help-13

[hot crag]

that is, as you get closer to 0

the graph continues to get closer to neg inf

and as you get closer to x = inf

the graph continues to get closer to 0

that is

[signal ibex]

alr thanks

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[signal ibex]

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✅

[signal ibex]

@pastel vault if you read the graph left to right wouldn't that mean that x is always going to be getting bigger?

[pastel vault]

This graph?

[signal ibex]

yea and the second one too

[pastel vault]

No, there's a turning point, so after x = 2 the graph gets smaller

Well we should look at the absolute values, the distance from y= 0

This graph has closer to 0, further from 0, then closer to 0

[signal ibex]

no I mean x

[pastel vault]

This graph has further from 0, closer to 0, further from 0

[pastel vault]

Oh right, yes x always gets bigger for any graph

That's how you should read any graph

[signal ibex]

so only focus on y when graphing this stuff?

[pastel vault]

Yes

[signal ibex]

ok thx

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[sturdy rover]

Consider the vectors a and b such that a = 12i - 5j and |b| = 15
(a) Find the possible range of values for |a + b|

[sturdy rover]

What i've tried so far is

i defined b as:

b = <x, y>

This means that |b| = sqrt( x^2 + y^2 ) = 15

and x^2 + y^2 = 225 (i will use later)

so the vector a+b = <12+x, y-5> has magnitude |a+b| = sqrt[ (12+x)^2 + (y-5)^2 ]

expanding the magnitude expression: sqrt [144 + 24x + x^2 + y^2 -10y + 25]

i substituted x^2 + y^2 for 225: sqrt[ 144 + 24x + 225 + 25 - 10y]

= sqrt[24x - 10y + 394]

so I need to find the possible range of values for the expression i got to, with the condition that x^2 + y^2 = 225

I'm a bit stuck here actually

I thought of substituting for one of the variables, say y: since y = sqrt [225-x^2]

sqrt[24x-10y+394] = sqrt[24x-10(sqrt(225-x^2))+394]

so there's a square root under a square root, which kind of bothers, since i'm not used to dealing with expressions like these. However I had the idea that I can treat it like a function, as there was is only 1 variable, so f(x) = sqrt[24x-10(sqrt(225-x^2))+394]

and the function does actually have a global minimum and maximum:

I just don't know if this is correct, lol. Can someone check it? (no pun intended by the checkmark graph)

It is a 2 mark question on a 110 mark exam, that is why i'm skeptical

PLEASE PING ME IF YOU RESPOND, TYSM!!

[dawn junco]

yeah there's much easier than that, think about the situation geometrically

[sturdy rover]

1 sec

don't give hints for now pls

maximum for sure would be 13+15 (if they're the same direction

so 28

and min, if they're the exact opposite direction, so 2?

is this correction? @dawn junco

*correct

LMAO what is that typo

[dawn junco]

yep it is indeed correction

the formal version of that reasoning is the triangle inequality

[sturdy rover]

hahaha. OK WTF. this is so annoying though, i love vectors because I like to visualize especially in 3d, quite annoying i didn't get this

[dawn junco]

https://en.wikipedia.org/wiki/Triangle_inequality#Normed_vector_space

[sturdy rover]

ooooooo can you teach me please

[dawn junco]

there's also the reverse triangle inequality, which gives you the minimization

[sturdy rover]

alright, i'll read this. Thank you

[sturdy rover]

ok. I have to go out apparently, but i will check these out! TYSMMMM

[dawn junco]

aight

[sturdy rover]

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[crimson sedge]

What does this exactly mean? Like what am I supposed to look at in order to get the answer?

Ye

Then

…….

Ye

Huh?

if x is less than 0

Then that’s negative

Then?

Ye

[keen python]

more of a "just look at it" problem

[crimson sedge]

That’s exactly what I said

But like what am I supposed to analyze in order to get the answer

[keen python]

start with the choices tbh i think thats the easiest way

[crimson sedge]

……

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[keen python]

did i do something wrong?

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[upbeat pond]

uhh whats teh derivative of tan3x

[void glen]

chain rule

do you remember how to differentiate tan(x)?

[upbeat pond]

nope

should i convert it to sinx/cosx

[void glen]

you can but that’s more difficult

just remember that $\frac{d}{dx}tan(x)=sec^2(x)$

[wraith dagger]

y0shi

[upbeat pond]

oh ok

[void glen]

so we differentiated tangent

but now we have to multiply it by the derivative of the inside

which is?

[upbeat pond]

sec^2x

[void glen]

well that was our outer function

our inner function is 3x

[upbeat pond]

wait what

[void glen]

do you remember what the chain rule is?

[upbeat pond]

wouldnt it be this?

[void glen]

nope

we didn’t need to square tan

$\frac{d}{dx}tan(3x)=sec^2(3x)\cdot 3$

[wraith dagger]

y0shi

[void glen]

see how secant is the derivative of tangent

then we multiplied it by the derivative of the inside function, which is 3

[upbeat pond]

ohhhh

i thought i could do this

[void glen]

yeah that’s not true

[upbeat pond]

would it be true if 3 was an exponent and not a coefficient of x

[void glen]

only if it was $tan^3(x)$

yep

[wraith dagger]

y0shi

[upbeat pond]

oh ok

tyty

[void glen]

yw!

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[silver vapor]

Hello, I am studying for a test and need some help

[silver vapor]

I have a triangle that is part of a bigger square, I know the length of all faces but none of the angles, and need to find the middle angle

[pastel vault]

Do you have an example question?

[silver vapor]

Yes

[pastel vault]

Okok

We need the picture then

[silver vapor]

One sec

I have more information if you need

I need to fine the o angle

Why isn’t it sending

Did you get it?

@pastel vault

[pastel vault]

Yeah sorry I was helping other people

Okay so for the acute angle in the middle

Divide the triangle into two smaller right angled triangles

The base has length sqrt((8 - 4)^2 + (5 - 3)^2)/2

And the height is sqrt((0 - 4)^2 + (11 - 3)^2)/2

So if the acute angle is theta

tan(theta/2) = opp/adj = base/height

This gives you theta and then the obtuse angle is just 180 - theta

[pastel vault]

Ok I've answered

[silver vapor]

What’s theta?

[crimson sedge]

Variable, name.

[silver vapor]

Ok thanks a bunch

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[silver vapor]

No I just became more confused

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[frigid sinew]

Is my overrall thing good

[opal schooner]

yes; I simplified the 140/14(...) when I tried it though

[delicate elm]

im wondering if you really have to do l hopital

[opal schooner]

remember to put it back in terms of the regular limit

[delicate elm]

i was thinking that you could convert it into the form (1+a/f(x))^(g(x) where f and g are some linear functions

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[dawn lion]

How can i say that 2 is true

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[dawn lion]

<@&286206848099549185>

[mellow moth]

?

[dawn lion]

WDYM WITH ?

[dawn lion]

.

[mellow moth]

oh ok

sorry

ill try to do it

[mellow moth]

hello

1 doubt

are k and l in the two pictures same

nothing

leave it

[dawn lion]

I mean try only the first picture

[mellow moth]

hmm ok

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[dawn lion]

Help

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[analog sorrel]

did i do anything wrong and if not, how do i sec^2(10x)tan(10x)?

[crimson sedge]

help

Help

[analog sorrel]

<@&286206848099549185>

[runic garnet]

honestly im not too sure what u did

but i can help u out if u want, from the start

it involves tabular integration, which is just a more compact form of integration by parts. are u familiar with that @analog sorrel ?

i dont know if my method is the most simple, but it works so

[analog sorrel]

i only bacically know intergration by part, for this question, you need to do by parts twice

[runic garnet]

is it required that you do it by using integration by parts twice? i did it only once and got the answer

[analog sorrel]

ye apprently

[analog sorrel]

how did you do it doing it once?

[runic garnet]

i started off by doing u = 10x

then i performed tabular on u^2 *sec^2(u)tan(u)

which gave the answer

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[cerulean oracle]

someone just solve this my teacher is a brainlet and never taught our class this

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[cerulean oracle]

<@&286206848099549185>

[iron jackal]

Do you know what a graph is (in this sense)?

[cerulean oracle]

ye

[iron jackal]

ok

So a weighted graph is like that but each edge has a "weight" assigned to it

[cerulean oracle]

ye

[iron jackal]

So you can use the table to see which node connects to which other node

for example, Fort Worth connects to Plano, and then just label the weight (54) of the edge

[cerulean oracle]

ah got it

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[paper warren]

hi i know the form will me x = a+wv but still can get my head around how to find them

[paper warren]

thank you

[ripe mirage]

I can try to solve it. I think that you should make those an equation and solve when true for both planes

[silver forge]

yeah, convert to cartesian form then perform gaussian elimination

u will have one free variable since the intersection is true over an entire line

[paper warren]

i convert both plane to cartesian form or ?

[silver forge]

convert both

[ripe mirage]

It is quite simple, I think

Notice that

[paper warren]

3*x1 + (-3)*x2 + (-3)x3 = -15
4x1 + (-5)*x2 + (-3)*x3 = -21
the last one is the free varible ?

[ripe mirage]

$x_1$

[wraith dagger]

mlysikowski

[ripe mirage]

component is identical

[paper warren]

right

[ripe mirage]

thus it will be along x_1

[paper warren]

<-1,1,3>

wdym by along x1?

[silver forge]

yeah last one is free

u eliminate it

and get a single equation

then u let x1 = omega or x2, ur choice

[paper warren]

you mean like 3x1 + (-3)x2 +15 = 4x1 + (-5)x2 +21

[silver forge]

no get rid of the x3

by subtracting both equations

[paper warren]

yeah so we gonna have
3x1 + (-3)x2 =-15
4x1 + (-5)x2 =-21
but then like what next

[silver forge]

u shud get -x_1+2x_2 = 6

by subtracting both equations

u r basically using elimination method to get rid of x3

[silver forge]

then let x2 = w

so -x1 + 2w = 6
x1 = 2w-6

[silver forge]

now go back to the first equation here.
let x1 = 2w-6
x2 is just x2=w
now u find x3 in terms of w

[paper warren]

sorry but i didnt understand know we got this yet

[silver forge]

3x1 + (-3)x2 + (-3)x3 = -15
4x1 + (-5)x2 + (-3)x3 = -21
subtract equations 1 with equation 2

[paper warren]

oh right

[silver forge]

so like 3x1 - 4x1 -3x2+5x2...

[paper warren]

i thought subtracting both equation is mean like we subtract both to something else

[paper warren]

I got w-1 = x_3

then may i ask what the next step?

[silver forge]

so u get
x1 = 2w-6
x2 = w
x3 = w-1

just write that as a vector

<-6, 0, -1> + w<2, 1, 1>

thats ur vector equation of a line

[paper warren]

right thank you

[silver forge]

np

[paper warren]

they also talk about the second method as well

i try to sub them in but idk how to like solve them

still trying but

like how do you convert them to the x = a+wv again

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@paper warren Has your question been resolved?

[paper warren]

<@&286206848099549185>

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@paper warren Has your question been resolved?

[paper warren]

i got -21=-24

.close

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.close

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[spice phoenix]

I am stuck on this problem, any help?

I tried to find the length of y and x, but I need an angle to do that

[livid hound]

you don't need the numerical values of the angle or x or y
consider soh cah toa

are you able to write out the ratios for
sin,cos and tan theta

[spice phoenix]

yea

sin would be x/y

cos would be 2/y

tan would be x/2

so ySin = x?

y sin theta = x

[livid hound]

first one has both x and y so you don't want that
also don't be lazy and include those thetas

since you want x and y in terms of theta only

solve cos(theta) = 2/y
for y

[spice phoenix]

2cos(theta) = y

wait no

one sec

would it be

2Sec(theta) = y

?

[livid hound]

yeh

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[misty nova]

I am trying to differentiate y and i dont really know to to proceed

[flint plinth]

can you paste a screenshot

ugh, that's rather gnarly looking

[misty nova]

yes very im not having fun

i found all the components but combining it all is making me dizzy

[flint plinth]

is that first term in the denominator (cube root of 2) times x? or does the cube root include the x as well

[misty nova]

included the x aswell

[flint plinth]

ah that's annoying, haha

i would consolidate the x^3 and cube root of x in the first term on the third line

x^3 times x^(1/3) = x^(10/3)

not a huge simplification but better than nothing

[misty nova]

wont that make the 3e^-4x more complex when expanding that out aswell?

[flint plinth]

no leave that part alone, you'll have x^3 (3e^(-4x)cos(3x))

[misty nova]

ah ok i see

[flint plinth]

no getting around doing a nested product rule there i guess

grunge work like this is why computers exist 😁

[misty nova]

is there a way to do it in matlab?

[flint plinth]

i dunno, they might have a symbolic package

could do it in wolfram alpha i'm sure

[misty nova]

i might try just doing it via that then cause it just makes me dizzy trying to pen and paper it

[hollow trail]

https://www.mathworks.com/help/symbolic/differentiation.html

[misty nova]

ty

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[crimson flower]

how would i solve this intergreal I tried doing u substitution but im not sure thats the right approch

[silver oar]

it is

u = 6cosx or u = 6sinx should work

[crimson flower]

oh alr thanks

[runic garnet]

just look at it geometrically

its just a semicircle

[crimson flower]

how did you get im not sure ive learned how to do this yet

[silver oar]

do you know how to solve the integral of sqrt(1-x^2)

[crimson flower]

we didnt learn that explicitly that's why im confused

but the teacher put it on the practice sheet so i think we are supposed to know how to do this

[warm crescent]

you know the area for a circle

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[steady plover]

i tried using u substitution

is this correct?

[livid hound]

no

u and x are related so you can't factor 1/(3x^2) out of the integral like that as though it were a constant

approaching this with sub isn't ideal

instead just expand
then power rule each individual term

[steady plover]

OHHHHHHHHHHH

gotcha

thanks

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[twin wave]

If i take 2 cards from a standard deck. what are chances they are king of hearts and ace of spades

[twin wave]

is it 1/52x51 or do you need to multiply that by 2

[drifting marlin]

the latter

[twin wave]

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[elder elm]

Could someone help me with this problem, 3A?

[buoyant latch]

What are you stuck on?

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[fervent vortex]

2B)
How do you use 2a) to complete this?

[fervent vortex]

I can see the resemblence of RHS (b) to RHS of (a) but not sure how to use it still

Though I orginally thought that the lhs of b was meant to be some binomial expansion but cos2kx doesn't work like that

[cedar kiln]

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[tiny geyser]

I need the derivative of e^(1/x), can anyone show me with the steps please?

[dusky coral]

Have you tried anything yet?

[tiny geyser]

Well Im learning derivatives pretty new, like 1 weeks so if I do the basics I can have the answer e/x^[(1-x)/x]

Which doesnt seems like right answer

[humble marsh]

no, you can't apply the power rule in that way

that's only for x^constant

not constant^x

do you know what the derivative of e^x is?

[tiny geyser]

Aint that still e^x?

My knowledge is pretty small right now

Sorry bout that

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[tiny geyser]

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[twin panther]

O is the centre <ABO=30 <ACO=24 <BAC=?

[stone mountain]

what have you tried

[cedar kiln]

@twin panther Has your question been resolved?

[pastel vault]

Draw a line connecting BC

Then let angles B and C be x

Fill in all the angles and you'll get an equation to solve x

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[crimson sedge]

Hello, my main problem is with part a and i don't understand the mark scheme. So i manipulated the equation to get this:

Then I decided to say that a=4 and b=4 as the question said nothing about a and b not being allowed to be the same thing.

I ended up getting a result of 5 and since 5>=4 then the statement has been proven to be true (for part a at least

So my answer is technically correct but I don't know if that's how they wanted it to be proven ? If my answer is wrong could someone please explain what they did in the mark scheme to prove the statement ?

Also, I have no idea where they came up with this:

[pastel vault]

Prove that for all positive values of a, b

You cannot substitute arbitrary numbers for a and b

[pastel vault]

So a correct proof needs to start from something that is true for all positive a, b

The order you write the proof is the exact opposite order to the order you come up with each step, when you're figuring stuff out

Luckily you can use $\iff$

[wraith dagger]

south

[pastel vault]

That means that you can read the lines top down or bottom up, both are fine

If you read the solution from bottom up, then the steps follow

4a^2 + b^2 = 4ab implies 4a^2 - 4ab + b^2 = 0

Compare this with 4x^2 - 4x + 1 = 0 where b = 1
or 4x^2 - 8x + 16 = 0 where b = 2

Each one of those gives you a perfect square

[crimson sedge]

How do you get 4a^2 + b^2 = 4ab ?

[pastel vault]

You have 4a^2/(ab) + b^2/(ab) = 4ab/(ab) right

Then just multiply both sides by ab

[crimson sedge]

so that gives you 4a^2 + b^2 right ? Can we factorise that ?

[pastel vault]

No

You get 4a^2 + b^2 = 4ab then

[crimson sedge]

Wait so 4a^2 +b^2 / ab =0 if we times both sides by ab we get 4a^2 + b^2 = 0 so how do we get it equal to 4ab ?

[pastel vault]

Read the MS again

You're missing the entire right hand side

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Oh wait so we can manipulate this ?

So we get (2a-b)^2 > 0 but how does that prove that it's >=4 ?

[modern sparrow]

the strategy with this kind of question is to basically assume what you want to prove is true, then manipulate/simplify/rearrange into something that has to be true, and then go backward in the exact reverse order.

For example, say I want to prove that
$x(x-6) \ge -9 for all x real$

Then I'm gonna assume that is true, and expand/simplify it

and I will get that

[wraith dagger]

lgkoo

[modern sparrow]

$x^2-6x+9 \ge 0$

[wraith dagger]

lgkoo

[modern sparrow]

which leads to

$(x-3)^2 \ge 0$

[wraith dagger]

lgkoo

[modern sparrow]

which we know has to be true for all x real, since a real number squared has to be >= 0

Then, you would start by stating this fact, and go backward to prove that $x(x-6) \ge -9$

[wraith dagger]

lgkoo

[modern sparrow]

So for your case, assume $\frac{4a}{b} + \frac{b}{a} \ge 4$ is true, and you will basically end up with $(2a-b)^2 \ge 0$ which has to be true. Then you propose that we start with this fact and work all the way backward. The mark scheme only showed the backward direction, which is why it looks like $(2a-b)^2 \ge 0$ seems like it came out of nowhere

[wraith dagger]

lgkoo

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[sharp cradle]

Hi, Does anyone know how to determine the image, zeros and signs of this function? sg is sign function

[sharp cradle]

I was able to do this

zeros are -2 and 3

I don't know how to determine the image and the set of positivity and negativity.

[cedar kiln]

@sharp cradle Has your question been resolved?

[cedar kiln]

@sharp cradle Has your question been resolved?

[sharp cradle]

I was able to extract the intervals of positivity and negativity.

I just need help to get the image please

<@&286206848099549185>

[sharp cradle]

😭

[crimson sedge]

Ah ok I get it now, thanks for your help !

[cedar kiln]

@sharp cradle Has your question been resolved?

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@sharp cradle Has your question been resolved?

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@sharp cradle Has your question been resolved?

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[small dust]

I feel like I didn't do this surjective proof correctly...

[small dust]

What do I need to prove to prove surjectivity here?

[upper abyss]

What's the input that gives the output b?

[small dust]

a_1

[upper abyss]

How do we define a1?

[wraith dagger]

Juke | ping me if no response

[upper abyss]

Is that an element of H?

[small dust]

hmm

h inverse is but...

b should be since b is in the coset of aH in H

right?

so bh^{-1} is in H by the above ?

[upper abyss]

So yeah, that's a proof of surjectivity

[upper abyss]

This makes sense to me, but worth seeing if there's a theorem from your class you can lean on

[small dust]

lemme check

I know I proved it once but I can't remember if that was before or after I wrote this one

the definition of cosets uses the word subset

so would that be fine?

I believe it's fine

thanks

.close

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[half isle]

How do i answer this using special triangle rules?

[drifting marlin]

you can use one of the rules to find the length of MP

[small dust]

^

and then since M is the midpoint, just double it to find the entire length of that side

[half isle]

so is the answer for PR 20√3?

[small dust]

what is tan(30)

[half isle]

umm root3/3

but is that answer right?

[small dust]

where did ur over 3 part go

should be 2(10*tan(30 deg)) which is 20sqrt(3) /3

or 20/sqrt(3)

both are the same

[half isle]

well i used this

and so i did 10*2

to get the h = 20

and then 20/1 * root3/2

[small dust]

where do u get 10*2 from

the 10 is the sqrt(3)h/2 side btw

[half isle]

from the 10 at the bottom

OOOOOH

[small dust]

notice the orientation of this vs ur problem

yea

[small dust]

mb for mentioning tan here, but same thing tbh

[half isle]

shiii thx lol i make this mistake of not looking at the triangle properly

[small dust]

ur good

anything else?

[half isle]

umm

well there is a b and c to this question

ill get back to u once ive done them

if thats okay?

[half isle]

thx again btw

[small dust]

alr

and yea np

[half isle]

what do i use to solve this?

resending the diagram aswl'

would i still use the special traingle rule but change it to the 45 45 one?

[half isle]

?

[small dust]

how do you know whether it is 45-45-90 or 30-60-90 though

what was part b btw

[half isle]

part b was proof

ill send it now

[small dust]

Ah

and PRG isn't 45 deg though

[half isle]

yea i realised now xD

[small dust]

sooo

Are you allowed to use trig?

or just the rules ur given

[small dust]

bc this has trig in it ...

[half isle]

im allowed to use trig just wanted to see if its possible ot do with special traingle rules

which i dont think it is

[small dust]

It isn't possible here

[half isle]

im guessing its ust simple trig then

[small dust]

yeah

lmk if u need help with that

[half isle]

aight

[small dust]

well

I assume you wouldn't

[half isle]

i also assume i wouldnt but my brain is working on like 2 brain cells rn

[small dust]

it's just pythagoras theorem lol

[half isle]

oh yea it is xD

[small dust]

right triangle, u have both legs

ye

[half isle]

mb lol my brain aint working well

[small dust]

ur good

same here

it's like 3am and i'm still doing hw

[half isle]

its 10 am for me but i got the worse sleep possible and this is due today so i had to get up try and atleast

[small dust]

fair enough

[cedar kiln]

@half isle Has your question been resolved?

[half isle]

so im rewriting the questions more clearly now just to make sure for the first one i do 10 * 2/root3 and then double it?

[small dust]

u doubled it with the *2 on the 10/root3

10/root3 is the length of half PR (which is the length of PM)

[half isle]

oh yea i meant to say the same thing but divide by 2 after to make it the right answere but thats quicker

[half isle]

sry for asking for alot of help but how do u do this?

[small dust]

ur fine

hmm

can you use a calculator

[half isle]

uh gimme a sec lmme check

yea u are allowed to

[small dust]

hmmmmmmmmm

lowkey, I'd graph it and find the intersection but we can do some algebra to it too

let's see.......

I can't see any useful manipulations immediately tbh

Maybe replace sin^2 with 1-cos^2

that might help?

it'll at least make the equation only invole cosine

[half isle]

aight ill try that

[small dust]

sounds good

i should get to 1-8cos^2(x)=2cos(x)

but honestly, idr how to do this without graphing it

[half isle]

i wouldnt know how to do this with graphing it or algebraically lol

[small dust]

felt

[dreamy void]

Use sin²x = 1-cos²x

Then you can substitute cos²x = u

You get something quadratic.

[small dust]

[small dust]

my suggestion was correct lets gooooo

haha

ty @dreamy void

[dreamy void]

oh didn't read what you wrote lol

[small dust]

ur good

[dreamy void]

you were just missing out that you can substitute cosine afterwards but you're good👍🏻

[small dust]

Idk how i missed that tbh

good stuff tho

[cedar kiln]

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[eager flame]

why can't ı solve the derivative like this.

[sonic robin]

,rotate

[sonic robin]

you can

[eager flame]

ı solve the equation for y and take the derivative instead of using implicit differentiation but it didn't work out

[sonic robin]

do it again

[eager flame]

yeah ı figured it out ı made some dumb misatke

thx

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[round elm]

Helloo

[round elm]

I don;t know how to solve it since it says more than 3 accidents

the formula and examples given to us only showed how to solve it when it's equal to x number

[rigid lava]

take 1 - P(X <= 3)

[round elm]

Is it like this?

[rigid lava]

that's just P(X = 3)

P(X <= 3) = P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)

[round elm]

that would be 0.56083

1-0.56083 is equal to 0.43917

so it should be 43.917%?

[rigid lava]

,w 1 - sum_{i = 0} ^ 3 poission(i, 2.7)

[wraith dagger]

Wolfram Alpha doesn't understand your query!
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Requested by m8of48

[rigid lava]

,w 1 - sum_{i = 0}^ 3 (2.7)^i/i! * e^-2.7

[wraith dagger]

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[rigid lava]

,w 1 - sum_{i = 0}^ 3 (2.7)^i/i! e^-2.7

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[rigid lava]

,w 1 - sum_{i = 0}^ 3 ((2.7)^i/i! ) e^-2.7

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[rigid lava]

uhh

nvm it was parsing it right

[rigid lava]

show work

[round elm]

I used this formula

then I added it with the values of 1, 2, and 0

so 0.22047 + 0.22496 + 0.18145 + 0.06721

making it 0.56083

[rigid lava]

,w sum_{x = 0}^ 3 2.7^x/x! * e^-2.7

[wraith dagger]

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[rigid lava]

,w 21251/(2000e^(27/10))

[wraith dagger]

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[rigid lava]

one of us is doing somethign wrong

double check your work maybe

[round elm]

I don't really know, this is the formula shown to us

[rigid lava]

no i mean

when you plugged it into your calculator

[rigid lava]

dbl check these are correct

[round elm]

oh

alri

[round elm]

I dunno, it seems good to me

[rigid lava]

ok hang on

ok

first of all

0.22496 should be 0.24496

second of all

,w 0.22047 + 0.22496 + 0.18145 + 0.06721

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[rigid lava]

you just added them up wrong

[round elm]

oh

[rigid lava]

,w 0.22047 + 0.24496 + 0.18145 + 0.06721

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[rigid lava]

1 - that is your answer

[round elm]

ohhhhhhh

Now I get it

alright, thanks!

[rigid lava]

np

[round elm]

Wait, so

[round elm]

will it be ok to know why I had to add up those numbers?

[rigid lava]

yes

so

you are asked P(X > 3)

this is equivalent to 1 - the opposite

which is 1 - P(X<= 3)

what values are less than 3?

[round elm]

ahhhhh

2, 1, and 0

[rigid lava]

er

[rigid lava]

less than equal to 3 *

but yes, 0,1,2,3

if you have a continuous rv

then you just integrate

or if you know the cdfs

you can just evaluate those as well

[round elm]

I see, I hope I'll learn about those in the future

Thanks a bunch! really!

[rigid lava]

npgl

[round elm]

.close

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[rigid lava]

just 2 no?

unless i am misinterpreting the question, just choose {3,99}

[prime bison]

I guess the choosing is random?

[rigid lava]

oh i see

[crimson sedge]

Someone good at trigonometry:))

[rigid lava]

should be |{1,3,..., 51}| + 1

[prime bison]

Well think about the pairs you could have, 99 and 3. 97 and 5, so on until they meet in the middle. 1 doesn't have a pair in your set

[crimson sedge]

?

[prime bison]

To guarantee you get one of those pairs you'd have to pick enough numbers to fill one half of your set plus 1 to account for the unpaired 1

[prime bison]

This is not your help channel

[crimson sedge]

O

Sorry

[rigid lava]

if i have all the numbers from {1,3,5,...,51}

and then i add any number into it

what will happen

ok

choose any number not in the list

that comes from {1,3,...,99}

so you can choose any number from {53,55,57,...,99}

that is you choosing the number that comes from {53,55,...,99}

because you choose 1 number

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[crisp flint]

What is the least value when sum of 3 values is 39 but all 3 value are chained

[crisp flint]

I submitted 12 but wrong

And i used brute force to find 12+13+14 = 39

[crimson sedge]

the minimum of abc?

[crisp flint]

It just has to be sum of 3 chained values is 39

Also it means least value of one of these chained value

[crimson delta]

what is chained supposed to mean

consecutive?

[crisp flint]

Something like 1,2,3

If the least value is a then the rests are a+1 ,a+2

[chrome delta]

what's the implication of least then?

there's only one triple that works right?

,w a + a+1 + a+2 = 39

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[chrome delta]

what's the actual question?

[crisp flint]

The a value when sum of chained numbers is 39

And the actual question is "연속하는 세 자연수의 합이 39보다 작을때, 합이 가장 큰 세 자연수중 가장 작은수는?"

[cedar kiln]

@crisp flint Has your question been resolved?

[cedar kiln]

@crisp flint Has your question been resolved?

[chrome delta]

oh

the question asks you for consecutive numbers such that their sum is less than 39

and which triple has the highest sum

still a bit confusing nvm

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[sleek lake]

I have 8 vectors of 2 3D rectangles and I already use SAT to find if they are colliding but I need to find their contact points. Does anyone know how I might find them given the vectors of the cubes?

[cedar kiln]

@sleek lake Has your question been resolved?

[sleek lake]

<@&286206848099549185> do you know how I could do this?

[uncut quarry]

vectors of 2 3D rectangle?

what dose that mean?

[sleek lake]

vectors of 2 3D rectangles

[uncut quarry]

of two?

[sleek lake]

yes

the vertexes

but they are just vector 3s so thats why I called them that

[uncut quarry]

so if i understand u correctly, we have two 3D rectangles, and we know their vertices

and what we want is to know where do they collide?

[sleek lake]

their contact points

but yes

where they collide

[uncut quarry]

are they parallel to the xy plain or they can be in any angle?

[sleek lake]

they rotate

like the purple points here but in 3d

[uncut quarry]

like they can be tilted?

[sleek lake]

yes

I need to know where the vertexes touch the other cube basically

[sleek lake]

they can freely rotate

[uncut quarry]

so, they are solid object, they cant go into each other

they can only touch each other

[sleek lake]

yes

[uncut quarry]

i see

[sleek lake]

I think there is a max of 4 contact points

since its not possible to get more

[uncut quarry]

well, maybe mathematiclly its hard to think of them in 3d, i was thinking about looking at the projection of their vertices into the 3 plains: xy, xz, yz

their projection into each plain

and if they touch each other in 2 of these projection, i can find the coordinates when they touch

[sleek lake]

I have the SAT info

so like the normals

where it collides

[uncut quarry]

sorry, but i need to go, u can mention the helper role to see if somebody can help

[sleek lake]

ok

[uncut quarry]

i might come back after 2h or so, i hope ur problem will be solved befor then

[cedar kiln]

@sleek lake Has your question been resolved?

[cedar kiln]

@sleek lake Has your question been resolved?

[sleek lake]

<@&286206848099549185> anyone?

[sleek lake]

I need to find the contact points of 2 rectangles in 3D space given their vertices

[uncut quarry]

ahh, more than 2h

[uncut quarry]

so, what dose this SAT gives u again?

[sleek lake]

Well

It gives me the normal of the collision

It gives me the depth of the collision

Sat is a algorithm for detecting collision

I have a working physics engine but I need to make it so it adds rotational force upon collision

And to get that I think I need to get the contact points. Let me know if there is any easier way to add rotational force based on a collision tho

[uncut quarry]

oh yeah, thats why u need to know here they touch

[sleek lake]

Yes

[uncut quarry]

but what do u mean the normal of the collision, is it like that?

[sleek lake]

I’m pretty sure yes

[uncut quarry]

and the depth is like the magnitude of that vector i think?

[sleek lake]

Yes

How much the object is in the other one on that normal

I think

[sleek lake]

I assume I will have to do something with projecting the vectors on the normals?

Or how will I get the contact points

The normal of the collision itself wont really help. I think it’s the normals of the cubes themselves

[uncut quarry]

well, when one object touch the other there will be a responce force from the collision, if we assume the second obj dosent move, then the full force will affect the first obj making it rotate which is what called torque

[sleek lake]

Well since the other object does move

Do we split it in half

[uncut quarry]

now it depend on the mass and velocity of each one of them

[sleek lake]

I’m using quaternion btw if that matters

[sleek lake]

I have that data

But we are getting ahead of ourselves

We need to first get the contact points

[uncut quarry]

i can see noway other than looking at the projection of the objects into the three plains

[uncut quarry]

for example, lets say the xy plain is the ground, and z is the hight

[sleek lake]

ok

[uncut quarry]

then the projection into the yz plain will be something like that,

now here thats a simpler problem to work on

we can check if a point is on the line between two other points

actually we already know if they collide or not, than we just want the coords of this point

[sleek lake]

yes

[uncut quarry]

and here would be like what u draw before

[sleek lake]

what is the one on the left

[uncut quarry]

left?

[uncut quarry]

thats an edge

i just marked it with red

[sleek lake]

ok

but we are doing it in 3d

what if its like this

[uncut quarry]

can u move the cam to look at it from the side?

and send the pic

[sleek lake]

edge to edge

[uncut quarry]

the side?

[sleek lake]

this cant happen in 2D

[uncut quarry]

i want the projection

[sleek lake]

you mean this?

[uncut quarry]

yes

[sleek lake]

so basically from the projections

I need to get where the vertexes

are touching the other cube

anywhere

[uncut quarry]

yes i know, using one projection will give me two coords of the touching point

[uncut quarry]

like here, i can know the y and z coords

but i wanna see it from down if u can

[sleek lake]

this is -z

[uncut quarry]

can this help u?
https://dyn4j.org/2011/11/contact-points-using-clipping/

[sleek lake]

that is 2D tho

[uncut quarry]

maybe try move it into 3D, cuz i think this is not a pure math problem anymore, i search about the SAT algorithm and it also use the same technique as i did which is the projection of the objects onto a line

[uncut quarry]

and since u can also implement it in 3D, maybe u can try this with 3D

or searching the internet about "contact points of collision game development" i can find some resources that talk about the same thing

dont get me wrong,im just trying to help, if u wanna u can still reopen ur channel to move it top since its now down between other channels and see if somebody can help u further

[sleek lake]

.close

[cedar kiln]

Channel closed

Closed by @sleek lake

Use .reopen if this was a mistake.

[sleek lake]

.reopen

[cedar kiln]

✅

[sleek lake]

like that?

[uncut quarry]

yes, now meantion ur original question up top

or reply to it

[sleek lake]

I have 8 vectors of 2 3D rectangles and I already use SAT to find if they are colliding but I need to find their contact points. Does anyone know how I might find them given the vectors of the cubes?

[sleek lake]

.close

[cedar kiln]

Channel closed

Closed by @sleek lake

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

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[heavy shoal]

Hello, what the symbol at right and left of 1,n ?

[surreal cave]

I presume it's interval notation

[1,n]

so this is saying all number in between 1 and n, including those two, where n is some natural number {0,1,2,3,4...}

[heavy shoal]

yes but it not [

it two [[ very near

I've read posts by the same author before and he set intervals normally, not with the symbol shown

[main needle]

yea it's just [ for integers, wonky notation

[surreal cave]

huh, interesting didn't know that :)

[heavy shoal]

What is the notation in LaTeX ?

[main needle]

uh I see it written as [\![ but idk latex

[heavy shoal]

$[![ [$

[wraith dagger]

Voltaire

[heavy shoal]

Ok, thanks you for the help

[cedar kiln]

@heavy shoal Has your question been resolved?

[cedar kiln]

Channel closed

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[ionic wolf]

how to do 10^log(9)

[frigid escarp]

Assuming logBASE 10?

[frigid escarp]

Can you convert to exponential

[ionic wolf]

oh wait

is it 10^9

[upper abyss]

10^ and log are inverse functions, so they cancel. You're left with 9.

[ionic wolf]

o

.close

[cedar kiln]

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[hollow totem]

Use similar triangle

[exotic sundial]

can I do it without similar triangles,

[hollow totem]

Yes

But its much more complicate

Similar triangle is the best option here

[exotic sundial]

I only need help with b

i already had a

[hollow totem]

Hmm

U can try to reverse it

[exotic sundial]

wdym

should i use pythagoras

[exotic sundial]

<@&286206848099549185>

[cedar kiln]

@exotic sundial Has your question been resolved?

[exotic sundial]

<@&286206848099549185>

[exotic sundial]

<@&286206848099549185>

[cedar kiln]

@exotic sundial Has your question been resolved?

[cedar kiln]

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[feral heart]

I kinda need help on letter c.

[crimson sedge]

set C = 20 i guess

and solve for t

[frigid escarp]

So then 1=6(1/2)^t/6

[feral heart]

ok I got 15.50977500.

[feral heart]

wait why this?

[frigid escarp]

Reduction

[feral heart]

oh ok. 1=6(1/2)^t/6 what do I with 15.50977500?

[frigid escarp]

Set that equal to t and check if it equals 1

[feral heart]

ok

no it doesn't equal 1

should I have used 15.51 hours equal to t?

it still doesn't equal 1

I did 20=120((1)/(2))^((t)/(6)), and got t=15.50977500 but rounded nearest hundredth of an hour. It's 15.51. Is the answer 15.51?

[crimson sedge]

,w 20 = 120(1/2)^(t/6)

[feral heart]

it doesn't equal 1 when I plug in 15.51 into 1=6(1/2)^t/6

[crimson sedge]

,w evaluate 6log(6)/log(2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by qylooo

[crimson sedge]

ok yes

that sounds correct

[feral heart]

15.51?

[crimson sedge]

yeah

[feral heart]

I will put that in

Let's go!

[crimson sedge]

ayee

[feral heart]

Can I ask another one?

[crimson sedge]

sure

[feral heart]

I am bad at math~

[crimson sedge]

ok let me read

[crimson sedge]

well its like saying

if you had 100 mg at the beginning after 800 days u woulod have 100*0.30 = 30 mg left

so u can use that to set up your equation of $\ds Ae^{kt} = B$

[feral heart]

ok

[crimson sedge]

hint: 100 is ur starting value, the time is 800, and the final amount is 30

[feral heart]

That doesn't seem something I know but I will try that equation.

I got

ah~

460 maybe?

[crimson sedge]

for what

[feral heart]

for question a, 460 days

oh wait I tried it, it is 460 days

hang on I am solving b

oh I solved it!

I love green!

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[nocturne furnace]

On my homework, I've been asked to make a proof for the properties of a given relation. However, I'm not exactly sure how I'm supposed to answer this.
I don't understand how the situation m-m could occur in any other scenario except for when m=n.

[nocturne furnace]

this is a discrete math problem

[upper abyss]

The wording of the proof is very weird

[nocturne furnace]

I hate webassign hw problems

always so annoying

[upper abyss]

If not m = n, it's gotta be m = m. I have no idea why you'd say that, but there's no other option

[nocturne furnace]

I also tried that

I'm on my last attempt

unfort

I don't get it though because they couldn't substitute m in for n unless they were equal

[upper abyss]

Well okay, we can choose a specific counter-example. The box probably takes any integer then?

[nocturne furnace]

I suppose it could but wouldn't that just be true for any arbitrary m

[upper abyss]

Yes, all integers can be used for a counter-example

But yes if you're constructing a counter-example you'd pick an integer

For example, 1O1 is false

[nocturne furnace]

Yeah I think I understand. I wish our professor spent more time talking about common practices for proofs. He reads directly from our textbook every class lol