help-26

[ivory wave]

that's not how this works

if you're saying isolate y then switch the x too

[spring tide]

Wdym switch x

[ivory wave]

the question is asking you to rearrange it

I think just make it in vertex form

[spring tide]

Yeah rearrange it

So isolate y

That’s how u get 3/2

[ivory wave]

well weird I've never heard something like this before

sorry can't help

[spring tide]

Oh

What’s the weird part

[ivory wave]

this

[spring tide]

U have to make like 1y right

So if u time 3/2 it get 1y

[ivory wave]

I just thought you could switch 2/3x=f(-3y+12)+6 by rearranging

[spring tide]

But the rearrange form is y=af though

[ivory wave]

I don't think you have a

just move 2/3y to 6

[spring tide]

A is 3/2

[ivory wave]

no?

[spring tide]

Huh

It don’t have it

Right

But if u times by 3/2, that’s where the a come from

[ivory wave]

you can't

[spring tide]

Y

[ivory wave]

multiple 2/3 by 6

tbh idk how to explain it

[spring tide]

Wait ok

Let’s multiple by 6

So u got 4y right?

But if u divide by 4

U still get 3/2

[ivory wave]

are you trying to make f does equal to 0, right?

[spring tide]

I try to isolate y

[ivory wave]

then switch variables too

[spring tide]

Can u write down ur step and final answer

Cuz it confuses both of us

[ivory wave]

I'm feeling sick tbh so might doing something wrong sorry

ill leave it to someone else

[spring tide]

Oh that ok

Ty though

<@&286206848099549185>

<@&286206848099549185>

[granite sundial]

what is the problem?

[spring tide]

Here

Just check my work

[granite sundial]

i see. let me check

[spring tide]

Ty

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[spring tide]

@granite sundial

how’s going on

[granite sundial]

How to got the m and n?

[spring tide]

I am not sure

But I think there’s a equation

M/b+c,n-d/a

I think but I am not sure

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[boreal mulch]

how do i find the geometric interpretation

Two parallel planes, not identical.
Other plane not parallel.
^that is the answer

[rocky iron]

the geometric interpretation is three planes that never meet or intersect

[boreal mulch]

looking at line 1 and 3, its clear that the LHS have a multiple of 3/2 difference

but RHS not

does RHS not matter? only LHS matters

if line 1 and 3 LHS have a multiple but RHS not -> parallel not identical planes
if line 1 and 3 LHS have a multiple and RHS too -> parallel and identitical planes?

[rocky iron]

yes, line 3 is a factor of -3/2 from line 1

if the RHS of line 3 was also a -3/2 multiple of line 1, then they would be the same plane

the RHS matters, because that is the reason that plane 1 and plane 3 are parallel and never meet

[boreal mulch]

ohhhh

while plane 2 is clearly independent of line 1 and 3 so it just intersects the parallel planes 1 and 3?

[rocky iron]

yes

ill show you the geometric interpretation as well

so just three planes that never have a common intersection

[boreal mulch]

ohh i see

tysm for clearing this up for me

[rocky iron]

you're welcome!

[boreal mulch]

👍

.cloes

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[autumn plaza]

how do I do dis

[rocky iron]

have you learnt logarithms?

[autumn plaza]

ermmm yes but like 2 years ago

..

[rocky iron]

you'll need logarithms to find x here

[autumn plaza]

r u sure?

[rocky iron]

another way is to change both sides to a base of 2

[autumn plaza]

Ok lemme try

[autumn plaza]

Did i do it right

[rocky iron]

yes

looks good

[autumn plaza]

Ok thx :=

:P

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[pearl fog]

idk how to start

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[brave fox]

Idk if that's the right start, but there is the property that the center of inscribed circle lies on angle bisectors

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[pearl fog]

what is it

[brave fox]

yeah this works

[brave fox]

C_2AO = OAB_1
AB_1O = OB_1C_1
AB_2O = OB_2C_2
B_1C_1O = OC_1A
B_2C_2O = OC_2A

then i just wrote expressions for literally every angle in the picture, and the equality just fell out

Idk if there is a less verbose way

[pearl fog]

oo

im too fucking tired for geonetry man wtf is wrong with geonetry

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[dry roost]

Can someone explain to me why this was solved with the equation being linear in x?

[dry roost]

Why cant it be linear in y? I have no clue

[sudden temple]

It can be I think

[dry roost]

The correct answer is 1/y according to the answer keys

Idk why

Because both 1/y and 1/x is in the choices

[sudden temple]

but mathematically A and B both are like correct

maybe there could be some error in question

[dry roost]

Oh wait i think the -y^2 violates linearity?

Like if y becomes the dependent variable it wont be linear anymore?

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[neon iron]

confused on 2nd half of the proof by contradiction

can sum1 explain

[cinder sequoia]

starting with which sentence

[neon iron]

if p is not prime

[cinder sequoia]

if P isn't prime, then it can be divided by some prime factor p_i. P = p_1 ... p_n + 1, so P - p_1...p_n = 1

if P is divisible by p_i, and p_i is in that list of primes, then P - p_1...p_n is divisible by p_i

but that would mean that equivalently, 1 is divisible by p_i

[neon iron]

sorry still cant understand lol im kinda stupid. confused on this part.

[cinder sequoia]

P - p_1...p_n is divisible by some prime factor p_i, right?

[neon iron]

yep

[cinder sequoia]

since P is divisible by some prime, based on P being composite, and p_i is a factor in that list

[neon iron]

we dont know which 1 but its 1 in the list

[cinder sequoia]

P - p_1...p_n is equivalently 1

since P = p_1 ... p_n + 1

now if two numbers are equal, then if some number divides one then it must divide the other

since they are necessarily the same number

p_i is some prime dividing P - p_1...p_n

P - p_1...p_n is identically the number 1

and therefore p_i would have to divide 1

[neon iron]

oh i see

ty

that helped

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[shell orbit]

FR tho 😭

Anyways, im guessing I use the distance formula somehow

Sqrt(a^2 + b^2) right?

Ah yeah nvm

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[serene quiver]

how do i get that equation?

[woeful drift]

Which equation?

[serene quiver]

the equation of the plane through those points

[dry quest]

u eat a pie

[woeful drift]

One way would be to compute the cross product of two of the vectors in the plane (you have three points so many vectors to choose from), that'll give you the normal vector of the plane

[serene quiver]

ill give it a go.

[woeful drift]

There are lots of 0s so the algebra should be simple

[serene quiver]

wait you mean the cross products of the points?

[woeful drift]

That's a meaningless operation

If you label the vertices A, B, C for example, one possible vector is A-B

[serene quiver]

oh

[serene quiver]

uhhh so i did from (1,0,0) to (0,-2,0) and (1,0,0) to (0,0,16)

that would make the first vector <1, -2, 0> and the second (<1, 0, 16> right?

i did cross product and got <-2, -16, 2> :/

[woeful drift]

Second is slightly off, check your signs

Err wait, both seem off

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[serene quiver]

i do not comprehend

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[hazy cosmos]

I would like to know how the fraction can be expanded like that

[opal vault]

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

[hazy cosmos]

ohh is it just basically changing it to the difference of powers raised to p?

[opal vault]

yep you recognize a difference of real numbers to the power of p

and then use x^p - y^p identity

[hazy cosmos]

alright thanks

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[vivid saffron]

I am learning new things with advanced exams for my level so i would like to know what i need to know for the second and third question, what do i need to study that will help me to answer these questions

I managed to do the first one

[quick dragon]

Where did "w" come from in part A

[vivid saffron]

sorry, i translated wrong

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[quick dragon]

Oh I see

Basically you can show that the distance from the center of a big circle to the nearest corner of the square is also "c"

idk if thats how you did A

[vivid saffron]

exactly, i made like it

[quick dragon]

but if it is then you can make an isosceles right triangle with legs of length "a" and hypotenuse "c"

which means you can solve for c in terms of a

so substitute that into your original expression for the diagonal in part A

since the square has side length of 1, you can find what the diagonal actually equals

[vivid saffron]

it is the question a) right?

[quick dragon]

when I wrote lowercase I meant the lengths in the picture

uppercase I meant the question letter

[vivid saffron]

alright

and about the b) and c), what would i need to know to do these questions? I want to study the content needed for making this exam you know

[quick dragon]

I showed how to do b

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[ripe shale]

what do i add in the circled area

[ripe shale]

so

[clear nova]

try using the given values and input them into that formula

replace the blank with x or some other constant, use the first set of data (shape number 1, number of matches 4) cuz itll be easier

[ripe shale]

oh ok

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[copper fossil]

Am I doing these correctly?

[copper fossil]

when it's not stated which side it's approaching from, it wants you to take into account both sides, right?

[cedar wagon]

Idk, there is a value for f(-4) and f(4) in the first graph

[copper fossil]

I think it's just asking for the limit, not the value of f(-4) or f(4)

[rigid ivy]

Yes

[cedar wagon]

Ok, then it is from the two sides

[copper fossil]

And DNE or undefined are the answers given when you have two different y values being approached despite approaching the same x value?

[cedar wagon]

Undefined is where it doenst approach a finite value, and DNE is for no limit, as some undefined approach Either +inf or -inf

[copper fossil]

alright thanks!

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[mortal oxide]

What is 808 days after august 15th 2022

[mortal oxide]

That’s it

Someone please answer I’m too lazy

[rigid ivy]

!noans

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[barren zodiac]

you can take 2x365 days out of that number as you'll get the same day just in a different year, and then it'll be easier to calculate the remaining days (hopefully there are no leap years though)

[rigid ivy]

You'll get into 2024, so there will be a leap year

[barren zodiac]

ah mb, we don't use that calendar

[barren zodiac]

so make sure to take that extra day into account as well

[crude pivot]

just minus 365 first then minus 366

then it becomes 77 days after August 15 2024

[mortal oxide]

Ok

I did it

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[polar blaze]

840x = pi mod 2pi. How many solutions are there given 0 <= x < 2pi?

[icy sky]

it would be easier to write it as 840x= pi+2npi

you can find bounds for n from there

[polar blaze]

oh cuz it has to be less than 1680pi

ohh ok its 840 ty

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[stone hare]

Phew, I’m trying to keep up with my professor’s simplification here and it is very confusing

I get lost after

[stone hare]

Oops wait, there is a 2nd part, 1/3 can be ignored

His keep change flip here

-h/1+h * 1/h

Shouldn’t it be

-h/1+h * h/1

Because that is the true reciprocal

[turbid sable]

no bc on the bottom is h by itself which is h/1 so the reciprocal is 1/h

[stone hare]

Okay..

[turbid sable]

mb for bad explanation

[stone hare]

No it’s not you,

When I look at it I see keep (-h/1+h)

Multiply instead

If I could picture my confusion, I see h as 1/h | x /h because it’s on the denominator

@turbid sable but I’m confusing myself the further I think on something like this and should just Move on from my mistake

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[vivid saffron]

Why sqrt(2) = 2a( sqrt(2) + 1 ) -> a = sqrt(2) / [ 2 ( sqrt(2) + 1 )]

[verbal moon]

divide both sides by 2(sqrt(2)+1)?

[vivid saffron]

well, i am trying to find the value of a

but i didn't understand what hapen there

[cedar wagon]

They just divide as 187 said

[vivid saffron]

but how sqrt(2) / 2(sqrt(2)+1) turned into a

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[neon iron]

.

[grave berry]

nvm

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[stray wave]

can somemone please tell me if im correct

[stray wave]

My work is shown in this picture

[stray wave]

These are the two questions

[craggy haven]

it looks like you calculated the surface area of a box and a cylinder and added them together, is that right?

[stray wave]

Yes it's a composite object then I subtracted the overlap so I subtracted a pi r^2

[craggy haven]

from both of them

since it covers up that area on both shapes

[stray wave]

So Im supposed to subtract 2 pi r^2

[craggy haven]

yeah

do you see why?

[stray wave]

Oh ok

Yes

It overlaps two times

[craggy haven]

yeah

the volumes just add together though like you did

[stray wave]

So I'm correct on the volumes?

[craggy haven]

i don't want to check your arithmetic :P but your method is right

[stray wave]

Ok

Thanks

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[languid dagger]

If the function has a local maximum end point that belongs to the x-axis, find the value of c, then find the equation of the tangent at its inflection point.

My only problem here is the y

Here y is supposed to equal c

So why is it positive

sorry if the translation isnt that clear

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[lofty coyote]

I need to find the angle theta in the above diagram..
Given: AB = 10 and BC = 10

What i have tried..

1. I find the hypotenuse AC using pythagorean theorem, which is 14.142
2. I find the length of MC by dividing the AC by 2, Which is 7.071
3. Now, i tried using tan inverse as i know MC(Opposite) and BC(adjacent), which came out 35.264 degrees..
   But the answer is 45 degrees (after round-off).. IDK WHERE I Gone wrong..

[lofty coyote]

Ping me.. While replying pls!

[fierce bluff]

The angle B is a right angle

You don't need to calculate the length of the hypotenuse

[lofty coyote]

Yeah.. Sorry not to mention earlier

[lofty coyote]

why..

[fierce bluff]

Because M is the midpoint of AC, while AB and BC are of equal length

Let's see if I can come up with some straightforward complete argument for it

The vague rationale is to try and justify that it's an isosceles triangle

It really isn't actually known unless you know the triangle is a right triangle

Ah

If they tell you B is a right angle, then it's clear

[lofty coyote]

And to mention.. The AB will not always equal to BC.. The mentioned problem is one of the cases.. In other cases (i have several one).. the two of them will vary.. So.. i need to find the genreal algorithim

[fierce bluff]

That's correct

AB=BC only for a 45-45-90 triangle

[lofty coyote]

ok.. But.. WE need to find angle MBC!

[fierce bluff]

It's 45

[lofty coyote]

how.. sorry..

[fierce bluff]

But I can see that from intuition, without any thought

I am trying to think of how to justify it properly with an algebraic argument

[lofty coyote]

...

[fierce bluff]

B is 90 degrees, you are told. The line BM is the bisector of the angle B, because AM and MC are of equal length

So MBC is 45 degrees

[lofty coyote]

Oh.. Now it makes sense!

[fierce bluff]

To see why BM is the bisector more clearly, note that MBC (triangle) and ABM (triangle) are congruent triangles

the whole construction is symmetric across the BM line

like, you can fold it over and it's the same

[lofty coyote]

hm.. Yeah..

[fierce bluff]

the reason your solution here doesn't work is because mc is not the "opposite" side of a triangle other than MBC, but BC is the hypotenuse (not adjacent side) of that MBC triangle

[lofty coyote]

oh.. So.. BMC is 90 degrees?

[lofty coyote]

Ok..So for the triangle MBC.. BC is hypotenuse.. but.. MC is adjacent or opposite..

Oh.. do MC is opposite in the triganle MBC as it is opposite to the angle theta?

[fierce bluff]

Yeah

[fierce bluff]

MC is indeed the opposite relative to theta

MB is adjacent

[lofty coyote]

So.. I need to use sin inverse right..instead of tan inverse

[fierce bluff]

Well, firstly you need to know that BMC is 90 degrees, and have to prove that 🙂

But assuming you do know it, then yes you could do sine inverse

[lofty coyote]

😅 .. Ok.. We can assume..

I got the answer for this case.. But when AB is 20 and BC is 10.. I can't get the answer!

I tried using asin but it is out of domain..

[fierce bluff]

I guess if I wanted to write a "clean" answer to this problem, I'd say something like this:

Because AB = BC and B = 90 degrees, we know this to be an isosceles triangle, so the line segment from the midpoint M of AC to the angle B is a bisector of B. Hence theta is 1/2 the angle of B, i.e., 45 degree.s

No trig needed

AB isn't 20

it's 10

This argument won't work if AB = 20

[lofty coyote]

in the other case, it is 20

[fierce bluff]

Try drawing the triangle, i have no tools at hand to draw with currently

Things will be clearer that way

[lofty coyote]

yeah.. i know

[lofty coyote]

AB is 20, in this case!

[lofty coyote]

I can understand your above argument.. as you are proving that BM is an angle bisector, so we can conclude that angle theta is 45, right?

[lofty coyote]

But in this scenario, where AB != BC, we can't use the formula AM/MC = AB/BC to prove that the line segemnt is an angle bisector.. So how can i use the trig to find the theta

[fierce bluff]

Yeah

[lofty coyote]

But what can we do in this case!

I try using sin inverse

[fierce bluff]

So in this case, you can fill in a bit with three pieces of information:

Use arctan to get A and C.

Use pythagorean identity to get the length of AM = MC.

that's just some extra data to work with to maybe make things easier

[lofty coyote]

i can't get it.. sorry!

[fierce bluff]

tan(A) = 10/20 = 1/2

tan(C) = 20/10 = 2

[lofty coyote]

oh ok

so we can find the angles A and C

oh wait.. If we know the angle C.. we can calutlate theata as.. theta + C + M = 180!

where M is 90 degrees

am i correct?

[fierce bluff]

Then you can get the length of MB using law of cosines

Yes

But you don't know CMB angle

[lofty coyote]

M is 90 degress

[fierce bluff]

No, it's not 🙂

It's acute

[lofty coyote]

ok sorry..!

[fierce bluff]

CMB is

AMB is obtuse

[fierce bluff]

This is the only way to do it I would think

[lofty coyote]

ok then.. What can i do with the length MB?

[fierce bluff]

Well, that's the magic

Once you have MB, you also have C, which is the opposite angle. And you have MC, the length which is opposite to the angle MBC!

So you can apply law of sines then.

MB/C = MC/MBC

So to solve MBC (theta), you just calculate MC*C/MB

[lofty coyote]

ok sounds cool.. I will try using magic!

[fierce bluff]

this should work even if you change the side lengths

it only relies on B being a 90 degree angle

[lofty coyote]

cool.. that's what i need!

But. Why i can't use sin inverse.. @fierce bluff

[fierce bluff]

Of what?

[lofty coyote]

of BMC..

😅

in triangle BMC

[fierce bluff]

it's not a right triangle

[lofty coyote]

oh.. So i use trig functions only in right triangle?

[fierce bluff]

yes

[lofty coyote]

to identify all the three sides..

ok.. thank you bro!

[fierce bluff]

np

[lofty coyote]

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[harsh shell]

whats the point of this stpe

[verbal crater]

just a list of all the information given

helps you in the solving steps

[harsh shell]

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[lofty coyote]

Since.. the traianlge ABC is right angle triangle.. we can conclude that A and C is 45 degrees each right?.. or not?

[prisma mesa]

ABC is right angle and isosceles

thanks to that, you can conclude that angle A = angle C

[fierce bluff]

only because AB = BC

[prisma mesa]

and since angle B = 90°, and B + A + C = 180, you can conclude that A,C = 45°

[lofty coyote]

Oh.. So not every right angle triangles has 90 + 45 + 45 = 180..?

[prisma mesa]

no, but since its also isosceles it's true

[lofty coyote]

But not when.. AB != BC right?

[prisma mesa]

yes

in that case we only know that A + C = 90

but A and C can be e.g. 60 and 30 respectively

[lofty coyote]

hm.. yeah making sense.. Thanks guys

[prisma mesa]

btw BM = MC, so BMC is also isosceles

and thanks to that, you can conlude what theta is immidiately

[lofty coyote]

BM = MC.. how>? @prisma mesa

[prisma mesa]

do you know about triangle congruence?

[delicate dragon]

Excuse me, could any of you explain to me what a limit value is?

[prisma mesa]

Try looking at ABM and CBM

[topaz sinew]

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

[lofty coyote]

yeah.. i gues..

[prisma mesa]

what can you tell about these 2 triangles then?

[lofty coyote]

they are isoceles

no.. they are equilateal

[prisma mesa]

nope

[prisma mesa]

that's right

but what about their relation to each other?

are they congruent, similar or neither?

[prisma mesa]

I just filled in the info we already know

[lofty coyote]

why.. They have all three sides equal right?

[prisma mesa]

they don't

BC is longer than BM and MC

and AB is longer than AM and BM

[lofty coyote]

ok sorry..

so they are similar?

[prisma mesa]

that's also true, but moreover, they are actually congruent

note that BC is same as AB, angle C is same as angle A and MC is same as AM

hence they are congruent by side - angle - side

[lofty coyote]

similar != congurent? sorry.. i forgot about those..

[prisma mesa]

similar can be scaled

the red shapes are both similar and congruent
the green shapes are only similar, but not congruent (since they have different size)

[lofty coyote]

Oh.. Making sense.. Awesome bro..

[prisma mesa]

so now that we know that BMC and ABM are congruent, we also know that all their corresponding angles and sides are equal

[lofty coyote]

So by this how can you conlude that BM == MC?

[prisma mesa]

meaning that e.g. angle CBM = angle ABM

[lofty coyote]

cool!

[lofty coyote]

But.. This doesn't add up!

is this a typo?

[prisma mesa]

we actually won't need it, but it will be a conclusion of what we are gonna do now

im using a slightly different approach rn, because i think it will be easier for you to understand

[prisma mesa]

okay so we know that the green angle is same as blue angle

[lofty coyote]

ytes

[prisma mesa]

and also that green + blue = 90°

[lofty coyote]

hm\

[prisma mesa]

what does this tell us about green and blue angles?

[prisma mesa]

this is because they add up to the angle B, which we know is right

[lofty coyote]

so they are 45 each

[prisma mesa]

correct

meaning that theta is also 45°

since theta is just the blue angle

[lofty coyote]

yeah..

[prisma mesa]

we can also note that BMC is isosceles, because those 2 angles are both = 45°

[lofty coyote]

so m is 90

[prisma mesa]

right

and BM = MC

since the 2 other angles are equal and so BMC is isosceles

[lofty coyote]

oh.. You are proving that BMC is isosceles to conclude that BM == MC

hmm.. making sense

[prisma mesa]

yeah, that's one of the ways to do it

there are thousands of other ways though

[lofty coyote]

So.. is it one of the critiria of isosceles triangle that if two angles are equal then it is isosceles?.. Sorry.. To ask this silly one!

[lofty coyote]

oh..

[prisma mesa]

yes, either 2 sides or 2 angles must be equal

[lofty coyote]

cool.. bro..

BUT.. This isn't gave me the solution of my actuall problem @prisma mesa

😅

[prisma mesa]

what was the actual problem?

[lofty coyote]

This is my problem

Where AB != BC

So.. We cannot use the isosceles triangle logic here!

[prisma mesa]

oh, i thought you needed to find theta

so now you need to find theta here, right?

[lofty coyote]

yep!

[prisma mesa]

one of the ways we can do it is by calculating MC and then using the triangle BMC to calculate theta

[lofty coyote]

We can calculate the MC easily using pythagorean theorem then..

[prisma mesa]

or by chance, do you know thalet's theorem?

[lofty coyote]

oh no..

[prisma mesa]

wait do you know thalet's theorem or not?

[lofty coyote]

or.. i can learn it now..

[prisma mesa]

*thales

[lofty coyote]

wait..

[prisma mesa]

this one

this theorem would allow us to conclude extremely useful fact

[lofty coyote]

Yeah.. I know it.. BPT

[prisma mesa]

what about this one?

[lofty coyote]

yeah..

[prisma mesa]

cool

so lemme draw one circle rq

[lofty coyote]

ok bro..

[prisma mesa]

so now M is the center of the circle

[lofty coyote]

hmm

[prisma mesa]

and MB is just a radius of the circle

as is MC and AM

and all of them should be equal

[lofty coyote]

ok...

[prisma mesa]

but since BM = MC, we also know that angle MBC = angle MCB

[lofty coyote]

isosceles criteria.,. cool

[prisma mesa]

red and green are equal, because BMC is isosceles

so it actually suffices to find the red angle

and then we will get the green one immidiately

and finding the red angle is just simple trigonometry

do you know how to find the red angle?

[lofty coyote]

Bro.. Since BM = MC.. BMC is isosceles.. SO it contains two equal angles... But we don't know which two among the three will be equal?

[prisma mesa]

we do, if the red sides are equal, then the purple angles are equal

[lofty coyote]

How you conclude that.. angle MBC = angle MCB? it can be.. angle BMC == angle BCM.. or other possibilites

[prisma mesa]

the 2 equal angles are the ones that are between the base and one of the equal sides

[lofty coyote]

oh.. oh.. Now i am making my fundamentals stronger.. Thank you So much bro.. You are awesome\

ok bro.. Give me some time to read the other details you gave in the diagram!

[lofty coyote]

Then.. I need some more minutes pls

lol.. easy.. by using tan inverse!!

[prisma mesa]

correct

[lofty coyote]

cool bro.. But in exacatly where we include Thales theorem?

[prisma mesa]

in here

without thales theorem, we wouldnt know that M is the center of the circle

[lofty coyote]

drawing a circle?

[prisma mesa]

we can draw a circumcircle for any triangle, but not in every triangle the center lies on the hypotenuse

that's specific to right triangles. And we know that thanks to thales theorem

and only thanks to the fact that M is the center, we were able to conclude that AM = BM = MC

[lofty coyote]

bro.. It seems quite confusing.. Because Thales theorem just says that if a line drawn parrelle to a side of a truiangle and if that line intersects other two sides, then that line segment divides those two sides in the same ratio..

[prisma mesa]

do you know the one with circle and right triangle

[prisma mesa]

this image basically describes it

[lofty coyote]

sorry.. i can't complety understand this one

[prisma mesa]

here it is with text description

and the converse to that holds too

[lofty coyote]

ok.. So we know that in our case.,. AC is a diameter.. But how we conclude that M is the midpoint of the diamter!

Oh yeah.. sorry..

Now i am clear..

Again.. A Big thanks for your time bro @prisma mesa

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[marsh roost]

I need help with sketching and finding the x-intercepts with: y = 2 cos 3 (x − π/4) when the domain is [-π, 2π]

[robust turtle]

for x-intercept, put y=0 and solve for x

and for the graph, you can put of values of x and y separately and plot the graph, it will be tedious tho

[marsh roost]

idk what to do after

because of the brackets

[robust turtle]

it means cos 3 (x − π/4) =0

[marsh roost]

what ab the 2?

[robust turtle]

ah

0/2

is 0

got it?

[marsh roost]

yeahh

[robust turtle]

consider x − π/4= P

it means cos 3P =0

when is cos 0?

in the given domain

[marsh roost]

idkk

[marsh roost]

π/2

[robust turtle]

yes

so, cos 3P = cos pi/2

now, 3P = pi/2

you can do now

try it

[marsh roost]

x = π/2, 2π - π/2 ?

[robust turtle]

no

3 (x − π/4) = pi/2

[marsh roost]

oh okay

5π/12

[robust turtle]

yes

that is one of the x-intercept

[marsh roost]

do i keep adding the period 2π/3 to find the other intecepts

why?

[robust turtle]

sorry

it is not

[marsh roost]

its okay

whats the period then?

[robust turtle]

okay

2pi/3

it is

i was calculating wrong

[robust turtle]

but from that you wont find the other intecepts,

lemme share the grap

[marsh roost]

okk

[robust turtle]

https://www.desmos.com/calculator/p9zjydr7kl

https://www.youtube.com/watch?v=R1iGQXyjaAc&t=12s
i found this video, it might help
it is about sine function, but the basis is same

[marsh roost]

thank youu

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[south pasture]

Can someone help me with this?

I don’t understand where did the sqrt2 / 2 came from in the first step

I did 1/2(1+i) but can’t move forward because sin theta = cos theta = 1

[coarse slate]

They’ve replaced 1 by sqrt2*(1/sqrt2)
Where 1/sqrt2 is cospi/4 & sinpi/4

[south pasture]

why though

[south pasture]

they do sqrt 2 / 2 though which is not equal to one

how did they find that value?

[coarse slate]

1/2 is already in the question. The value inside the bracket becomes (1/sqrt2 + i*1/sqrt2). For the bracket to become (1+i) they’ve multiplied by sqrt2

[topaz sinew]

@south pasture Has your question been resolved?

[topaz sinew]

@south pasture Has your question been resolved?

[south pasture]

i dont get it where did the 1/ √2 came its not even in the question

<@&286206848099549185>

like

1/2(1+i) is what we have

not √2/2

also i think ur trying to say we did √2/2 to multiply it by the inside things and get 1/2 but like

√2/2(cos45+ i sin45)

and yes we do get 1/2 + 1/2 i but the thing is

1/2(1+i)

and cos 45 is not equal to 1.

shouldn't we be sure that we have this form?

[nocturne flume]

well
any complex number a + bi can be expressed as r(cos(theta) + i*sin(theta)
where r is the length from the origin to that complex number as a point in the complex plane, which is sqrt(a^2 + b^2)
and theta is the angle of the line you draw to get to that point from the origin
so we got it since sqrt(1^2 + 1^2) = sqrt(2), and we divide it by 2 since we've multiplied by 1/2 at the start

but if you havent learned about the complex plane yet, we just multiply by sqrt2 / 2 to make sure it ends up equaling 1/2 + 1/2 i when you multiply it all out

heres a more visual explanation

[topaz sinew]

@south pasture Has your question been resolved?

[south pasture]

i learned about the complex plane but idk the 1/2(1+i) confuses me

I get it that we are trying to make it equal to 1/2 + i1/2 by doing sqrt 2 /2 ( cos 45 + i sin45 ) but i dont underdtand why we did 1/2(1+i) in the first place 🤷🏻

[nocturne flume]

because it was 0.5+0.5i so we factored out the 0.5 and it became 0.5(1 +1i)

are you asking why we thought of doing the first place? if so i think its because it makes converting into the form r(costheta + isintheta) easier since youre working with the common 1 1 sqrt(2) right triangle

well actually now that i think about it a 0.5 0.5 sqrt(2)/2 triangle also isnt hard to derive, but some people prefer to do it the other way

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[sour carbon]

I’m stuck and don’t understand what to do

tried drawing AD and calling the midpoint of AD R. I found some similar triangles from this as NR is parallel to BC

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[sour carbon]

.reopen

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✅

[clear lodge]

is ABC a right triangle?

[sour carbon]

?

what do you mean

[clear lodge]

yeah, i've seen the original pictures. Are those the full instructions of the problem?

[sour carbon]

yes

[reef adder]

try using reverse mid point theorem... draw a line parallel to AB from point N

So NP will be half the length of AB (by mid-point theorem)

I tried proving MP = NP

Then draw NT line parallel to MP... you will get a rhombus

[reef adder]

Diagonal of rhombus bisect the angle at vertices...

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[sour carbon]

Thank you so much

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[lean pawn]

alternatively; if the measurement of 1 is congruent to 4, then measure of angle 2 is congruent to angle 3?

[mint crescent]

What’s the question

[lean pawn]

this itself

[mint crescent]

If you keep the complement conditions, then yes.

[lean pawn]

aightt

tyy

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[desert agate]

how should i go about solving this sequence

[frail reef]

i would look at the terms of the series

it seems to be the sum $\sum_{n , = , 1}^{\infty} \frac{e^n}{n^3}$

[thorny flame]

eugene_krabs_has_cake

[frail reef]

so for this sum to converge, these terms have to tend to 0

[desert agate]

so because they don't tend towards 0 it would just diverge instead right

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[sharp dew]

The solution provided by the site, looks a bit strange

They use the definition of derivative

Bro

[haughty ember]

mb it didn't show in the thingy

[sharp dew]

It is okay

[sharp dew]

If the equation holds, then the function is differentiable at x=0

[raven sparrow]

That equation only means that the limit of the function exists at x=0. It doesn`t even guarantee continuity at x=0, so certainly not differentiability.

For differentiability you want the derivatives to agree.

[sharp dew]

I see

I have to prove that h’(x) at x=0 is continuous

No

[sharp dew]

Yes

Is it valid?

Can someone check it for me

[raven sparrow]

Not necessarily continuous. Actually, checking with the definition is probably better, but in this case the left and right handed limits will end up the same as the derivatives.

I think that if the limits are the same, then it is a bit stronger (having continuous derivative is good), but it works fine in practice.

[sharp dew]

Help me

[raven sparrow]

It`s fine, yes.

[sharp dew]

Thank you

.close

[topaz sinew]

Channel closed

Closed by @sharp dew

Use .reopen if this was a mistake.

[raven sparrow]

It is sufficient to show that the limits of f'(x) around 0 are the same, since we know it's differentiable everywhere away from 0 (it's a polynomial in those places).

The reason why you'd want to use the definition is that it's possible for f'(0) to be defined despite the limit of f'(x) about 0 is not.

[topaz sinew]

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[spice tundra]

Can I get some help with this problem. I am struggling to remember what I am supposed to do.

[mint crescent]

Do you remember the conditions for a sequence to converge or diverge?

If not, look them up.

[spice tundra]

The conditions for it to converge or diverge? Like if it’s converging it eventually goes towards a number and if it diverges it goes to infinity?

[mint crescent]

Finite limit vs a limit that’s doesn’t exist/is infinite

Yeah

So let’s start with (a) - how would we start?

[spice tundra]

I don’t remember how to prove that it converges or diverges I can plug in numbers and see it converges but I don’t remember besides that.

[mint crescent]

Just set up the limit for now

[spice tundra]

Like Lim n-> infinite (1/n)^1/n

[mint crescent]

Yup

$\lim_{n \to \infty} \left(\frac{1}{n} \right)^{\frac{1}{n}}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

Do you remember the trick where you rewrite using natural logs if you have stuff in the exponent?

[spice tundra]

I do not remember that

[mint crescent]

$a^b=e^{\ln(a^b)}=e^{b \ln(a)}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

This is the main idea

[mint crescent]

If we apply that to this, what do we get?

[spice tundra]

e^((1/n)ln(1/n) )right?

[mint crescent]

Yeah

$\lim_{n \to \infty} e^{\frac{1}{n} \ln \frac{1}{n}}=e^{\lim_{n \to \infty} \frac{1}{n} \ln \frac{1}{n}}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

Any ideas now?

[spice tundra]

Would you be able to use a test there then?

[mint crescent]

What are you thinking of?

[spice tundra]

Am I thinking of the wrong thing here? I was thinking of like divergence tests like ratio test or something but that might be wrong for this.

[mint crescent]

That’s for series convergence

You’re just calculating a limit rn

[spice tundra]

Right just calculating a link right now

Sorry I’m mixing things up cause there is a bunch of different problems in here and some are like that.

As n got bigger and bigger there though it would approach 0 though meaning it would converge to 1 right…?

[mint crescent]

Yeah, the sequence converges to 1.

[spice tundra]

Would that be enough to prove it converges to 1?

[mint crescent]

Yeah, you showed the limit is 1

[mint crescent]

Specifically this one

[spice tundra]

Ok thank you

So for these problems do I not need to do any convergence tests or anything like that?

[mint crescent]

Nah

[spice tundra]

So I can just simplify it like that and plug in numbers or solve the limit basically?

[mint crescent]

Do not plug in numbers

1. It’s not rigorous even for continuous functions

2. Sequences don’t have a discrete domain, so it’s gonna be even worse

Just calculate the limit like a normal person

[inner oracle]

Is the second point true?

[spice tundra]

I think the other two problems also converge to 1

[coarse tusk]

You’re right

@mint crescent why?

[mint crescent]

?

Oh wait

Wrong emoji

[coarse tusk]

Oh lol

[spice tundra]

I think I have those problems good then if they all converge to 1

Thank you

Could I also get some help with these ones then?

These I think actually use the divergence tests which was why I got it mixed up.

[coarse tusk]

Just the limit test works here

[spice tundra]

Just the limit test works on the first one?

[coarse tusk]

Yes

[spice tundra]

I will have to look the tests up again because it’s been a while but finding which test to use is a lot of it for me

Would I then use alternating series test for the second one?

[coarse tusk]

That might work

[spice tundra]

Would there be another better option?

[mint crescent]

I’d use alternating too

[coarse tusk]

Yeah alternating is pretty easy here

For the last one the integral test is the most suitable

[spice tundra]

Alright thank you.

For the alternating series test would I have to convert the function the the ( -1)^n form

[coarse tusk]

Yes

But notice that cos(n*pi/2) is only 1 or -1 at all even values of n

[spice tundra]

What would that mean?

Sorry it’s been a while since I have done this I don’t quite remember what I have to do.

[mint crescent]

Wait we might be certified idiots - isn’t this the same as $$\sum^{\infty}_{n=1} \left(\frac{1}{4n}-\frac{1}{4n-2} \right)$$

[thorny flame]

Civil Service Pigeon

[mint crescent]

,w sum of 1/(4n) - 1/(4n-2) from n=1 to infty

[thorny flame]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[mint crescent]

,w sum of 1/n * cos(n*pi/2) from 1 to infty

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[mint crescent]

@coarse tusk idt we cooked with alternating lol

[spice tundra]

For which question?

[mint crescent]

b

[spice tundra]

For 2

Wait how would that be the same as 1/n cos(npi/2)

[mint crescent]

Write out the first few (nonzero) terms and it becomes obvious

[spice tundra]

0,-1/2, 0,1/4

First 4 terms

[mint crescent]

A few more than that

[spice tundra]

Wait but even with those I don’t think they equal do they?

[mint crescent]

??

[spice tundra]

Cause like 1/8-1/6 does not equal 0

Or I mean -1/2 not 0

But still

Actually looking at it I think it is basically correct

[topaz sinew]

@spice tundra Has your question been resolved?

[mint crescent]

?

Just use limit comparison now

[spice tundra]

Yeah you can use limit comparison at that point

[spice tundra]

But also how do you get to that equation from the first one?

[mint crescent]

Did you not confirm it yourself here

[spice tundra]

Yeah I did but then my brain was like wait actually idk

I was like it seems kinda right but also I just don’t know how to get to the equation.

Sorry I am struggling with brain rn been doing this stuff for the past few days straight and my brain doesn’t want to work right anymore 😂

[spice tundra]

Like would I be able to show any work as to how you could arrive at that equation or is it kinda just seeing relations?

[mint crescent]

Uh

Pair consecutive terms in the expansion

$\left(\frac{1}{4}-\frac{1}{2} \right)+\left(\frac{1}{8}-\frac{1}{6} \right)+\left(\frac{1}{12}-\frac{1}{10} \right)+\cdots$

[thorny flame]

Civil Service Pigeon

[mint crescent]

The larger denominators are increasing by 4

The rest falls into place after that

[spice tundra]

Ohhh wait I get it now actually

For some reason my brain did not like that that eqaution did not have the zeros from the series however the zeros don’t matter.

Ok that makes sense now.

Only other question. For integral test I just take the integral of function and if integral converges to a value and not infinity it converges correct?

But also would I even need to use integral test for the last one since ln of infinity is still infinity?

So function would end up being 1/infinity right?

[topaz sinew]

@spice tundra Has your question been resolved?

[topaz sinew]

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[topaz sinew]

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[eager timber]

Could anyone help me solve this differential equation?
$x \cdot \frac{dy}{dx} + 2y = 1 - \frac{1}{x} \$
$\cdot \frac{dy}{dx} + \frac{2y}{x} = \frac{1}{x} - \frac{1}{x^2} \$
$\frac{dy}{dx} +\frac{1}{x}2y = \frac{1}{x} - \frac{1}{x^2} \$
$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{2} \cdot (\frac{1}{x} -\frac{1}{x^2}) \$
$\frac{d}{dx}(xy) = \int \frac{1}{2x} - \frac{1}{2x^2}dx \$
$\frac{d}{dx}(xy) = \int 2x^{-1} -2xdx = 2\ln x - x^2 \$
$xy = 2\ln x -x^2 \$
$y = \frac{2\ln x -x ^2}{x} + C$

[thorny flame]

Tomi

[eager timber]

what is wrong with this? (hope didnt anything)

[coarse slate]

You didn't divide 1st term by 2

There's no need to divide tho

[coarse slate]

How did you get the 4th step?

[topaz sinew]

@eager timber Has your question been resolved?

[eager timber]

shouldnt the y be on its own?

[thorny remnant]

4thstep is wrong

[eager timber]

well its like the product rule so I end up with x and y

[coarse slate]

There's a little mistake I assume

[eager timber]

should I just leave it as xy= the integral of whatever is there

[thorny remnant]

there is no need of xy here

[eager timber]

well since the integrating factor is x

I multiply that out

and after I take the integral of that side I end up with xy

[thorny remnant]

if you are applying IF then there is no need to to have $\dv {xy}{x}$ in the LHS

[thorny flame]

convergence

[eager timber]

then it should have been
$xy = \int \frac{1}{2x} - \frac{1}{2x^2}dx$

[thorny flame]

Tomi

[thorny remnant]

yes

no wait thats wrong

IF is not x

[eager timber]

$x(\frac{dy}{dx} \frac{1}{x}y)$

[thorny flame]

Tomi

[eager timber]

here i end up with dy/dx(x) and y

that is the product rule with the derivative of x being 1

[thorny remnant]

iintegrating factor is not x!

[eager timber]

$e^{\int \frac{1}{x}}$

[thorny flame]

Tomi

[eager timber]

then it is e^{ln x}

and x

[thorny remnant]

$e^{\int \frac{2}{x}}$

[thorny flame]

convergence

[eager timber]

here P(x) is 1/x right?

[thorny remnant]

no P(x)=2/x

[eager timber]

well it depends, if I calculuate P(x) at the beginning then yes, if I divide everything by 2 it is one over x

so I could calculate P(x) with 2 over x so I don't need to divide by 2.

[thorny remnant]

no it doesnt the case you did is wrong

[eager timber]

let me see what I got if p(x) is 2/x

[thorny remnant]

$\frac{1}{2} \cdot\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{2} \cdot (\frac{1}{x} -\frac{1}{x^2}) $

[thorny flame]

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

[thorny remnant]

this what you will get if you divide by 2

[eager timber]

but now the term front of y is 1 over x, no?

well yes I forgot

to multiple

by 1 over 2

in the left term

LHS

[thorny remnant]

yes so we cant apply the IF in this case

it should be in the form of
y'+yP(x)=Q(x) to apply IF

[eager timber]

just realised by IF you mean "integrating factor"

[thorny remnant]

yes

im too lazy to type that

[eager timber]

so I should divide by 2 and then do the IF part

[thorny remnant]

no need to do that just take P(x) =2/x makes your life easier

[eager timber]

alr let me try this again and see what i get

[thorny remnant]

yeah

[eager timber]

$(2 \ln |x| (\frac{dy}{dx} + \frac{2y}{x} = \frac{1}{x} - \frac{1}{x^2})$

[thorny flame]

Tomi

[eager timber]

so I just multiply through by that

[thorny remnant]

yeah its wrong

[eager timber]

integral of 2/x is that

ah my bad

i forgot to deal with e

[thorny remnant]

if you are using IF then the eq will be in
$y*IF=\int Q(x)\cdot IF$

[thorny flame]

convergence

[eager timber]

you gotta multiply each of the terms

[thorny remnant]

well its the same thing with differnt notation mb

[eager timber]

so i end up with
$\frac{xdy}{dx} + 2y = x - 1$

[thorny flame]

Tomi

[eager timber]

is this correct then @thorny remnant

[thorny remnant]

almost correct

[eager timber]

x2y

[thorny remnant]

its $\frac{x^2dy}{dx} + 2xy = x - 1$

[thorny flame]

convergence

[eager timber]

now I just have to integrate both sides?

[thorny remnant]

yeah

you should get
$\dv{x^2y}{x}=x-1$
then you know what to so

[thorny flame]

convergence

[eager timber]

wait, how did you get this?

[thorny remnant]

apply derivative rules on the LHS you will find that its the same thing
$$\frac{x^2dy}{dx} + 2xy =\dv{x^2y}{x}$$

[thorny flame]

convergence

[eager timber]

what about the right handside?

[thorny remnant]

multiply dx on both sides

[thorny remnant]

for this

[eager timber]

i totally lost this now

[eager timber]

I got it until this point.

[thorny remnant]

did you undersatnd this?

[eager timber]

not exactly sure what you are trying to do there.
now you would need to take the integral of both sides after multiplying by IF and simplify it down to y = smt + c.

[thorny remnant]

yk what ill show wim
$\frac{x^2dy}{dx} + 2xy =\dv{x^2y}{x}\
\dv{x^2y}{x}=x-1\
\dd{x^2y)=(x-1)\dd{x}\
\int \dd{x^2y)=\int (x-1)\dd{x}$

[eager timber]

$\frac{x^2dy}{dx} + 2xy =\dv{x^2y}{x}$

[thorny flame]

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

Tomi

[eager timber]

how did you get \frac{dx^2y}{dx} from x - 1?

arent you supposed to take the integral what does that have to do with derivatives

[thorny remnant]

i never got it from x-1

[eager timber]

okay, so it relates to the first 2 terms in the LHS but can you tell me why its valid? like i dont know if ive seen derivatives written up like this before

[thorny remnant]

do you kmow derivative rules?

[eager timber]

ofc

this just confuses me, maybe im just being really stupid here or cant focus enough

[thorny remnant]

do you know the derivative of the product of two functions?

[eager timber]

yeah y'x + x'y

[thorny remnant]

now let x=x^2 and y=y in that eq

[eager timber]

ah get it
x = x^2 x' = 2x
y = y y' = dy/dx

[thorny remnant]

yes

[eager timber]

so it is just the derivative of x^2 multiplied by y with respect to dx

[thorny remnant]

yes

[eager timber]

oh okay, thank you

I will continue with this problem later

.close

[topaz sinew]

Channel closed

Closed by @eager timber

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[topaz sinew]

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