help-39

[tawdry wing]

no

[wispy quail]

🥹

the answer should be this

[tawdry wing]

maybe i made mistake due to speed

i'm sry

[wispy quail]

oh okay

i forgot to subtract

at the beginning

[tawdry wing]

i just see it

ahahahahahah

speeed

ahhh

Here it is!

Hope it helps! @wispy quail

[wispy quail]

thank you so much

!!!!

[tawdry wing]

you get it?

[wispy quail]

yes, i just needed the algebra on this

[tawdry wing]

🙂

[wispy quail]

btw, where did you do this?

[tawdry wing]

In high school.

[wispy quail]

oh . i meant. is there a site? application...

[tawdry wing]

???

[tawdry wing]

By myself.

If you meant that.

[wispy quail]

ok, idk what i meant.
thanks a lot!

[tawdry wing]

no problem.

my english isnt that good, sry

🙂

[wispy quail]

do you have instagram? could help to bother you with a few things

[wispy quail]

your english is perfect

[tawdry wing]

you can write me in DM on Discord.

If you want.

Or here.

As you want.

[tawdry wing]

Thanks a lot! 😉

[wispy quail]

i don't use discord much. but thanks

appreciate it

.close

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[tawdry wing]

see ya!

[pearl pond]

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[karmic osprey]

ive done part a but im not very sure of my answer, and i dont really know how to start with part b

am i supposed to parametrize the contour? i think that would be annoying and that its not the point of the exercise

I understand that the argument somehow jumps from C to E as well as B to F so having this contour allows f(z) to move smoothly, but what about A and D?

<@&286206848099549185>

[pearl pond]

@karmic osprey Has your question been resolved?

[midnight haven]

Can you send an image of your work and maybe some of the notes you have used for doing part a? .
Thanks

@karmic osprey

[karmic osprey]

heres what ive managed so far

actually i think a) should be this and now it really doesnt make sense

[pearl pond]

@karmic osprey Has your question been resolved?

[midnight haven]

I will try to look, at it tomorrow(morning) . The calculation seems fine, I just want to sketch the function for myself. Cause it really confusing. Is there any reference (topic for part B), part A seems intuitive one.

[pearl pond]

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[edgy jasper]

can someone help me?

[summer imp]

[edgy jasper]

Determine the length of BC to the tenth. With angle ABC= 39 degrees, Angle ADC= 90 degrees, Angle
ACD= 65 degrees, side AD= 12.7 m. (Note that C is in between B and D)

can u check this

[pearl pond]

@edgy jasper Has your question been resolved?

[edgy jasper]

no ones gonna help?

usually i see 30 people helping one kid

[harsh orbit]

It's correct. You used the sine law properly; however, in the future make sure you're using recommended formatting for it. Like you know how you label using a^2+b^2=c^2 when you're using the Pythagorean Theorem, you should use a,b,c for labelling side lengths and A,B,C for angles.

[harsh orbit]

But what you're doing isn't wrong, but it just makes it look cleaner when you're using sine law!

[edgy jasper]

@harsh orbit ty bro and next time ill use sine law

[harsh orbit]

your using it already

[edgy jasper]

i meant to say . Like you know how you label using a^2+b^2=c^2 when you're using the Pythagorean Theorem, you should use a,b,c for labelling side lengths and A,B,C for angles. lol

are u in high school

or uni or college

[harsh orbit]

Oops, sorry, I was rushing. I didn't read it fully properly. I don't think it's 27.2 metres.

It should be 9.8 metres

[harsh orbit]

high school

here is a diagram

You use tan(x) to figure out DC, which you can use then to figuree out AC (pythagorean theorem). Then you can use the 180 angle rule to figure out angle BAC, and then sine law to do (14)/(sin39) = (a)/(sin26).

[pearl pond]

@edgy jasper Has your question been resolved?

[edgy jasper]

wait so

its 20.2 right

@harsh orbit

[harsh orbit]

huh

is it not b to c

9.8

[naive hemlock]

Yo

Here for the help

@harsh orbit

Try similarity of triangles

@edgy jasper

[harsh orbit]

Won't work bc their angles don't match up. Easiest to use trig. functions + pythagorean theorem & then sine law.

[naive hemlock]

No listen

The small triangle

And big triangle

Not middle one

Yus

[harsh orbit]

theres only 2

triangles

[naive hemlock]

Bro

[naive hemlock]

See in this

3

One the stretched one

One small

And one which contains both of them

3 fr

@harsh orbit tell what we have to find

[harsh orbit]

thats already solved

we aren't looking for a scale factor or anything it's solved

i was helping the other guy

[naive hemlock]

Okok

[harsh orbit]

its the diagram i made for his problem

[pearl pond]

@edgy jasper Has your question been resolved?

[pearl pond]

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[kindred reef]

Dikshant writes down 2k+1 positive integers in a list where k is a positive integer. The integers
are not necessarily all distinct, but there are at least three distinct integers in the list. The
average (mean) of all 2k + 1 integers is itself an integer and appears at least once in the list.
The average of the smallest k + 1 integers in the list and the overall average of the list differ
by less than 1/2025. Likewise, the average of the largest k + 1 integers in the list and the overall
average of the list differ by less than 1/2025. Determine the smallest possible value of k.

Clarification: If m is the overall average, m1 is the average of the smallest k + 1 integers, and
m2 is the average of the largest k + 1 integers, then the conditions are |m − m1| <1/2025and
|m − m2| <1/2025.

[kindred reef]

i dont even know where to start

[viscid sierra]

me neither

[vital estuary]

play with some lists of numbers and try to see how to get the average of the low numbers close to the total average

[kindred reef]

😭

[vital estuary]

what

[kindred reef]

would it work

[vital estuary]

im giving you somewhere to start

[kindred reef]

o ok

ill try

[viscid sierra]

I guess you can try with a limiting case

m-m1=1/2025

and m2-m=1/2025

The
average (mean) of all 2k + 1 integers is itself an integer and appears at least once in the list.

This should definitely be of some help

[kindred reef]

k thx

[vital estuary]

heres a pretty big hint: mean of 6 and 12 is 9, mean of 6, 12, and 12 is 10

[kindred reef]

ykw

i think imma give up on this problem 😄

its fine cuz it didn't rlly matter

thx for the help tho lol

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[rain turret]

45?

[pearl pond]

@rain turret Has your question been resolved?

[rain turret]

. close

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[dense iris]

Can i get help on this question

This is the first part of it

[naive hemlock]

Yo

Boy

@dense iris

[dense iris]

yo

yee

[naive hemlock]

Me back

[dense iris]

sup

[naive hemlock]

What we have to do

Btw

Matching?

[dense iris]

This

[naive hemlock]

Oh

Oh oh

[dense iris]

I sent first part of the question above

cause it's related

Idk how to use that info to solve this tho

[naive hemlock]

Okok

What if we expand

Like sin(A-B)

And tan(A-B)

[dense iris]

I need to expand?

[naive hemlock]

Wait they have given above

[dense iris]

yeah

[naive hemlock]

Got it

See

It is

Given that

We have to solve

5sin(2x-50)=9tan(x-25)

We can set x-25 as t

So we have

5sin(2(t))=9tant

Hello you there?

[dense iris]

ye

[naive hemlock]

So now we expand sin 2t

So we have

[dense iris]

2sintcost?

[naive hemlock]

10sint.cost=9tant

[dense iris]

ok

[naive hemlock]

So we get

10sint.cost=9sint/cost

Sint cancel

We get

Cos 2 t=9/10

So we get cost=+-√9/10

[dense iris]

oh crap

I was asking for part b

soz

shoulda made it clear

[naive hemlock]

Bro listen in b it i same

Just take x-25 as t

And it convert to cos^2t=9/10

so u get two values of t

Hello?

[dense iris]

yeah

[naive hemlock]

Did u understand?

[dense iris]

so I do arc cos root 9/10

and divide it by 2?

[naive hemlock]

no

This is t

[dense iris]

ohhh

[naive hemlock]

You need x

[dense iris]

yaeh

[naive hemlock]

t=x-25

So x=25+arccos(+-√9/10)

One time plus one minus

Ok

[dense iris]

uh divide by 2 as well right?

[naive hemlock]

Why?

[dense iris]

oh no

ok lemme check

[naive hemlock]

Listen they have done what

We got answer by cos^2t=9/10

So two answers

If we do by sin

We get two answers

🙂

That's why they having 4 answers

[dense iris]

Thing is

they get -18.4 here

and I get positive 18.4

when taking arc cos of root 9/10

[naive hemlock]

Bro listen

We have arc cos (+-

Understanding

Cos we did root

So we got cos t=+-

Now try urself

[dense iris]

yeah if i do positive arc cos

I get 18.43

negative arc cos returns me 161.545 tho

[dense iris]

how dyu know ur meant to use the positive value or negative value

[dense iris]

they only show the neg value here

[naive hemlock]

Listen bro

This is because bro

If we use -ve

@dense iris

They have only used -ve?

[dense iris]

lemme just see if I got it right

I'm supposed to do arc cos root 9/10 right?

that gets me positive 18.4

but they have negative 18.4

in the mark scheme

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[dense iris]

.reopen

[pearl pond]

✅

[dense iris]

Why is the length here not just 6?

[sterile turtle]

?

length of OD?

use your length formula

[dense iris]

l = r theta

I'm so confused why half of the diamter

[sterile turtle]

wait why do you think its 6

[dense iris]

doesn't get me it

cayse it says the length is 12 in AOD

and O is the centre

[sterile turtle]

theyre two different circles

so the radius is different

[dense iris]

😱

ok

so l = r theta

[sterile turtle]

ye

[dense iris]

I forgot

how to do this

fk

theta is ofc 0.4

[sterile turtle]

lmao

[dense iris]

radius is what

[sterile turtle]

we trying to figure that out

[dense iris]

ohh right

[sterile turtle]

so thats just a pronumeral

[dense iris]

so 3 = r theta

[sterile turtle]

yep

[dense iris]

3 = 0.4 r

r = 1.2

l = 1.2 x 3

[sterile turtle]

oooo mate

thats not a times

thata a divide

[dense iris]

fakkkk

[sterile turtle]

l/theta = r

[dense iris]

yeah r = 12

[sterile turtle]

uh-

[dense iris]

7/5

i mean

7.5

[sterile turtle]

ye

nice

[dense iris]

oh thats the final answer

[sterile turtle]

yup

[dense iris]

ok different circle

damn that confuzzled me

thanks

.close

[sterile turtle]

np

[pearl pond]

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[dense iris]

I got the answer to this with ratios

but

I kept get it wrong when trying to form an equation of a line

why?

y - 16,000 = -1750t

basic algebra mistake

thats why

fuck

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[wild fable]

how do i do this?

[ruby monolith]

Is it option B

?

[wild fable]

idk

i dont have the answer

[ruby monolith]

@wild fable

[wild fable]

thanks ill have a look

[midnight haven]

blud doing trigonometry 🙏🙏

[wild fable]

shush

go study for jee adv

[midnight haven]

nah

i quit

[ruby monolith]

Same I am also s

Studying for jee adv

[wild fable]

bruh

how cnan u quit

[midnight haven]

what's the prob

[wild fable]

why quit

[cinder flower]

water beam

pun pun

[midnight haven]

its pain man i dont wanna do it

[wild fable]

slayla

[midnight haven]

sharp

[wild fable]

but you still have to do it

[cinder flower]

do it now

[ruby monolith]

@wild fable have you got your answer

[wild fable]

yeagh im looking thru it 👍

[wild fable]

you tell him

[cinder flower]

@midnight haven

[midnight haven]

there must be other options too

it was just a waste of money after all

🙏

[wild fable]

if slayla tells u to do something u do something

how are u gonna become the first indian astronaut lawyer doctor

[ruby monolith]

@cinder flower please don't talk here

If you wanna talk go in #discussion

[wild fable]

[ruby monolith]

Or #chill

[cinder flower]

ok welcome to my opp list

enjoy your stay

[ruby monolith]

@wild fable have you got?

[pearl pond]

@wild fable Has your question been resolved?

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[midnight haven]

,rotate

[midnight haven]

What property is used in solution?

[sharp smelt]

$i+i^2+i^3+i^4=0$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[midnight haven]

No the summation one

r=4 to r=1

[spare lark]

Its decomposition of the original one

[midnight haven]

Decomposition means?

[spare lark]

I mean the same as in vectors : AC = AB+BC

Or in integral

A sort of chasles relation

[midnight haven]

Ok

Thanks

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[tawny tiger]

X and Y are 2 independent variable. X is with uniform density in [-1,1= and Y is exponential with parameter 1. I need to find P(X>Y|XY>0)

[tawny tiger]

i know that fx = 1/2 for x in [-1,1]

and Y = e^-y

but i am not sure what to do from here

[pearl pond]

@tawny tiger Has your question been resolved?

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[visual canyon]

i think it isnt possible because both sides are not the same after simplifying

[visual canyon]

does that seem right?

[sharp smelt]

!show

[pearl pond]

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

[visual canyon]

sorry my caption is all my work

i think that it isnt possible

when you simplify it is different

[sharp smelt]

$1+cos(x)=2cos^2(x/2)$

yes?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[cursive wraith]

why are you using this identity?

[cursive wraith]

? you sure?

last time I checked up with you I asked if you found the domain, i.e the values of theta for which the LHS is even defined

[sharp smelt]

I planned on writing sin^2(x) as 4sin^2(x/2)cos^2(x/2)

[cursive wraith]

ok but it's using half angles for no real reason

where as you could just write sin^2 = 1 - cos^2...

in any case

algebraically verify means:

[visual canyon]

i am not sure no

[cursive wraith]

"let theta be in (this domain)
sin^2(theta)/(1+cos(theta)) = 1 - cos(theta) <=> ...
<=>...
<=> Obviously true statement
Thus the original statement is true"

[pearl pond]

@visual canyon Has your question been resolved?

[pearl pond]

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[dense iris]

How do I work this out

[naive hemlock]

Yo

[hazy solar]

you can subs y=ax and work it out algebraically

just manipulate inequalities

[naive hemlock]

Correct

[hollow gull]

and mindful about abs value tho

[dense iris]

yeah I tried and got it wrong

ig maybe i messed up the algebra

[dense iris]

yh p sure it has to be negative

[dense iris]

did i misunderstand

cause I set ax equal to the other equation

and it doesn't get me the right answer

[hazy solar]

show your working

[dense iris]

ax = 2mod x+4 - 5

ax = -2(x+4)-5

ax= -2x+3

ax +2x = 3

a = 3/x -2

[hollow gull]

abs value can really give negative or positive depends on x tho

[hazy solar]

what happened to your abs

[dense iris]

i thought i use a negative

[hazy solar]

apply your definition carefully

[hollow gull]

not really

can be 2 scenarios

[hazy solar]

\begin{aligned}
|x| := \begin{cases}
x \quad \text{ if } x> 0 \
-x \quad \text{ if } x < 0

\end{cases}
\end{aligned}

[hollow gull]

yep exactly

[dense iris]

yeah and x is less than 0 here

so isnt it negative

I don't see the problem with what i've done

[jolly parrot]

normalAtmosphericPa=101,325
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hollow gull]

it doesnt say negative in the text tho

[dense iris]

so ok

I have ax = 2mod x+4 -5

[hazy solar]

why is this not rendering properly

anyways

[dense iris]

where do I go from there

[hazy solar]

you do it by cases

[dense iris]

wdym by that

[hazy solar]

case 1: x > 0 ...

case 2: x < 0 ...

[dense iris]

how do I know which one is right

ok lemme j expand it for now

[hazy solar]

it's not exclusive either or

[dense iris]

so when x > 0

[hazy solar]

yep

[dense iris]

ax = 2x + 8 - 5

ax = 2x +3

[hazy solar]

"suppose x > 0" ... then we have that ...

"now suppose x < 0" ... then we have that ...

[hollow gull]

|x-4|=0 gives two cases , -x+4 when x<4 and x-4 when x>4

[dense iris]

oh

ok

so

ax = 2(-x+4) -5?

ok cool

I got it now thanks

.close

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[hollow gull]

yr welcome

[dense iris]

.reopen

[pearl pond]

✅

[dense iris]

Dyu know why for this

the negative is on the outside

and its not negative x +4

but rather -2(x+4)

[sudden heath]

For |x+4|, there are 2 cases.
Case 1, x+4>0, then |x+4|=x+4
Case 2, x+4<0, then |x+4|=-(x+4)

[dense iris]

oh rite ok

thx

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[dense iris]

.reopen

[pearl pond]

✅

[dense iris]

Is the range of the inverse function the same as the range of the original funciton?

[inland ivy]

No

The range of the inverse function is the same as the domain of the original function

[dense iris]

ah I get confused with teh two terms

[inland ivy]

$f:A\to B$ has an inverse $g=f^{-1}:B\to A$ iff. $f$ is a bijection from $A\to B$

[dense iris]

so

[jolly parrot]

kheerii

[dense iris]

this is the range of the original function right?

and the domain would be greater than or equal to 16

[dense iris]

idk what a bijection is 😭

[inland ivy]

No, you’re confusing the two terms

For $f:A\to B$ the domain is A and the codomain is B

[jolly parrot]

kheerii

[inland ivy]

The range is the set of possible outputs of f(x)

Range$={f(x)\ |\ x\in A}$

[jolly parrot]

kheerii

[dense iris]

oh right

so then the range for g(x) is 0 to 16

[inland ivy]

Have you heard the term “one-to-one mapping”?

[dense iris]

yeah

[inland ivy]

Yes

[inland ivy]

That’s what a bijection is

[dense iris]

and the domain would be what?

[inland ivy]

Have you heard the terms “injectivity” and “surjectivity”

[dense iris]

noperino

[inland ivy]

What are the possible values of x?

The input

[dense iris]

0, 1, 4, 9, 16

thats the domain?

those values?

[inland ivy]

What?

Why?

[dense iris]

those are the possible values of x

oh

[inland ivy]

What

How did you even get those values

[dense iris]

I subbed in the values of 0-4

into the x^2 function

oh i mean

Ig values of x

are

0,1,2,3,4

[inland ivy]

Does it state anywhere that x must be an integer?

[dense iris]

ohh

no it does not

[inland ivy]

So what would the domain be?

[dense iris]

x <16

but x >0

[inland ivy]

No

It’s quite literally written in your question

[inland ivy]

This

Is the domain

[dense iris]

I thought that was the range

ahhh

[inland ivy]

No

The values of the input to the function form the domain
The values of the output from the function form the range

[dense iris]

oh right

[dense iris]

would this be the range then

[inland ivy]

No

[dense iris]

less han or equal to 16 tho

[inland ivy]

The range is the set of outputs

[dense iris]

oh so all the outputs

[inland ivy]

Yeah, so state it in one inequality

That will be the range

[dense iris]

o

so 0<x<=16

right?

[inland ivy]

Range of $g(x)=[0, 16]$

[jolly parrot]

kheerii

[inland ivy]

Domain of $g(x)=[0,4]$

[jolly parrot]

kheerii

[dense iris]

so is this right?

[cursive wraith]

(small precision : range of g and domain of g, not g(x))

[inland ivy]

Yeah my bad

[inland ivy]

No, you can’t write it in inequality form like this

[dense iris]

oh ok

[inland ivy]

x doesn’t lie between 0 to 16

[dense iris]

ah I get it now ok

[inland ivy]

g(x) lies between 0 to 16

[dense iris]

its like newton raphsons method interval questions

ok thanks

i got it now

[inland ivy]

But we usually use intervals to represent range and domain, not inequalities

[dense iris]

ty

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[tulip sequoia]

Hi, I was wondering if anyone could help me solve 17bii

[sharp smelt]

write $tan^2(x)$ as $sec^2(x)-1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[tulip sequoia]

Ok

[brisk steeple]

break tan^5 x

[tulip sequoia]

Into what

[brisk steeple]

This is a hint.

[tulip sequoia]

I understand that and I could do it for 17bi but I’m not really sure how to do it for the 2nd one

[dapper dawn]

youve just done tan³x

so how do u think u can split tan⁵x

[tulip sequoia]

Tan3x and tan2x?

[dapper dawn]

exc

then u can use ur expression for tan³x

multiply theough and see if u can now solve the integral

[tulip sequoia]

Oops

Oops meant

[dapper dawn]

id keep tan²x as jus tan²x

[tulip sequoia]

Right

[dapper dawn]

u could prolly do it this way tbh

[brisk steeple]

You once used to be cool. f(Why am. I here) = I don't Know

[tulip sequoia]

I’m still a bit unsure

Do I need to know integration by parts?

Hello?

@dapper dawn ?

[odd blade]

Are you here @tulip sequoia

[tulip sequoia]

Hi

Yes

[odd blade]

Hi

[tulip sequoia]

I’m here

[odd blade]

ok so I would start by breaking up tan^5 into ((tan^2)^2)*(tan)

[tulip sequoia]

Ok

.

[odd blade]

yep

then, do you know the trig identity for tan^2?

[tulip sequoia]

Yep

Sec2x - 1

Right?

[odd blade]

correct

so we can replace the tan^2 with that

[odd blade]

right

now can you spot a potential u-sub that would help simplify this problem

[tulip sequoia]

Um

No I can’t

[odd blade]

thats ok

so the common pattern when we do a u-sub is we have a lot on the left side, in this case (sec^2 -1)^2 and something simple on the right, in this case tanx
the tanx does not really fit the problem and so we usually look for a usub to get rid of it
in this case u=secx helps us
this is because when we find du = secxtanxdx this will help divide out the tanx

[tulip sequoia]

So

What exactly should I do next

[odd blade]

this is what it should look like

does this make sense @tulip sequoia

[tulip sequoia]

So

Like that

[odd blade]

yes

[tulip sequoia]

Then you integrate with respect to u right?

[odd blade]

correct, id recommend first multiplying out the numerator

[tulip sequoia]

So this?

[odd blade]

yes very good

what would be the last step from here

[tulip sequoia]

Sub u for sec x

Right?

[odd blade]

right

[tulip sequoia]

Is this correct?

[odd blade]

should be a negative on the sec^2x

[tulip sequoia]

Oh oops

[odd blade]

otherwise yes

[tulip sequoia]

Right

Thank you so much

I’m sorry I took such a long time and everything

[odd blade]

So overall for these types of problems with high exponent trig function, try to follow the pattern:
-Break down using identities and try to get it into two trig functions
-Use one of these two trig function to do a u-sub and try to get it to cancel out one of the trig functions
-Integrate as usual
-Plug back in u

[odd blade]

no worries it didnt take long at all, let me know if you have more questions im happy to help

[tulip sequoia]

Thanks so much

[odd blade]

No problem

[tulip sequoia]

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[visual canyon]

ok trying more of these, are they correct?

[brisk steeple]

,w cos 15

[jolly parrot]

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[brisk steeple]

,w tan 105

[jolly parrot]

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[brisk steeple]

,w (root 6 - root 2) / 2

[jolly parrot]

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[pearl pond]

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[distant dock]

How do I find the point of infliction?

[distant dock]

Its the 2nd derivative right?

[fallen tartan]

yes when it passes the x axis

[distant dock]

so i plug in 0 for f''?

[fallen tartan]

yep

.close

!close

lol nvm

[distant dock]

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[edgy jasper]

Determine the length of BC to the tenth. With angle ABC= 39 degrees, Angle ADC= 90 degrees, Angle
ACD= 65 degrees, side AD= 12.7 m. (Note that C is in between B and D)

[pearl pond]

@edgy jasper Has your question been resolved?

[spare lark]

Have you drawn it ?

[edgy jasper]

na i just need someones help witht he answer

what would the final anser be

[pearl pond]

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[hybrid basin]

Hello

[hybrid basin]

$\left(-ab\right)^{2}=\left(-a\right)^{2}\left(b\right)^{2}$

[jolly parrot]

Chaewon

[hybrid basin]

Is this statement true?

[regal herald]

yes

[tough copper]

Comes from the exponent rule (ab)^n = a^nb^n

[pearl pond]

@hybrid basin Has your question been resolved?

[ruby atlas]

can someone help me

[naive orbit]

!status

[pearl pond]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pearl pond]

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).

[ruby atlas]

oh

[hybrid basin]

Oh wait that was a dumb question

Thanks though!

.close

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[tranquil root]

Hey

I couldn't understand how the guy from a vid I watched solved this division, particularly the reducing step.
I understood that we can reduce 4 from the first fraction but how can we reduce the 9 and the 13 if it's from different fractions???

Thanks

[maiden badger]

heya

here to help

so, we know how to multiply factions right?

[tranquil root]

yeah

[maiden badger]

and how dividing fractions work

right?

[tranquil root]

yeah

[maiden badger]

ok

if I have $ab \times c$ its the same as $a \times bc$

[jolly parrot]

Max-Cat

[maiden badger]

this makes sense right?

[tranquil root]

yes

[maiden badger]

so what he did is that he turned the two fractions in one

$\frac{9 \times 4 \times 13 \times 5}{13 \times 4 \times 9 \times 3}$

[jolly parrot]

Max-Cat

[tranquil root]

ohh make sense, but how does this related?

[maiden badger]

then he canceled the 9, the 4 and the 13 as usual

you could turn the previous fraction into $\frac{9 \times 4}{9 \times 4} \times \frac{13 \times 5}{13 \times 3}$

[jolly parrot]

Max-Cat

[maiden badger]

its just re-organizing the numbers

after all, multiplication is comutative

[tranquil root]

ok

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$

[jolly parrot]

𝓣𝓪𝓼𝓴

[tranquil root]

is what you mean?

[maiden badger]

yes

thats how you multiply fractions

[tranquil root]

ok now that's making more sense

thanks

[maiden badger]

np

[tranquil root]

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[violet bronze]

            for x_train, y_train in zip(X, Y):
                activations = [x_train]
                for layer in self.layers[:-1]:
                    activations.append(layer.forward(activations[-1]))

                Z = self.layers[-1].forward(activations[-1], False)
                A = relu(Z)

                error = cost(y_train, A)
                deltas = [error[k]*relu_derivative(Z)[k] for k in range(len(Z))]

                for l in reversed(range(1,len(self.layers))):
                    current_deltas = deltas if l == len(self.layers) - 1 else next_deltas

                    for i in range(len(deltas)):
                        self.layers[l].neurons[i].bias -= learning_rate * deltas[i]
                        for j in range(len(self.layers[l].neurons[i].input_weights)):
                            self.layers[l].neurons[i].input_weights[j] -= learning_rate * deltas[i] * activations[l][j]
                    
                    if l > 0:
                        next_deltas = [
                            sum(current_deltas[k] * self.layers[l].neurons[k].input_weights[i] * relu_derivative(activations[l])[k] for k in range(len(current_deltas))) for i in range(len(activations[l-1]))
                        ]

def cost(y_true, y_pred):
    return [(y_pred[i] - y_true[i]) ** 2 for i in range(len(y_true))]

layers = [2, 3, 1]
nn = NeuralNetwork(layers)

X_train, Y_train = ([[0,0],[0,1],[1,0],[1,1]], [[0],[1],[1],[0]])

nn.train(X_train, Y_train, 0.01, 1000)

is there a problem with my code's logic? im trying to use gradient descent to improve the accuracy of my neural network. for some reason, it outputs [0] for all training samples.

[violet bronze]

i know this isnt a programming server, but i guess yall could help with the logic

[pearl pond]

@violet bronze Has your question been resolved?

[pearl pond]

@violet bronze Has your question been resolved?

[violet bronze]

<@&286206848099549185>

[jagged ore]

yup

[pearl pond]

@violet bronze Has your question been resolved?

[violet bronze]

<@&286206848099549185>

[real crater]

When you start study Radiation?

[pearl pond]

@violet bronze Has your question been resolved?

[pearl pond]

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[midnight haven]

If A is a non-empty set bounded from below in R, show that the set -A={-a/ae A} is bounded from above and sup(-A) = - infA.

[midnight haven]

Sorry if it sounds weird i used Google translate

I need to prove that sup -A = -inf A

[warm current]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[midnight haven]

1

[warm current]

<@&268886789983436800> @radiant terrace

[warm current]

Do you know how inf and sup are defined?

[midnight haven]

Yes

[warm current]

So how would inf A be defined?

[midnight haven]

For every a part of A inf A <a

And

For every ə>0 inf A +ə is not a lower bound of A

[warm current]

So you first need to show -A is bounded above

[midnight haven]

It is ?

Ohh right it is

So how do i do that

@warm current

[warm current]

You have a hint right here

[midnight haven]

I can't see it

[warm current]

sup -A, by definition is an upper bound of -A

And what is it equal to?

I must sleep now

[midnight haven]

That -sup -A = A

[midnight haven]

Ohh ok

Can you just give me the solution really quick?

I guess not

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[midnight haven]

If A is a non-empty set bounded from below in R, show that the set -A={-a/ae A} is bounded from above and sup(-A) = - infA.

[midnight haven]

Pls some help 🙏

<@&286206848099549185>

.Close

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[distant stream]

Sorry I just cannot figure out what the method for this would be

[distant stream]

I’ve lost my answer from my original attempt

It’s not homework, just a summer practice excercise

[uneven smelt]

rewrite f into (x+p)^2+q and graph it

[pearl pond]

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[distant stream]

Thank you. Will give that a shot.

[pearl pond]

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[brittle onyx]

What does it mean by what values can you take?

[brittle onyx]

Is there like a restriction or smthn?

[uneven smelt]

what values can y take such that it satisfies the previous equations

[pearl pond]

@brittle onyx Has your question been resolved?

[brittle onyx]

So 3.5<=y<=4 ?

[pearl pond]

@brittle onyx Has your question been resolved?

[uneven smelt]

how did you get 3.5?

[brittle onyx]

Part c

Asks when y is 3.5

[uneven smelt]

why can't y be smaller

[pearl pond]

@brittle onyx Has your question been resolved?

[pearl pond]

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[lucid moth]

Im really far into this question but idk how to solve it

I have 2 fractions, 1 of which I already integrated (you can see it in the answer box) and another one that needs to be integrated

4/(x^2 - 2x + 7)

Im not really sure of how to integrate this second fraction

I know im supposed to do something with arctan

But not sure exactly how to go about it

.close

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[rain vessel]

is a matrix, A, times column j of another matrix, B, the column j of the product of A & B?

[pine jay]

what do you think?

or rather, what have you have done so far

[rain vessel]

I'd think yeah, but I'm struggling with lin alg

so I'm unsure

[pine jay]

So that is infact true (if im not missing anything)

for example, if B = [ b1 ... bj ... bn]

then AB = [ Ab1 ... Abj ... Abn ]

b1,...,bn are column vectors

it also depends on the def. of matrix multiplication

but you can see this quite directly using the def.

[rain vessel]

I see, thank you

[pine jay]

when in doubt, just go back to the def.

[rain vessel]

👍

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[acoustic furnace]

does the derivative cancel out the antiderivative or do I need to do the integral first

[sharp vigil]

you can apply the fundamental theorem of calculus here

[acoustic furnace]

wait whats that

[acoustic furnace]

ohh ok i think i get it

so the answer would be d then?

since it plugs in x into f(t)?

ok i get it thank you!

.close

[pearl pond]

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[acoustic furnace]

why is this wrong

[cursive wraith]

h' = f, it's the fundamental theorem of calculus

so h' is the function that is graphed

[acoustic furnace]

o

[cursive wraith]

does this look like a relative max?

[acoustic furnace]

so it literally graphs h'

[acoustic furnace]

na

[cursive wraith]

yes, you have the graph of h'

[acoustic furnace]

o ok

ty

[pearl pond]

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[dense iris]

How am I to find the lengths of the sides of this triangle?

[dense iris]

I did this for all the lengths

then I used this and got it wrong

.close

[pearl pond]

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[dense iris]

.reopen

[pearl pond]

✅

[dense iris]

rly dunno what im doing wrong ere

[regal herald]

this doesnt look like AC

it looks like AB

[dense iris]

ah yeah true

Im gonna redo it

niceee

ok

silly mistakes

.close

[pearl pond]

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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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[lucid moth]

ln(9x)/x

u = 9x
du = 9
x = u/9

ln(u)/81

1/81(u)

1/729x

Which is wrong too lol

[regal herald]

?

[lucid moth]

?

[regal herald]

thats not what your sub would leave you with, and after that you didnt integrate, you differentiated

the sub your using also wouldnt really simplify the problem in any particular way

it would leave you with ln(u)/u

[lucid moth]

Ah oke, what method would you use?

method of integration

[regal herald]

id still do a substitution, just of u=ln(9x), should deal with the x in the denominator

[lucid moth]

ln(9x)/x

u = ln(9x)
du = 1/x
x = (e^u)/9

[regal herald]

last line, no

dont do that

and its du=1/x dx

[lucid moth]

Ah yeah

so its just u

[regal herald]

yeah

[lucid moth]

haha

(u^2)/2

((ln(9x))^2)/2

[regal herald]

+c

[lucid moth]

Thank youuuu

❤️

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[pearl pond]

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[pearl pond]

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[ancient ether]

A recycling company in Colorado collects three types of recyclable materials: paper, plastic, and glass. The company has a limited capacity to process each type of material and wants to maximize the total amount of material it can recycle while minimizing the transportation and processing costs. Data:

The company can process a maximum of 100 tons of paper per day.
The company can process a maximum of 80 tons of plastic per day.
The company can process a maximum of 60 tons of glass per day.
The cost to transport and process one ton of paper is 20USD.
The cost to transport and process one ton of plastic is 30USD.
The cost to transport and process one ton of glass is 40USD.

[ancient ether]

It's a problem about linear programming

[pearl pond]

@ancient ether Has your question been resolved?

[pearl pond]

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[pearl pond]

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[teal python]

can somene check

I think I did it right

radii of circles are 15 and 6. With that I connect them and made this bottom line segment thingy which is 21. I seperated the uadrilateral into 2 shapes, a rectangle and triangle, which split the 15 on the side into a 6 and a 9.

[vital estuary]

looks good

[teal python]

ok cool

thanks

.close

[pearl pond]

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[pearl pond]

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[undone yoke]

.ask

.ask

[junior vapor]

The solution was (6!*2)/7! but i don't get why you have to multiply by 2

[undone yoke]

thanks

can u help me?

[junior vapor]

u need to go to another channel

[undone yoke]

oh

[junior vapor]

ye

[undone yoke]

why?

[junior vapor]

cus im using this channel for my question

[undone yoke]

sorry

[junior vapor]

alg

[undone yoke]

idk how to ask

[junior vapor]

go to the "how to get help" channel

[junior vapor]

<@&286206848099549185>

[pearl pond]

@junior vapor Has your question been resolved?

[rancid lichen]

You have to multiply by 2 since you’re accounting for both books and where they are

They could be ordered in two different ways

[stoic pecan]

u already opened a help channel

#help-5

go back here

ask questions there

[pearl pond]

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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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.