help-27

[jaunty mantle]

How do you judge it based on gradient

Is 1/16 good?

Is 1/2 impossible?

Is 1/30 too inefficient?

You first need to define what is “good” in terms of those metrics

If we have a test and 0 marks means the worse and 100 marks mean the best, only then does the mark you get have any meaning

[fiery thorn]

my very rough idea was going to be something like a simulation but then I found sources saying that it should be 1/12 to 1/16 which I then shifted my goal a bit to try to maintain the gradient in this boundaries instead of judging it

[jaunty mantle]

So given that the gradient doesn’t change, the length also wouldn’t change since it’s a function of the gradient

Then you’re left with just the cost

I guess you could graph the cost of concrete against gradient ranging from 1/12 to 1/16

It’ll probably be one of the boundary cases though if you look at the area of the triangle it makes

And not really a derivative = 0 kind of deal

[fiery thorn]

wouldnt the price of concrete be constant?

[jaunty mantle]

But the volume required to fill the ramp changes with gradient

I assume that’s what you meant by the cost of concrete

[fiery thorn]

aahhh i see

so I could do it just like a normal right angled triangle times the depth of the ramp to have a 3d shape then change the gradient, or actually the angle would be more straightforward i think

[jaunty mantle]

Let’s say the gradient is m then m=h/k where h is the height and k is the run

Solving for k we get k = h/m

The area is rise * run /2

So h/m * h /2 or h²/(2m)

So the volume would actually be proportional to the square of the height it seems

Which makes sense, we want to find the area under a linear line

That would be the same as integrating f(x) = mx

[fiery thorn]

it is still a linear relationship right?

[jaunty mantle]

Nope

[jaunty mantle]

.

[fiery thorn]

i didnt understand this last part

[jaunty mantle]

Ok

Do you know calculus

Like basic integration

[fiery thorn]

yeah

[jaunty mantle]

Ok a triangle is just a diagonal line on a graph

f(x) = mx

If you just integrate this from 0 to k you get a right angled triangle of base length k

And gradient m

f(x) = mx is a linear function but when we integrate it for area we get that the area is proportional to the square of x

Which we know is directly proportional to the height

Which makes the area proportional to the square of the height as well

[fiery thorn]

yes i understood that just not the part where it is proportional to the square of x

[jaunty mantle]

Ok

f(x) = mx

f(x) is the height of the triangle yeah?

[fiery thorn]

yes

[jaunty mantle]

Ok integrate it for the area under the curve, aka area of the triangle

We get some quadratic in x

F(x) = 1/2 mx^2

[fiery thorn]

yeah

[jaunty mantle]

F(x) is the area

[jaunty mantle]

So F(x) 𝛂 x²

[jaunty mantle]

And we know from here height 𝛂 x

[fiery thorn]

yes

[jaunty mantle]

So F(x) 𝛂 [f(x)]²

Or, the area is proportional to the square of the height

[fiery thorn]

i see

then I can use that for the concrete price and multiply that by the depth for the volume?

[jaunty mantle]

Well that’s just area

You need volume

So you need how wide the ramp is as well

[fiery thorn]

that value I had just defined as 2 meters

[jaunty mantle]

Right

You could also have a look at the area used by the ramp

That’s simple just the base of the triangular prism of concrete

[fiery thorn]

yeah from the models I had in mind they have a very big different in the space they use

and then I could use solids of revolution to make a protection

something like this but round

@jaunty mantle thank you so much for the help and for being so attentive!

[jaunty mantle]

👍 good luck with IB

[fiery thorn]

thanks lol

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[karmic hound]

Hello

[karmic hound]

So I’m doing dilations on a scale

And the dilation is 1 half. Like 1/2 - would that just be like for example 4 times 1/2

And would that be just be 2?

So I just, I don’t know if I’m over complicating it in my head? Is the same as 1/2 * 4 ?

Maybe I’m just over thinking it haha

Hm well that’s all I needed for now, how to close

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[vast finch]

can someone help me

[vast finch]

if someone can help id appreciate it

[vast finch]

<@&286206848099549185>

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[stone jay]

hi yall

[scarlet sequoia]

u know cosine law?

and sine

[stone jay]

yea-

i did used it

but it went negative

[scarlet sequoia]

show work

[stone jay]

i am unable to photo

but

nvm deh

[stone jay]

is this right?

[scarlet sequoia]

thing is

it interprets 25 as radians

[stone jay]

i see

[scarlet sequoia]

,w sqrt(150^2 +120^2 - 2(150)(120)*cos(25 deg))

[woven radish]

Results provided by WolframAlpha

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

Requested by Modus#7591

[upper schooner]

[if you want, on your one, select here where it says "deg" to make sure you get degrees rather than radians]

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[ancient dust]

Solve sin x >= -1/2

[ancient dust]

do i give a range for this?

[sick fulcrum]

,tex \factoid original

[ancient dust]

Solve this trigonometric inequation sin x >= -1/2

[sick fulcrum]

,tex \factoid original

[ancient dust]

this is it

[sick fulcrum]

Is there no picture

[ancient dust]

its all text

and another lanugage

[sick fulcrum]

better to post it anyway

even if you cannot translate

[ancient dust]

1 sec

7th quest

[sick fulcrum]

they probably want a range then

[ancient dust]

yes

[sick fulcrum]

yup

[ancient dust]

its x E {-pi/6, -pi/2}

?

[sick fulcrum]

question should really state it imo

ask wolframalpha

[ancient dust]

so its asking in that format?

.close

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[thin gulch]

how do i do c) There are 75 cars in the parking lot
of a car dealership. Two-thirds of the cars are new. How many of the cars are new? is the answer 10

[thin gulch]

.close

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[lavish canopy]

Could someone plz help me w #4?

What proportion should I set up?

@dry robin

Nvm I got it

Could someone help me w #5 part a? <@&286206848099549185>

[devout snow]

@lavish canopy Has your question been resolved?

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@lavish canopy Has your question been resolved?

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[split walrus]

Can someone explain how to solve the integral of e^4x sinx dx using integration by parts?

[hybrid snow]

Let $K = \int e^{4x}\sin(x),\mathrm{d}x$. If you use IBP twice, then K will reappear, allowing an easy algebraic manipulation

[woven radish]

Umbraleviathan

[split walrus]

bro im sorry but i just learned it today and the teacher taught the concept in a minute and just went onto practice problems

apparently theres the formula integral Udv = uv-integral vdu
but then the teacher just skipped out on that and went to draw a U DV table where she differentiates one side and then integrates the other?

[hybrid snow]

Pretty much yeah

[split walrus]

but then that doesnt make sense in some cases like these where you can just keep on differentiating it and you dont knwo when to stop?

like sinx would just switch back and forth with cosx SO HOW WOULD U KNOW WHEN TO STOP MAN

😭

[hybrid snow]

You label one factor "u", and then another factor "dv". You'll have to find du and v

[hybrid snow]

Usually it should resolve itself

An easier integral will appear

[split walrus]

um?

[hybrid snow]

But for cases like this one, it won't, but it will reappear

Which allows for easy algebraic manipulation

[split walrus]

if i set e^4x as u then wont it just keep on getting to like

1/4 e^4x

1/16 e^4x

etc etc...

[hybrid snow]

No you don't keep on doing it

You only care about the first order derivatives

Start with doing the first IBP instead of trying to do both at the same time

[split walrus]

Well this is what I got so far

[hybrid snow]

The derivative of e^4x is not 1/4 e^4x

[split walrus]

but then we get stuck at the other integral again

man did i mental math my usub wrong

oh wait

[hybrid snow]

You'll have to do IBP again for that new integral

If you do it correctly, the og integral should reappear

[split walrus]

so you're saying if i do IBP again for the integral of -cosx * 4e^4x dx

then somehow magically it'll dissolve and all my problems will be gone?

[hybrid snow]

If you do it right, yes

[hybrid snow]

And then this applies

[split walrus]

okay i integrated it again

and got -16e^4x sinx + 16 integral sinx e^4x dx

so that thing did appear

what do i do with the k...

[hybrid snow]

Well you gotta collect like terms

Gather the K's together

iirc it'll end up being 17K = something

[split walrus]

okay so im seeing what symbolab is telling me

so you just combine the e^4x sin x dx whatever crap

but how'd you get 17....

thats such a wack number

how do they even get 17....

if it's K = soomething - something + 5K

so then that's not possible to get 17 out of the blue

oh wait

the -4 is inside the parenthesis

so it's k = -16k essentially

finally i solved it out

thank you

now i got another 5 problems to do after this one...

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[neon wigeon]

How do I do this?

[waxen harness]

What to find?

[versed basin]

Can someone help me with this

Where did the 144 come from?

[polar chasm]

#❓how-to-get-help

[versed basin]

Ok thanks

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@neon wigeon Has your question been resolved?

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[devout snow]

@copper hatch Has your question been resolved?

[flat raptor]

ii is 3/10 i think

i think iii is 3/10 as well

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[lethal rose]

Given a polygon A, how does one get the polygon B that is defined as the polygon that is made out of all the points that are farther than x units inside A

[devout snow]

@lethal rose Has your question been resolved?

[lethal rose]

A conceptual answer or the actual mathematical solution could both help me

[devout snow]

@lethal rose Has your question been resolved?

[timber pebble]

the question doesnt really mak sense

are you sure the polygon youre looking for isnt just scaled version of the original

[lethal rose]

no, i'm pretty sure that a scaled down version doesn't always work for that problem

[timber pebble]

youd have to be more clear about the problem i think

[lethal rose]

imagine a very big square with a small bevel in the corners

the inner polygon with a big enough x would be a perfect square if I'm not mistaken

[timber pebble]

right

this is what makes me think its just scaled down version

[lethal rose]

this is an example of that not working

because by your logic the inner would still have the bevel

maybe i'm mistaken

[timber pebble]

then i don understand the problem

[lethal rose]

i'm not convinced a scaled down version satisfies my prerequisites

and i'm not sure where to go from here

[timber pebble]

you need to more clearly define the problem.

IMHO

think about like

if you start at some point on the shape

and head directly away from the shape one unit

[lethal rose]

Get polygon b so that every point in B is inside A and at a minimum x distance away from A

[timber pebble]

then repeat this process for every point on the shape

youll just get a scaled version

wait

youll get a bunch of lines

well it depends on the shape

[timber pebble]

minimum?

[lethal rose]

yeah so every point of B is like atleast 3 meters from any side of A

[timber pebble]

so filled in

hrm

well you could think of a circle

thats what its gonna be

like what you need is to

for every point on A

draw a circle of radius whatever

and delete that neighborhood

then repeat this over all of A

[lethal rose]

surely there is less computionally intensive way

[timber pebble]

which you could replicate using a finite number of circles

[lethal rose]

cuz my lines gonna look wobbly

[timber pebble]

are you coding this?

[lethal rose]

yes

[timber pebble]

ah

maybe constructively is easier

lemme think

im almost positive its just a scaled down version

still

[lethal rose]

for weird shapes it doesn't really work

if some part is thinner than the required x

then you can never scale it down enough to work , even though other parts of the shape could be wide enough to have it

or some shapes are gonna become two polygons

like two squares connected by a thin band

I guess i'll go circle for now and try some smoothing algorithms if it's too ugly

tho I believe math has a solution for me

seems like it would be a common problem

I guess if I do just one circle per corner it will be ok

[timber pebble]

if its just to create like an image

then that should be fine right

you could look into blurring, too

like if youre just creating something you need visually to look like that, and it doesnt need a precise description

just blur the original, then take a difference

[lethal rose]

sadly it is computational mapping

in real time

surely an algorithm exists

[devout snow]

@lethal rose Has your question been resolved?

[devout snow]

@lethal rose Has your question been resolved?

[devout snow]

@lethal rose Has your question been resolved?

[violet wind]

If A is convex you can just move all the sides inwards by x units

handling points of intersection might get complicated though

if the previous side intersects some side after the next side does, remove it ig

[devout snow]

@lethal rose Has your question been resolved?

[lethal rose]

I need to work with any type of polygon

[agile swan]

Visualize please

[lethal rose]

what I would b given is the big hexagon, and I want to get the smaller inner one where each of the point of the red one is atleast x units away from any side of the outer one

[agile swan]

you only require area?

[lethal rose]

no, I want the coordinates of every edge

or enough information that defines it

[agile swan]

for finding the coordinates of any polygon, you shall use sarrus rule

and, you have to define, how much it shall be far away in both axes which results in such

[lethal rose]

bro this seems very complicated

especially to apply to my specific problem

[violet wind]

That's a pentagon

If it's not convex, the shape you get isn't a polygon.

[lethal rose]

I needed a simple example, but you get the idea

it can be many smaller polygons

I just want a solution and work from there

[violet wind]

?

you want to approximate a curve with polygons?

[lethal rose]

in what case would there be a curve ?

[lethal rose]

and even if it was possible yes but I can manage that part myself

[violet wind]

like this

whenever you have an inside corner it will be like this

[oblique panther]

wait

its true

[lethal rose]

I don't see how the bottom diagram represents anything from the top one

oh I understand

it's flipped

I guess even the non rounded version would be fine by me

I just need basic visualization

[lethal rose]

[fossil chasm]

if your only restriction is for points to be > than some unit x, you could probably find the bounding box for your polygon and conform your most "concave vertex" to that

[craggy dagger]

so if you have concave, that's one of the issues - it's not a polygon

[craggy dagger]

so it's generally writing a polygon-cutting routine, do you know how to write that?

then all we need to do is figure out which lines to cut by

that's for the convex case

then concave case, if x is small you can see where the two lines meet

but if x is large you might get some degenerate case

[devout snow]

@lethal rose Has your question been resolved?

[woven wave]

Can someone help me with this

[languid fossil]

You need to put this in an available channel

The ones that Don't have usernames next to them

[woven wave]

okay thanks

[hardy tartan]

People I hope y’all having a nice day and fun and enjoyment learning math! 😃

[trim vine]

I think it was mentioned before, but I think a possible procedure is create a gray region of no-touchy-touchy by dragging a circle of radius x along the line of the polygon A. Any point inside or outside that region is fair game for polygon B.

[lethal rose]

The problem is there can be multiple polygons B

[past scaffold]

@lethal rose ask the problem in math stack exchange... Ig someone there can give u a solution

[lethal rose]

Good idea

[fossil chasm]

polygons can be concave, not sure wht you mean by that

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[cunning narwhal]

no clue how to do part ii

ive done i.

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[willow quartz]

so I know that du = pi/12, and du = pi/12 dx

[willow quartz]

how do I find dx, it says its supposed to equal 12/pi but that doesnt make any sense to me

[restive river]

rearrange du=pi/12 dx to make dx the subject

[willow quartz]

yeah

and then i need to find dx

though

[restive river]

you would get dx=12/pi du, which is what you want?

[willow quartz]

yeah, but why did the pi/12 change to 12/pi?

[restive river]

divide both sides by pi/12

[willow quartz]

OH

that gives you this though right ?

dx/du = 12/pi

[restive river]

i showed u this like 10mins ago lol

[willow quartz]

oh and then u multiply du

[willow quartz]

yeah srry I made a mistake

I get it now

thanks

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[wind rune]

how can i simplify (5b^3)(-2b)^3?

[winter brook]

u can distribute the exponent of 3 into the ()

and then mutiply that to the 5b^3

[wind rune]

ok ok thanks

[winter brook]

u can distribute the exponent bc the inside is multiplication

(ab)^x = a^x*b^x

[devout snow]

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[fierce ruin]

@wind rune

are u finished?

[devout snow]

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[dusk sand]

Where did I mess up?

[odd badger]

U can also use d²=4at. This is also a quadratic equation.

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[devout snow]

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[dusk sand]

.reopen

[devout snow]

✅

[dusk sand]

How?

[red sierra]

Without subbing in 3 points to find the quadratic equation, you can notice that
4.9=0.1×7^2
19.6=0.1×14^2
44.1=0.1×21^2, and so on
Therefore you can conclude that
y=0.1×(7x)^2
=0.49 x^2

[devout snow]

@dusk sand Has your question been resolved?

[dusk sand]

where did you get ^0.1 and 7^2? at the top

@red sierra

i graphed 4.9x^2 and got the answer. but how?

[red sierra]

4.9=0.1×7^2
19.6=0.1×14^2
44.1=0.1×21^2
78.4=0.1×28^2
122.5=0.1×35^2
You understand this, right?

[dusk sand]

i understand but how did u come up with those numbers

did it pull it out of a magic hat

i wanna learn this trick

[red sierra]

I just noticed the patterns

The given numbers seemed like squares of multiples of 7

[dusk sand]

ok so you just found a pattern

is there a more reliable way to find this equation?

that simply observation? in a future test the professor will want us plug in 3 points but tbh i dont know where i messed up

[dusk sand]

maybe somewhere at the bottom?

[red sierra]

The most reliable way is to just plugg in 3 points

[dusk sand]

i can retake the photo with flash if that helps

sorry for my awful handwriting, im tipsy. jk jk

[red sierra]

You got these right

You can multiply 2 at the top one and 3 at the bottom one to eliminate b

[dusk sand]

are you pulling my leg

l0l

what about the rest of the problem, do u also multiply there? something feels off

omg uwere right!

LOOOOOOL

WOW

@red sierra

[red sierra]

Wait what

[dusk sand]

nothing

ty

.close

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[devout snow]

Please don't occupy multiple help channels.

[supple knot]

!onechannel

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Please stick to your channel.

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[restive river]

.close

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[restive river]

.close

[wooden kiln]

you needed help?

why closing so much now

[restive river]

I figured it out

[wooden kiln]

Ah good good

[restive river]

@wooden kiln I figure it out

[wooden kiln]

ya

[restive river]

It a lot of second guessing that my issue

I get it right but I second guess

Bc I’m scared to fail the assignment

[wooden kiln]

well this is a straight line right?

[restive river]

Yea

[wooden kiln]

like y=ax+b?

[restive river]

YEPS

mango you’re the g

I wish the best for you

But I’m gonna be leaving this server

[wooden kiln]

so take the x numbers and the corresponding y numbers

[restive river]

Have a amazing life

Mango the g

[wooden kiln]

and put em into y=ax+b

ok?

[devout snow]

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[copper trout]

Can someone please explain why the answer is not -2?

[final laurel]

The way that I like to think about these problems is that you have a polynomial of degree 1 (in absolute values) in the top and a polynomial of degree 1 in the denominator. So if you divide x/x = 1, you end up with a line. But since it's |x|/x you end up with a split when x = 0. In this problem you know |2x-26| > 0 for any x, and 13-x < 0 for x > 13 and 13-x > 0 for x < 13. So as you approach 13 from the left side (x < 13) 13-x takes on a positive value meaning the answer must be 2. Alternatively approaching from the right side gives (x > 13) so 13 - x < 0, which is your result of -2.

[devout snow]

@copper trout Has your question been resolved?

[copper trout]

gotcha I'm gonna try my best to digest this lol.

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[nocturne cape]

In the following figure, O is the center of the circle, PBAˆ=PBQˆ and PQ is tangent to the circle at P.

Given that |AB|=3 and |AP|=1, what is |QB|?

Can u help me please

[devout snow]

@nocturne cape Has your question been resolved?

[red sierra]

Did you learn cosine rules

[jovial mauve]

You will skip this problem as you do not have margins at all

[red sierra]

Wdym?

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[onyx nymph]

3a

[onyx nymph]

it’s very obvious that the answer is 5 and it goes up to 8 but i’m really confused why when i tried to solve it algebraically it gave me 8.21 (correct) and 9.21

what did i do wrong so the 9.21 solution is there?

.close()

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[eager wave]

if
5b+7=A
2c+4 = B
what is the minimum value we can give to A

[eager wave]

so far what i did is this;

But i am not sure if this is correct

I tried this now but i don't think this is correct

[devout snow]

@eager wave Has your question been resolved?

[eager wave]

i tried doing this but the answer is none of the options

<@&286206848099549185>

but i can not give C even lower number than 3

to more clarify what i have been doing

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[fossil moth]

.rotate

[main gull]

,rotate

[fossil moth]

Yes that

[jaunty mantle]

it's a comma before the word rotate

[fossil moth]

I was wondering why my answer is wrong or different? I got cosx and the textbook got sinx

[jaunty mantle]

you're integrating cos x's with du

those are functions of x not functions of u

[fossil moth]

Alright lemme try it again

Oh I get it

Thanks

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[golden wren]

Hey guys

[golden wren]

If a+b+c=6

Where a b and c are non negative real numbers

How do I find the maximum value of a^2+b^2+c^2

?

[devout snow]

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[devout snow]

@golden wren Has your question been resolved?

[devout snow]

@golden wren Has your question been resolved?

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[native bobcat]

prove that the below afirmations are true or false and show if its true or prove its negation if its false

[native bobcat]

par = even impar = odd

[devout snow]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[native bobcat]

1

[restive river]

recall that odd + even = odd

[native bobcat]

ok

but 5n² + 2m how i know if they're even or odd?

[restive river]

2 is even

even * anything is even

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[safe tulip]

I'm confused because It usually tells you the point they want you to find the radius of curvature for

so how should i choose a point?

[lusty sapphire]

You're 11 years late turning this in btw

[lusty sapphire]

It's asking you to find the point

The point where the curvature is smallest

[safe tulip]

haha

[safe tulip]

it just sounds like 2 parts

1. find the radius of curvature
2. find the point where the radius is smallest

so i should only find the point?

[lusty sapphire]

Find radius for any point in the graph

Then determine the global minimum

[safe tulip]

i see

i think i got it

thanks

.close

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[restive river]

If we have the equation of a circle and the equation of a line that intersects the circle in two points A and B, what's the fastest way to calculate the length of AB?

[restive river]

We could find the coordinates of the two points and use the distance formula, but that's slow. We could also find the distance of the line from the center of the circle and then use Pythagoras with the radius of the circle, then double what we get

But is there a formula for this? Or a faster method?

[devout snow]

@restive river Has your question been resolved?

[restive river]

Bloop blip blop

I'm calling the <@&286206848099549185> 'cause I'm about to go to bed

Oh wait maybe this is not the place to ask something like this

[restive river]

Bdoof bdaf bdof

Goodnight

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[fading ether]

Hey, I’m confused on how the zeros of a function can be a square root
And I’d understand if that would be the answer after using the quadratic formula, but even the quadratic formula doesn’t give a proper answer for x
I got the sqrt of 17/2

[scarlet peak]

You need to use the factor theorem. This allows you to get a linear term and a quadratic. That’s where the sqrt comes from

[hollow pollen]

you cant really use the quadratic formula for a cubic

besides you can just plug in all the options and see which works

thats probably the quickest

[fading ether]

I thought that if I simplify the equation to a quadratic one, it might work

[hollow pollen]

good insight

how did you simplify it into a quadratic exactly?

[fading ether]

I divided the whole equation by x to simplify, and moved the -4 to the other side
Or just kept it there outside the parentheses
I don’t think that’s right though

[hollow pollen]

not quite since you would get 4/x

but you can do polynomial long division

[fading ether]

Would long division get me a real number as a zero (or, answer)?

[hollow pollen]

it would allow you to factor the equation

into (x+1)(a quadratic)

and then you can solve for the zeros of that quadratic

you can factor (x+1) = (x - (-1))

since x = -1 is a root

[fading ether]

Okay, I see where you’re coming from there

But about the -4

Would that stay in the quadratic?

[hollow pollen]

im not sure what you mean by that

[fading ether]

Oh nevermind I just figured that out lol

Putting that into a quadratic, I got sqrt of 33/2

[hollow pollen]

after polynomial long division?

[fading ether]

Oh sorry, then I did that totally wrong

Aight I got it, thanks 👍

[hollow pollen]

nice

[fading ether]

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[deft terrace]

I dont know what w-y is

[warped relic]

Hmm

Do you know your special triangles?

@deft terrace

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@deft terrace Has your question been resolved?

[warped relic]

Welp Mijo waddled away

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[short iron]

Why is it wrong to solve c and d like this?

[short iron]

,w 7/10 * 6/9 * 3/8

,w 7/10 * 3/9 * 2/8

[woven radish]

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[devout snow]

@short iron Has your question been resolved?

[short iron]

<@&286206848099549185>

[neat night]

which one is your C

[misty copper]

I think it’s the first one

[misty copper]

well the probability of picking a man two times is (7/10)^2, not 7/10*6/9

[short iron]

My bad

[short iron]

c

[short iron]

d

[misty copper]

and the probability of picking a woman is 3/10

[short iron]

I guess I don't understand then. Are the people being replaced after being picked? Why would the number of people stay the same after each pick?

[misty copper]

they aren’t being replaced, they are being picked at the same time

so you don’t see the outcome of each pick

plus, if the events were dependent, the multiplication rule wouldn’t apply

you would have to use P(A and B) = P(A|B)*P(B) = P(B|A)*P(A)

but you can just assume that each pick is independent

[short iron]

,w (7/10)^2 * 3/10

[woven radish]

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[short iron]

,w (3/10)^2 * 7/10

[woven radish]

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[misty copper]

it’s really close funnily enough

[short iron]

I still don't understand, the book is saying neither of those are correct?

[misty copper]

hmm

then I guess my assumption was wrong

are you working with conditional probability?

or maybe combinatorics

[short iron]

The instructions were to use counting principles to find the probability

[misty copper]

okay so I think I got what the problem wants you to do

count the number of ways that you could have 2 men and 1 woman, and divide it by all possible groups

should work

[short iron]

All possible be groups being 10 choose 3?

[misty copper]

yeah

(7C2)*3/10C3 maybe?
I’m solving the problem with you at this point haha

[short iron]

,w ((7/10)^2 * 3/10)/binomial(10, 3)

[woven radish]

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[misty copper]

,w binomial(7,2)*3/binomial(10,3)

[woven radish]

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[misty copper]

great

[short iron]

Okay wait

Logically why are we doing that?

[misty copper]

so we are using 2 things:

7C2 is how many combinations of 2 men we can have

3 is how many combinations of 1 woman we can have

you then apply the multiplication rule

and get how many combinations of 2 men and 1 woman you can have

and then it’s just dividing by 10 choose 3

[short iron]

Gotcha

So for the future how do I know if I have a problem of this type?

[misty copper]

hmm probably if it involves the probability of seeing a certain combination of a larger set of elements

in this case the elements being 7 men and 3 women, and the combination being 2 men and 1 woman

[short iron]

Okay, thanks for help!

[misty copper]

no problem! I learnt something too!

[short iron]

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[cinder pewter]

hi! may i get help solving this problem please

so i thought that you were supposed to use the equation 1/2 integral from a to b (r)^2 d(theta)

but then i got stuck so i wasnt sure where to go next

[hybrid snow]

Well for starters, a = 0

Makes life easier

All you need to do now is find when the function closes itself

[cinder pewter]

may i ask how i could find that please

[hybrid snow]

I mean if you think about it, if you want it to close:

$$4 + \sin(b) = 4 + \sin(0)$$

But $a = 0$, and $a ≠ b$.

[woven radish]

Umbraleviathan

[cinder pewter]

would that be at pi?

[hybrid snow]

Well I can see you get that, but no

What I would do is if you're able to graph it, graph it

It's basically some kind of round structure

You wanna take into account that it's a polar equation, so it's gotta circle back. When theta = pi, r = 4 but it's not at the right place/quadrant

[cinder pewter]

would i be graphing the r = 4 + sintheta equation? because if so, the boundaries seem to be -3 and 5 so that wouldnt be right i dont think

[hybrid snow]

No... you need to look at theta not the x and y

As a round object you'll find it'll close itself up when t = 2pi

It's not quite a circle

But it'll act like one

You cannot mix up polar coordinates with Cartesian coordinates

[cinder pewter]

ohh i understand what you mean now, thank you sm!

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[frail lake]

could someone please help me with 10e

[frail lake]

is it "Ex (D(x) or F(x) or C(x))"

(the Ex is supposed to be the existential quantifier)

please @ me if anyone responds

[devout snow]

@frail lake Has your question been resolved?

[frail lake]

<@&286206848099549185>

<@&286206848099549185>

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[icy depot]

My answer comes out to be {f(6.88)}

But correct answer given tells it is {f(7.88)}

I don't know what to do now

I don't understand why is correct answer {f(7.88)} and not f{6.88)}

[magic pine]

is that period only 2

im so confused

[icy depot]

Period is 4

[magic pine]

where is the rest of the function then

they only defined it for x in 0,2

[icy depot]

This is what the graph i draw

The rest of the function can be found out by the information that f(x) is even and have a period 4

[magic pine]

oh it's even

i missed that i see

how did you get {f(6.88)}?

[icy depot]

That means f(-1 + 4) = f(-1) that is f(3) = f(-1) and soo one

On*

[magic pine]

i understand

but how did you get {f(6.88)}

[icy depot]

Okay let me tell you

[bright burrow]

I assume ur answer has to be one of the given 4 options

[icy depot]

If we go by graph , than f(6) = 2 and than decreases linearly and reaches f(7)= 1 that means somewhere near left part of 7, the value of f(x)= 1.12...and we can find it with analytical method also

[icy depot]

I feel option given does not include correct option..

I maybe wrong

[magic pine]

f(5.12) = f(5.12 - 4) = f(1.12)

what is f(1.12)?

[bright burrow]

No does a works

A does work*

What is wrong with my english today

[icy depot]

f(1.12) equal to 1.12

[icy depot]

How?

[magic pine]

so what is {f(1.12)}

[magic pine]

follow along with me and you'll see

[icy depot]

.12

[bright burrow]

Find f(7.88)

[magic pine]

great

now find f(7.88)

use the fact that the period is 4

[icy depot]

7.88 - 4 equal to f(3.88)

[magic pine]

do it once again

well

no

do 3.88-4 again

[icy depot]

It is -1.12

[magic pine]

what is 4 - 3.88

[icy depot]

Oops sorry lol ...it's .12

[magic pine]

yes so 3.88 - 4 = -.12

use the fact that it's even

to compute {f(-.12)}

[icy depot]

.12

[magic pine]

yes

[icy depot]

And f(3.88) = f(7.88) = f(.12)

That means {f(7.88)} = .12

I guess i draw graph wrong?

Thank you

[icy depot]

.?

Graph that i draw Is wrong, right?

Oh wait... that's right ....

It's just i forgot that 0.xx can also have fractional parts

Thanks

.close

[devout snow]

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[devout snow]

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[lean onyx]

I was stuck in a test question today
the question was "what is V if a = 4 and b = 8"
V= a(5-b)^2
can someone state the answer?

[neat night]

https://tenor.com/view/cat-mad-angry-calculator-gif-10034459

[lean onyx]

Because
4(5-8)^2
4*((5-8)^2)
4 x 9=36?

i got the question wrong

[devout snow]

@lean onyx Has your question been resolved?

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[warm dew]

The perimeter of an isosceles trapezoid is 64, each leg has a length that is 6 less than the lenght of the shorter base, and the longer base has a length that is 4 less than twice the length of the shorter base. Find the length of an altitude of this isosceles trapezoid.

[devout snow]

Please don't occupy multiple help channels.

[supple knot]

!onechannel

[devout snow]

Please stick to your channel.

[warm dew]

.reopen

[red sierra]

show you're work

I can't just tell you the answer

[warm dew]

Let x = length of shorter base Length of longer base = 2x - 4 Length of legs = x - 6 Perimeter = 2x + 2(x - 6) + (2x - 4) 64 = 4x - 8 + 4x - 4 8 = 8x x = 8 Length of altitude = (1/2)(8 + (8 - 6) + (2(8) - 4)) = (1/2)(20) = 10

[red sierra]

why 2x?

[warm dew]

I asssumed you would hav eto doulbe the shorter base

[red sierra]

x = length of shorter base Length of longer base = 2x - 4 Length of legs = x - 6

x + 2(x-6) + (2x-4)=64
5x-16=64
5x=80
x=16

[red sierra]

no

there's only one short base of an isoceles trapezoid

[warm dew]

hmm

.close

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[ashen vault]

I learn about continues sets and continues order.
I've got a definition:
set X us continuously ordered if
$$\underset{\emptyset \neq X_0 \subset X}{\forall} \left{x\in X : a < x \underset{x \neq a \in X_0}{\forall}\right}$$
is either empty, or has minimal element.

Does it mean that natural numbers have continues order, while reals and Q doesn't have one?

[woven radish]

redve

[ashen vault]

Sets theory / topology question

we assume order relationship <

[devout snow]

@ashen vault Has your question been resolved?

[ashen vault]

<@&286206848099549185>

May I ask just for "yes" / "no" ?

[devout snow]

@ashen vault Has your question been resolved?

[ashen vault]

soup soup tasty soup

[devout snow]

@ashen vault Has your question been resolved?

[ashen vault]

2 hours in btw

[devout snow]

@ashen vault Has your question been resolved?

[devout snow]

@ashen vault Has your question been resolved?

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[quaint hearth]

how would i go about proving this?

induction i know

this is what i have so far but i dont know what i am doing

this is Applied Combinatorics btw

Another approach

[wicked turtle]

suggestion, the left hand side looks a lot like the derivative of $\sum_{i=0}^{n}C(n,i)x^i$ evaluated at $x=1$

[woven radish]

Bungo

[quaint hearth]

i dont think my teach wants us using derivatives

[wicked turtle]

ah, that's too bad, it makes this problem so simple

[quaint hearth]

https://tenor.com/view/cry-cute-gif-25985192

[slender portal]

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that
∠ B = ∠ C

[wicked turtle]

that doesn't seem relevant here

[quaint hearth]

yeah idk how triangles can prove my proof

[slender portal]

im new idk how to ask a doubt

[wicked turtle]

#❓how-to-get-help

[slender portal]

thank u

[quaint hearth]

my prof only gave us the hint of "oh its just like how we did the binomial in class!!!"

[wicked turtle]

i think this approach works!

copying again for easier reference

[quaint hearth]

AYO

ok so

[wicked turtle]

assuming it's all ok up to that point:

you know what the first sum is, by the induction hypothesis

[quaint hearth]

my thought next was to then substitute i-1 for i

[wicked turtle]

the second one is the sum of C(n,0) through C(n,n-1), which is the same as the sum of C(n,0) through C(n,n), minus C(n,n)

and you know the sum of C(n,0) through C(n,n) is...?

[quaint hearth]

2^n

[wicked turtle]

yep

and C(n,n) = 1

try making those substitutions and simplify

i think it works

[quaint hearth]

i=1 though

[wicked turtle]

sure

[quaint hearth]

from c(n,0) -> c(n,n-1)

oh

[wicked turtle]

no, if i goes from 1 to n, then i-1 goes from 0 to n-1

[quaint hearth]

youre saying instead of i->n

yeah

[wicked turtle]

so you're getting all of C(n,0) + ... + C(n,n-1)

and you're missing C(n,n)

[quaint hearth]

oh wiat

[wicked turtle]

(you could make this more formal by letting j = i-1 if you like)

[quaint hearth]

can i just make a for loop

[wicked turtle]

that's basically what a finite sum is

[quaint hearth]

wait

but what about the i in the second sum

[wicked turtle]

that's the one i'm referring to

[quaint hearth]

im not specific

one sec

[wicked turtle]

oh are you concerned about the first sum starting at i=1 instead of i=0?

[quaint hearth]

im concerned about the one multiplying C(n,i-1) each time

[wicked turtle]

yeah the second sum

[quaint hearth]

$\sum_{i=1}^{n}iC(n,i-1)$

[woven radish]

Splash

[wicked turtle]

$$\sum_{i=1}^n C(n, i-1) = \sum_{i=0}^{n-1}C(n,i)$$

[woven radish]

Bungo

[quaint hearth]

the inside is being multiplied by i

[wicked turtle]

ohh, i missed that, one moment

$$\sum_{i=1}^n iC(n,i-1) = \sum_{i=0}^{n-1}(i+1)C(n,i) = \sum_{i=0}^{n-1}iC(n,i) + \sum_{i=0}^{n-1}C(n,i)$$

[woven radish]

Bungo

[quaint hearth]

the 3rd sum is the assume

[wicked turtle]

and both sums on the RHS look do-able

[quaint hearth]

and the 4th is 2^n

[wicked turtle]

the 4th is 2^n - 1

because it only goes up to n-1

[quaint hearth]

yep

[wicked turtle]

so it's missing C(n,n), which is 1

[quaint hearth]

3rd if you start the sum at i=1 and go to n thats still the assume right? $2^{n-1}n$

[woven radish]

Splash

[quaint hearth]

so then we are at $$2^{n-1}n+2^{n-1}n+2^{n-1}+n+1$$

[woven radish]

Splash

[wicked turtle]

yea hm, why isn't this working..

are we sure that the work you had in the screenshot is right? i'll double check

[quaint hearth]

oh hell no

no way i know its right

[wicked turtle]

oh your third line is wrong

oh wait no it's not

haha

[quaint hearth]

$$(n+1)C(n+1,n+1)=n+1$$

[woven radish]

Splash

[quaint hearth]

i didnt write it

then i use the expansion rule for C(n+1,k) = C(n,k)+C(n,k-1)

[wicked turtle]

oh wait

the first sum on the RHS is not the induction hypothesis

it only goes to n-1

[quaint hearth]

can we

[wicked turtle]

so it's missing the n'th term, which is n

[quaint hearth]

WE ADD IN THE N+1 from THE OUTSIDE

FUCK

im wrong

[wicked turtle]

that first sum on the RHS should be $2^{n-1}n - n$

[woven radish]

Bungo

[quaint hearth]

that deals with the n on the outside then

[wicked turtle]

and the second sum is $2^n - 1$

[woven radish]

Bungo

[quaint hearth]

that deals with the 1

$$2^{n-1}n+2^{n-1}n+2^n $$

[woven radish]

Splash

[wicked turtle]

ah yes it works!!

lemme type

$$2^{n-1}n + (2^{n-1}n - n) + (2^n - 1) + n+1$$

[woven radish]

Bungo

[wicked turtle]

do you agree with that part

[quaint hearth]

yep

[wicked turtle]

ok good

[quaint hearth]

lets se

[wicked turtle]

i'm gonna rearrange it a bit:

there are two terms of the form $2^{n-1}n$, add them together and you get $2n2^{n-1}$, which is $n2^n$

[woven radish]

Bungo

[quaint hearth]

makes sense

[wicked turtle]

so then we have $n2^n + 2^n$ (after canceling the $n+1$)

[woven radish]

Bungo

[wicked turtle]

and $n2^n + 2^n$ is just $(n+1)2^n$

[woven radish]

Bungo

[wicked turtle]

which is exactly what you want

[quaint hearth]

oh

wow

thats it

wow

thats amazong

so i was on the right track at least

[wicked turtle]

it was so confusing going back and forth between the screenshots haha

[quaint hearth]

and my bad handwriting

[wicked turtle]

and mentally making substitutions

[quaint hearth]

how did you connect the dots

ive been trying for the past few hours

[wicked turtle]

after finally getting the right initial expression, the simplifying part wasn't too hard

[quaint hearth]

ah okay

i can never see the simplification

[wicked turtle]

mainly spotting stuff like 2 x 2^n = 2^(n+1)

[quaint hearth]

time for latex wooooo

[wicked turtle]

btw, now that you solved it the "combinatorics" way, can i show my derivative way that i was mentioning earlier?

[quaint hearth]

please do

i hate calculus, more of a finite man myself

[wicked turtle]

ok:

$$(1+x)^n = \sum_{i=0}^{n}C(n,i)x^i$$ by the binomial theorem

oops hang on

missed the C(n,i)

[woven radish]

Bungo

[wicked turtle]

now take derivatives of both sides with respect to x:

$$n(1+x)^{n-1} = \sum_{i=0}^{n}iC(n,i)x^{i-1}$$

[woven radish]

Bungo

[wicked turtle]

and plug in x=1:

$$n2^{n-1} = \sum_{i=0}^{n}iC(n,i)$$

[woven radish]

Bungo

[quaint hearth]

that is ridiculous

3 lines

[wicked turtle]

yep

now spotting a trick like that does require having seen similar tricks before

unlikely that you would just come up with it out of the blue

[quaint hearth]

for the binomial theorem he did it totally differnet

[wicked turtle]

ha, figures

[wicked turtle]

"do it just like we did with the binomial theorem"

[quaint hearth]

so he takes the s+1 out of the 3rd step and moves that +1 out of the summation

and substitutes k for k+1 so he can take out a y

[wicked turtle]

maybe he just meant to remember to use the identity C(s+1,k) = C(s,k) + C(s,k-1)