help-23

[unborn terrace]

I can help

You need y value, yes?

[loud cedar]

Yes

@unborn terrace Cemirul

[unborn terrace]

Please try search the y value from quadratic equation of 3y²+3y+6=0

[mint raven]

would it be +3y or -3y

[unborn terrace]

It was -3y but I prefer to make the whole equation positive in the end

[mint raven]

i got 3y²-3y-6=0 as the final eq

[loud cedar]

solution pls

Idk how to write it in paper step by step

[unborn terrace]

The final value does not seem clear

[loud cedar]

Yes

Like whats the value of y

[unborn terrace]

It was in friction

[loud cedar]

Value of y??

Just equated it to 0 but the value of y?

im so sleepy

Im gonna sleep now Ill just wake up early tomorow

.close

[safe radish]

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[lean otter]

This is the problem I am stuck on and my work. It seems simple and i get 69 no matter what I do but the computer says it’s wrong.

[pastel sinew]

@lean otter that is weird, it's just a simple geometric series

it should be $\sum\limits_{n=0}^\infty 23\cdot\left(\frac{2}{3}\right)^n = \frac{23}{1-\frac{2}{3}}=69$

[flat frigate]

🇵🇸Mína🔆

[gentle herald]

Think carefully
Some distances will be travelled twice

[pastel sinew]

OH WAIT

[pastel sinew]

you're right, i forgot about that part xd

[safe radish]

@lean otter Has your question been resolved?

[pastel sinew]

then it should be $\frac{23}{3}+\sum\limits_{n=1}^\infty 23\cdot 2\cdot\left(\frac{2}{3}\right)^n=\frac{23}{3}+\frac{46\cdot\frac{2}{3}}{1-\frac{2}{3}}=\frac{276+23}{3}=\frac{299}{3} [EDIT : THIS IS WRONG BTW]$

[flat frigate]

🇵🇸Mína🔆

[pastel sinew]

which is almost a hundred instead of 69, upsetting

@lean otter do you see why it's like this?

[lean otter]

no sorry i am still a little confused

[pastel sinew]

Alright, let's go over this again

[pastel sinew]

the ball starts at height 23, which we can interpret as distance traveled = 0.
at a certain point it will have fallen a distance of $\frac{23}{3}$.

[flat frigate]

🇵🇸Mína🔆

[pastel sinew]

distance $\frac{23}{3}$ means the ball is exactly $\frac{2}{3}$ of $23$ above ground

[flat frigate]

🇵🇸Mína🔆

[pastel sinew]

from that point on we can start counting our infinite sum

i will draw a picture, that may be more clear.

but i forgot one thing actually

so this will be two infinite sums

total distance traveled will have been $\sum\limits_{n\in\mathbb{N}}23\cdot\left(\frac{1}{3}\right)^n+\sum\limits_{n\in\mathbb{N}}2\cdot23\cdot\left(\frac{2}{3}\right)^n$

[flat frigate]

🇵🇸Mína🔆

[pastel sinew]

sorry for not thinking it through properly before @lean otter

before each bounce-back of the ball, it needs to travel to the peak of the upcoming bounce first. That is the term of the first sum. After passing that peak, we can start counting the distance of the bounce twice. That is the term of the second sum.

We can simplify the sums above as such:\
$\sum\limits_{n\in\mathbb{N}}23\cdot\left(\frac{1}{3}\right)^n+\sum\limits_{n\in\mathbb{N}}2\cdot23\cdot\left(\frac{2}{3}\right)^n=23\left(\sum\limits_{n\in\mathbb{N}}\left(\frac{1}{3}\right)^n+\sum\limits_{n\in\mathbb{N}}2\cdot\left(\frac{2}{3}\right)^n\right)=23\left(\frac{1}{1-\frac{1}{3}}+\frac{2}{1-\frac{2}{3}}\right)=23\left(\frac{3}{2}+6\right)=23\cdot\frac{15}{2}=172.5$

[flat frigate]

🇵🇸Mína🔆

[pastel sinew]

And THIS, I believe, should be correct. Someone right me, if I'm wrong

@lean otter I hope this is satisfactory.

[lean otter]

i have not tried typing in the answer in my computer yet but that does make a lot more sense

thank you so much

i also wanted to say i appreciate your solidarity with 🇵🇸 btw as i am palestinian

[safe radish]

@lean otter Has your question been resolved?

[pastel sinew]

Oh holy shit, my heart goes out to you all.
I am in utter disbelief from the amount of hate and denial everywhere, especially my super-zionist country

Palestine will be free, insha Allah

And of course, none of us are free, until we're all free 🏴

[lean otter]

yes inshAllah!! ❤️❤️❤️

[safe radish]

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[frail blade]

I don’t know where it’s wrong, it’s apparently supposed to be another answer, I need help.

[frail blade]

I hope you can read my handwriting btw

[warped lotus]

what have u done with this part

[frail blade]

Yhhh I just saw It 2 sec ago

[warped lotus]

i got 4x^2-20x-24

[frail blade]

I’m stupid

Thx btw

[warped lotus]

dw, everyone makes mistakes

[frail blade]

Ik, but that was a stupid mistake honestly

I was trying to understand , it literally took me 10 minutes to see it

[warped lotus]

aight, in the end u have to get 3x^2-38x-105=0 if i am not wrong

[frail blade]

I don’t think so, I got 3x^2-2x+345

I’m probably wrong actually

[warped lotus]

can u show me how u got it

[warped lotus]

wait a min

[frail blade]

Np

[warped lotus]

i think u multiplied with 11 by mistake

ur supossed to get 6 there not 66

it s x+1, right? not x+11

[frail blade]

I don’t understand

[warped lotus]

here

[frail blade]

Oh yhhh

Imma check it 2sec

It’s wrong but where?

Ohhh

I forgot to square the -x-9

Lemme fix that

[warped lotus]

did u square -x-9?

[frail blade]

Nah

I’m doing it

There

Now I need to check

And then I’m done

I’m a 100% sure that’s right

The first one works

But not then second one

The only answer is 15

[frail blade]

Am I right?

[warped lotus]

well

i got 3x^2-38x-105=0

same thing

didn t solve the equation tho

wait a min

[frail blade]

Can ya?

Thx

[warped lotus]

-7/3

and

15

didn t check them tho

[frail blade]

I probably got them right

Thanks for your precious help dude

Have a lovely day!

[warped lotus]

cheers

[frail blade]

.close

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[shrewd flint]

Hello, would anyone be able to tell me what the answers to c and d are? I've tried solving them but my answers end up larger than the total amount of combinations without any special requirements

[shrewd flint]

For a) I got 42504 possible hands

For b) I got 9900 possible hands

For c) I tried C(6,1) x C(6,1) x (22,3) but that didnt work

For d) I tried C(6,1) x (4,2) x (21,2)

[stray steppe]

the question says at least one heart and club

so it can be more than 1

[shrewd flint]

Yes i figured that since i was only taking away 2 from the total 24 in C(22,3) that would leave room for decks with any other number of hearts or clubs to be the remaining 3 cards

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@shrewd flint Has your question been resolved?

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[limpid basin]

Hello, i will send in a minute what i did with this, i want to solve this with L'Hospitals rule

[limpid basin]

Should i do L'Hospital rule now?

[forest gust]

you can but it would be painful

[limpid basin]

Whats the other solution (i can't use substitutions)

[forest gust]

(it's substitutions)

[limpid basin]

well...

hahahahaha

[forest gust]

lol

[limpid basin]

Wish me luck then

[forest gust]

🫡

[limpid basin]

🤝

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[polar plinth]

How would I find if one of these constraints is redundant?

[safe radish]

Please don't occupy multiple help channels.

[mortal sandal]

The easiest way would probably be graphing

Ah 3 variables

[polar plinth]

yeahhh, I tried with proof and believe (1) is redundant but dont feel confident in it

[mortal sandal]

Multiply inequality (2) by 2

The extra b can only make it even more redundant

(as in, add to b≥0)

oh 3m though

[polar plinth]

Yeah I've been stuck on this one for awhile, the second part of the question asks to solve the contrained problem for b,m,g lol

any help for just this part woul be great

[mortal sandal]

Well, desmos has a 3d calculator now

You could graph it with that

there's a linear programming way to do it, apparently

Minimize b+3m+6g subject to the bottom 2 constraints (and b,m,g≥0)

[polar plinth]

I think I have to do by hand with lagrangian to solve

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[spare dawn]

can a linear transformation be one to one and onto?

[spare dawn]

So can the linear transformation T map Rn onto Rm and also be one to one at the same time?

[fickle pendant]

id is bijective and also linear

[spare dawn]

what is id?

[fickle pendant]

the identity function

[spare dawn]

and what does bijective mean

[fickle pendant]

bijective = injective and surjective

= has an inverse

[spare dawn]

okay

[fickle pendant]

= 1-to-1 and also onto

[granite idol]

only if it's from Rn to Rn I believe

[spare dawn]

when you say identity function does that mean your mulitplying the vector by an identity matrix

[fickle pendant]

I mean the function defined as f(x)=x

for all x in...anything

[spare dawn]

ah ok

[fickle pendant]

I'm being handwavey as to the specifics

but it's certainly possible for a linear map to be bijective but it doesn't mean every linear map is bijective

[spare dawn]

why does being bijective mean you are both onto and one to one

[fickle pendant]

yeah

bijective = injective and surjective
injective = 1-to-1
surjective = onto

same thing, different words

[spare dawn]

ohhhh

so you can only be both if A is a square matrix

[spare dawn]

so in other words this

which also means you can have inverse

thanks for the help :))

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[lean otter]

How would I do Q8

[crisp mural]

Firstly, we need to find the value of b

and you're given that the lines FE and ED are perpendicular

if two lines are perpendicular, then what can you say about them?

@lean otter

[lean otter]

I did it already

My@bad

Forgot to close it

I clocked afterwards

[crisp mural]

oh alright

[lean otter]

I can prove it’s 1 by showing that DE’s perpendicular gradient includes F, thus proving

[crisp mural]

yup

[lean otter]

Thank you tho

I appreciate it brother

[crisp mural]

I didnt do anything, really

<3

[lean otter]

:))

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[safe radish]

@pure ingot Has your question been resolved?

[safe radish]

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[ionic stirrup]

What is the meaning of perfect square

[ionic stirrup]

.close

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[safe radish]

@dreamy monolith Has your question been resolved?

[safe radish]

@dreamy monolith Has your question been resolved?

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[lean otter]

The problem is :

My method was to find A(t) for amount of water in the tank per minute, and then calculate A(9)

This is my answer

But somehow omitted the -16/15 according to the answer sheet. I don't know where I got wrong because I think my calculations are right.

[plucky elk]

You didn't evaluate A(0)

You just assumed it was 0

Where did you justify that?

[lean otter]

The solutions assume that A(0)=0 too. It's ok to assume that in the standards of this course.

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@lean otter Has your question been resolved?

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[rancid dagger]

Guys, i'm a little out of braincells today, can any1 explain why -2-19 is -21? in easy terms pls

[rancid dagger]

<@&286206848099549185>

[lean otter]

take away the negative sign

what's 19+2?

[rancid dagger]

21

[lean otter]

right

as an analogy

I lost 19 dollars

but I then lost 2 dollars

how many dollars in total did I lose

[rancid dagger]

all?

oh wuit

21

thanks

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[rancid dagger]

hey guys

[rancid dagger]

I don't get it, can som1 explain to me?

[mortal sandal]

Kinda looks like they plugged in 3 for x and -2 for y

[rancid dagger]

yeah

let me check my paper

oh

but I got the answer +30

hw to do?

[mortal sandal]

What did you do?

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[safe radish]

Please don't occupy multiple help channels.

[royal mist]

hi! how do i solve this problem?

[safe radish]

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[lean otter]

Let f(x) = (a-b-c)x^2 +ax+ (b+c) where a-b-c not equal to 0.

1. show that the roots are real of the equation f(x)=0
2. if the roots of f(x) = 0 are double of other one then show that, b+c = a/3 or b+c = 2a/3

[old prism]

Where are you up to

[lean otter]

sorry didnt get u

[quasi bison]

show your progress so far, if any

or if you have none, then say you don't know how to begin

[lean otter]

i dont know how to begin

[quasi bison]

btw this looks translated from another language

did you translate?

[lean otter]

yeahhh

[quasi bison]

from what language?

[lean otter]

sinhala

[quasi bison]

ok nevermind

i asked in hopes it would be one of the languages i speak but it's not

ok

so do you know how to tell whether or not the roots of a quadratic are real?

[lean otter]

yepp

[quasi bison]

ok, how do you do it

[lean otter]

its real if the discriminant is equal or greater than 0

[quasi bison]

greet

great*

find the discriminant of your quadratic

try to show it is ≥0

[old prism]

So just b^2 - 4ac

Need to be +

[lean otter]

compare the quadratic with the standard format of quadratic: ax^2+bx+c=0
if b^2-4ac>=0 then the roots r real

if b^2-4ac=0 roots r real and equal

[lean otter]

yep

i get stuck here

can you send the whole problem?

like the original question?

[lean otter]

btw why r u trying to expand it?

[lean otter]

how to show its ≥0 without expanding

[quasi bison]

no actually you're on the right track

now notice that this is 4(b+c)^2 - 4a(b+c) + a^2

which is [2(b+c)]^2 - 2 * 2(b+c) * a + a^2

= (2(b+c) - a)^2

[lean otter]

ahh yeahhh

ooh so u had to prove that from the standard format itself?

my bad

[old prism]

Hold in is a-b-c > 0 or just not equal to 0

Nvm

I forgot that squaring is always positive 💀

[lean otter]

so when it comes to the second q, the answer is a/2 rytt

or did i mess up somewhere

.close

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[split fulcrum]

Verify that for all real numbers, $\left(\cosh t\right)^{2}-\left(\sinh t\right)^{2}=1$

[flat frigate]

water beam

[split fulcrum]

not sure how to start

[forest gust]

$\cosh\left(x\right)=\frac{e^{x}+e^{-x}}{2}\\$
$\sinh\left(x\right)=\frac{e^{x}-e^{-x}}{2}$

[flat frigate]

Combustion

[split fulcrum]

yes

[forest gust]

just square them and subtract them

[split fulcrum]

oh is that it

[forest gust]

yeah lol

[split fulcrum]

wow lol

okay

thanks

.close

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[frozen ledge]

Help

[frozen ledge]

Hello, I wanna know how did we get 12 roots?

.close

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[woeful hemlock]

I need to determine whether A1, A2 and A3 lie on the same line

[woeful hemlock]

Their equations for obtaining y through x are different, so I thought they’re supposed to not be?

But the answer list says that they are??

[umbral swan]

well can you first create a line using the two points A1 and A2?

[woeful hemlock]

To draw them on a graph?

Well, every two points can create a line

[safe radish]

@woeful hemlock Has your question been resolved?

[woeful hemlock]

Okay nvm I found out how to do it

.close

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[urban eagle]

What do I sub into $\int_{0}^{\frac{\pi}4} \frac{2}{\sin^2x+1} dx$

[flat frigate]

ЖѲTЇЇC

[urban eagle]

@lean otter did you find what I could sub into it

Accidentally let the other channel close

[urban eagle]

Or is there a simpler way to do $\int_{0}^{\frac{\pi}4} sin(x)artanh(sin(x)) dx$

[flat frigate]

ЖѲTЇЇC

[lean otter]

ignore the theta part on the left

[safe radish]

@urban eagle Has your question been resolved?

[urban eagle]

Ok ty I would’ve never thought of anything like that

Probably an easier way to do the original q idek

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[lean otter]



[umbral swan]

pls occupy another :)

[safe radish]

@umbral swan Has your question been resolved?

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@umbral swan Has your question been resolved?

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@umbral swan Has your question been resolved?

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@umbral swan Has your question been resolved?

[plucky elk]

You treat A like your L in |f(x) - L|<eps and find a delta such that |x-(-1)|< delta implies the above inequality

More jargon here
https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

Although your function is much more complicated, the pattern is the same

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[jade lotus]

Radical 63- 2 radical 28 + 5 radical 7

[jade lotus]

Number 6

[lean thorn]

break down each radial into it's prime factorizations, then simplify

just like how you did for #3 but you'll just have to do them separately then combine like terms

[jade lotus]

Ok

I need help solving 19

[finite dust]

the domain is all real numbers

for the roots you just gotta solve the quadratic equation (-x^2-6x-5=0)

[jade lotus]

What do I put in for x?

@finite dust

[finite dust]

you gotta solve for x

[jade lotus]

How would I solve this one?

Number 16

<@&286206848099549185>

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[cinder pagoda]

how do i proceed to finding ẞ?

[austere parrot]

if you take the inverse sine of both sides, you're left with $2\beta = 3\beta + 2 \pi k \text{ where } k \in \mathbb{Z}$, then you just solve it

[flat frigate]

kahve

[cinder pagoda]

in degrees

[austere parrot]

well, it has infinite solutions. a notable one is 0

[cinder pagoda]

beta is an angle in an isosceles triangle

so it cant be more then 90

[austere parrot]

in that range, it seems like there are 2 solutions

one is 0.63 radians, and the other one is 1.9 radians

edit: i took the range as 0 to 180, from 0 to 90 it should be 0.63

[cinder pagoda]

😐

[austere parrot]

well you can do the conversion yourself, right

[cinder pagoda]

no

I havent learnt about radians yet

[austere parrot]

okay, so 360 degrees = 2 pi radians

[cinder pagoda]

this is what Im thinking but its not true

[austere parrot]

thats the solution where they meet at 0

there's one more, and that is
$2 \beta = \pi - 3 \ \beta + 2 \pi k , k \in \mathbb{Z}$

[flat frigate]

kahve

[austere parrot]

solving that leads to $\beta = \frac{2 \pi k}{5} + \frac{\pi}{5}$

[flat frigate]

kahve

[cinder pagoda]

36

thanks got it👍

.close

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[drifting ridge]

do you only multiply the 2 with the numerator

[empty gyro]

yes

then you simplify

[drifting ridge]

why not both

[empty gyro]

by both, you mean both 1/4 and 1/3, or do you mean both 1 and 4?

[drifting ridge]

b

both 1 and 4

[empty gyro]

1/4 (one-fourth) is a representation for part of a whole. It is to be understood that, if you had a whole of something, and needed to break it into four equal parts, each part would be designated as "one fourth of the whole".

A popular analogy is a cake or pizza. Take your pick. Cut it into four equal slices. Each slice represents 1/4 of the whole. When you multiply a number by 2, you add that number to itself 2 to times.

In the food slice example, that's like taking to slices. You know have two "one-fourth" slices. In words, you could say you have "two-fourths" of the whole. You wouldn't say you had "two-eights".

[drifting ridge]

ah i see

[empty gyro]

yay

[drifting ridge]

thanks for the help

.close

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[scenic bough]

How do you verify if a function is an inverse?

[peak estuary]

an inverse of what

[scenic bough]

just 21

i got the inverse already

but how do i "verify" it

[peak estuary]

lets call the function you found g

check g(f(x))=x and f(g(x))=x for all x

[scenic bough]

Thank you

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@cerulean birch Has your question been resolved?

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[runic sail]

I honestly have no clue how to do this

I know how to take partial derivatives but the notation is really tripping me up for some reason and I don't understand how to set it up

[granite idol]

make those substitutions and see what you get

[runic sail]

so what am I taking the partial derivative of?

[obsidian oracle]

This quantity

With respect to r

Shouldn't be too much of a hassle

[runic sail]

is that just

?

[obsidian oracle]

Yes

[runic sail]

oh

that's easy

[granite idol]

yup

[runic sail]

thank you

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[crystal yarrow]

Could someone help me with the question? It’s worded awkward. I thought that the company would be lying because they said it was 1 in 6 chance you get a winners cap but 3 out 7 friends got it.

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@crystal yarrow Has your question been resolved?

[crystal yarrow]

<@&286206848099549185>

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@crystal yarrow Has your question been resolved?

[crystal yarrow]

.close

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[hearty jungle]

I have a quick trignonometry question, its a word problem involving the law of cosines, which is pretty easy on its own, however...

question #8

ill draw a quick diagram of it quickly

im in particular having trouble finding the angle which is located within the triangle at B

with that I can find the angle at F and solve the triangle relatively easi;ly

quick correction

that heaidng 345 extends all the way to the edge of the 15

mb

ok so ive come to the conclusion that the angle at b would end up being 95

which i think seems a lil wierd

but could def be a possibility

yep

was right

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[unborn venture]

yo

[unborn venture]

for marices

augmented matrices

does everyone follow it a specific way because i feel like i could solve it a diff way then someone else and still get the way it’s supposed to be like three zeros and the 1’s

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@unborn venture Has your question been resolved?

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[rigid path]

Number 56

[rigid path]

It's kind of confusing

I don't rlly know where to stsrt

[snow haven]

whicj

[rigid path]

im trying to make it clearer lol

[snow haven]

which

[rigid path]

56

ye

is it clear?

[snow haven]

id say simultaneous equation

construct two different equations

[rigid path]

yeah id say making a system aswell

but thats what im confused about

its more or so idk where to start in making one

[snow haven]

it says for the same distance

[turbid geyser]

You have to half 7 3/4

[snow haven]

why

[rigid path]

right i wrote something along the lines of like

[turbid geyser]

And multiply each by 16 and 15

[rigid path]

16x+15y=73/4

*7 3/4

[turbid geyser]

It says both are the same length of time

[rigid path]

yeah so wouldnt like

[snow haven]

no

[snow haven]

distance

[rigid path]

uh yeah same distance travelled

[turbid geyser]

Oops sorry

[rigid path]

and since we are trying to find the distance

would the equation iw rote work?

[snow haven]

thats wh

[granite idol]

stop giving wrong answer

[snow haven]

dont half anything

[rigid path]

.

uhhhhhhhhhhhhhhhhh

interesting

anyways

[rigid path]

its an honest mistake lol

its ok

:D

so uhmm back to the system lol

oh well

that died fast

[turbid geyser]

Could you solve it like 16/x + 15/y = 7+3/4

[rigid path]

hmm thats what i was thinking but idk about the 16/x

i was thinking like

15x+16y=7 3/4

[snow haven]

wait no

[rigid path]

oh

[snow haven]

distance are the same remember

so

[rigid path]

well x = y

ig

uhh

15x + 16y

equals to 7 3/4

?

[junior smelt]

(to be clear, what are you representing with x and y?)

[rigid path]

distance @junior smelt

[turbid geyser]

Variables

[rigid path]

since the distances are the same

i guess we could use the same variables

or variable

lol

the only issue is that like

[junior smelt]

fair enough - might want to do the time(s) and write distances in terms of the time?

[rigid path]

we need to figure out the distance

so uh,

[rigid path]

uhhhhh that is 15 km per hour

is half the distance

[snow haven]

aint it 15x + 16x = x/7.75

[rigid path]

and uhhh 16 km per hour is like

wut the fawk

idk how that works lol

[junior smelt]

Alright, maybe let the time for the first "trip" he does be t, and the time for the second trip be s

[rigid path]

its the same trip tho

[junior smelt]

If you travel at 16km/h for t hours, how far, in terms of t, do you travel?

[rigid path]

16t

[simple oak]

Can you use speed is distance over time for this

[rigid path]

no

there is no rate

[turbid geyser]

16t+15s=7.75?

[rigid path]

uhmmmmmmmm

i said that but if you look at it more it doesnt sound right tbh

cuz uhm

[junior smelt]

Speed is a rate

[turbid geyser]

The rate is 16/kph and 15/kph

[rigid path]

oh

im smart :D

rate is distance over time

right>

?

[granite idol]

be careful with the units, as well

[rigid path]

sobbing tears of confusion

ok

[turbid geyser]

Speed

[simple oak]

So we have two speeds so just make two equation

[rigid path]

yeah a system

right?

so uhhh

16t+15t=73/4

[simple oak]

If total time is 7.45hours can we say that t1+t2=7.45

[rigid path]

that could be the first one

uhhhhh

[granite idol]

no, units don't work

[rigid path]

oh ic

can i just use x and y

lol

[turbid geyser]

It doesn’t matter

[rigid path]

ok

[granite idol]

think of what you need to do so that "km / h" just becomes "h", if you understand

[junior smelt]

Remember that you don't necessarily take the same amount of time for each "portion"

[rigid path]

wait why is it 7.45 hours tho

[simple oak]

7 and 3/4 of an hour

[rigid path]

thats 75...

[simple oak]

Good job

[rigid path]

😭

[simple oak]

[rigid path]

gaslamping me rn

ok so

:OIHEWPSIUHREWOULISFJKMNFLSIU

how should i write my second equation

wouldnt the second one be likeeee

uhhhhhhh

nvm

[simple oak]

Umm

I set up two speed formulas

No idea if right but here we go

[turbid geyser]

Can we just talk about how this guy was able to cycle at 15-16/kph for 7.75 hours?

[rigid path]

....

[simple oak]

t1+t2=7.45

[rigid path]

yeah

[turbid geyser]

He must have the most stamina of anyone else in the world

[simple oak]

16=d/t1

[rigid path]

oh

[simple oak]

15=d/t2

[turbid geyser]

He should try out for the Olympics

[rigid path]

im in pain

[junior smelt]

thats 75...

[rigid path]

so many conflicting things

i also said 75

[simple oak]

what's 75?

[rigid path]

😭

[granite idol]

7 3/4 = 7.75

[rigid path]

7.75

1/4 is 25

[simple oak]

3/4 of an hour

Guys

[rigid path]

25*3 is 75

no we're not talking in minutes tho

are we?

[simple oak]

3/4 of an hour is 3/4 times 60

Oh

Umm

Pardon me

😄

[rigid path]

100 percent

[turbid geyser]

HOW WAS THIS GUY CYCLING AT 16/kph FOR 7.75 HOURS

[rigid path]

😭

ac in confusion

:D

[turbid geyser]

WHAT THE HECK DUDE

[simple oak]

ignore that

umm

[rigid path]

who actually knows what they're doing

[simple oak]

So it is t1+t2=7.75 hours

[turbid geyser]

I think he was on a e-bike prob

[simple oak]

isolate t1

[rigid path]

uhhh

[simple oak]

t1=7.75-t2

Then plug that into the speed formula

Do some algebra

And boom

Answer

[rigid path]

ehmmmmmmmm

how did you get the t1 and t2 tho

[simple oak]

Well there's two guys cycling

One is faster than other

[rigid path]

no theres 1 guy cycling

...

[simple oak]

They ride the same distance

Wait wht

[rigid path]

did you read the problem

he speeds up midway

[simple oak]

It says another

[rigid path]

he goes for 7 3/4 hours

oh

shoot

no wonder

[simple oak]

Lmao

[rigid path]

i just read it wrong

holdon i think i can do this

maybe

[simple oak]

👌

[rigid path]

so technically

like uhm

jfeipahjfeiwaphrewaipfeapfehuaip

i hate life

so

[simple oak]

well first just umm have those two speed formulas

[rigid path]

i dont understand what the problem is asking then

its asking us to find the total distance traveled right?

[simple oak]

yah

[rigid path]

but since both traveled at the same distance

and assuming they traveled at the same time

then the slower one has to finish later right?

[simple oak]

Yh

[rigid path]

so then

uhh

why is the one going 16 km/h important

[simple oak]

?

[rigid path]

if the one going 15 km/h is finishing after him

[simple oak]

Beats me

[rigid path]

.....

[simple oak]

Don't think about that

Do the algebra

[rigid path]

wut

so then wouldnt you just do r= d/t for just the 15 km/h

??

[simple oak]

ok heres the equations

wot

umm

Okay then try

Don't think itll work

What is your time

[rigid path]

uhhhhhhh

lemme do the thing

[simple oak]

Thing is

I don't think you can find the time…

cause you have two unknowns

[rigid path]

oh yeah

ur right

[simple oak]

So you need a system

To eliminate an unknown

[rigid path]

yes but the thing is that like

is the other guy traveling at the same time

or at a different time?

im rlly confused rn

[simple oak]

I don't think it matters

Just add up the times

[rigid path]

uhmmmmmmmm

[simple oak]

Umm

Actually

Lets

The question is unclear

Soo

Assume different

I'd try both ways if this was my homework question

But meh

[rigid path]

sobbing rn fr

[simple oak]

Since it wants total time I assume they travel not at the same time so times can be added up

[rigid path]

right

[simple oak]

Anyhow just do the math

1. t1=7.75-t2

2. 16=d/t1

3)15=d/t2

[rigid path]

what how did you get those tho 😭

oh

so the two equations are uhh

x+y=7.75

and then 16x=15y?

[rough storm]

yes

[simple oak]

Do you understand why that is

[rigid path]

wait why did that suddenly get like

cut out

lol

[rigid path]

oh ye

so uhh judging from the work

you have to multiply 16 by the time because uhh

every hour you multiply by 16

so then uh

the time for person 1 and the time for person 2

would be 7.75

and then uhhhh

idk uhhh

holdon

[simple oak]

Show your work using algebra

Hint you use the two speed formulas I showed and you rearrange it to equate each other

[rigid path]

yeah d = rt

right

so uhh

im trying to wrack my rbain rn lol

[simple oak]

what's rt

[rigid path]

so uhhhh the rate is 15 and the time we dont now but that woiuld equal the distance

rate times time

[junior smelt]

rate * time

[rigid path]

yes

[simple oak]

oh I see

[rigid path]

so we know that they go the same distance

so we can do uhh

16 and the time 1 equals 15 and the time 2

so the two equations are x + y = 7.75 and 15x = 15y

right?

if we switched the variables out

[junior smelt]

16x but yea

[rigid path]

:D

[simple oak]

16x

[rigid path]

oh oops

[simple oak]

Ok yes

You know how to go from here?

[rigid path]

yes

you solve the sytem right?

[simple oak]

Yes

[rigid path]

ok i can solve it

just give me some time

:D

[simple oak]

Gtg now but just remember that what you get from this is not your answer yet, you will have to use speed formula to find distance

From this you find time so you only have one unknown so you can solve

Gn

[rigid path]

okk

ty

ill ask someone to check for me then

gnn

.close

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[fleet condor]

perimeter is the sum of all the sides

[safe radish]

@digital bay Has your question been resolved?

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[rigid path]

Number 57 is also quite confusing

[rigid path]

i know we need to make a syste

syste,

system

breh

but idrk what to do asside from that

lo

lol

[plucky elk]

Show

[rigid path]

i have like

uhh holdon

lemme send it

actually there isnt even enough to do that struggle i only have busuits = x and days = y

i know that likee theres like 16 busicuits a day in 1 case

and 12 biscuits a day in the other

and thats about it

[plucky elk]

Use the part where it says "6 days less"

[mighty berry]

inequality

[rigid path]

uhhh

[rigid path]

Like this

?

[mighty berry]

If she eats 16 biscuits a day, the total number of biscuits will last for x/16 days.

If she eats 12 biscuits a day, the total number of biscuits will last for x/12 days.

[rigid path]

how did you even come to that

[plucky elk]

@rigid path you should get used to thinking of problems in terms of rates like Blue ice showed you here

[rigid path]

uhm

oh ic

[mighty berry]

now just solve for x

[rigid path]

wait what would the x be

days or busuits?

[mighty berry]

biscuits

[rigid path]

oh uhh

let me try to understand that first lol

[mighty berry]

cuz i think thats what we're trying to find

[rigid path]

hmm

thats slightly interesting

of what

[rigid path]

indeed

ok so ur saying that uhhh

i should solve this equation now right>

[mighty berry]

288 biscuits

[rigid path]

?

[mighty berry]

thats the answert

[rigid path]

what the fawekkwejkkwakwkwkawkwakwkakwkwk

holdon

😭

[plucky elk]

Can you not

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[rigid path]

ye so wait

[mighty berry]

my bad

[rigid path]

is x/12 -x/16=6 the only equation?

[mighty berry]

yesd

[rigid path]

i assumed we needed a system

[mighty berry]

then find the common denominator

[rigid path]

right so uhh ill send an ss of my work

lemme jsut solve it

[mighty berry]

ok

[rigid path]

uhh what

i did it differently

ish

uhmmmmmm

let me show u what i have so far

[mighty berry]

the common denominator isnt 16

it's 48

[rigid path]

what

no i mean you can multiply both by 4/3

i mean not both

uhh the top and the bottom fraction

of x/12

because 4/3 over 4/3 is 1

so ur just manipulating it so that the bottom is 16

[mighty berry]

just subtract the fractions like normal

you should get this ☝️

you're making it complicated

[rigid path]

well the answer i got was 288 in the end

i think

[rigid path]

ye mp

mb

lol

[mighty berry]

lol

[rigid path]

tysm for the help

[mighty berry]

as long as you get the right answer

[rigid path]

😭

[mighty berry]

also you can use chatgpt or something idk how good it is for math but i use it a lot to teach me something i dont know

[rigid path]

oh ok

lol

im in pain lol ty :D

.close

[safe radish]

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[safe radish]

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[wind jasper]

How do you know when to use absolute value and when to not?

[median vigil]

$\sqrt{x^2} = |x|$ is generally true. but, if $x \ge 0$ ($x$ is always positive), then we can drop the absolute value because it's not needed

[flat frigate]

cloud

[wind jasper]

.close

[safe radish]

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[spare dawn]

Im sorry if my question may be vague, but can someone help me see how thos vectors form the basis for that octahedral lattice? I firstly dont understand how a basis could be used as a description of the unit cell of a lattice

[safe radish]

@spare dawn Has your question been resolved?

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[stiff aurora]

hlo my fellows i want u to answer this if u respect mathematical rigour, myquestion is , In linear algebra the determinant is that area of parallelopied in R^n space

[stiff aurora]

but then why cross product is also solved using determinant

shouldn't cross product be a completely different thing since thaey give us a special vector perpendicular to two given vectors

<@&286206848099549185>

hloo

r u guys here?

plz reply if get what i m saying

<@&286206848099549185>

[median vigil]

a * (b x c), the triple product, gives us the area of the parallelopiped given by the 3 vectors, with the magnitude of the cross product being the area of the parallelogram at its base

[stiff aurora]

can u simplify plz

imean now y is scaler triplet is introduced here

[safe radish]

@stiff aurora Has your question been resolved?

[stiff aurora]

hlooooooooo

<@&286206848099549185>

[safe radish]

@stiff aurora Has your question been resolved?

[safe radish]

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[clear crest]

if a limit approaches infinity and its function is a sin function +1, do i just consider +1 and completely discard the sin?

[clear crest]

since a sin function is not limited

on the x axis at least

[quasi bison]

question unclear as stated

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.