx

[ruby cape]

and the diagonals

why are you assuming thats the only right triangle to try?

i see lots and lots of right triangles

😉

[visual wolf]

hmmmm

this?

But I don’t know if it’s possible to say that the side is 25:2

[ruby cape]

but is that a right triangle?

if its not a right triangle you cant use Pythagorean theorem

[craggy veldt]

it can be if ABCD is a rhombus

[ruby cape]

except its not

[visual wolf]

this?

[ruby cape]

does that feel like a right triangle?

[visual wolf]

not really

[craggy veldt]

actually i think we can work something out with this triangle

using trig laws or something

because CD is 8x

[ruby cape]

how about try triangle BFD

[visual wolf]

I think this can be solved more simply, but how?

[visual wolf]

this?

[ruby cape]

yep!

this is a right triangle, is it not?

[visual wolf]

but in it we only have BD

[ruby cape]

true, but are you scared of making more variables?

check this out

let BF = h, since its the height

AF = d just because i like that letter here

can you write an expression for FD?

using x, d, h

[visual wolf]

FD=AD-d

[ruby cape]

good, but what is AD

[visual wolf]

19x

[ruby cape]

perfect

[visual wolf]

FD=19x-d

[ruby cape]

also dont forget you can drop altitudes outside the shape too

drop an altitude from C

down to the ground next to AD

extend AD

[ruby cape]

yeah give that bad boy corner a name

[visual wolf]

that?

[ruby cape]

eyy

now look at triangle ACE

[visual wolf]

we only have AC

[ruby cape]

oh really?

BF = CE, no?

arent they both heights?

[visual wolf]

ohhhh I forgot that we decided that it was h

[ruby cape]

aha

how about AE

remember we have x, h, and d now

[visual wolf]

19x+d

[ruby cape]

bingo

so now we have three triangles

BFA, BFD, ACE

all right triangles

write out Pythagorean for each one, you get 3 equations

but look, you have 3 variables!

isnt that promising?

[visual wolf]

hmmm

25^2= h^2+(19x+d)^2

?

[ruby cape]

state Pythagorean theorem

[visual wolf]

?

[ruby cape]

after you have it stated correctly, we can check to see if we applied it correctly

what is the Pythagorean theorem?

what does it say?

[visual wolf]

the square of the hypotenuse is equal to the sum of the squares of the legs

[ruby cape]

yep

but i dont see any squaring in your equation

i see the edit

dont sqrt yet, just keep it as squares

because you are going to be manipulating the equation

its going to be way harder to do stuff when its inside the sqrt

good

now do the other two equations and solve for x

unfortunately i gotta go so

good luck, sorry i cant see this through

[visual wolf]

But I’m thinking, isn’t there a theorem (The sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its two adjacent sides)?

.close

[marsh citrus]

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[lavish granite]

Easy question here, but I don't understand how the answer is C instead of B, I did got 2/5 but apparently I have to add 1/3 and I don't understand why

[dry prawn]

what do you mean "add 1/3"

what method are you using to compute x?

[iron meadow]

did you maybe compute the distance between 1/3 and x? then you would have to add 1/3 to reach what x actually is, if that's what you mean

[marsh citrus]

@lavish granite Has your question been resolved?

[marsh citrus]

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[lavish granite]

i did (7/9-1/3)*9/10

.reopen

[marsh citrus]

✅

[lavish granite]

and got 0.4

[iron meadow]

yeah so do you see why that isn't the answer?

your calculation is correct but it's just the distance between 1/3 and x, now what is the coordinate of x?

[lavish granite]

oh yea i understand that

[lavish granite]

but i dont get why i need to add 1/3

[iron meadow]

so do you understand or do you not

you just said you understand but then you asked a question showing you don't =\

i don't know which to believe

ok, say i have the coordinate x = 2 and another coordinate 5 to the right of it on the x axis, what coordinate is the second one?

[lavish granite]

7

oh

yea

ok

im just not used to doing it with fractions

sorry

[iron meadow]

so same idea, does that clear it up?

[marsh citrus]

@lavish granite Has your question been resolved?

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[red sierra]

May someone check question 13 to see if I did it right

[marsh citrus]

@red sierra Has your question been resolved?

[red sierra]

<@&286206848099549185>

[marsh citrus]

@red sierra Has your question been resolved?

[cunning fiber]

what's your answer?

If I'm not wrong, rotating by pi/2 clockwise means you add pi/2 to the angle

[marsh citrus]

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[faint hare]

Let us call a whole number "lucky" if its digits can be divided into two
groups so that the sum of the digits in each group is the same. For
example, 34175 is lucky because 3 + 7 = 4 + 1 + 5. Find the smallest 4-
digit lucky number, whose neighbour is also a lucky number (i.e. the
whole number next to it is a lucky number as well).

[faint hare]

Let us call a whole number "lucky" if its digits can be divided into two
groups so that the sum of the digits in each group is the same. For
example, 34175 is lucky because 3 + 7 = 4 + 1 + 5. Find the smallest 4-
digit lucky number, whose neighbour is also a lucky number (i.e. the
whole number next to it is a lucky number as well).

<@&286206848099549185>

[sinful thistle]

!15m

[marsh citrus]

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.

@faint hare Has your question been resolved?

[marsh citrus]

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[fringe prairie]

i have to find the zeros of this but I don’t understand

[novel juniper]

zeros of what?

could you type in in TeX

trying to solve $e^{\sin\left(x\right)}\cos\left(x\right)$?

[elfin berry]

ƒ(Why am. I here)=I don't know

[fringe prairie]

noo wait

I have to do it like that

the green box

@novel juniper

[marsh citrus]

@fringe prairie Has your question been resolved?

[marsh citrus]

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[rigid kelp]

how do you simplify this? ((1/x) - (1/8))/(x-8)

[rigid kelp]

$((1/x) - (1/8))/(x-8)$

[elfin berry]

wakamole

[rigid kelp]

how do i simplify this?

[arctic thunder]

you can multiply the top and bottom by x

you could also multiply the top and bottom by 8 to get rid of the fraction

[rigid kelp]

?

how

[arctic thunder]

a/b = 8a/8b

[rigid kelp]

$\frac8/8x - \fracx/8x$

damn that fdidnt work

i got a common denominator

8/8x - x/8x

so thats 8-x / 8x ?

[arctic thunder]

those two things are equal yea

[rigid kelp]

ok then that over x-8

[arctic thunder]

yea

then simplify that

[rigid kelp]

how do i simplify it

[arctic thunder]

(a/b)/c = a/bc

right

[rigid kelp]

idk

that one

[arctic thunder]

ok well think about it

if i divide a by b, and then divide that by c

thats the same as dividing by b * c

right?

[rigid kelp]

b times c?

[arctic thunder]

yea

[rigid kelp]

no

[arctic thunder]

um

are you sure

[rigid kelp]

(4/2)/1

= 2

[arctic thunder]

= 4/(2*1) = 2

[rigid kelp]

ok wait

yea

[arctic thunder]

you can see that its true by multiplying both sides by bc

((a/b)/c) * bc = a/b * b = a

and (a/bc) * bc = a

[rigid kelp]

ok so how do i simplify my equation

i dont get that

[arctic thunder]

at any rate, you can use (a/b)/c = a/bc to simplify your equation

since you have something of that form

[rigid kelp]

i do?

[arctic thunder]

you have ((8-x)/8x)/(x-8)

right

thats what u told me a minute ago

[rigid kelp]

ohhh

i see it now

[arctic thunder]

good

[rigid kelp]

so bring down the 8x

[arctic thunder]

yep

[rigid kelp]

so its 8-x / (8x/x-8)

[arctic thunder]

no

[rigid kelp]

i dont get it then

[arctic thunder]

reread this

[rigid kelp]

oh mulitply

[arctic thunder]

and try again

yep

[rigid kelp]

sry

[arctic thunder]

no worries

[rigid kelp]

8-x / 8x(8-x)

so thats 8-x / -8x^2 - 64

or 8 -x / 8x(-x-8)

?

im actually trying to find the limit as x approaches 8

[arctic thunder]

thats correct

[rigid kelp]

i did this calculation already but wolfram is saying it is -1/64

idk how it got that

i already got that equation but thought i did something wrong

[arctic thunder]

i think so too

i dont think the limit of what you sent exists as x -> 8

wait

maybe im being silly

[arctic thunder]

ok fair

[rigid kelp]

this is fromwolfram

[arctic thunder]

oh

divide the top and bottom by x

perhaps?

[rigid kelp]

from this eq or my simplified

[arctic thunder]

ill think for a sec

[rigid kelp]

ok this is the actual problem

[arctic thunder]

oh

[arctic thunder]

you can factor this out

there is an error here

it should be -x+8 on the bottom

that allows you to cancel it out

[rigid kelp]

oh

wai t a sec

[arctic thunder]

so you have -1/8x

so if you take x = 8, you get -1/64

[rigid kelp]

but it is 8-x on topp

[arctic thunder]

yeah and 8-x on the bottom

[rigid kelp]

and x-8 on bottom

[arctic thunder]

yeah

they differ by a fcator of -1

[rigid kelp]

it is reversed signs

you can still cancel?

[arctic thunder]

so cancel it out and put a -

yea

[rigid kelp]

oh

put a neg on top?

im confuesed how they got the 64

then it would be -1/8x

[arctic thunder]

no

yes

then put x = 8

[rigid kelp]

ha

ok

hold up

omg

that was so confusing

i didnt know you can cancel out if they are revrerse signs

like

for ex

oh ok i get it

[arctic thunder]

well (x-3)/(3-x) = -(x-3)/(x-3) = -1

[rigid kelp]

its not like this (x+4) / (x-4)

its like (4-x) / (x-4)

then u can cancel

right

[arctic thunder]

yea that doesnt simplify

[arctic thunder]

yes

[rigid kelp]

okk

my b

that makes way more sense

ty

[arctic thunder]

np

[rigid kelp]

ok im gna close it now

.close

[marsh citrus]

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[rigid kelp]

i guessed 1 but i have no idea why that was correct

[undone bane]

why are you guessing? you have the hint, work with it.

[rigid kelp]

so i make it $e^x(1+e^-x)) / e^x(1-e^-x))$

[elfin berry]

wakamole

[rigid kelp]

you can cancel those out??

[shadow warren]

it should be $\frac{e^x(1+e^{-2x})}{e^x(1-e^{-2x})}$ instead

[elfin berry]

y0shi

[plain geode]

take e^x out u will have 1+e^-2x/1-e^-2x substitute limit e^-infinity is 0 ur answer is 1

[shadow warren]

since if you distribute what you have up there, it wouldn’t go back to your original result

[rigid kelp]

thank you and why -2x?

[marsh citrus]

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.

[shadow warren]

well try distributing the e^x back in

[rigid kelp]

oh right exponent mult is addition and subtraction

ok but still

how can you cancel them out thoufh

its positve e on top and negative on bottom

[shadow warren]

you don’t cancel out those terms

you cancel out the e^x

and then take the limit

[rigid kelp]

ok so how would you take the limit of whats left over

[shadow warren]

well what’s the $\lim_{x\rightarrow \infty} e^{-2x}$?

[elfin berry]

y0shi

[rigid kelp]

i dont know how to do that yet i just startefd learnin limits last week in class and our teacher didnt teach us with 'e'

[shadow warren]

okay so we can rewrite that as $\frac{1}{e^{2x}}$

[elfin berry]

y0shi

[shadow warren]

by exponent rules

and as x approaches infinity

what happens to the denominator?

[rigid kelp]

gets bigger

[shadow warren]

well no

yep that’s right

[rigid kelp]

i meatn thee fraction getssmaale

[shadow warren]

yep

and as it goes to infinity

[rigid kelp]

so how is that 1 then

[shadow warren]

the value of the fraction would go to what

[shadow warren]

we’ll get there

[rigid kelp]

it would go to 0

oh

ok

[shadow warren]

yep

so if you plug that in whenever you see e^-2x

it will all turn to 0

which leaves only 1 in the denominator and the numerator

which evaluates to 1

[rigid kelp]

what exponent rules did we use here?

to rewrite that

[shadow warren]

oh if we have $a^{-b}$ that’s equal to $\frac{1}{a^b}$

[elfin berry]

y0shi

[rigid kelp]

i dont get how we went from (1+e^-2x)/(1-e^2x) = $\lim_{x\rightarrow \infty} e^{-2x}$? = $\frac{1}{e^{2x}}$

[elfin berry]

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

[shadow warren]

well it’s not necessarily equal to that

we’re just observing what happens to that e^-2x

[rigid kelp]

ok instead of = ->

[shadow warren]

when x goes to infinity

[rigid kelp]

ok

but what happened to the equation

its just e^-2x in the limit all of a sudden

ill have to ask my tutor at school in person might be easier i have to go to the bathrooo,

[shadow warren]

alright

[open kayak]

We factored out e^x

[rigid kelp]

ok that leaves us with (1+ex^-2x)/(1-ex^-2x)

not just e^-2x

[open kayak]

Yes

[rigid kelp]

so why is the lim -> inf = e^-2x @open kayak

[open kayak]

So we have $\lim_{x \to \infty} \frac{1 + e^{-2x}}{1 - e^{-2x}}$ after cancelling out

[rigid kelp]

yep

[open kayak]

Well now think about what happens to those + e^(-2x) and - e^(-2x) terms

As x goes to infinity

Will those get large, will those get small, ...

E.g. is e^(-100000000000000) a big or a tiny number

[rigid kelp]

small

[open kayak]

Yes

[open kayak]

They both go to 0

So this term goes against 1/1

That's 1

[rigid kelp]

but why is it just lim -> inf = e^-2x

#help-33 message

[open kayak]

It's not, he just meant to ask you exactly what I asked you

[rigid kelp]

oh

ok

[open kayak]

"Will it become huge, will it become tiny, ..."

[rigid kelp]

thats why i was confused cause it just left the building

so the equation is still lim -> inf = (1+e ^-2x) / (1 - e ^ -2x)

[open kayak]

e^, not ex^

[rigid kelp]

and you just care about the e / e part

my b

[open kayak]

$\lim_{x \to \infty} \frac{1 + e^{-2x}}{1 - e^{-2x}}$

[elfin berry]

Kepe

[rigid kelp]

yes that

so you only care about the e... / -e... parts

[open kayak]

No, we of course care about the whole thing

[rigid kelp]

cause they have highest order

[open kayak]

We just looked at them to see what happens to them

[rigid kelp]

yea

[open kayak]

And we concluded they go against 0

So they don't matter at all

[rigid kelp]

ok

thx!

i think i get it more now

i still have some questions but im too tired

but i will figure it out

thanks

[open kayak]

np

[rigid kelp]

closing time

.close

[marsh citrus]

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[rugged wolf]

do anyone know how to do these types of questions

[plain geode]

use middle-term splitting method example 1st ques 6x^2-7x+2 multiply Coefficient of x^2 and constant( say some a ) and express coeffecient of x as product of a , so 6 x 2 = 12 and -7 can be written as -4+-3 ,-4x -3 = 12

[rugged wolf]

huh?, can you explain it more i still don't get it

[rugged wolf]

ohhh

i see

tHANk you

[rugged wolf]

but how do you do question24 then

[plain geode]

treat y as constant

do the same thing

x^2 +5xy - xy -5y^2 , here coeffecient (5y) x( -y ) = -5y^2 which is constant term

[rugged wolf]

oh

[rugged wolf]

i do the same process for all questions right

[plain geode]

yea

[rugged wolf]

ok ok THAnk you

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[fallen yew]

could someone help me with the second part?

[devout mauve]

how did your argument go for the first part

for the second part you should carefully look at how you used positivity of the weights

[fallen yew]

First, we show that the subset of edges of minimum total weight that connects all the vertices is a tree. To see this, suppose not, that it had a cycle.
This would mean that removing any of the edges in this cycle would mean that
the remaining edges would still connect all the vertices, but would have a total
weight that’s less by the weight of the edge that was removed. This would contradict the minimality of the total weight of the subset of vertices. Since the
subset of edges forms a tree, and has minimal total weight, it must also be a
minimum spanning tree.

[fallen yew]

I think i have alreadt figured it out

If you have a graph with 3 vertices that is a cycle. and all edges have weight -1 that means the minimum weight would be -3 but this would mean that a tree with 3 vertices would have 3 edges which means it cant be a tree.

I still have to word it better

but this is the general idea

[devout mauve]

yes that works

[fallen yew]

Great 🙂

sorry for wasting your time haha.

have a nice day

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[marsh citrus]

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[true bluff]

how did 4w become 4w^3

[runic temple]

multiply by w^2

[true bluff]

where did they get the w^2?

[runic temple]

to get rid of the w^-2

[true bluff]

u can do that??

[runic temple]

whay is (w^2)(w^-2)

[true bluff]

w^-4?

[runic temple]

no

[true bluff]

oh wait its 0

[runic temple]

no

[true bluff]

oh

[marsh citrus]

@true bluff Has your question been resolved?

[marsh citrus]

@true bluff Has your question been resolved?

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[merry mulch]

Hello. I can't figure out what I'm doing wrong with this 3D summation approximation.

[marsh citrus]

@merry mulch Has your question been resolved?

[marsh citrus]

@merry mulch Has your question been resolved?

[marsh citrus]

@merry mulch Has your question been resolved?

[marsh citrus]

@merry mulch Has your question been resolved?

[marsh citrus]

@merry mulch Has your question been resolved?

[molten moss]

@merry mulch are you sure you did the calculation right?

I'm not a pro on numerical integration but the formula looks right to me?

[marsh citrus]

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[drowsy sundial]

Asked to deduce how therea = 360sin(a/2) (where a= vertical angle) of a cone, while considering the arc length of the sector (l, l behind theta/360 * 2pir)from which the cone is formed, hitting a roadblock

[drowsy sundial]

previous question asked to deduce how r (r being the radius of the base) is equal to s * sina/2 like this (picture) but unsure how to do the other one

[marsh citrus]

@drowsy sundial Has your question been resolved?

[marsh citrus]

@drowsy sundial Has your question been resolved?

[merry mulch]

I may have not been in the correct mode for angles

[molten moss]

lmao sorry fam

i just got suspicious since your answer was so close to the total volume of the 5x5x5 box itself even though cosine should be sufficiently smaller than 1 most of the time

[merry mulch]

Apparently no matter how high of a math level I make it to, the wrong mode is still going to haunt me

[molten moss]

What is theta?

maybe you can draw up a quick picture in paint, that would probably help you and any helper

[drowsy sundial]

yeah sure bear with me

[drowsy sundial]

not the best artist but lmk if this is what you need

[molten moss]

ye i think that helps

Oh so the idea is that you make the cone out of this sector by pushing the edges together?

so the radius of the sector is equal to the side s of the cone, right

[drowsy sundial]

yeah exactly, it's investigating the dimensions that are changed/added when this happens

[drowsy sundial]

yup

[molten moss]

Now I'm just wondering if you're making the cone out of the big sector (with angle theta) or the small sector

[drowsy sundial]

Big one, small one is just discarded

[molten moss]

gotcha! never seen that problem before but i'll think about it a bit

it definitely clears things up

[drowsy sundial]

yeah, pretty interesting 😁, it's an investigation/assignment so not just a problem, this is literally the last part i haven't done, fucking with my head 😹

[drowsy sundial]

cheers

[molten moss]

Were you able to do anything with the "by considering the arc length" comment of the exercise?

[drowsy sundial]

I was trying to do something with it, as part of the investigation was using l = θ/360 * 2pi * r but couldnt get anywhere

and the arc length of the sector is equal to the circumference of the cone's base (C=2piR) but couldnt really get anywhere

[molten moss]

awh you're so close man

[drowsy sundial]

😭 really

[molten moss]

what do you get when you set these two equal?

[drowsy sundial]

uhh θ/360 * 2pir = 2piR

right

yeah i think so

[molten moss]

solid, so getting rid of the 2pi you have theta/360 = R

uuuh with another r

theta/360 *r = R

can you think of any way you can bring your angle alpha/2 into this equation

[drowsy sundial]

uhhh

hm wait

bro i just know im so close i was looking at it like this before

i did something similar to this before i think on another question

fuck

[molten moss]

it's always the same with trigonometric problems

u can do it fam, give it a couple minutes

[drowsy sundial]

well sin(a/2) = R/s

oh wait

hm

shit man i know i can do this i can feel it

[molten moss]

get rid of s

you know what s is

[drowsy sundial]

hmm

[drowsy sundial]

what do you mean by get rid of it

[molten moss]

you already know that s is equal to some other variable

and it's easier if you have less variables in the system, so just replace s with the other variable and forget about s

but i might be confusing you so don't take what i'm saying too seriously hahaha

[drowsy sundial]

nah its alright lol

ohhh

oh i get what you mean

replace s with r?

[molten moss]

aye

meaning yes

[drowsy sundial]

haha alright alright

idk man right now i'm at θ/360 R = R/r

[molten moss]

oh sorry for ghosting you

wait how did you get that

[drowsy sundial]

dw about it

[molten moss]

you had theta/360 *r = R and R/r = sin(alpha/2)

right?

[drowsy sundial]

no yeah youre right i did have it like that

[molten moss]

you just need to bring those two equations together now

[drowsy sundial]

holy fuck yeah i think i got it

wait give me one sec

okok so

theta/360 * r = r * sin(alpha/2)

[molten moss]

based

[drowsy sundial]

therefore theta/360 = sin(alpha/2)

then theta = 360 sin (alpha/2)

right?

[molten moss]

yes sir

[drowsy sundial]

goddamn

that was a fun one 😜

[molten moss]

the geometric construction of this 3d-object out of a 2d circle sector might seem complicated mathematically

but all you needed is to rely on this construction keeping the radius of the original disk the same, and the circumference of the original sector

b/c that translated to r = s and theta/360 = R

super cool & super nice

so you don't always need to understand everything about a complicated system, it's enough to understand what things don't change

the math pros call that the invariants of the system

that was fun, good luck with your math future

[drowsy sundial]

yeah yeah, i have trouble with visualising stuff like this which makes it a little hard but thanks for the nudge, understand it really well now man cheers

[drowsy sundial]

i will 👍

[molten moss]

that's why the drawing also helps!

math problems can't always been drawn easily, but when they can, definitely make use of that