help-27

[supple knot]

Did you try king's rule

[devout snow]

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[glossy dew]

back

[glossy dew]

like pi-x and adding both integrals?

[grand edge]

\begin{align*}
I &= \int_{0}^{\pi}x^{3}\ln\left(\sin\left(x\right)\right)dx \
&= \int_{0}^{\pi}\left(\pi-x\right)^{3}\ln\left(\sin\left(x\right)\right)dx \
&= \int_{0}^{\pi}\left(\pi^{3}+3x^{2}\pi-3x\pi^{2}-x^{3}\right)\ln\left(\sin\left(x\right)\right)dx \
&= \frac{1}{2}\int_{0}^{\pi}\left(\pi^{3}+3x^{2}\pi-3x\pi^{2}\right)\ln\left(\sin\left(x\right)\right)dx\
&= \frac{1}{2}\int_{0}^{\pi}\left(\pi^{3}\right)\ln\left(\sin\left(x\right)\right)dx+\frac{1}{2}\int_{0}^{\pi}\left(3x^{2}\pi\right)\ln\left(\sin\left(x\right)\right)dx-\frac{1}{2}\int_{0}^{\pi}\left(3x\pi^{2}\right)\ln\left(\sin\left(x\right)\right)dx \
\end{align*}

[woven radish]

Roy

[grand edge]

Then I guess they used the results $-\frac{\pi^{2}}{2}\ln\left(2\right)=\int_{0}^{\pi}x\ln\left(\sin\left(x\right)\right)dx$ and $\int_{0}^{\pi}\ln\left(\sin\left(x\right)\right)dx=-\pi\ln\left(2\right)$

[woven radish]

Roy

[grand edge]

So $$I = \frac{3\pi}{2}\int_{0}^{\pi}x^{2}\ln\left(\sin\left(x\right)\right)dx+\frac{\pi^{4}}{4}\ln\left(2\right)$$
$$= \frac{3\pi}{2}\int_{0}^{\pi}x^{2}\ln\left(\sin\left(x\right)\right)dx+\frac{3\pi}{2}\int_{0}^{\pi}x^{2}\ln\left(\sqrt{2}\right)dx$$

[woven radish]

Roy

[grand edge]

Which is a pretty neat way to force out a factor of the denominator once you combine them with log laws

[grand edge]

You can derive these two with repeated applications of kings rule and symmetry

Or with fourier series

[glossy dew]

yikes what an insane compression

thanks

.close

[devout snow]

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[grand edge]

Yeah their working of it is not very nice

[devout snow]

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[quaint valley]

hi can someone help me with question 2e and 2f please

[undone dirge]

what have you done so far?

[quaint valley]

this is what i have done for e

but idk am i in the right track or not

[undone dirge]

just looking at the start, you've differentiated sinx correctly but the xsin(2x) term requires the product rule

you have two functions in x so we need to use $\frac{d}{dx}uv = u'v + v'u$

[woven radish]

melon

[undone dirge]

can you redo the differentiation of xsin(2x) ?

[quaint valley]

ahhh so then i need to use the product rule for the second term

okok

[undone dirge]

correct

[quaint valley]

oh and the derivative of sin(2x) is 2xcos(2x) right

[faint hearth]

No, use chain rule

Derivative of 2x is 2

[grand edge]

Same thing

[undone dirge]

as stated already, when we differentiate sin we need to chuck the derivative of the insides to the outside

more formally $\frac{d}{dx}\sin{(f(x))} = f'(x) \cos{(f(x))}$

[woven radish]

melon

[quaint valley]

is it looks better now

[undone dirge]

yes perfect

now we can continue with the question

[quaint valley]

then i will use this as a gradient for the tangent

then i will find the equation using y - y1 = m(x-x1) right

[grand edge]

Yes

[undone dirge]

yes very good

[quaint valley]

ahhh ok TYSM

<333

.close

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[devout snow]

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[unique canyon]

Someone Help me prove this

[devout snow]

@unique canyon Has your question been resolved?

[unique canyon]

<@&286206848099549185>

[sage ibex]

do u know what lcm*gcd=

[unique canyon]

Product of numbers?

[azure parrot]

It looks really weird to me, isn't n a common multiple of a_i ?

since gcd(ai,bi)=1 , k_i/b_i should be an integer, or I am being dumb here

[devout snow]

@unique canyon Has your question been resolved?

[round raptor]

Why do you say

[azure parrot]

uh say what

[round raptor]

n is rational =aiki/bi so ki= nbi/ai, since bi and ai are coprime ai|n

as ki is natural number

[azure parrot]

n is natural number

[round raptor]

n belongs to q+

[azure parrot]

What

Okay lmao it say i=1,2,...,n so I assumed n is integer

But smth like a_{1/2} is weird as hell

[round raptor]

Right

Question is badly framed

But integers lie in q

[unique canyon]

.close

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[devout snow]

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[quaint valley]

can someone help me with this please

[pseudo basin]

do you know log laws

[devout snow]

@quaint valley Has your question been resolved?

[devout snow]

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[wise violet]

I need help with 87 and 88

These are the trig identities I’ve learned so far

[tiny kestrel]

what books that

[wise violet]

Keep in mind that my teacher doesn’t use the unit circle much btw yall

[wise violet]

Pre calculus with limits 8e graphing approach

[tiny kestrel]

dam

[tiny kestrel]

sin( pi -t) = sin t , sin (pi +t) = -sin t , cos(pi-t)=cos(pi+t) = -cos t

[wise violet]

Is that for 87 ?

Or both

[tiny kestrel]

both

[wise violet]

How is sin (pi+t)=sin t

Like why

I don’t understand why it is equal

[tiny kestrel]

-sin t

[tiny kestrel]

unit circle

[wise violet]

😭

How so

[tiny kestrel]

its

how do i explain

[wise violet]

Does this have anything to do with parallel lines intersected by a line

That’s what a teacher told me

[grand edge]

Grab any point on the unit circle. Rotate it 180 degrees or π radians around the origin. Notice that its y-value is the negative of its original y-value as it's been reflected about the origin.

[wise violet]

Idk the points of a unit circle

😭😭😭😭

[grand edge]

I'd recommend you watch a video on that, then

[tiny kestrel]

it defines trig girl

trg starts from unit circle

[wise violet]

No like what are the points of a unit circle

Is it the thing they tell us to memorize

My teacher told me not to memorize that 💔

He rlly said “we ain’t doing that in this math class”

[grand edge]

The unit circle gives a great intuition for trigonometric functions

[wise violet]

This right?

Or something else?

[steady kettle]

i think its right

[wise violet]

Oh okay

[hollow ice]

thats the unit circle, but the one thats useful is this one

[wise violet]

Wait i still don’t understand why sin (pi - t) = sin t

[wise violet]

What the 😭😭

[hollow ice]

try to plot the angle t and the angles pi+t and pi-t using this

[wise violet]

Bro I think I’m gonna go to sleep and just ask my teacher tmrrw 😭

[steady kettle]

lol

[hollow ice]

ehh, its pretty simple tbf

[steady kettle]

kinda

[hollow ice]

you got a circle, and you mark an angle on it

[wise violet]

Bc it’s super late and not worth it atm, I’ll just go over it tmrrw when I have more time

Wait no I see what that image is showing lowkey

[hollow ice]

then the x coordinate is cos and y one is sine

[wise violet]

I see it now

[hollow ice]

and you gotta then track the angles and sides for those various configurations asked in your problem

[wise violet]

☹️

Okay I’ll try that tomorrow

Thank you guyssss

.close

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[devout snow]

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[paper dawn]

Hai..
Can someone help me how to solve...

• Probability equation, Union And Intersection, and the dice probability thing: 1/4 + 1/4 - 1/4 = 1/4

• Distance formula

• Permutation and Combination

• General form

[stoic lotus]

What?

[paper dawn]

Wiat no, wrong wording.

How to do a probability equation?

[stoic lotus]

In what context?

There are a lot of equations for probability

[devout snow]

@paper dawn Has your question been resolved?

[paper dawn]

Wait

P(AUB) = P(A) + P(B) - P(A (upside down U) B)

[stoic lotus]

So it's the probability of one or the other happening, but not both

That is, if events A and B can happen at the same time, we need to account for that

Probability of A and B both happening is just probability of A multiplied by probability of B

So if P(A) = a, P(B) = b
P(A and B) = a*b

[devout snow]

@paper dawn Has your question been resolved?

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[winged hearth]

!rotate

how does this work

[stoic lotus]

, rotate

[winged hearth]

oh thanks

[stoic lotus]

So consider this

If I want to choose someone for the first position, how many ways to do that are there?

You have 15 people

So there are exactly 15 ways to choose the first player of the team

[winged hearth]

15^10?

[stoic lotus]

No

[winged hearth]

or 10^15?

[stoic lotus]

If you chose someone for the first spot, how many players unchosen are there?

[winged hearth]

so 10^15,9^14?

...

[stoic lotus]

Not at all

Again

Think about it logically

Forget the math

You have 15 players for the first spot in the team

[winged hearth]

first position we can get is 15, then 14 ,then 13?

[stoic lotus]

Yes!!!

[winged hearth]

14 for 2nd position

aah

[stoic lotus]

13 for the third

[winged hearth]

so half knowledge is bad

[stoic lotus]

15
14
13
12
11
10
9
8
7
6

And at each step there's a branch

So we need to multiply

Makes sense?

15 * 14 * ... * 6

[winged hearth]

uuh

why multiply

oh got it

[stoic lotus]

For each of the first 15 picks you can choose 14 picks in the second spot for which you have 13 picks in third and so on

[winged hearth]

k

[stoic lotus]

It's called the number of permutations

Permutations are ordered. That means it matters whether John is in the first or 10th spot

But in our case we only care whether John is in the team or not

Doesn't matter which spot

So we essentially just overcounted

[winged hearth]

is this an example?

[stoic lotus]

So if team is (player 1, John, etc) it's different from (John, player 1, etc) in this case

[winged hearth]

k

[stoic lotus]

But

We don't want that

We need to account for that

[winged hearth]

k

[stoic lotus]

How many ways are there for 10 team members to be in the team?

Again, 10 for the first spot, 9 for the second, etc, 1 for the last one

[winged hearth]

oh

why 10 tho

[stoic lotus]

Because there are 10 spots for the team members

[winged hearth]

wait I'll try and example

[stoic lotus]

Imagine you have 10 people. How many ways are there to arrange them? It's 10 * 9 * 8 * 7 * 6 * 5 * 3 * 2 * 1

This is called 10 factorial

10!

[winged hearth]

ywa

1098

ik that

[stoic lotus]

Okay. So that's the factor by which we overcounted

We have 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 permutations

If we don't care about the order, we must divide this by 10!

[winged hearth]

one sec

[stoic lotus]

We get what's called combinations

Combinations don't care about order

[winged hearth]

so we get 12 possible outcomes when position doesn't matter. and we have 4 players and 3 slots

no I'm wrong

we get 4

when position doesn't matter

and 24 when we count position

so 4!/3!

4* 3* 2/4*2
not counting 1

in this case we get 15!/10!

?

[stoic lotus]

No, not 15!

15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6

[winged hearth]

we had 151413... right

oh

isee

[stoic lotus]

First 10 terms of 15!

[winged hearth]

15!-5!?

[stoic lotus]

Not minus, divided by

[winged hearth]

aah

so the final is

[stoic lotus]

And then we divide that by 10!

[winged hearth]

15!/5!?

then /10! wow

[stoic lotus]

Yes. Because again, 15!/5! is permutations

That means it matters how players are chosen, in which order

1 2 3 4 5 6 7 8 9 10 is not the same as 1 3 2 4 6 5 7 9 8 10

[winged hearth]

yea

[stoic lotus]

But we don't care about that

[winged hearth]

permutations are the total positions counting the place?

[stoic lotus]

Because the same 10 players in different order is the same team

[winged hearth]

got it

[stoic lotus]

So we want to exclude the number of ways in which we can order 10 players in the team

[winged hearth]

how will we calculate the probability that player one is in position 6

[stoic lotus]

That's why we divide 15!/5! by 10!

So the final answer is 15!/(10! * 5!)

[winged hearth]

so 15 is the number of players=N
10 is the number of slots or position which =M
the formula will be N!/M!*(N-M)!

[stoic lotus]

Yes!!!!!

[winged hearth]

ooooo

[stoic lotus]

You're a genius

[winged hearth]

no

i am dumb

people my age are learning this

and I don't even know the chapter's name

btw when we learn this

and how will we calculate the probability that "John" is in position 6 when position matters

[devout snow]

@winged hearth Has your question been resolved?

[round raptor]

which question are we on?

[winged hearth]

umm not the question

idk permutation and combination

so I am trying to solve but I am learning

i don't like textbooks

they don't make me understand

[round raptor]

ok, what do you know ?

[winged hearth]

ik permutation formula

combination formula

[round raptor]

ok

do you know what permutation means?

[round raptor]

you are referring to nPr = something and nCr

permutation is arrangement

where as combinations is choosing different set of things

eg you have 9 people and you have to seat them at 9 seats in a row, the total number of ways you can seat them is permutation

[winged hearth]

shit

i forgot

[winged hearth]

yea

[winged hearth]

ik that

[round raptor]

ok good

are you in high school?

[winged hearth]

10th grade

yea

[round raptor]

why do you have undergraduate role ?

[round raptor]

ok

india ?

[winged hearth]

yea

[round raptor]

ok

so what doubt you have ?

[winged hearth]

rn

if there r 15 options

10 slots

and John a player has to be there

in 1 of the 10 slots

[round raptor]

ok, how many players

[winged hearth]

15

total

what is the probability that John is in 6th position out of 10

[round raptor]

so you have 15 players to be seated in 10 seats and find probability jon is in 6th spot ?

right ?

@winged hearth

[winged hearth]

yea

btw a question first

is this hard or the probability question

[winged hearth]

, rotate

[devout snow]

Channel closed

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[winged hearth]

.reopen

[devout snow]

✅ Original question: #help-27 message

[winged hearth]

@round raptor

[round raptor]

wher is probability question

[winged hearth]

thats my question

that i made

[round raptor]

oh

they are both different situations

[winged hearth]

ik

what maths wise what will be harder

i want to go for easy ones

to get my basics clear

[round raptor]

this

[winged hearth]

harder or easier

[round raptor]

not harder this will help in thinking your question and that, i would say both easy but that better

[winged hearth]

k

let's do that one then

[round raptor]

tbh you should not focus on difficulty rn

[round raptor]

do both

[winged hearth]

i mean if I am doing a hard question at the start without knowing what I'm doing then no point ig

[winged hearth]

yea I meant first that

[round raptor]

right

[round raptor]

ok

[round raptor]

so first this ?

[winged hearth]

yea

[round raptor]

what do you think for a) 1)

[winged hearth]

no restriction means the same people again in different positions or positions doesn't matter

[vapid bluff]

hey I have a math qsn can i post it here?

[round raptor]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[vapid bluff]

alr

[winged hearth]

bt different i think position doesn't matter

by*

so it means no matter John is number 1 or 2

?

[round raptor]

no restriction means we can do operations normal way

[winged hearth]

what do we mean by normal

[round raptor]

however position does not matter here

to form a team you are choosing 10 people

[winged hearth]

so combination?

[round raptor]

so does not matter you choose jon first or last the team formed will be same 10 people

[round raptor]

yes

[winged hearth]

shoul I use calc

[round raptor]

eg if they applied restriction that they want john and tom in every team then we need to choose john and tom everytime , every team will have them but other 8 members will change

[round raptor]

no, for small numbers

[winged hearth]

is 15! small. idk first time working with factorials for a question

[round raptor]

so you have to make team of 10, so choose 10 out of 15

[round raptor]

no, but it will cancel out with denominator

[winged hearth]

k

[round raptor]

so in nCr

put n and r

[winged hearth]

what is n and what is r

is n 15

and r 10

[round raptor]

yes

[winged hearth]

so after manual calculation I am getting 3 7 13 * 11

[winged hearth]

that's some work

[round raptor]

right

[winged hearth]

yea it's 3003

[round raptor]

yes

[winged hearth]

but it's 3 people are the same

so it is 15!/7!*8!

[round raptor]

?

[winged hearth]

no

[round raptor]

where are 3 same ?

[winged hearth]

3 sisters must not be seperated

it's wrong

[winged hearth]

this

[round raptor]

oh that is next part

applying restrictions

so, 3 sisters are always selected

[winged hearth]

oh shit

i thought that 3 sisters part is for 1a

mb

k

[vapid bluff]

i think the answer will be very big

[round raptor]

it is but there two parts for 1

[round raptor]

for?

[round raptor]

now you know 3 always selecte

[vapid bluff]

for the pinned qsn

[round raptor]

a 1?

[winged hearth]

calc is allowed

[vapid bluff]

this one

[winged hearth]

yea

[vapid bluff]

alr

[round raptor]

you have to select remaining 7

[winged hearth]

yes

but there will be 12 people to select from

[round raptor]

right

so 12C7

[vapid bluff]

you know the ans arsh?

[winged hearth]

k

[winged hearth]

i have answer key but i have not checked

792

[vapid bluff]

its coming out to be 19508428800

💀

[round raptor]

dw we'll see if its big we will leave it without calculating

[round raptor]

lmao no way

[winged hearth]

not possible

[vapid bluff]

alr maybe i did some mistake

[winged hearth]

e u using calc

r u

[round raptor]

right

[winged hearth]

that's the answer

?

[round raptor]

yes

[winged hearth]

k

I'll check

[round raptor]

,calc 1198

[woven radish]

Result:

792

[winged hearth]

it says 858

?

[round raptor]

huh

[winged hearth]

answer key says 858

[round raptor]

oh i misread the question

they say sisters must not be separated

[winged hearth]

uh then?

[round raptor]

if you select none of3 then they would not be separated

[winged hearth]

oooooo oooooo

wow

[round raptor]

so you need to make cases where you always select them and you select none of them

and the cases

[winged hearth]

i was so happy

k imma try

wait

[round raptor]

we have answer for 1st case

find 2nd

out of 15 you don' have to select 3

[winged hearth]

got it

12C10

[round raptor]

pnc is like that

[round raptor]

correct

[winged hearth]

792+66

[round raptor]

right

[winged hearth]

yaaaaaaay

[round raptor]

that would be 858

[winged hearth]

yeye

[round raptor]

https://tenor.com/view/qawe-asd-gif-26050335

[winged hearth]

imma post again

, rotate

[round raptor]

can you send image

this is blur

[winged hearth]

, rotate

[winged hearth]

k

i believe this is harder

[round raptor]

what do you think

fo a number to be even

[winged hearth]

last place should be 0,2,4,6,8

0 is not an option for the first place

we find permutation for all conditions

• when 0 is 1st place(idk how

then - when 0,2,4,6,8 is not the last place

[round raptor]

divide in two cases where 0 is in last place and it is not

[round raptor]

0 can't be in 1st

[winged hearth]

we - the answer from the main thing

[round raptor]

case 1 0 is in last place

[round raptor]

oh

ok

so consider you can place 0 in first place

how many ways can you arrange them all

[winged hearth]

so idk how

[round raptor]

how many ways can you fill first digit

how many numbers can you put

[winged hearth]

10

but actually 9

[round raptor]

right

[round raptor]

we consider 0

[winged hearth]

so 10

[round raptor]

2nd digit

[winged hearth]

9

no

we have to consider that even number

[round raptor]

ah wait

[winged hearth]

ig

[round raptor]

yeah

let's fill last digit first

[winged hearth]

k

[round raptor]

so how many ways for last digit

[winged hearth]

5 of the numbers can be in the last place

[round raptor]

yes

now let's fill others

how many ways

[winged hearth]

so assuming last is even

then 5odd and 4 even

so 9

[round raptor]

right

so till 4 more digits

9*8*7*6*5 multiplied by last 5

[winged hearth]

huh

[round raptor]

the* gets formatted

[winged hearth]

ik that

but why 9

ok i understand 9

but why 8

and. not 9

[round raptor]

you have filled 2 digits with 2 numbers how many remain

10-2

[winged hearth]

so 9 can be 1st place bcz 0 is not there so we have 9 numbers technically for 1st place but we have 10
but when we go to 2nd position we have that 0 so it is 9?

[round raptor]

we consider 0 at first place

[winged hearth]

why

to subtract?

[round raptor]

your method

[winged hearth]

k

got it

so 1 st is 9

thinking we have 1 even for last

[round raptor]

right

[winged hearth]

so

9* 8* 7* 6* 5* 5

[round raptor]

yes

now

[winged hearth]

this is when it is 0

when it is and it isn't 0 then it will be

10!/6!

[winged hearth]

but this isn't including that last is even

[round raptor]

this is when first place is 0 or not zero

[round raptor]

how ?

[winged hearth]

it is 0

[winged hearth]

10 total possible outcomes
but only 6 can be used

[round raptor]

we include both cases where first place canbe 0 and not, now subtract where 1st place is0

[round raptor]

?

[winged hearth]

the normal formula

[round raptor]

no, we have to put even number as last digit

[round raptor]

.

subtract cases where 0 is in 1st place

[winged hearth]

other way aroun

?

can not be 0- can be 0

[round raptor]

no
you said we - arrangements where 0 is in 1st place
so we consider we can place 0 in first place and calculated arrangemets
5numbers of last digt
then 9 for 1st, here if we put 2 as last digit
then 0....2 and 1...0...2 is also possible so this contains both cases where 0 is in 1st place and not

so we just subtract the cases where 0 comes 1st like 013542

[winged hearth]

yea that's what I said ig

so 102746-023138

eg

[round raptor]

ok, instead of cannot you should write can and cannot

[winged hearth]

oh yea my bad

[round raptor]

so, how many arrangements with 0 at first

[winged hearth]

i was thinking this and wrote that that's y I lose marks

[round raptor]

i feel you

[winged hearth]

75600

[round raptor]

,calc 87654

[woven radish]

Result:

6720

[round raptor]

hm..

[round raptor]

what did you do?

[winged hearth]

98765*5

[round raptor]

ok

[round raptor]

and for this

[winged hearth]

no

wait

[winged hearth]

i am wrong with this

i can't get - can i

[winged hearth]

151200

[round raptor]

you are rght but that is total arrangements

[round raptor]

what did you do?

[winged hearth]

oooooo my god

i am so dumb

109876*5

[round raptor]

mm.. no

put 0 in first place, 1st digit fixed

now last digit

how many numbers can you put ? (2,4,6,8)

[winged hearth]

4

umm

[round raptor]

right

so 0 x x x x (4ways) x represent slots

[winged hearth]

this is for when it is 0 and even?

[round raptor]

right

now you have fixed two numbers 0 at first and even at last

how many ways can you fill other 4 slots

[winged hearth]

so out of 4 even we used 1 we have 3 remaining and we have 1 removed already (0)
so 8 numbers can be in the remaining 4 slots

8

[round raptor]

yes

so how many ways?

[winged hearth]

1680

8!/4!

[round raptor]

right

but 1680* 4

last digit

[round raptor]

so this

[round raptor]

- this

[winged hearth]

uuh

why *4

[round raptor]

last digit has 4 ways

even

[winged hearth]

yea

ok

got it

yes yes

[round raptor]

yay

[winged hearth]

6720

[round raptor]

turned out to be harder this way

[round raptor]

right

[winged hearth]

oh

[round raptor]

but we reached

[winged hearth]

waah

[round raptor]

ye

it's late

do you have boards ?

this year

[winged hearth]

yes 10th boards

who cares

[round raptor]

oh

gl

[winged hearth]

igcse so doesn't matter

[round raptor]

hm..

i was icse

[winged hearth]

u r in?

[round raptor]

11 th

[winged hearth]

college?

[round raptor]

yeah school

[winged hearth]

k

r u doing any competitive exams

[round raptor]

yeah

[winged hearth]

which ones

[round raptor]

engineering

[winged hearth]

jee?

[round raptor]

yes

[winged hearth]

oh

which coaching

[round raptor]

i can't tell here

dm

btw !done

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[winged hearth]

.close

[devout snow]

Channel closed

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[devout snow]

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[neon wagon]

can someone help with the d) so

[neon wagon]

can i just do p(x<=92) + (1-p(x<=113)

or

for the d)

[polar chasm]

are you sure u should interpret it as x <= 92?

[neon wagon]

oops

[polar chasm]

shouldnt it be just x <= 93 (since height is "continuous")

[neon wagon]

but it says less than 93

ik what u mean by continious but

[polar chasm]

yeah, but its quite unlikely that anyone has exactly 93cm

[neon wagon]

uhh

i dont understand

[polar chasm]

you already have to count 92.9999999999999999cm people

x <= 92cm doesnt count those

x <= 93cm is practically the same as x < 93cm if x is distributed among reals

[neon wagon]

hmm okay

yeah its confusing cuz it says less than

ohhh

i get what u mean now

shits confusing

[polar chasm]

it makes more sense that way, because now 93 and 113 are same distance from the mean

[neon wagon]

hmmm

[polar chasm]

so u can do it even more simply by 2*p(x<=93) (by symmetry of the bell curve)

[neon wagon]

huh

yeah the bell is symetric but

the mean is 103

but my way is correct too?

[polar chasm]

yeah

[neon wagon]

also what do u refer to

like lets say ur doing normal distribution but a height unit or something

you dont have a p then right

a probability

[polar chasm]

wdym?

[neon wagon]

how do you refer to the given like

wait let me show u an example

[polar chasm]

here's what i meant btw, both 113 and 93 are 10 units from mean, so the areas of the tails are gonna be the same, due to symmetry

[neon wagon]

so, you see the 2500 parts, theyre different from when you use a discrete probability

[neon wagon]

you cant call them p

[polar chasm]

are u asking how to refer to the parts?

maybe just call x the number of scrap parts produced, then its asking about p(x=0)

[neon wagon]

yes yes

okay so

i already did that i need help on another thing

wait

oh wait btw

its "r“

i saw

[polar chasm]

what is r?

[neon wagon]

the 2500 parts

like

if i would use the fi function

r - mu / sigma

is the Z value

if im not allowed to use a calculator

[polar chasm]

fi function?

oh do you mean phi?

[polar chasm]

the integral?

[neon wagon]

this stuff

[neon wagon]

i guess yeah

[neon wagon]

this is another exercise tho

[polar chasm]

yeah, so whats ur question?

[polar chasm]

Are u asking about what r should be here? (in the machine parts Q)

[neon wagon]

oh no wait

let me show u

so here

it translated poorly but its still there

[neon wagon]

so the 100 km are they relevant in any way

no right

[polar chasm]

not really

the only important thing is the rate of consumption (and the deviation)

[neon wagon]

yeah

okay so

uhh

wait let me think

so for the d)

[polar chasm]

i dont see d

[neon wagon]

i just have to like

oops

i mena c

c)

uhhh

yeah wtf do i do there 😭

is it something with the deviation

[polar chasm]

I'd figure out the real probabilities

[neon wagon]

so 2% of 18 and 10% of 18

[polar chasm]

whats the probability that a vehicle is between 2% below and 10% above the mean?

[neon wagon]

omg

its like

wait

p(mu-2% of the consumption <= x <= mu + 10% of the consumption)

?

[polar chasm]

yep

[neon wagon]

holy shit

[polar chasm]

and the salesman claims that that probability is at least 50%

[neon wagon]

so if the probability isnt above or 50 then its wrong

but wait why is <= x < = correct

[polar chasm]

"...deviate from the standard fuel consumption by a maximum of 2% below and 10% above"

[neon wagon]

oh

yes

can i keep this channel open ? im studying rn and i get stuck sometimes

[polar chasm]

hmm actually it might be better if u open a new one every time

ill probably not be able to respond soon and opening a new channel will bring u more attention

[neon wagon]

okay ty

[devout snow]

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[coral forum]

Hi! I need help with letter (b).

[coral forum]

Thanks in advance!

Do i need to find v first?

[vapid veldt]

you need help with b) of problem 3.23 or 3.24?

for 3.24, it helps to remember that cosine is the ratio of the adjacent and the hypotenuse of a right triangle

so, for a ratio to be positive, the numerator and denominator BOTH have to be positive or negative

now, since the hypotenuse is always positive, the adjacent must also be positive, which means that angle v points in a direction where the horizontal (x) axis is greater than 0

[devout snow]

@coral forum Has your question been resolved?

[coral forum]

Ohh problem 2.34

[devout snow]

@coral forum Has your question been resolved?

[hollow jolt]

you mean 3.24?

[woven vale]

we know a few things about sin(90°), cos(90°), and cos(v) on their own, right?
is there an identity that would "split" (90° - v) into ratios of 90° and v, by themselves?

[devout snow]

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[marble hearth]

Divide −4𝑥^2+6𝑥 by 3𝑥

When dividing algebraic expressions, we can write the division as,

−4𝑥^2 + 6𝑥 / 3𝑥 or (−4𝑥^2+ 6𝑥) ÷ (3𝑥)

[marble hearth]

We use the fact that division is the same as multiplication by the mutiplicative inverse:

(−4𝑥^2 + 6𝑥) * 1/3𝑥

And now we use the distributive property to evaluate the expression,

−4𝑥^2 /3𝑥 + 6𝑥/3𝑥 = −4/3𝑥 +2

Why does the reciprocal matter here?

aren't we just dividing each one by 3x? I don't see any multiplication in the distributive property

[young spade]

tbh it doesnt really matter

its just another way to write it

in the reals and rationals

Division is literally multiplication by the inverse

[void fox]

the first term should be written as -4x/3 instead of -4/3x

[young spade]

oh yeah, didnt see

[marble hearth]

but how would we even multiply say 6x * 1/3x

6x/1 * 1/3x = 6x/3x?

[jade bobcat]

maybe it's helpful to latex the fractions so where the x is is unambiguous

$6x\cdot \frac{1}{3x} = \frac{6x}{3x} = 2$

[marble hearth]

ahh I need to familiarize myself with latex

[devout snow]

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[woven haven]

!help

[devout snow]

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

[violet wind]

that should probably be different if you open a thread with it

anyways this is the right place, post your question and people will see it and hopefully help

[devout snow]

@woven haven Has your question been resolved?

[devout snow]

@woven haven Has your question been resolved?

[woven haven]

.close

[devout snow]

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[wise violet]

Can someone pls check if I did this right? I’m in AP Precalc and we’re working on trig functions for this chapter

For the work above, I ignored the 9 degree angle bc I didnt see a use for it. Pls lmk if I’m wrong

[azure parrot]

that's a sign that you might be doing smth wrong

And yah it's wrong

[wise violet]

Nvm I did it wrong

I don’t have the full base length oops

Gimme a moment to figure this out 😭👍

Lowk dunno how to do it

😭

I redrew the diagram tho

Hopefully this is at least right

[round raptor]

yes

[wise violet]

Idk what to do next ☹️

[azure parrot]

you should be able to relate d and the whole base length to h right?

[wise violet]

Oh

[azure parrot]

similar to what you did here

[wise violet]

Tan of both angles
And then rearrange so both equal to h

Then set them equal to each other?

[azure parrot]

yeah

[azure parrot]

or you rearrange them so both equal to d, so you can find h directly

[wise violet]

Ohh

[azure parrot]

this's right, I think, after you get d you can compute h

[wise violet]

Oh

I think I’d rather get to h directly

Would this be the right process so far

[azure parrot]

looks good to me

[wise violet]

Idk how to isolate h 😭😭😭

[azure parrot]

say you have 5h=3h-13

How would you find h

[wise violet]

Oh subtract 3h from both sides

[azure parrot]

well, 1/tan(3.5) is just a number

approximately 2.6696

you can just do the same thing here

[azure parrot]

factor out the h

[wise violet]

Oh
I keep forgetting that I can factor in situations like this 😭

Yay

How come I can’t do this btw

In the box

Instead of dividing -13 by a fraction, I flip the fractions in the denominator? It doesn’t work tho and idk why

[azure parrot]

$\frac{1}{\tan(9^\circ)}-\frac{1}{\tan(3.5^\circ)}$ is not A fraction

[woven radish]

Alexis_Fx <- sucks at Math

[azure parrot]

There're 2 fractions

If we make them a single fraction

$\frac{1}{\tan(9^\circ)}-\frac{1}{\tan(3.5^\circ)}=\frac{\tan(3.5^\circ)-\tan(9^\circ)}{\tan(9^\circ)\tan(3.5^\circ)}$

[woven radish]

Alexis_Fx <- sucks at Math

[azure parrot]

then can flip the rhs

[wise violet]

😭

Okayy

Tysm for your help 🙏

[azure parrot]

np

[wise violet]

Have a nice day

.close

[devout snow]

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[wanton tiger]

i tried to square it then subtract by 4b both sides to try and get a workable eq but i didnt

i honestly dont have any clue how to solve with powers when u cant simplify it

[vagrant skiff]

is this a multiple-correct question?

[wanton tiger]

yea

[gloomy aurora]

$a^2 - b = (a+ \sqrt{b})(a-\sqrt{b})$

[woven radish]

Annie Maqionde

[proud perch]

This doesn't even make sense. Roots in which variable?

[vagrant skiff]

x

[wanton tiger]

what do u mean?

[vagrant skiff]

it does make sense

[wanton tiger]

mb a and b are constants'

[proud perch]

Oh I see

[gloomy aurora]

take $y = (a+\sqrt{b})^{x^2-15}$ and its a quadratic in y

[woven radish]

Annie Maqionde

[wanton tiger]

how would i express a - root b?

in terms of y

uhh

1/y

mb 😭

so

im getting the roots of the quadratic in y as (a + rootb) and (a - rootb)

so from that

im getting options 1 and 3

[vagrant skiff]

correct

[wanton tiger]

dang

its kinda hard to think in that way

tysm yall

.close

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[viral lynx]

So for f_4(x)

Just so that I understand, is this what they did

[smoky gyro]

odieee

[fossil locust]

there's no asymptotes

[viral lynx]

Hi beeeeeam!!!! 🤗 ❤️

[viral lynx]

Wait so its wrong?

[fossil locust]

yeah, cause for x < -1, f(x) exactly equals -2

and for x > 1 f(x) exactly equals 2

[fossil locust]

yep

[viral lynx]

Damn crazy

So f_2(x) is the only answer?

[fossil locust]

yeah

[viral lynx]

Ok thanks!

❤️

.close

[devout snow]

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[fossil locust]

no worries!

[devout snow]

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[sand girder]

Given a, b, c > 0 so that a+b+c=3, find min of $\frac{a^2+67}{a+1} + \frac{b^2+67}{b+1} + \frac{c^2+67}{c+1}$

[woven radish]

Thomas

[sand girder]

I think that we can transform it to $a - 1 + b - 1 + c -1 + 68*(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}) \rightarrow 68*(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1})$

[woven radish]

Thomas

[steady kettle]

yes i just did it

[sand girder]

which means that we only need to find min of $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

[woven radish]

Thomas

[azure parrot]

Titu's Lemma should be able to do this

[sand girder]

which gives 68*3/2 = 102

so that should be it

can anyone send q query to draw this out?

[steady kettle]

thats correct

[sand girder]

onto the next question in the practice test that i have problems with: geometry

[fossil locust]

who tf made that practice test

67

[sand girder]

Given two circles (O) and (O') with equal radius R. Draw O'A as a tangent to (O), and draw OB as a tangent to (O'). On the ray opposite to AB and BA, respectively, take points C and D such that AC = BD. Construct the rectangle EFCD whose length is greater than 4R. Draw (I) with diameter EF. Choose point L on (I) such that triangle ELF is an isosceles right triangle with the right angle at L. Draw EH and EN as tangents from point E to (O); draw EK and EM as tangents from point E to (O') (with N lying inside (O'), and M lying inside (O).

Prove that HMNK is an isosceles trapezoid.

[rough nova]

Diagram please

[sand girder]

Wait

am redrawing it on geogebra

my original one looks 3x more confusing just from the scrub marks alone

[devout snow]

@sand girder Has your question been resolved?

[sand girder]

<@&286206848099549185>

[frail summit]

What part are you stuck

[sand girder]

I'm just rlly bad at geometry in general

I just can't see where the pieces connect

[frail summit]

That so much going on

[sand girder]

I kinda need help w this: "draw EK and EM as tangents from point E to (O') (with N lying inside (O'), and M lying inside (O)" (original ques but i don't know how to draw that so i just let N inside (O) and M inside (O'))

[frail summit]

You draw this?

Oh I thought it was given

[frail summit]

I think you got confused here it should be " (with N lying inside (O), and M lying inside (O')."

[devout snow]

@sand girder Has your question been resolved?

[sand girder]

but that doesn

't give me trapezoid

[devout snow]

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[iron chasm]

stuck

.close

[devout snow]

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[whole hornet]

looking for a good book to practice math analysis from start
(proving axioms and then reals etc)

[supple knot]

#book-recommendations

[whole hornet]

.close

[devout snow]

Channel closed

Closed by @whole hornet

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

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[raven kayak]

\begin{Theorem}
Assume $A$, $B$ and $C$ are sets. Then, \ $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
\end{Theorem}

\begin{proof}
Let $A$, $B$ and $C$ be sets.

First, we will proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.
Suppose that $$x \in A \cap (B \cup C).$$
Then, by the definition of the intersection, $$x \in A \text{ and } (x \in B \cup C).$$
And, by the definition of the union, $$x \in A \text{ and } (x \in B \text{ or } x \in C).$$
For this to be true, $x$ must be in $A$, and it can be either in $B$ or $C$.
Said differently, $$(x \in A \text{ and } x \in B) \text{ or } (x \in A \text{ and } x \in C).$$
Then, by the definition of the intersection, $$(x \in A \cap B) \text{ or } (x \in A \cap C).$$
And, by the definition of the union, $$x \in (A \cap B) \cup (A \cap C).$$
Since $x \in A \cap (B \cup C)$ implies $x \in (A \cap B) \cup (x \in A \cap C)$,
we have $$A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C).$$
Next, we will proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.
Suppose that $$x \in (A \cap B) \cup (A \cap C).$$
By the definition of the union, $$x \in (A \cap B) \text{ or } (x \in A \cap C).$$
Then, by the definition of the intersection, $$(x \in A \text{ and } x \in B) \text{ or } (x \in A \text{ and } x \in C).$$
For this to be true, either $x$ is in $A$ and $B$, or in $A$ and $C$.
In both cases, $x$ must be in $A$, and then it can be either in $A$ or $B$.
Said differently, $$x \in A \text{ and } (x \in B \text{ or } x \in C).$$
Then, by the definition of the union, $$x \in A \text{ and } (x \in B \cup C).$$
And, by the definition of the intersection, $$x \in A \cap (B \cup C).$$
Since $x \in (A \cap B) \cup (x \in A \cap C)$ implies $x \in A \cap (B \cup C)$,
we have that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.
Therefore, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
\end{proof}

[woven radish]

Mor Bras

[raven kayak]

Hello, could someone check if this proof looks good please?

[obsidian torrent]

seems fine to me but im no expert

[raven kayak]

Yay! ...I guess?

[obsidian torrent]

oh u did misspell "and" in ur fourth last line

[raven kayak]

\begin{Theorem}
Assume $A$, $B$ and $C$ are sets. Then, \ $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
\end{Theorem}

\begin{proof}
Let $A$, $B$ and $C$ be sets.

First, we will proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.
Suppose that $$x \in A \cap (B \cup C).$$
Then, by the definition of the intersection, $$x \in A \text{ and } (x \in B \cup C).$$
And, by the definition of the union, $$x \in A \text{ and } (x \in B \text{ or } x \in C).$$
For this to be true, $x$ must be in $A$, and it can be either in $B$ or $C$.
Said differently, $$(x \in A \text{ and } x \in B) \text{ or } (x \in A \text{ and } x \in C).$$
Then, by the definition of the intersection, $$(x \in A \cap B) \text{ or } (x \in A \cap C).$$
And, by the definition of the union, $$x \in (A \cap B) \cup (A \cap C).$$
Since $x \in A \cap (B \cup C)$ implies $x \in (A \cap B) \cup (x \in A \cap C)$,
we have $$A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C).$$
Next, we will proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.
Suppose that $$x \in (A \cap B) \cup (A \cap C).$$
By the definition of the union, $$x \in (A \cap B) \text{ or } (x \in A \cap C).$$
Then, by the definition of the intersection, $$(x \in A \text{ and } x \in B) \text{ or } (x \in A \text{ and } x \in C).$$
For this to be true, either $x$ is in $A$ and $B$, or in $A$ and $C$.
In both cases, $x$ must be in $A$, and then it can be either in $A$ or $B$.
Said differently, $$x \in A \text{ and } (x \in B \text{ or } x \in C).$$
Then, by the definition of the union, $$x \in A \text{ and } (x \in B \cup C).$$
And, by the definition of the intersection, $$x \in A \cap (B \cup C).$$
Since $x \in (A \cap B) \cup (x \in A \cap C)$ implies $x \in A \cap (B \cup C)$,
we have that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.
Therefore, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
\end{proof}

[woven radish]

Mor Bras

[devout snow]

@raven kayak Has your question been resolved?

[raven kayak]

<@&286206848099549185> 👋

[woven vale]

looks good to me!

[raven kayak]

are you an expert? 😆 (look earlier messages)

[woven vale]

well, i did all these set theory problems 2 years ago :) but i dont think i'm an expert >.>

[raven kayak]

It's a joke, thanks for your response

.close

[devout snow]

Channel closed

Closed by @raven kayak

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• 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.

[trail cypress]

Not sure how to proceed. I was thinking of inverting the second fraction which would then be first fraction multiplying the second fraction but I don't think it will lead me any further...

[scarlet sequoia]

Maybe firstly raise the powers and then do it

You'll be able to cancel out something and it'll become simpler

[trail cypress]

Ok brb