help-38

[real herald]

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[wraith hinge]

how can I do this

[wraith hinge]

all I could get to was this

[north shale]

You know that these are 3 regions of equal area

[wraith hinge]

oooooooooooooh right

how do you calculate the area of equilateral

[north shale]

Half base times height

[wraith hinge]

no thats for right angle triangle

we dont know height here

[north shale]

An equilateral is two right triangles

And you found the height to be root 12

[wraith hinge]

is it correct?

[north shale]

Yes I think

Do u get it

[wraith hinge]

yeah let me try

are these two equal?

[sharp heart]

Not necessarily.

Did you get the area of the equilateral triangle?

[wraith hinge]

yes

square root of 48

I got the answer correct if I assumed that they were equal

[sharp heart]

OK, so the area of each trapezoid is equal to that.

[wraith hinge]

okay then how do I get the small line

[sharp heart]

OK, so the triangle just above the equilateral triangle is half the area of the equilateral triangle, so you were correct.

[wraith hinge]

ooh thats why it works

[sharp heart]

So, the area of the smaller triangle is sqrt(12).

[wraith hinge]

so divide by 2 you get square root of 3

[sharp heart]

Right.

[wraith hinge]

alright thank you

[sharp heart]

No problem.

[wraith hinge]

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[supple sun]

z^2=xy x+y=4 x+y=6

[supple sun]

Calculate the volume of the block

So I started with sth like that

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[supple sun]

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[supple sun]

So I have to calculate the volume of a figure

Thst is limited by oxy oxz z=3x and circle x^2+y^2=4

What do I do wrong?

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[vocal oracle]

if i have to compute the integral of
sqrt( 1-x^2 - y^2 ) dy
i wonder if i could
substitute x^2+y^2 as r^2
and then use trig substitution

[vocal oracle]

and if i were to do so, when i do the first change of variable in to r, i should also compute what dr is, right?

it would be
dr = 2y dy

but i don't see this helping me. i donìt think it's working

[clear cloud]

Its just dy

?

You can consider x² as a constant in that case

[vocal oracle]

wait i made a mistake

hmmmmmmm my logic doesn't work i think

x^2+y^2 is r^2 not r
can i do the substitution with r^2 and then.. find r and compute dr?

it won't help me with trig subst

:/

[dusky thunder]

as the helper above said, dy means integrating wrt y

[clear cloud]

The real questions is, is it dy dx or just dy as you write

[vocal oracle]

just dy

[clear cloud]

So theres no polar

[vocal oracle]

oh ok sorry

there will be dx later

[clear cloud]

!xy

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[vocal oracle]

but my shape is not polar

[wraith hinge]

Is it a double integral or not?

In the original problem?

[vocal oracle]

this is the problem

so i was incorrectly setting the plane as z^2 + y^2 +x^2 = 1

then i figured that it would simplify. But then i wanted to know how to ingetrate that instead

so deviating from the original problem

so yea in that case i could just switch to polar i guess...

so in that case

it would become
sqrt(1-r^2) r dr dtheta

where r is the jacobian really

the scale factor

[wraith hinge]

How are you getting this?

[vocal oracle]

as i said it was a MISTAKE 😄

my brain decided yeah square all terms

[wraith hinge]

Sorry, my bad

[vocal oracle]

np

so if it were just dy, x is a constant.. i could directly use trig subst right?

i could set 1-x^2 = a^2

and go from there

[wraith hinge]

Why will that even be needed here? Are you still exploring how to solve the double integral √(1-x²-y²) dxdy apart from the given problem?

[vocal oracle]

no

i'm just playing with what i have in front of me, trying different things

if i had sqrt(1-x^2-y^2) dy

how would i solve that?

[wraith hinge]

same as integral for sqrt(a²-y²)dy where a \in R

x is a constant

[wraith hinge]

And to start to solve the problem, my suggestion would be to find the limits of integration first

with that, you can either solve through triple integration (or an easier way would be to solve using only geometry by figuring out the vertices)

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[vocal oracle]

thanks !

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[lean kraken]

Here I am thinking about two cases one case is if I take 7

Second is if i take double 7

5^4+6=731?

[fleet bear]

I think you should consider using nCr

[lean kraken]

Same thing

5C4.4!+ 4c2.4!/2!

5.24+6.12

120+72=192?

[fleet bear]

My bad, we should be thinking about permutations not combinations in this case

[lean kraken]

Check it

Is this correct?

[frail heron]

How about no seven?

[lean kraken]

Huh?

Why?

[frail heron]

Form a 4 digit number with just 3 5 6 8

[lean kraken]

If I am taking four digits from 356 and 7 will it not count

[frail heron]

Oh, is your idea choosing 4 none 7 and a seven?

That makes sense

[lean kraken]

I am choosing four from 5 digits

In my first case I am only taking 1 7

[frail heron]

5C4 x 4!+ (4c2 x 4!/2!)

What’s the bracelet about?

You pick 4 numbers out of 3 5 6 7 8, and permute the four numbers you pick

[lean kraken]

Yes

5C4 is picking four numbers out of 5

And then permutations

4c2 is when we are taking two's 7

So we have only two place to choose

[frail heron]

Pretty neat, the work looks fine to me

Mb, I thought a line only stands for one condition

[lean kraken]

I see

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[solar wagon]

I forgot what I need to do with this

I did 2 ^12 got 4096

[trim lichen]

good question, what DO you need to do w that

[solar wagon]

And did 2^(3^4]

[trim lichen]

its full value is big and monstrous

!xy

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[trim lichen]

2^(3^4) is the correct reading

[frail heron]

Mb

[trim lichen]

but again like

we gotta know what they want from you exactly

if it really is "Find the full value of 2^81" then show us where it says that

[trim lichen]

oh so you need to compare them against each other.

[solar wagon]

Yes

[dull pilot]

[trim lichen]

but NOT necessarily to find any of their values in full

[dull pilot]

this is why !xy is important

[trim lichen]

ok so

[frail heron]

[trim lichen]

who are you comparing 2^(3^4) with

[solar wagon]

3^4^2

[trim lichen]

ok so 2^81 vs 3^16

who do you think is bigger

[solar wagon]

2

[trim lichen]

don't abbreviate like that

[dull pilot]

im think we might wanna be a bit more rigorous lmao

[trim lichen]

well yes sure but my next question would be why he thinks 2**^81** > 3^16

[solar wagon]

Big numbers

[trim lichen]

...

[dull pilot]

[trim lichen]

nope

not gonna cut it

[dull pilot]

$0^{120043030430} > 1^1$?

[solid kiln]

Percy

[trim lichen]

you realize this is not an argument for anything right

[solar wagon]

Yes

[trim lichen]

ok so

[dull pilot]

try and be a bit more serious then

[trim lichen]

should we take it that you have no argument

except a vague and vibes-based feeling

i will hope that we can do something more serious here

[dull pilot]

mfw 0^{120043030430} > 1^1

[trim lichen]

@solar wagon are you prepared to be serious here

yes or no

[solar wagon]

Yes

I’m writing notes

[trim lichen]

so you are trying to cook something up

ok, show us when you're done

[solar wagon]

2^9 =512

3^4 =81

[dull pilot]

...?

[solar wagon]

That’s the only argument that I have other than 2^81 being bigger the the other numbers

[dull pilot]

how is that an argument for anything

are you thinking that $2^{81} = (2^9)^2$?

[solid kiln]

Percy

[solar wagon]

No

2^9 and 3^4 are two completely different numbers

[dull pilot]

what are you thinking

[solar wagon]

One just has a greater value

[dull pilot]

and thats not the numbers you were asked about

[solar wagon]

^

cant i just divide

[dull pilot]

divide what

by what

[solar wagon]

2^81 and 3^16

[dull pilot]

@trim lichen you still here?

[dull pilot]

good luck doing that

[solar wagon]

4^9 can be (2^2 )^9

or 2^18

and 4^8 can be 2^16

[dull pilot]

sure

what are you gonna with the $3^{whatever}$ stuff

[solar wagon]

and we know that 2^81 is greater than all the 4's and 2's

[solid kiln]

Percy

[solar wagon]

i have no idea about that

[dull pilot]

can you compare the rest

[solar wagon]

2^81 2^64 2^18 2^16

and then 3^16

[dull pilot]

can you compare the powers of 2?

[solar wagon]

3^16 = 4.3 x10^7?

[dull pilot]

how.

you aren't allowed calculators are you

and even if you are, this isn't supposed to be done like that.

[solar wagon]

what do you want me to do?

[trim lichen]

playing chess irl sorry

[solar wagon]

Any YouTube explanations for this

[solar wagon]

All good, I was watching the 2.2 lesson for this

And the video didn’t show any problems likes this

But an example that made a bigger number into smaller parts

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[winged hinge]

Sus link

[marble wharf]

<@&268886789983436800>

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[drowsy moss]

Hi.

[drowsy moss]

Can't we generalize the quadratic reciprocity law?

[clear cloud]

What is it

[drowsy moss]

The quadratic reciprocity laws says this.

I say that let p and q be any odd integers integers so that at least one of them is of the form 4k + 1. Then (p-1)(q-1)/4 = an even integer. Therefore, (-1)^even = 1 = (p/q)(q/p).
Note: (p/q)(q/p) can be rewritten as (p/p)(q/q) = 1 x 1 = 1

Hence, we have the generalized quadratic reciprocity law.

<@&286206848099549185>

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[drowsy moss]

<@&286206848099549185>

[carmine spade]

yes, see here https://en.wikipedia.org/wiki/Jacobi_symbol#Properties

[drowsy moss]

Yes?

oh

Thanks

[carmine spade]

nws

is that all @drowsy moss ?

[drowsy moss]

I think so. Thanks.

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[lunar stirrup]

first i did 1/f=1/v-1/u
u=-15,f=30
then i got v = -30
so first image at 45 cm from mirror so it will be formed on 45 cm to left of mirror
now again that image will produce rays acting on the convex lens.
1/30 = 1/v-1/u

now my question is should i take u as -60 or + 60?

because its on the right side of the lens?

and also can someone help me understand nature of all 3 images formed

[undone gorge]

The sign convention says that if the measurement from the pole is in opposite direction to the incident ray, then we have to take the sign to be -ve

[lunar stirrup]

one second

[undone gorge]

Another que?

[lunar stirrup]

yes but its part of the same thing

u have to take object distance positive for this one

since image is on right of mirror

which i dont understand

[undone gorge]

Wait a min

[lunar stirrup]

yeah

[undone gorge]

Mirror? You mean concave lens?

[lunar stirrup]

yeah

[undone gorge]

The object distance will be +ve only

Drawing the ray diagram, all the rays are going from left to right

[lunar stirrup]

yea but isnt it the same for the first q?

[lunar stirrup]

wait wait and isnt the ray coming from the image formed going from right to left?

[undone gorge]

Let me draw it one sec

[lunar stirrup]

oh ok

[undone gorge]

It isn't the best ray diagram possible, but you get the idea that the rays always move forward

Like no ray is going to "come back"

[lunar stirrup]

what about the rays from i1?

isnt that coming back?

[undone gorge]

I1 doesn't form in reality

It's just what would form if only convex lens was there

Which now acts like an object for concave lens

[lunar stirrup]

yeah but dosent it act like a real object for the concave lens?

[undone gorge]

Yup it does

But think about it does the i1 completely forms then concave lens does it's stuff?

[lunar stirrup]

or maybe i would understand if u could tell me the difference in why we take negative object distance here and positive in the other question

[lunar stirrup]

im sorry i didnt understand that..

[undone gorge]

Well i1 is kind of an illusion we create to solve the question easily

[lunar stirrup]

oh right yes

[undone gorge]

I1 doesn't form in reality cuz the rays which were going to form it get diverged by concave lens

[lunar stirrup]

oh ok yes

but even in my original question isnt it a virtual image being formed there as well?

[undone gorge]

Let me check

Ray diagram, if you can't understand it I will draw it again

As you see the ray going to form the final image is gonna be opposite to the distance so -ve

[lunar stirrup]

yeah got it

[undone gorge]

Trick to most of this questions is to continuously draw the ray diagram, it really helps

[lunar stirrup]

huh but isnt that happening in the other one too? with the concave lens?

sorry to trouble u so much about this 😭

[undone gorge]

No man no problem

[undone gorge]

No cause mirror reflects but lens refract, so in mirror the direction(in sense complete 180 kind) of light does change but in lens it doesnt

[lunar stirrup]

isnt it kinda changing here as well? or does it have to be full 180?

[undone gorge]

That's imaginary line cause the lines won't meet on the right side of concave lens ever

[lunar stirrup]

oh thats an imaginary line

[lunar stirrup]

right because image would be real then

okay i think i got it now bro

thank u so much

[undone gorge]

Yh my drawing wasn't great lol

[lunar stirrup]

it was way better than mine lol

[undone gorge]

Lol

[lunar stirrup]

ty again

.close

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[wraith hinge]

how would you even calculate the triangle part

[wraith hinge]

you cant know the height of the bent triangle

im guessing the answer is B or D since 1 is from the square then the triangles add up to the square root of 2 or 3

[steep knot]

the height of the cube is 1

[trim lichen]

you can!

[steep knot]

so now you just need to find the height of the pyramids

[trim lichen]

make a pyramid with a square for its base and equilateral triangles for its faces

the center will be directly above the center of the square

you can do some Pythagoras magic to get the height

[steep knot]

Hint: use Pythagoras

[trim lichen]

imagine saying that just after i mentioned the very same thing in more detail

[steep knot]

lmao using Pythagoras is also my idea

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[wraith hinge]

sorry I went to do something

how would pythagoras help

[dull pilot]

you want the height.

you have the hypotenuse and the base.

pythagoras!

[wraith hinge]

it dont work

is the base squareroot 2 / 2 and the height 1?

[hidden dew]

draw a perpendicular line through the middle of the triangle

[wraith hinge]

that wont work since in the 3d shape its bent

lets say I get 1 as the height for this in the shape it could be 0.7

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[tired shoal]

over what interval does the graph of the function have a slope that is positive but the slope of the function around it decreases with as you move

slope is negitive in that interval

nuh uh there the slope is positive but

its increasing

try using your hand as tangent and keep it at different points

and move along the function as your hand as tangent

the angle your hand makes with horizontal is slope , if its above your horizon AND Less than 90 deg then slope is positive

i think they mean the slope of the function itself is decreasing. the question asks
where the slope is positive but the value of slope is getting smaller (like from 8 to 6 to 2 to 1 ) but it stays positive

yes

the slope at 2 is large, and then slope at all other values is positive, but less than 2

the slope is decreasing

but its positive

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[haughty condor]

Hello!
How do I get x_a and x_b here?

[humble forge]

lambda are equal

[haughty condor]

But what to do after I go: -200+2x_a=-220+2x_b?

2x_a = -20+2x_b

[humble forge]

wait didn't see that

you get Xb - Xa = 10

[haughty condor]

How?

[humble forge]

$2X_a = -20 + 2X_b \$
$20 = 2X_b - 2X_a\$
${dividing both side by 2}\$
$10 = X_b - X_a\$

[haughty condor]

But 10=x_b -x_a? Right?

[humble forge]

yeah my bad....

[haughty condor]

So after that? We have 10 = x_b - x_b how do we get xa and xb?

[solid kiln]

alphaX

$2X_a = -20 + 2X_b \\$
$20 = 2X_b - 2X_a\\$
$\{dividing both side by 2}\\$
$10 = X_b - X_a\\$
```Compilation error:```! Extra }, or forgotten $.
l.51 $\{dividing both side by 2}
                                \\$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.


Underfull \hbox (badness 10000) in paragraph at lines 49--53```

[humble forge]

you have other eqn too

$20 = X_b + X_a$

[solid kiln]

alphaX

[haughty condor]

But when 10=xa-xb how can xa+xb be 20

[humble forge]

why not,
one is adding and another is subtracting

like 10+5 = 15
10 - 5 = 5
they are not same

[haughty condor]

Is there a way for equations where you can't find it out in a puzzle?

[haughty condor]

I heard think this is also possible to solve in a linear equation model? @humble forge

[humble forge]

exactly

[haughty condor]

Could you explain to me how please?

[humble forge]

just add the two equation.....

[haughty condor]

What do you mean?

I tried to solve for xa : 10xa = xb but this is wrong

[haughty condor]

Thank you very much!

[humble forge]

I hope u understood ....

[haughty condor]

No really, but I found some videos for the linear equation. Thanks a lot anyways 🙏

.close

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[mystic iris]

How exactly are you supposed to do this question I tried similar triangles but nothing seems to work

[trim jolt]

@mystic iris Has your question been resolved?

[mystic iris]

<@&286206848099549185>

https://tenor.com/view/angry-birds-angry-bird-red-gamble-numbers-gif-16030169547675327124

[trim jolt]

@mystic iris Has your question been resolved?

[mystic iris]

similar triangles

but there is no technique i know that can solve this

i can tell both the small triangles are similar

[umbral hazel]

How do you know that the triangles are similar?

@mystic iris Are you still here?

[mystic iris]

yes

[umbral hazel]

Ok

[umbral hazel]

I would say let’s not jump to that conclusion

But let’s break down the problem by going backwards

[mystic iris]

let me think

[umbral hazel]

So start by finding the equation to the Triangle area, and then find out how to calculate each required component

This may help you determine what to find

[mystic iris]

just a question do you yourself know how to solve it?

[umbral hazel]

Not off the top of my head

But I normally go with this approach of tutoring

[mystic iris]

how do you find an equation for the area

i know its b*h*1/2

[umbral hazel]

Oh, that ones easy

[umbral hazel]

That’s right

[mystic iris]

but i dont know how you find the length PA or BQ

which i need for the triangles area

[umbral hazel]

Those are not the sides you need for calculation

[mystic iris]

but i need sides AC and BC which i get by finding those two sides

[umbral hazel]

Ok

@mystic iris The good news is that we know that all sides of the square are equal to 3

[mystic iris]

yeah but i dont even know how that helps to solve the problem

[umbral hazel]

Well, we’re going to use Pythagorean theorem a lot

Knowing the sides will help us effectively calculate what we do need to know

[mystic iris]

BQD is similar to APD because angle BDQ= angle DAP as they are corresponding angle and they both share a right angle so its angle angle similarity

[mystic iris]

how do we use pythagorean theorem here

[umbral hazel]

Well, for starters, (AC)^2 + (CB)^2 = (BA)^2

(4sqrt(7))^2 = what??

We can also deduce the Following:
(QB)^2 + (QD)^2 = (BD)^2
(PD)^2 + (PA)^2 = (DA)^2

Oh, right

@mystic iris What’s the square of the product of 4 times the square root of 7

[umbral hazel]

Or this

[mystic iris]

sorry i went offline

[mystic iris]

112

[umbral hazel]

Good

Keep this in mind. You may or may not need it later

[mystic iris]

do you have any idea of how to solve it though

i guess it made sense to use pythagoras theorem but there are two unknown parts

how do you even find them

[grave lynx]

triangles BQD and DPA are similar right?

[umbral hazel]

I don’t think it necessarily is

Although there was no indication either way

[grave lynx]

wait can i do my solution and send it here

[umbral hazel]

It would be preferable not to just say the answer without walking him through the problem.

But if you can help him find the answer, that would work

[grave lynx]

kk

[mystic iris]

Please do

[leaden trellis]

I would set up 4 equations (two with pythagoras, BD+DA=4\sqrt(7) and one with ratios DQ/PA = BQ/DP = BD/DA) and solve from there

[mystic iris]

corresponding angles they definitely are similar

[leaden trellis]

im not sure if ure still trying since the chat died, but the area should be 21
try and see if u can get that result by setting up the 4 equations
I think thats the easiest way

[grave lynx]

im trying but i think i gtg rn

[mystic iris]

well thanks i guess

i went off trying to create a formula for the primes

[leaden trellis]

[mystic iris]

.close

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Channel closed

Closed by @mystic iris

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[knotty oriole]

I think you can angle chase to show the big triangle and both smaller ones are similar?

Oh closed

[mystic iris]

.reopen

[trim jolt]

✅

[mystic iris]

i just got scared when i saw a mod typing

[knotty oriole]

Lmao it's fine

[mystic iris]

what does angle chase mean

[knotty oriole]

You have two parallel lines given by the squares and those parallel lines tell you about the angles of the bottom right corners of both triangles.

https://www.mathdoubts.com/pics/transversal/corresponding-angles-parallel-right.gif

Not sure if this is a good gif

[mystic iris]

no no it helps

[knotty oriole]

Yeah like this basically. The marked angles are the same. Your bottom right triangle corners have this setup.

They are also right triangles.

The remaining angle is determined

[mystic iris]

I already said that though

[knotty oriole]

Yeah you probably did. Your thinking for why they are similar is good.

[mystic iris]

I just dont know how to make any progress

[knotty oriole]

What chelrelleren said here makes sense.

Name the missing lengths

Spam pythagoras and similar triangle relationships to get a bunch of equations

Then solve the system of equations you get

[leaden trellis]

ong

[mystic iris]

Okay thanks ill try that now

[leaden trellis]

once u get one of the missing sides the others become very easy to find

[knotty oriole]

Yeah ideally you can whittle it down to two equations and two unknowns.

Then solve by substitution or something.

[mystic iris]

This seems like an incredibly hard equation I got a quartic

I give up lol

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[burnt mulch]

@mystic iris Note that you want the area of ABC - you don't need the value of AD individually.

For convenience, I'll let BQ = x and AP = 9/x.

[mystic iris]

wait what

[burnt mulch]

Then, we need to compute the value of $$\frac{1}{2}(x+3) \left(3+\frac{9}{x} \right)$$ given that $$(x+3)^2+\left(\frac{9}{x}+3 \right)^2=112$$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

At this point, consider expanding out the quantity you need to find and the equation from the Pythagorean theorem and writing both in terms of some other quantity

See where you get with that

(spoiler alert: pls don't solve for x outright)

[burnt mulch]

?

[mystic iris]

im going to see if substitution works

[burnt mulch]

It could, depending on what you're substituting

[burnt mulch]

That's kinda what I was getting at here

[burnt mulch]

What exactly are you substituting?

[mystic iris]

nvm that doesnt make sense

[burnt mulch]

Did you do what I said here?

[mystic iris]

i'm gonna be honest I dont even understand what you mean by this

[mystic iris]

.

[burnt mulch]

Basically expand out both

and think of something that you can substitute

[burnt mulch]

I'd focus on the quantity we're trying to find

and substituting the "ugliest" part of it

[mystic iris]

?

[burnt mulch]

nvm just expand out both

and we'll work from there

[mystic iris]

this =112

[burnt mulch]

It's $\frac{9}{x}+3$, not $\frac{9}{x+3}$.

[solid kiln]

Civil Service Pigeon

[mystic iris]

bruh

im so ass

[burnt mulch]

nah reading is hard sometimes I feel you lol

(this isn't meant to be sarcastic snark)

[mystic iris]

cut off

[burnt mulch]

?

wdym

[burnt mulch]

=112

that's our equation from pythag

what about the quantity that we need to find

[mystic iris]

should i expand that as well?>

[burnt mulch]

yeah

[grave lynx]

i made my progress till here

[mystic iris]

I still dont see how this relates to the last equation

[burnt mulch]

I'm rewriting this as $\frac{3}{2} \left(x+\frac{9}{x} \right)+9$ cuz I'm lazy

[solid kiln]

Civil Service Pigeon

[burnt mulch]

So if we can find $x+\frac{9}{x}$, then we're chilling

[solid kiln]

Civil Service Pigeon

[burnt mulch]

So we want to rewrite $x^2+6x+\frac{81}{x^2}+\frac{54}{x}=94$ in terms of $x+\frac{9}{x}$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

Any low hanging fruit that you can see here?

[grave lynx]

this is such a good problem whoa

[mystic iris]

OH SHIT

[burnt mulch]

so u good now?

[mystic iris]

if i square (x+ 9/x)^2 i see a lot of the terms match up

[burnt mulch]

[burnt mulch]

[grave lynx]

imma solve this again from scratch tonight

[burnt mulch]

[burnt mulch]

@mystic iris ?

[mystic iris]

LETS GO IT FACTORS NICELY

[burnt mulch]

ggs

[mystic iris]

HOLY SHIT IT IS 21

but it seems like there is 3 other cases

[burnt mulch]

elaborate

[mystic iris]

though you told me to not directly solve for x

i kinda ended up getting to it

[burnt mulch]

um -

try plugging in each of those values of x

and you should be able to figure it out

[mystic iris]

yeah i did the first one and i got 21

[burnt mulch]

did you do the other 3

cause what's rlly happening is that the area is fixed based on x + 9/x

aka each of the two values of x s.t. x + 9/x achieves a given value will yield the same final area

[mystic iris]

so the area is the same in all the cases?

[burnt mulch]

test it out for yourself

[mystic iris]

ok so the 4 - sqrt(7) also gives the same area

-12

what the actual

this makes no sense

how is the area -12

[burnt mulch]

exactly

this case is ridiculous

so you can ignore it

[mystic iris]

so the only true solution is 21

holy thank you so much

ive been confused for this problem for a while

and i was about to give up

[burnt mulch]

Yeah I saw

then scrolled up and was like

wtf is going on here

😭

[trim jolt]

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

[mystic iris]

alright im going to sleep now lol

.close

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[warm sedge]

Is this correct?

[warm sedge]

I’m not allowed to use a calc so that’s why it’s like that

[shy needle]

does anyone know why this is wrong? these are previous answers and a prasctice question. (the 390 is the one i need to do)

[warm sedge]

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[vague citrus]

do you know what rolle's theorem states

[harsh siren]

yeah

continous in close interval

diff in open interval

and there is a point where diff is 0

[zinc ginkgo]

Show the entire context

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@harsh siren Has your question been resolved?

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[wary schooner]

what value of k would make the following trinomial a perfect square. my teacher got k=4 but i dont think thats right.

[simple panther]

Yes, it's not right

What did u get when u tried?

[solid plank]

we know the property that $(a+b)^2=a^2+b^2+2ab$

[solid kiln]

Aria

[wary schooner]

I got 24 but honestly i dont know how I got that

[drowsy moss]

For a quadratic polynomial to be a perfect square, its discriminant must be 0, otherwise, it would have two distinct roots, which would contradict the fact that the quadratic trinomial is a perfect square.

[simple panther]

$k\cdot(x)^2-2(12)x+(\sqrt{6})^2$

[solid kiln]

Suika

[simple panther]

From comparing with this, we'll get b = √6

Now try to find the value of a

[wary schooner]

okay i understand

[drowsy moss]

It's correct.

,w 24x^2 - 24x + 6

[solid kiln]

Results provided by WolframAlpha

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Requested by philosophizingphilosophy

[drowsy moss]

A single root, verily.

D = b^2 - 4ac = 0
b^2 = 4ac
576 = 4 * k * 6
k = 24

[wary schooner]

ohh okay

so my teacher just confused me lol

thanks for the help!

[drowsy moss]

No problem.

[wary schooner]

.close

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[little glen]

D = 0

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[harsh siren]

@zinc ginkgo rest are options

[solid plank]

it satisfies rolle's theorem, thus f(3)=f(5)

[harsh siren]

ohhh then integration will be 0

thanks

[stark sandal]

no

[solid plank]

no no no

[harsh siren]

.close

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[harsh siren]

.reopen

[trim jolt]

✅

[harsh siren]

why not?

[solid plank]

lmao it wont be zero because we have to INTEGRATE the function

[harsh siren]

but limits are same?

[stark sandal]

f(3) = f(5)

3 is not equal to 5

the limits are 3 and 5

[harsh siren]

ahh

suppose the integration is F(x)

so it wil be F(5)-F(3)

[solid plank]

so here we'll assume that f(3)=f(5)= 0 for simplification

[harsh siren]

sure

[solid plank]

yeah but we dont know F(5) or F(3)

[harsh siren]

mm

right

[solid plank]

thus we can form a polynomial with roots x=5 and x=3 right?

[harsh siren]

not sure

[solid plank]

lets say f(x)=a(x-3)(x-5)

[harsh siren]

i see

a(x^2-8x+15)=a(x^3/3-8x^2/2+15x)

now limit time

[solid plank]

wait wait

let me check what you did

yeah okay

also, to simplify it, we can choose a=1

[harsh siren]

sure

is it necessary to make that polynomial

[solid plank]

if we dont get the exact answer, then we can just choose the one with its multiple

[solid plank]

i mean that's how my teacher taught me and how i do it, not sure about any other method

[harsh siren]

( \left(x^3/3-8x^2/2+15x\right) ) okay

( 125/3-100+75-9+36-45 )

[solid kiln]

Tom

[solid plank]

simplify the equation first

make it $x^3/3-4x^2+15x$

[solid kiln]

Aria

[harsh siren]

-4/3

[solid plank]

it matches the option (d) so we dont need to multiply

[atomic seal]

im a little confused here. f(3)=f(5) so the function can literally be anything you want right? it can even be a constant or a polynomial or anything. so integration from 3 to 5 can take any values right? like it can be 0,-ve or +ve

[light trout]

here f(3) = f(5)

[harsh siren]

i see

thanks

.close

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[harsh siren]

ohh guys you wanna say something

.reopen

[trim jolt]

✅

[atomic seal]

like i can assume y=0, it would satisfy the conditions ig. but the integration would be 0.

[light trout]

yeah

[atomic seal]

its confusing

[light trout]

area under the curve between x=3 and x =5 could be zero if the function is symmetric around some point or balanced positive and negative in that interval.

[atomic seal]

exactly

[light trout]

yess

[atomic seal]

so arent all options possible?

[light trout]

no i dont think so

[atomic seal]

please explain

[vapid grove]

y = 1, answer is 2
y = 0, answer is 0
im also confused

[atomic seal]

there is no mentioning of anything symmetric or so. i can assume any function i want which satisfies those condition for rolle's theorem.

[light trout]

f(x) is symmetric or has equal area above and below the x - axis so it might suggest integral of f(x) dx with limts 5,3 be zero ?

[atomic seal]

yes why not

given that f(3)=f(5)

[vapid grove]

this problem doesnt make sense lol

[atomic seal]

it can have any number of points in (3,5) where f'(x)=0 so area can vary

[light trout]

yeah although the function is not given right?

[atomic seal]

thats what im tryna say

[stark sandal]

we cant say, given these condition you can make the integral whatever you want by choosing the right f(x)

[atomic seal]

yes its not given

[vapid grove]

is there more before it? it looks like this is a part of a multi step problem

[stark sandal]

technically all 4 are correct

can be*

[harsh siren]

actually no

problem is bad

thanks all of you

[atomic seal]

yeah so the question is not specific

[stark sandal]

we need more context

[harsh siren]

i need to ask my next question

[stark sandal]

sure

[harsh siren]

so i am trying to close the channel

geatest integer function is increasing or decreasing?

and how can I apply derivative thing over it?

[light trout]

ques should be more specific 😦

[harsh siren]

because non integers derivative is zero but it is undefined at intgers

actually someone is asking me about it

[atomic seal]

at non integer since derivative is 0, its neither increasing nor decreasing

[stark sandal]

non decreasing

[harsh siren]

like how?

[stark sandal]

you can say its increasing ig

i like to use non decreasing

[harsh siren]

and what is the different

[stark sandal]

well non decreasing means that it doesnt decrease

increasing means that it at least stays constant

same thing as non decreasing

and strictly increasing means its increasing for all x

[vapid grove]

how can it increase if it stays constant

[outer kiln]

it’s a definition thing

[stark sandal]

in terms of derivatives f'(x) >= 0 is the condition for non decreasing and f'(x) >= 0 (but with a caveat that it can only be 0 at discrete points) for strictly increasing

[stark sandal]

yeah its not. thats why i prefer non decreasing. though the commonly understood definition for increasing includes constant functions

at least from what ive read and what we were taught

[vapid grove]

i was not taught that

[stark sandal]

thats why you have a distinction between strictly increasing and increasing

[stark sandal]

ig it varies place to place

gif is non decreasing

[harsh siren]

constant function is decreasing or increasing

[stark sandal]

could be both

its a constant function

we just say its constant

[light trout]

are you talking about floor function ?

[stark sandal]

yes

gif = floor

[light trout]

It never decreases as x increases. But it stays constant between integers (like from 2.1 to 2.9, it's still 2).It jumps at integer values (like from 2.9 to 3.0, it goes from 2 to 3) ?? right

So it's not strictly increasing ,just non-decreasing stepwise

[light trout]

1

[harsh siren]

how?

[light trout]

sqr bs of given eqn you will get sinθcosθ = 1

use tanθ trig ratio formula and same for cotθ

[harsh siren]

great

thanks

[light trout]

done?

after solving tanθ+cotθ you will get sin^2θ+cos^2θ/sinθcosθ and since sinθcosθ is 1 and sin^2θ + cos^2θ = 1 it will now become 1/1 finally you will get 1

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[lean kraken]

Limit points set is derived set and how it is closet set

[lean kraken]

Because derive set content some points which are not on our boundary

I didn't understand this statement how it is close

[trim jolt]

@lean kraken Has your question been resolved?

[willow urchin]

it may help to use the "contains all limit points" definition of closed

[lean kraken]

but limit point can be outside of that boundary no?

[willow urchin]

yes but i think ur not being careful about the sets ur talking about

[lean kraken]

I didn't understand??

[willow urchin]

why r u talking about boundary?

[sudden mist]

What’s ur definition of closed

[lean kraken]

To me of we have a circle

Then all of its boundary points and inside points

[sudden mist]

That’s a bit unclear

What about the circle?

[lean kraken]

|z-4|<=1 it is closed

[sudden mist]

Mhm

Can you state your definition of closed that you’re using

That’s a bit too vague, its only a specific type of set

For example are you working through a book or lecture notes?

[lean kraken]

Random youtube video

[sudden mist]

What did they say?

You won’t get far without a proper definition, and you’ll get entangled in confusion

But in that case I guess youre instead curious in getting a feeling for why the set of limit points is closed?

[lean kraken]

@sudden mist

[sudden mist]

Cool

So if you want to show that D’ is closed in your case

Then you want to pick an arbitrary boundary point from D’ and show it’s in the set itself.

Then by definition it’s closed.

@lean kraken

[lean kraken]

I guess definition is not clear to me I am thinking about some limit points which are not on boundary they are out of boundry

@sudden mist

So the can be limit point

??

[sudden mist]

You might be confusing the boundary of the set D for the boundary of D’

[lean kraken]

I guess yes

[sudden mist]

They are different

[lean kraken]

Please explain?

[sudden mist]

Why they’re different?

Well if say D was just a singleton set = {a}, then the boundary of D is just {a}.

However D’ is empty (check why!).
So the boundary of D’ is empty too!

So they’re different

The reason I’m mentioning this is because for a set S to be closed, by your definition, then we have to take boundary points from S and show they’re in S.

Where S in our case is D’, not D.

[lean kraken]

I think you are proving that if complement of D is open then our set D is close

Am I right

@sudden mist

[sudden mist]

I’m not, but yes that is also a valid way of going about it

That is, if you can show that. Then you can if you desire show that the complement of D’ is open instead.

For clarity I’m assuming this is the end goal for you right? As you were wondering why D’ was closed.

[lean kraken]

For every del >0 they should contain elements of D

To be limit point

??

[sudden mist]

Oh wait this is wrong for the definition of a limit point. Why are they writing exists delta > 0. It should be for every delta > 0…

[lean kraken]

So if you make neighbourhood of that point and if we get any common points then it will be limit point

So points other than boundary cannot be close because we can make many small delta

I mean outside of the disc

I understand now

Thanks

And sorry for taking your time for so long

[lean kraken]

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[hollow wedge]

if B is the midpoint of PR then how is BRS half the area of PRS????

[woven python]

it halves the base

hence half the area

$\Delta \propto b$

[solid kiln]

parabolicinsanity

[woven python]

feels weird using this in a maths context

[hollow wedge]

wdym

[woven python]

you usually dont see the proportional sign in maths mostly in physics and engineering

anyways

the base is halved so the area is halved

[hollow wedge]

oh

wait that makes sense

thanks

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[valid hamlet]

Why is it 8 x 10^4 and not 5 x 10^4.

[valid hamlet]

The net force is 500 N.

[vagrant prism]

the forward thrust is still 800 N

the output power doesn't care about the net force

[valid hamlet]

So they want the power of the engine and not the net power?

[vagrant prism]

so, yes

[valid hamlet]

Well I didn’t know there was even a difference before but now i know i guess

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[livid bolt]

Use the inequality in between the geometric and aritmetic ¿mean? (Dunno if that's the right word) in between 3 real positive numbers to demonstrate this

[livid bolt]

I am not sure how to proceed, this is my attempt at a translation

[wraith hinge]

It asks you to lower bound LHS by AM-GM inequality

Your attempt is on the right path

Now just apply the inequality

[livid bolt]

This is the part where I'm kinda lost

[wraith hinge]

Do you know the AM-GM inequality?

[livid bolt]

The aritmetic mean vs geometric mean

A friend that was passing by explained it to me that the exercise could be resolved like this

But I am kinda lost on the AM vs GM, I can replace the values with anything that fits the shape so long as I adapt the formula? What is the theory behind?

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[wraith hinge]

What your friend did is also correct but I do not get the last part of your question sorry

The inequality is valid for all non-negative a,b,c

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[livid bolt]

How does the AM GM inequality work?

[livid bolt]

I was used to the form (a + b)/2 >= root(a•b)

But he changed it to this

Could someone explain a bit just so I make sure I understand

Is the essence of it that the aritmetic mean of X amount of numbers will always be larger that the Geometric Mean of the same amount of numbers?

[frail heron]

Greater than or ** equal to**

Don’t forget that a, b, and c should be positive

[buoyant sluice]

Or 0

Trivially

[livid bolt]

Oh okay important data, what you just said

[buoyant sluice]

What

[livid bolt]

And what exactly is the geometric mean by the way

I'm kinda lost at what that thing is meant to be

[frail heron]

No, zero is not the case

unless you’re referring to all of them being zero

[buoyant sluice]

Any can be 0

Am gm will still hold

[frail heron]

Mb

[buoyant sluice]

GM doesn’t have any specific meaning that’s important I’d say

You can interpret it using a semicircle

But it’s just the byproduct an algebraic manipulation

[livid bolt]

Okay thanks

That is all

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[hot moth]

I need to prove using the pigeonhole principal that for a 3x7 grid of dots where each dot is either orange or purple, that there is a rectangle (whose sides are parallel to the sides of the grid, no diagonal lines) with all four corners being the same color

[hot moth]

I managed to prove this for 3x9 but I can't seem to do it for 3x7

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@hot moth Has your question been resolved?

[hot moth]

Either I'm blind or the following disproves it:
⬜ ⬜ ⬛ ⬛ ⬛ ⬜
⬜ ⬛ ⬜ ⬛ ⬜ ⬛
⬛ ⬜ ⬜ ⬜ ⬛ ⬛

Nvm that's 6x7, solved it

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[short hollow]

Help

[short hollow]

I understand what it’s asking me but what do I put in quotient and in product

[vapid grove]

copy what they did above with different numbers

[main sigil]

you can pick any numbers, its up to you

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@short hollow Has your question been resolved?

[short hollow]

Yes but what do I put in quotient or product

Both of them but my own numbers

[vapid grove]

copy what they did man and change the big number

[vapid grove]

yep

[short hollow]

What do I put in quotient

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[little shuttle]

4

[short hollow]

Wdym just the number 4

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[azure dagger]

if I get "find dy/dx of (x^2+y^2)/(xy) = 5" can I just find dy/dx of x^2-5xy+y^2 = 0 where x,y≠0?

multiplying both sides by xy gives:

x^2+y^2=5xy

and subtracting both sides by 5xy gives:

x^2-5xy+y^2 = 0

[vagrant prism]

yes

[azure dagger]

what confuses me though

is that

for the first function

implicit differentiation gives dy/dx =

but the second function implicit differentiation gives dy/dx =

[austere cedar]

They may not look the same, but they're the same

Because y is some complicated function of x

[azure dagger]

is there a way to prove this

[austere cedar]

We basically just did

[vagrant prism]

moreover, $\frac{y^3 - x^2 y}{xy^2 - x^3} = \frac{y(y^2 - x^2)}{x(y^2 - x^2)} = \frac{y}{x}$

[solid kiln]

artemetra

[austere cedar]

But you can sub y in for something else in an attempt to show they can transform into the same thing.

art might actually be able to do that, and I'm going to be impressed

[vagrant prism]

i tried but this is what i ended up with lol

it's 1 am here

[austere cedar]

It's often hellish to transform like this, as y can be pretty complicated

[azure dagger]

what is transform 😭

[vagrant prism]

you can probably replace x^2 + y^2 as 5xy somewhere

[azure dagger]

can i say (5y-2x)/(2y-5x) = y/x where x,y =/ 0

[vagrant prism]

ehhhh

[vagrant prism]

on its own, no

[vagrant prism]

if you do this, sure

[azure dagger]

right but if i input values for y and x do u think it will be equivalent

im gonna try this

[vagrant prism]

$\frac{y^3 - x^2 y}{xy^2 - x^3} = \frac{y^3 - (5xy-y^2)y}{x(5xy-x^2) - x^3} = \frac{y^3 - 5xy^2 + y^3}{5x^2 y - x^3 - x^3} = \frac{2y^3 - 5xy^2}{5x^2 y - 2x^3}$

[solid kiln]

artemetra

[vagrant prism]

hm

this is.. similar

but not the same

[austere cedar]

You can, if you already know that (x² + y²)/xy = 5

[azure dagger]

what if i dont

[vagrant prism]

as long as (x^2 + y^2)/(xy) = 5, yes

[vagrant prism]

then no

[austere cedar]

Then it's no longer true

[azure dagger]

oh wait what am i askng

ok ur right

im going to input (0.20871, 1) and see what happens

[vagrant prism]

$\frac{y^3 - x^2 y}{xy^2 - x^3} = \frac{y(5xy-x^2) - x^2 y}{xy^2 - x(5xy-y^2)} = \frac{5xy^2 - x^2 y - x^2 y}{xy^2 - 5x^2 y + xy^2} = \frac{5xy^2 - 2x^2 y}{2xy^2 - 5x^2 y} = \frac{5y-2x}{2y-5x}$

[solid kiln]

artemetra

[vagrant prism]

@austere cedar be impressed

[azure dagger]

how did u do this

[vagrant prism]

x^2 + y^2 = 5xy

so y^2 = 5xy-x^2, and x^2 = 5xy-y^2

[azure dagger]

oh so

u just manipulated the

equation

[vagrant prism]

yep