help-27

[balmy quest]

calculate i^1,i^2,i^3,i^4,i^5......, then you will find the regulation

[still pendant]

That's 101i

I don't understand what to do next

@balmy quest

[balmy quest]

you know $i^1=i, i^2=-1, i^3=-i, i^4=1, i^5=i,i^6=-1, i^7=-i, i^8=1, \dots$

[woven radish]

秋水

[still pendant]

No idk that

[balmy quest]

do you learn complex number?

[still pendant]

Yes

Oh

That's i

Ok

Ok

So what now

Uhhmm so i^100 + i^1

[balmy quest]

no, just i^101

101 = 4*25+1

[still pendant]

100/4 = 25 .. the remainder is 0 therefore it's i^0 which equals 1

• i^1 which is 1+ √-1
  So 1+i

So a= 1

[still pendant]

b= 1

(1,1)

[balmy quest]

no, the answer is (0,1)

[still pendant]

Explain pls

How

[balmy quest]

The period of the cycle is 4, and the sum of these four is 0. right?

we calculate $i^1+i^2+i^3+i^4+\dots+i^{101}$

[woven radish]

秋水

[balmy quest]

101 = 25*4+1

so we have 25 periods, and their sum is 0

the left is $i^{101}=i$

[woven radish]

秋水

[still pendant]

How did u calculate the number of cycles

Oh ok

100/4

[balmy quest]

a cycle has 4

101 = 25*4+1

[still pendant]

Alright so 4 * 25 means that there are 25 cycles ?

100 cycle 😭

Wait

I did it right tho

[balmy quest]

I mean i+i^2+i^3+i^4 is one
i^5+i^6+i^7+i^8 is another

[still pendant]

i^100 is 1
i^1 is i
1 × i

[balmy quest]

sum of every 4 terms is 0

yes, i^101 = i

[still pendant]

Yes my method is simpler ig lol

[still pendant]

Yes my method is simpler ig lol

Or is it just wrong lmao 😂

[balmy quest]

where is wrong?

[devout snow]

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[devout snow]

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[light nova]

Hello

[merry seal]

hi

[light nova]

Question is sending

Omg it won’t send

[merry seal]

do you understand this part?

[light nova]

No

I’m lost Ngl

Wait

Are the b’s

For the angle

[merry seal]

it said angle in question

[light nova]

Ye

But like what u drew

[merry seal]

yeah

b is angle

[light nova]

Are u drawing those to show what angle it is like obtuse

I just need to know the measure of b

<@&286206848099549185>

[shut brook]

You need help with his?

[light nova]

Yes

I need my memory refreshed

[shut brook]

Ok

Do u know about alternate interior angles in a transversal?

[light nova]

I kinda got a understanding of it

[shut brook]

Ok... So basically, alternate interior angles in a transversal are equal, right?

[light nova]

Yes

[shut brook]

And <b is the alternate interior angle to the 125° angle, isn't it?

So <b=125°

[light nova]

So

Do I not have to like solve anything else

[shut brook]

I don't think so... It's a pretty direct answer

[light nova]

Okay

I thought there was more to it

[shut brook]

You can take linear pairs and do it if you want to elaborate

The angle to the left of 125 will b 55 cz they're supplementary
And since b is on the same side as the 55 angle, they're supplementary too

So you can get b that way

Or the angle corresponding to 125° is vertically opposite to b

[light nova]

Ahh

I think I’m starting to get it

[shut brook]

👍

And for 5, they're vertically opposite angles, so they're equal... Or u can again take a linear pair as said before

[light nova]

.close

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[light nova]

.reopen

[devout snow]

✅

[light nova]

Okay

.close

[devout snow]

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[ashen vault]

I tried to make a simple integral, and i got weird results

[ashen vault]

Where is a mistake here?

[stiff ruin]

@ashen vault Chain rule

[devout snow]

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[merry stream]

howdy

[merry stream]

these two expressinos are asymptotic to each other right?

I was told that (given bottom expression) you can find something asymptotic to it by taking leading term of top / leading term of bottom

or is this a better asymptotic expression (asymptotic to the big one)

[torn vessel]

no?

[torn vessel]

that's better but still not correct

[merry stream]

okay

so like

how can I make it better?

[torn vessel]

notice that there are two terms in the denominator that have x^3? They are the same 'degree'

[merry stream]

yes

[torn vessel]

so they are both part of the leading term

there's 2 of them

[merry stream]

oh so 2x^3?

wait

[torn vessel]

2root(x^3) yes

[merry stream]

(2x^3)1/2

and the numerator leading term is 3rootx^2 yes

double checking

[torn vessel]

cuberoot(x^2) yes

[merry stream]

coolio thanks

and btw do you happen to know the corrolay to the nth term test?

[torn vessel]

i don't think so

[merry stream]

no worries thanks again

.close

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[pale crater]

what should i do after x^2 = 4/3y -7/3

[tiny carbon]

$g(x) = \frac{4}{3}x - \frac{7}{3}$

[woven radish]

᲻᲻

[tiny carbon]

This?

[pale crater]

yes

but it was in parentis

[tiny carbon]

Yes that is the same thing

[pale crater]

brackets i mean

oh okay

[tiny carbon]

So you can say g(x) is y, since thats really what it means

$y = \frac{4}{3}x - \frac{7}{3}$

[woven radish]

᲻᲻

[tiny carbon]

Then swap x and y, and solve for y

$x = \frac{4}{3}y - \frac{7}{3}$

[woven radish]

᲻᲻

[pale crater]

is it y = 4x-7/3

[tiny carbon]

$y = 4x-\frac{7}{3}$

[woven radish]

᲻᲻

[tiny carbon]

Like this?

[pale crater]

no like 4x is on numerator

[tiny carbon]

Then you need to use parenthesis...

$y = \frac{4x - 7}{3}$

[woven radish]

᲻᲻

[pale crater]

yea like that

[tiny carbon]

Let me check

I got a different answer.

Can you show your steps?

[pale crater]

ok

[tiny carbon]

To getting (4x - 7)/3

[pale crater]

then

after that

y = 4y-7/ 3

[tiny carbon]

What

Why is there an x^2

[pale crater]

shouldnt you do it?

[tiny carbon]

No?

[pale crater]

oh

[tiny carbon]

That makes no sense

[pale crater]

thats sqrt

right?

[tiny carbon]

No?

You don't need square or square root

In this

This is pure +, -, *, /

Go back to the beginning

$x = \frac{4}{3}y - \frac{7}{3}$

[pale crater]

4/3 -7/3 = 1

[tiny carbon]

Oops

[pale crater]

-1

[woven radish]

᲻᲻

[tiny carbon]

Solve for y, get it by itself

[pale crater]

you switch side and then it becomes y= 4x-7/3

[tiny carbon]

Nono

What I wrote is after switching

Solve for y in that equation I just put

[tiny carbon]

.

this

[pale crater]

ok

x= -3y/3?

or maybe x= 4y-7/3

[tiny carbon]

Seriously, brackets are needed

4y-7/3 is NOT THE SAME AS (4y - 7)/3

Also, I don't know how you are coming to these conclusions because you aren't showing your work

[pale crater]

.close

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[swift rampart]

Hey

[swift rampart]

How can I use vietta law if that's the final answer for my question

Vieta's law/formula

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[low nexus]

hello

i'm new to logarithmic equations

i kinda skipped over it when calc came along

can someone instruct me on how to go about solving these?

please ping me when you have a respons

[wooden zodiac]

did u look at log laws

[low nexus]

nope

i should right

[wooden zodiac]

yes

[low nexus]

these good?

[wooden zodiac]

yes

[low nexus]

so uhh

we can move log_9(6x-1) to the other side of the equation

is that the first step?

[wooden zodiac]

ummm I don't think so

u can simplify left term

look at last law

[low nexus]

ah i see

uhhh

should it should be

1/2log_9(x)

$1/2log_9(x) -log_9(6x-1)=0$

[woven radish]

aguaman

[wooden zodiac]

no

M is x^(1/2)

p is 2

[low nexus]

shouldn't p be 1/2?

[wooden zodiac]

no the form u see is p*logb(M)

so u have x^((1/2)*2))

[low nexus]

huh???

ok nvm i'll figure this out myself

.close

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[uneven wigeon]

I am once again asking

[uneven wigeon]

how do i even start tackling this

i have been trying to figure it out for a while now and i think im doing it wrong

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[pallid tusk]

In statistics, if I measure as y a quantity b that will be offset by a number within the range -v to +v, and I can take many measurements of y all the while knowing what v is, how can I determine when to stop taking measurements and approximate b as the average of the various y's I measured?

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@pallid tusk Has your question been resolved?

[pallid tusk]

c.lose

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[cedar schooner]

Let R be integral domain but not a body and

[cedar schooner]

$a \in\ R \ ({0} and R*)$

[woven radish]

Eichhorst

[cedar schooner]

show that a and t are coprime in R[t]

.close

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[restive river]

how did you get b - c = k (sin B - sin C) in blue box from red ?

[pallid vector]

From the {\color{red}{red box}},
$$ \frac{b}{\sin B} = k : \text{and} : \frac{c}{\sin C} = k$$

[woven radish]

Si Arya

[restive river]

what did they do to get b-c ?

[pallid vector]

@restive river
To get b, multiply both sides by sin B from the left equation.

And similarly, multiply both sides by sin C from the right equation and you get c.

Then substitute the value of b and c

[restive river]

k thx

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[brazen hill]

I need to proof this

[steel sage]

Prove it by definition

[brazen hill]

I am stuck, I am supposing that there is a x in the first intersection but then I dont know what i should do

[restive river]

What does the definition say?

[brazen hill]

{x | x ∈ Aα for all α ∈ I}

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[brazen hill]

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[fleet apex]

Can anyone please explain what is parameters is in linear programming

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[vagrant crest]

How do u do this gain I forgot

Question 7

[merry seal]

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

[golden vale]

@vagrant crest difference of 2 squares, like

$(x+1)(x-1)=x^2-1$

[woven radish]

Hyperlix

[golden vale]

and then just cancel them

[merry seal]

using 1 might confuse them because they have a 9

[golden vale]

same concept, it's just one example

[devout snow]

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[vagrant crest]

Oh so I should just times the numerator and denominator?

Ohhh

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[lone sky]

hello can someone help me with this series convergence question? I’ve included my work but im so so so lost hfjdkfghj idk how to do these :,)))

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[lone sky]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[restive river]

what u need help for

[lone sky]

How to integrate with the substitution they give, and integrate by parts twice

I don’t understand how they’re getting that solution

I tried integrating by substitution as above, but I’m not getting anywhere

If I’m not queuing right please tell me

No one is answering my question

If no one knows or I’m not asking right please tell me

[restive river]

whart

wahta

what

im here

lets see

[gentle estuary]

$\int_{1}^{\infty} \frac{u}{e^u} \cdot 2u \ \dd u = 2 \int_{1}^{\infty} u^2 e^{-u} \ \dd u$ after the first substitution

[woven radish]

castroploiin

[gentle estuary]

Applying integrating by parts once leaves you partly with the integral of u e^(-u). This is where you apply integration by parts again

[restive river]

first u need to do this

ull get this

u got it?

@lone sky

[lone sky]

Yes wait I’m trying it (sorry I write slowly)

[restive river]

here u can quick understand it

just by replacing

[lone sky]

but there’s no (x)^-1/2 to put in du though? that’s my issue

[restive river]

u want step by step?

[lone sky]

That would be super helpful :,) im so sorry idk why im not getting this

[restive river]

dont be sorry

[lone sky]

hfjdkfh thank you :,))

[restive river]

@lone sky u got it?

[lone sky]

wait I am processing

ohhhhhhhhhhhhhhh

I GET IT I understand what I was missing

[restive river]

tell me

[lone sky]

Yes let me write show you

I needed to write it in terms of dx (here, it’s dn), instead of writing it in terms of du

Because then I was missing the step in the blue arrow, and manipulating the integral wrong (my last line)

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[proven acorn]

I have to find the length of CB but i do not even know where to start.

I dont know what strategy/steps to use to solve the problem

[small field]

what's the question

the original one

[proven acorn]

Thats the original

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[light nova]

Ello

[light nova]

Can some one help me find the distance between (0,4), (-6,6)

[topaz axle]

6.325

[stuck field]

Don't do people's work for them.

[stuck field]

Do you know distance formula?

[untold marlin]

The difference in the x coordinate is the horizontal distance, the difference in the y coordinate is the vertical distance and then that makes a right angled triangle you can use Pythagoras on

[topaz axle]

offer declined

[stuck field]

What offer?

No one's giving you any offer.

[topaz axle]

you offered advice?

[merry seal]

visuals!

also note that $|a|^2 = a^2$

[woven radish]

Doggo

[merry seal]

heyy, the answer is 3

[stuck field]

You might wanna review what offer means, that wasn't an ”offer”. It's not something you get to choose, it doesn't make sense if someone posts how to do a sum and you don't tell them how rather give the answer, that was my point, that's all.

[merry seal]

does that mean anything to you?

I just said 3

woaah it's a number, where's the work though

[stuck field]

Exactly.

[topaz axle]

i don't get it

[stuck field]

You would if you knew how the server works.

[untold marlin]

👁️

[topaz axle]

usually there's no implied rule that you need to keep server working the way it works

[stuck field]

When exactly did they ask for the answer?

You tell me.

[topaz axle]

it's against the rules to do that if you didn't know

[stuck field]

Do what?

[topaz axle]

like you can't expect an answer

[stuck field]

Exactly

That's my point

Giving answer isn't much issue, you didn't help at all though.

[topaz axle]

i know

[stuck field]

Nice then we're fine here.

[topaz axle]

but what was it if not advice or an offfer?

it all started when you said it's not an offer

[stuck field]

I mean it doesn't have to be an offer.

It was just a statement I threw. I don't expect you to change what you did. Just a statement you read being neutral.

[topaz axle]

ohh

you were irritated and you asked me to not do it in the future

thanks i get it

[stuck field]

No you don't. But it's fine. Not everyone can understand everything after all.

[topaz axle]

true that

[stuck field]

Anyway @light nova sorry for bickering in your channel.

[topaz axle]

sorry ryan

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[merry seal]

$a = \left(\frac{1}{3}\right)^{n+1}$

[woven radish]

Doggo

[merry seal]

then what do you have?

[shell cosmos]

1^n+1/3^n+1

[merry seal]

can you write with latex or send image of that?

[shell cosmos]

uhm

like you put the n+1 over both the 1 and 3

.close

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[tacit cape]

anyone know how i get the perimeter of the square?

[rich heron]

P=20

.close

@tacit cape

[tacit cape]

wait i still dont understand

[rich heron]

Omg you use algebra on the area rule A=a*b/2

[tacit cape]

but i dont know b

[rich heron]

You sub in A=12 and a=6 and solve for b

12=6b/2

12=3b

12/3=b

B=4

Do you get it

Then you use c^2=a^2+b^2 rule to find the hypotenuse

[tacit cape]

hang on food

,close

,close

.close

[devout snow]

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[tacit cape]

.reopen

[devout snow]

✅

[tacit cape]

ok i have food now

[tacit cape]

where do i sub it in

[rich heron]

Sub A=12, a=6 into A=ab/2

Which is 12=6b/2

Solve for b, year 7 algebra

[tacit cape]

ohh i see

hang on im going to try it

one sec

[rich heron]

Then you find hypotenuse

Using Pythagoras

[tacit cape]

did i do it

[elder yarrow]

hello friends can you guys help with 2b i could not prove that the function is odd

[tacit cape]

this is an occupied server

go to the available section and click on one of those

and then post your question

[rich heron]

Yes

[tacit cape]

ok thank you

.close

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[vast abyss]

How do I do this?

[spark sleet]

Develop first

[vast abyss]

wdym?

[spark sleet]

(3a+1)^2

[vast abyss]

expand?

[spark sleet]

Yes

[stone stump]

well 4^2 is not 1 mod 9

[vast abyss]

ok now what xd

[supple knot]

a = 1

1 is a whole number

[vast abyss]

ok but lets say i were to go about proving it

what would i do

[spark sleet]

Haha right

[graceful cosmos]

True that you can expand, and get rid of any terms with a 9. So this does simplify to:
6a + 1 = 1 (mod 9)
6a = 0 (mod 9)

[spark sleet]

This is false

[supple knot]

i don't think it's a good use of time to prove false things

[vast abyss]

how did 6a become 0?

[graceful cosmos]

Reread

Oh, and don't forget that I did expand (3a + 1)² to get my first line

[vast abyss]

wouldnt you reduce 6a+1 by any multiple of 9

and see if u get 1

[graceful cosmos]

That's the same as reducing 6a by any multiple of 9, and seeing if you get 0

(Subtract 1 from both sides)

[vast abyss]

but 6a+1 isnt \equiv 1 right

nor 0

[graceful cosmos]

6a + 1 is equivalent to 1, yes

(mod 9)

[vast abyss]

what

[graceful cosmos]

Don't forget that I did expand (3a + 1)² to get my first line

[stone stump]

well under the assumption (3a+1)^2 = 1 mod 9

which is false

[vast abyss]

oh ok

[graceful cosmos]

(Give that a shot, if you're lost)

[vast abyss]

but I still dont get how you went from 6a+1 to 0

[stone stump]

because we prove (3a+1)^2=1 mod 9 wrong by reducing it to 6a = 0 mod 9 which is wrong

[graceful cosmos]

6a + 1 = 1 (mod 9)
6a = 0 (mod 9)

Subtract 1 from both sides

[vast abyss]

wait so you write 6a+1=1(mod9)

not the equiv 3 lines sign

[graceful cosmos]

Because lazy

[stone stump]

= or equiv is more or less the same. I didnt wanna write equiv

[graceful cosmos]

Interpret as \equiv

[vast abyss]

ugh idk if im stupid but i dont get whats happening here

6a + 1 = 1 (mod 9)
6a = 0 (mod 9)

[graceful cosmos]

I subtracted 1 from both sides

[stone stump]

we assume that (3a+1)^2 = 1 mod 9

then we expand to 9a^2+6a+1 = 1 mod 9

[vast abyss]

yeah I get that part

[stone stump]

9a^2 = 0 mod 9 so we can remove it

[vast abyss]

and then 9a^2 goes because its a multiple of 9

[stone stump]

then we are left with 6a+1 = 1 mod 9

then we subtract 1 from both sides and get 6a = 0 mod 9

and this is wrong, so our assumption (3a+1)^2 = 1 mod 9 must have been wrong

[vast abyss]

1 mod 9-1=0mod9?

oh yeah it is

[graceful cosmos]

Why is there a mod 9 in the middle like that? Haha

[vast abyss]

i just subtracted 1 both sides

[graceful cosmos]

Now, correct me if I'm wrong, I believe I can divide both sides by 2.
3a = 0 (mod 9)

[vast abyss]

I've no idea

[stone stump]

yeah, 2 is invertible mod 9

[graceful cosmos]

And that's as simple as this gets, I think

[supple knot]

,calc 2 * 5 mod 9

[woven radish]

Result:

10

[graceful cosmos]

Technically correct

[supple knot]

damn you calc

[vast abyss]

1

[stone stump]

technically we are multiplying by 5 instead of dividing by 2

[graceful cosmos]

,calc 2*5 % 9

[woven radish]

Result:

10

[graceful cosmos]

Nop

[supple knot]

,calc (2 * 5 % 9)

[woven radish]

Result:

10

[supple knot]

(╯°□°）╯︵ ┻━┻

[devout snow]

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[livid musk]

hey, can anybody help me? I drew the graphs but cannot make it.

[devout snow]

@livid musk Has your question been resolved?

[devout snow]

@livid musk Has your question been resolved?

[wooden veldt]

You haven't shown what the question is

[livid musk]

Question is about polar curves's area

[devout snow]

@livid musk Has your question been resolved?

[supple knot]

.close

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[devout snow]

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[restive river]

How do i find the angles in this particular question?

[devout snow]

Please don't occupy multiple help channels.

[restive river]

nothing drawn to scale/accurately

[merry seal]

I suck at geometry, but I think you can only find $x + y$?

[woven radish]

Doggo

[long kettle]

Yeah I don't think there's enough info

[barren gorge]

Yea

You can't find x and y

Just x + y

[restive river]

then

how

does it say

[merry seal]

is anything parallel or whatever?

[restive river]

x = 85

y = 110

in the correct answers

i dont know how

[long kettle]

Must be you're leaving out key info

[barren gorge]

^

[restive river]

wait

ye

i missed

that an entire cirlce

is around it

ty

[barren gorge]

[restive river]

but

still

how

[wild creek]

is one of the lines the diameter?

[barren gorge]

I hope

[restive river]

no

[long kettle]

Neat property about quadraterals inscribed in circles

Opposite angles add to 180

[restive river]

nvm

ty

makes sense now

.close

.cloes

.cloes

.close

[devout snow]

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[devout snow]

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[past matrix]

Anyone know any good website where I can get exam questions for a given topic?

[past matrix]

I can only find websites with textbook style questions

[devout snow]

@past matrix Has your question been resolved?

[past matrix]

<@&286206848099549185>

[past compass]

You should specify what exam

[past matrix]

Edexcel AS level

[past compass]

Have you tried past papers

[normal crater]

Pmt

Physics and maths tutor

Has PPQs topic specific

[past matrix]

Is that a website or something?

I’ve done whole past papers, but I’m looking sir specific topic related questions

[past matrix]

Will try that thanks

[devout snow]

@past matrix Has your question been resolved?

[devout snow]

@past matrix Has your question been resolved?

[past matrix]

Any more websites? <@&286206848099549185>

[supple knot]

This really isn't the right place to ask for vague resources. Go to #book-recommendations or do your own googling

.close

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[past matrix]

Did loads of googling, and I’m looking for websites not books

[devout snow]

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[fervent jewel]

I found U(20) = {1,3,7,9,11,13,17,19}

ANd I know the identity is 1

is there a quick way to find the order of the elements other than, say for 3

$3^1 = 3 \ne 1$

[woven radish]

Kurama

[fervent jewel]

$3^2 = 9 \ne 1$

[woven radish]

Kurama

[fervent jewel]

etc..

and checking until I hit 1 multiple of 20?

Actually nvm. They mantually tested the powers in the textbook

...

.close

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[lime linden]

for f(x, y) = (x^2)(y^3) if we are taking the derivative with respect to x shouldnt we treat y as a constant, and that would make fx = 2x?

[stone stump]

d/dx f = 2x, yes

or f_x

or whatever your notation

[lime linden]

on the vid

the guy doin the prob had 2x * y^3

but the question was asking me to find the slope for fx and fy

given a point

would that be the reason why?

[stone stump]

wait addition or multiplication

either way we treat y as a constant

when differentiating with respect to x

[lime linden]

multi

[lime linden]

yeah but if we are tryna find the slope of fx given a point, will this still stand?

[stone stump]

well first we differentiate and then sub in the given coordinates into the derivative to get the slope

[lime linden]

so it wouldnt stand?

y wouldnt be a constant?

since we need it to find the slope given a point?

[stone stump]

well that's the same thing as in one dimension. to find the slope of f(x) at x=2 we calculate f'(x) and then sub in x=2

first we just calc the derivative in general

not at a specific point

but then to find the slope at a specific point we have to sub in specific values

and while we are calculating the derivative we are treating y as a constant. just like we would treat a 2 or whatever

[lime linden]

yeah so keep y, and not get rid of it right?

<@&286206848099549185>

[stone stump]

well treat it like a constant while differentiating with respect to x. if that means getting rid of it (for example if we have something like x^2 +y) then get rid of it, otherwise leave it just like you would leave a constant

after you have done it, to actually use it to calculate the slope at a specific point, you have to substitute in the relevant y value of the point you are looking at. same as having to substitute the relevant x value

[devout snow]

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[leaden aurora]

Could someone explain to me what the difference between upthrust and drag is? When travelling through a liquid. And do both forces act simultaneously when you travel through a liquid

[devout snow]

@leaden aurora Has your question been resolved?

[leaden aurora]

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[celest tide]

How do I solve $\log_{x}\bigg({\frac{5}{2} - \frac{1}{x}}\bigg) > \bigg(\frac{5}{2} - \frac{1}{x}\bigg) \newline$
I know that the domain is $x > \frac{2}{5}, x ≠1 \newline$
I have to take cases for the base, when $x > 1$ and $\frac{2}{5} < x < 1 \newline$
I am stuck at this point

[woven radish]

kinglacto

[devout snow]

@celest tide Has your question been resolved?

[celest tide]

<@&286206848099549185>

[night cosmos]

I actually tried this problem, it's a bit hard

[celest tide]

yea

[maiden bridge]

I am trying rn

give me some time

[devout snow]

@celest tide Has your question been resolved?

[normal crater]

need to approximate with newton-raphson

intersection do not have closed forms

[maiden bridge]

https://www.sarthaks.com/2045155/which-of-the-following-is-not-the-solution-log-x-5-2-1-x-gt-5-2-1-x

@celest tide

I have found a solution online

but can't really how they started

see if anyone can explain the steps

[normal crater]

you can only approximate

[celest tide]

yeah i checked that out too but didnt understand 😓

[celest tide]

hmm dunno what that means

ill try looking it up

[normal crater]

you won't be able to get an exact answer

you will need to use numerical methods such as iteration or newton-raphson to find answer

[celest tide]

but the answer key states $x \in (1, 2) \cup (\frac{2}{5}, \frac{1}{2})$

[woven radish]

kinglacto

[celest tide]

which seems exact?

[celest tide]

.close

[devout snow]

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[devout snow]

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[main gull]

Please read #❓how-to-get-help
You have to wait at least 15 minutes before pinging helpers

You don't get special privileges to ping early. Rules exist for a reason

No need to be hostile

[neat coral]

bro is probably at an exam lol

[restive river]

what a jerk

i blocked them dw i block cringe

[neat coral]

can't even solve a first grade question and starts insulting people

[main gull]

And all I said, was they needed to wait at least 15 minutes before pinging helpers

[tawdry trellis]

<@&268886789983436800>

Hate this stuff

[wicked rover]

they left but ty

[small field]

$$\int_{\pi}^{3.1}\left(\cos\left(x\right)^{2}+\tan\left(x^{0.5}\right)+\frac{3x^{2}}{2x}\right)dx$$

[woven radish]

Chunkin

[small field]

I think this is the answer

[small field]

ban them incase they come back?

[wicked rover]

discord cant do that by default

i think the mathbot can do it, ill do that

[small field]

man pulled up in calculus

💀

[small field]

🇼

[devout snow]

@trim island Has your question been resolved?

[neat coral]

yes his question has been resolved

[wicked rover]

.close

[devout snow]

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[tepid aspen]

Let's say Bob sells 3 million balls at price of $30 each while Ann sales 25 million balls at price of $10. so in total they both make $340 million. (Bob $90 million and Ann $250 million). By how many more balls should Bob sell in order to get total earning from $340 million to $400 million? worth to mention, let's say if Bob sells extra 1 million balls, then Ann will sell not 25 million, but 24 million.

[tepid aspen]

help please

[hybrid snow]

That's a very confusing statement

Are you saying that if Bob sells 1M more balls, then Ann will decrease her sales by 1M balls

So no matter what, they end up selling the same number of balls?

@tepid aspen

[tepid aspen]

@hybrid snow yeah exactly

[hybrid snow]

Okay so

Set up a system of equations

[tepid aspen]

B + X = A - X = 400 million?

[hybrid snow]

Let $b$ be the number of balls, in millions, that Bob sells; and let $a$ be the number of balls, in millions, that Ann sell.

$$a+b = 28$$
$$30a + 10b = 400$$

[tepid aspen]

should i now combine those two equations?

i'm very bad at linear equations

😦

[hybrid snow]

Just solve it like a normal system of equations

Solve for one variable in terms of the other, substitute it

[tepid aspen]

30a + 10 B = 28 + 340?

i replaced a + b with 30a + 10b

[hybrid snow]

It doesn't work like that

[tepid aspen]

oh, sorry for that. just got stuck

[hybrid snow]

Also@my bad

I did a funny

Hold on

[woven radish]

Umbraleviathan

[hybrid snow]

So what I'm going to do is because I want to find out what $b$ is, I need to rearrange the equation:

$$a + b = 28$$
To:
$$a = 28 - b$$

I am going to substitute it into the other equation:
$$30(28-b) + 10b = 400$$

[woven radish]

Umbraleviathan

[hybrid snow]

And you should be able to solve it from there

[tepid aspen]

oh how did u do that?

840-9B= 400

9b = 1240

[hybrid snow]

Right here

Goes back to knowing equations and their properties

[tepid aspen]

by 150k

[hybrid snow]

Wdym by 150k

[hybrid snow]

Check your work

This isn't correct

[tepid aspen]

oh wait a second

840 - 30b +10b = 400

840 - 20B = 400

1240 = 20B

62

[hybrid snow]

No

What is 400-840

[tepid aspen]

-440

[hybrid snow]

What is that divided by -20

[tepid aspen]

21

[hybrid snow]

Check again

[tepid aspen]

22

my bad

[hybrid snow]

Yeah

So Bob needs to sell 22 million

HOWEVER

the question asks "by how many more balls does Bob need to sell"

So 22-3 = 19

Bob needs to sell 19 more balls

[tepid aspen]

but 22 * 30 gets us = 660 million

[devout snow]

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[devout snow]

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[devout snow]

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[restive river]

If I had an integral
$(\int(f(x)dx)^(-1)$

[restive river]

Is there any way to bring the exponent inside?

[woven radish]

7aman

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

?

Is there a general rule for this or is it case-by-case?

[devout snow]

@restive river Has your question been resolved?

[supple knot]

No

[supple knot]

No general rule

[restive river]

🗿

thanks

.close

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[terse haven]

Translate;

Ada wanted to write the integral question written on the paper in her notebook, but could not see the whole solution of the problem because the left side of the paper was torn.

It cannot exactly see the boundaries of the region up to the point 3 on the x-axis under the curve.

According to this, what is the integer value that m can take?

[merry seal]

you solve it like you'd solve any other integral, then evaluate from m to 3 and you get a cubic in terms of m I think?

[terse haven]

I tried but couldn't find the result

[merry seal]

mhm so where did you get stuck exactly?

[merry seal]

2(3)^3 + 2(3) = 54 + 6 = 60

not 55

other than that it seems fine

[terse haven]

3^3 27

Lolll

İ though 27x2 49 lmao

İ am idiot

Thank you bro

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[quiet mulch]

any help with the answer?

[fluid elbow]

You can use the identity $a^2 - b^2 = (a - b)(a + b)$, and note that $9x^2 - 36 = (3x)^2 - 6^2$.

[woven radish]

pi over four

[quiet mulch]

i do not understand 😅

[supple knot]

have you heard of FOIL?
https://www.cuemath.com/foil-formula/

[quiet mulch]

is the answer the last choice? (3x+6)(3x-6) not sure

[fluid elbow]

Yes

[quiet mulch]

alright now i understand

thank you for explaining it

[quiet mulch]

can u explain more

.help

[devout snow]

Commands:
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help: .help

Type .help <command name> for more info on a command.

[quiet mulch]

.close

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[quiet mulch]

.delete

.open

.reopen

[devout snow]

✅

[quiet mulch]

i need help with this question!

[devout snow]

@quiet mulch Has your question been resolved?

[cursive smelt]

When factorising you look for a common factor so what do those both have

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[hybrid snow]

Bucko we can't help if you don't even post a question

[eternal delta]

[hybrid snow]

@ocean valley what do you need help with

Yeah if you're just gonna troll

<@&268886789983436800>

Third troller

[wicked rover]

ty

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[hybrid snow]

I feel bad having to ping you guys 2 times today lol

[wicked rover]

no need, its our duty

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[restive river]

How do i solve part C?

[restive river]

i tried solving this way:

but i keep getting the wrong answer

[vernal yew]

g(g(x)), right? When you put g(x) inside itself, expand it out entirely

[restive river]

like this ?

[vernal yew]

How did you get that?

(a+b)²=(a²+2ab+b²)

[left robin]

$g(x)=x^2+1$
$g(g(x))=(x^2+1)^2+1$

[woven radish]

~Martin

[left robin]

so
$g(g(x))=x^4+2x^2+2$

[woven radish]

~Martin

[left robin]

basically you take g(x) and substitute x with another g(x) where x has been

[restive river]

oh alrightt

thanks

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[bright ice]

So lets say I have 3 numbers, each with a number probability from [0-1], for example [0.1, 0.5, 0.2]
How can I make a random selection wherein the highest probability number has the highest probability of being selected? What i've been doing is generating a random number r and testing the absolute difference between r and the probability which I just now realized is not the right way to do it

[bright ice]

Basically how to do accumulated probability interval calculations

[graceful cosmos]

What is a "selection" here?

Like a subset of your three numbers?

[bright ice]

The result given a generated random number [0-1]

[bright ice]

Yep a single return val

[graceful cosmos]

Oh I see. Split your choices into their own intervals

So with your example above,
If r is between 0 to 0.1, return #1
If r is between 0.1 to 0.6, return #2
If r is between 0.6 to 0.8, return #3
Aaand I'm just realizing your example doesn't sum to 1, so that's possibly a problem you might be running into?

[bright ice]

Okay I think I can do that, I can restrict the sum to max 1 which is something I also forgot to do

[graceful cosmos]

I can fix your example by multiplying all probabilities by 1/0.8

Where 0.8 is the current sum, and 1 is the new sum.

[bright ice]

Are you pretty familiar with coding? I'm tring to think of a way I can write that nicely

[graceful cosmos]

Some. Go ahead

[bright ice]

Yeah I know what the max sum would be so I can normalize accordingly

[graceful cosmos]

Another way, which might be simpler from a coding perspective, would be to subtract intervals off until your number is negative

[bright ice]

closest = abs(probability - r) for each number wouldn't work right?

[graceful cosmos]

So let's say r = 0.4.
Subtract off the first number. Now r = 0.3
Subtract off the second number. Now r = -0.2
Since r is now negative, return #2

[bright ice]

I'm not sure I'm following that exactly

do you subtract to find the greatest difference?

between r and the number

[graceful cosmos]

You don't want the closest probability, no.

So let's say the probability of selecting between three outcomes is given by
[0.2, 0.5, 0.3]

And your random number generator hands you r = 0.3

[bright ice]

alright

[graceful cosmos]

With the interval method above:
If r is between 0 to 0.2, return #1
If r is between 0.2 to 0.7, return #2
If r is between 0.7 to 1, return #3

[bright ice]

Ahh alright

[graceful cosmos]

Will accurately simulate the probability of each outcome

In other words, [0.2, 0.5, 0.3] should give the size of each interval

[bright ice]

How does the subtraction method work with that? Seems like the right approach for coding it

[graceful cosmos]

Subtract off the sizes of each interval until you get a negative number

r = 0.3 with outcomes [0.2, 0.5, 0.3]
Subtract the first, r is now 0.1
Subtract the second, r is now -0.4
Since r is now negative, return the second outcome

This is just another way to test which interval r falls into, just a bit easier from a coding perspective, I think

[bright ice]

Agh that's not really clicking for me, surely this would result in

a bias towards the start of the array right?

[graceful cosmos]

If r is smaller than the first element of the array, then you'd correctly return the first outcome

If r is larger, then you'd continue to the next element

[bright ice]

Ohh okay I think I'm getting it

let me whiteboard it for a second

@graceful cosmos Awesome works great 👍

I went for like a week on my project doing it the wrong way not knowing why everything was broken, hopefully this is the reason haha

[graceful cosmos]

Cool! Feel free to ask if you have any other questions on it.

[bright ice]

will do thanks!

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[teal gust]

Hi. I have a test on factoring and inequalities coming up, and atm I still have no idea how to do this, especially the first one since it's a parabola I believe? Please help out. Thank you. :)

[bold prism]

y < x^2 means it'll be under the parabola

try to draw the parabola first then pick any point outside it

if that point satisfies that inequality

that means that area needs to be shaded

[teal gust]

Wait, so how do I find the x-intercepts (or y) of the parabola?

[bold prism]

set y = 0

solve for x in x^2 = 0

[teal gust]

so x is 0?

[bold prism]

thats your x intercept

yes

can you draw the parabola now?

[teal gust]

Do I need anything else?

maybe another point, Ig?

[bold prism]

nope

as long as you know the graph of x^2

[teal gust]

Ah, I see.

Thank you.

Let me try this out real quick.

:0 I think I have it, tysm!

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[ruby shell]

hu

[ruby shell]

anyone there

[bold prism]

Yes ffs

post your question

[merry seal]

nah noone

[ruby shell]

wait lemme type

[bold prism]

dont open multiple channels, ping helpers without sending your question and waiting

you've broken multiple rules in less than 10 seconds

read #❓how-to-get-help 🗿

[ruby shell]

tan-1(log4/5(5x^2-8x+4)

domain and range of the above fn

[bold prism]

did you try anything?

[merry seal]

I'd start with the denom. on inside

[ruby shell]

4/5 is the base and rest is argument

[ruby shell]

unfortunately no can't

[merry seal]

aah

[ruby shell]

no idea how to proceed

[merry seal]

well you want to make sure the inside of log is positive, assuming no complex numbers are involved

[bold prism]

start with every composition and determine its domain

[ruby shell]

yes

so that quad>0

[merry seal]

the domain of inverse tan is all real IIRC

the range should just be 0 to 2pi I think..?

[bold prism]

$f(x) = tan^{-1}(log_{\frac{4}{5}}(5x^2-8x+4))$