help-19

[acoustic yacht]

honestly i just raw dogged it not using a formula but indeed it is correct

how do you come with formulas so easily

[dense moth]

That's stated to you

[dense moth]

So that must be the logical approach!

[acoustic yacht]

but eitheir way with formula you would have to do trial and error

[dense moth]

Not actually

[acoustic yacht]

how so

[dense moth]

Here, we have only one restriction

B = 9

So a can be anything logically

[acoustic yacht]

can you give an example

xor-11

.close

[odd edge]

Channel closed

Closed by @acoustic yacht

Use .reopen if this was a mistake.

[odd edge]

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.

[surreal summit]

Currenty trying to sniff around in the economic field, and I for some reason can't wrap my mind around the math. Honestly you can skip the context parts if u want, but I added them for clarity.

Context:
Equation 2.1 is a general functional form—it does not specify exactly how Q var-
ies with the explanatory variables, p, pt, and Y. The estimated demand function that
corresponds to the demand curve D1 in Figures 2.1 and 2.2 has a specific (linear)
form. If we measure quantity in millions of lbs per month, avocado and tomato prices
in dollars per lb, and average monthly income in dollars, the demand function is

Q = 104 - 40p + 20pt + 0.01Y. (2.2)

When we draw the demand curve D1 in Figures 2.1 and 2.2, we hold pt and Y at
specific values. The price per lb for tomatoes is $0.80, and average income is $4,000
per month. If we substitute these values for pt and Y in Equation 2.2, we can rewrite
the quantity demanded as a function of only the price of avocados:
Q = 104 - 40p + 20pt + 0.01Y
= 104 - 40p + (20 * 0.80) + (0.01 * 4,000)
= 160 - 40p.

AND

Answer (important part) (still text from the book)

1. Express the price that consumers are willing to pay as a function of quantity.
   We use algebra to rewrite the demand function as an inverse demand function,
   where price depends on the quantity demanded. Subtracting Q from both sides
   of Equation 2.3 and adding 40p to both sides, we find that 40p = 160 - Q.
   Dividing both sides of the equation by 40, we obtain the inverse demand function:
   p = 40 - 0.025Q

QUESTION:
By subtracting Q from both sides I should get 0 = 160 -40p -Q
-> Then by adding 40p to both sides I should get 40p = 160 - Q
-> Finally by dividing both sides I should get p = 4 -0.025Q.

HOWEVER, the book says that it equal p = 40 -0.025. Am I missing something?

[odd edge]

@surreal summit Has your question been resolved?

[surreal summit]

<@&286206848099549185>

[sour notch]

to me it looks like the book made a mistake with saying p=40-0.025Q instead of p=4-0.025Q

cause 160/40 is just 4

[surreal summit]

Problem is that it goes on using it

So it can't be a mistake

[sour notch]

the book itself already uses the equation 40p=160-Q which would definitely result in p=4-0.025Q so i do believe it is a mistake
either way, the anser to that question in the picture equals the same because 40-40=4-4

in this case i would say it´s okay to assume the book made a mistake, unless there are multiple excercises with the exact same problem

[surreal summit]

But they keep on using those numbers. Ngl the lack of explanation this book provides makes me not want to study economics. I'll close the channel though. Thanks for the help denise.

.close

[odd edge]

Channel closed

Closed by @surreal summit

Use .reopen if this was a mistake.

[odd edge]

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.

[shadow pilot]

they want me to determine the domain of this (f/g)

i got 1 -sqrt2 and 1+ sqrt 2 when i did quadratic

the correction sheet says -1 and sqrt 2

did i do something wrong

i did quadratic with g(x) since its denominator

[south plume]

you are correct

[south plume]

idk what they are on

[shadow pilot]

so just a problem in the correction

[south plume]

maybe

show what they wrote

[shadow pilot]

wait i misread it it says 1 plus minus sqrt 2

i read it differtently

im the one on smth

[south plume]

lmao

[shadow pilot]

ty

,ckise

'.close

.close

[odd edge]

Channel closed

Closed by @shadow pilot

Use .reopen if this was a mistake.

[odd edge]

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.

[mystic saffron]

I’m dumb

.close

[odd edge]

Channel closed

Closed by @fast oar

Use .reopen if this was a mistake.

[odd edge]

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.

[heavy wraith]

Is this function right differentiable in 0

My solution comes out to 0, but the solution should be 1/2, but I don't see my mistake

[vale vapor]

whats in front of the tan?

[heavy wraith]

Bgtan

I mean Bg is in front, so it is Bgtan

[vale vapor]

whats that?

arctan?

[hexed hinge]

Arctan

wow

[vale vapor]

i looked it up and i have never seen it written like that in my life

[hexed hinge]

same

[forest sky]

can you show your solution?

[heavy wraith]

Where I live we always use Bgtan instead of arctan

Yeah

That is what I though I should do

This is what I should get

[hexed hinge]

dutch? boogtangens?

[vale vapor]

it should just be x in the denominator and not x-1

[heavy wraith]

yessir

[vale vapor]

never heard of it in my life wtf

[hexed hinge]

because you are not dutch unfortunately

[heavy wraith]

don't you use this formule lim(x->a) (f(x)-f(a))/x-a with a = 0

[vale vapor]

i am dutch though

lol

[hexed hinge]

really?

damn

[vale vapor]

jazeker

(de hypotheker)

[hexed hinge]

how does it feel like riding those bikes over there

[vale vapor]

yes you do use that

but you write yourself there that /(x-a) with a=0

[heavy wraith]

ja das dom

[vale vapor]

yet what you wrote down was /x-1

[heavy wraith]

domme fout

[vale vapor]

kan gebeuren

[heavy wraith]

Merci

[vale vapor]

feels really good

[heavy wraith]

jullie zijn echt legends man

.close

[odd edge]

Channel closed

Closed by @heavy wraith

Use .reopen if this was a mistake.

[odd edge]

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.

[whole anvil]

Can I do anything with the right side?

[vale vapor]

nope, thats the most simplified form

[whole anvil]

Damn

That's sad

[vale vapor]

it is what it is

[whole anvil]

Well thank you

.close

[odd edge]

Channel closed

Closed by @whole anvil

Use .reopen if this was a mistake.

[odd edge]

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.

[whole hazel]

How did we get the red box?

[sand horizon]

Tan² + 1 = sec²

[royal herald]

notice only the numerator changed. a pythagorean trig identity is tan^2 = sec^2 - 1

[sand horizon]

Its an identity

[frosty badge]

you can prove the identity from sin^2(x) + cos^2(x) by diving with cos^2(x)

[vale vapor]

$\tan^2=\frac{\sin^2}{\cos^2}=\frac{1-\cos^2}{\cos^2}=\sec^2-1$

[clever fjord]

Bonk

[royal herald]

@whole hazel whos message was the most helpful?

[whole hazel]

This one was most helpful

[frosty badge]

you'll get $tan^2(x) + 1 = 1/cos^2(x) = sec^2(x) -> tan^2(x) = sec^2(x) - 1$

[whole hazel]

Ty

[clever fjord]

alex <3

[royal herald]

we got bonked by Bonk i guess

bye

[sand horizon]

Bye

[frosty badge]

when you start from sin^2(x) + cos^2(x) = 1 and divide by cos^2(x)

[whole hazel]

ty guys

.solved

[odd edge]

Channel closed

Closed by @whole hazel

Use .reopen if this was a mistake.

[odd edge]

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.

[stone ermine]

how do i get to the yellow step? it's not a gaussian integral is it?

[vale vapor]

physicists and notation.....

im not 100% positive, but i think its an odd-evenness trick

the integral of an odd function from -inf to inf is 0

so $\int (x-a\cos(\omega t))^2xdx$ reduces down to $\int -2ax^2\cos(\omega t)dx=-2a\cos(\omega t)\int x^2dx$

how they get rid of the last integral im not quite sure

[clever fjord]

Bonk

[vale vapor]

oh wait, its in the exponent

disregard my messages then

[stone ermine]

hahaha sorry

[vale vapor]

who tf puts dx at the front of the integral

[stone ermine]

my professor lol

I found it weird at first but I see it so much now that it’s oddly comforting

[odd edge]

@stone ermine Has your question been resolved?

[quasi sparrow]

You need limits to do the integral

And looks like you wrote the wave function squared wrong

Show the full steps

[stone ermine]

hmm okay i just copied it off my prof's answer and he didnt show the full steps thats why i got confused

nevermind i believe its a gaussian integral and he used an identity, ill look it up

[odd edge]

@stone ermine Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[weak oriole]

How do I find x?

[boreal crag]

Can you use the fact that the lines are parallel?

[weak oriole]

Idk

I hope so that’s literally all it gives me

Just find x and thst

[fair hemlock]

Ur allowed to use calculator right

Cuz im pretty sure u need sin to do this

[weak oriole]

Yeah

It’s sohcahtoa chatpter

[narrow hound]

Even with sinus I don’t think it would be possible

[fair hemlock]

U can see that the top triangle is congruent to the bottom triangle

[weak oriole]

Yeah

[fair hemlock]

Well just mirrored

[weak oriole]

Would that mean the other side is also 4 cm?

[fair hemlock]

Yes

Now just use sin(18)

[narrow hound]

How you’re getting at this conclusion

[weak oriole]

Ok one sec lemme get calculator

[weak oriole]

Cause the two triangles are the same

[fair hemlock]

SAA

[weak oriole]

Ok so 0.309

[narrow hound]

No proof of that

[fair hemlock]

No sin is opposite over hypotenuse

[weak oriole]

Yeah but ig you just have to assume

[fair hemlock]

AAS

Two angles are same and one side is shared

[weak oriole]

So 4 over x?

[fair hemlock]

Yes

Yes

Solve for x

[weak oriole]

0.309 = 4/x

Then multiply by x divided by 0.39 right?

[fair hemlock]

Wait

Wrong one

[narrow hound]

But the side isn’t the same as the other one, one is hypothenus and other one is adjacent

[fair hemlock]

Its tan

Not sin

[weak oriole]

OOoH

[fair hemlock]

Opposite over adjacent

Cuz u need x

[weak oriole]

Yeah

So it’s 0.324

[fair hemlock]

So tan18=4/x

Solve for x

[weak oriole]

1.3?

🙏 for help

[fair hemlock]

I meant 4/x

Mb

[weak oriole]

Alr how do I close the channels again

[fair hemlock]

.close

[narrow hound]

.close

[weak oriole]

Thx

.close

[odd edge]

Channel closed

Closed by @weak oriole

Use .reopen if this was a mistake.

[odd edge]

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.

[near hazel]

Trying to check these over

I don’t know if I am absolutely right though

[royal crystal]

The algebra for part a looks correct

[raw gazelle]

Everything looks right

[royal crystal]

Part b as well

[raw gazelle]

As long as you calculated the zeroes for the quadratic in part a correctly you should be good

Looks good 👍

[near hazel]

Hey, also, is graphing a log count as algebra?

[raw gazelle]

Wdym?

Like graphing a log function?

[near hazel]

Because I was wondering if my graph looks to bad

I have all the points right, and everything, but it seems like I extended the arrow outside the boundary. Do teachers penalize this usually?

[raw gazelle]

No

[near hazel]

Ok thank you!

[raw gazelle]

As long as you draw in the asymptote you’re being pretty clear with what you’re saying

[royal crystal]

All that really matters is that you have points and the correct shape

Oh yeah and asymptote

[raw gazelle]

So it doesn’t really matter if the arrow goes a little bit past the asymptote line

[odd edge]

@near hazel Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[whole anvil]

I don't get how to do this one

[whole anvil]

Is it even seperable?

[karmic lotus]

hi again

[whole anvil]

Hey 😭

[karmic lotus]

its a linear ode

have you learnt that

[whole anvil]

I have

I just don't remember

I know you use integrating factor

[karmic lotus]

yeah

[whole anvil]

Let me try to do that

How do you identify if it is linear when it looks like this?

Is it due to the +xy?

[chilly ravine]

erm

You just factorise it haha

(1+x)(1+y)

[whole anvil]

How does it multiply?

[karmic lotus]

holup

you might be cooking

[chilly ravine]

factorise rhs

[whole anvil]

Who?

[whole anvil]

I get the y(1+x) part

[whole anvil]

But the (1+x)(1+y) doesn't make sense to me

What happens between 2 and 3 😭

[chilly ravine]

huh?

[whole anvil]

Here

[chilly ravine]

2 is just factoring out y+xy

[whole anvil]

Yeah

[karmic lotus]

ika pls

[chilly ravine]

3 is noticing y(x+1) +(x+1) has the same common factor, ie (x+1)

take out common factor (x+1) and you get (x+1)(y+1)

[whole anvil]

Aaaaaaah

[chilly ravine]

it's a basic common factoring expression so that's why I saw it lol

[whole anvil]

No no ofc

Idk why I didn't see it

[karmic lotus]

same

[whole anvil]

I was so stuck on the + 😭

[chilly ravine]

all the math comp prep memorisation came in clutch 😎

[whole anvil]

I wish I tried harder at math as a kid

I have so many holes due to not caring 😭

[chilly ravine]

this question wasn't really a hole imo it's just memorizing

kinda like not knowing your time tables (yes I don't know them 😬)

[whole anvil]

(Me neither)

[karmic lotus]

heresy

[whole anvil]

I know the basic ones

5, 9, 10 (oooh crazy Ik), 2, 4, 1, 6 (partialy 😭)

But 7 and the later parts of 8 mess me up 🥲

[chilly ravine]

I know my squares and the rest I just do mentally (for example if I have 9×8 is just do 8²+8 or 9²-9)

[whole anvil]

Wait so is it seperable?

[whole anvil]

I just count down from 90 😭😭😭😭😭😭

Kinda

[chilly ravine]

?wdym

[chilly ravine]

Had to have those memorized since they were a common thing in comps

[karmic lotus]

im glad the timetable has a catchy flow in my language so its easy to remember

[karmic lotus]

you seperated it

[whole anvil]

Yeah

[karmic lotus]

gg

[chilly ravine]

still not quite sure what you meant

[whole anvil]

Like seperable or linear

[karmic lotus]

both

[whole anvil]

:(

[karmic lotus]

theyre not distinct qualities

[whole anvil]

Now what does that mean 😭

Huh

What about this?

[karmic lotus]

never saw this lemme read

who are you to say that theres no "linear" bubble under yes seperable

it just isnt significant cuz separating is many times easier

[whole anvil]

I thought it had to be either or

That's how I understood that slide 🥲

[karmic lotus]

just choose a method

they all work

[chilly ravine]

I think it assumes they cannot factor out or something

[karmic lotus]

all it says is that theres another option if you cant separate it

did they teach a third method tho

whats under good luck

[whole anvil]

It's a joke I think

Like

If neither works

Good luck

[chilly ravine]

yeah I thought so too

but it's pretty easy still ngl😭

[karmic lotus]

nono my uni taught a third method

[whole anvil]

🥲

[karmic lotus]

homogenous odes

[whole anvil]

Yeah they're a part of this

[chilly ravine]

That's 2nd order no?

[karmic lotus]

nah 1st

[chilly ravine]

I kinda forgot what the names meant

[karmic lotus]

my 2nd order slides only cover linear

[whole anvil]

Both

I think

[karmic lotus]

every term has the same power or sth

[whole anvil]

Homogenous is if it ends in 0

[karmic lotus]

so if you turn x and y into tx and ty you can factor out t^n nicely

[whole anvil]

Inhomogenous is if it has shit going on after =

[whole anvil]

I'ma continue on my questions 😭

[karmic lotus]

.

[whole anvil]

Thanks y'all

.close

[odd edge]

Channel closed

Closed by @whole anvil

Use .reopen if this was a mistake.

[odd edge]

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.

[shell finch]

pls help me im stuck for more than hour

[vale vapor]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shell finch]

1

[vale vapor]

there are multiple ways to approach it

y=mx+c is in a form called slope-intercept form

where m is the slope (how steep the line is)

and c is the intercept (where it crosses the y-axis)

to find c, we simply need to look where the line crosses the y-axis

what is c?

[shell finch]

at 20?

[vale vapor]

yep

the intercept (c) is 20

and now, what is the slope?

i.e. if x moves by 1, how much does y go up or down

[shell finch]

3

[vale vapor]

careful now, the y-axis is not 1 every square

[shell finch]

i can only see that when x moves by 1 y moves by 3

[vale vapor]

but look at the numbers on the y-axis

it goes 5,10,15,20,25,...

[shell finch]

oh ok

[vale vapor]

so yes, it moves 3 squares

but every square is actually 5 units

(in the y-direction)

[shell finch]

so when x moves by 1 y moves by 15

[vale vapor]

yes!

[narrow hound]

wonderful

[shell finch]

thanks

[vale vapor]

to conclude, what are m and c?

[shell finch]

y=15x+20

[vale vapor]

perfect!

[shell finch]

les go finally i got it

thanks

.close

[odd edge]

Channel closed

Closed by @shell finch

Use .reopen if this was a mistake.

[odd edge]

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.

[tepid quarry]

$ax+bx+ay+by$ we need to factor by grouping terms.

[clever fjord]

Simon James B

[tepid quarry]

if we group the first 2

$x(a+b)$ if we factor the last 2 $y(a+b)$

[clever fjord]

Simon James B

[tepid quarry]

so now a+b is also a common factor here

$x(a+b) + y(a+b) = (a+b) + x + y$ is what i said but it is not right

[clever fjord]

Simon James B

[tepid quarry]

and i have no diea why

[vale vapor]

try replacing a+b with c

and see if you can do it then

$xc+yc$

[clever fjord]

Bonk

[tepid quarry]

same thing tho just different common factor

[rapid reef]

x(a+b) + y(a+b)
Factor (a+b)
(x+y)(a+b)

[vale vapor]

!nosols

[odd edge]

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.

[tepid quarry]

i don't see why it's multiplication between them

[vale vapor]

can you factor this?

[tepid quarry]

c(x+y)

[rapid reef]

Shush

[vale vapor]

exactly

now replace c with (a+b) again

then what do you get?

[viscid flint]

no, he's right. please let him cook in this case.

[tepid quarry]

x(a+b)

[vale vapor]

not quite

$c(x+y)$ and $c=(a+b)$

[clever fjord]

Bonk

[tepid quarry]

a = ax nad b = bx

the common factor is x right and a+b would be x(a+b) i do not understand thatt

[vale vapor]

here we get (a+b)(x+y)

do you see why?

[tepid quarry]

no

i just started this lesson today with grouping terms and i am stuck very very very stuck

got the idea of grouping terms to help us ig but the rest not so much

[vale vapor]

okay, lets work backwards

assume (a+b)(x+y) is the solution we want

[tepid quarry]

right

[vale vapor]

try to distribute the terms

[tepid quarry]

ax + ay + bx + by

[vale vapor]

can you show your steps?

[tepid quarry]

you multiply each from the first with each of the second

what is there to show

(a+b)(c+d) = ac + ad + bc + bd

[vale vapor]

its nice that youre able to see it that quickly

do you notice how that right hand side is what is being asked in your current problem?

[tepid quarry]

yea

ig i got it

[vale vapor]

so then do you see how its equal to (a+b)(x+y) and not (a+b)+x+y

[tepid quarry]

yea

.close

[odd edge]

Channel closed

Closed by @tepid quarry

Use .reopen if this was a mistake.

[odd edge]

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.

[ocean nest]

Hi, how to show that a function is good for derivation?

in the correction they use limits as well which i didn't understnad

[quasi sparrow]

Use the limit definition of derivative

[ocean nest]

i was not aware of that

[lapis rain]

They should have made you aware of it if they're giving you exercises whose solutions use it ...

[ocean nest]

i am catching up with missed classes

is that this formula: $\tau(x) = \frac{f(a+h)-f(a)}{h}$ ?

[clever fjord]

Badr

[ocean nest]

lim for h \to 0?

[quasi sparrow]

Should be f'(a) on the left side

[weary pelican]

and missing the limit

[odd edge]

@ocean nest Has your question been resolved?

Channel closed

Closed by @ocean nest

Use .reopen if this was a mistake.

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.

[winged bear]

How would i calculate the GCD of 9 9 digit numbers where each number has the digits from 0-9 but no repeats

[winged bear]

is it 3?

i think its 3

[nimble blaze]

3 would be a common factor but it won't be the greatest

[winged bear]

woah moderator

an honour

[nimble blaze]

disability rules indicate a higher common factor, but that isn't necessarily the greatest either

[winged bear]

i think i explained it really badly

[nimble blaze]

also if it has digits from 0-9
those would be 10 digit numbers

[winged bear]

also the digits can't repeat so if one number ends in 1 none of the others can

my fault

its 9 digits and you choose 9 digits from 0-9

[nimble blaze]

do you have a pic of the original problem

[winged bear]

yeah

but the original is a bit differeny

ive started by trying to work out the gcd

[nimble blaze]

if you don't use 9, they'd all be divisible by 9

[winged bear]

hmm

thats good to know

[nimble blaze]

but there could potentially be a higher possible value

is this supposed to be a coding question?

[winged bear]

I don't think so but I'm not sure

which is why I was hesitant to put the full question here

cause ik you can work out te gcd mathematically

[nimble blaze]

don't think there's an easy way to do this

[signal yacht]

do you the full question?

oh wait nvm it's just above

really sorry

[winged bear]

no worries

[winged bear]

I think it can be done logically

cause i got it as a puzzle

but no solutions 😦

[signal yacht]

this genuinely feels like something id read in like leetcode lol

[nimble blaze]

who gave the puzzle

[winged bear]

should i ask a coding discord maybe

no its not supposed tobe done with coding

[nimble blaze]

if the GCD is 9, the answer won't be unique
so it should be something higher

[winged bear]

I think 27 could work as well

or any other higher power of 3

[odd edge]

@winged bear Has your question been resolved?

[winged bear]

.close

[odd edge]

Channel closed

Closed by @winged bear

Use .reopen if this was a mistake.

[odd edge]

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.

[ember pelican]

answer needs checking

[odd edge]

Please don't occupy multiple help channels.

[signal yacht]

i mean it looks fine

[odd edge]

@ember pelican Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[deep willow]

So I have the answer for this which is ( C ) - pi/18 +( pi/9)n

I really need help with the steps and how this works, it is trigonometry, I am good at math but I forgot how this question gets solved, please help ❤️

Question 11 btw

[wooden gorge]

1. cot x = tan (π/2 - x)

2. tan x = tan y => x = y + nπ

[deep willow]

What

[mental wharf]

first you should convert the cot into tan using the formula

then you can use the second to find a general form instead of a single solution

[deep willow]

Yea

Tan 5x = 1/tan4x

Hm

[mental wharf]

dont write it like that tho

write cot(4x) as tan((pi/2)-4x)

[odd edge]

@deep willow Has your question been resolved?

[odd edge]

@deep willow Has your question been resolved?

[odd edge]

@deep willow Has your question been resolved?

Channel closed

Closed by @deep willow

Use .reopen if this was a mistake.

[deep willow]

.close

.Close

[odd edge]

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.

[sturdy geode]

may i get help

[warped glacier]

just ask!

[sturdy geode]

isnt smth missing here

[warped glacier]

what does that mean?

show the original question please

[sturdy geode]

i am supposed to use pythagorean / euclid theorem, the original question is in my language, lemme translate it:

In the right triangle ABC (γ=90°), a=4cm, cb=6cm. Calculate side b side c.

but i feel like smth is missing there

[warped glacier]

a = 4cm?

[sturdy geode]

yep

[warped glacier]

that notation isn't right

[sturdy geode]

yea i was wondering if i am dumb

[warped glacier]

like a is the side opposite angle A

but then CB is already 6cm, so it's impossible

[sturdy geode]

that line drawn from point C is the height to its side and divides it into Ca and Cb

so in conclusion, C will be longer than 6cm but like

[odd edge]

@sturdy geode Has your question been resolved?

[woven pier]

What?

We can only conclude that 'a=4cm' implies that the side opposite to angle A = 4cm, which is BC, but again, CB = BC = 6cm is given in the question. So the question is not properly communicated. Also, was any diagram given originally, or did you just out of the blue drew one?

[sturdy geode]

Cb is not BC

i solved it

[woven pier]

Ok, then are you done?

If you are, then .close it.

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[odd edge]

@sturdy geode Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[carmine sentinel]

Help please

[carmine sentinel]

How do I do this?

[odd edge]

@carmine sentinel Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[wooden gorge]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pastel matrix]

Is it x = -3/4 and then for the terms -3/4, 2, -13/4

Is that the answer?

[wooden gorge]

That's right.

[pastel matrix]

Thanks

would the answer to part b. od the question be $750

[wooden gorge]

Right

[pastel matrix]

Ok

Thank you

[odd edge]

@pastel matrix Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[quasi sparrow]

uh not enough information

[royal aurora]

Lemme grab some

[quasi sparrow]

Oh yes the x axis width is inconsistent

Each square should represent the same spacing

[royal aurora]

So what would I put down as the 2 answers?

Sorry im really confused 😖

..?

[quasi sparrow]

This should tell you one of them

[royal aurora]

So the scale on the horizontal axis should go up om equal-size?

[quasi sparrow]

Yes

[royal aurora]

What about the other mistake

[quasi sparrow]

Were you taught histograms should start at 0 on the x axis

[royal aurora]

I wasn't in for the first few lessons

[quasi sparrow]

Or that there should be gaps between the bars

Go borrow some notes then

[royal aurora]

I'll wing it with this

.close

[odd edge]

Channel closed

Closed by @royal aurora

Use .reopen if this was a mistake.

[odd edge]

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.

[timid sky]

What exactly is the mathematical rigour behind these "theorems"? X and F are Laplace Transforms

[odd edge]

@timid sky Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[cosmic obsidian]

Hello! I'm working on a project involving atmospheric dispersion modeling, especially related to particle dispersion (e.g., radioactive fallout). Can someone recommend any key papers, books, or resources on the topic? Computational methods and practical applications are of particular interest.

[odd edge]

@cosmic obsidian Has your question been resolved?

[deep willow]

Hm

Did you try asking chatgpt

It could help sometimes

[cosmic obsidian]

yeah, it gave me some ideas but i prefer real people opinions lol

[deep willow]

Fair

[odd edge]

@cosmic obsidian Has your question been resolved?

[odd edge]

Channel closed

Closed by @cosmic obsidian

Use .reopen if this was a mistake.

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.

[whole hazel]

What’s a way I can remember the cos tan relation?

[whole hazel]

Is there a way I can get it geometrically w/ a right triangle r sum?

[lethal spoke]

Do you know the trig identity
Cos²x + sin²x=1

[whole hazel]

Yes

[whole hazel]

Yes

[lethal spoke]

Divide through by cos²x
And you get 1+tan²x=1/cos²x

Then express cos x in terms of tan²x

To find out if cos x is positive

[whole hazel]

Ty

.solved

[odd edge]

Channel closed

Closed by @whole hazel

Use .reopen if this was a mistake.

[odd edge]

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.

[mystic saffron]

In the following graph, AC=AD and BE=9, EC=7.
Find DE.

[meager juniper]

Let F be the intersection of AC and ED. Triangles AFD and EFC are similar. This should be enough information to find a and x using a system of equations.

@mystic saffron

[mystic saffron]

thanks

[mystic saffron]

can you give me please another hint

[meager juniper]

!show

[odd edge]

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

[mystic saffron]

theta + aplha = 90

mm

7k = 16

?

above

is not useful

right

no

16

[wooden gorge]

Chasing angles is no help

[mystic saffron]

s0

so

maybe

k is 16/7

and a = 16

[wooden gorge]

You get two equations:

AC • DE = AD • CE + DC • AE
DE² = DC² - CE²

Now the question is, how to explain it without ptolemy's

[meager juniper]

Let b = FC, let c = EF, let k be the constant of similarity.

7^2 + x^2 = 2a^2
(a-b)^2 + a^2 = (x-c)^2
c^2 + 7^2 = b^2
c = k(a-b)
b = k(x-c)
a = 7k

5 unknowns, 5 equations.

There's probably a nicer approach though

[mystic saffron]

mm yeah, I don't think they want me to use matrices or sm like that

idk

[mystic saffron]

idk that

[meager juniper]

Can't, not linear

[wooden gorge]

Here, CX = 7/16 a, XA = 9/16 a --- (by bpt)
Now,

DE² = DC² - CE²
DE = DX + XE = √(a² + XA²) + √(CX² - 49)

That's the simplest it can get.

Solving wouldn't be too tough if you plug value of (2a² - 49) from equation 1, to equation 2

[mystic saffron]

I'll return in a coupole of hours

I'm in class

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[rose herald]

i need help with this please

[rose herald]

i know how to grid the normal ones, but i dont know how to solve these and grid them, all i know is to isolate the Y and nothing after that

[static nacelle]

Solve for y, and then grid as if it was a normal one

[rose herald]

what do you mean?

can you just help me?

bruh

please someone

<@&286206848099549185>

[olive needle]

What is confusing you?

[rose herald]

like

when i isolate the y

i js dont know

after that

[olive needle]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[rose herald]

shut up

please

[olive needle]

<@&268886789983436800>

So do you know what a dependent and independent variable are?

[rose herald]

how would that corelate to the question

but ys

yes i do

x and y

[olive needle]

Because that will allow you to begin plotting some points for each equation.

[rose herald]

right

i just need a follow through of the steps

cause im stuck

[olive needle]

If you put each equation into y = mx + b form, you will make x the independent variable and y the dependent variable.

From there, it's just a matter of choosing any value for x and solving for y. That will give you an (x,y) ordinal pair which you can plot.

[lavish slate]

??

[odd edge]

@rose herald Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[scenic cypress]

okay so im currently doing a practice test my teacher gave me, its about working with formulas concerning parabools for the most part, there is a question on the practice test that goes as following:

from the parabool p: Y = -1/6X^2 + bx + 2, the Y-top is 2.2/3
calculate the intersects of the X-axis algebrically

usually i would do a ''stelsel vergelijking'' but im not sure if thats possible, and they dont really use it in the answer sheet, this is the answer sheet for anyone wondering

[outer coyote]

what does oftewel and geeft mean

😭

[scenic cypress]

LMAO OH YEAH ITS IN DUTCH

en = and
geeft = gives
oftewel = aka/in other words

dus = so

[outer coyote]

wait so the parabola's (think uve been calling them parabools) equation is $y = \frac{-1}{6}x^2 + bx + 2$, right?

[clever fjord]

poudel

[scenic cypress]

true

[outer coyote]

and $y$ top just means that the turning point is at $y = 2 \frac{2}{3}

[clever fjord]

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

[outer coyote]

ok

[scenic cypress]

ik it dont make a difference but the negative sign is infront of the fraction not on the 1

yeah basically

[outer coyote]

and we need to work out where it intersects with the x axis

ok got it

are you familiar with this form for a parabool

$y = a(x - h)^2 + k$

its known as the vertex form

[scenic cypress]

yea for the top point

[clever fjord]

poudel

[scenic cypress]

h is the x coardinates for the top and k is the y coardinates for the top

[outer coyote]

yeah

[scenic cypress]

mhm

[outer coyote]

dont think u need it tbh I was going to say rearrange it into that form

but its better to just

sub in $y = 2 \frac{2}{3}$

[clever fjord]

poudel

[outer coyote]

into the parabool

and solve for x like that

$y = \frac{-1}{6}x^2 + bx + 2$
\ \ $\frac{8}{3} = \frac{-1}{6}x^2 + bx + 2$

[clever fjord]

poudel

[scenic cypress]

how can i solve for x if b is unknown?

[outer coyote]

hmm

[scenic cypress]

it looked simple but the more i thought about the less sense it made

[outer coyote]

oh wait

did you know this asw

[scenic cypress]

fuck he means w ytop is 1.5 of b squared + 2

[scenic cypress]

asw?

[outer coyote]

as well sorry lmao

[scenic cypress]

where did he get that from?

[outer coyote]

$x_{top} = -\frac{b}{2a}$, when $y = ax + bx + c$

[scenic cypress]

dude i finished the whoel unit and i thoguht i was a 1000 precent ready but this odnt make sense

for this xt = 3b

i got that

[clever fjord]

poudel

[scenic cypress]

I KNOW

[outer coyote]

then you just sub x top into the parabool

[scenic cypress]

b

is unkiwng

unkown

i need to find b

and then solve

[outer coyote]

listen

solve for y top in terms of b

[scenic cypress]

wym?

[outer coyote]

do you know how to do that?

[scenic cypress]

fuckj no

[outer coyote]

yeah alr

you know that x top is 3b right

[scenic cypress]

yea

[outer coyote]

thats what you were just showing me

yeah aight

now

if you sub x top into the parabool

what do you get

you get y top

i hope you knew that

[scenic cypress]

yea but idk what the xtop is, i know its 3bs but no idea what the vlaue of b is

[outer coyote]

yeah we're getting to that

you don't need to know what b is yet

just do what i'm saying rn

and you'll understand trust

[scenic cypress]

so do i substitute x for 3b?

[outer coyote]

yes

[scenic cypress]

alr 1s

[outer coyote]

since the coordinate for the vertex is (x top, y top)

[scenic cypress]

i got b = 1.86666667

[outer coyote]

,ask what is 2/3 in decimal

[clever fjord]

Results provided by WolframAlpha

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

Requested by poudel

[outer coyote]

according to the answer sheet you're wrong

what did you do after I told you to work out y top in terms of b

[scenic cypress]

yeah i know

idk how

ok so

i substitued x for 3b

[outer coyote]

$y_{top} = -\frac{1}{6}(3b)^2 + b(3b) + 2$
\ \ $y_{top} = -\frac{1}{6} \times 9b^2 + 3b^2 + 2$
\ \ $y_{top} = -\frac{3}{2} b^2 + b^2 + 2$
\ \ $y_{top} = -\frac{1}{2} b^2 + 2$

is what you should have rn

[scenic cypress]

so -1/6 of 9b is -1.5b + 4b ( bx for x is 3b is 4b no?)

huhh

wait

ok makes sense that its 3b squared

OOOOOHHH

okay

wait

[clever fjord]

poudel

[outer coyote]

you should've gotten this

what mistake did u make lmao

i see your mistakes

[scenic cypress]

it just brainfarts icl

but its really wierd ngl

[outer coyote]

you turned (3b)^2 into 9b instead of 9b^2
and you turns b x 3b into 4b instead of 3b^2

lmaooo

[scenic cypress]

i thought of substituting x for 3b

[outer coyote]

massive brainfarts

[scenic cypress]

but i got a completely different anwer 😭

i dont blame mytself tho

i slept at 7pm and woke up at 12 am

yea

its very confusing

ill go with the rest of the answer sheet

hey uh can i like somehow go back to this afte ri close it

[outer coyote]

yeah allg

gl

[scenic cypress]

.close

[odd edge]

Channel closed

Closed by @scenic cypress

Use .reopen if this was a mistake.

[odd edge]

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.

[viral tapir]

B is a brain teaser and im having trouble wrapping my head about how to do it

[ocean bramble]

,rotate

[viral tapir]

Can you pls help me🙏😭

[ocean bramble]

you know that 114 perform in band.
you know that 56 perform in chorus.
You also know that 75% of the 212 perform in something, which are 159

i'm assuming "either band or chorus" includes playing band, playing chorus, or playing both

[viral tapir]

Wait hold on

[ocean bramble]

the best way to visualize it is probably drawing a venn diagram

[viral tapir]

So 75 percent of 212? Why not the 170?

[ocean bramble]

because that assumes that you can only perform one of those

you're never told that. In fact, you're explicitly asked how many do both

[viral tapir]

So is that 75 percent of 212 how many do one or the other not both?

[ocean bramble]

no, im counting that 75% of 212 (159) as they do band, they do chorus, or they do both band and chorus

[viral tapir]

It says either tho not both

[ocean bramble]

yeah, because english fucking sucks

[viral tapir]

Ik thats why im not going to be an english teacher

[ocean bramble]

i guess you can do it assuming it does not include the both as an option

[viral tapir]

So we have the 170 which is all the band and choir and we have the people that do one or the other which is 159 so we subtract it and we end up with 11 doing both?

[ocean bramble]

you'd have something like this:

A would be only band, C only chorus, and B would be both

[viral tapir]

k

[ocean bramble]

you know that A+B = 114, and that B+C=56

and that A+B+C = 159

so as you said, B = 11

you could understand the 159 as doing only A+C=159, bud you'd notice than then you'd have B=6.5, which obviously cannot be

and thus why english suck

[viral tapir]

Ik thats why i focus double the time in math

Anyways tysm for your help

[viral tapir]

I got one more thing how do you find the fractions on paper without a calulator speciffically 159?

Nvm got it

[odd edge]

@viral tapir Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[silent zephyr]

how do i find the x-intercepts

[silent zephyr]

without the graph

or am i supposed to make a graph on my own

[real musk]

you can find the x intercepts when letting y to be equal to 0

so your equation would be -x^2+6x+7=0

Now try solving for x

[silent zephyr]

uhh

i got

5x^2=-7

@real musk

o dot think thats right

[real musk]

hmmm

[real musk]

try to factor this equation

can you do that?

[silent zephyr]

i cant factor

[real musk]

oh, I understand

In this type of problem, its really best of you are able to factor equations

I'll help you how to factor

so our equation is: -x^2+6x+7=0

First we need to multiply the coefficient of the first term and the coefficient of the last term

[real musk]

Our first term here is -x²

The coefficient is -1. You can think of it as -1x²

And the last term is 7

And its coefficient is 7 itself

Can you follow up to here?

[silent zephyr]

UMMM

ur using big words

[real musk]

ok ok hahaha

I'll try to simplify things for you

Ok so our equation is -x²+6x+7=0 right?

[silent zephyr]

yes

[real musk]

now -x², 6x and, 7 are what we call terms

and coefficients are the numbers in those terms

so for example, the coefficient for 4x is 4 because thats the number of it

another example, the coefficient for x is 1 because you can think of it as 1x

Did you get that?

[silent zephyr]

okok

ik

yes

[real musk]

Ok for -x², what do you think our coefficient is?

Clue is that we can think of -x² as -1x²

[silent zephyr]

1

yes

-01

-01

-1

[real musk]

Yes youre correct

How about the second term, 6x?

What is its coefficient?

[silent zephyr]

6

[real musk]

Nice

And how about the last term 7?

[silent zephyr]

7 is just 7

[real musk]

Yes youre correct

Now, to factor, we need to multiply the coefficient of the first term and the coefficient of the last term

[silent zephyr]

7

[real musk]

remember that the coefficient of the first term is -1

And the last term is 7

[silent zephyr]

oh

-7

[real musk]

Correct!

Now remember the coefficient of the second term which is 6

Now, we need two numbers that if we multiply, it will result to -7 and when we add it, it will become 6

Try to think of those two numbers

[silent zephyr]

uhhuhuhuhhuuh

-7 an d 1

wati

wiat

7+-1

[real musk]

Nice !!

So we have the two numbers 7 and -1

Now, you've factored the equation!

Did you get it?

Wait, Im sorry I made a mistake

[silent zephyr]

yse

OKay

[real musk]

Okay so this is our two numbers

Technically you already got the x intercepts which is -1 and 7

If you want to continue to learn factoring I can help you

[silent zephyr]

I got it now

Thankx

so do i just have to factor

to find all the x's

[real musk]

Yess

First you just have to let y=0 and then solve for the value of x, it can be solved using factoring or any other methods

[silent zephyr]

sighh

okay..

.close

[odd edge]

Channel closed

Closed by @silent zephyr

Use .reopen if this was a mistake.

[odd edge]

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.

[tawdry pagoda]

How to understand the concept of limit, continuity and differentation in better ways.

[mellow tundra]

Graphs

How have you learnt it tho?

[odd edge]

@tawdry pagoda Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[woeful rivet]

Hi! Im going into 12th and i was wondering what would be the most complex topic here so i can learn ahead

[wild hull]

jee?

[woeful rivet]

Whats jee

[wild hull]

nvm

[wild hull]

go for functions then integration

[copper quarry]

bum chicken

[woeful rivet]

HI WATER BEAM

why u still here

[copper quarry]

why not

[woeful rivet]

ur graduated.

[copper quarry]

ye

im chillin

[woeful rivet]

this is not chilling

[copper quarry]

integration is easier than vectors

[celest acorn]

not rlly

[woeful rivet]

vectors are so hard

😭😭😭😭😢😢😢

[copper quarry]

they are for a highschool level

ive seen their topics

[celest acorn]

highschool = jee level, right?

[copper quarry]

no

[woeful rivet]

WTF IS JEE

[copper quarry]

they are not jee level

[celest acorn]

oh

yeh

than vectors harder > integration

[woeful rivet]

waterbeam whats a jee

[celest acorn]

exam

[copper quarry]

india exam

[woeful rivet]

erm

[copper quarry]

u dont need to knwo it but

integration u can learn ez vectors u should prob focus more on that

[woeful rivet]

i like integration

at least in year 11 methods

omg

WTF IS THIS

[forest sky]

jee is a university entrance examination given in india, which appears to give people studying for it the impression that it's the only thing worth studying for ever

[copper quarry]

those are all calm lol

u know how to do them

[woeful rivet]

no i dont

[forest sky]

derivatives are really all very formulaic, it's just a matter of learning the appropriate formulas

[woeful rivet]

i know how to use power rule

[copper quarry]

im pretty sure ive seen u do them

[woeful rivet]

...