help-10

[timid silo]

i dont think you really get how it works

[valid crown]

welp, I guess I'll check it on my own

plugging some coefficients to see whether that 3 will transform into e

[kind hawk]

I mean the difference between 3^x and e^x is just some coefficient in the exponent

[valid crown]

the best way of maximizing (non including local maxima) is to choose to assign once every 3 growths

dunno why

[kind hawk]

how what works. recursion?

or does the cpp compiler automatically catch cases like that? that could be the case I guess. I don't know cpp that well

[timid silo]

because it is not like fibonaci, it doesnt call previous values to calculate value needed now

[kind hawk]

ah you mean your code. yeah I admit I didn't look into it long enough

not really sure what it does

[obtuse pebble]

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[wise kernel]

I need help with this question

[worldly heron]

do you know how to multiply 2 matrices

[wise kernel]

i think i do

[worldly heron]

then do the right side of the equation

what do you get?

[wise kernel]

I thought i would take away the first matrice from the one on the left

[worldly heron]

yea but you can't subtract a 2x2 from that product yet

you need to multiply it out first

[wise kernel]

okay let me try to do it

i got

0 8
11 8

@worldly heron

@warm canopy

so i took it away

and got

-2 4
9 5

<@&286206848099549185>

[obtuse pebble]

@wise kernel Has your question been resolved?

[twilit ivy]

the 9 should be 13

[wise kernel]

Ok

So after that

Do I get rid of the 2 by dividing x by 2

@twilit ivy

[twilit ivy]

yes divide by 2 on both sides

[wise kernel]

Ok

so would the answer be x=
-1 2
7.5 2.5

[twilit ivy]

13/2 is 7.5?

[wise kernel]

6.5

sorry

[twilit ivy]

yeh then that would be the answer

[wise kernel]

thanks

[warm canopy]

Don't ping people in the future

[wise kernel]

sorry bro

[obtuse pebble]

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[peak roost]

Hi, just learned Lagrange multipliers and coincidently a question in the weekly maths challenge by kings college just posted this one here. Just wondering if you can even use Lagrange multipliers and if so where do I begin, I have x+y<=2 and y^2<=2.

[obtuse pebble]

@peak roost Has your question been resolved?

[peak roost]

<@&286206848099549185>

[peak roost]

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[pallid flame]

bruh he closed as soon as i got a sol.

[pallid flame]

basically start with difference eq x - y = a (we know x is non-negative and having a negative y value would increase difference while changing nothing else from its positive counterpart, so a >= 0)

so y = x - a

sub this back we get x + (x - a) <= 2 and x >= (x - a)^2

we solve these for a

a >= 2x - 2 and a <= x + sqrt(x)

so we want maximum a for x + sqrt(x) <= 2x - 2

meaning sqrt(x) <= x - 2 thus x <= 4

a increases as x increases so x = 4 is max

going back to x + y = 2 we get y = -2

thus 4 - (-2) = 6

max difference 6

@peak roost

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[peak roost]

Got 6

[peak roost]

Ah

Didn’t mean that

Sorry

But nice one

.close

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[normal wind]

what am i doing wrong ?

[normal wind]

this is the question

can't find the answer

[obtuse pebble]

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[obtuse pebble]

@normal wind Has your question been resolved?

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[hushed marsh]

Hi

[hushed marsh]

im wondering why this isnt right

may the math overlords come to my aid :3

[brave bramble]

f(t) = f(t) = t/(4+t)

You've got one too many f(t)s

[hushed marsh]

OH

lmao. this aint a math question no more. I'm just dumb

thanks

👍

I switched this around so many times lol

.close

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[foggy peak]

Would like to be walked through this problem and recieve tips on surface area

[foggy peak]

so 2pi(7)(5)?

[naive owl]

no

read the formula lol

[foggy peak]

isnt 7 a radius

and 5 the height

[naive owl]

yup

read the entire formula

[foggy peak]

ohhh

i thought it said or

2pi(7)(5)+2pi7^2?

[naive owl]

yes

now solve it

remember order of operations 😳

[foggy peak]

im being dumb rn

its taking me forever to solve the equation

.close

[obtuse pebble]

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[teal python]

do these look right?

[nocturne minnow]

You have to wait at least 15 minutes before pinging helpers

[teal python]

mb

<@&286206848099549185>

[obtuse pebble]

@teal python Has your question been resolved?

[teal python]

<@&286206848099549185>

bro someone

hllo?

its been 40 minutes bro someone

.close

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[lost totem]

I have all the calculations done, I was just wondering if anyone could graph this on a calculator since I do not have one.

[obtuse pebble]

@lost totem Has your question been resolved?

[lost totem]

<@&286206848099549185>

[hazy sedge]

I'm having trouble seeing the image because my computer is bad but if you look up demos and use that graphing calculator you can graph it your self to check

[lost totem]

it's all good, I managed to find a website that would graph it, for reference if you want to try and calculating it, null: p1 = p2, alt: p1 < p2 , x1 = 0.3, n1 = 100, x2 = 0.4, n2 = 100, and it's 2-tailed

[kindred spoke]

Try Desmos

Also your image is kinda blurry

[lost totem]

apologies

the problem has been solved, thank you for putting up with me 👍

[kindred spoke]

Okay!

.close to close thread please

[lost totem]

.close

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[bitter ridge]

--

[gilded needle]

can you work out the position as a function of time, for the two particles?

[bitter ridge]

here's what i got so far

[gilded needle]

,rotate

[bitter ridge]

did i do it correctly?

[gilded needle]

yeah, looks right as far as i can see

so now just set the the two s(t) formulas equal to each other and see if there is a solution

[bitter ridge]

thank u!

.end

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[worn saffron]

Hi, I'm trying to convert some code into a function I'm able to transpose to work out values,

the code's purpose is rudimentary projectile motion, i've already done all the maths with the code however in this instance I wish to make a range chart that allows users to plug in values to determine range, effect of wind speed and such.

Direction = (Barrel.lookVector*420) - (Vector3(0, (60), 0) * dt)
bulletPos = bulletPos + Direction * dt

lookVector is a is a 3d vector value as well as the the Vector3(0,60,0)

[restive pilot]

No stress ^^

[worn saffron]

there

dt represents delta time, in context; time lapsed since last last calculation

please ping me on any responses

[worn saffron]

should be all done

[restive pilot]

Im not sure what your question is now

[worn saffron]

I wish to work out the range of the projectile using the above formula, as my code uses raycasting, instead of a set formula

probably worded it terribly, sorry been at this for a while and its burnt me out

[restive pilot]

Not sure what this means 😬

[worn saffron]

it might help if i produce a diagram

of how the code works

[restive pilot]

Yes 👉 👈

[worn saffron]

does that make more sense?

however the product is in 3d, but as the mortar fires roughly in a straight line (spread not needing to be calculated into distance), i simply need to work out x distance

and the reason i said a formula is because

the 420 value is just a numerical value in this example, but will be interchangeable as it represents the projectile velocity and same with 60 as it represents the drop per portion

@restive pilot

[restive pilot]

It should work like this based on my coding experience

[worn saffron]

no it does

so it moves the projectile and such

but i want a way for the user to be able to know the distance of their projectile

as the origin look vector

will change

[restive pilot]

Okay how often is dtbapplied

[worn saffron]

its every frame stepped

[restive pilot]

So 60 times a second?

[worn saffron]

yes

i assume

[restive pilot]

Okay then you need the start height

Okay for simplicity dt=1/60

And then its an if case for range.
You say max distance is when it touches the ground

Which is the last value before you go negative

Just lookisng at y variable in vector (x, y ,z)

[obtuse pebble]

@worn saffron Has your question been resolved?

[restive pilot]

@worn saffron got something let me write it down

[worn saffron]

alright thank you

um no in this case, simply the Z vector

[obtuse pebble]

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[worn saffron]

.reopen

[obtuse pebble]

✅

[worn saffron]

im sorry some of its not too legible

[restive pilot]

Sorry 😬

[mighty ether]

would someone be able to help me with this?

i am so confused

ah crap

my bad

channel taken

[restive pilot]

@worn saffron shouldn't the length be just 2*y?

I might how overthought it

[worn saffron]

yeah

[restive pilot]

Cause you go y steps until you reach midpoint

[worn saffron]

so i need to work out the end vector

[restive pilot]

And the same amount to get to 0 again

[worn saffron]

no, it'd be z steps

[restive pilot]

💀

[worn saffron]

as Y is vertical plane

[restive pilot]

Okay

[worn saffron]

the distance between the origin vector

and then and projectile landing

(if on a flat surface)

[restive pilot]

Now you confusend me again

[worn saffron]

if the origin was on a flat plane

and the trajectory was not effected by an unflat surface

what is the point of intersection between y = 0

[restive pilot]

Yeah should be 2*height

[worn saffron]

x = 2*height?

[restive pilot]

If you do this the need to aubtract the variable that is responsible for the height dimension

Like you got x,y,z in your coordinate system

[worn saffron]

yeah

[restive pilot]

And let x and y be the ground

The z is your height

[worn saffron]

im getting confused

oh i might've forgotten to state

the projectile could orientate left and right

so it would be the x and z value, y value wouldn't matter as thats the height

[restive pilot]

So in (x,y,z) y is your height correct?

[restive pilot]

The range is = 2 * y

Wont change

Not dependent on z

As z is not change throughout your algorithm

[worn saffron]

so this graph is a result of

y = -x^2 + 10

how do i find out the height

what about this @restive pilot distance = ((initial_velocity^2)(sin(2 * angle))) / drop

thats the function for actual ballistics

[restive pilot]

The +10?

[restive pilot]

Ugh

[worn saffron]

thats just to show it above the line

because if it doesnt have it

its just intersecting with x=0 at y =0

actually

dont worry i think i worked it out

thank you for the help

.close

[obtuse pebble]

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[static kiln]

Hi. I'm searching for the name of this curve. The idea behind it is : the average of mark of the group " terminal " is ~4 and the average mark of the group " polytech " is ~9, so it's like the curve ascend when they have many peoples at this moment

[static kiln]

I don't know if it's clear sorry

[obtuse pebble]

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@static kiln Has your question been resolved?

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[kind nexus]

There are two cars going towards each other from city A and city B. After they met, they spent 27 and 12 min to go to city B and city A. How much time did each car spent to go from city A to B or city B to A

[kind nexus]

This is what i have tried so far

[clear cloak]

City a to b takes 27minutes after they met?

[kind nexus]

Yes

For the first car

City b to a takes 12 min for the second car

After they met

[timid silo]

,rotate

[valid crown]

check those

[obtuse pebble]

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[spare current]

Wondering if i messed this derivative up or is it good?

[timid silo]

Why don't you cross multiply the equation?

[spare current]

what do you mean? with what?

[timid silo]

You have,
y = (1+y)/(1-y)

On cross multiplication,
y(1 -y) = 1 + y
y - y² = 1 + y
1 + y² = 0

Now differentiate it

I think the appropriate question might be to differentiate f(y) = (1+y)/(1-y)

What's the actual question btw?

It's just to differentiate 1+y/1-y. right?

[spare current]

well the problem is to find the 2nd derivative but i think i gotta get the first one and then the 2nd.

[timid silo]

@kind nexus
this is wrong though

[valid crown]

where was an error?

oh I see it

[spare current]

this just seems a bit unconventional for me, dont even know where i would getting the 2nd derivative from there

[valid crown]

that's not neutral element

[timid silo]

the speed of the cars are 30km/h 45km/h the distance between AB is 22.5 km

[valid crown]

I can only multiply both denominator and the upper one

[timid silo]

This channel is occupied

[timid silo]

well atleast this works as there can be other options too

[timid silo]

who cares his question was last

we point out error

[timid silo]

What's the actual question?

[spare current]

the question is whether that approach is right

[timid silo]

What do you want to differentiate?

Only right side?

Of equation.

[spare current]

yes

[timid silo]

Last step is wrong

-(1+y)(-1) = 1+y

i.e.
(1-y)(1) - (1+y)(-1) = 1-y + 1 + y

[spare current]

alright, thanks.

.close

[obtuse pebble]

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[timid silo]

i did it like this:
S=AB
v1 - car going from A speed
v2 - car going from B speed
t - time it took them to meet

t=S/(v1+v2) > S=t(v1+v2)
27=(S-tv1)/(v1) > v1=S/(27+t)
12=(S-tv2)/(v2) > v2 = S/(12+t)
S=t((S/(27+t)) + (S/(12+t)) divide by S
1=t((1/(27+t)) + (1/(12+t)) from here t=18
so t1=18+27=45, t2=18+12=30
@kind nexus

[obtuse pebble]

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[tacit barn]

i don't quite understand this question

[tacit barn]

i can start it off but i don't know how to continue it

[timid yacht]

how did u start it?

[tacit barn]

i basically multiplied everything buy un+1

so i got

Un+2/Un+1 = Un+1/Un+1 + Un/Un+1

[timid yacht]

we need the limit

so let x= lim u_n+1/u_n n->oo

and we can split u_n+1

[tacit barn]

why is it n- ??

[timid yacht]

?

n goes to infinity

its an arrow

[tacit barn]

oh but u placed a minus

so i was wondering why it was negative

[timid yacht]

so what is the next step?

[tacit barn]

Hahah that’s the part I’m not quite sure??

[timid yacht]

did u split it?

cuz then you would see some hint in the fraction

@tacit barn

[tacit barn]

sorry im here

[tacit barn]

okay let me try

so U_n+2/ U_n+1 --> oo

but for the question

it would be

that symbol lol

idk what it is

[obtuse pebble]

@tacit barn Has your question been resolved?

[tacit barn]

<@&286206848099549185>

[compact shadow]

You can solve u_n

The characteristics polynomial is p(λ)=λ^2-λ-1 which has roots (1+sqrt(5))/2 and (1-sqrt(5))/2

So $u_{n}=a(\frac{1+\sqrt{5}}{2})^{n}+b(\frac{1-\sqrt{5}}{2})^{n}$ for some constants a and b

[warm shale]

Cogwheels of the mind

[compact shadow]

Use u_1 and u_2 to solve a and b

The rest is clear, limit can be calculated by definition

[tacit barn]

ohhh okay

i got the answer needed

thank youu

.close

[obtuse pebble]

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[tough hornet]

help

[tough hornet]

<@&286206848099549185>

[tough hornet]

<@&286206848099549185>

[timid silo]

If you have this product of binomials, if my english terminology is correct then im guessing it looks something like (ax + c)(bx+d) with x the variable. Multiplying ax by bx should definitely get you 4x^2, so you know they should both be? Thats right, 2x. Now its just a matter of choosing c and d.

[tough hornet]

i don't get it

[timid silo]

I think i get it, i could be wrong

First and foremost, do you get all the terminology used to ask the question?

[timid silo]

Of this one

[tough hornet]

a little

[timid silo]

If you dont know what it means for it to be factored then its gonna be somewhat harder of course

[tough hornet]

i did this on, can i show u

[timid silo]

Nice, can you send the work?

Or do u first wanna work through the first picture

[tough hornet]

Just wanna know if it's right

hello

[timid silo]

Hmmm

[tough hornet]

wrong?

[timid silo]

Think so

The forth line

[tough hornet]

im kind of rushing so im forgeting things

[timid silo]

Or no

Nvm

Brb

So u solved for

When is the volume zero right

How would you determine the length width and height from that? Or did u just wanna solve for the roots and didnt care about the question

I mean idk much about rectangular prisms tho

But i can defo talk u thru the first picture

[tough hornet]

im supposed to answer the question

[tough hornet]

ok please

[timid silo]

Ahh

Aight

[tough hornet]

only have 6 mins lol

[timid silo]

So what terminology is still fuzzy?

Oof

So in short:

We have this polynomial and to say that it is factored as a product of two binomials (i could be totally wrong cause im not taught in English) means that theres two binomials (ax+c) and (bx+d) and their product is the polynomial.

Clear on that?

Can I go on?

[tough hornet]

yes

[timid silo]

So the product of two binomials is the polynomial like this (ax+c)(bx+d) = polynomial

If im not wrong

And then, for sure a and and b gotta be 2, right?

Cause if you multiply then 2x times 2x is 4x^2

[tough hornet]

yea

[timid silo]

Another thing is the product of c and d is 5

Now what should k be

We know that if we multiply the things we get axc + bxd

Uh

Ax c + bx d

It keeps messing w the * but u get the picture

So we wanna know k

And axc+bxd = kx

And if we factor the left we get

x(ac+bd)

Doesnt that look familiar

Its x times (something), or just kx

So k should be ac+bd i guess

A and b are definitely both 2, remember?

And the product of c and d should be 5. Well, its easy to find two numbers whose product is 5.

Wait actually sorry

Nvm its good lol

Yeah so

In other words

k = 2c+2d (because k=ac+bd)

And cd=5

We can factor out the two

K= 2(c+d)

cd = 5 so c=5/d

We can substitute that in

$k=\frac{5}{d} + d$

[warm shale]

jb

[timid silo]

And ofc d could be any ol number

I think

Did you get it? :)

A bit of discovery and playing around was involved on my part so i hope u can follow

First and foremost did u get the key things? Like why c*d = 5?

[timid silo]

And how we discovered this?

I guess time ran out :,

So your answer should read: all possible values of k are determined by $k=\frac{5}{d} + d$ with d any number at all. Or something like this

[warm shale]

jb

[obtuse pebble]

@tough hornet Has your question been resolved?

[obtuse pebble]

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[shell tundra]

125x^3-64

i need help factoring this.

[warm canopy]

are you trying to find the roots?

[shell tundra]

I need to factor this by putting it in 2 brackets. so should look like this

()(___)

[warm canopy]

is the question exactly "put this into two brackets"?

[shell tundra]

It is "factor this completly"

[compact shadow]

a^3-b^3=(a-b)(a^2+ab+b^2)

a=5x, b=4

[shell tundra]

oh..... okay i get it.

just realised that 64 is a perfect cube.

thanks

.close

[obtuse pebble]

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[timid silo]

Hi! I need someone to explain the steps to me of this problem. The answers are shown but I am confused on which step goes where and what to do in each step

[timid silo]

Currently I'm confused on how you would get 1.0485

[drifting wraith]

first column then second

divide by 103

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

oh okay thanks

[obtuse pebble]

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[lean spear]

why does the 81 go to the bottom>

[timid silo]

$3^{-4}=\frac{1}{3^4}$

[warm shale]

Pi Creature

[timid silo]

Which takes it to the bottom

[lean spear]

ooooo

.close

[obtuse pebble]

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[timid silo]

I'm slightly unsure on how to work this out

[subtle adder]

volume of pyramid: 216 cm^3
volume of prism: 504 cm^3
volume of composite figure:720

[warm shale]

☙ Kiwwi ❧

☙ Kiwwi ❧

[subtle adder]

total volume of composite figure =

[warm shale]

☙ Kiwwi ❧

[timid silo]

I done the formula and get 37.70

but it's wrong

[subtle adder]

for this?

[timid silo]

yep

[subtle adder]

volume of a cylinder is pi*r^2*h

and volume of prism = length*breadth*height

volume of composite figure = volume of cylinder + volume of prism

[timid silo]

so 37.68

[subtle adder]

how though?

oh wait

yes

37.68

[timid silo]

Yes

[timid silo]

now i gotta solve the thing under

[subtle adder]

yes

easy enough

6*5*7

[timid silo]

so that is this which is easy

[subtle adder]

210

210+37.68

247.68

[timid silo]

yep

I got like 120 questions to go through lol

Do you know what the formula is for these 2 shapes?

I think it is volume = width x height x length

@subtle adder do you know the formula for this one because it keeps coming up wrong.

[subtle adder]

they are asking the SURFACE area

not volume

[timid silo]

So the top shape is 150
and bottom is 404

so it'll be 554

damn it's wrong

[subtle adder]

you mustn't count it like that

you can't simple add them up

because there is a side that is covered

you must count the surface area as in the visible part, or uncovered area

[timid silo]

How do I minus the covered area? or only count the covered area. Can you show me the steps if you got time.

[subtle adder]

take the blue cuboid

add up the surface area of the visible sides ONLY

not the bottom side

your generic 2(ab + bc + ac) won't work

[timid silo]

I'm stuck 😫

[subtle adder]

or you could just subtract the covered area by the total surface area

[subtle adder]

alright i'll solve it

let's take the top cuboid alone

[timid silo]

Can you how me how you did it after you solve it

[timid silo]

yep

[subtle adder]

find the total surface area of the top cuboid

what do you get

[timid silo]

150

[subtle adder]

now find the surface area of the covered area

which is a normal rectangle

of dimensions 11*5

we have the length of the cuboid because; length of top cuboid = length of bottom cuboid

by length i mean this

[timid silo]

Is that 55 for one side?

[subtle adder]

this part

[subtle adder]

yes

55 for the covered side

150-55?

95

[timid silo]

yep

[subtle adder]

95 is the area of the cuboid WITHOUT the covered side

[timid silo]

so i minus 55 again for the bottom one that's covered

[subtle adder]

for the bottom cuboid, notice the covered area is the same

[subtle adder]

yes

404-55

444

[timid silo]

so in total 444?

yes

[subtle adder]

is it the correct answer?

[timid silo]

Nope

😭

lol

[subtle adder]

what is it then

[timid silo]

11 x 3 = 33

[subtle adder]

hold on i'll calculate surface area of each shape

[timid silo]

It is 488!

I got it!!!!!!

Thank you!

[subtle adder]

im getting the surface area of the top cuboid as 206

[subtle adder]

alright then

[timid silo]

Yep I'm trying the next one now on my own to see if I get it right

66.6 is the pyramid

• 16 to remove the square covered

so should be 50.6

nope it's incorrect

.close

[obtuse pebble]

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[main cargo]

for a 3x3 plane

[main cargo]

is there a standard direction that the z axis goes

so

the x axis increases as you go right

the y axis increases as you go up

for the z axis, is it increasing as you go into the page, or out.

aka

is the one on the left correct, or the one on the right

[nocturne minnow]

Right hand rule

[main cargo]

which finger is which axis?

you can make both correct

[nocturne minnow]

You can't make both

[main cargo]

uh how can i make both

[nocturne minnow]

But I'm pretty sure index is x, middle is y, thumb is z

[main cargo]

i can literally make both

[nocturne minnow]

You can make both but arrows would be different

[main cargo]

uh

what

well it dont help then does it

is the one on the left correct or right?

if u just tell me which one ill prolly remember forever

[nocturne minnow]

Using the right hand rule, the right image is the proper way

[main cargo]

but if you make index point away, thumb go right, and middle go up

then you make the one on the left

[nocturne minnow]

For the z axis to point like the left, y axis is down and x is left

As I mentioned, x is index finger, y is middle, and thumb is z

So having your index and middle point, the only other option for z is the right image

[main cargo]

ah i get it, thanks for the help 🙂

.close

[obtuse pebble]

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[timid silo]

I don't get this, please help.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[heady kiln]

Could anyone help me with substituing simultaneous equations ? I can do them pretty easily but once there's fractions involved I get stuck. Like for example 4x-3y=5 and x^2 +3xy=10

[heady kiln]

Sorry but... <@&286206848099549185>

[obtuse pebble]

@heady kiln Has your question been resolved?

[long sinew]

$y = -\frac{5 - 4x}{3}$

[warm shale]

Breeziboi

[long sinew]

Doesn't the denominator cancel out when you substitute it into the second equation?

[obtuse pebble]

@heady kiln Has your question been resolved?

[obtuse pebble]

@heady kiln Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[dark vapor]

if i have a function with a domain (-inf, inf) would this function's domain be open or closed?

[dark vapor]

i'm thinking open since it uses parentheses, but infinity might be an exception to that rule?

[neat valley]

what is your definition of openness

[dark vapor]

if i have a set like [8, 10] that's closed since 8 and 10 are included in the set, but if it was (8, 10) it would be open since 8 and 10 are not in the set

but with infinity i guess infinity is in the set? thats why i'm thrown off

or i guess to be more specific, my domain is the entire xy-plane, this is multivar calc lol

[obtuse pebble]

@dark vapor Has your question been resolved?

[dark vapor]

no 😔

.close

[obtuse pebble]

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[neat valley]

.reopen

that didnt work lol

[dark vapor]

.reopen

[obtuse pebble]

✅

[neat valley]

but it's actually both open and closed technically

there's a formal definition of openness you'll learn in analysis

[dark vapor]

ahh so infinity is an exception

[neat valley]

clopen

if you're thinkin about the xy plane it's probably better to think about $\mathbb{R}$

[warm shale]

μ₂

[neat valley]

the set of all real numbers is probabky what you mean by infinity and that set is both open and closed

[dark vapor]

that does make sense

thank you. one more question to go with that one if you dont mind? about bounded vs unbounded

bounded being if the domain is bounded by some function, bounded would mean everything inside the function or below or above or smth, literally bounded

and unbounded would be outside the bounded area right?

so I think all real numbers would be unbounded. is that the right way to think about it?

[neat valley]

yeah i dont think you need to think about boundedness too hard for what im assuming is early college calc

though these are worthwhile questions you'll circle around to if you take analysis

[dark vapor]

okay, thank you so much! I think I get it now. I think I was just thrown off because my domain is all real numbers

.close

[obtuse pebble]

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[timid silo]

can someone help me with this im kinda confused
Jacob draws a line that passes through the point A(-2,8) and has a steepness of -3. What is the equation of the line that Jacob draws? Show your calculations.

y=-3x-2, y=-2x-3, y=-3x+2, y=2x-3

[naive owl]

use point slope form

cause u have a point

and a slope

https://tenor.com/view/clash-royale-mini-pekka-dance-awesome-emote-gif-22234774

what have you tried tho

[timid silo]

im actaully a dumbass

tysm man

.close

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[teal blade]

#3 pls

[naive owl]

what have you tried lol

[teal blade]

im braindead

[obtuse pebble]

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[obtuse pebble]

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[naive owl]

can the function x! be approximated or matched by a^b

[naive owl]

when the domain is limited to 2 and above

why or why not

just curious lol

[brave bramble]

See Stirling's approximation

It can't be approximated with only an exponential, it grows too fast

[naive owl]

woah that's cool

,w plot \frac{x!}{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}}

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chunkin#4952

[naive owl]

it converges to 1 interestingly

[obtuse pebble]

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[coral shale]

Hi can someone answer number 3 with complete solution while I answer 1-2

Deadline is approaching so I'd really appreciate it if you would help me

[obsidian isle]

1-2 = -1

[obtuse pebble]

@coral shale Has your question been resolved?

[coral shale]

<@&286206848099549185>

[obtuse pebble]

@coral shale Has your question been resolved?

[coral shale]

<@&286206848099549185>

[cedar lichen]

We don't just give answers

[coral shale]

Well... I'm done with the problem. Thanks for letting me use the channel tho?

.close

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[native basin]

Can anyone explain to me how to write an equation for a hyperbola given this?

[obtuse pebble]

@native basin Has your question been resolved?

[tardy epoch]

https://www.mechamath.com/precalculus/asymptotes-of-a-hyperbola-formulas-and-examples/

[obtuse pebble]

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[real mist]

i got all the factors of 2000

and im kind of stuck there

[royal basin]

you want to count those factors of 2000 which are strictly greater than 8

[real mist]

oh

then its 15 right?

[royal basin]

(otherwise N isn't big enough for a remainder of 8)

15... can you list them?

just to ensure you haven't forgotten anything by mistake

[real mist]

ok

10,16,20,25,40,50,80,100,125,200,250,400,500,1000,2000

these are it right?

[subtle adder]

correct, it's 15

[real mist]

oh thanks :)

so now i have to close this channel?

[royal basin]

well you can either ask some follow-up questions if you have them

or if not, then type .close to close the channel

[real mist]

not right now but i might ask them in future.

.close

[obtuse pebble]

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[meager osprey]

Hello is this proof correct or do we need to expand further?

[obtuse pebble]

@meager osprey Has your question been resolved?

[meager osprey]

<@&286206848099549185>

[obtuse pebble]

@meager osprey Has your question been resolved?

[real mist]

damn my guy is not gettin any helpers

[compact shadow]

You wrote it yourself already

|(sqrt(a)-sqrt(a_n))(sqrt(a)+sqrt(a_n))|=|a-a_n| <ε

and for n large enough you can let max{a-1,0}<=a_n<=a+1

So sqrt(a_n)+sqrt(a)<=sqrt(a+1)+sqrt(a)=C is a constant

So |sqrt(a)-sqrt(a_n)|<ε/C

You can replace ε with Cε at the beginning

[meager osprey]

Thank you

I was just wondering if the way I wrote it and what I wrote is okay and isn’t missing anything vital to make this proof stand it’s ground

I’m not very good at writing proofs and knowing what to put in and what to leave out

I forgot a conclusion to make it complete I guess

[compact shadow]

No problem

[meager osprey]

.close

[obtuse pebble]

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[cyan thistle]

Evaluate in scientific notation using three significant figures: (5.04)^-4

[cyan thistle]

i have no idea how to do this

if someone can write a step by step solution

[nocturne minnow]

Calculate the number, put it in scientific notation, then round to 3 sig figs

[cyan thistle]

.close

[obtuse pebble]

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[timid silo]

Im trying to solve for the derivative so far this is what i’ve done

[strong vale]

ok

lets see

[royal basin]

before you continue theres a big issue with what youve written here

[timid silo]

Yeah i had a feeling it was wrong but I continued on bc I didn’t quite get it

[royal basin]

your first line looks a lot like $f(x) = 1 + \sqrt{x^2 + 3x} + 6$, which is wrong bc it denotes a different function than what you're working with.

[warm shale]

Ann

[royal basin]

this is purely a handwriting issue

you should make sure that the radical symbol extends to cover everything that goes under it

it may be good to put a little hook at the end to make it not want to retract

[timid silo]

Ohh ok i’ll remember that next time

[royal basin]

other than this, you have set up the limit correctly so far

[timid silo]

Do i just work it out how i usually do and solve?

Oh wait it wont work nvm

[royal basin]

well plugging in x=2 won't work as is

you need to do some simplification

maybe some conjugate fuckery

[timid silo]

Yeah i just realized

Im trying to use 2 examples our teacher gave to solve

But i have no idea how she got some of the things in the equation

So im stuck where i left off in that photo

[strong vale]

so u havent learned any differentiation techniques? only limit definition

[timid silo]

Nope

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[silk pond]

which part is causing you trouble

[timid silo]

Idk how to continue what i started on

[timid silo]

Here

Like idk the next step

[obtuse pebble]

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[silk pond]

.close

[obtuse pebble]

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[obtuse pebble]

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[meager sorrel]

How do I know what value to use for ii

[night igloo]

You can try x = 10 I guess

The expression will be -10/729 but then you can multiply the approximation by another -10

Which should be easy enough if you're asked to perform a decimal approximation

Though x = 10 isn't really that small...

Yeah, not a good idea, nvm

Ah yeah, x = 1/10 works

Which is also perfect for doing decimal calculations 😌

[obtuse pebble]

@meager sorrel Has your question been resolved?

[meager sorrel]

alright thanks

[obtuse pebble]

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[grave halo]

When integrating by parts integral(u dv) = uv- integral(v du)) what is the elegant way to solve when the original integral reappears in the integral part of the result

[grave halo]

I remember there was a way to solve this but i have forgotten it

for example when integrating e^x*sin(2x)

[warm canopy]

Call the integral I

And rearrange for it

[grave halo]

ok will try thanks

works cheers

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[rain socket]

Hi, how do i solve a question like this?

[rain socket]

the solutions to the equation: sin(2θ) = sin(θ) for -Π/4 < 0 < Π/4

[haughty coyote]

Use double angle identity

Or the fact that sin is injective on that interval, if you've seen what that means

[rain socket]

yeahh ill just stick with the double angle identities lol

[haughty coyote]

Culture point: injectivity means no different inputs map to the same output. This in particular is true or strictly monotonuous (i.e. increasing or decreasing) functions, which is the case of sin here.

[rain socket]

Well we've never went over that deep into functions

but i do know sin(2θ) = 2sinθcosθ, so does that mean 2sinθcosθ = sin(θ) in the question?

[haughty coyote]

That's why I said that as a bonus. The more theory, the less computations needed. When you don't have the theory, you just show it "by hand"

[haughty coyote]

Yes, and you solve for theta

[rain socket]

ah i see

so is the question asking for the solutions in that specified range?

-Π/4 < 0 < Π/4

[sonic fog]

yes

[haughty coyote]

Theta IS the solution

Finding theta is solving here

[rain socket]

ah i see, then my issue is not knowing how to solve for theta

[haughty coyote]

Two cases: sin theta = 0 or not

[sonic fog]

I suggest you subtract sin(θ) from both sides and then factor it

[haughty coyote]

Sounds like a more complicated way of doing it, though only slighlt

[sonic fog]

Nope, that is not how it works

[rain socket]

ah i figured

oh wait

[haughty coyote]

Yes

[rain socket]

Alright thanks for the help guys i think i know what to do

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[obtuse pebble]

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[zealous sable]

I was just wondering if this was congruent and if so does anyone know if it's SSS, Asa, Sas, AAS, AAA, SSA

[narrow hawk]

it is... by ASA property

[zealous sable]

Alright, Thank you so much, have a great day!

[narrow hawk]

👍

[zealous sable]

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[cursive solar]

I don't understand what the question is asking me to do. the translation is "Find the partial sum of the following series and find the limits of said sums"

[cursive solar]

It's practice tasks before an exam - What would be a partial sum in this context?

[royal basin]

the sum of the first N terms

[cursive solar]

isn't that given?

Maybe you can help me out with a)?

[royal basin]

no, what you're given is the terms themselves

not their sums

[cursive solar]

ohhh its a latex error, right?

its not actually sum of n equals

[royal basin]

you need to find $\sum_{n=1}^{N} \frac{n-1}{n!}$ in terms of $N$

[warm shale]

Ann

[royal basin]

and yes part a has a latex error even though it is trivially fixable

[cursive solar]

Right! I didn't udnerstand it lol

Okay

Thank you

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[obtuse pebble]

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[cursive solar]

Actually, I could use some more help on it. This is what I've arrived at

[cursive solar]

Don't know any neat trick to simplify the denominator, the numerator just turned into an arithmetic sequence

So I guess what I'm looking for is $\sum_{n=1}^{N}\frac{1}{n!}$

[warm shale]

Meduza3

[cursive solar]

I relaize my huge mistake

[royal basin]

you are confusing factorials with squares

[cursive solar]

I am. Let me quickly fix that up

[royal basin]

2! is 2, not 4
3! is 6, not 9
4! is 24, not 16
5! is 120, not 25

also you can't add fractions the way you tried to

[cursive solar]

Haha, oh my god so many mistakes

Thanks for pointing that out

[royal basin]

anyway (n-1)/n! = n/n! - 1/n! = 1/(n-1)! - 1/n!

so you will have a telescoping series

[cursive solar]

So its recurrent?

I can solve it using the Recurrence relation?

[obtuse pebble]

@cursive solar Has your question been resolved?

[obtuse pebble]

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[rain coral]

I don’t rly get 2c

am I supposed to integrate it twice

so I can set u=1+sin2x

then find the du/dx which is 2cosx?

[timid silo]

Try u=2x

[rain coral]

ohhh

then I can expand the bracket?

wait last question

to integrate sec^2usinu

should I use integrate by parts

or is there a short cut

oh nvm

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[dark vapor]

can someone help me with this? this is the problem I have and how I'm answering it. I'm not sure where I'm going wrong

[dark vapor]

the first answer is the answer to when it hits the ground, the second is how far it travels

wait a second

i think

omg

i messed up

quadratic formula

-b

that was

stupid

[dark vapor]

that was rlly it wow

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[obtuse pebble]

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[shadow phoenix]

anyone know how to find the measurements listed below?

[shadow phoenix]

I'm pretty confused with it

[nocturne minnow]

Do you know what $\cong$ means?

[warm shale]

dldh06

[nocturne minnow]

@shadow phoenix

[shadow phoenix]

no I dont

wait

nvm i do

[gloomy trellis]

Why can't u used sine law here?

[nocturne minnow]

No need to

It's congruent shapes

[gloomy trellis]

Oh I see

I didn't see that

[nocturne minnow]

Meaning if you line up the shapes, properly, for example (because this may be wrong), angle R = angle T

[shadow phoenix]

so how would I find the different measurements

[nocturne minnow]

It's all given, you just need to match it up properly

The only one is just one angle, and that's found, by using the concept of sum of angles in a triangle equals 180

[shadow phoenix]

so I could go like UV=PR

to line up

[nocturne minnow]

Yes

[shadow phoenix]

Okay so I'm not sure if this is correct but could the measurement of r be 75?

thank you guys

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[obtuse pebble]

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[wide oriole]

hi

[grizzled eagle]

That's right.

[obtuse pebble]

@wide oriole Has your question been resolved?