help-28

[ornate bobcat]

Yeah this is called
Determinate of the quadratic formula
Because it determines the type of solutions the quadratic equation will have

[edgy shore]

I need help with a level functins transformation

[full forum]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[cedar mulch]

This channel is occupied

[cedar mulch]

Ahhh okay

So what's going to happen ??

[ornate bobcat]

When it is negative it means that it has no real solutions
When it is zero it means it has only one real solution that is repeated twice
When it is a positive number it means it has 2 different real solutions

When it has no real solutions it means it doesn't intersect the x-axis
Which means it either always positive or always negative

No what do you think
How do you know if it is always positive or always negative?

The numerator of our question

[cedar mulch]

IDK sorry huhu

[ornate bobcat]

Doesn't get into your mind to just substitute any value in x and see if it will output positive or negative @cedar mulch

[cedar mulch]

For example, x=2. Then it's -88

Now what

[ornate bobcat]

So it is always negative
Got it ?

The easiest to substitute is zero so use it instead

But 2 works as well

Now we knew that the numerator is always negative

Check when the denominator is positive and it will be the solution to the question

@cedar mulch
Got it ?

[cedar mulch]

Aight thanks

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[pearl mural]

So, even more combinatorics homework that I simply dont grasp. So this question consists of an A and a B part. The first one I kind of understood and got the right answer, C(11, 5) being correct, but I cant for the life om me understand the second one. Here goes the question:
In a city there exists a street system that is totally rectangular. You're at position A and want to go to position B. There are many ways through he city that have the same length. One of these is already marked out in the picture below. This path has a lenght of 11.

a) How many different paths exist between A and B that has the length of 11?

b) A road work is impeding(?) all the traffic at the marked X. How many fewer paths now exists between A and B?

I simply have no idea how to even begin with the b) question.

[rare dock]

for (a), my first observation is that the 11 length paths are exactly the ones that use only "right" and "up" moves

if there are any other types of moves, you can't reach B in 11 moves

and any path that uses only right and up moves will get you to B

moreover, you need to use exactly 5 up moves and 6 right moves

soooooo

you can view paths to B as permutations of uuuuurrrrrr

[pearl mural]

Yeah so I figured that there are 6 "side-moves" and 5 "up-moves"

So the answer would be C(11, 6)=C(11, 5)=462

Which was right

My main issue is with this b) question

[rare dock]

oh ic

have you tried counting the paths that cross through X?

(to subtract from the total)

[pearl mural]

I began but the I realised that there was quite the amount

The answer is supposed to be 105 fewer than without the X

[elder field]

yo

[pearl mural]

Which is quite a lot

[rare dock]

began how?

it looks very possible to count them to me

[pearl mural]

I mean I just started counting

[elder field]

guys i have a question,
does this server support lgbtq or no?

[rare dock]

yes

[elder field]

uhh

but the logo

[elder field]

isnt the lgbtq flag

[pearl mural]

Homie what

[rare dock]

yes it is an lgbtq logo

why are you asking about this here tho lmao

[elder field]

oh so the rainbow thing is for supporting lgbtq

ty

[steady nexus]

[elder field]

i joined 2 mins ago 😂

i dont know the channels

[pearl mural]

Its fine 🙂

[rare dock]

ok can we go back to this now lol

[pearl mural]

Yeah haha

[rare dock]

sooo

[elder field]

yea ty
bte

bye

[pearl mural]

I just dont understand how we get to 105 using combinatorics

[rare dock]

passing through the X road requires you to start with one of these 3:
urr
rur
rru

then r to go through the X road

[pearl mural]

Im with you so far

[rare dock]

then count the paths after that point to B

[pearl mural]

Wait isnt it just the same as the first question now

When we have passed the x

[rare dock]

yea same idea now

[pearl mural]

There is 4 up and 3 right

[rare dock]

yep exactly

[pearl mural]

So 3*C(7, 4)

Gives 105

Thats fantastic

[rare dock]

yay ^_^

[pearl mural]

Wohoo

Thank you so much

[rare dock]

no problem!

[pearl mural]

.close

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[raven flume]

I have to simplify this expression. Any idea which law I should use to start? Can I use the distributive law?

[wild sleet]

the channel is taken do it again

[full forum]

@raven flume Has your question been resolved?

[raven flume]

The question says it has to be equivalent with one of these three

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@raven flume Has your question been resolved?

[timid canyon]

any one can help me?

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@raven flume Has your question been resolved?

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@raven flume Has your question been resolved?

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[grand herald]

If the perpendicular of the curve y = f(x) at the point (1,1) is : x +4y = 5

[grand herald]

Then f(1) = ?

[upper estuary]

! status

[full forum]

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

[grand herald]

1

If only I had the slope then that question wouldn't be too hard

[upper estuary]

You can solve for the slope.

[grand herald]

How

[upper estuary]

Just put the equation into the form y=mx+b

[grand herald]

What is m and b

[upper estuary]

Those are both constants. m is the slope and b is the point where the curve intersects with the y-axis.

[grand herald]

So should I move y and 5

[upper estuary]

I would subtract x from both sides and divide by 4.

[grand herald]

Thats y

Wait my bad

So y = -(4/5)x + 5/5

[grand herald]

What's beside x is the slope ?

Perpendicular slope

[upper estuary]

-(1/4)x+(5/4)

[grand herald]

Oh nvm

Keyboard mistake

So slope of perpendicular is -(1/4)

[upper estuary]

The slope of the perpendicular is 4

[grand herald]

You mean the tangent

[upper estuary]

The tangent to the line you are tying to find, yes.

[grand herald]

Ook

So now what

[upper estuary]

We have the slope, all we need now for the equation for the line is the y-intercept.

[grand herald]

That's 5/4

[upper estuary]

That was the y-intercept of the other line.

[grand herald]

-4/5?

[upper estuary]

To find the y-intercept of this line, we can notice that the perpendicular line passes through (1,1), as stated in the question.

So, if we want to find the y-intercept, we can think about where a line with slope 4 and passing through (1,1) will intercept the y-axis.

Basically, we need to set x to 0, and find out how much lower the y value is than what it was at (1,1).

[grand herald]

We set x to zero in the line equation we just made right ?

[upper estuary]

Um. Think about it like this: by knowing the slope, we know how much y has changed, given some change in x.

Because we know the value of the function at x=1 (y=1), we know the y value at some number of units in the x direction. In this case, we want to find the y value at x=0 (the y-intercept), which is one unit to the left of the point we know.

Because we know the slope, we know that a change of -1 units in the x means a change of -4 units in the y, because the slope is 4.

[grand herald]

I need to prove this mathimaticly

[upper estuary]

Prove what?

[grand herald]

Prove how I got the y intercept

[upper estuary]

How rigorous?

[grand herald]

Eh I'll figure this one out

Anyways

When we have y intercept and slope

We still don't have the curve equation

[upper estuary]

The y intercept is equal to 1-4=-3.

m=4, b=-3, so y=4x-3

[grand herald]

That's the curve?

[upper estuary]

Yeah.

[grand herald]

Well I'm sorry but there's one more question that confuses me

[upper estuary]

Ok. Ask away.

[grand herald]

If the line : y+x-1 = 0 is the tangent of the curve f(x) = x² - 3x + a , then find a

[grand herald]

.

[upper estuary]

Does it specify which point it is the tangent of?

[grand herald]

Nope

That's what's really confusing about it

It is also given the answer which is 2

So I just have to prove that x = 2

I mean a

[upper estuary]

Do you know about derivatives?

[grand herald]

The whole lesson is about applying derivatives to tangent equations 💀

But it always asked about the tangent equation those are the only 2 questions that don't ask for them

That's why I'm confused

[upper estuary]

Ok. Then all we have to do is find out which point the derivative of the first function is equal to the second. This is where they intersect (because the second is the tangent of the first)

[grand herald]

Oooh

[upper estuary]

Then we can set the two equations equal to each other and solve for a.

[grand herald]

Well what if We get the slope of the line using the same way we did the first time , then we get the slope of the curve using derivatives

And we solve for a

Wait

That is exactly what you just said

[grand herald]

Am I right?

[upper estuary]

Yes. Take the derivative of both functions to find the point where they're equal and solve for x.

[grand herald]

Then solve for a

Nice

[upper estuary]

Yup

[grand herald]

Thanks now I can solve the rest of hw cause it's all repeated

[upper estuary]

Ok. Cool.

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[oblique skiff]

is this right?

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[stiff swan]

does this apply if the inside is a sum

[stiff swan]

(this is matrix btw, i am noob at linear algebra)

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[stiff swan]

wow

.close

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[buoyant heath]

why is integrating this easier than

and can you integrate the integral on the above integral using normal means? integration by parts doesnt seem to do anything

[formal scarab]

you can use normal substitution here u=x^3

,ask integrate xexp(-x^3)

[glossy valve]

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[formal scarab]

way different

[buoyant heath]

3x^2dx = du,
becomes integral of (-1/3)e^-2u du ?

[formal scarab]

yes

[rough tundra]

with just x instead of x^2 we get an exponential integral situation with u=x^3 (which is very garbage)

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[molten flare]

hey, how would i do lim[x->0] (cos^2(2x)-1+sin^2(2x))/x

[placid oar]

do you know trig identities?

[queen salmon]

think about it this way:
we know the identity sin²(θ)+cos²(θ) = 1, therefore the numerator becomes exactly 1 - 1 or 0. So the overall expression becomes 0 multiplied by 1/x = 0.

We can apply this shortcut because x only tends towards 0 but the numerator remains exactly zero throughout its whole domain so the overall effect is the same as multiplying the expression by 0.

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[molten flare]

oh i understand

thank you

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[fallow sage]

why is this partially right

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@fallow sage Has your question been resolved?

[fallow sage]

<@&286206848099549185>

yo someone help me pls

[sullen mantle]

What quadratic did you end up with, and how did you find it?

[fallow sage]

x^2+x-12=0

and then factored that to get -3,4

[sullen mantle]

I'm getting a different quadratic. How did you find that one?

Expanding the determinant, we should wind up with det(...) = x(x+1) - 4*1 = x^2 + x -4

[queen salmon]

yes but we have to equate that to 16

x(x+1) - 4*1 = 16

[fallow sage]

uh i did x(x+1)+(4)(1)=16 x^+x+4=16 minus 16 on both sides and i got x^2+x-12=0

[queen salmon]

because the value of the determinant is given to be 16

[queen salmon]

correct

I think your answer is complete

I can't find more than two solutions either

must be an error then

[fallow sage]

was i supposed to subtract the (1)(4) or add

[queen salmon]

oh wait

yes you were supposed to subtract

[fallow sage]

-20

omg

[sullen mantle]

Yeah

[queen salmon]

yes

my bad

[fallow sage]

damn

[queen salmon]

made the same mistake

[fallow sage]

i made a lil boo boo

[queen salmon]

answer should be 4, -5

[fallow sage]

yea i see what i did now

thanks

can i get help on this

im reviewing sum notes rn

was i supposed to put them in brackets

like

domain = [-oo,7] range = [-6,oo]

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[fallow sage]

<@&286206848099549185>

[formal scarab]

k(x) doesn't end in the corner there

there's no circle that indicates that like for (-3,7)

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[fallow sage]

.close

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[sharp nebula]

Is this supposed to be
$y_{2}$ ?

[glossy valve]

Kai Funaba

[formal scarab]

yes

[sharp nebula]

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[full mist]

Hi, I'd like help with solving the following equation:

[full mist]

It's my first time seeing an equation with both an exponent variable and a coefficient variable.

[fast peak]

well you can guess one solution. which you are likely supposed to do

actually solving it requires the lambert W function which you don't know

[full mist]

Could you elaborate on this?

Also, from looking at the answer key, it definitely wants two solutions

[open quail]

you could also use a logarithm I believe

[full mist]

I tried that, but I wasn't sure how to get the x from 1 over 2 out of the log

This is what I had

[mighty ridge]

there's no way to solve it cleanly, it necessarily involves still having log in it in the end

the solutions are related to the product log function exactly because you can't just take the log on both sides

like, obviously, for x = 2

you get 2^2 - 3 = 1 which is true

but 2 isn't the only solution

but it is the only one that cleans itself

there is another solution around -5.96

for which you won't have exact form

[full mist]

That's one of the solutions for it on my answer key

How did you get to that?

[fast peak]

cheating with WA

[full mist]

WA?

[fast peak]

wolfram alpha

[full mist]

ah

[fast peak]

you can find it numerically

with eg newton

if you know what that is

[full mist]

Unfortunately no, I don't.

I'd be happy to learn about wolfram, but I was hoping to find out the expected method for my class

For reference, I've just started my first year of IB Maths HL

[fast peak]

show the origin problem statement and your solution sheet

[full mist]

This is the answer key

scroll to question 8c

Sorry by the way if it doesn't just open in chrome. I hate having others download things

[fast peak]

ask your teacher. properly solving this is a bit outside the scope of the rest of this

[full mist]

Will do then, thanks for the help.

[fast peak]

I genuinrly don't know what the hell they expect

[full mist]

By the way, how do I close this chat?

So others can use it

[fast peak]

write .close

[full mist]

.close

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[torn jolt]

oops

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[frigid phoenix]

hi
i'm trying to understand the limit formula of the geometric series

[frigid phoenix]

a/(1-r)
the common ratio should be less than 1.

but this means as r approaches 1, the result get closer to infinite, right?

[wide sundial]

"closer" to infinite is very wack

infinity is infinity far away from any finite number

a finite number will always be infinitely far away from infinity so if it converges it will be nowhere near infinite

[fathom saddle]

If your ratio approaches 1 (from the left), then the limit formula approaches inf, yes.

[frigid phoenix]

so, if it approaches infinite, it means the series diverge?

or maybe we are not playing that game here?

we just pick a number r, we are not taking the limit. right?

but still ... 🤨

[fathom saddle]

Even if r = 0.99, that formula will still output a real number. "Outputting a real number" is what a converging series does.

[wide sundial]

well...

convergence is a bit more than that

[frigid phoenix]

but in my mind, for this to hold, it should hold even if we take the limit

[wide sundial]

we do take limits

[frigid phoenix]

of r ?

[wide sundial]

$\lim_{n\to \infty}\sum_{i=1}^n a_i$

[glossy valve]

Frosst

[frigid phoenix]

but that is of n

i mean of the formula a/(1-r)

[fathom saddle]

Note this formula is just the value the series converges to and is not the series itself. There are no limits placed on the formula.

[wide sundial]

yes, the $r$ is a property of the sequence ${a_i}_{i=1}^n$

[glossy valve]

Frosst

[wide sundial]

the formula works only for things that converges

so if r = 1 the formula isn't correct

[frigid phoenix]

but the formula should tell me what it converges to right?

[wide sundial]

well in a sense it's correct that it diverges to infinity but i think you could just add infinity to the range

[next sail]

yes, but we don't usually consider different r's; we consider a fixed r, and then let n go to infinity

geometric series

1 + 0.5 + 0.25 + 0.125

r = 0.5

it's 0.5^0 + 0.5^1 + 0.5^2 + 0.5^3 + ...

the same r for all of them

[frigid phoenix]

yea

[next sail]

ok

[frigid phoenix]

ok i think i've got this

and when we talk about POWER series, the differente is we have a variable in there ?

or the harmonic series is a power series? and the variable doesnt matter?

because when we need to check the radius of convergence, we need to find the x basically

that makes the convergence <1

is the radius of convergence something that you do when you have a variable ?

ok thanks anyway!

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[frigid phoenix]

hey mee again..
i'm following the MIT course and at this point there is this video

[frigid phoenix]

how am i supposed to know which function is represented by those?

[onyx glen]

recognize common taylor series

namely exp and 1/(1-x)

these are modifications of those two

[frigid phoenix]

she does it by heart but i dont understand where she plug the from

oh ok, so it's not in order in their videos, since taylor comes next.

so makes sense i am a bit lost here

thanks for the hint

[onyx glen]

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

[glossy valve]

Ann

[onyx glen]

did you see this yet

[frigid phoenix]

no

[onyx glen]

hm

[frigid phoenix]

https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-5-exploring-the-infinite/part-b-taylor-series/session-97-power-series/

it's here in case you are curious

net chapter is on taylor

🙂 thank you

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[quick cairn]

what's the difference between these 2 statements ??

[fast peak]

well in the first the eta can depend on x and epsilon

in the second it can only depend on epsilon and has to hold for all x

[quick cairn]

kk ty

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[long quarry]

Hello. I have no clue why my answer is wrong here, I've worked on the problem for a while now and maybe I'm overlooking something really small here but I just have no clue

[full forum]

@long quarry Has your question been resolved?

[deft hinge]

Given the model your trying to solve, I don't believe it makes sense to have a negative result for any food group.

you can't include negative 10 carrots in a soup

so all the values in your result must be 0+

[long quarry]

Ah, gotcha, that makes sense

It's the number I get though, maybe if I put zero it will accept it?

Changing it to zero was not the answer

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[faint salmon]

(8,-2), (r-6),m=-4

I solved it but it’s not correct

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[dull lagoon]

The graph presented corresponds to a curve y = f(x). Select the alternative that contains the graph of y = f'(x).

[dull lagoon]

How can I find the correct graph for this y=f(x)?

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@dull lagoon Has your question been resolved?

[lethal mirage]

use your intervals of increase, decrease and your inflection points

and your relative extrema on the parent graph

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[barren badge]

What’s the derivative of 6xy and why ?

[sand stone]

with respect to what

[barren badge]

Ik product rule (uv’)+(vu’)

[barren badge]

Y

i think

[sand stone]

what does x represent

[barren badge]

u?

[sand stone]

a constant? a function?

[barren badge]

function

[sand stone]

in terms of y?

[barren badge]

yea

[sand stone]

alright

so we want d/dy(6xy)

[barren badge]

yep

[sand stone]

first you can take out the constant

[barren badge]

we’ve been doing either y’ or dy/dx

[sand stone]

if you know that rule of derivatives

[barren badge]

i don’t think i do

[sand stone]

are you sure this isn't with respect to x?

[wide sundial]

y’ and d/dy is not the same

[barren badge]

idk man

[wide sundial]

Show us the original question

[sand stone]

dy/dx is the derivative of y with respect to x

[wide sundial]

!original

[full forum]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[wide sundial]

Show us the instructions

[barren badge]

my fault it wasn’t sending

[sand stone]

dy/dx

is the derivative of y with RESPECT TO X

[barren badge]

okay

[sand stone]

okay

so where are you confused

for taking derivative of 6xy

[barren badge]

why does derivative of 6xy= 6x + 6y

[sand stone]

it doesn't

why do you think that

[barren badge]

oh?

I have the answer sheet

[sand stone]

that's not correct

show the answer sheet

[barren badge]

i meant

6xy= 6y+6xy’

my fault

so like why does it

i get the y’

but not the 6x +6y

[sand stone]

how so

what do you get when you apply the product rule

[barren badge]

i’m confused on how to apply the product rule

in this situation

cuz there’s 3 things

[wide sundial]

Where does it even say 6x + 6y

[barren badge]

6y+6x is what i meant

[wide sundial]

That’s the same thing

Where does it say that

[barren badge]

^

we are ignoring the y’

[wide sundial]

Why are we ignoring the y’

[barren badge]

cuz i know that

[wide sundial]

You know what

[barren badge]

we can keep it in but i alrdy know why it’s there

that’s not what i’m confused on

[wide sundial]

So what are you confused about

[barren badge]

bro😭😭

Explain it to me like i am 5 please
Why does 6xy= 6y+6xy’

[wide sundial]

d/dx 6xy = 6y + 6xy’

It’s just the product rule with chain rule

It’s not really even chain rule

[barren badge]

Idk the chain rule in this circumstance

[wide sundial]

The chain rule is just the y’ part

You might not even call it chain rule at that point

It’s simply just product rule

[barren badge]

what are the two items for the product rule

what is u

what is v

[wide sundial]

well do we have 2 functions of x

Yes

6x is a function of x

y is a function of x

[barren badge]

So 6x and y?

[wide sundial]

d/dx (6x)(y)

Doesn’t have to be

[barren badge]

why not 6 and xy

[wide sundial]

That’s perfectly fine

[barren badge]

Oh

[sand stone]

because you need the product rule to differentiate xy

you CAN do that

it's just not very productive

[wide sundial]

Yes but it doesn’t stop you from splitting it like that

Like

[barren badge]

(6x)(1)+(6)(y)y’
6x+6yy’ that makes sense lol

i’m slow 💀

[wide sundial]

$\frac{d}{dx} f(x)g(x) = \frac{d}{dx} 1 \cdot [f(x)g(x)]$

[glossy valve]

Frosst

[wide sundial]

You can keep product ruling this

But you won’t get anywhere

[sand stone]

not quite

[barren badge]

is 6x and 6y swapped

so the order of the product rule matters yea?

[wide sundial]

How did you get (6x)(1)

[sand stone]

yes.. but saying that doesn't show any understanding

[wide sundial]

No, addition is commutative

[barren badge]

(6x)(derivative of y)

[sand stone]

not at all

[sand stone]

what's the derivative of y

with respect to x

[barren badge]

derivative of y is 1, no?

[sand stone]

with RESPECT TO X

remember it's d/dx

[wide sundial]

y is a function of x

[barren badge]

so how does that change anything

[wide sundial]

Think about y = 3x

[barren badge]

so that’s where i add y’?

[wide sundial]

What is dy/dx

[barren badge]

idk

[wide sundial]

What do you know you don’t know

[sand stone]

exactly, y is a function of x, i.e y(x), that could be anything we don't know what it is so the furthest we can go is just write it as dy/dx or y', these are both notations for the same thing

[wide sundial]

You don’t know how to differentiate 3x wrt x?

[barren badge]

i’m def overthinkin bro

idk what with respect to means dawg

[wide sundial]

That means what’s on the bottom of the derivative operator

d/dx means to differentiate wrt x

[sand stone]

using frosst's example y(x) = 3x if we differentiate with respect to x it means X is the variable that's changing, it is the one we are "differentiating" here the derivative with respect to x will be 3 certainly not 1

[wide sundial]

d/dy means to differentiate wrt y

d/dt² means to differentiate wrt t²

[sand stone]

of consider y(x) = x^2, here the derivative of y with respect to x will be 2x certainly not 1

[barren badge]

so y=3??

[wide sundial]

No, y’ = 3

[barren badge]

right

[sand stone]

the derivative of y with respect to x is 3

[barren badge]

yea

[sand stone]

however

in this question

[barren badge]

ok i copy

[sand stone]

we do not know what y(x) is

[wide sundial]

You claimed here that y’ = 1 for all y

[sand stone]

so when differentiate it

[wide sundial]

We are trying to show you this is not true

[sand stone]

we just write it as y' or dy/dx

[barren badge]

gotchu

[barren badge]

right

[sand stone]

the latter is clearer, loses less information but you can generally deduce it is with respect to some variable with the context of the question

[wide sundial]

This is why we call it implicit differentiation

Because we haven’t quite said what exactly dy/dx is

Just that it’s related to the other stuff in some way

[barren badge]

Okay

i partially understand now 💀💀

[wide sundial]

What don’t you understand?

[barren badge]

naw i do it just is abt remembering

y’all explained it fine

appreciate it

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[nocturne trench]

So I suck at problem solving in math and precalc in general
I uhh very much struggle in uhh limit proofs epilson and delta
would it be wise to go back and do more basic limits problems?
Since it takes me like 2-4 hours to fail at a problem
after 2-4 hours on a pricise limit problem I succeed like only barely sometimes mostly failing?
I should go back and do basic problems of limits right?

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@nocturne trench Has your question been resolved?

[nocturne trench]

<@&286206848099549185>

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@nocturne trench Has your question been resolved?

[nocturne trench]

.cloe

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[rugged trench]

how do i do this?

[wanton thistle]

what?

[rugged trench]

How do i go about solving this questions^^

[wanton thistle]

sorry maybe there's a problem with my laptop cause i can't see anything

[rugged trench]

oh ok

.Find the positive value of a,correct to 4 decimal places,such that cos⁡(ax)+x^2 has exactly 5 stationary points.

thats the question

[wanton thistle]

ohhh i see now, sorry i had to restart discord

ah sorry i haven't done those problems before so i won't be of any help

[rugged trench]

ok no worries

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[lofty wigeon]

What does he wants in this question? I don't understand

[formal scarab]

ordered samples of size 3 means you pick 3 balls in order

as opposed to taking them all out at once without arrangement

[lofty wigeon]

The wording got me. Thank you for the help

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[cunning fractal]

Need help solving to find given answer

[ornate bobcat]

Do you know the simple interest rate formula

[cunning fractal]

Yea

So since the payments are the same can I assume he’s paying 50 grand each time

Hello?

[rough plaza]

hello~

i wish to help, but i dont know what is focal date

[cunning fractal]

It means end date

[rough plaza]

oh i see

[cunning fractal]

Yea I’m stumped

[rough plaza]

sigh .. this feels so weird

simple interest

usually when i was calculating simple interests

i just do 100,000*(1+24%) for one year

so it's 100,000*(1+12%) for 6 months

but these choices for the answers ain't that simple

[cunning fractal]

Yeaa

It’s ok I’ll figure it out

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[sullen fiber]

this picture is the quesiotn

not sure where I went wrong

[onyx glen]

0.4% per annum compounded quarterly means you apply 0.4%/4, or 0.1%, every quarter

not 0.4% every quarter

[sullen fiber]

OHH

i thought It said 0.4 per quaterr

you have opened my eyes

I appreciate you immensly :0

I completetly skipped that one thing my lord smh

thank you so much

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[normal thicket]

x=1/(x-3)
x+1/x=?

[onyx glen]

is lowercase x different from uppercase X?

[normal thicket]

Nope both are same

[onyx glen]

ok then this looks like a simple quadratic equation

[normal thicket]

Yes it is but i am getting complex roots

[onyx glen]

show work

[normal thicket]

,w x^2-3x-1=0

[glossy valve]

Results provided by WolframAlpha

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

Requested by arjunn5589

[onyx glen]

not so complex now are they

i think these might give different values for x + 1/x tho.

[normal thicket]

I got value of x

But need to find x+1/x

Looks complicated

[hot herald]

did you try doing it?

or made the assumption that it might be complicated and stopped

[normal thicket]

After solving more i found this

@onyx glen

@hot herald

No idea so? Are you stuck too?

No problem i am closing

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[normal thicket]

.reopen

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✅

[normal thicket]

I solved it by not solving complete roots

[rotund birch]

Multiply denominator by the conjugate

[normal thicket]

,rotate

[glossy valve]

Couldn't find an attached image in the last 10 messages.

Sep 30, 2023, 5:35 AM

[rotund birch]

(Numerator as well)

[normal thicket]

,rotate

[normal thicket]

@onyx glen @hot herald see

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[normal thicket]

4 men work equals to 6 women, 4 women work equal to 6 boys. If a boy completes a work in 60 days. 1 man and 1 woman together will complete the work in how many days?

[normal thicket]

,rotate

[normal thicket]

Ohh i solved it by ratio

Total work=4×60=240

240/(9+6)

=16

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[primal pewter]

Guys, why is this true? By the power rule exponents this wouldnt be true or am I wrong?.
$\left(\frac{a}{b}\right)^{\frac{1}{n}} = \frac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}}$

[glossy valve]

Gabe

[primal pewter]

Ohh shit

I realised it

[shrewd hamlet]

The equation is. True

[primal pewter]

It is same as saying but with 1 over n instead

$\left(\frac{a}{b}\right)^{2} = \frac{a^{2}}{b^{2}}$

[glossy valve]

Gabe

[shrewd hamlet]

That equation is also true

[primal pewter]

Yeah I was trying to understand it in general. Although this still doesnt fully prove the previous equation

[shrewd hamlet]

Wdym by prove

[primal pewter]

Like to make it obvious not to just accept it

Perhaps I'm overcomplicating it

[shrewd hamlet]

$\left(\frac ab \right)^c = \frac {a^c}{b^c}$

[glossy valve]

Stephen

[primal pewter]

I mean yeah that's a good enough proof, I assume c can be a fraction as well then.

[shrewd hamlet]

Yea

it can be anything

Anything real at the least

[primal pewter]

Yep, thanks for helping

[shrewd hamlet]

Np

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[molten tundra]

Hi

[formal scarab]

ok

what about it

[molten tundra]

waittt

ok

theres that thing

are these two

the same?

[white smelt]

yes

[molten tundra]

how tho

[rough tundra]

if you divided both top and bottom by sqrt(x^2+y^2) then yes

[molten tundra]

i dont get it

[white smelt]

multiply top and bottom by (x^2 + y^2)^{1/2}

[molten tundra]

but its to the power of minus half

[rough tundra]

You can rewrite this as: $\frac{\frac{x}{\sqrt{x^2+y^2}}-y}{x-\frac{y}{\sqrt{x^2+y^2}}}$

[glossy valve]

FancyBredFries

[molten tundra]

there should be a y

ok yeah

there

[rough tundra]

do you see now where it came from? :)

[molten tundra]

wait wiait

should it be the opposote

like denominator in the numeriator

[white smelt]

what

[rough tundra]

[white smelt]

no

[molten tundra]

ok yeah

i get it

[rough tundra]

I think you're getting confused about the negative exponents

[molten tundra]

i get this one

then what do we do after

[rough tundra]

$\frac{y}{\sqrt{x^2+y^2}}=y\left(x^2+y^2\right)^{-\frac{1}{2}}$ :)

[glossy valve]

FancyBredFries

[molten tundra]

what...

oh

yeah

[molten tundra]

i still dont get how this

[molten tundra]

and this are equal

[rough tundra]

gimmie a sec to type of this LaTeX

$\frac{x-y\sqrt{x^2+y^2}}{x\sqrt{x^2+y^2}-y}=\frac{x-y\sqrt{x^2+y^2}}{x\sqrt{x^2+y^2}-y}\cdot\frac{\frac{1}{\sqrt{x^2+y^2}}}{\frac{1}{\sqrt{x^2+y^2}}}\=\frac{\frac{x}{\sqrt{x^2+y^2}}-\frac{y\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}}{\frac{x\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}}=\frac{\frac{x}{\sqrt{x^2+y^2}}-y}{x-\frac{y}{\sqrt{x^2+y^2}}}$

[glossy valve]

FancyBredFries

[molten tundra]

wow

mathematics is a hell of a drug

thanks for the help

[rough tundra]

[rough tundra]

also RIP Stan Lee

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[timid canyon]

.

[torn jolt]

Hi, is this true or false?

[hot herald]

are you cutting out info?

[torn jolt]

no

[hot herald]

well then the x = tan(a)
doesn't seem that relevant

the part after the arrow is a pythagorean trig identity

x being tan(a) doesn't really imply that

[simple ridge]

But it is a true statement implying a true statement so it is true ?

[hot herald]

its not really an implication though

[torn jolt]

I am not sure, the indications say just to put if it's T or F
It's and old exercise, and I remember the answer was F

What I also thought is what if tan(a) is like 1 or -1

-1 or 1 = x

would that be relevant

[hot herald]

i am on discord → its the year 2023

[torn jolt]

m so it's not relevant

[simple ridge]

But yes they aren't related

The second isn't a consequence of the first

[torn jolt]

I agree
I was just saying that I thought before if x = 1 or -1, then alpha could be 45° and it would be "right"
but I understand that they are not related

in this case

?

I put it as True

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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[lime cedar]

why -3 / 0.75 = -4 ? i dont get it

[lime cedar]

dude if i knew how to simplify i woudnt be here

[sharp flame]

multiply the top and bottom by 100

It should becomes easier to simplify then

[lime cedar]

ty

got it

.close

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[cosmic maple]

How do i use MPE to solve algebra. Example question: -d = 10

[cosmic maple]

<@&286206848099549185>

anymoe

anyone

[torn jolt]

referyt

burh

[fast sigil]

gonna start learning calculus, what should i learn before it or can i judt start straight up

[reef dust]

#❓how-to-get-help channel occupied

[cosmic maple]

can anyone help?

[reef dust]

!15m

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[cosmic maple]

sorry

[azure grove]

please read guidelines before requesting help

[runic bloom]

if you have -1, what can you multiply it by to make it 1

[cosmic maple]

-1 x -1

[runic bloom]

right

so you can apply that here

to turn -d to d

[cosmic maple]

so -1/1 x (-d) = 1/1x (10)?

[cosmic maple]

am i correct?

[cosmic maple]

im sorry

[azure grove]

All good 🙂
I more meant that to the other guy...

[cosmic maple]

oh okay

[runic bloom]

uhh i think you are overthinking this

you just need to multiply by -1 on both sides

[cosmic maple]

so -1 * -d and -1 * 10?

[runic bloom]

ya

[cosmic maple]

ok

so -1 x 10

is -10

how do u get -1 x -d

[runic bloom]

-d just means -1 * d

[cosmic maple]

oh ok

so d = -10?

[runic bloom]

yep

[cosmic maple]

oh nice

tysm

[runic bloom]

np

[cosmic maple]

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[cosmic maple]

how do u use MPE to get -121= -11c

[cosmic maple]

so

first.

its -1/121(-121) = -1/121(-11c)

yes?

<@&286206848099549185>

the -1/121 has to be in both sides

oh why u deleted

[sullen mantle]

My bad, I made a mistake.

[cosmic maple]

oh ok

am i correct?

[sullen mantle]

That is a valid way to use MPE, however it doesn't get us closer to finding the value of c

We want to isolate c on the right hand side (RHS)

[cosmic maple]

oh

so it should be

-1/11 (-121) = -1/11(-11c)

yes?

[sullen mantle]

Exactly!

[cosmic maple]

so that means -1/11 (-121) = c?

[sullen mantle]

Exactly

[cosmic maple]

then whats next

[sullen mantle]

Next we just need to simplify the fraction. We can easily divide a -1 from the numerator (top) and denominator (bottom), so we have 121/11

Since 121 = 11 * 11, 121/11 = (11 * 11)/11

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@cosmic maple Has your question been resolved?

[torn jolt]

@cosmic maple just memorise the technique. If you are having a hard time understanding it now, it'll come to you as you continue to do algebra

There is no use in keeping the ticket open, especially when you have the solution

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[atomic vortex]

Idk what to do next

[onyx glen]

erase all this

you did way more than asked for

you were not asked to solve the system

only to verify that [2; -1; 1; 0] is a solution of it

[atomic vortex]

how would I do that then

[onyx glen]

well

if i gave you an equation and a number and told you to check if that number is a solution of the equation

what would you do?

[atomic vortex]

plug it in ?

[onyx glen]

yes exactly

do the same thing here

just plug x = [2; -1; 1; 0] in

that's it, no complicated system-solving required

[atomic vortex]

oh alr lemme try that brb

[atomic vortex]

Is this correct so far?

oh wait

[onyx glen]

no you are STILL doing more than asked for

you don't need any fucking row reduction crap

[atomic vortex]

ye idek why I put that there I never rlly used it

[onyx glen]

you just need to calculate $A \cdot \bmqty{2\-1\1\0}$

[glossy valve]

Ann

[onyx glen]

that's it. that's it!!

[atomic vortex]

ah

got it

ty

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[fallen jungle]

Hello, need help on existence and uniqueness if u available should only take a min pls

[formal scarab]

[fallen jungle]

This is the IVP

I assume i just need nominator = 0 and demoninator to be not be zero, so y = 1 or y = -1 whch means any T is a solution except t = 0 or t = - 2

?

something like this?

but answer given was this

Which doesn't sound right, because it would mean 0/0 which is undefined and not a solution?

any thoughts would help, thx

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@fallen jungle Has your question been resolved?

[rough plaza]

I can only recall a little of such problems.
I am guess it's about "has more than one solution", that is, when it's the case 0/0, it will be indeterminate such that it can have more than one solution.

without further context, i cannot conclude anything about this IVP.

Hope that helps!

dy/dt may exist for such values that makes the fraction 0/0

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[hoary lava]

so i understand that when lim(x), and x goes towards 3 for example, then lim(x) ≈ 3 (as far as i understand it). but what do i do when x goes towards negative (or positive for that sake) infinity?

[elfin stream]

,rotate

[grand scaffold]

this would solve your problem

[hoary lava]

gotcha, but i don’t get how this turns out to be 1. if x ≈ 0, this should be 5000/1, which is 5000. then it would be 5001, no?

[grand scaffold]

any rational function thats a constant over infinity is basically 0

[grand scaffold]

its the whole fraction that becomes 0 not just x

[hoary lava]

misunderstood that then, think i get it now. thank you

[grand scaffold]

np :)

[hoary lava]

in this case, would this also apply?

[grand scaffold]

nope

cuase the numerator isnt a constant

[hoary lava]

right

[grand scaffold]

for that case you would need to use l'hopitals rule

since its an indeterminate form of infinity over infinity

[hoary lava]

yeah nah i don’t know that unfortunately

unless that is a different name for something i know

[grand scaffold]

what'd they teach for indeterminate forms?

[hoary lava]

no idea honestly, we use norwegian terms. this is however how the book expects me to solve it

[grand scaffold]

ohhh yeah you can also use that method

[hoary lava]

but wait, wouldn’t 1/x = 0 here though?

wouldn’t we just multiply everything with 0 then

[grand scaffold]

your multiplying by 1/x to make the numerator a constant in example c

[grand scaffold]

yes but the equation in the parentheses is still infinity, giving (infinity)0/ (infinity)0 which is still indeterminate

thats why you distribute the 1/x

do you know how to do derivatives?

[hoary lava]

uh, that was a while ago now

but somewhat yeah

definitely a bit rusty

[hoary lava]

i still don’t get where they pulled the 1/x from though

[grand scaffold]

since they're trying to make the numerator a constant they need to divide it by x to remove the x

[grand scaffold]

l'hopitals rule is just getting the derivative of the numerator and denominator separately

which would make it easier for less fractions if thats what your struggling with

[hoary lava]

oh, so if i’m correct then, (2x)*(1/x) = 2x/x = 2?

[white smelt]

1/x isnt 0, it just approaches 0, so no you're not multiplying by 0, you're multiplying be 1

i know it sounds weird at first but that's how it is

[grand scaffold]

yes

[hoary lava]

if i’m not mistaken that would be lim(2)/x, which is 0?

ahh

[grand scaffold]

2/2x*

but yes

[hoary lava]

certainly does, takes some time to grasp it

[hoary lava]

oh yeah, my bad

[grand scaffold]

all good

[hoary lava]

appreciate it

how would i write it to show that i have derived it? lim’(2/2x)?

[grand scaffold]

usually you would just write Dx(2x+3)/Dx(x^2+1) = 2/2x

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[sudden yew]

if A={1,2,3} then
is {1} a subset of P(A)
and is {{1}} a subset of P(A)

[wide sundial]

{1} is an element of the powerset of A

the elements of the powerset of A are themselves sets of elements of A

[sudden yew]

so if {1} belongs to P(A)
then {1} is a subset of A
but what about {{1}} is it a subset of P(A)?

@wide sundial if you dont mind tagging..

[wide sundial]

but {1} is not a subset of A