help-19

[frozen stream]

Ok ty

.close

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[mystic saffron]

i need help

[mystic saffron]

i have it due in 30mins

@tough torrent can you help me pretty please

@keen talon can you pretty please help me

[keen talon]

?????

[mystic saffron]

help me on a math problem

please

i have it due in 30 mins

[keen talon]

Sorry busy now

[mystic saffron]

@proud swift hello i need help please i have it due in 30 mins

[cold sage]

is it assessed work or something

[mystic saffron]

wdym assessed work?

[cold sage]

something that contributes to a grade

also stop pinging people

[mystic saffron]

no it isnt assessed

[mystic saffron]

sorry

[cold sage]

which part do you need help on

[mystic saffron]

the hole thing

spent a rough 3 hours on it

[cold sage]

how much fabric in total is needed for one apron

[mystic saffron]

asked help for my friends

5/8?

[mystic saffron]

5 over 8?

[cold sage]

how did you come to that?

thats just how much is needed for the tie

[mystic saffron]

oh then 1 1 over 4?

[cold sage]

not quite

its 1 1/4 + 5/8

[mystic saffron]

it is

7 over 8?

[cold sage]

do you know what 1 1/4 is as a real fraction

i said real, i dont accept mixed fractions, repulsive things

[mystic saffron]

im sorry i dont know what you mean bro im dumb im sorry

[cold sage]

1 1/4= 5/4
so youre doing 5/4 + 5/8

[mystic saffron]

oh

then i dont know what to do

[cold sage]

add the fractions

get a common denominator

[mystic saffron]

11/4 and 5/8

?

[cold sage]

what?

[mystic saffron]

you said add

add what

[cold sage]

5/4 and 5/8

5/4 + 5/8

[mystic saffron]

15 over 8

[cold sage]

yeah

15/8 yards of fabric are needed for one apron to be made

so how much is needed for 3

[mystic saffron]

45/8?

[mystic saffron]

45/8?

[cold sage]

yeah

[mystic saffron]

but i gotta show very step

[cold sage]

i told you every step pretty much

[mystic saffron]

where did you get 5/4?

[wind flame]

1 + 1/4 is 5/4

since 1 can be rewritten as 4/4

[mystic saffron]

but it doesnt say plus

[wind flame]

its written differently

i just wrote it with a plus for clarity

so you didn't think it was 1 * 1/4

[mystic saffron]

you know what forget bout it imma get a F anyways that bitch wont help me

[wind flame]

maybe try practicing online? if your teacher isnt a help you can take it upon yourself to get a good grade

[mystic saffron]

not you

the teacher

[mystic saffron]

how tho?

plus its due in 5 minutes

[wind flame]

oh i thought this was practice for a coming test

[mystic saffron]

no no no its due

[wind flame]

alright

[mystic saffron]

i dont care no more bro

[wind flame]

i dont think i can explain adding and subtracting fractions to you within 2 minutes to leave you enough time to finish the assignment

i also have to go, sorry about that

[mystic saffron]

i mean it will just be marked as overdue

ok

[odd edge]

@modest summit Has your question been resolved?

[odd edge]

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[jagged bramble]

hey! how would i prove that the relation x mod y = 0 is transitive for all positive natural numbers?

[low locust]

x mod y = 0 means that there exists some integer k with x=ky

now do the same for y mod z = 0

[jagged bramble]

ah so, say we have a, b and c

a mod b = 0 -> there is a y with a = yb
b mod c = 0 -> there is an x with b = xc

and so we have a = xy c and therefore the transitivity is fulfilled. My question is now, how would I show this formally

[timid siren]

That looks like a formal proof to me

[jagged bramble]

okay, thank you!

.close

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[mystic saffron]

first i need to find the centre but idk how to

[stark coyote]

look, two points share the same y value. that should be a big hint.

[mystic saffron]

yh the midpoint of the points r 1,0

[stark coyote]

do you know the standard form of a circle

[mystic saffron]

yh

[stark coyote]

what do we need to fill it out?

[mystic saffron]

x^2 + y^2 = r^2

yh

[mystic saffron]

umm

we need to find the radius

and the centre

but i dont think we can find the radius using those points because we don't know if its a diameter

[stark coyote]

$\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}$

[clever fjord]

b0ngl0rd

[stark coyote]

whats the diameter? the length from one point on the circle to the opposite point.

[mystic saffron]

yh

[stark coyote]

visualize those points A and C on a graph

what does AC represent?

[mystic saffron]

just a chord?

[mystic saffron]

ummmm i dont think its the diameter because we dont know if the y axis r the line of symetrical

however

[stark coyote]

each point is on the circumference. we can create a system of circle equations and solve for h and k and r.

[mystic saffron]

isnt h and k the coordinate of the centre?

[stark coyote]

yes.

[mystic saffron]

but i think ik what the h is

becuase when we found the midpoint between the points

it gives the line of symetrical

which is x=1

(1, k)

this is very hard 😭

[lavish osprey]

here's another way thinking about this. What does the radius represent? What can we say about the radius in relation to these 3 points?

[mystic saffron]

a triangle?

[lavish osprey]

no

the radius represents an equidistant line between the center and the points on the circumfrence

in other words, each point shares the same length

with the center

[mystic saffron]

true

[lavish osprey]

so what can we say about the distance between points and the radius?

[mystic saffron]

idk what can we say about them?

right triangle?

[lavish osprey]

no don't think about triangles

the istances are equal

by definition

[mystic saffron]

yh

[lavish osprey]

so in other words, if you do the distance formula between a known point and the unknown radius, it should be the same between another point and the radius

[mystic saffron]

true

[lavish osprey]

in other words the distance between every point on the circumfrence and the radius is equal

is there a formula we know we can use?

hint: The name of the formula is in the previous sentence

[mystic saffron]

wait i dont get that

[lavish osprey]

ok so the points lie on the circumfrence right

[mystic saffron]

yh

[lavish osprey]

all points of the circle

the radius is a line that represents the distance between the center and that point

[mystic saffron]

yh

[lavish osprey]

Now, if we move the radius around, does the value of the radius change?

[mystic saffron]

no

[lavish osprey]

bingo

I'm not sure why you're showing me that lol

[mystic saffron]

so they r all the same

so what does that tell us

[lavish osprey]

that the distance between a point and the radius is equal

say if we have points A, B, and C

then the distance from A to the radius is equal to the distance from B to the radius

same goes for C to the radius

in other words, we can equate the distances and solve for our unknowns

what is the distance formual?

*formula

[mystic saffron]

pythagorus

[lavish osprey]

$\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$

[clever fjord]

MellowDramaLlama

[lavish osprey]

that's the answer I was looking for

the distance formula

then we can do the following

[mystic saffron]

AB => 8root2

[lavish osprey]

hey can you do me a favor

can you stop jumping the gun?

just follow along

[mystic saffron]

oh ok

[lavish osprey]

I'm not looking for the distance between AB

I'm looking for the distance between teh radius and A

AND

the distance between teh radius and B

ok so watch this

[mystic saffron]

ok

[lavish osprey]

Let $(h, k)$ be the center of our circle.$\\$ The distance between our radius $r$ and our point $A$ is as follows: $\sqrt{(-4 - h)^2 + (0 - b)^2}\\$ Similarly, the distance between our radius $r$ and our point $B$ is: $\sqrt{(4 - h)^2 + (8 - k)^2}\\$ Finally, the distance between our radius $r$ and our point $C$ is: $\sqrt{(6 - h)^2 + (0 - k)^2}$

[clever fjord]

MellowDramaLlama

[lavish osprey]

Since they're the same distance, we can set them equal to each other. Let's do the first two:
$\\ \sqrt{(-4 - h)^2 + (0 - b)^2} = \sqrt{(4 - h)^2 + (8 - k)^2}\\$
Then we get $(-4 - h)^2 + (0 - b)^2 = (4 - h)^2 + (8 - k)^2$

[stark coyote]

try $$

[clever fjord]

MellowDramaLlama

[lavish osprey]

Now expand those out and find an equation with both h and k

think you got that?

are you following the logic here?

[mystic saffron]

yes

[lavish osprey]

ok expand and simplify as much as possible and get back to me

[lavish osprey]

Thanks I'll do that next time. I'm still learning LaTex (and am too lazy to look things up haha)

[stark coyote]

i think its more to do with the latex bot. i think double $ is for multi-line.

[lavish osprey]

aaaah ok

Since they're the same distance, we can set them equal to each other. Let's do the first two:
$$ \sqrt{(-4 - h)^2 + (0 - b)^2} = \sqrt{(4 - h)^2 + (8 - k)^2}\$$

[clever fjord]

MellowDramaLlama

[lavish osprey]

aaaye look at that

much cleaner

[stark coyote]

🅱️

[lavish osprey]

🅱️ 🕛 🇳 🇬 👑

[stark coyote]

i wanted to bring attention to the third variable b in your latex

[lavish osprey]

oh whoops yeah your're right

my bad lol

ok hold on

[stark coyote]

i think he gets what you're saying though lol

[lavish osprey]

Since they're the same distance, we can set them equal to each other. Let's do the first two:
$$ \sqrt{(-4 - h)^2 + (0 - k)^2} = \sqrt{(4 - h)^2 + (8 - k)^2}$$

Then we get
$$(-4 - h)^2 + (0 - k)^2 = (4 - h)^2 + (8 - k)^2$$

[clever fjord]

MellowDramaLlama

[lavish osprey]

there we go

[lavish osprey]

yeah I think so too

[mystic saffron]

ngl i think i found more efficient way to solve it

[mystic saffron]

@lavish osprey u here?

what if we just find the midpoints of the each points

and draw a symetrical line to find the centre?

using this method, i got the centre of 1,3

[stark coyote]

technically thats what you were doing with his method of equalities. you were finding the point where all 3 circle equations intersect.

you just did it geometrically

[mystic saffron]

oh

[stark coyote]

you also could've found the circle through some fancy geometry using the chord AC

or any of the chords honestly

can you find the radius though?

[mystic saffron]

yh i can just find it using the centre and any points

using the pythagrous

[stark coyote]

use the distance formula

[mystic saffron]

yh

[stark coyote]

(x-h)^2 + (y-k)^2 = r^2

solve r

[mystic saffron]

yh

so i just got 34

and the mark scheme said its correct

[stark coyote]

nani

[mystic saffron]

typo

[stark coyote]

should be $\sqrt{34}$

[clever fjord]

b0ngl0rd

[mystic saffron]

yh i meant in the formula

thanks for u guys help

$\sqrt{34}$

[clever fjord]

rella

[lavish osprey]

I guess there is another way with using chords that you were picking up on @mystic saffron .

Do you want to stick around and see?

[mystic saffron]

sure

[lavish osprey]

if not no worries, I just need a minute to get my bearings in order

[mystic saffron]

no plz

i want to expand more knowledge

[lavish osprey]

this is super cool I never thought about doing this before

maybe it works only for this problem I'd need to test it out

[mystic saffron]

oh please tell me

[lavish osprey]

ok so we have just our 3 points to start out with, yeah?

[mystic saffron]

yh

[stark coyote]

actually the fact they gave 3 points might allude to them wanting you to do this with chords.

[lavish osprey]

so we find the formula for the line between two points

say if we did (-4, 0) and (4, 8)

the equation would be y = x + 4

[mystic saffron]

yh

[lavish osprey]

now let's do teh same between the points (4, 8) and (6, 0)

[mystic saffron]

yh

[lavish osprey]

we get the equation y = -4x + 24

now, like you said, we find the mid point between the two

[stark coyote]

to make it cleaner you can set boundaries for your lines like {-4 <= x <= 4} for that line from (-4,0) to (4, 8)

[lavish osprey]

so we get (0, 4) and (5, 4)

updated photo, thanks bonglord ❤️

now we find the perpendicular line of our midpoints

for our point (0, 4) we get y = -x + 4

(I didn't limit this one because wedont' have an intersection yet

repeat for the same for (5, 4)

[mystic saffron]

k

[lavish osprey]

we get y = 1/4 x +11/4

now, my claim is that the center of the circle is at the intersection of these two perpindicular lines

so we set -x + 4 = 1/4 x + 11/4

and we solve

and we get x = 1

we plug that back into one of the perpendicular equations (y = -x + 4) and we get y = 3.

so our center should be (1, 3)

[mystic saffron]

👍

thats beautiful

[lavish osprey]

now we just find the distance between a circumfrence point and teh radius and square it to get r^2

and we get 34

$\left(x-1\right)^{2\ }+\ \left(y\ -\ 3\right)^{2\ }=\ 34$

[clever fjord]

MellowDramaLlama

[lavish osprey]

Plugging that in we get this

sorry I'm terrible at geometry so I'm super proud. This is most likley pretty well known

but there ya go

that's how to use two chords to find the center

you were on the right track

so really nice work

[mystic saffron]

very beautiful work

thanks for showing me this

❤️

[lavish osprey]

yeah of course

give yourself credit too

[lavish osprey]

you did like 75% of the work with this

the other method would work too btw, but I agree with you this is easier

[mystic saffron]

true but i love how u showed that there're always another way to solve it

[lavish osprey]

I felt like we grew together

lol

did you have any other problems?

[mystic saffron]

yh i do

[lavish osprey]

ok shoot

[mystic saffron]

i found the distance between the 0,12 and the tangent but idk how to find the formular

[lavish osprey]

hmm ok lemme look

[stark coyote]

hmmm

[lavish osprey]

oh

oh yeah okay

[stark coyote]

y = mx + c, keep m as a variable, sub in the point 0,12

[lavish osprey]

so we have a linear line y = mx + b. Can you tell me what the b is here?

ah yeah bong is already on it lol

[mystic saffron]

12

[lavish osprey]

good. Now you can substitute y = mx + 12 into your equation of a circle for where you see y. You'll get (x-6)^2 + (mx + 12 - 5)^2 = 17. Now you can expand and solve

[mystic saffron]

but we get x and m

as the variable

[lavish osprey]

oh yeah hmmm

thinking

oh well m is just a constant right

[mystic saffron]

yh

[stark coyote]

think of it like a coefficient

[lavish osprey]

if we expand we get the following:
$$ x^2 - 12x + 36 + m^2x^2 + 14mx + 49 = 17 $$
Simplifying and grouping we get:
$$ m^2x^2 + x^2 + 14mx - 12x + 36 + 49 - 17 = 0$$
Finally we can factor out x terms and simplify our constant:
$$ x^2(m^2 + 1) + x(14m - 12) + 68 = 0 $$

[clever fjord]

MellowDramaLlama

[lavish osprey]

Now we can use the quadratic formula

which will be messy lol

[stark coyote]

not even

you need the double root

[lavish osprey]

OOOOOOH

[stark coyote]

so we just need to make sure b^2 = 4ac

[lavish osprey]

yeah you're right

[mystic saffron]

speak english plz?

[stark coyote]

wait let me double check this

hmmmmmmmmmmm

i dont think that = 2m^2x^2 lol

[lavish osprey]

lol

[stark coyote]

but we do still need the double root

[lavish osprey]

just curious why we're ignoring the other factors of x?

[stark coyote]

we are not im being dumb, excuse me

[lavish osprey]

no you're fine!

[mystic saffron]

we only need to find the coordinates of the tangent

hmmm

can we use the radius to find the other cords?

[lavish osprey]

I think the idea that bonglord is trying to say is that since our lines are tangents, there is only one solution for when the lines intersect with the circle. For this to be true, the discriminant must be 0 (aka b^2 - 4ac = 0). The only way for that to happen is if b^2 = 4ac.

[mystic saffron]

oh

[lavish osprey]

So if that's the case, then we let $a = (m^2 + 1), b = 14m - 12, c = 68$. Thus:
$$ b^2 - 4ac \implies (14m - 12)^2 = 4(m^2 + 1)(68) $$
Then we can solve for m, our slope

and that's all we care about. We don't have to worry about what x actually is

we just used x to get to m lol

[clever fjord]

MellowDramaLlama

[stark coyote]

yes! damn why was that so hard for me. my brain is fried from linear algebra

[lavish osprey]

lol no worries at all.

I still have COVID brain so I'm moving a mile a day

[mystic saffron]

ok so whats m

[stark coyote]

gradient/slope of the tangent

[mystic saffron]

we can use that to find m?

[lavish osprey]

we have y = mx + 12, so we're looking for m so we have our actual lines

and yes we can 🙂

We get the following:
$$ 196m^2 -336m + 144 = 272m^2 + 272 $$

[clever fjord]

MellowDramaLlama

[lavish osprey]

now we just need to solve for m

[mystic saffron]

wow!!

[lavish osprey]

we'll need a calculator for this for sure

[mystic saffron]

i got

-4 and

-0.421

[lavish osprey]

ok let's graph and see real quick

[mystic saffron]

$$ -76m^2 -336m -128 = 0 $$

[clever fjord]

rella

[mystic saffron]

but which gradient do we use tho

[stark coyote]

-4 and -8/19

i believe we have to substitute

[mystic saffron]

so we r gonna get 2 equation??

[stark coyote]

to check

[mystic saffron]

oh

[stark coyote]

we have to check which satisfies our equations

[stark coyote]

maybe sub back into

[lavish osprey]

aaaye look at that

[stark coyote]

OHH

its both of them

its the two possible tangents on the circle intersecting 0,12

[mystic saffron]

oh yhhhh

THIS IS CRAZY

[lavish osprey]

pretty cool right?

[stark coyote]

makes sense

[lavish osprey]

isn't math exciting?

[stark coyote]

so the answer would be each linear equation using those slopes

exact is -8/19 for the decimal you guys got

[mystic saffron]

U GUYS DID IT

OHHHH MYYYT

FINNALY I FINISHED 5/23 of my hw

[lavish osprey]

oh geez haha

you still 18 problems to go? 😮

[mystic saffron]

yh haha

i can do them on my own now

u guys can sleep

[lavish osprey]

awesome, we believe in you ❤️

best of luck!

[mystic saffron]

thanks all❤️

[stark coyote]

we all learned together haha

🫡

[lavish osprey]

[mystic saffron]

👍

.close

[odd edge]

Channel closed

Closed by @foggy pebble

Use .reopen if this was a mistake.

[mystic saffron]

what an absolute legends

[odd edge]

Available help channel!

Send your question here to claim the channel.

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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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• Be polite and have a nice day!

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[ripe ether]

Hi, can someone point me in the right direction?

[narrow forum]

which part

[ripe ether]

The whole thing

like how do i find an equation that is hard to solve, in backthought of the newtons method?

[narrow forum]

transcendental equations?

find an engineering example of when u need to solve one

[ripe ether]

Yes but what is considered an engineering example?

[narrow forum]

osciliations n stuff idk?

im not an engineer lol. what have u done in ur classes?

[ripe ether]

First year student here

our teacher shows us how to solve equations, like normal equations. Then we got 4 tasks, this is the first one.

[narrow forum]

I never took physics past like AP but im guessing most forms of engineering have something similar

whats ur major?

[ripe ether]

Data

[narrow forum]

data engineering?

[ripe ether]

yes

[narrow forum]

lol idk then

[ripe ether]

But

this task doesnt have anything to do with my major

[narrow forum]

yeah but I'm trying to think of examples

that maybe u would know

u can just google this if u wanna find examples

[ripe ether]

Yeah, idk where to start, cuz it needs to be a hard equation to solve

The problem isnt to find an example, it needs to be difficult

Then i have to show

Like ik if i get the newtons method to get onto the maximum it ends. or that i can get into "newtons loop" or that i can get the zero point of an unintended point

But finding a so called difficult equation to solve is what im trying to find. Cant find much on the internet

[narrow forum]

just find any transcendental equation

if u dont need a specific example just say

[leaden widget]

any polynomial that you don't know how to factor

[narrow forum]

x=cos(x) appears in osicilations or smthing

then talk about solving w/ newtons method etc

[ripe ether]

if you cant factor it, then there isnt any zero point, that wont work?

[leaden widget]

your teacher isnt gonna be like "i've asked 10 engineers and 9/10 say they wouldn't use this equation' just come up with anything that isnt immediately obvious how to solve

[narrow forum]

just do cos(x) = x

[ripe ether]

i guess, got an equation u think would be a good one?

[leaden widget]

ok so can you factor this for example?

[ripe ether]

cos(x)=x okey

bruh

hell nah

[leaden widget]

well it has roots

[narrow forum]

I dont think thats what they want anyways

[ripe ether]

degree 4

[leaden widget]

yeah exactly, but newton's method would work great for it

[ripe ether]

This doesnt even have real roots

[ripe ether]

think ill try that one,

you get a straight line with it tho

[leaden widget]

i believe it does since i literally multiplied some slightly larger factors to get it 😂

[ripe ether]

my calculator is fed then ahhaa

said no real roots

[leaden widget]

well i think i made my point well then that the roots are hard to find

if starting with the expanded version

[ripe ether]

nvm i didnt put "-" in the 4th term

but yeah

[ripe ether]

My brother in humanity, this is hard, I'll give you that, thank you. But when im going to put a screenshot of the graph, its going to look fed

Do i have your permission to use that exact example you gave? @leaden widget

[leaden widget]

sure, but also you can literally come up with anything you want by just multiplying a couple ugly looking factors lol

[ripe ether]

true true

[leaden widget]

maybe a cubic example would do instead of a 4th degree

there IS in fact an analog of the quadratic formula called cubic formula for cubic equations but it is one ugly looking mofo

so enough motivation to avoid using it and use newton's method to approximate instead 😂

[ripe ether]

yeah

okey perfect i have a start now. Thank you.

.close

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[rough hound]

I can't seem to figure out the volume

[odd edge]

@rough hound Has your question been resolved?

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@rough hound Has your question been resolved?

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[scenic schooner]

Total points 0, 1, 2, 3, 4, 5
Probability 0.05, -, -, 0.1, 0.35, 0.4

You have created a test designed to screen job candidates for an interview. There are five questions on the test and each correct response is awarded one point. The table below shows the expected probability of achieving each total score. There is an equal probability of scoring a total of one or a total of two points.

What is the probability of scoring more than one point on the test?

[scenic schooner]

Is the probability is 0.85?

[odd edge]

@scenic schooner Has your question been resolved?

[odd edge]

@scenic schooner Has your question been resolved?

[odd edge]

@scenic schooner Has your question been resolved?

[boreal crag]

!show

[odd edge]

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

[odd edge]

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[lusty swift]

Hello, sorry not really a question but just wanted to see if this was correct

[lusty swift]

if that's allowed

[fervent pewter]

how did you get c?

[lusty swift]

I did c^2=4^2+2(4)(8)cos95

[fervent pewter]

what formula is this?

is this cosine rule?

[lusty swift]

This is law of cosines, so c^2=a^2+b^2-2ab cos C

and yeah

[fervent pewter]

where is b^2 and why is there no negative sign infront of cos?

[lusty swift]

oh shoot

oh wait yeah no I did 4^2+8^2 I just didn't type it

[fervent pewter]

this isn't the correct calculation..cosine formula is perfect but place the values correctly

[fervent pewter]

still you did a^2+b^2+2abcosC..it should be a^2+b^2-2ab cosC

[lusty swift]

ah shoot, my mistake yeah I just mistyped, I wrote down - cos though

[fervent pewter]

ok fine then

[lusty swift]

My bad

[fervent pewter]

what did you use for A and B calculation?

[lusty swift]

for A and B calculation I used the same law just moved things around to isolate Cos A and cos B

[fervent pewter]

hmm.. that is fine as well

but sine rule would have been a bit easier in this case..you already have c and C and b, a are already mentioned

your process is alright!

[lusty swift]

haha well that's a relief at least, and yeah that makes sense figures I'd take the harder route

bad habit of mine I guess not intentional lol

[fervent pewter]

I haven't done the calculation part..just ensure you don't make any mistakes in it..just recheck that part and you are good to go!

[lusty swift]

I'll try law of sines and see if I get the same thing?

[stark coyote]

sum of internal angles = 180

[fervent pewter]

yep sure

[lusty swift]

ah shoot, I'm off a bit

I got .5 leftover somehow, let me try again

[stark coyote]

only round for the answer, not for calculations

[lusty swift]

hm okay give me a sec bare with me and thank you guys

[versed junco]

if your calc has a store feature, use that

[lusty swift]

I'm just usin my phone D:

[versed junco]

i would not recommend that

[stark coyote]

should have a scientific mode with basic memory functions

[lusty swift]

I lost my Ti 84 awhile back it's been a struggle

[versed junco]

phone calculators are a recipe for disaster, i recommend investing into a scientific calculator if you don’t yet have one

[versed junco]

big problem

[stark coyote]

you can get a basic non graphing calc at walmart for $15

a TI as well

[versed junco]

by having a proper calculator, you can do id say 5x more questions, which means you’ll get better 5x faster

[lusty swift]

All great points, I'm just a little short on dough these days

I've been saving up to try and get a good one like my ti84 I had since I'll probs be needing it for future classes anyway

but anyway gimme a sec lemme try law of sines, ty guys

[stark coyote]

use a loaner calculator in class, but for now use something like desmos as a calculator? even windows has a calc with a memory

[lusty swift]

Alright will do, the other thing I use is calculator.net not sure if that's up to par

[stark coyote]

desmos is perfectly fine

[lusty swift]

oh so they're not just a graphing calculator?

sry am actual noob

[stark coyote]

nope. it is a super fancy normal calculator as well

[lusty swift]

sweet thanks

[stark coyote]

but only use it to check or get decimals. it is a bit advanced and can do a lot of things

[lusty swift]

hm, how can I use law of sines with only 2 values

[stark coyote]

after finding c you have one angle and all the sides

[lusty swift]

ah right

ty

Shucks, okay my new answer is 59.5 for B and 25.5 for A

That makes more sense to me

cause that adds up to 180 right

[stark coyote]

looks good, it adds up to 180 now

[lusty swift]

Woohooo! You guys are so freaking amazing

[stark coyote]

you did most of it, just needed a push

[lusty swift]

haha well you guys are amazing pushers then

[stark coyote]

[lusty swift]

Really appreciate yall, Noob, and A4 as well

that's all from me for now, I'll most likely be here tomorrow haha

Take care ❤️

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[lethal willow]

i dont think this is right, as trying an, a= 3, b= 4, c= 5 triangle, im getting x = -3/16, which makes no sense

nvm i got it

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[hazy hornet]

I solved the differential equation for To = -40 and got T=-40+e^(kt+c). how do I get e^c=64 which is what the question is saying?

[fervent pewter]

T0 = -40

now solve the differential equation to get an expression in T, t and k

[hazy hornet]

I solved it already and got T=-40+e^kt*e^c but their answer is -40+e^(kt)*64 how do I get e^c=64?

[fervent pewter]

hmm

you tell me

you have to find c

there is another additional information present at some time t with some value of T

use it

[hazy hornet]

The only other information is that the initial temperature is 24 so T=24 but you still can’t solve for c if you plug it in?

[fervent pewter]

yes you can

it means at t = 0 => T = 24

I hope you will get your answer now

[hazy hornet]

Oh, right I forgot thx for ur help

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[lethal willow]

letting a = 3, b = 4 and solving for x, is not getting the same area for a1 and a2

whats wrong with this

[hoary marsh]

I'm confused on what you did

Why is b in the numerator?

[lethal willow]

Where

Wait wrong pic

one sec

nvm i figured it out

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[leaden widget]

$\int\atan\sqrt{x},dx}$
completely blanking on how to approach this kind of integral

[clever fjord]

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

[leaden widget]

$\int\tan^{-1}\sqrt{x},dx}$

[clever fjord]

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

[narrow forum]

probably by parts?

[toxic monolith]

nods

[leaden widget]

oh right good idea : )

ok ill give that a go thx

[hoary marsh]

That looks nasty

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[copper quarry]

how can I solve for x in: -cosx + 3 = x-1

[copper quarry]

dunno how this would work

[tall veldt]

Are you explicitly being asked "solve this equation"

[copper quarry]

yea idk my friend asked me this and it looked really odd so its either that or he changed the original question

i dont know how this would even work tho

[tall veldt]

Yeah an equation like this is really only solved with numerical methods to get an approximation

There's no closed form

[copper quarry]

i thought as much. it looked really odd

thanks

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[mystic saffron]

Since a and b are natural numbers,

a^2=b^2+92

Gcd(a,b)=2
a-b what is the difference?

[mystic saffron]

a^2-b^2=92

I got this but

I don't know how to solve the equation

[low locust]

can you factor this?

[mystic saffron]

No

I domt know the factors

[low locust]

a^2-b^2 is a difference of squares

it can be factored as (a+b)(a-b)

good idea to remember that, it comes up from time to time

[mystic saffron]

a=2K b=2m

a^2-b^2=92

2K^2 -2M^2=92
4K^2-4M^2=92
K^2-M^2=23
(K+M).(K-M)=23
K+m=23 k-m=1
2K=24 K=12 M=11

24-22=2

I valued

Maybe some equetions wrong cause
Cant write here good

Valid?

[low locust]

well you really should write (2K)^2 instead of 2K^2

but yes its valid

[mystic saffron]

Ik

But

Its hard to write

[low locust]

but using brackets is very hard, I know

[mystic saffron]

Anyways cya

[low locust]

having to type two additional symbols

[mystic saffron]

Real😭

I m gonna close

.close

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[odd edge]

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[wild crater]

hi! i'm practicing implicit differentiation and i can't figure out how to further factor/simplify this and come up with y' on one side

[modern sundial]

multiply xy(x+y) on both sides and expand the left hand side after that

[wild crater]

like so? idk if signs are right, but where do i go from here

[modern sundial]

move everything that doesn't have a y' to the other side, then factor out the y' from what remains

[wild crater]

like this?

[modern sundial]

yes

[odd edge]

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[tiny vigil]

hello I need help here

[tiny vigil]

how can I list the 2hours and 30minutes like this in the video cuz the my problem if 2hours and 30 minutes

[viscid latch]

i don’t understand your question

[tiny vigil]

ok let me mark it

[tiny vigil]

this how can I list the 2 hours and 30 minutes if I want to divide it

[tiny vigil]

the problem is in here 2nd pic letter a

[junior junco]

would dividing the time in minutes by 60 work?
that should result in the time in hours

[viscid latch]

oh

(2 x 60) + 30

[tiny vigil]

thats how I list it?

[viscid latch]

yea

[tiny vigil]

ok thx

[tiny vigil]

I tried to calculate the one from yt using this method but it didnt work 305/5x60=1.02

[tiny vigil]

idk,I think it wants km/h

[viscid latch]

ah

then 30/60

so its 2.5 hours

sorry sorry i was wrong

mb

[tiny vigil]

how?

[viscid latch]

this should be correct

[tiny vigil]

ok thx so much

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[odd edge]

Please don't occupy multiple help channels.

[viscid latch]

.close

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[gleaming hedge]

I'm trying to understand the definition of linearly independent vectors

[gleaming hedge]

is it true that if you have a set of vectors, say,

a1, a2, a3..., an

they are linearly independent only and only if their coefficients are 0

and it all equals 0

this channel is taken

seek an open one

[lean willow]

yes

[gleaming hedge]

oh ok tysm,
But so for linearly dependent, it's so that each can be expressed as a linear combination of the others

[grizzled tide]

if the only solution for adding them is all 0s, obviously that will always be a solution

[gleaming hedge]

yes

[grizzled tide]

linearly dependent is if any vector can be represented by a combination of any other set of vectors, yes

[gleaming hedge]

so for them to be linearly dependent, there's at least one coefficient that's not 0

and their sum equals to 0

so when put in a equation

each can represent the others are a linear combination

that's true yes?

[grizzled tide]

no not each

[1 0] [2 0] [0 1]

this set is linearly dependent, but V3 can't be represented by a combination of V1 and V2

[gleaming hedge]

ok

so only the vectors that do not have a coefficient of 0

can represent the others

as linear combinations

[grizzled tide]

I guess kinda

wait no that's still not true, because of mt example

[gleaming hedge]

I don't understand your example

[grizzled tide]

2V1-2V2+0V3 = 0

[gleaming hedge]

ngl

yeah?

2V2 and 2V1

can express

[grizzled tide]

sure can't express V3

[gleaming hedge]

right

so another question

there was one property of linear spans

of b is in l(a1, a2, ..., an)

meaning

l(a1, a2, ..., an, b)

then it's true that

l(a1, a2, ..., an, b) = l(a1, a2, ..., an)

or without it

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[ocean pendant]

how to do this?

[low locust]

have you figured out what R^3x3 is?

and what its dimension is?

[ocean pendant]

i would think it was 3 but bunny said it wasnt

[grizzled tide]

The implication was that you reference your notes/slides/internet to figure out what R^3x3 is

R^3 is 3 dimensional, this is not the same

[ocean pendant]

there was nothing else to the question if thats what youre looking for

[low locust]

unlikely that this question is the first time your course has ever used this notation

[grizzled tide]

I'm not looking for anything, I'm waiting for you to figure out what R^(3x3) is

[low locust]

somewhere in your notes this notation was introduced

we want you to look for that

[grizzled tide]

(and on the very off chance they dropped this out of nowhere, it's easily googleable)

[ocean pendant]

ive seen it with matrices if thats the same?

so V could be 3 vectors (1,0,0), (0,1,0) and (0,0,1)?

[grizzled tide]

you know how in R there's numbers, in R^2 there's (x,y) in R^3 there's (x,y,z)?

R^(3x3) extends this pattern

just with 3x3 matrices

[ocean pendant]

but would that matrix be made up of the vectors i listed?

[grizzled tide]

you mean the identity 3x3 matrix?

{1 0 0, 0 1 0, 0 0 1}

[ocean pendant]

i guess it could be any values no?

[grizzled tide]

yes, R^(3x3) contains all (real) 3x3 matrices

[ocean pendant]

so i just gotta find another matrix that has the dimension 6?

alr i think i understand the dimension thing a bit more now

i dont see how i can prove that another matrix W is a subspace without having numbers for W

alright would {1 1 1,1 1 1,1 1 1} not be a subspace?

[low locust]

no

thats just a single matrix

[ocean pendant]

how not?

[low locust]

well for starters, a subspace should be a set

[ocean pendant]

yeah a set of vectors?

[ocean pendant]

if im supposed to give a set of vectors how do i make sure its dimension is 6??

[low locust]

do you know what dimension means

[ocean pendant]

apparently not

[ocean pendant]

please clarify

[low locust]

the dimension of a vector space is the number of basis vectors in a basis of that vector space

[ocean pendant]

so {v1,v2,v3,v4,v5,v6} has a dimension of 6?

[low locust]

no

thats a set of 6 vectors

the span of those might have dimension 6

depending on if they are linearly independent

[ocean pendant]

so what i have to find is 6 basis vectors that are in V and then group them in a set?

[low locust]

and then the span of those works, yes

[ocean pendant]

wdym by the span of those works?

[low locust]

do you know what a span is

[ocean pendant]

dont think so

[low locust]

the span of a set is the smallest vector space which contains that set

equivalently, its the set of all the linear combinations from vectors of your set

[ocean pendant]

doesnt this set fulfill everything?

[low locust]

thats not a vector space

also its not a subset of R^3x3

[ocean pendant]

which vector space property isnt defined?

[low locust]

basically most of them

sum of two vectors isnt again in the set

scalar multiples arent again in the set

zero vector isnt in the set

additive inverses arent in the set

[ocean pendant]

alright how would you find a subspace of V?

and make sure its dimension is 6

[low locust]

find 6 linearly independent vectors in V

then take their span

[ocean pendant]

yeah but how do you just do that?

[low locust]

well you managed to find 6 linearly independent vectors in R^6

now do the same but this time the vectors arent one column with 6 entries and instead 3 columns with 3 entries each

[ocean pendant]

isnt that a matrix tho?

[low locust]

yes

and?

vector in this context means "element of vector space"

its not mutually exclusive with the word matrix

[ocean pendant]

like this?

[low locust]

no

thats a set with three row vectors

[ocean pendant]

you want this?

[low locust]

thats a matrix, yes

now find 6 linearly independent ones

[ocean pendant]

6 matrices that are linearly independent? have not learnt that

[low locust]

do you know what linearly independent means

[ocean pendant]

for vectors not for matrices

[low locust]

whats the difference

you can still add matrices

the same way you add vectors

if you have a 3x3 matrix and a vector with 9 entries, except for the shape, how is there a difference when its just about adding them

[ocean pendant]

idk man. ive spent almost 3 hours with this and im not making progress

[low locust]

the thing is that in the context of it being a vector space, there is no difference between R^3x3 and R^9 except for how the entries are arranged

[ocean pendant]

alright this right?

[low locust]

yes those are 6 linearly independent matrices

btw just a single 1 in each would be easier

[ocean pendant]

ight cool

whats the quickest way to argue the dim=6 and its a subspace?

[low locust]

well the span of them is by definition a subspace

and dim is 6 cause its 6 linearly independent vectors which generate it

again, by definition

[ocean pendant]

alr

ty for the help

.close

[odd edge]

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[median fable]

I hope everything in the image is clearly visible.
The purple line is drawn so that one point is always inside of the circle, but the other one is outside. The purple line's points are arbitrary whilst following the said rules.
I need the vector OA, which means getting its X and Y components(green lines).

[median fable]

As you can see, I have tried using a ratio but I have no way to prove that the green and orange triangles are similar.
There's a very specific detail that I don't know if anyone will get as it involves game dev, but it basically lets me test if I have correctly gotten the result

And I have tried using the formula as if the triangles are similar, and it works!
But I have no idea why, because as you can see the orange and green triangle are clearly not similar

[gleaming hedge]

similar means in ratios

why aren't they similar?

[median fable]

Because the angle Alpha is different for both of them

The only thing that I can see they have in common is the Hypothenuse and the right angle

[median fable]

same goes for Beta

[gleaming hedge]

well, if OC and OD lie on the ox line, there's one sign

they also both have 90 degree angle

then CA is proportional to DB as they both go the same direction

another sign

wasn't it one theorem that if you have 2 sides proportional to each other

and an angle between them

making them similar?

[median fable]

the SAS? side, angle, side

[gleaming hedge]

yeah

[median fable]

but how can you tell that OC:OD is the same proportion as AC:BD?

[gleaming hedge]

they lie on the same line, no?

then you have parallel lines for ac and bd

[median fable]

OC and OD are on the same line, but AC and BD are parallel

[gleaming hedge]

well yeah, doesn't it make them proportional in one sense?

they have the sam estarting point

when you rearrange it

then you multiply AC by some number to get BD

more or less

fraction or not

[median fable]

Wait does this mean that, Angle AOC = Angle OBD?

[gleaming hedge]

I have kind of lost experience with proving similar triangles ngl haven't done that in at least over a year

[median fable]

so that Alpha of the green triangle is proportional to the Beta of the other triangle

[gleaming hedge]

might be

I can't tell for sure tbf

you might use some circle properties

[median fable]

such as?

[gleaming hedge]

I just haven't done this in quite a long time

idk

just a guess

[median fable]

fair fair

[median fable]

But yeah there is a possibility that this is true

Because you can clearly see the Alpha for both triangles is not the same

The orange triangle has a bigger angle

[gleaming hedge]

yeah

[median fable]

I'm going to try and test it a bit more

[median fable]

If this turns out to be correct I'll close the question

and thx for the help

[gleaming hedge]

welcome!

some orientation is better than no help at all

[median fable]

yep

[odd edge]

@median fable Has your question been resolved?

[median fable]

Nope, I don't think that this is the case

I've tried using numbers and nothing adds up

In this example you can basically see that almost none of the angles are coorelated in any way

<@&286206848099549185>

[odd edge]

@median fable Has your question been resolved?

[median fable]

welp

what now

[odd edge]

@median fable Has your question been resolved?

[odd edge]

@median fable Has your question been resolved?

[novel finch]

THIS DOESNT SEEM RIGHT

UNLESS HYPOTENUSE IS DEFINED WEIRDLY AS 1/D

[median fable]

Ah

I seem to have been blind

While writing those formulas

[novel finch]

IT BE LIKE DAT SOMETEIMS

[median fable]

It's y/d not d/y

Wait but how did I write the first part of the formula wrong, but then when I rearranged it, it became right

The hell was I smoking when I wrote that

[novel finch]

AND X/D

[median fable]

Yeah that too

[novel finch]

71.166.126.0 , 39°43′N 140°07′E

[novel finch]

IF YOU EVER FIND OUT, YOU KNOW WHERE TO CONTACT ME

[median fable]

What is uh, what is this?

[novel finch]

IP ADDRESS AND COORDINATES

ID BE A MADMAN TO GIVE A STRANGER MY PHONE NUMBER

[novel finch]

R U GETTING THROUGH WITH YOUR QUESTION

[median fable]

I'm calling it a day. Hopefully tomorrow someone'll be able to help out

[median fable]

What's really funny/terrifying is that I pasted this whole message(including the ip) into google maps and it was pretty damn close to where I live

[novel finch]

RESTATE THE QUESTION, BECAUSE I AM ALSO LOST AS TO WHAT UR ASKING

[median fable]

The goal is to find the lines OC and AC if that's what you're asking

Both of which are x and y components for point A

I think I've explained most of what's important already

[median fable]

Don't mind that

[novel finch]

IF YOU FIX THESE UP THEN THE QUESTION IS DONE

[median fable]

They're not that important because I don't have alpha anyways

[median fable]

Wdym?

How

[novel finch]

YOU WANT TO FIND ALPHA IN RELATION TO WHAT

[median fable]

Alpha in that formula is the angle AOC

Which is not known. I have just labelled it

The formulas are correct, but they don't do much for me

Well the first part of the formula is wrong but the end formula is somehow correct

[novel finch]

SO YOU WANT TO FIND THE COMPONENTS OF OA, GIVEN A POSITION VECTOR OB ?

[median fable]

Yes, and the radius of the circle

Oh and the coordinates of the center of the circle

It should be possible right?

Given that the hypothenuse of the green triangle is the same as the radius

[novel finch]

||I DIDNT INTEND TO STAY THIS LONG SO UH,|| <@&286206848099549185>

[median fable]

I'm going to sleep. All the necessary data is in the history of this question

[odd edge]

@median fable Has your question been resolved?

[median fable]

One more note: The sharp angles of the Orange triangle are also given IF they are even of any importance

[wind flame]

May i ask why you have been typing in capslock this entire chat

[median fable]

.close

[odd edge]

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[fierce ore]

both q1 b and q2 please

[odd edge]

@fierce ore Has your question been resolved?

[narrow forum]

for q1b just plug into quadratic formula

[fierce ore]

Oh

[narrow forum]

actually you can do it with factorizing too

probs easier

$\frac{4}{3}x^2 - 8x + 12 = 0$

[clever fjord]

Abyss

[narrow forum]

$4x^2 - 24x + 36 = 0$

[clever fjord]

Abyss

[narrow forum]

$x^2 - 6x + 9 = 0$

[clever fjord]

Abyss

[narrow forum]

and thats simple to factor to solve for x

as for q2 you havent shown the full question so...

[fierce ore]

Hm

Butnlike

[fierce ore]

Oh oops

I did thw quadratic formula

It worked

I think i did 2 right

[fierce ore]

.close

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[leaden light]

Can someone help me with the second question :
Determine if exist non-vertical asymptote to the right of the graph, If it exist , indicate it.

[tall veldt]

You want to study the behaviour of the function as x tends to infinity

[leaden light]

I don’t think that I got you right

[tall veldt]

That's not what the function looks like as x tends to infinity

For x > 3 it is (x+2)/(x²+1)

[leaden light]

Key if this function gonna tends to the infinity , it will infinitely approach to the zero isn’t it ?

[tall veldt]

It tends to 0 yes