help-10

[alpine raven]

this ?

[timid silo]

well, my confusion is that i don't see this in what he wrote

like 1*10 to the power of 3

[alpine raven]

its the weight of each number

[timid silo]

where did the 3 come from?

[alpine raven]

the rank

[timid silo]

ohh right

and in the first example he gave (1011 in binary), why is every number multiplied by two?

[alpine raven]

because 2 is the base

or at least, thats a representation of a number in binayr

binary*

[timid silo]

but what determines the base? or is that something that's bound to a set of rules, like a formula or such?

so any conversion of binary must always include 2 as its base, and it's to the power of its weight

is that correct?

that makes a lot more sense. thank you

one last thing

i've been really meaning to start self studying math. i've had an interest in doing so for the past 4-5 months. what's a good starting point? (only resource i know of is khan academy)

right

so that's the only resource i'll need? how far will it take me before i have to start searching for books on my own?

well, i graduated high school but never went to college

i'd say i need a refresher on my math though

[alpine raven]

just study what you need for

[timid silo]

i'm not really doing it for a course or program. i just want to self study out of interest, so i don't exactly need math for anything specifically

[alpine raven]

do all algebra courses, geometry and calculus

i think thats enough ?

[timid silo]

novice programmer. math can help with computer science and some data structures and algorithms concepts

[timid silo]

awesome. how far would you say i can go with khan academy? i'm wondering when i need to start searching for books on my own

awesome, those'll be first on my list

[alpine raven]

I dont really know the limit of Khan academy, ive never used it

they go to calc 3 at least

i dont know if they do linear algebra

what do you know so far in maths

[timid silo]

@timid silo if you have experience with khan academy, i'd like to know your input on this

[timid silo]

well i'd need to brush up my math since it's been a bit but i know whatever the australian curriculum teaches up to 12th grade

no problem

haha small world

[alpine raven]

and what is their limit in high school maths ?

[timid silo]

if you're talking about extension classes, i didn't take those

i remember taking advanced maths classes but that's about as far as i went with it

[alpine raven]

like, which notion

[timid silo]

i'm not sure what you mean. i remember learning certain topic such as logarithmic functions, exponential functions and some calculus

but as i said, i've forgotten a lot of it

i never used to have much of an interest in math when i was in school but that's changed years after

[alpine raven]

thats basically what i was asking
what is the curriculum 🤔

you said exponential function and logarithm

I guess you did complex number too

[timid silo]

no, i think that was above my level

[alpine raven]

ah

[timid silo]

i wasn't the best at math though. that was my biggest thing

i was always that guy who barely studied and just managed to pass his exams

[alpine raven]

well im not asking if you are the best or not lmao
I just want the syllabus so I can at least tell you where to start

[timid silo]

which is why i want to go back and fill in missing gaps in my knowledge

[timid silo]

yeah well, that was my level

[alpine raven]

so you only did calculus in 12th grade ?

[timid silo]

it was never listed as methods or specialist in my school but looking at it online, i guess i did methods. here, i'll link it

https://au.ixl.com/standards/maths/year-12?documentId=2007000022&subsetId=2007000096

i did a lot of this stuff

but i probably need to go back and redo a lot of it

[alpine raven]

ok, if you remember at least a bit of everything you did in 12th grade, you can start with abstract algebra (the basic)

monoid, group, ring, field at least

after that you can start linear algebra

going through all linear algebra maybe

well thats on you

[timid silo]

thank you both, you've helped out a ton

.close

[obtuse pebble]

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[alpine raven]

ah dont forget arithmetic too

[timid silo]

i won't. thanks again

[obtuse pebble]

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[hoary idol]

Need help with 2 and 4

[spring trail]

from the top?

[hoary idol]

Yes

[spring trail]

!show

[obtuse pebble]

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

[hoary idol]

Buf not sure where to start but what to do

Im supposed to use this I think

[spring trail]

are you aware of the product formulas?

[hoary idol]

this?

[spring trail]

not these ones

[hoary idol]

can i not do it with thees ones?

[spring trail]

you can

@hoary idol

try following me okay?

[hoary idol]

i need to do it with this

ok

[spring trail]

u are comfortable with the idea of variables?

[hoary idol]

is it neccecery?

[spring trail]

here look at the two formulas of sine

there is alpha and beta; i need you to understand that they are variables and represent something

and we can make new variables to represent new things

[hoary idol]

ok

[spring trail]

ok lets define two variables c and d

c= alpha + beta

d = alpha - beta

so what is c + d

[hoary idol]

2alpha

[spring trail]

so alpha is (c+d)/2

similarly what is d

(take ur time)

[hoary idol]

c-d/2

[spring trail]

yep

do u have a notebook w u rn?

[hoary idol]

yea

[spring trail]

so write down those 2 formulae of sine

[spring trail]

these ones

only the sine ones tho

remember the way we added these two?

add those two formulae together as well

now u can replace the alpha + beta with c

alpha - beta with d

and also make use of the formulas we made for c and d

[spring trail]

these ones

[hoary idol]

ok so what do i do?

[spring trail]

ahhh

show ur work again

[hoary idol]

do i start doing the equation or what?

[spring trail]

if u have been doing all that ive told till now

send me a pic of it

we are arriving at a new formula using the old ones

to easily solve ur problem

i actually gtg rn

if u still need help

!15m

[obtuse pebble]

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.

[timid silo]

Whoever is giving u these questions is a demon

[hoary idol]

are they hard?

.close

[obtuse pebble]

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[timid silo]

Holy can you wait

Are you this impatient

Q4 otw

[hoary idol]

oh i thought you left

[obtuse pebble]

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[timid silo]

here

its correct

[timid silo]

Ofc not

First one is just tedious

[boreal jay]

Wtf is tg

Tan?

[timid silo]

yeah i guess

tangent

[hoary idol]

tg

tan yea

thanks a lot birdwatcher

[timid silo]

nah i did it in like 3 minutes

just do more

trig is super easy

[hoary idol]

ok

how long have you been doing it?

[boreal jay]

It's tedious if you expand it using sin(a+b) and sin(a-b)

[hoary idol]

i get it better that way i prefer it tbh

[boreal jay]

Like their individual expansions

[spring trail]

dang i forgot u can just plug in

[boreal jay]

Then yeah go for it

But I want to ask

U know the formula for sin a + sin b?

[spring trail]

they dont

[boreal jay]

Ah

[spring trail]

he

[boreal jay]

In that case u do need the individual expansions

[hoary idol]

i havent lerned that yet

[spring trail]

tho if u wanna explore, have fun or just make ur maths life easier

try seeing through what i told u

and getting to that formula

[hoary idol]

ok thanks ill think about it tonight

[obtuse pebble]

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[brisk arrow]

I’m making a note, is it true

[meager glade]

yes

if you want something intuitive

you can make a graph

of sin(x)

and sin(-x)

[brisk arrow]

Thank you

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[vocal pollen]

In the context of a kernels (SVMs), a kernel is valid if it is:

1. Symmetric
2. Positive semidefinite

What is the point of the constants c in condition 2?

[obtuse pebble]

@vocal pollen Has your question been resolved?

[obtuse pebble]

@vocal pollen Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[sullen sparrow]

i am struggling with this exercise, because i dont know whether this is a subspace or not

[sullen sparrow]

if i want to consider U and V belonging to the subspace, i dont know how to set them

[royal basin]

(a-1, -2a+2) and (b-1, -2b+2) perhaps.

[sullen sparrow]

like, its U = (a,a) and V = (b,b)?

[royal basin]

no.

[sullen sparrow]

uhm

[royal basin]

.

[sullen sparrow]

okok

alright i got this

one more question tho

in this exercise, how am i supposed to do this

U = (x1,y1,z1) and V = (x2,y2,z2) ?

[timid silo]

ye

should be clear that it is closed under multiplication and contains the zero vector

verify that it is closed for addition though

hint: find a counterexample

[sullen sparrow]

yeah thats where i am struggling

[timid silo]

consider liie

the vectors (1.0, 1) and (1,1,0)

do those two work out for what you have

clearly they hold true individually, but what about when summed?

[sullen sparrow]

because u + v = (x1 + x2, y1 + y2, z1 + z2) , so then we should do something like (x1+x2)(y1+y2)(z1+z2) ??

[timid silo]

yeah

exactly

and that should be equal to 0

[sullen sparrow]

but if we only work with x y z (and not numbers) how do i justify it

because i get somewhere in the equation x1 multiplying by y2 and z2 or z1

so i cannot say that it is = 0 because i dont have enough proof that it wont be 0

[timid silo]

A lot of people have this misconception that you need to be direct about proofs and such always, and contradictions/counterexamples are not sufficient

but it really isnt

showing one counterexample makes the whole thing collapse

and that's what you are trying to show. So don't trouble yourself by doing it generally or whatever

[sullen sparrow]

ok so when do i know whether i should do a general case or get an counterexample?

[timid silo]

Really hard to generally answer that , but for subspace questions counterexamples are a good choice generally

as shown above

also like it should be obvious to say that x1 + x2 isn't always equal to 0, even if either terms are in fact 0,

[sullen sparrow]

yeah yeah

thank you so much!!

.close

[obtuse pebble]

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[wind sphinx]

help me to solve please 🙏

[obtuse pebble]

@wind sphinx Has your question been resolved?

[tropic terrace]

n is a positive integer?

[timid silo]

You could use integration by parts

@wind sphinx

[wind sphinx]

i think yes

[timid silo]

@wind sphinx , you could differentiate the $sin^n(x)$ term

[warm shale]

Palahoo

[wind sphinx]

okay ill try

.close

[obtuse pebble]

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[rough marten]

how to do part c)

[rough marten]

is part c inverse of part b?and how do i connect bth of them

[obtuse pebble]

@rough marten Has your question been resolved?

[meager glade]

what happens when in b, you substitute x with e^y?

[rough marten]

e^e^y?

[meager glade]

mb, y= e^x

[rough marten]

ya get x

[rough marten]

and then?

[meager glade]

then what does the integral become

[rough marten]

a straight line?

[meager glade]

i meant the terms in the integral

the integral changes

when you use substitution

[rough marten]

get x

[meager glade]

?

rewrite the integral

for y

the whole integral

everything about the integral

[rough marten]

whats integral

[meager glade]

$\int$

[warm shale]

Flappie

[meager glade]

thats an integral

[rough marten]

o lmao..

[meager glade]

maybe language issue

thats okay

[rough marten]

so rewriting the question..?

[meager glade]

yes

[rough marten]

let y = e^x right

[meager glade]

yes

[rough marten]

i cant integrate x, can i

[meager glade]

substitute the rest aswell

right now you have only subtituted one value of y

there are 4

[rough marten]

4??

wtf

[meager glade]

yes, 4

[rough marten]

theres like only 2 y

[meager glade]

4 places you need to replace

the bounds aswell

they go from y=2 to y=3

right?

[rough marten]

how do you change the bound

[meager glade]

you have your substitution

y=e^x

[rough marten]

oh

[meager glade]

and we have the bounds y=2 and y=3

so to get the lower bound

we fill in y=2 into y=e^x

so we get 2=e^x

but we want our new integral to be in terms of x

so we get x=ln(2)

for the lower bound

similar for the upper bound

[meager glade]

yes

but d(e^x) does nothing for us now

so transform it into a dx

[rough marten]

and how do u do that

[meager glade]

we have our substitution y=e^x

[rough marten]

e^x to x... add a ln?

[meager glade]

we differentiate y to x

so we get dy/dx = e^x

thus dy=e^x dx

[meager glade]

into this

[rough marten]

oh

[meager glade]

we replace dy with e^x dx

[rough marten]

okok

then

[meager glade]

exactly

have you had integration by parts?

[rough marten]

lemme check back my notes

this?

[meager glade]

yes

that last line

[rough marten]

oh alright den its doable now

[meager glade]

$\int u dv = [uv] - \int v du$

[warm shale]

Flappie

[rough marten]

actually why must u change the ln y to in terms of x tho

[meager glade]

look at question b

there is a reason why they asked us to calculate this

its beacuse we are gonna need it in the next step

so in the ln(y)dy integral we wanna go to something of that form

we couldve also substitued x=ln(y)

and then we get dx/dy=1/y, dx=1/y*dy

hmm actually

then we wouldnt go to the form of ln(y)dy that they ask

[rough marten]

how do you even know to sub y=e^x tho

is it to juz make the intergrand in terms of x

[meager glade]

we see e^x in the first integral

and we see ln(y) in the second one

so we need some kind of substitution that makes us go from e^x to ln(y) or vice versa

the only thing i can think of is y=e^x or x=ln(y)

this is also kind of a experience thing

if you see this more often

[rough marten]

[meager glade]

you'll see what you need to substitute for

also we see the bounds

the bounds in b are of the form ln(something)

and in c they are something

which also indicates we need some e^x substitution thing

[rough marten]

alrite thx for the help

[meager glade]

can you do it with integration by parts?

[rough marten]

i think so , i will follow the guide notes

the questions jumped from intergrating trigo to this type of question so i was so lost

thx again! :)

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[obtuse pebble]

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[meager glade]

no problem

good luck

[obtuse pebble]

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[sacred trail]

Need help with this problem

Would it be the original answer divided by 5?

[fossil crag]

Do you know how to make the first integral appear?

[sacred trail]

Nope

[fossil crag]

What is different between the first and second integral, not looking at the bounds?

[sacred trail]

The first one is 0 to pi and the second is 0 pi/5

[fossil crag]

not looking at the bounds

[sacred trail]

?

[fossil crag]

Like, is there another difference apart from the bounds of integration?

[sacred trail]

There’s a 5x in cos

[fossil crag]

Yes

But to find the first integral, you want there to be just "x" in cos

So how do you switch from 5x to x?

[sacred trail]

Multiply by 1/5?

[fossil crag]

Wdym?

[sacred trail]

I’m not sure how you switch

[fossil crag]

What properties allow you to change from 5x to x?

[sacred trail]

I’m not sure

Divide by 5

[fossil crag]

I'll repeat in a more obvious way

What allows you to change from the variable 5x to the variable x?

[sacred trail]

Well don’t you divide both sides by 5

[fossil crag]

Ok let's try again

You want to turn ln(a+bcos(5x)) into ln(a+bcos(t))

How do you do it?

[sacred trail]

U substitution?

[fossil crag]

Yes

[sacred trail]

Oh

And then what do you do

[fossil crag]

Did you do the u sub?

Bc I'm pretty sure you haven't gone through it yet

[sacred trail]

Idrk what I’m suppose to do after I u sub

[fossil crag]

What do you get from u sub?

[sacred trail]

Does it become the original answer

Ln(a+bcos(u)du/5

[fossil crag]

Well then

If you find the bounds of integration, the answer becomes pretty clear

[sacred trail]

Idk the bounds of integration though

[fossil crag]

u = 5x no?

So if x = 0, then u=...

And if x = pi/5, then u = ...

[sacred trail]

Pi

[fossil crag]

Then you know the new bounds of integration

[sacred trail]

So the integration would be the same

Thing

?

@fossil crag

[fossil crag]

Is this the same thing as what the first integral is?

[sacred trail]

Well no

But when you plug in for x you get the same thing as the other integral

[obtuse pebble]

@sacred trail Has your question been resolved?

[sacred trail]

<@&286206848099549185>

[tropic terrace]

just sub 5x=t

the limits become 0 and pi, and dt/dx=d(5x)/dx=5 so there's a factor of 1/5

answer is 2/5 pi ln 2

[sacred trail]

Ok thank you

.close

[obtuse pebble]

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[loud grail]

sure, what are you struggling with?

[tardy epoch]

Oh are you taking a quiz right now?

[timid silo]

More importantly matrices In high-school??

Precalc and calc are understandable but I never so someone to early linear in hs

Also we can't help with graded quizes

It's against the rules

[meager glade]

!noans

[obtuse pebble]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[slim cove]

Is this something you're supposed to be working on your own for?

[obtuse pebble]

@oblique jacinth Has your question been resolved?

[long plinth]

.close

[obtuse pebble]

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[mint nebula]

Through the middle M of the arc AC of the circumscribed circle of the triangle △ABC, a chord (MN) parallel to AB is drawn.

Show that the arcs BNC and MCN are congruent.

[mint nebula]

I was thinking to prove that angle BOC and angle MON are congruent

Please kind the choppy drawing.

mind*

[obtuse pebble]

@mint nebula Has your question been resolved?

[mint nebula]

<@&286206848099549185>

Haven't done much

[obtuse pebble]

@mint nebula Has your question been resolved?

[steady lake]

Try proving BC = MN which would prove that the angkes are equal

[mint nebula]

Well.

OB = OC = ON = OM

We've also got quite a lot of isoscelles triangles

[steady lake]

I gave it a try and was only able to prove that the 2 triangles zre similar but i need them congruent

ok so we need BNI and MCI congruent, ive proved that theyre similiar which can be easily done. for proving one of the sides equal i wanna make use of the same slope and M being on the middle of the arc info together

@mint nebula you there?

[mint nebula]

Yeah, sorry.

[steady lake]

any idea how we can use that together?

[mint nebula]

Not really.

Ill look into it tomorrow tho

Thanks anyways!

Have a good one

[steady lake]

alright

[mint nebula]

.close

[obtuse pebble]

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[little vault]

Is the following true? How can I prove it? : p | a => (p-1) | (a-1)

[little vault]

nvm

.close

[obtuse pebble]

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[vague isle]

Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram

[timid silo]

need help on this one guys

not good on these types but i can do the rest like what

[vague isle]

ill free the server for you

.close

[timid silo]

i just need the answer

[vague isle]

i had a question as well but we wrote at the same time, i'll pick another one

[timid silo]

.close

[obtuse pebble]

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[polar fossil]

draw a tree diagram

should be three cases i think

h1, ~h1 and h2, ~h1 and ~h2

yea

[obtuse pebble]

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[restive fog]

doing linear algebra: isnt the whole point of gaussian elimination that its the same as gaussjordan without the last part? how can you solve a system with one or the other? doesnt gaussian not always produce an incomplete solution, and if it doesn't, then theres no more steps for you to do with jordan?

[obtuse pebble]

@restive fog Has your question been resolved?

[obtuse pebble]

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[deep ravine]

How do I find the right answer?

[unreal musk]

The purple bits seem fine (maybe excluding integral limits) - what did you do after that?

[deep ravine]

I tried taking antiderivative of u

Which would make it (-e^(9/2))/6+1/6

But the actual answer was ((-e^(-3))/6+1/6) somehow?

[unreal musk]

Could you show what you got as that?

[deep ravine]

I crossed it out but you can still see it

[unreal musk]

Also remember the comment about integral limits and them not being changed

[deep ravine]

Yea i didnt change them

[unreal musk]

so you did it of u by itself rather than of the new integral in terms of u

[deep ravine]

?? What would that look like?

[unreal musk]

Once you get this, you want to replace all x’s with u’s

[timid silo]

Wait I forgot the 1/6

Hahaha

[unreal musk]

Also answer is here

[deep ravine]

But i set u to -3x^2?

[unreal musk]

Exactly, replace that so the integral is in terms of stuff that has u’s in it only

[timid silo]

@deep ravine

[unreal musk]

That’s the whole point of substitution

[deep ravine]

Hang on let me look at this

[deep ravine]

Is the limit supposed to change?

[timid silo]

Yes

[deep ravine]

What is the condition that elicits this change?

Bc i havent seen it anywhere else

[timid silo]

Wdym with that?

You take a look at your u sub

In this case it’s u=-3x^2

Now you see the first limit, it’s 0, what happens when you sub in 0 for -3x^2, correct, also zero, so we write that down

What happens when you sub in 1 for -3x^2? @deep ravine

[deep ravine]

Becomes -3

[deep ravine]

But dont we have to take the antiderivative of u first?

Or is that a different method

[timid silo]

Antiderivative? What do you mean?

You just sub in the old limits to the u sub and get the new limits

And you don’t need to sub back after that

It stays u

[deep ravine]

How I did it, you raise the power of u by 1 and divide by the new exponent

So u becomes (u^2)/2

At least thats how my teacher told us to do it

[timid silo]

No that would be if you integrate u @deep ravine

But we want to differentiate it so we get du to make our sub and solve for dx

This will help you to understand it better @deep ravine

https://youtu.be/tM4RWc9ryx0?si=VuvqKoLqr-HxH39p

[deep ravine]

Ok ill take a look

[deep ravine]

Do you think its possible to solve this without changing the limit?

[timid silo]

Hmm perhaps there’s another method but the most obvious and easy one is for sure the u sub

It’s easy to understand the concept, just take a look at a few examples and you‘ll get it

[deep ravine]

Yea i understand it now, its just that my teacher never mentioned such a method

Anyways thanks @timid silo for ur help

@unreal musk

[timid silo]

Maybe you interchange this with integration by parts

Because there you integrate

[deep ravine]

Does that include u substitution?

[timid silo]

Nope it’s a bit different you just form your integral to a new integral

But this also doesn’t work always

[deep ravine]

Ah nvm then

.close

[obtuse pebble]

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[obtuse pebble]

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[fallen inlet]

Hiiii I need help with this because I only just found out that 125^1/3 would be 3 radical 125 so idk which one exactly makes that true

[fallen inlet]

Does it matter where the x is

Isn’t is useless ???

Idk tho

[timid silo]

$\sqrt[a]{x^b}=x^{\frac{b}{a}}$

[warm shale]

Şêro

[fallen inlet]

Is it always like that?

Like a rule

[timid silo]

Yes it’s true for the real numbers

But in this case

You must use another rule too

[fallen inlet]

Why

My problem doesn’t have real numbers ??

[timid silo]

It does 125 is for example a real number

Ignore this now

What can you write $\sqrt[3]{125}$ as with the rule mentioned?

[warm shale]

Şêro

[fallen inlet]

I can 125^x/3

But what is x/3

[twin sapphire]

@fallen inlet in fact the more general rule is that the definition of $\sqrt[n]{a} = x^{\frac{1}{a}}$

[timid silo]

where did you got x?

125=125^1

[fallen inlet]

Because in the problems there’s x in all equations

[timid silo]

.

[warm shale]

Benjamin

[timid silo]

See this, your x is 125

[twin sapphire]

lets not make things confusing

x already exists in the exercise

[timid silo]

Yes

[twin sapphire]

so did you get this dia?

[fallen inlet]

Yes

[twin sapphire]

ok so apply that to our case

to get rid of the root

[timid silo]

You can rewrite the $\sqrt[3]{125}$ as $125^{\frac{1}{3}}$

[warm shale]

Şêro

[fallen inlet]

It would be 125^1/3 ?

Yeah

[twin sapphire]

yes

so now

do you know your rules on exponents?

[fallen inlet]

Not really

[timid silo]

Now you need to use this rule $(a^b)^c=a^{b\cdot c}$

[warm shale]

Şêro

[twin sapphire]

do you get why we are using this rule?

[fallen inlet]

No

I don’t know how to use it because I have fraction now

[fallen inlet]

This

[timid silo]

Well tell me what’s your a,b, and c in this case

[twin sapphire]

so we have $(125^{1/3})^x$ right?

[warm shale]

Benjamin

[fallen inlet]

Yeah

[twin sapphire]

1/3 =b ?

[timid silo]

It doesn’t matter if it’s integers or fractions, a,b and c can stand for anything

[twin sapphire]

1/3 is a number right?

[fallen inlet]

Kind of yeah

Ohh ok

[twin sapphire]

wdym kind of

its a number

[fallen inlet]

Bc it’s a fraction

Yeah still a number

Lol sorry

[twin sapphire]

its not an integer right

[fallen inlet]

Nooo

[twin sapphire]

but its a number

?

[fallen inlet]

Yes

[twin sapphire]

so use the rule

[fallen inlet]

125^1/3timesX ?

[twin sapphire]

yes

[fallen inlet]

Ohhh ok I got it now thx

But there’s another kind of similar can u also help me ???

It’s this and I only know that 81^1/3 is 3 radical 81

Should I simplify that

[timid silo]

You need to find the factors of 81

[twin sapphire]

you can do it from one way or the other

nvm there are 4 choices

[timid silo]

$81=9\cdot9=3^2\cdot 3^2=3^3\cdot 3$

[twin sapphire]

use \times sero

[warm shale]

Şêro

[timid silo]

Not used to that

[fallen inlet]

Oh

Okay

What about the extra 3 s

And stuff

In the answer options

I only ended up with 3 radical 3

And 3 inside the radical used to be 81

Or is 3^3 times 3 the same as 3^3 radical 3?

[twin sapphire]

now take this equation to the 1/3 exponent

and apply the power rules

[fallen inlet]

Ok

[twin sapphire]

what do you get?

[fallen inlet]

Which power rules exactly ?

Imnot sure

[twin sapphire]

there are only 3 power rules

a^-1 = 1/a

and $a^{b+c} = a^b \times a^c$

[warm shale]

Benjamin

[fallen inlet]

Oh ok

.close

[obtuse pebble]

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[obtuse pebble]

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[polar bramble]

i need help with this trig identity

[polar bramble]

i tried to do the left side first

and i am struggling

i just did the sum and difference identity

[robust raven]

very well

and next

just divide numerator and denominator

by

cosalpha * cosebta

you shud see , them , right side at once

[polar bramble]

you mean cos(a)cos(b)/cos(a)cos(b) ?

[robust raven]

numerator / cosacosb and denominator / cosacosb

such thing:

$\frac{cos\alpha\text{}cos\beta-sin\alpha\text{}sin\beta}{cos\alpha\text{}cos\beta+sin\alpha\text{}sin\beta}=\frac{\frac{cos\alpha\text{}cos\beta-sin\alpha\text{}sin\beta}{cos\alpha\text{}cos\beta}}{\frac{cos\alpha\text{}cos\beta+sin\alpha\text{}sin\beta}{cos\alpha\text{}cos\beta}}$

[warm shale]

Joanna Angel

[robust raven]

and that is almost all

[polar bramble]

what

it would cancel each other out

[robust raven]

do what i say

[polar bramble]

oh wait

i see the factoring

yeeah

i did it

1-tan(a)tan(b) / 1+tan(a)tan(b)

[robust raven]

$=\frac{1-\frac{\sin\alpha}{\cos\alpha}\cdot \frac{\sin\beta}{\cos\beta}}{1+\frac{\sin\alpha}{\cos\alpha}\cdot \frac{\sin\beta}{\cos\beta}}$

[warm shale]

Joanna Angel

[robust raven]

yes! 🙂

[polar bramble]

trig identities are hard for me ngl

there are so many

[robust raven]

they require experiences and practises

[polar bramble]

but i have to revise them since i am doing calc next sem

[robust raven]

ok)

[obtuse pebble]

@polar bramble Has your question been resolved?

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[golden sierra]

Ok I have a system of equations that I have to solve.

p^2+2p+q=24
2q+p=4

[golden sierra]

I’m not really sure what to do

!status

[obtuse pebble]

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

[golden sierra]

I don’t know where to begin

[azure anchor]

do you know any method at all for solving systems of equations?

[golden sierra]

Yes, I know a few. I believe you can add the equations, subtract them, or substitute

Or you could graph them (but I’m not familiar with this)

[azure anchor]

okay so we have 2 equations and 2 unknowns, p and q

you're right that we can kind of add and subtract with them

so with that in mind, what operation would you do here with the two equations?

[golden sierra]

Addition?

[azure anchor]

well, i think it'll be easier if we just try and work with one variable, don't you?

how can we work with only one variable here?

[golden sierra]

Each equation has multiple unknowns

[azure anchor]

yes, but my point is that with some creative addition or subtraction we can derive an equation of only one unknown

then solve for one, and use that to solve the other.

[golden sierra]

I get that

[azure anchor]

can you see any way of getting that single variable equation?

[golden sierra]

Would it be good if we got one variable on each side of the equation?

Or no

[azure anchor]

imagine the first equation is a variable, A, and the second is B. how can we add or subtract multiples of these equations together like xA + yB = C

there's 2p in the first equation, and only p in the second equation

so if we wanted to get rid of 2p, what should we do?

[golden sierra]

Could we multiply the second equation *2 and then subtract the two equations?

I feel like that’s not what you said but I’m not sure what else

[azure anchor]

no you are on the right track

so we multiply the second equation by 2 and subtract.

how would you set this up?

[azure anchor]

ugh im sorry, i wasn't paying attention. notice if we get rid of p we are still going to work with 2 variables because p^2 remains.

[golden sierra]

That doesn’t work

Wait

[azure anchor]

so we should get rid of q

[golden sierra]

Lmao it’s ok

Ok

[azure anchor]

similar logic though

[golden sierra]

Ummm

Multiply the first equation by 2?

Will it become like a quadratic

[azure anchor]

yes

[golden sierra]

Ughhh

Hate quadratics

[azure anchor]

$$x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a},$$

[warm shale]

b0ngl0rd

[golden sierra]

What is X

[azure anchor]

p in this case

[golden sierra]

Oh

[azure anchor]

there will be two solutions

[golden sierra]

I have 2p^2+3p-44=0

Is this right

[azure anchor]

its what i got

[golden sierra]

Ok

I got a weird number

[azure anchor]

what did you get

[golden sierra]

-3+-sqrt-343

Wait no

I got

-3+-sqrt361

Ok so My final answer is p=(4,5.5) and q=(0,4.75)

[azure anchor]

thats what i got

they might want exact answers though

not decimals

[golden sierra]

Idk what’s an exact answer

[azure anchor]

fractions, radicals

[golden sierra]

Sorry, what’s a radical?

[azure anchor]

root signs

[golden sierra]

Ohh

Got it!

[azure anchor]

but hold upo

your second p answer should be negative

[golden sierra]

Ok

Oh sorry

You are correct

[azure anchor]

can you write it out in exact form?

[golden sierra]

Ummm

I don’t know

What that looks like

[azure anchor]

the answer before you made them decimals

[golden sierra]

9.5/2

[azure anchor]

decimals are an "approximate" answer when working by hand

[golden sierra]

Uh

[azure anchor]

you should've gotten something like that while working the quadratic formula

[golden sierra]

Yeah that’s exactly what I got

[golden sierra]

Should I just write that for p

[azure anchor]

thats what i mean

by exact answer

[golden sierra]

Oh

Ohhhh

Ok

[azure anchor]

its a better answer in nearly all cases unless it asks you specifically to use decimals

but you had it right, so good job

[golden sierra]

How did u get 19/4

[golden sierra]

Oh well I’m not that smart

I just used a calculator

[azure anchor]

you should practice this if you plan on taking college algebra

[golden sierra]

I’m a junior right now

[azure anchor]

[golden sierra]

My

school system is strict with taking classes

U can take AP calc only senior year

OK anyways thank you for

The help

So much

And being patient

[azure anchor]

well, what i mean is, if you plan on going further. itll help you a lot to work with pen and paper

no problem, good luck with your class

[golden sierra]

OK IM gonna close the ticket if that’s ok

.close

[obtuse pebble]

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[slow pier]

$$
l1 : \frac{x-2}{n}=\frac{y-5}{-9}=\frac{z+4}{-5}
$$

[warm shale]

kisek

[slow pier]

$$
l2 : \frac{x+3}{-1}=\frac{y-1}{n}=\frac{z+2}{1}
$$

[warm shale]

kisek

[slow pier]

theese lines will never be parallel for any n?

right?

$$
\frac{n}{-1}=\frac{-9}{n}=\frac{-5}{1}
$$

[warm shale]

kisek

[slow pier]

so if we look at 1 and 3, n must be 5, but -9/5 is not -5

am i missing something?

[obtuse pebble]

@slow pier Has your question been resolved?

[obtuse pebble]

@slow pier Has your question been resolved?

[obtuse pebble]

@slow pier Has your question been resolved?

[obtuse pebble]

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[kind gust]

The base of the right prism is a triangle and its side has lengths a=6 cm and b=5. Calculate the surface area of ​​that prism, if your height isH 4 cm

[kind gust]

ive done up until finding everything

it just wont

[fathom flicker]

what do you mean "finding everything"

!show

[obtuse pebble]

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

[kind gust]

yes

wait

c=H so thats 4

ive found the base thats a times b

i dont have it on paper

let me write it

[fathom flicker]

What is a "right prism"

[kind gust]

wait

since english isht my firdt

but the bases in this case

are triangles

ive found a b and c

then found the base

it still wont come out

right

i csnt seem to do it

[royal basin]

write it on paper

we can't tell what you're doing wrong without seeing it in writing

[kind gust]

this is the drawing

now how do i find perimiter of base

[obtuse pebble]

@kind gust Has your question been resolved?

[royal mulch]

hello @kind gust I can help if you are here

[slow pier]

scheme looks kind of...

weird

is that triangle prism?

triangle base, two bases are equal shape and their surfaces parallel?

[royal mulch]

yeah they will need two more rectangles to solve this bad boy

[slow pier]

i dont like how only 2 lines connect base vertices

also triangle on top must be same as on bottom

unless perspective, but perspective isnt very common in math schemes

actually perspective is bad as i know, distortion can be confusing

[obtuse pebble]

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[surreal bone]

Hi

[obtuse pebble]

Please don't occupy multiple help channels.

[surreal bone]

I cant find mu channel

My

[alpine bison]

this is your channel

[surreal bone]

O ok

Im not rlly sure what the question is asking

And i would like some help

[silk mauve]

I have a binomial question and I just need to know if its correct, the question I have is "Six people are asked to choose a number between 1 and 20, what is the probability that two people choose the number 9", would I use the formula nCr(20,0)(9)^0(11)^9 + nCr(20,1)(9)^1(11)^8 + etc, until I reach nCr(20,9)(9)^9(11)^0??

[surreal bone]

@silk mauve i think ur in wrong channel

[silk mauve]

is there a binomial channel

[surreal bone]

No u jus gtta start an available one

[silk mauve]

oh ok

[surreal bone]

👍

I am stuck on This

<@&286206848099549185>

<@&286206848099549185>

[tropic terrace]

!15min

[obtuse pebble]

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.

[tropic terrace]

x+y=9 and xy=20
write y=9-x in the second 20=xy=x(9-x)=-x^2+9x => x^2-9x+20=0

[obtuse pebble]

@surreal bone Has your question been resolved?

[surreal bone]

Got it!

Thx

[obtuse pebble]

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
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[chrome talon]

I want to make help channel

[chrome talon]

Can anyone help me solve this?

[tropic terrace]

sqrt(5+sqrt(24))=1/(sqrt(5-sqrt(24)))

[chrome talon]

There is also exponential x

[tropic terrace]

set a=(sqrt(5+sqrt(24)))^x then the equation becomes a-10=1/a

[chrome talon]

Ohh

[tropic terrace]

this is a quadratic equation

[chrome talon]

Oh yeah?

Wow thank you

[tropic terrace]

welcome!

[chrome talon]

.close

[obtuse pebble]

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[chrome talon]

.reopen

[obtuse pebble]

✅

[chrome talon]

One sec could you tell me how we got 1/a I know there is - but how we understand its going to be 1/a?

ahh now I see

After I solve the a-10=1/a I get x = 5+- sqrt(26)

from that part how do we solve it?

[warm shale]

moriaritie

[tropic terrace]

,w solve a-10=1/a

[warm shale]

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[tropic terrace]

ok a is positive because 5+√24 is positive so a=5+√26

[plucky sentinel]

Ghhidhdxkxi

[chrome talon]

mhm

[tropic terrace]

so (sqrt(5+sqrt(24)))^x=5+√26 use log

[chrome talon]

Alright

its should be about 2,01745?

Right

?

.close

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[urban topaz]

Is the answer to this B?

[light raft]

yeah think so

[obtuse pebble]

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[spring steeple]

How do I find the sum of the smallest elements of r elements subsets of the set {1,2,3,...,n}?

[royal basin]

let k be a number between 1 and n. in how many sets of r numbers does it appear as the smallest?

[spring steeple]

I guess in one subset only?

Oh wait

[royal basin]

incorrect.

[spring steeple]

(n-k) choose (r-1)?

[royal basin]

yes

[spring steeple]

Oh ok and any element greater than or equal to (n-r) cannot be a smallest element? thus k ranges from 1 to (n-r-1)

[royal basin]

n-r itself can be the smallest

even n-r+1 can

anything higher can't

[spring steeple]

Yeah yeah right n-r+2 and higher can't be

F(n,r) = $\frac{\sum\limits_{k=1}^{n-r+2} k\binom{n-k}{r-1}}{\sum\limits_{k=1}^{n-r+2} \binom{n-k}{r-1}}$?

[warm shale]

Normed

[spring steeple]

How do I simplify this to (n+1) over (r+1)?

[obtuse pebble]

@spring steeple Has your question been resolved?

[obtuse pebble]

@spring steeple Has your question been resolved?

[obtuse pebble]

@spring steeple Has your question been resolved?

[obtuse pebble]

@spring steeple Has your question been resolved?

[spring steeple]

<@&286206848099549185>

[spring steeple]

.close

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[candid rock]

express...what lol

oh

[timid silo]

lol

im just a bit confused on the slope part

ive expressed it as finding the roots

and then also finding the coordinates of inflection in terms of k

but i dont understand how to do

part b and c

idk if u here

@candid rock sry for ping would you help or are you busy

ive got a couple more too after

that i need a bit of help w

[candid rock]

yeah sorry, I'm fairly clueless 😦

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[timid silo]

so x>0

[final thunder]

Yes

Wait

x^2-52x-52>0

[timid silo]

oh yh s

so i think we can use qudratic formula here

i got 26+/- 4root39

[timid silo]

im not sure the follow up

would it be just 51?

[final thunder]

,w solve x^2-52x-52>0

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[final thunder]

Well apparently that’s wrong

,w 26-2sqrt(182)

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