help-13

[vagrant elbow]

Any idea on how I'd eliminate p and q from the resulting expression?

[tropic oxide]

uhhhh

okay so hold on

f(x) = a + 2b arccos(x)

isnt the domain of this unambiguously [-1,1]

[vagrant elbow]

oh

I am

💀

Thanks Ann

.close

[cedar kiln]

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[sullen quest]

need help with law of sines no. 5-7

[south tundra]

!status

[cedar kiln]

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

[sullen quest]

1

[south tundra]

According to the law of sines, $\frac{BC}{\sin{\angle{BAC}}} = \frac{AB}{\sin{\angle{ACB}}}$, right?

[wraith dagger]

A Lonely Bean

[south tundra]

Ah, you are stuck at 5th too?

[sullen quest]

yeah

[south tundra]

What's the sum of angles in a triangle generally?

[sullen quest]

arent they suppose to sum total of 180degree?

[south tundra]

Yes, do you think that helps us solve for the angle ACB?

[sullen quest]

im not sure because theres the D angle thing

thats where im confused

[south tundra]

That is a problem when considering the triangle ABD

But when you consider the triangle ABC

Then you have 2 known angles and 1 unknown angle which sum up to 180

Meaning you can solve for that unknown angle, namely, the angle ACB

48 + 55 + ACB = 180
ACB = 77, right?

[sullen quest]

isnt theres a 2 unknown angle C and D?

[south tundra]

We will get to angle D later

[sullen quest]

yes thats corrrect

[south tundra]

Another thing is that the angles ACB and BCD should add up to 180 as well, so BCD is supposed to be of measure 107

Which is the answer to no.5

[sullen quest]

uhm i dont think i was able to comprehend

[south tundra]

This?

[sullen quest]

yes

[south tundra]

ACB and BCD are complementary angles here, right?

[sullen quest]

yeah

[south tundra]

So they should add up to 180
ACB + BCD = 180

We already know the measure of ACB, it is 73

73 + BCD = 180
BCD = 180 - 73
BCD = 107

[sullen quest]

how did it become 103 for the final answer?

if its 107

[south tundra]

I made a mistake, it's 107

Wait there is no 107

One moment

[sullen quest]

theres also no 107 in the choices

[south tundra]

Ah, ACB is 77 instead, not 73

[sullen quest]

oh

[south tundra]

Because 180 - (48 + 55) = 180 - 103 = 77

And BCD is hence 103

[sullen quest]

ohh wow okay i understand now

[south tundra]

Moving onto no.6

In order to solve for BC, we need to apply the law of sines in the triangle ABC, which yields [ \frac{BC}{\sin{\angle{BAC}}} = \frac{AB}{\sin{\angle{ACB}}}]
Right?

[wraith dagger]

A Lonely Bean

[sullen quest]

yeah

[south tundra]

We know the values of AB, BAC and ACB

So, why not solve for BC using that equation

Can you do that?

[sullen quest]

uh i'll try

[sullen quest]

uhm is this the formula for BC?

[south tundra]

Yes, we can solve for BC starting with that equation

Make sure to plug in the known values and then isolate BC

[sullen quest]

might take me more than 5minutes huhu wait

[south tundra]

Shouldn't be too hard

[sullen quest]

i cant im lost again😭

[south tundra]

Alright, let us start by plugging some values in

[south tundra]

That yields $\frac{BC}{\sin{48}} = \frac{18}{\sin{77}}$

[wraith dagger]

A Lonely Bean

[south tundra]

Now, we need to isolate BC by getting rid of that sin(48) on the left hand side

Which can be done by multiplying both sides by sin(48)

Finally getting us the answer for $BC$: $BC = 18\cdot\frac{\sin48}{\sin77}$

[wraith dagger]

A Lonely Bean

[sullen quest]

wait uh how did it become BC/sin48? isnt its like all of angle BAC?

[south tundra]

We can't manually calculate the values of sin(48) and sin(77),
(Well, we rather are not expected to do that*)
So we put that expression into a calculator

[south tundra]

BAC = 48, so BC/sin(BAC) = BC/sin(48)

Similar thing happened on the right hand side, ACB = 77, so AB/sin(ACB) = AB/sin(77)

We also know that AB = 18, so that turns into 18/sin(77)

[south tundra]

Ultimately, we get this

[sullen quest]

is sin48 from angle A? because thats where im kinda confused

[south tundra]

Yes, it's coming from the value of angle A

[sullen quest]

oh okay i understand now

is that answer 13.7?

[south tundra]

,w calculate 18 * sin(48)/sin(77)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by A Lonely Bean#7998

[south tundra]

Yes

[sullen quest]

ohhh i understandd

my mind keep going blank

[south tundra]

Now, moving onto no.7

Let us apply the law of sines once again, but instead in the triangle BCD this time

It yields $\frac{\sin\angle{D}}{BC} = \frac{\sin\angle{BCD}}{BD}$, right?

[wraith dagger]

A Lonely Bean

[sullen quest]

is angle D still unknown or have we already solved it that i just dont notice?

[south tundra]

No, D is still unknown and we are on the way to solving it

[sullen quest]

oh wait uh i think i'll try this one

[south tundra]

Alright

[sullen quest]

uh is BD still uknown?

[south tundra]

BD is given as 25

[sullen quest]

uhm i think theres something wrong with my equation

im not sure how to proceed now

is there something im doing wrong?

[south tundra]

Should be 13.7 instead of 17.3

[sullen quest]

i didnt noticed that lol uhh this is what i get

[south tundra]

Right, that's sin(D)

Now you put that into arcsin()

[sullen quest]

the sum i got?

[south tundra]

This, yeah

[sullen quest]

so like arcsin 0.5339

[south tundra]

Yes

[sullen quest]

the closest answer in the choices is 32degree16'

uh is this right?

what is the 16' for in the choices?

[south tundra]

16' means 16minutes

1min = 1/60deg

,w 16/60

[wraith dagger]

Results provided by WolframAlpha

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

Requested by A Lonely Bean#7998

[south tundra]

Yup, matches up

[sullen quest]

oh i see

aaaa tyy very muchh

[sullen quest]

have a good dayy

.close

[cedar kiln]

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[pastel sonnet]

Got a question,

ABCD is a Kite

[pastel sonnet]

A = (-2,10)

B = (-27/5, 4)

C = (4, -5)

AB = AD

CB=CD

find coordination of D

So I tried

setting BC = AD

[tropic oxide]

do not drop the parentheses from a coordinate specification

[pastel sonnet]

and find intersection

[dreamy sleet]

do you know distance formula

[pastel sonnet]

Yes I tried that too

I got a circle at the end

But it looks very weird

[dreamy sleet]

Can you show all your steps?

[pastel sonnet]

sure, it is very messy tho

I will do it again

[dreamy sleet]

And by the way, always write a point’s coordinates as (for example) (-2, 10), not -2, 10

[pastel sonnet]

ok gotcha

[dreamy sleet]

did you swap them?

[pastel sonnet]

nope

I just added the brackets

[dreamy sleet]

you can send it first

[pastel sonnet]

I am redoing it, because it is wayyyy too messy

It’s all around the place

Sorry

oh

I think I got where I went wrong

I forgot to squareroot

On finding AD

I skipped a step ahead at the end

I set them equal together and got rid of the sqrt

Nvm I got it

n.close

close

n’éclose

.close

[cedar kiln]

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[graceful ether]

Hey, I'm 8th grade and need help with this structured question, it's really the whole thing because I don't really understand squat. So please if anyone willing to help me with this please help

[zenith nymph]

take a guess at what y was at t=50

then subtract that from 200

[graceful ether]

Y and t?

do you mean Y and x axis?

[zenith nymph]

when x=50 then

i usually see time as t

[graceful ether]

so x = 50 right

[zenith nymph]

yes

[graceful ether]

would y be 120?

[zenith nymph]

no

[graceful ether]

hm

so if I choose 120

subtract that from 200?

[zenith nymph]

well, yeah but no

you're wrong about 120

[graceful ether]

💀

[graceful ether]

100 ?

[zenith nymph]

draw a line up from where x=50

[crimson sedge]

how'd you get 120

[graceful ether]

ok hold on

[graceful ether]

done

do i draw the line straight up?

[crimson sedge]

yes

[graceful ether]

dont mind the line that exceeding the page

[zenith nymph]

where does it intersect the curve

[graceful ether]

right here

uhm, do i say 199 ?

[crimson sedge]

it isnt 199

slightly lower

[zenith nymph]

roughly 196

and then you subtract that from 200

to get the number of students stil taking the test

[graceful ether]

how did you know it was 196 ?

[crimson sedge]

the scale goes up in 4

its on the line below 200, therefore, 196

[graceful ether]

oh.

So x = 50 and Y = 196

so 200 - 196 = 4

so it'll be 4 students ?

[crimson sedge]

indeed

[south tundra]

Yes

[graceful ether]

holy it was that easy?

so guys how do I solve the medium?

do i add everything up and divide it by 2

[zenith nymph]

no

at what time have half the students (100) solved the problem

i.e. where does y=100?

[graceful ether]

33 minutes

yes?

o

[south tundra]

Draw a horizontal line crossing the spot where 100 is on the y-axis

[graceful ether]

ok 🫡

[crimson sedge]

isnt that 33 minutes

or more precisely between 33 and 34

[south tundra]

Let me check more precisely

Ah ye it's 33min my bad

[graceful ether]

there guys

it intersect at 33 ?

[crimson sedge]

yeah 33 mins

[graceful ether]

33 minutes 🫡

[zenith nymph]

yep

[graceful ether]

wait that's the answer ?

[zenith nymph]

yep

[graceful ether]

oh waw am stupid

[zenith nymph]

that”s when half the students have solved

[graceful ether]

So how do I solve probability?

random took at most = 28 minutes

[zenith nymph]

what number of students had solved by 28 mins

[graceful ether]

i will do that now 🫡

oh boy its somewhere in 80 and 60

63 ?

[zenith nymph]

It’s lower than halfway

yeah

64 is probs what theyre looking for

[graceful ether]

hm is there a way to show this in step wise manner ?

[zenith nymph]

no

its just increments of 4

and the closest is 64

so 64/200 = 8/25 is your probability

[graceful ether]

yeah I got that

but my brother,

the hardest part has come

[zenith nymph]

da?

[graceful ether]

last time I did this I tryhard but got 0

[zenith nymph]

ez

[graceful ether]

nani

😲

[zenith nymph]

just subtract f(10) from f(20)

[graceful ether]

alright hold on,

[graceful ether]

i got

-10

[zenith nymph]

how

[graceful ether]

10 - 20

= - 10

[zenith nymph]

no

what is y when x = 20

that is f(20)

[graceful ether]

ok w h a t 😶

[graceful ether]

let me break this down

x = 20

ok brb when i get it

y would be

27?

[crimson sedge]

yeah

wait this is stats

[graceful ether]

so my brothers

f(20) would be 27? @zenith nymph @crimson sedge

[zenith nymph]

no

[crimson sedge]

x is the midpoint, not time

[zenith nymph]

?

[graceful ether]

what

[graceful ether]

hm

26?

[zenith nymph]

about 34ish

[graceful ether]

BROO

I MIXED UP 30 WITH 40

[crimson sedge]

sorry didn't look at the table

[graceful ether]

so f(20) is 34

[graceful ether]

do u want me to find for f(10)?

[zenith nymph]

yes

[graceful ether]

lets be real

its definitely 18

[zenith nymph]

its literally 12

anyways i gtg

[graceful ether]

.

so subtract f(10) from f(20)

34 - 12 = 22

[graceful ether]

ay before you go would it be 15.5 x 22 ?

oh he gone gone Don't worry depreeestri.

I will solve this and make u happy

[cedar kiln]

@graceful ether Has your question been resolved?

[graceful ether]

f(50)

f(40)

f(50) = 196

f(40) = 160

f(50) - f(40)
196 - 160 = 46

[cedar kiln]

@graceful ether Has your question been resolved?

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[median burrow]

If a particle collides with a wall and has coefficient of restitution e, do you multiply the initial velocity of said particle by the coefficient of restitution (e) in order to find the rebound velocity?

[median burrow]

I understand that coefficient of restitution = velocity of separation / velocity of approach

Wait if it’s a wall then it’s constant

So e x velocity of approach = velocity of separation

Ah okay that make sense

Okay so if

The particle is moving

And it collided with another particle moving

And you knew the initial velocity of both

Then let’s say they coalise

Wait then the velocity of separation = 0

So there is no coefficient ah that makes sense

Okay well what if the two particles do not coalesce

So both particles collide while moving

You’d have to know more information than the initial velocities of both

So let’s simplify the e equation

So it’s (v2-v1) / (v1-v2)

Oh rightttt

No you don’t need more information

Because you can form an equation using momentum

As the momentum before = momentum after using vectors

Giving you an equation for both velocity of 1 and 2 after

Then you can form an equation using e

As you know the initial veclotieis

*velocities

This setting up a simultaneous equation

So all you need is the coefficient of restitution and the initial velocities

Wait but that only applies if e = 1 as otherwise surely there would be a loss to the objects

So the momentum’s are not consistent

Apparently that is not true

So it works

.close

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[clear gyro]

I keep getting 148 for this

[cedar kiln]

Please don't occupy multiple help channels.

[clear gyro]

But im not sure what im doing wrong

[crimson sedge]

What does it say Z is?

3A?

[clear gyro]

z=3

A= what its solving for

[median burrow]

I mean do Z^X first

Then W*Z

Then just carry through with the answers from that left to right

[crimson sedge]

^

How'd you get 148 can you show your process @clear gyro

[clear gyro]

basically I did 5 * 3-3 = 12^2= 144+4= 148

[crimson sedge]

What on earth

This is how I did it

You're reading it wrong

Tbh I don't blame you, questions like these shouldn't exist - they test more on your reading skills than actual math

(w * z) - (z^x) + y

[cedar kiln]

@clear gyro Has your question been resolved?

[cedar kiln]

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[dull oxide]

I evaluated these sets correctly, right?

[finite raven]

Isn't (1,A) = {{1,A}, {1}}

{1} X {A} = {(1,a)}

[untold torrent]

{1} x {A} should be {(1, A)}

Yeah

[finite raven]

yeah

[dull oxide]

Righto

Okay thanks.

.close

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[white radish]

I am currently working on a little project that involves real world coordinates and its giving me some issues as I am not very good with math. I am attempting to draw a circle of points with radius r around a given coordinate with latitude and longitude. To do this I have found two formulas online, one that helps produce evenly spaced points on a circle:
(xk,yk)=(x0+rcos(2kπ/n),y0+rsin(2kπ/n))for k=0 to n−1. (pasted from the internet so styling is a little off)
The other formula is to basically add meters to a coordinate:
latitude + (dy / r_earth) * (180 / pi) for latitude and: longitude + (dx / r_earth) * (180 / pi) / cos(latitude * pi/180) for longitude
where r_earth is the approximate radius of the earth (I.e. 6378137) and dx/dy are the distance want to shift the point.

The way I have currently combined them is as follows.
y = origin.Y + ((radius / r_earth) * (180/ π)) * sin(2kπ / n)
and
x = origin.X + ((radius / r_earth) * (180 / π) / cos(origin.Y * π / 180)) * cos(2kπ / n)
where origin x/y are the center point and radius is in meters.

This is very close to working, but the result is always an oval instead of a circle as show in the attached image. I have checked and the latitude seems to be the correct distance away from my center point according to google maps so I am assuming something is going wrong in the longitude calculation but I do not know what.

What am I doing wrong here and how do I correctly adjust for the curvature of the earth in a calculation like this?

[cedar kiln]

@white radish Has your question been resolved?

[fading narwhal]

I'm not 💯 % sure about whether this is the problem, but the Earth isn't perfectly spherical. If you didn't account for this in your calculations then it might be stretching the circle by some factor along an axis.

[lethal jackal]

nah it's almost perfectly spherical, so it doesn't explain that amount of stretching

[fading narwhal]

yeah it's a small area

[white radish]

I would assume it has to do with the fact that coordinates don't equal the same amount of meters depending on how far along the x axis you are, right? But from what I gathered, the / cos(latitude * pi/180) part of the longitude formula was meant to account for that.

[lethal jackal]

What programming language are you doing this in?

[lethal jackal]

yeah it is

[white radish]

C#, but I didn't copy past my current code into the original question so as to keep it purely math. Here is what the actual code looks like. Ignore the return statement, I'm just attempting to have it print the correct values first.

[lethal jackal]

My guess, and this is just a guess, is that because you're doing so many degrees-to-radians conversions, somewhere you're not multiplying it correctly idk

also the order the factors are arranged is a little weird

[white radish]

Thanks for the insight. I'll go make sure that everything is happening correctly and in the correct order as well. I'll just leave this open while I do it.

[white radish]

How so?

[lethal jackal]

Like the straightforward way I would imagine computing it would be:

angle = 2 pi k / n
dx = r cos(angle), dy = r sin(angle)

point.x = latitude + (dy / r_earth) * (180 / pi)
point.y = longitude + (dx / r_earth) * (180 / pi) / cos(latitude * pi/180)

also there's a slight small angle approximation here for some reason

it's supposed to be arcsin(dy/r_earth)

but that doesn't change too much

[white radish]

Could you explain what an angle approximation is?

[lethal jackal]

small angle approximation means that one takes sin(theta) = theta for small angles theta

[white radish]

And what is theta in that instance (I'm not english by birth so I used different like names for stuff in highschool trig haha).

[lethal jackal]

$\theta$

[wraith dagger]

Saccharine

[lethal jackal]

variable name

$\sin \theta \approx \theta$

[wraith dagger]

Saccharine

[lethal jackal]

I guess this is a first-order approximation anyway so it doesn't matter

[white radish]

Oh right so you're saying it doesn't make much sense to even do the sin since the angle is so small?

[lethal jackal]

the proper way to go about it is considerably more annoying

actually wait no it's exact

it's not supposed to be arcsin(theta)

because you're actually talking about traveling the great-circle distance

in any case, it doesn't matter

but yeah try computing the intermediate quantities like I listed and see if any of them don't make sense

the formulae look mostly correct

[white radish]

Yeah I'm on it in the mean time. Haven't discovered anything yet but I'll keep on it.

One weird thing I'm noticing that, to me, indicates I'm doing it wrong is when I completely remove the / Math.Cos(origin.Y * Math.PI / 180) part from the x/longitude calculation the only thing that actually changes in my output is that the oval becomes slightly less tall even though that part should be adjusting the width, right?

[lethal jackal]

depends on which direction is north lol

you didn't indicate it on the map

[white radish]

North is up.

[lethal jackal]

yeah I can't see all of your code or understand how it works, so that would be strange

but there's also a possibility that you've just changed the zoom level; the division by Math.cos is supposed to make the thing fatter

[white radish]

I'm a little confused. It is possible that the tool I'm using might be interfering with results here. I'm using this website the check the polygon I've drawn: https://www.keene.edu/campus/maps/tool/ but for some reason requires coordinates to be written latitude, longitude instead of longitude, latitude. When I input the results of my function as x, y (longitude, latitude) it does result in a circle, but not in the correct location on earth (it doesn't recognize the location at all) while if I input the results as y,x it does display it at the right location but as an oval.

[lethal jackal]

latitude, longitude is the standard way of writing it

isn't that what your code is supposed to print too?

[white radish]

Yeah that's why I'm printing it that way.

[white radish]

Really, why does google maps do it the other way around then?

Oh shit

I think I fixed it.

I changed from the first to the second.

[lethal jackal]

the tool asks for longitude, latitude

[white radish]

I thought Y is latitude?

[lethal jackal]

not latitude, longitude

[white radish]

You're right... How come when I changed the two around then it was all the way off near afrika?

I clearly made some base level mistakes here in x/y long/lat which messed it up.

[lethal jackal]

not sure, but you probably transposed the two somewhere wrong

I suggest renaming variables to latitude/longitude instead of x/y

because x/y doesn't actually make a whole lot of sense on a sphere

[white radish]

Yeah, that would probably make it clearer. I'm also just going the document the shit out of this so no one else makes my mistakes xD.

Thanks so much for your help!

I'm glad I got this cleared up. Been working on it all day.

Haha

[lethal jackal]

In particular, these lines are fairly bugged in that your circle is rotated if you think of "y" as north

[white radish]

In my mind Y is the same as latitude. So that would be correct, would it not? The higher Y is the further up from the equator, right?

[lethal jackal]

If that's the case, then this code makes no sense, because the first line computes the longitude of the point

[white radish]

Yeah, that's why the cos was on the X first haha. So I guess that assumption of mine is wrong actually. I think I should just play around some more and print the values until I'm 100% on which value is actually what.

[lethal jackal]

I spent some time reproducing the correct code before I went to paste it in the tool and saw the order was different

import math
def compute_points():
    r_earth = 6371 * 1000
    n = 50
    r = 100
    start_latitude, start_longitude = 42.92541974089839, -72.28108600708926
    for k in range(n+1):
        angle = 2*math.pi*k / n
        dx = r * math.cos(angle)
        dy = r * math.sin(angle)

        lat = start_latitude + (dy / r_earth) * (180 / math.pi)
        lng = start_longitude + (dx / r_earth) * (180 / math.pi) / math.cos(start_latitude * math.pi/180)
        print(f"{lng}, {lat}")
compute_points()

your problem is that the variable names in your code (at least if you use y to represent latitude, x to represent longitude) indicate that you're printing latitude, longitude

but you've changed the code around so what you're really printing is longitude, latitude

and there's the whole rotational symmetry of the circle

and somehow you've made all of the mistakes cancel out

the tool requires you to print the stuff out in longitude, latitude format

[white radish]

Hahaha, leave it to me to make so many mistakes that it comes full circle and is almost right again.

That's actually crazy btw because origin is from a geospatial library and it stores its long/lat as x/y so it is also doing it wrong like me haha.

[lethal jackal]

It's really common in a lot of code, because if you want to reuse some vector libraries or whatever, you're stuck with that

but would highly not recommend confusing yourself with stuff like that

[white radish]

Yeah I'll remember this for sure. I aint ever making this mistake again haha.

Hey I'm just going to brush this up a little and then quit working on it for the day, but thanks so much for your help again! I wouldn't have figured it out alone. This has been great :).

[lethal jackal]

good luck

[white radish]

.close

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[eternal knot]

How Do I solve the system of equations to find the critical points?

im stuck on the algebra

[granite eagle]

🐸

[inland gust]

You can multiply them both together to get $25sin(x)cos(x)sin(y)cos(y) = 0$ which is equivilent to $sin(2x)sin(2y) = 0$

That might help getting solutions

[wraith dagger]

WeAreIngram

[eternal knot]

@inland gust does that look right?

also -pi/2 since the range is from -pi to pi

im also confused on how to match them up to find actual (x , y) pairs

[inland gust]

Look good

[inland gust]

You could manually check them all 🙃

[eternal knot]

yes perhaps

.close

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[tall mica]

.

Need help with a PDE question

[tall mica]

what I have up to now

Those are my ideas for how to go on

[cedar kiln]

@tall mica Has your question been resolved?

[tall mica]

<@&286206848099549185>

[cedar kiln]

@tall mica Has your question been resolved?

[tall mica]

<@&286206848099549185>

[dire geode]

What's your question

[tall mica]

this is "the" question

dont know how to use the hint abt the Fourier coeffs of sin^2

[cedar kiln]

@tall mica Has your question been resolved?

[cedar kiln]

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[glossy shoal]

I have no idea how to start this

[rose swan]

do you understand what information the problem gave you?

[glossy shoal]

Yes

[rose swan]

you might want to start labeling sides with the lengths they gave you

[glossy shoal]

I just write the ratio to prevent messy

[rose swan]

that's a good start

can you extend the plane to visualize where it intersects AD?

[glossy shoal]

Ok

its not drawn to cale tho

s

[rose swan]

yeah it's tricky, but that should be good enough

you can label L now

[glossy shoal]

OK I labelled

[rose swan]

i'm trying to get a nice picture on geogebra, you can give it a try yourself: https://www.geogebra.org/m/aY75dEkf

[glossy shoal]

Thx

[rose swan]

yeah, your picture looks pretty good

[glossy shoal]

Yeah

I can't make this

[rose swan]

so some triangles you can draw now include LAB, LHD, and LEH

[glossy shoal]

what theyre for?

[rose swan]

they might help you find some side length relationships

since ultimately you want an expression for AL:LD

[glossy shoal]

Yes

Let me see

[rose swan]

there's more triangles you can make which may be useful, but i haven't solved this yet

i'm guessing it's just spamming the pythagorean theorem and using similar triangles

[glossy shoal]

I still can't see anything

[rose swan]

keep playing around with the different triangles you can make: there are more than the three i suggested

[glossy shoal]

I still have no idea how to do it

There are too many variables

[rose swan]

yeah i'm struggling with this too, maybe see if anyone else can help

i think the side lengths all have to be (1+r)(1+s)(1+t) though

[glossy shoal]

Ok

Thank you for your help

[rose swan]

i'll let you know if i figure it out later

[glossy shoal]

👍

[cedar kiln]

@glossy shoal Has your question been resolved?

[cedar kiln]

@glossy shoal Has your question been resolved?

[cedar kiln]

@glossy shoal Has your question been resolved?

[cedar kiln]

@glossy shoal Has your question been resolved?

[cedar kiln]

@glossy shoal Has your question been resolved?

[glacial sentinel]

is this proving?

you better start by writing out the equations

$$ \frac{AI}{IB} = \frac{1}{r} $$ $$ \frac{DJ}{JH} = \frac{1}{s} $$ $$ \frac{FK}{KG} = \frac{1}{t} $$

[silent finch]

The synthetic solutions seems difficult, it seems best to place the cube on the 3D cartesian plane and find the equation of the plane in terms of x,y,z,r,s,t and find where it intersects AD

[wraith dagger]

Bring Back Beatrix

[glacial sentinel]

and you know if you draw a line from K to perpendicular to BC, you know that for example name this point K_1, then BK_1:K_1C=FK:KG

but wait

since it uses r, s and t

you could write this

$$ \frac{IB}{AI} = r $$ $$ \frac{JH}{DJ} = s $$ $$ \frac{KG}{FK} = t $$

[wraith dagger]

Bring Back Beatrix

[glacial sentinel]

then its just a matter of substitution

you just reverse engineer the equation

fuck latex

ms word equation is much better

[cedar kiln]

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[worn pulsar]

Does anybody know how to solve this? It's database task, but someone maybe know it? It says "Find keys: relation"

[crimson sedge]

there's a cs server in #old-network

[dawn junco]

still here ?

[worn pulsar]

Yep

[dawn junco]

have you tried anything or not ?

[worn pulsar]

trying to find solution on yt

idk nothing yet

[dawn junco]

lol

alright

you know what the C -> D notation means ?

[worn pulsar]

c determines d

[dawn junco]

yeah alright

could you draw the graph of dependencies in this relation ?

at least it's gonna be visual it's more fun

[worn pulsar]

yes i'll do it

Hahahahahaha

[dawn junco]

yeah p much indeed

I'll say the cycle has shocked me a little at first lol

so now I guess the thing you forgot is, what does it mean to be a key from the dependencies ?

I also forgot tbh had to look it up

[worn pulsar]

HAHAHAHAHA

bro honestly i don't know really, im watching indians on yt

it's explained so bad in school presentation so im pretty much hopeless without any help

[dawn junco]

I mean the idea behind a key, is that you can identify which tuple you have only knowing the key right

[worn pulsar]

aight

[dawn junco]

i.e. if you know the key, you can deduce all the other attributes in the relation

[worn pulsar]

yea i get that

[dawn junco]

(the key can have multiple attributes inside of it btw)

now let's have fun with the graph a little bit

if we suppose we know AB, can we deduce all the other attributes from that ?

[worn pulsar]

C?

and D?

[dawn junco]

yeah you can know C cause AB->C

and since C->D and we know C we know D

so we won

AB is a key

[worn pulsar]

Oh

it was that simple? xd

[dawn junco]

we're not finished tho

there could be other possible sets of attributes which are also keys

[worn pulsar]

hmm

[dawn junco]

have fun with this cycle mate

[worn pulsar]

honestly

i think they are all keys

xd

[dawn junco]

exactly lol

[worn pulsar]

HAHAHHAHA

wow

[dawn junco]

that cycle thing makes it fun lol

[worn pulsar]

i'm sleepy so i didn't notice

YEA exactly lol

sorry for taking your time bro

[dawn junco]

nah it's ok

[worn pulsar]

i was so stupid

can you help me with one more task? short one

[dawn junco]

let's see that thing

[worn pulsar]

it says

find shutters

or how i can translate that to english

[dawn junco]

sounds like transitive closure to me

i.e. what can I deduce from B

[worn pulsar]

maybe, i dont know how to translate it correctly

[dawn junco]

what can I deduce from C

etc...

[worn pulsar]

exactly yea

so we have written here, that:

we look left side of arrows

so we look b, e, c ,d

and we check is the letter part of a) B, b) D...

you get me?ž

am i doing it right

[dawn junco]

i get you roughly

we start from a state where we know only B

[worn pulsar]

yea

[dawn junco]

from there we try to see if there's dependencies we can apply

for example B -> DE is usable here

now we know new stuff

[worn pulsar]

yea right

[dawn junco]

we know B D and E

etc..

and you continue this process until your state doesn't grow anymore

the final state is your answer

[worn pulsar]

so for a) would answer be BDE or should i continue procces

[dawn junco]

can you apply some rules knowing B D and E ?

[worn pulsar]

yea ok, so first step is to see left side of B -> DE, E -> C.. so i look for first letter, which are B, E and others

then i check if B is part of Bf

B is part of Bf

and then i check if there is DE there, if there is not, i add DE

that's written in my presentation, so i HOPE it's right

[dawn junco]

sounds about right yeah

[worn pulsar]

thank god :D

so, is for a) BDE right answer?

[dawn junco]

nah

you can add stuff still

with these newfound D and E

[worn pulsar]

Oh, so i can add C also, because E -> C, then i can add A, couse C -> A and finally add B couse D -> B

?

[dawn junco]

yeah exactly

[worn pulsar]

so it's pretty much all the letters? :)

[dawn junco]

yeah

so B is a potential key here

[worn pulsar]

oh yea B is potential

so i don't write it twice

BDECA then?

[dawn junco]

the order doesn't matter

ABCDE is fine

[worn pulsar]

you are legend mate

thank you so much

[dawn junco]

you can do the others, i can check if you want

[worn pulsar]

imma do b) D, and then i'll need your help with d) AC couse i don't know with 2 letters

[dawn junco]

that just means you start with A and C

simple

[worn pulsar]

so with D, it's all numbers again, because D -> B, B->DE, E -> C, C->A

So ABCDE, right?

[dawn junco]

yeah

[worn pulsar]

<3

[dawn junco]

you have D->(a key) so you can deduce everything

[worn pulsar]

i'mma try with E

with E, only ECA

couse E -> C, C->A and A -> nothing xd

right?

[dawn junco]

yup

[worn pulsar]

thank you bro you are savior

really thanks for explaining me in 20mins which this guy couldn't in his 2000 words document

[dawn junco]

more like 30mins

cheers mate

[worn pulsar]

yup sorry for that xd hope you do well bro wish you best

[dawn junco]

good luck with your exam ig

[worn pulsar]

!close

how do i close thiis

xd

[dawn junco]

.close

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[mint moth]

I'm not sure what to do or where to start solving...

I first did 18 + 4.50 : 4
And I didn't know what to do from there and that's probably wrong 😅

[scarlet garnet]

do both equations first

Money(Time)

[mint moth]

?

I'm a bit confused on what u said there

[worn pulsar]

do (18 + 4.5) / 4 = 15 : 8 so it would be 8 x (18 + 4.5 ) = 15 x 4

[mint moth]

Ohhh

Okok then I solve from there ^

[worn pulsar]

good luck :)

[mint moth]

Hmm I got 180 = 60

how would I use that to find out how many weeks

[worn pulsar]

ok so let weeks be x

so 4.5x

it would be 8 x (18 + 4.5x) = 15 x 4x

then yu have 144 + 36 x = 60x

then you have 60x - 36x = 144

24x = 144

x = 6

you get it? xd

[mint moth]

I'm trying to understand how u got that answer by trying to do it myself with ur guidance - might take some time

.close

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[hasty dagger]

The question circled in green and pink,
how would you go about solving it?
It’s 1 mark so it should only take a minute according to exams but I’m not too sure how 500 is the answer.

[dawn junco]

just look at the recursive definition straight in the eyes

[hasty dagger]

Sorry?

[dawn junco]

it relates very closely to the sum you have

[dire geode]

Solve for your summand

pretend n=r

[slate lintel]

that's a cute problem

[cerulean star]

funny indeed

@hasty dagger you need to substitute

certain given information

as one does in math

[hasty dagger]

Alright, thank you!

Another, question, how would you resolve the problem? Currently written the forces but not sure how to find speed, would you calculate GPE loss and then KE gain?

[cedar kiln]

@hasty dagger Has your question been resolved?

[slate lintel]

that's how I'd do it yeah

[hasty dagger]

Lovely, thank you

.close

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[crimson sedge]

I am stuck in a

[crimson sedge]

Express the following

5x-35

-12mn+2m

[grand forge]

hm?

[crimson sedge]

So, i am stuck in a

[dire geode]

You're missing a lot of information

[crimson sedge]

ohh

sorry

[dire geode]

Read what you sent again and see if anyone besides you can understand

[crimson sedge]

what i meant

was that i am stuck in finding out the solution to A or 5x-35

Sorry again!

[grand forge]

can u send a picture of the question

[crimson sedge]

how to?

[grand forge]

use your phone

[crimson sedge]

ok

i did it

but i dont know how to send it to my compyuter

[slate lintel]

sign into discord on your phone

or email the picture to yourself, that's a trick I've always used

[crimson sedge]

ok

sorry if its blurred

[iron saffron]

as a what

[crimson sedge]

product of two or more terms

[iron saffron]

So, do you know what product means?

[crimson sedge]

yes

[iron saffron]

Do you see any product there?

[crimson sedge]

i do not think so

[iron saffron]

what does 5x means to you?

[crimson sedge]

x+x+x+x+x

[iron saffron]

isn't that the definition of product?

[crimson sedge]

or 4x+1x

[iron saffron]

3+3+3 = 3 * 3

x+x+x+x+x = 5 * x

right?

[crimson sedge]

yes

[iron saffron]

you have a product there then

and 35 is another product

which one

[crimson sedge]

i think we just need to factor the term

[iron saffron]

which product is inside 35?

[crimson sedge]

7x5

[iron saffron]

ok, what's in common then

between 5x and 35

[crimson sedge]

5

[iron saffron]

do you know how to factor out?

[crimson sedge]

no

[iron saffron]

read this carefully

a * b + a * c = a(b+c)
a * b - a * c = a(b-c)

What do you have?

first or second?

[crimson sedge]

second

[iron saffron]

write everything replacing with your numbers

[crimson sedge]

like everything everything

or just the secind

[iron saffron]

a * b - a * c = a(b-c)

this one

[crimson sedge]

5x35-5x7=5x(35-7)

[iron saffron]

that's not what you have

you have 5x + 35

we agreed 5x = 5 * x

and we agreed 35 = 5 *7

also, never use "x" as multiply symbol

use asterisc *

[crimson sedge]

ok

[iron saffron]

so write again

5x = 5 * x and 35 = 5 * 7

a * b - a * c = a(b-c)

[crimson sedge]

535 - 57=5x(35-7)

oops

[iron saffron]

?

[crimson sedge]

535 - 57=5x(35-7)

[slate lintel]

use x instead of star

wait don't, but use uhhh

[iron saffron]

leave a space between asteriscs

[crimson sedge]

5 * 35 - 5 * 7=5x(35-7)

[iron saffron]

ok but read the exercise

The second thing you typed is correct

• 35 is equal to - 5 * 7

but the first thing is 5x

5x is NOT equal to 5 * 35

5x is equal to 5 * x

x is a number you don't know

it's not a symbol

[crimson sedge]

5 * x - 57=5x(35-7)

[iron saffron]

i guess you meant 5 * x - 5 * 7

sorry

this

but after the equal you are wrong

this is incorrect

remember what I wrote

a * b - a * c = a(b-c)

Tell me who is "a", who is "b", and who is "c" in your exercise

[crimson sedge]

a=5 b=x c=7

[iron saffron]

perfect

so now you must write this

a(b-c)

this means a * (b-c)

[crimson sedge]

5(x-7)

[iron saffron]

that's correct

when you put a parenthesis after a number, you don't need to use multiply symbol, it's there even if you don't put

so that's it

now you do it with part b

find what's in common

and do the same

[crimson sedge]

-2m(6m-1)

oops

[iron saffron]

that's close

[crimson sedge]

-2m(6n-1)

[iron saffron]

perfect

[crimson sedge]

thank youu

.close

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[cedar kiln]

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[pallid ether]

Can somoene please tell me if i have done this problem correctly ? my prof gave us examples to do but didnt give us an answer key.. here is the problem: https://preview.redd.it/z4gkgu5lx25b1.png?width=1640&format=png&auto=webp&v=enabled&s=eec1cd94d5f8df0cb8ec5b23a1c0bea851844ecb

[pallid ether]

and this is what ive got so far: https://flic.kr/p/2oGcy4s

[slate lintel]

looks right to me. I don't think you needed to actually do the computation, but that can help to prove that it is indeed what you think it should be (dollars per month)

[pallid ether]

thank you for answering !

.close

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[dense basalt]

Are both x's and angles equal?

[scarlet widget]

They have the same name...

[dense basalt]

I have used the name to signify

I can redraw it with different names

is alpha = beta

and x = y ?

[livid hound]

do x and y represent the lengths?

[dense basalt]

Yes

Just the name

[livid hound]

then yes

[dense basalt]

Is this the reason why we can say that 120 = 90 + 30

Therefore, use 30 degrees

In calculations?

[scarlet widget]

Yes they are necessarily equal sorry

[dense basalt]

(with appropriate sign of course)

[slate lintel]

you can show the triangles are the same via ASA (one of them is a right angle)

[livid hound]

well you don't really use
30 because of 120=90+30

[dense basalt]

Why do you do so?

I mean

Yeah just explain whatever xd

[scarlet widget]

Beta represents 180° + alpha actually

[livid hound]

the angle marked is congruent to alpha

[dense basalt]

Yeah it does, and I removed the 180 degrees part for the very reason to ask if they are the same

So to check whether that's why they are used

[dense basalt]

Congruent?

They are the same

ALright xd

[livid hound]

the use of references/related acute angles simplifies the values in the problem making it easier to evaluate

[dense basalt]

Yes

It does

But why is it mathematically / logically correct

To use them?

[livid hound]

congruent triangles, vertical angles, angle sums

[scarlet widget]

@dense basalt from your example you can say that sin(alpha + 180°) = -sin(alpha) and cos(alpha + 180°) = -cos(alpha)

[dense basalt]

Again, that doesn't "prove" what I asked

[dense basalt]

This does

I suppose

Thank you very much!

.close

[cedar kiln]

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[viral mural]

I'm confused why different expressions are returned?

[viral mural]

impDiff(y=f(x), x, y) returns the derivative in terms of x

come on man

[cedar kiln]

Please read #❓how-to-get-help

[slate lintel]

my guess is that those are equivalent based on the (implicit) definition of y(x) but I haven't checked

[viral mural]

could you check? I don't think they are

oh

yeah no

[slate lintel]

what do you get if you multiply the numerator of the first one by $x^2(y^2+2y-1)$, which equals 1?

[wraith dagger]

kitten.in.a.teacup

[viral mural]

cough

fair enough

thank you!

.close

[cedar kiln]

Channel closed

Closed by @viral mural

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[edgy heart]

Hi can someone help me with an easy thing please

[edgy heart]

I have (x+1)(x-2)<0 (inequalities and stuff)

And the answer is -1<x<2

Can someone explain it to me?

[scarlet widget]

It's a multiplication between 2 factors

[edgy heart]

Idk why the x is bigger

Than the -1

Because i separate the two parentheses

And i have x<-1 and x<2

[scarlet widget]

positive × positive = positive
positive × negative = negative
negative × positive = negative
negative × negative = positive

[edgy heart]

But the answer says that the x is bigger than the -1

[slate lintel]

i'd recommend sketching the parabola and marking where it crosses the x-axis

[scarlet widget]

(x + 1)(x - 2) is negative, so we have to see the bold cases

[edgy heart]

Ye ik but you only have to substrate or add the 1 in the two sides right?

[edgy heart]

Oh fuck

YOU'RE A GENIUS MAN

[scarlet widget]

lol

[edgy heart]

But wouldn't be there

Two possible answers?

If a is positive and b negative

And viceversa

[scarlet widget]

Let's try to solve it first

Let's start with positive × negative = negative

Which means that (x + 1) is?

[edgy heart]

POSITIVE

And x-2 negative

Right

?

[scarlet widget]

Yeah

x + 1 > 0
x > -1

[edgy heart]

So it's has to be positive

Oh shet

[scarlet widget]

And

x - 2 < 0
x < 2

In the first case, x is between -1 and 2

[edgy heart]

Yeah u right

[scarlet widget]

But you get it?

[edgy heart]

And in the other case I have x<-1

And x>2?

[scarlet widget]

Let's see

x + 1 is negative and x - 2 is positive

[edgy heart]

Holy molly

What's that

[scarlet widget]

x + 1 < 0
x < -1

And

x - 2 > 0
x > 2

But x can't be less than -1 and greater than 2

Second case is invalid

[edgy heart]

Why it can not

Oh shit

Stupid question

Sorry

My bad

[scarlet widget]

No problem lol

[edgy heart]

Yeah I understand now