help-10

[sharp heath]

decrease

but

[ruby finch]

But?

[sharp heath]

why does r! reflect the number of combinatinos overlapped or whatever

hmm thats not worded so well is it

[ruby finch]

You should probably search a derivation of the formula

[sharp heath]

okay its probably not worth my time rn i have an assignemnet wit this tmrw

i dont need to know details i just like math

[ruby finch]

Do that some time later then lol

[sharp heath]

yea

so

would it be

6!/4!2!

[ruby finch]

For this order does matter

[sharp heath]

yes

im aware

[ruby finch]

Yeah

[sharp heath]

im not using any formula

was i wrong

if they were all different colours

it would be 6!

but since 4 overlapping red and 2 overlapping blue

6!/4!2!

correct?

[ruby finch]

Correct

[sharp heath]

okay thanks

so

it would be

8!/2!2! + 8!/2!2! ?

[ruby finch]

remove the duplicate letters

[sharp heath]

i thought i did

[ruby finch]

except e

[sharp heath]

yea

2 m 2 t

/2!2!

[ruby finch]

9 letters total

so 9! permutations

[sharp heath]

well

if you lock an e at the beginning

its 8 letters

and then you repeat with an e at the end

[ruby finch]

Yeah you're right with that logic

[ruby finch]

And this should be correct

Wait a minute

[sharp heath]

im just wondering

[ruby finch]

You can't add them

[sharp heath]

wouldnt u have to subtract something at the end

[ruby finch]

Subtract what and why?

[sharp heath]

because some permutations with e at the start may overlap with permutations of e at the end

so maybe like

[ruby finch]

I think only one of them will

[sharp heath]

8!/2!2! + 8!/2!2! - 6!/2!2!

no

im pretty sure im right acc

because

assuming

no

8!/2!2! + 8!/2!2! - 7!/2!2!

because

the first part makes up for

e - - - - - - - -

the second part makes up for

• • • • • • • • e

so the third part removes

[ruby finch]

Just a minute

[sharp heath]

e - - - - - - - e

which would be 7 different permutations

and 2 doubles

7!/2!2!

i think i did it

[ruby finch]

Right, sorry

I had to go for a minute

[sharp heath]

np

oh wait theres answers at the bottom

[ruby finch]

What does it say?

[sharp heath]

the answer for this question is apparently just 8!/2!2! + 8!/2!2!

20160

[ruby finch]

Hmm

[sharp heath]

hmm indeed

can i ping advanced 👀

[ruby finch]

Not sure

but you can ping helpers

[sharp heath]

thats a lot of people tho

[ruby finch]

It's allowed I think

every 5 minutes

[sharp heath]

<@&286206848099549185>

answer on the page says 8!/2!2! + 8!/2!2!

[stoic trout]

Yeah one where it starts with e other where it ends with e

[sharp heath]

yes but what about the ones that overlap where it starts and ends with e

[ruby finch]

I am pretty sure there will be common cases

if you pin e and e to the end and start

those are going to be the common cases

[sharp heath]

yea

[stoic trout]

Hmm

[sharp heath]

im convinced my teacher just did it wrong

[raven spire]

um, you can just do 8!/2!2! + 8!/2!2! - 7!/2!2!?

[sharp heath]

i was thinking that

[ruby finch]

Yeah that's what we think

Yeah you're right

[raven spire]

ooo the answer must've been wrong or smth... confirm with your teacher?

[sharp heath]

i will

[ruby finch]

The answer doesn't take the common ones into account

[sharp heath]

yea

thanks guys

i will close for now

maybe reopen

[ruby finch]

👍

[sharp heath]

.close

[obtuse pebble]

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[past solar]

School A has Increased 10% of Male students and increased 30% of Female Students. and the Total students increased by 25%, how many are Male?

[raven spire]

You just need to frame correct equations, let x = number of males initially and y = number of females

the question asks you the percentage of males finally

[past solar]

nvm the question asked what percent of the total number were male

[raven spire]

Yeah

[past solar]

hmm im stuck on constructing the equation

[ruby finch]

take x as the amount of males and y as the amount of females

[past solar]

Ok Ok

[ruby finch]

Then increase them by 10% and 30%

Are you sure the question is correct?

The total amount of students aren't mentioned

[past solar]

yes the question was School A has a 10% increase of Male students then 30% of Female students then there is a 25% increase of total student, What percent of the total number is a male

[ruby finch]

1.1x + 1.3y = 1.25(x+y)

The problem is how do you do anything with this

[past solar]

hmmm

[obtuse pebble]

@past solar Has your question been resolved?

[ornate dove]

Common factor?

I'm not totally sure though

[past solar]

wdym

[ornate dove]

I personally feel that there might be integers at which have smth like a common factor

Wait nvm

Idt it works

Becoz of the 25%

[past solar]

hmm it looks like 2x + 6y = 5x + 5y

[raven spire]

y = 3x yes

:o

[past solar]

does it only have 1 solution?

[raven spire]

mhmm

1.1x/1.25(x+y) = %age of male finally

[ornate dove]

Wait

[past solar]

22%

[ornate dove]

Is the answer in alg. or no.

[past solar]

nope

an integer

[ornate dove]

H.

Hm*

[past solar]

y=3x has only 1 sol right and its 6 and 2

which means that 1.1 x 2 / 10 = % of male

2.2/10 = 0.22 which is just 22

is it though

[raven spire]

whaat? y = 3x has only one solution?

why not x = 1, y = 3? why not x = 3, y = 9?

[past solar]

still .22

I forgot I watched i vid with y=3x then 3y= x vid

[obtuse pebble]

@past solar Has your question been resolved?

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[rich holly]

where are these from?

[alpine raven]

they just multiply by 1

(x+3)/(x+3) = 1
you should just ask yourself, why they did that

[ruby finch]

Common denominator

Oh

Yeah, common denominator

So you can subtract the fractions

What you do to the numerator should be done to the denominator and vice-versa

otherwise you would change the fraction

[rich holly]

thought you had to apply it to everything

like 4

to make it equal

[obtuse pebble]

@rich holly Has your question been resolved?

[obtuse pebble]

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[shadow trellis]

why’s this method wrong:

so the answer = 0 bc 0/tending to 0 = 0

[raven spire]

what

No

[hazy marlin]

There is no zero tending to zero

they both tend to zero

therefore you have an indeterminate form

[shadow trellis]

lol okay, thanks

so sinx/x as x tends to 0 is not exactly = 0

?

[hazy marlin]

yea

it actually converges to an actual number

[shadow trellis]

wdym?

[hazy marlin]

sinx/x as x tends to 0 is 1

[raven spire]

Also, I think what you've done is tantamount to: $\lim \limits_{x \to 0} \frac{1}{x^2} \frac{\tan x - \sin x}{x}$

[warm shale]

Ansh

[hazy marlin]

but your intuition is right, when both the nominator and denominator tend to zero, it's never actually zero therefore it's an inconclusive form

[raven spire]

which is wrong because by product law of limits, you can only multiply lim PQ = lim P lim Q, iff P and Q are finite limits

[shadow trellis]

umm so wdym by finite limit?

is the 1/x^2 the problem here?

[hazy marlin]

a number

Limit arithemtics can be applied only if the limits exist

[raven spire]

I think

[hazy marlin]

I mean

if one limit is a number, and the other is divergent

[shadow trellis]

sorry wdym by divergent? never heard that before

[hazy marlin]

divergent? approaches infinity?
w/e
then you can conclude the limit approaches infinity

[raven spire]

then the product .. need not be divergent right? You have a clear example here

0 \times infty - ...? does lim PQ diverge though 👀

[hazy marlin]

well.. 0 can be excluded and it gets infinitely small

but if it's a real number

say 3

[raven spire]

hmmmmmmm

Right

[raven spire]

Here :D

[ruby finch]

Not exactly sure but you can make it into the subtraction of two limits

[raven spire]

not here kirby, he needs to use the maclaurin series

[ruby finch]

Ah ok

Sorry

Just ignore what I said

[hazy marlin]

omg I completely forgot maclaurin is a thing

[raven spire]

Lmao

[hazy marlin]

I think l'hopital is also viable tho

it might.. get just a bit messy but not too bad

[raven spire]

Basically, you use @shadow trellis

upto x^3 cause that's the degree of denominator anyways

If in future you come across say, a limit that goes something like (a sin x + b tanx )/x^5 or something

you can always use higher powers

[shadow trellis]

right

[hazy marlin]

l'hopital works too
verified

[shadow trellis]

wait so my soln was wrong on these terms:

1. there was no 0 and tends to 0
2. I couldn’t just multiply bc for 1/x^2 the limit did not exist

[hazy marlin]

yesssssssssssssss

there was no 0 tends to 0, everything was tending to zero
thus indeterminate

[shadow trellis]

okay, thank you

.closed

.close

[obtuse pebble]

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[simple grail]

Can someone verify if I did this correctly? The question said sec(θ) = -13/5 and θ is in the 3rd quadrant. I have to find sin(2θ), cos(2θ) and tan(2θ). I got 120/169 for sin(θ), -119/169 for cos(θ), and 600/595 for tan(θ)

[obtuse pebble]

@simple grail Has your question been resolved?

[obtuse pebble]

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[sudden spindle]

I need some help verify the answer of this question

[sudden spindle]

Like I am not completely sure if I did it correctly

[royal glacier]

You did it right

[sudden spindle]

do u mind explaining how it work tbh i kinda forget the process

[alpine root]

oo chinese

I finally understand something 🥲

but not the math

🌝

[west sierra]

5x+5 is incorrect

[sudden spindle]

the button or upper

[west sierra]

What is button

[sudden spindle]

Buttom

uh

The lower

[west sierra]

Both bottom and top seem to have 5x+5 as answer

[sudden spindle]

yea

[west sierra]

Which is incorrect

The question is $(5x^2+5x+1)/x$

[warm shale]

Ji Ning

[sudden spindle]

yes

[west sierra]

Then did you forgot the +1

[sudden spindle]

i think i just took 5x+5x out

[west sierra]

Because only when $(5x^2+5x)/x$ the answer would be $5x+5$

[warm shale]

Ji Ning

[west sierra]

The remainder is not 0

If remainder is still 1 so you can't take out the quotient

[raven spire]

Oh lol, Ji Ning you might be misunderstanding something

[west sierra]

What?

[raven spire]

The question is asking for remainder and quotient when (5x² + 5x + 1) is divided by x using long division method

[west sierra]

You can still multiply by 1/x

[sudden spindle]

🗿

[raven spire]

So quotient = 5(x+1)

remainder = 1

[west sierra]

x(5x+5+1/x)

[raven spire]

...

[west sierra]

Seems to be perfect quotient

[raven spire]

Quotient is supposed to be a polynomial 🗿

...

👀

[sudden spindle]

🗿

[west sierra]

I can't read chinese

[sudden spindle]

idk anymore

[west sierra]

Did you read the question

[raven spire]

Do you understand the long division process ?

question for Classified ..?

[west sierra]

Yeah

[sudden spindle]

my guy i just learnt this today and wrote whatever the teacher did

[raven spire]

Ahha

hmm

Do you know algebraic multiplication?

[sudden spindle]

Yes

[raven spire]

Like what is x times 5x+5?

Okay!

Do you remember long division process for dividing numbers? 👀 we used to do back when we were kids

I mean, we still do 🤦‍♂️

[sudden spindle]

Give me an example 🗿

[raven spire]

🗿

Ah

Waito

[sudden spindle]

Chinese school doesnt teach english term

I have to purposely google translate

[west sierra]

When you divide 15 by 2

[sudden spindle]

Ah

So it would be 7 and 1

[west sierra]

Yes

[raven spire]

Exactly !

[west sierra]

The same process here

[raven spire]

The polynomial long division uses the same process

[sudden spindle]

but how do i get the thing to divide

[west sierra]

For 5x+5 divided by x, at x=2

[raven spire]

for example... you see the first term 4x² in the dividend

and your divisor is (x-3)

so what do you multiply with (x - 3) to get a term 4x²?

[sudden spindle]

how do i get x-3, do the question always provide it

[raven spire]

Yes !!! exactly like how they always provide the divisor in the long division process for numbers

The question is always given like ( 4x² - 5x - 21 ) ÷ (x - 3)

[sudden spindle]

ah that make sense

[raven spire]

Hmm!

[sudden spindle]

4x or something

[raven spire]

Exactly!

4x

perfect

you're a genius 🗿

[sudden spindle]

no

[raven spire]

so we multiply (x - 3) with 4x and get

(4x² - 12x)

[sudden spindle]

yes

[west sierra]

Something is your unconfidence

[sudden spindle]

I consider it as a fact

[raven spire]

You multiply (x - 3) with (4x) and get (4x² - 12x)

[sudden spindle]

where did the 7 come from tho

[raven spire]

but the initial expression which we were supposed to divide had 4x² - 5x instead of 4x² - 12x

so, we subtract to get whatever extras we're getting from our multiplication different to our initial dividend

Which here, would be (4x² - 5x) - (4x² - 12x) = 7x

and we bring down the ( - 21 ) from the dividend just like we do in long division process for numbers

[sudden spindle]

ah right

[raven spire]

So now you're left with (7x - 21)

and the divisor is (x - 3)

Again the same question

what do you multiply with (x - 3) to get the first term of (7x - 21), that is 7x

[sudden spindle]

So the top one is made from 2 answers

In combined

[raven spire]

answer this first 🥺

[sudden spindle]

didnt u answered it yourself 🗿

[raven spire]

no,

the question is

what do you multiply with (x - 3) to get something like 7x

[sudden spindle]

7 ues

ues

yes

[raven spire]

Yes!

so when you multiply (x - 3) with 7

you get 7x - 21

which is exactly the expression left, and we have no remainders to bring down to the next step in the long division method

Hence, what is the final quotient we get???

👀

[sudden spindle]

google translating the meaning of quotient

right lets see

0 it says there maybe

[raven spire]

Lmao, what's it called in chinese? maybe I can make better sense of it

[sudden spindle]

Its called 商

X-3 is the 除

[sudden spindle]

yes

[raven spire]

Hmm

No then

商 is (4x + 7)

[sudden spindle]

oh i thought the 0 at the buttom is the 商

[raven spire]

which is the answer to, what should be multiplied with (x - 3) to get (4x² - 5x - 21)

[sudden spindle]

so the 4x+7 is the answer to that

[raven spire]

And the bottom one is 余 = 0

[sudden spindle]

🗿

x-3 doesnt need to write ok

[raven spire]

Yes lol

x-3 was given in question

why would they ask for something that's already given

[sudden spindle]

anyways

[raven spire]

anyways

[sudden spindle]

how come u tell

if it have to be positive or negative during the uh

4x^2 -5X

4X^2 -12

Phrase

Or is it have to be negative

[raven spire]

ooo

it's exactly the same reasoning as for the long division of numbers

for example take 279 ÷ 5

first you do 5 * 5 to get 25

[sudden spindle]

55.8

[raven spire]

and then 27 - 25

you get 2, bring down 2 with the 9 from 279

and you have 29

again 5 * 5 = 25

29 - 25

4

See?

[sudden spindle]

🗿 what

Did u simplified the divide

[raven spire]

No I just did the long division method

[sudden spindle]

uh

[raven spire]

hmm?

[sudden spindle]

I guess i have a different doing that but anyways

This part

[raven spire]

Yes

[sudden spindle]

It was negative 12x right

how do u know that it have to be negative

Like this

[raven spire]

oh

because

You were given an initial expression and (x - 3)(4x) accomplished it's task of removing the 4x² term from the dividend in the next step... so you're left with (4x² - 5x - 21) - (4x² - 12x) = (7x - 21) to divide by (x - 3)

basically, each step in the long division is meant to reduce the highest power terms remaining from the previous step

[sudden spindle]

ahhhh

I got it

So it have to be negative

[raven spire]

Genius >_< !!

SO? think you got a grasp on long division in polynomials?

[sudden spindle]

meh, i still suck in math but

At least i understand how this work so

thanks anyways

.close

[obtuse pebble]

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[azure lake]

Hey in the solution, why is it specifically written the the function is continuous at 0 when it’s continuous everywhere?

[spiral maple]

cause at x != 0, they're obviously continuous

at x=0 you had the chance of a discontinuity, so they're empathsizing that it is continuous at x=0

[azure lake]

But why was there a chance of discontinuity at x=0?

[brave bramble]

Because there's a piecewise there. One can easily create discontinuities this way. You'd have to check, and they did.

[spiral maple]

cause the function changes at 0

[azure lake]

Ohh got it

Thanks…

.close

[obtuse pebble]

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[obtuse pebble]

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[unborn owl]

i sovled it and plug into my calc

and i get the bottom answer

im using sin sum and diff formula

and i am adding because its the + formula

a+b

i changed it from adding to subtracting and i got the correct answer

why are you subtracting

in other wrods i did

(7/25)(2sqrt(13)/13) + (24/25) (3sqrt(13)/13)

whereas the answer was

(7/25)(2sqrt(13)/13) - (24/25) (3sqrt(13)/13)

wait is it cause it can be positie or negative

[humble wind]

why don't u use the sin(a+b ) formula what do u get by doing that

[unborn owl]

i did but i got the wrong answer now

but its fine i see why

i didnt see they gave the domain at the top

.close

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[meager osprey]

Hi, I have a question about injectivity

[meager osprey]

So for the following function I am asked to find f^-1

we find this

Which leads to this conclusion

I am a bit confused on the intervals though?

How did we find the intervals for this?

thanks

Also I am confused why it says {x,-x/2} if x=+-2 for example

Should it not rather be {x,-x/2} if y=+-2?

[raven spire]

You start with the definition of injectivity.
f(x1) = f(x2) => x1 = x2
And so, you equate f(x1) = f(x2) to check whether this function f(x) is injective or we need to break it into intervals to make it into injective.
Also, notice that f^-1 exists iff the function is surjective in the codomain defined and it indeed is that way.

Now, next step is continuing with f(x1) = f(x2), factorize a bit and the expression you get is:
( x1 - x2 )( (x1)^2 + (x1)(x2) + (x2)^2 - 3 ) = 0
Assume, we're in an interval where f(x1) = f(x2) has solutions other than x1 = x2... so right now, we're interested in ( (x1)^2 + (x1)(x2) + (x2)^2 - 3 ) factor and so, equate it to zero... to obtain an equation
(x1)^2 + (x1)(x2) + (x2)^2 - 3 = 0
which is quadratic in any of x1 or x2... Here, I am taking this is quadratic in (x2)
Check discriminant - D = (x1)^2 - 4((x1)^2-3) = 3(4-(x1)^2)

Now, for real solutions for (x2), by which we mean that, for f(x1) = f(x2) to have more than one solutions in our part of the interval chosen from the codomain, we'll have: D ≥ 0, or 3(4-(x1)^2) ≥ 0 or (x1) lies within (-2, 2)

So, you figure out that okay, the function f(x) has the property that f(x) is not injective within (-2, 2) or rather that f(x1) = f(x2) has more than one solutions only in the interval (-2, 2)

Also, notice that you're not really asked for f^-1 but rather f^-1({f(x)}) which is actually the image of the inverse of the function... which in the interval (-2, 2) will follow the quadratic (x1)^2 + (x1)(x2) + (x2)^2 - 3 = 0
and hence, the part f^-1({f(x)}) = those scary roots and x in the interval (-2, 2)

put x1 = 2 to check for equal roots case in the quadratic obtained earlier to get f^-1({f(x)}) = {x,-x/2} for x = ±2

However, the Discriminant D < 0, for x1 not lying in [-2, 2] and so the only possible image of the inverse of f(x) in the interval [-2, 2]^c is {x}

@meager osprey

Hope that fills you in on everything you want to know smh
also, sorry for leaving you half way through the last time

[obtuse pebble]

@meager osprey Has your question been resolved?

[obtuse pebble]

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[wooden bay]

how would i go about solving arcsec(sec(-2)) = 2

[wooden bay]

i know the answer i just need some help getting the method down

[unique solstice]

Do you mean evaluating?

[wooden bay]

yes

idk first time doing double sec(sec(x)) and im kinda stuck

usually there is a sin/cos/tan in there

yes, but how does that help me solve it

[grave thistle]

wdym

secx is 1/cosx

cosx can give negative outputs

why cant secx do the same

[wooden bay]

maybe because the range for cos is 0 to pi

[grave thistle]

nuuU

heres a graph for secx

and yes it does have negative values

[unique solstice]

arcsec doesn't

[grave thistle]

just wait

[wooden bay]

ok so how do i solve my thing 😭

[grave thistle]

you need to know that secx is an even function

meaning sec(-x) = sec(c)

arcsec is the inverse function of secx

=> arcsec(sec(-2)) = arcsec(sec(2))

= 2

[wooden bay]

wait

[grave thistle]

whaa

[wooden bay]

ok then
let a = arcsec(sec(-2))
sec(a) = sec(-2)
a = -2, why does here a not equal -2

[grave thistle]

it is…

sec(2) is equal to sec(-2)

its an even function

🥲

[wooden bay]

how do u properly indicate it

[grave thistle]

wdym

[wooden bay]

when ur showing proof of work

is there like a way to say it

[grave thistle]

secx is equal to 1/cosx

cosx is an even function

hence secx is also an even function

there are probably more rigorous ways of proving it

[wooden bay]

hmm ok

.close

[obtuse pebble]

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[wooden bay]

my bad for closing and reopning kinda the same question, but I :
arcsec(sec(-2)) = 2 does not fit the parameters

[wooden bay]

because you do:
let a = arcsec(sec(-2))
sec(a) = sec(-2)
a = 2; but 2 does not fit the ranges

what should i do to make it fit in the ranges

[grave thistle]

you could extend the definition of arcsec

into the imaginary plane

where it has no ranges

[wooden bay]

@grave thistle wdym

[grave thistle]

nvm that

anyways

wdym it doesnt not fit the ranges

[wooden bay]

because ya know 2 is not between o and 1/2pi and also not between pi and 3pi/2 @grave thistle

[grave thistle]

well neither is -2

[wooden bay]

yea so thats my problem

[grave thistle]

well

math is weird

[wooden bay]

cant i like 2pi + or 1pi+

[grave thistle]

try to find out why those ranges

[wooden bay]

or maybe pi - a

i know in cos functions you can 2pi - a, and u have the same cosine

can i do something similar here

[grave thistle]

sort of

arcsec(-x) = pi - arcsecx

[wooden bay]

so would that work

pi-2

so it respects the boundaries

[grave thistle]

ykno those ranges only apply for triangles right

[wooden bay]

hmm, pi-2 fits into the first range

[grave thistle]

just like how cosx seem to have limitless range

so does arcsecx

and basically all of the trigonometric functions

we just give them ranges when dealing with a geometric problem

involving triangles

[wooden bay]

okay but my teacher wants us to respect the ranges 0 -> 1/2pi and pi -> 3pi/2

[grave thistle]

oh

[grave thistle]

tell him thats hes sleeping on the job

no im jk

[wooden bay]

this dude is lowkey cracked tho fr

[grave thistle]

anyways

do you know that the pi here is in radians

[wooden bay]

yes

[grave thistle]

yeah

pi = 180 degrees

correct

[wooden bay]

yes

[grave thistle]

exactly this range only applies for triangles

secx = 1/cosx

whats the range for cosx

[wooden bay]

ok i get it now, man this shit is hard af then, how do i make arcsec(sec(-2)) respect the range

[grave thistle]

t^t

the range is irrelevant

[grave thistle]

istg

im dying with you here

[wooden bay]

alright bro its over for me

.close

[obtuse pebble]

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[grave thistle]

there are no ranges

[obtuse pebble]

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[timid silo]

How would I do question (ii) and (iii)?

[obtuse pebble]

@timid silo Has your question been resolved?

[spice rune]

ok, tell us how you solved the first one

and i'll explain how you can do the same with 2 and 3

[timid silo]

well -5 divided by 2 is -(5/2) so I multiplied 8 by it and got a as -20

I think I got it

I turned the vectors into

8/(b+3) and (1-5b)/(4b-6)

Cross multiplied

and factorised

to get b = 1 and b = -(51/5)

[balmy mortar]

?????????

vectors aren't fractions, what?

Just use the definition of parallel for vectors:

They are multiples of each other

[timid silo]

you turn them into fractions to find the parallel vectors

it worked perfectly so

[balmy mortar]

If I was presented with working like that, I would think you'd misunderstood how vectors work

There is no need to 'cheat' by turning them into fractions

Same math, but looks improper.

[timid silo]

That’s what I was taught to do

If you had the vector 4i + 2j

it can be shown as 2/4

[balmy mortar]

That feels terrible, but if thats how your teacher wants it.

[timid silo]

Nick does want it like that, and he’d kill me if it wasn’t

[timid silo]

How would you do it then?

[spice rune]

the problem is you can't use the same technique to solve the 3rd one

[spice rune]

this is what you want to do instead

[timid silo]

I did

But I cross multipled the top and bottom line

@timid silo what would you do?

id use the same thing idk how else youd do it

bcs they have the same gradient so the fractions are equal to each other so it makes sense

[spice rune]

ehhh

listen

[timid silo]

That wouldn’t work for something complex though

I cant just divide one by the other, find the factor and use it

[balmy mortar]

I'll write it.

$$\begin{pmatrix}a\b\c\end{pmatrix}$$

$$\begin{pmatrix}x\y\z\end{pmatrix}$$

[spice rune]

vectors are parallel when there exists such $k$ that
[
\mqty(a_1 \ \vdots \ a_n) = k \mqty(b_1 \ \vdots \ b_n)
]

[warm shale]

rept1d

[spice rune]

which means that you just need to solve the following system of equations:
[\begin{cases}
a_1 = k b_1 \
a_2 = k b_2 \
\vdots \
a_n = k b_n \
\end{cases}]

[warm shale]

rept1d

[timid silo]

which is similar to what I did, but I just solved it simultaneously instead

[spice rune]

yes, if you have only two equations, you can divide left and right sides
[
\frac{a_1}{a_2} = \frac{k b_1}{k b_2} = \frac{b_1}{b_2}
]

[warm shale]

rept1d

[spice rune]

and if you understand that, it's perfectly fine

but for more equations you usually don't have these "hacks"

you could actually do that trick for your 3rd task (by first considering only lines 1 an 3) just because there are only cs (and no ds) there

but in general (when every line contains n - 1 variables), you still need to solve that systen

[timid silo]

I got the answers right with my method so idk

[meager osprey]

hey im sorry I just now am seeing this. I really appreciate you spending so much time and energy and really no worries for last time you've helped me more than i could ever ask for. thanks a lot. big appreciate

[raven spire]

Hope you understood it though

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[toxic elm]

can someone help me with this question please?

[timid silo]

so obviously youd have to assume there are 100 people

French is 78/100 and german is 84/100

It doesnt state whether they are independent events really

[outer quail]

Two different tests would be independent events

[timid silo]

But if you do badly in one, you're more likely to do bad in the other

[outer quail]

[short spruce]

not what the question says

[timid silo]

okay so for independent events

P(FnG) = p(F) x p(G)

[toxic elm]

I'm confussed

[toxic elm]

I got 0.6552

can someone help me please?

<@&286206848099549185>

[tardy epoch]

How'd you get this

[toxic elm]

78/100*84/100
@tardy epoch

I don't really understand the logic behind this

[tardy epoch]

That's not even the question

The question tells you to draw the Venn diagram

You don't know enough to calculate the exact value of the intersection.

[toxic elm]

how to solve this question?
@tardy epoch

[obtuse pebble]

@toxic elm Has your question been resolved?

[obtuse pebble]

@toxic elm Has your question been resolved?

[obtuse pebble]

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[silent bramble]

I need to find the range of y=f(x) if f(x)=2-4SQRT5x-3.

[neat valley]

if you're ever in doubt with stuff like this it doesnt hurt to try brute forcing it first

[silent bramble]

Lol what’s that mean

[neat valley]

plugging in values to try and guess the range

[silent bramble]

Yes but I’d need to show how I got the value

[obtuse pebble]

@silent bramble Has your question been resolved?

[obtuse pebble]

@silent bramble Has your question been resolved?

[limpid sonnet]

write an quation of the line that is parallel to 7x+5y=35 passing through point 3,2

[obtuse pebble]

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[foggy sandal]

How can i solve this integral? I have been trying for an hour

[warm shale]

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

[foggy sandal]

sorry wrong latex text

$\frac{x²}{x⁴+1}$

[warm shale]

QuantumMeme

[foggy sandal]

now its right

[iron geyser]

You need to factor the denominator

Then use partial fraction decomposition and then you should be able to do the integral

I looked up online just to make sure that the denominator was factorable

[forest sinew]

the partial fractions looks horrible

is it

[iron geyser]

No

[foggy sandal]

I got x⁴+1=(x²+1)²-2x², but idk what to do whith the -2x²

[forest sinew]

almost looks like difference of square

[tardy epoch]

spoiler: ||wolfram alpha can do partial fractions||

[forest sinew]

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

so $x^2 \pm \sqrt 2 x + 1$?

[warm shale]

jan Niku

[foggy sandal]

Thanks!

.close

[obtuse pebble]

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[timid silo]

consider rolling a 6 sided die, what is the probability that it takes at least two rolls for the first 5 to be observed?

Attempt: (5/6)(1/6) = 5/36 but that's wrong. My thoughts: first roll is 1,2,3,4,6 and second roll needs to be 5, hence my attempt. Should I be using combinatorics here or am I misreading the prompt?

[forest sinew]

its a common distribution

geometric

[timid silo]

Uh sorry I don't understand those terms

[mental solstice]

you're misunderstanding 'at least 2' as 'exactly 2'

[forest sinew]

oh

[timid silo]

hmm

[forest sinew]

well i think

youll still need geometric

[mental solstice]

...no you don't, or at least not the word 'geometric'

[forest sinew]

okay

[mental solstice]

P(at least 2 rolls) = P(first roll is not 5)

[timid silo]

ah

that makes sense, so it's just 5/6 lol

[forest sinew]

oh

ye

[mental solstice]

[timid silo]

got it, thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[proper coyote]

Hi

[proper coyote]

The line L goes through O(0,0,0) and is perpendicular to the line k: λ(1,0,1). The line L makes an φ angle with the y-axis so that cos φ = 1/3

Give a vector equation of line L

So if it’s perpendicular to K i have this: a + c = 0

I have tried with inproduct but there are too many variables for me to have an answer

So i don’t know what to do now

[obtuse pebble]

@proper coyote Has your question been resolved?

[proper coyote]

<@&286206848099549185>

[jovial crane]

Idk it either lol

[proper coyote]

Lol my classmates also don’t and teacher is not responding to messages 🥲

[obtuse pebble]

@proper coyote Has your question been resolved?

[proper coyote]

.close

[obtuse pebble]

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[last crow]

hello

[last crow]

I am sort of confused as to what the word calculate in this sense means

[spice rune]

get rid of the infinite sum

[last crow]

so convert it into a function?

[spice rune]

it's a function even if it has an infinite sum

but the task is most likely about expressing it using elementary functions

[last crow]

so how does one get rid of the infinite sum

yea i guess that sounds about right

okay that makes a little more sense - i thought i had to find a numerical value or something like that and i was confused

[spice rune]

do you know what is the value of a series?

[last crow]

honestly i forgot all about series i studied this stuff 6 months ago

[spice rune]

by definition
[
\sum_{n=1}^\infty f_n(x) = \lim_{k \to \infty}\sum_{n=k}^\infty f_n(x)
]

[warm shale]

rept1d

[spice rune]

and there are also some helpful formulas

[last crow]

but is this problem specifically looking for the value of that series?

[spice rune]

yes, they just probably want you to use the given fact

the first one is obvious for example

[last crow]

yeah i gotta see how to apply this then

[spice rune]

[
\sum_{n=1}^\infty p (1-p)^{n-1} = p \sum_{n=0}^\infty (1-p)^n = \frac{p}{1-(1-p)} = 1
]

[warm shale]

rept1d

[last crow]

ohhh

yea i gotta refresh on these

thank you for your help @spice rune

[spice rune]

np

[last crow]

@spice rune sorry one more quesiton

would part b just be constant of integration + 1 then?

not sure if thats how you should represent your answer

im probably trolling nvm

[spice rune]

it'd be (1-x)^n, not n-1

so the sum is starting from n=1

you've probably missed that

[last crow]

yeah i noticed that

[spice rune]

did you get an answer?

[last crow]

Nah im stuck i’m trying to find resources to study from but i’m not in luck

Assuming this is a geometric series

[spice rune]

where exactly are you stuck?

[last crow]

I did the integral which was easy

So just the constant + the integrated series

And the bound is still at n = 1

But i’m stuck at what to do next

Do i just take the limit?

[spice rune]

that's another possibility (and is much easier in my opinion), but let's still discuss the method the book wants

I assume you know how the formula of the derivative of power series

[last crow]

Yeah

[spice rune]

in our case k=1, so it's very simple

do you agree that
[
\sum_{n=1}^\infty np(1-p)^{n-1} = -p \qty(\sum_{n=1}^\infty (1-p)^n)'
]

a mistake

wait

[last crow]

Positive p

[spice rune]

no

[last crow]

Oh

[warm shale]

rept1d

[spice rune]

now it should be right

[last crow]

Oh its because 1-p

[spice rune]

yes

[last crow]

Yes

I see that

[spice rune]

can you go from here?

[last crow]

Oh I accidentally took the integral lol

I guess there was no point in doing that

[spice rune]

kinda true

[last crow]

I got it yeah just use the derivative formula

[spice rune]

the answer should be 1/p

[last crow]

And then the bound will be n = 0

[spice rune]

i might have misunderstood you, but the derivative formula has been used already

[last crow]

Yeah im lost then

[spice rune]

i mean
[
\sum_{n=1}^\infty np(1-p)^{n-1} = -p \sum_{n=1}^\infty \qty((1-p)^n)' = -p \qty(\sum_{n=1}^\infty (1-p)^n)'
]

that's how we got there

[last crow]

I’m unable to understand how to evaluate with the n=1

We cant make the exponent n-1 right?

[spice rune]

evaluate with n = 0 and then subtract f_0(p)

[warm shale]

rept1d

[last crow]

Okay

But if you evaluate with n = 0

You must change the exponent to n-1

[spice rune]

do you?

look at the formula at the top

it's just x^n

[last crow]

It is but how do we start at n = 0 if its n = 1

We can just start at n = 0 and remove the first term?

[spice rune]

Yes, that's what I said

[last crow]

Ah right

[last crow]

Well i think i got the steps

I just used the general formula of the sum of the series though

I didnt understand the subtraction part

[spice rune]

:(

[last crow]

Can you explain the subtraction part?

Also the limit

[spice rune]

[
\sum_{n=1}^\infty (1-p)^n = \sum_{n=0}^\infty (1-p)^n - (1-p)^0 = \frac{1}{1-(1-p)} - 1 = \frac{1}{p} - 1
][
-p \qty(\sum_{n=1}^\infty (1-p)^n)' = -p \cdot \qty(-\frac{1}{p^2}) = \frac{1}{p}
]

[warm shale]

rept1d

[last crow]

Ohhh

Yea im dum

[spice rune]

i won't get into details but the main thing you need is that (1-p)^k converges to 0

[obtuse pebble]

@last crow Has your question been resolved?

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[mellow trail]

.reopen

What sum will amount to ₹53,240 in 3 years if the rate of interest is 10% p.a. and the interest is compounded annually?

[mellow trail]

<@&286206848099549185>

[scenic sundial]

what have you done?

we won't be able to give you answer. Show work first @mellow trail

[mellow trail]

Well, I don't know how to approch it

Little confused

<@&286206848099549185>

[timid silo]

Don't abuse the helper role.

Do you know formula for finding compound interest?

[scenic sundial]

^

[mellow trail]

Yes

[grave thistle]

do you know why the formula is like that?

[raven spire]

$\text{Amount} = \text{P.A.} \qty(1+\frac{r/n}{100})^{nt}$

[warm shale]

Ansh

[obtuse pebble]

@mellow trail Has your question been resolved?

[obtuse pebble]

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[steady lodge]

someone hel[ [ls

[mild ocean]

you haven’t tried at all have you

[steady lodge]

i did

[mild ocean]

try the answers given

[steady lodge]

i tried l'hopital

[mild ocean]

bruh

[steady lodge]

i tried taking log

but still didnt

[steady lodge]

i want method how to solve , not just the ans

eh its for understanding of how to solve such qs rather than just getting the ans

[deft hazel]

L'hopital?

[steady lodge]

l'hopital's rule

[deft hazel]

oh yeah

sorry

didn't read the message where you said you've tried

Yeah and didn't work out?

[steady lodge]

nah

[mild ocean]

idk how that would help since that would just make it more complicated

[steady lodge]

i got stuck , and it all just went on getting more n more complicated

[steady lodge]

exactlyy

idk wot to do now

[deft hazel]

hm idk i'd apply l'hopital

sub in x=a
simplify

solve

[steady lodge]

hmmm lemme try

ty

.close

[obtuse pebble]

Channel closed

Closed by @steady lodge

Use .reopen if this was a mistake.

[obtuse pebble]

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[nocturne aspen]

the answer is 1.5 but i need someone to break down the steps

[royal basin]

@nocturne aspen have you made a diagram yet

[nocturne aspen]

Ye

[royal basin]

show your diagram

[nocturne aspen]

Inside triangle is the pond

[royal basin]

that's not a triangle

[nocturne aspen]

Rectangle*

[royal basin]

okay

[nocturne aspen]

Yeah autocorrect

[royal basin]

so what are the dimensions of the outer rectangle?

[nocturne aspen]

X I guess

[royal basin]

why guess

x is the width of the path

it's not what i'm asking you

it should be clear from the diagram that the length and width of the outer rectangle are ||15+2x and 12+2x||

[nocturne aspen]

2x being both sides right

[royal basin]

"both sides" what?

[nocturne aspen]

Top bottom left and right sides of the outer rectangle

[royal basin]

you're overthinking it

look at your diagram horizontally

you have the path (x), the pond (15) and the path (x)

x + 15 + x

= 15+2x

[nocturne aspen]

Ok

[royal basin]

same for the vertical direction

are you able to continue from here?

[nocturne aspen]

So now I just multiply and factor right

Nvm factor just comes back

[timid silo]

Area = 90
Does this make sense -
(15+2x)(12+2x) - 180 = 90

The total area of the entire place including path

And subtracting off the area of the pond

[nocturne aspen]

Yeah

[timid silo]

He has 90unit^2

So you got it now?

[nocturne aspen]

Yup

.solved

.closed

[timid silo]

.close <

[nocturne aspen]

.close

[obtuse pebble]

Channel closed

Closed by @nocturne aspen

Use .reopen if this was a mistake.

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Remember:
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• 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.

[timid silo]

Nevermind

That was awfully stupid of me to ask I immediately just

Yeah sorry

.close

[obtuse pebble]

Channel closed

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[fiery apex]

did i do it right?

the variance, and standard deviation i mean

[obtuse pebble]

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[grave thistle]

x,x1, x2...x9 are integers satisfying the conditions below, what is the value of P

[raven spire]

what is x?

ohhh, I got it nvm

[raven spire]

Are x1, x2, x3, x4 ... factors of x or something? or just integers, that's it?

[grave thistle]

just integers i guess

[raven spire]

Huh? there has to be something else given as well...
Tried with two integers x1, x2: x = (1-(x1)^2) and x1 + x2 = 0 => P = -(x1)^2 (1-(x1)^2)

Can you ss the complete question?

[grave thistle]

oh its in a different language

[raven spire]

No worries

[grave thistle]

i translated it all

1 to 1

[raven spire]

send lol

[grave thistle]

aaaaa

fine

[raven spire]

👀

[grave thistle]

thats it

[raven spire]

hmmmm

Anything you've tried so far?

[grave thistle]

yeah

ive tried multiplying it by its conjugate

then dividing the conjugated 1st part with the 2nd part ^2

=1

im pretty sure P = 0

but i need za proof

[raven spire]

Tried setting a polynomial with x1,x2,x3...,x9 integer roots??

[grave thistle]

no

wait

isnt that just x1x2...x9=0

[raven spire]

how?

[grave thistle]

no idea

but x1=x2=x3=...=x9=0 and x=1 satisfy

[raven spire]

Lol 😂 do you know something?

[grave thistle]

idk

i thought i had it

so i came up with the x1=...