help-0

[stone garden]

I think i dont know what i think...

Do we need maybe the p-q formula now?

[pseudo ice]

Not really

If the one I’m thinking of is what you mean

There’s one obvious solution here

[stone garden]

Do i need to calculate them first?

Oh no wiaat

I think i know i must write only 0,01x^3 - 0,3^2 + 2,25x = 0 and i need to say that it starts from zero in a sentence in the text idk

And then i need to calculate it first? Or

[pseudo ice]

What do you mean by “it starts from zero”? I think you’re saying what I wanted you to say(!)

[stone garden]

I think that the calculate zeros start's from zero because we start there and when we have for example -5 it procceds to start before i think

Hmmmm

I don't know

[pseudo ice]

Alright for a moment forget the whole question

Let’s say I give you y^2 + 5y + 4 = 0, could you find what y could be for me?

[stone garden]

I think y is the time in minutes

Do i need to write it in minutes??

Hmmm

Is it then 2y^1 + 5y?

[pseudo ice]

No no noooooo

Ignore everything else, like the whole question and everything

Here y isn’t time or anything, it’s just a quadratic equation

,w solve y^2 + 5y + 4 = 0

[ocean seal]

Results provided by WolframAlpha

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Requested by chartbit#3171

[stone garden]

Hmm how do we find that we have -1 and -4?

[pseudo ice]

You can for example factor it as y^2 + 5y + 4 = (y+4)(y+1)

Or you can use the quadratic formula

[stone garden]

I think i know now because y is 1 bc there is y and so we need ^2 and make it ^1 and 1 is then left so it is plus i made the plus to minus and then the 4 i do the same and then its -4

But why especially the -4 can i say -5y too?

I don't understand rip

[pseudo ice]

Hmmm not too sure what you mean?

Also I have to go and get a bit of sleep

May be back later but if I don't see you then, hopefully I'll catch you around

[stone garden]

Oki

Good night 🌼

Thanks you still helped me a bit to understand it more😌🌼

See you around🌼

[pseudo ice]

[pseudo ice]

Hopefully I did

[lone heart]

@stone garden Has your question been resolved?

[lone heart]

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[tawny sonnet]

I’m not sure where to start graphing this ik that it’s an exponential function

[strong compass]

It’s a polynomial, exponentials are when x is in the power

Do you know what I mean when I say the roots of the equation?

[tawny sonnet]

Would it be the entire equations squat root

[strong compass]

Not quite, it’s when x^ -9= 0

You need to know what x values make that true

[tawny sonnet]

3^2=9

[strong compass]

Nice so 3 is a root

Do you know another root

Hint ||the sqrt(a^2) = +a and -a||

[tawny sonnet]

Hmmm I’m not sure is there another squat root that equal 9

[strong compass]

Are you aware there are 2 solutions to a square root?

[tawny sonnet]

Ohhh yeah because -2x-2 is 4

[strong compass]

There is only one type of square root, but let’s say for another example I have x^2 = 4

Of course one solutions is 2

But -2 is also a solution

Because -2 * -2 = 4 (because -1*-1= 1)

Exactly

[tawny sonnet]

Yeah

[strong compass]

So given that what’s the other solution

[tawny sonnet]

-3^2

[strong compass]

So x = 3 or x= -3

Do you know how to factorise it from here

[tawny sonnet]

Hmm I’m not sure

[strong compass]

Nvm then, so you know when x = 3, y= 0 and when x = -3, y=0

If you wanna plot those on your axis

We can begin to work out the points that this curve goes through

[tawny sonnet]

Ohh ok I think I’m starting to get it

[strong compass]

That’s good

So we need 3 points to work out how we should draw this, we have when y=0, any idea what we should try find next?

[tawny sonnet]

Would you take a random x point in either direction and plug it into the formula

?

[strong compass]

You could, but that might be a bit of a pain to work out

What happens if I put x= 0

[tawny sonnet]

Would it be y=9

[strong compass]

Yep

Oh wait not 9

Y= x^2 -9

[tawny sonnet]

-9?

[strong compass]

Yeh -9

So you know if goes through those points, now you jusy need to sketch a smooth ish line through those

Does that make sense to you?

[tawny sonnet]

Just to recap when x=-3 y=0 and when x=0 y=-9 and x=3 y=0

[strong compass]

Yeh

[tawny sonnet]

How did you know to flip the sighns on the equation ?

[strong compass]

What do you mean?

[tawny sonnet]

The original equation called for f(x)=-x^2+9

[strong compass]

My miss read sorry, it’s the same roots but you were correct earlier x= 0, y=9

[tawny sonnet]

Ok I think that clears things up

[strong compass]

It’s the same roots because 0 = -x^2+9
Adding x^2 to both sides: x^2=9

Does that make sense now

[tawny sonnet]

Yeah

[strong compass]

Ok good

[tawny sonnet]

Thank you

[strong compass]

For a short cut, it might be worth knowing, this type of quadratic is called a “difference of 2 squares” as you get comfier your might be able to recognise these and jump straight to the solution, it’s whenever you can write something as y=(x+a)(x-a) where a is a number, it means that there is no x term, just y=x^2-a^2

[tawny sonnet]

Yeah

.close

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[queen bear]

can someone help me figure out the rest of this problem?

[queen bear]

having trouble with D

[placid zinc]

a and c are right, but d is wrong?

[queen bear]

how so

i meant d second part and e

im not sure how to progress

@placid zinc

[placid zinc]

https://tutorial.math.lamar.edu/classes/de/linear.aspx

Basically this is a first order linear DE. There's a whole solution method to them you'll be tested on. Give that page a skim.

[queen bear]

will do, would you happen to know the answers to the ones i am missing so i can check the work?

[placid zinc]

If you just want answers, check out Wolfram Alpha, which can generate answers to basically anything for you

[queen bear]

for the same topic?

[placid zinc]

Easy in-browser math engine

Can do differential equations, yeah

[queen bear]

cool thanks for the help!

i didnt use the wolfram but i did end up getting 2.33 for final answer

the link you sent earlier helped actually, will have to check out more of their stuff

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[pulsar bane]

Hello, this may look like a dumb question but its 4 am and yeah... Ok so why's the first one wrong but the second one is correct (the 2 identities on the right)

[gusty gorge]

how are both of these correct?

[lone heart]

@pulsar bane Has your question been resolved?

[pulsar bane]

the first one isnt

but the second one is

[pulsar bane]

i've looked it up, my question is why is the first one wrong

and why's the second one correct

idk why the first one that i did is wrong

Oh god I'm stupid and sleep deprived, he moved the tan cuz you cant do the other stupid thing that i did

Nvm I'm sorry 🤣

.close

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[low zephyr]

can i get a step by step on what to do for these types of problems ?

[lone heart]

@low zephyr Has your question been resolved?

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@low zephyr Has your question been resolved?

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[alpine sable]

14048

[alpine sable]

How would I round that to 3 sig figs?

[tight locust]

depends on context

you can't just have sig figs in isolation

it depends on the accuracy of your measurement

[alpine sable]

101115.65401744

then I used this number to calculate return of investment

and got 14048

Teacher requires 3 sig figs... so I have no clue how to round this

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[alpine sable]

.close

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[ocean seal]

AwitNamanSayoPar

[stone plaza]

??

[last ether]

What about that

Solve for x?

[stone plaza]

its is x = -1,1

[last ether]

Take the square root of both sides

Yeah

[lone heart]

Channel closed

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[stone plaza]

what is $$x^2=-1$$

[ocean seal]

AwitNamanSayoPar

[stone plaza]

help

is the answer x = 1,-1 😭

[ocean seal]

Stephen

[long axle]

No

Try squaring those and see what u get

[stone plaza]

ahhh $$x= \pm{i}$$

[ocean seal]

AwitNamanSayoPar

[stone plaza]

??

wrong?

[long axle]

Yea correct lol I literally just said that

[stone plaza]

ah ok

thank you

[long axle]

Np

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[last ether]

i has cyclical properties that makes your solution incorrect

[last ether]

Shit

.close

[lone heart]

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[long axle]

Wait what @last ether

i is wrong?

[last ether]

No, -i is wrong @long axle

i is correct

They had x = +- i

[wary stream]

Did you need math help?

[last ether]

No

It was bad timing

[long axle]

But if u square -i don’t u get -1?

[last ether]

No

You get 1

I'm pretty sure

Am I being dumb

Yeah no (-i)^2 = -1

I'm not being dumb

Oh wait

I'm going to throw a fit

LOL

DID I JSUT THINK 1 = -1

[long axle]

$(-i)^2 = (-1)^2 \cdot (i)^2$

[last ether]

I'm done

[ocean seal]

Stephen

[last ether]

Naw naw naw I'm done

I just thought -1 = 1

[long axle]

Lol rip

[last ether]

Nah bro G I'm resigning

I can't just correctly do a Riemann summation and then do this shit

[long axle]

[last ether]

Screenshotting this

[long axle]

Lol

[fallen verge]

.coose

[modern sedge]

Does it count as bingo?

[fallen verge]

no

[lone heart]

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[high wolf]

Could someone help me solve this using partial fractions:

[modern sedge]

I gtg soon, so I wont help you with the partial fractions, but I would start by factoring out 3 and moving it out of integral

[high wolf]

so I'd be left with 3 integral of 1/x-2?

[modern sedge]

and also (2x+1)/(x^2+4)

[high wolf]

So I can then put it in the format of Ax/(x-2)+B/(x^2+4)?

[lone heart]

@high wolf Has your question been resolved?

[slate monolith]

thats unnecessary

you can already find the integral of 3/x-2

[high wolf]

But they ask me to use partial fractions..

[slate monolith]

for the second part yeah

pfd is for a single fraction usually

the first fraction though is really simple

seperate into two integrals

[lone heart]

@high wolf Has your question been resolved?

[high wolf]

What do you mean?

[slate monolith]

you have two fractions

make those two integral

[lone heart]

@high wolf Has your question been resolved?

[lone heart]

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[vale kayak]

To prove if a function is discontinuous at a certain point using epsilon-delta definition; what if i showed that |f(x) - f(c)| is less than some number, hence not true for all epsilon > 0, hence limit does not exist at c?

[mortal trellis]

do you mean bigger than some number?

[vale kayak]

i can only show its less,
take f(x) = x+1, and f(1)=0
then |f -f(1)| = |x+1| for some |x-1|<delta

[mortal trellis]

less doesnt show anything

either less than all epsilon or bigger than one epsilon

less than one epsilon doesnt give anything

[vale kayak]

then how do i go about proving the discontinuity?

[gusty gorge]

wait what

oh I see

you basically show that there exists an epsilon such that you can't find the delta

[vale kayak]

f(x):= 0 when 0<=x<=1
x+1 when 1<x<=2
i have to show its discontinuous at x=1

[gusty gorge]

in this case, epsilon = 0.5 works fine

[vale kayak]

so i choose epsilon instead of delta like in proving limit = infinity?

[gusty gorge]

huh?

epsilon-delta definition of continuity at c is that for all epsilon > 0, you can find a delta such that |f(x) - f(c)| < epsilon for all x within delta of c

[vale kayak]

yes

[gusty gorge]

suppose you say epsilon = 0.5

now show that |f(x) - f(c)| has to be greater than 0.5, no matter how small you make delta

[vale kayak]

got it, thx

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[glossy shuttle]

I don’t understand where it says “when two terms do not have any common factor except 1. Can someone please explain that to me?

[stray bloom]

those would be co-prime numbers

for example, 3 and 4 are co-prime, 2 and 6 are not (both even)

[lethal belfry]

That would mean that if you were to factorise them, the only common factor would be one

[glossy shuttle]

Okay. So how would I know exactly when to divide and multiply? I’m doing these problems and I’m answering them correctly but I think I’m just getting lucky because I don’t know when to multiply or divide.

[stray bloom]

in those cases you need to perform the "rule of 3" i think it's called

yeah, direct rule of 3

[glossy shuttle]

Okay thank you.

.close

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[stray bloom]

I'm looking for information on how to construct an ellipse given 5 points of it, graphically, I've seen a way or two but I don't know whether the way I found has a name, is documented, or is novel

[stray bloom]

In my case I start with 4 points and draw the regions where there can be a center of an ellipse that could pass through them, repeat that for all combinations of 4 out of 5 points and it should give 1 unique point or none if the shape is concave

for some interaction: https://www.desmos.com/calculator/qutb7ab0aa?lang=es

[lone heart]

@stray bloom Has your question been resolved?

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@stray bloom Has your question been resolved?

[prime crown]

tf

[stray bloom]

🤷‍♂️

[limpid spade]

jesus what

[stray bloom]

It's a bit obtuse

basically drawing parallel lines and middle points of segments

[lone heart]

@stray bloom Has your question been resolved?

[lone heart]

@stray bloom Has your question been resolved?

[heady cargo]

Hello>>

Hello??

[stray bloom]

yeah

[alpine sable]

Ax+b = 0 correct yes?

[stray bloom]

please use other channels

[alpine sable]

?

[stray bloom]

#❓how-to-get-help

[lone heart]

@stray bloom Has your question been resolved?

[slate monolith]

@stray bloomI'm guessing this is your own experimentation?

idk if you'll find someone here

[stray bloom]

correct

[slate monolith]

there might not be a solution

[stray bloom]

my guess is that those blue regions on the picture are the regions where you can put a center, starting with 2 points it's most of the plane, with 3 you can discard more, and then with 4 and 5, it should leave a single point or none

[slate monolith]

yeah I have no idea

[stray bloom]

maybe i should try to learn how to use the 3b1b software

[slate monolith]

thats just for visualization

not sure how that would help

[stray bloom]

at least for explaining the method

where could i find researchers that respond to messages?

[slate monolith]

the high high majority of people on here arent researchers

you could try #advanced-lounge or a specific advanced channel in #get-advanced-access

[stray bloom]

would this be algebraic geometry?

[slate monolith]

maybe? honestly I have no idea

what year are you even

[stray bloom]

i'm not studying, i've done baccaulaureate about 3 years ago

[karmic rapids]

???

this does not belong in the advanced channels

[slate monolith]

this person has already finished their bachelors

i think that is worthy of advanced

[karmic rapids]

what? just because someone has X degree doesn't mean every question they ask is on that level

[slate monolith]

I didnt say a research level

early uni is part of advanced

just not here

[karmic rapids]

no it's not

there's an early uni section

[slate monolith]

yeah but I thought that was unlocked when you got advanced access

[karmic rapids]

also this is far from being algebraic geometry

[slate monolith]

what is it then

[stray bloom]

I honestly don't even know the definition

[karmic rapids]

#geometry-and-trigonometry ?

alg geo is like

doing geometry with algebraic means

[slate monolith]

well yeah

[karmic rapids]

like comm alg

[slate monolith]

geo-trig is not enough

this is not something a highschooler can do

at least not a highschooler learning geo/trig

[karmic rapids]

geo trig is not only for high schoolers

this question belongs there

[slate monolith]

alr

[formal glade]

.close

u should do dat bc this is unused

[slate monolith]

we know how .close works lol

.close

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[serene mirage]

hey

[serene mirage]

\int (x)/(cos^(2)4x)dx

[last ether]

Use tabular method for IBP

[last ether]

Hm

Yeah I'd still use tabular/IBP

[serene mirage]

the x in the numerator threw me off

[last ether]

Well that's the same as integrating x • sec^2(4x)

Which is easy to do IBP/tabular on

[serene mirage]

oh wait

we got sec cause 1/cos

does the x act like a 1

[last ether]

Wdym does the x like a 1

No I just simply rewrote the integrand

x/cos^2(4x) = x • 1/cos^2(4x)

The x "doesn't act like a 1", that wouldn't make sense

[serene mirage]

ok

sorry it took me a while to process this i see now

so u = x and dv = sec^2 4x @last ether

[last ether]

Sure

[serene mirage]

would my answer be

[last ether]

Lemme check

[last ether]

You're fine

[serene mirage]

alright thanks

.close

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[naive python]

Hey, can someone help me with geometric calculations?

[slate monolith]

[naive python]

Can u help me find the area of 5?

[naive python]

Can u help?

[slate monolith]

nah

idk what is this lol

[naive python]

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[naive python]

Sorry, I am kinda in a time crunch

<@&286206848099549185> now it has been 15 mins

I found another side length btw

Please help y’all

<@&286206848099549185> I highkey need to keep spamming

Desperation rizz — pleaseeee

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[cerulean goblet]

Hey i’m relatively new to advanced maths and just wanted an easy way to estimate what a cubic equation would look like on a graph.

[slate monolith]

do you mean y = x^3

[cerulean goblet]

For example, a quadratic equation is easy to visualise

[slate monolith]

,w graph x^3

[ocean seal]

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[cerulean goblet]

I just mean any cubic equation and how to read the values, if you get what I mean

[slate monolith]

its the same thing as any function

plug in some x values

[limpid turret]

There are different kinds of cubics

[slate monolith]

trace it out

[cerulean goblet]

Oh, that makes sense

[limpid turret]

,w plot y=x^3-x

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[worn fox]

You should play around on desmos with sliders for the coefficients to see how they affect the shape of the graph

[naive valley]

you can make a few general observations

when x is large, in one direction the function must go to +infty and in the other direction it must go to -infty

[cerulean goblet]

will do!

[naive valley]

there can be at most 3 real roots

etc

[cerulean goblet]

alrighty! Thank you everyone for the help!!

.close

[lone heart]

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[junior warren]

how to go about solving this?

tried this didnt work

[surreal meadow]

you’re assuming the two ns are equal, they are not

[junior warren]

so two variables/

[surreal meadow]

i would list the values that are 3 mod 4 and 5 mod 7

and find the one that is in both

[junior warren]

would this process work?

actually i can really isolate for m or n

so probably not?

[surreal meadow]

you can’t solve for n and m with just one equation

i would do what i said above

[junior warren]

oh since the 3 and 5 is the remainder

which gets placed in front of the modulus?

[surreal meadow]

i would start with the mod 7 values and then go to the mod 4s, since there’s less mod 7 values

ok i seem to have misunderstood

actually no

i believe you are solving for a = 3 mod 4 and a = 5 mod 7

so i would list the numbers between 0 and 27 with remainder 5 when divided by 7

and then do the same with the values with remainder 3 when divided by 4, until you find a match

[junior warren]

oh ok, so theres no set algebraic way to solve it?

moreso just guess and check

[surreal meadow]

not that i’m aware of

but i’m not too well versed in this so there could be

[junior warren]

yeah i tried an algebraic solution but got 3mod4 congruent to 2

which it just isnt

so its 13?

[worn fox]

13 is not congruent to 3 mod 4 unfortunately

[junior warren]

hm

even though it has r5 when /7 and r3 when /4

isnt that what i am looking for?

i dont think it has to be congruent

it is asking for the number before the mod i think?

maybe i am wrong

[worn fox]

13 does not have remainder 3 when divided by 4

[junior warren]

it has r1

yeah

let me try agian

i believe 19?

[worn fox]

19 yes

[junior warren]

19 r5 /7 and r3/4

and lowest

ok i see the pattern

however in a next one i have same idea

but no number in common. ):

[worn fox]

Youre missing a number in each list

(Which necessarily will be the answer)

[junior warren]

possibly 1?

[worn fox]

Yes

[junior warren]

since 1=0/4 +1

[worn fox]

That's 1 more than a multiple of 4 and a multiple of 7 yes

[junior warren]

ok i didnt know i had to do that first one

i see

ok that makes sense

so quick example if i needed remainder 0

its all perfect multiples

AND 0

[worn fox]

And zero yes

[junior warren]

ok thank you

got all of them right

appreciate it

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[flat vale]

how to do 💀

[flat vale]

idk how to start

[lone heart]

@flat vale Has your question been resolved?

[worn fox]

Assume a is not zero wlog

What can you do to the equation ab = 0 to conclude that b = 0

[lone heart]

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[alpine sable]

Can someone define functions in simple terms

[lavish cedar]

sikeman this is my box?

could you go into a free one

please

thanks

Need help on probablity

just probobility

proabiltiy

[lone heart]

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[lavish cedar]

.close

[lone heart]

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@serene trellis Has your question been resolved?

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[molten pivot]

factor out 16 inside the radicand

[serene mirage]

i use trig sub for this

[molten pivot]

or just incorporate it into the sub yeah

[serene mirage]

the final page is where im at right now

am i doing anything wrong yet

[molten pivot]

looks good so far

[serene mirage]

ill continue and let u know again

@molten pivot

how about now

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[quiet mural]

i need help with probability

[vale wigeon]

image on its way?

[gusty gorge]

let her finish the fortnite game

[vale wigeon]

what fortnite game

[quiet mural]

hold on

sorry

[vale wigeon]

in the future, post your question right away, don't say "i need help" then go silent

[quiet mural]

im struggling with something ok but yes i understand

[vale wigeon]

tell us what you are struggling with!!!

show us the problem(s)!!

[quiet mural]

it was the ss lmfao

but i got it

[lone heart]

@quiet mural Has your question been resolved?

[vale wigeon]

did that mean "i understood your instructions" or "i know how to solve my problem"

[lone heart]

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[stray merlin]

im struggling with understanding on how to come to an answer with these problems

[chrome salmon]

tell me you already did the first

[stray merlin]

i did the first one

[chrome salmon]

use distribution of summation in 2nd

[stray merlin]

how do i do that

im lost

[chrome salmon]

summation (a+b) = summation a + summation b for finite sums

[stray merlin]

okay ty

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[woeful wind]

How do you calculate the most additions needed in dijkstras when given the number of vertices?

[woeful wind]

I know the most comparisons is given by 1/2n (n-1)

But I'm not sure what it is for additions

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[buoyant cipher]

Why is c the right answer here

[carmine reef]

there are 3 possible shapes

top left circle, top right circle, bottom middle circle

each one appears in each row and column an odd number of times

[undone falcon]

I see 4 different shapes

small dot, 2 small dots. small and big dot, 2 small dots and 1 big dot

[carmine reef]

those are patterns consisting of multiple shapes

[undone falcon]

ah alr

[carmine reef]

no idea why in the world everything is flipped vertically here for the answers

[buoyant cipher]

I see i see, am impressed how fast u understood the problem😭

[carmine reef]

knowing the answer already does help lol

actually first i saw that the top left circle was in every single square

the top right circle makes a T shape and so i suspect the bottom middle circle also makes a T shape

which it does by the answer

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[queen hazel]

If I know one angle is 10 degrees, and the adjacent side is 1410m, what is the length of the opposite side? (The triangle is a right triangle)
How do I find it?
I used the tan= opp/adj rule, but then I got 14100m for the opposite side...
I don't know if my calculations are wrong, but I'm not meant to get this number because I'm literally trying to calculate the height of a building...
If it is right, I probably got the angle wrong...

[vale wigeon]

how did you get 14100 for the opposite side?

[quasi vector]

,w tan10degrees

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[queen hazel]

uh...
tan = opp/adj so opp= tan x adj
Then i put in the numbers and did
opp= 10 x 1410m = 14100m

😭

[quasi vector]

opp = tan(10degrees) x 1410, not 10x1410

[vale wigeon]

tan(10°) is not 10

[queen hazel]

oh shit

what is it then

I literally havent learned this 💀

[quasi vector]

tan10° doesn't have a known value

you'd need a calculator to find that

[queen hazel]

o

[quasi vector]

maybe your angle is wrong

[queen hazel]

Yeah probably

what if it's 30 degrees?

[quasi vector]

do you know what tan30° is?

[queen hazel]

according to the calculator its 0.577350269

[quasi vector]

well

,w tan30°

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[quasi vector]

tan30° is 1/sqrt(3)

you just have to know these values

[queen hazel]

oh

Can I round it to 0.6 and try to calculate?

[quasi vector]

why would you need to round

[quasi vector]

tan = opp/adj so opp= tan x adj
Then i put in the numbers and did
opp= 10 x 1410m = 14100m
use this logic that you gave

[queen hazel]

So there would be sqrt in my answer?

[quasi vector]

if you wanted you can round the result after multiplying

,w 1410/sqrt(3)

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[quasi vector]

your answer would be around 814.06

whatever unit it is

[queen hazel]

ohhh

I get it now

I probably have the angle wrong because the actual height of the building is 234.371m lol

[quasi vector]

,w arctan(234.371/1410) in degrees

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[quasi vector]

that doesn't seem right.

can you send me the entire question?

[queen hazel]

and it's like "Find a building and calculate the height

from afar

something along those lines

so I used elevation and the distance between me and the building

to try and calculate it-

They said use ratio

The distance is 1.41km

and my not so accurate method of calculating the elevation without technology said the elevation was 10 degrees

so all I know is the adjacent and angle

I should recalculate the angle...

Thank you for helping me

[queen hazel]

But wait does this mean it actually is close to 10 degrees?

[quasi vector]

it does

with that height and distance, the angle is 9.4 degrees

[queen hazel]

Just a slight difference in the angle made the answer change so much, wtf

[queen hazel]

Thank youuuu

[queen hazel]

I'll probably record this in the homework lol

Thank you so much

[quasi vector]

no, you weren't calculating it properly.

,w tan(10degrees) *1410

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[queen hazel]

My dumbass thought tan 10 was 10 degrees 🤦

Guess not lol

But thanks a lot

I hope you have a nice life

.close

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[tepid mango]

how do you figure this out?

[lone heart]

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[robust oak]

Find HM

[rotund jolt]

have more information in this question ?

[robust oak]

no

[limpid turret]

AC=7 and AB=6?

[robust oak]

yes

[limpid turret]

Not enough info

[jovial rune]

is HM parallel to AC?

[robust oak]

idk

[robust oak]

what info added would make it solvable

[limpid turret]

Many things

There's no single answer

[robust oak]

it says H is As bisector to Bs projection or smth

[jovial rune]

i think if HM parallel to AC it could solve

[robust oak]

i dont think its parellel though

[limpid turret]

Is it not in English?

[robust oak]

yeah its not in english

[normal flower]

Can you translate?

We can say M is MP of BC

H is the angle bisector of A

AB is 6

AC is 7

[limpid turret]

Still not enough info. Many values could exist for <BAC. But perhaps HM would remain the same?

[normal flower]

True

[robust oak]

this is put through google translate

In the triangle ABC, AB = 6, AC = 7. H was the base of the projection from B on the bisector of A, and M was the midpoint of BC.

Find HM

[normal flower]

AH = AC

Can we say that?

[robust oak]

why

[normal flower]

Aren't the angles same?

HAC = BAH

Wait fk

[sour sleet]

Hi

[robust oak]

BAH=HAC?

[robust oak]

hello

[normal flower]

Yes

And BH will be 9.21

[robust oak]

that still doenst make AH=AC

[normal flower]

If angles are same the sides are same in a triangle

[normal flower]

BH is 9.2

[robust oak]

how

[normal flower]

This isn't enough

[normal flower]

AH = 7

[normal flower]

Not enough info

We might need HM value to solve

[robust oak]

In the triangle ABC, AB = 6, AC = 7. H was the base of the projection from B on the bisector of A, and M was the midpoint of BC.
Find HM

this was all of the problem

maybe ive drawn it incorrectly

[limpid turret]

Ok I looked into it

There's enough info

[robust oak]

rlly

[normal flower]

I think we can solve the entire question using angles itself

I am seeing angles part

Sides aren't enough but angles are

[robust oak]

is there enough angles provided

[normal flower]

Yep

Using angles you solve for sides

[limpid turret]

Hint : HM is parallel to AC, but you need to prove it

[robust oak]

what

how is hm paralle to ac

[normal flower]

Yep

[limpid turret]

!nosols

[lone heart]

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.

[normal flower]

Interesting method

I used angles

Good method capitan

[jovial rune]

thanks

[robust oak]

umm

HM would be parallel to TC if uh

wait

[normal flower]

Where's TC?

[robust oak]

?

[normal flower]

Converse Midpoint theorem interesting

What's TC actually?

[jovial rune]

ABD is an isosceles triangle and the bisector of it is perpendicular so it has to be median too, so the angle of ABD = angle of ADB, so AB=AD and DC = 1 use the midsole theorem and solve it

[normal flower]

What does T represent?

[robust oak]

T is just BH stretched out

For triangle BTC

H is the midpoint of BT?

@limpid turret

but i need to prove it right

[jovial rune]

yes

[robust oak]

ah

[jovial rune]

because AH is bisector of BAT and its perpendicular so it has to be median, and ABT has to be isosceles triangle too

[robust oak]

ahhhh

[jovial rune]

my english is not that good im using google translate sometimes, sorry about it 🙂

[robust oak]

thank you thank you

thank you all

finaly

[jovial rune]

your welcome

[robust oak]

ill close it now

.close

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[hidden musk]

Trying my hand at 3c.
This domain and range seems rather difficult like its a composite function or something.

[hidden musk]

Maybe?

Or...

I got this handout to help determine domain and range...

/

[alpine sable]

it's been a while, but doesn't the range of f(x) become the domain of gof(x) ?

[flat vale]

No need for composite fn or whatever

we know that x^2 - 5 must first be > 0

solve for the root

and that will be ur domain

[alpine sable]

(this is wrong nvm)

[flat vale]

and then the range is trivial from the domain

[flat vale]

So dom = R\(-sqrt(5), sqrt(5))

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[sudden compass]

So I was studying for my test when I came across this one on the sheet. We are given some formulas for but i'm stilk not really sure as for where to plug the numbers into and why

[lone heart]

@sudden compass Has your question been resolved?

[tepid mango]

.reopen

[tepid mango]

ohh this is the compound interest formula

one sec

so for the first question

you would do

A = 5000(1+6/4)^7x4

i thiiink

and then for the second one you would do 9000 = 5000(1+6/n)^nt and jus work backwards

@sudden compass

[sudden compass]

Ty

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[lone heart]

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[tidal ingot]

If S is a gamma random variable with rate lambda , then lambda S is a gamma random variable with rate 1

[tidal ingot]

why is that?

been at this for quite a while cant understand it

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[broken falcon]

kk

[broken falcon]

trying to resolve this limit but idk how to resolve from that point

i was thinking to cancel the n with the -n

.close

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[karmic osprey]

does anyone know if theres something wrong with my graph for (a) i.

i got 3 because at x=-1

both sides is approaching 3

i got DNE because x=1 has 2 values for (a) ii.

i got 1 because x=2 has a point at 1 for (a) iii.

for (b) i got dne because in the chart where the f(x) is it's approaching positive and negative

instead of being the same sign

[chrome salmon]

are these both different questions?

[karmic osprey]

and i got 0/8 because of the work shown

a b and c are 3 different questions yea

[chrome salmon]

no i mean the photos

[karmic osprey]

oh

yea

[chrome salmon]

can you send clear ones

[karmic osprey]

yea

[chrome salmon]

the one where you got limits from graph is right

not here

delete it

open your own channel

this channel is already occupied

I can't see properly but is the other limit $\lim_{x\to-2} \frac{x^2-4}{x-2}$

[ocean seal]

fäf

[karmic osprey]

yea

x-> 2

tho

[chrome salmon]

you got limit equal to 0?

[karmic osprey]

yea

[chrome salmon]

the limit is 4 tho

[karmic osprey]

oh

is (a) and (b) right though?

[chrome salmon]

x^2-4 = (x-2)(x+2)

a is right

what does b asks? it's not visible

yes DNE is right

[karmic osprey]

okay

how would i solve c

because the reason why i did what i did

is because i assumed i should multiply by the conjugate

since that was what was done to a similar problem

but im not sure what went wrong

[chrome salmon]

you may have done mistake there too

[karmic osprey]

thanks

i think i can figure the last one out

i was just trying to figure out if i got any wrong

.close

[lone heart]

Channel closed

Closed by @karmic osprey

Use .reopen if this was a mistake.

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• 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.

[barren wren]

Hello !
My question is:
when has this system no solutions(depending on alpha) ?

[barren wren]

I tried to use matrix but I don't remember the theorem

[limpid spade]

did u try row reducing

[barren wren]

I mean, the fast way

sorry for not beeing clear

and they gives us few possible answers

[lone heart]

@barren wren Has your question been resolved?

Channel closed

Closed by @barren wren

Use .reopen if this was a mistake.

[lone heart]

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>.
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• 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.

[steep moon]

.open

.reopen

ok

[steep moon]

uh

how do I find odd numbers from even numbers

so ok

i was asked how many 3 digit numbers can be formed from 0,2,3,5,and6 my answer is 243

so how do I find how many here are odd numbers and how many here are even numbers?

[lone heart]

@steep moon Has your question been resolved?

[peak bough]

How did you arrive at this number?

[steep moon]

nah that was my fault I confused the types of equations but my answer is actually just 60

I thought that one was one of those probability stuff I'm dumb

[peak bough]

I'm also a bit confused on how you got to this number

[steep moon]

well

it's like

5p3

I'm using the calculator so I just add up the 0,2,3,5,6 and then P 3

that's how they taught us this

[peak bough]

You have done permutations without repetitions here, and have assumed that a number like 222 can't be chosen

[peak bough]

I assume you did 3⁵ do get to this number?

[steep moon]

yes

[peak bough]

This is on the right track - using permutations with repetition - but you've done a mix-up on the way

And forgot to account for something

[steep moon]

what did I forget?

[peak bough]

Some times these digit-arranging questions can be taken much easier one digit at a time, instead of wondering which "formula" to use

First think about how many ways the first digit can be chosen

[steep moon]

6

nono

5

i miscounted

[peak bough]

No

[steep moon]

huh

[peak bough]

This is the tricky part

0 can't be chosen as the first digit

[steep moon]

oh

[peak bough]

Because then it will not be a 3-digit number

(At least by some definitions)

[steep moon]

nah for this one I think its ok to count it

[peak bough]

Maybe you do define something like 023 as a 3-digit number in your course?

When we did something similar in one of my courses, you had to account for 0 in this way

[steep moon]

man I shoulda asked her if we would count the ones that start with 0

[steep moon]

ok lets just say we do account for 0

[peak bough]

You can do both, and write it down

They're not that different in nature

[steep moon]

oki

[peak bough]

We can start with treating 0 as a no-no for the first digit

So for the first digit, there are 4 ways to choose this

[steep moon]

yes

[peak bough]

Then for every digit here, how manys way are there to choose the second digit?

[steep moon]

5

[peak bough]

Yes

And then for the third digit?

[steep moon]

5

[peak bough]

Exactly

[ocean seal]

Mikkel

[steep moon]

100

multiplied with each other

so what we doing with

3

[peak bough]

So there are 100 ways to make a 3-digit number

[steep moon]

oh

so we just multiply all the possible ways

[peak bough]

Yes

[steep moon]

I see

[peak bough]

If we didn't care about the 0-thing, the 4 would have been a 5

And there would have been 125 ways instead of 100

[steep moon]

then howd I get 60

[peak bough]

You did permutations without repetitions. So you calculated it like if 5 was chosen for the first digit, then it couldn't be chosen for the second or third

[steep moon]

I see

[peak bough]

Now you need to find the numbers that are even, and the ones who are odd

[steep moon]

and how do I do that

[peak bough]

Usually even numbers are defined as numbers than can be written on the form 2k where k is a whole number

[steep moon]

hm

[peak bough]

Odd numbers are usually either defined as simply whole numbers that are not even or a second definition that is often used is a number that can be written on the form 2k + 1 where k is a whole number (some prefer 2k - 1, which is equivalent)

The first definition for an odd number presented here would be the most useful to use

Since we just need to find the amount of even numbers and take that away from the total amount of numbers

[steep moon]

and how do we do that?

isn't this a sequence type of equation?

I think there was some type of arithmetic formula that could maybe help

[peak bough]

A neat trick to see if a large number is even is to see if the last digit is even

[steep moon]

ohh ok

[peak bough]

So something like 19474752 can easily be seen to be even, since the last digit, 2, is even

[steep moon]

yes

[peak bough]

If you want an explation for why this is, it follows from the fact that a number like 19474752 can be written as 1947475 ∙ 10 + 2 ∙ 1 which can be factored to be on the form of an even number

(And this logic follows for all numbers with 2 digits or more)

Okay, back to it

We know that the even 3-digit numbers have to end on an even digit

Which are our options for this?

[steep moon]

uh

[peak bough]

We just want the possible digits our 3-digit number can end on, for it to be even

From our set of numbers {0, 2, 3, 5, 6}

[steep moon]

yeah

like we gotta find how many 3 digit numbers are even?

[peak bough]

Right now we want to find which digits our number can end on

[steep moon]

let's say from 100-125? and 100 102 , 104, etc.. that stuff?

[peak bough]

Just the third digit of our 3-digit numbers

Can it end on 0?

[steep moon]

oh just

uh yea

yes

I think

[peak bough]

Yes

Because 0 is even

[steep moon]

yes

[peak bough]

(Can be written as 2 ∙ 0)

What other numbers as the last digit, would make the 3-digit number even?

[steep moon]

2,4,6,8,0

[peak bough]

4 and 8 are not part of the digits we can choose

But otherwise, yes

[steep moon]

oh

oh ok

[peak bough]

So we know that the last digit has to be either 0, 2 or 6 for the 3-digit numbers to be even

[steep moon]

yes

[peak bough]

Now we can do a similar procedure as earlier, but with out new knowledge of the third digit

How many digits can be chosen as the first?

(If we don't count 0)

[steep moon]

9

[peak bough]

No

[steep moon]

huh

[peak bough]

We have the numbers 0, 2, 3, 5 and 6 to choose from

[steep moon]

oh just those?

[peak bough]

Yes

[steep moon]

oh 4

[peak bough]

Yeah