help-0

[limpid turret]

Gotcha

And is y, the cooldown reduction, measured in seconds or percentage?

[next tundra]

Seconds

[limpid turret]

Okay

[prime badge]

that's what google says the formula is for yasuo

but that's from bonus

oh it's year old info

[next tundra]

The guy I'm trying to solve this for said this

❤️ — Today at 2:30 PM
yea that is what i need
The problem with this is you have to account for the asymptote or restriction at y = 4 while also account for the infinite plateau of y cd at some >= x value
my idea was a negative logistic growth equation
which is just a inverse
equation

Idk what it means

[limpid turret]

Sounds like nonsense tbh

Think they're just trying really hard to sound smart

[prime badge]

the point is it doesn't automatically go horizontal after the cap

they want to solve that

i don't know what they want to happen on the left side

it's not hard to do the zigzag with abs

[next tundra]

Ok

Thanks for the help

[prime badge]

it goes flat on both sides automatically

@next tundra

[lone heart]

@next tundra Has your question been resolved?

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[lone heart]

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[craggy hill]

Hello, I posted earlier today my question but I am not sure I understand. Basically I need to calculate Pokemon attack/damage/hp. My math was 44 - ((55/2) * 65/100) = 26.125 but apperently it is not correct. I was pointed out that I should 100-65 and that makes 44 - ((55/2) * 35/100) = 34.375. But I don't udnerstand, what if Pokemon has defense over 100, I will get negative result? Please help me formulate formula for this example. HELP

[lone heart]

@craggy hill Has your question been resolved?

[craggy hill]

Nope

[lone heart]

@craggy hill Has your question been resolved?

[lone heart]

@craggy hill Has your question been resolved?

[alpine sable]

Yes, if the pokemon has a defense over 100, it will get negative damage. But that shouldn't happen anyway, since defence is the percentage of damage that will be blocked. A pokemon should not be allowed to blocked over 100% of the damage

In this case, the formula should be quite clear. First, let's label the variables (you can use different labels for your purposes of course):
a is the attack strength of the attacking pokemon
d is the defense of the attacked pokemon
h is the current HP of the attacked pokemon

First, since the attack strength is lowered by half, we have a/2. Next, a certain percentage of this will be blocked. Here we're only interested in the percentage of the damage that isn't blocked, which is 1-d/100. Multiplying these 2, we get the actual damage dealt, which is a(1-d/100). Then, subtracting that from the current HP, we get h-a(1-d/100) as the new HP

[craggy hill]

Okay, than you kindly for your reply !

[alpine sable]

If that's all, you can .close this channel

[craggy hill]

.close

[lone heart]

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[real rune]

i have the answer already

but

i wanted to ask something about the domain

so basically

thats my work and i was wondering how the forced domain change works

like if the domain is already at -pi <= 2x <= pi

[gilded vessel]

,rotate

[real rune]

do i have to double the domain again?

[gilded vessel]

double the domain?

[real rune]

well like you know hjow normally it would be like

-pi <= x <= pi

like the x is just x

if its already given as 2x

-pi <= 2x <= pi

do i have to double it since my angle is a 2x angle

or does the 2x in the given domain already fulfill that requirement

like the forced domain change thing

[gilded vessel]

i dont understand what you are asking

theres nothing to double

having the domain for 2x works well since thats the input of the trig function

you know there is another solution for tan every multiple of pi

[real rune]

yeah sorry if its a little confusing

[gilded vessel]

but that will exceed the domain

[real rune]

yea

but

ok so for double angles

the domain must fit for the double angle right?

so like normally it would be like

-2pi <= x <= 2pi

right

[gilded vessel]

i think you are confused because the domain is usually given in terms of x

lets say the domain for this question was

0 <= x <= pi

[real rune]

ya

[gilded vessel]

this is just an inequality

you can multiply it all by two

0 <= 2x <= 2pi

then you will have another solution

its useful to have a domain for the body of the trig function

[real rune]

but if the angle im finding is a double angle

doesnt the domain have to be doubled?

like can you do tan2x = -1

in the

[gilded vessel]

double angle isnt some special angle

its just a word

[real rune]

-pi <= x <= pi

[ornate condor]

fancy~

[gilded vessel]

what are you asking

[real rune]

here

lemme send a picture of notes i took

the forces domain change thing

[gilded vessel]

i think ur teacher just confused u

they really did nothing special

they just multiplied the inequality by 2

the point of that is just to have the domain in terms of the input of the trig function

thats all

the "forces domain change" is just confusing

[real rune]

when would i multiply and when would i not

cus like

if i multiply the number of answers would change

[gilded vessel]

you just want to have the middle part be the input to the function

[gilded vessel]

no it wouldnt

its only for convenience

[real rune]

wait really?

interesting

[gilded vessel]

as long as you multiply all parts you will get the same inequality in essence

yes for example

[real rune]

mmmm

[gilded vessel]

this question

[real rune]

so the only important part would be

[gilded vessel]

lets say you multiply by 2

0 <= 4x <= 2pi

[real rune]

ya

[gilded vessel]

then lets say you get two solutions for 2x

2x = pi/4, 5pi/4

if you were to multiply both sides by 2

4x = pi/2 , 5pi/2

that first solution is still valid

but the second solution exceeds that domain

you can multiply as much as you want but the solutions stay the same

its just for convenience to have 2x in the middle

[real rune]

wait wdym the second one exceeds it

[gilded vessel]

so you know how many solutions to extract when you remove the trig function

[real rune]

5pi/4 is less than 2 pi

[gilded vessel]

5pi/2 > 2pi

[gilded vessel]

its 5pi/2

[real rune]

oh that one

mbmb

[gilded vessel]

since you multiplied both sides by 2

to get 4x

[real rune]

so in that case

when i did my answer

[gilded vessel]

you only multiply to have the domain in terms of the input

[real rune]

i got tan 2x = - 1

[gilded vessel]

its all for convenience

[real rune]

so if the domain given was say 0 <= x <= pi

i would not have to multiply

[gilded vessel]

wasnt it tan 2x = 1

[real rune]

oh yea mb

1

not -1

[gilded vessel]

theres no matter of have to

its just

[real rune]

if u want to

?

[gilded vessel]

easier to have 2x in the middle

[real rune]

mmm okay

[gilded vessel]

since thats the input

and we get the number of solutions

when we apply the inverse

[real rune]

so then when i apply inverse i get pi/4

[gilded vessel]

its just easier to see how many solutions we should get at that poi nt

[real rune]

but since its tan 2x i would have to divide by 2 to get x

[gilded vessel]

yeah

[real rune]

ah

[gilded vessel]

you could keep the inequaliyty as is

0 <= x <= pi

its just

its not as easy to tell

[real rune]

so the domain multiplication is purely based off comfort

[gilded vessel]

how many solutions you get

[real rune]

ahhhhh okay

[gilded vessel]

yeah

another method you could do

tan(2x) = 1

then 2x = pi/4+ pi*k

where k is some integer

then solve for x

x =pi/8 + pi/2 * k

oops

[real rune]

wait so i have a question then

[gilded vessel]

i pmesse dp

[real rune]

if the solution i got was pi/4

cus technically tan 1 is just root2/root 2

but isnt tan also positive in quadrant 3?

[gilded vessel]

yep

[real rune]

so would there be another solution?

[gilded vessel]

thats why a solution is every pi multiple

that would be quad 3

but its out of the domain

[real rune]

rn the only solution ihave is pi/8

[gilded vessel]

quad 3 is out of domain

[real rune]

oh yeahhh

ur right

because its 0 <= x <= pi

[gilded vessel]

anyway from here

[real rune]

i see

[gilded vessel]

we hae k =0

for x = pi/8

then k = 1

x = 5pi/8

then k = 2

x = 9pi/8

but then that third solution

exceeds domain

so you dont include that and stop there

[real rune]

wait

[real rune]

in this step what is the second pi doing

[gilded vessel]

this is for the domain u gave me

[real rune]

lke the pi*k

[gilded vessel]

0 <= x <= pi

[gilded vessel]

since

[real rune]

what do u put for pi

[gilded vessel]

solutions to tan

are every multiple of pi

k is just any integer

[real rune]

every multiple of pi?

[gilded vessel]

thats just a method u can do to not change the doman

[real rune]

oh wait nvm im dumb

mb

i see

[gilded vessel]

but its a lot easier for people to just do

0 <= x <= pi --> 0 <= 2x <= 2pi

then at this step

tan(2x) = 1

2x = pi/4, 5pi/4

adding pi again will be out of the domain

[real rune]

so the solutions would be pi/4 and 5pi/4?

[gilded vessel]

so its just for convenience to see how many solutions to extract at that step

[real rune]

for htat new domain

[gilded vessel]

for 2x

[real rune]

ye

[gilded vessel]

well its not a new domain

[real rune]

well the updated one

[gilded vessel]

but those are the solutions for 2x

[real rune]

i see

[gilded vessel]

and the domain is in terms of that

[real rune]

okay so

[gilded vessel]

yeah so you dont need to worry

about multiplying the domain

[real rune]

what happens if

[gilded vessel]

and certain times

[real rune]

i have 2 factors

[gilded vessel]

its all just for convenience

[real rune]

one thats cos3x and one thats sin2x

[real rune]

i see that makes more sense then i guess

like for the domain i just use original domain and multiply accordingly

depending on the x right

[gilded vessel]

yeah

[real rune]

ah okay

[gilded vessel]

its important to note the domain isnt actually changing for x

its like if you had x = a

multiply by 2

2x = 2a

its the same equation

[real rune]

yeah

i see

that makes a lot of sense

thanks so much B)

[gilded vessel]

np

[real rune]

okay well ill go back to doing more work B)

.close

[lone heart]

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[naive flare]

I'm not sure how to start this or how to solve it even

[gray isle]

approach it in the same way as you would when finding intersections of simpler lines

you'll have a quadratic equation in one variable

and you can use something like the quadratic formula

[naive flare]

to find the x-intercepts. But what do I do with the x-intercepts?

[gray isle]

sub back into either original equation to get the respective y coordinates and hence points

[naive flare]

sub back into the standard form?

to find the POI?

[gray isle]

wdym into the standard form

[naive flare]

like sub the x-intercepts back into standard form or what original equation you talking about?

[gray isle]

either original equation

you were given 2 equations

sub the x-coords you found into either one of them

[naive flare]

so any of the x-intercepts I found, I can put just one in into one of the original equations

[gray isle]

each x-coord you found gives you a point

subing first x-coord gets you its y-coord giving you one of the points
subbing the second x-coord gets you its y-coord giving you another point

[naive flare]

ok

thanks

.close

[lone heart]

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[crisp prism]

The resistance, R, to electricity of a cylindrical-shaped wire is given by the equation R = pL/πd^2, where p represents the resistivity of the wire’s material, L represents the length of the wire, and d represents the diameter of the wire. What happens to the resistance of the wire as the diameter approaches 0?
The resistance approaches 0.
The resistance approaches Pi.
The resistance approaches L.
The resistance approaches infinity.

[raven haven]

show work

[crisp prism]

I haven't written anything

I don't really understand what to do

[raven haven]

Can you try something first?

,w plot 1/d^2

[ocean seal]

Results provided by WolframAlpha

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Requested by DerpZ#9141

[crisp prism]

ok so

I know that when d approaches 0, it becomes undefined

[raven haven]

undefined?

or infinity

[crisp prism]

anything divided by 0 is undefined right?

[raven haven]

well

Here you don't see "the resistance is undefined"

lim x->0 1/x is undefined because the limit doesn't exist

But if you look at said plot what do you notice

[crisp prism]

the y values approach infinity?

oh

OH

[raven haven]

yep

[crisp prism]

ok thank you so much

[raven haven]

np

[crisp prism]

.close

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[quasi venture]

How do I open

[lone heart]

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[real rune]

how do i do this question?

[raven haven]

tan (2x) = sin2x/cos2x

Try to turn it into something that looks like a polynomial first

[raven haven]

Apply double angle identities here

[ocean seal]

Modus

Modus

[real rune]

ye rn i have

8cos^2x - cos/sin - sin2x/cos2x

can i cross out a sin and cos? im assuming no right

[raven haven]

Nope

Because they can be 0

[real rune]

mm ok

[raven haven]

$8(\cos x)^2 - \frac{\cos x}{\sin x} - \frac{\sin 2x}{\cos 2x}$ = 0

[real rune]

ye thats what i have rn

[ocean seal]

DerpZ

[raven haven]

Use these

[real rune]

do i really have to expand it all though?

[raven haven]

Get rid of the sin 2x and cos 2x

Well

[real rune]

theres prob a faster way right

[raven haven]

I don't know

[real rune]

cus like my immediate thought was to do that but i feel like theres a faster way to do it

[raven haven]

I mean why not do that then

[real rune]

and less tedious

[raven haven]

Well here's how I'll do it faster:

,w tan2x = 8 (cosx)^2 - 1/tanx

[ocean seal]

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Requested by DerpZ#9141

[real rune]

uh

[raven haven]

yeah lol

[real rune]

what

LMAO

[raven haven]

That's called "cheating"

Just expand it out

[real rune]

okay

[raven haven]

grr wolfie gave decimals >:(

[real rune]

yeah

k so i expanded them out

but like now we have to common denominator and everything right

there is 100% a faster way to do this

without all these steps\

[raven haven]

you just asserted it without any proof lol

[real rune]

no like

when an equation is like this

theres a faster way to do it

[raven haven]

how do you know that

did someone tell you there is

[real rune]

well i mean

its just math

[raven haven]

or is it just a gut feeling

[real rune]

its neither

its an educated guess

[raven haven]

I'm not aware of any faster way

You can check with your teachers

[real rune]

hm

well i guess i just have to multiply it out then

unfort

[minor needle]

I'm not sure you'll be able to solve for exact forms, maybe Weierstrass substitution will be required

[raven haven]

I think they wouldn't have learnt that lol

[real rune]

yeah

i hvae not

[raven haven]

well then

[real rune]

yeah no its def not correct

because it gives a huge fraction

[raven haven]

does not immediately mean it's not correct lol

[real rune]

not factorable

no like its

[raven haven]

oh well that's suspicious

can you show?

[real rune]

8sinxcos2xcosx^2 - cos2xcosx -sinxsin2x/sinxcos2x

[minor needle]

I'd do it on the paper, I'm curious

[real rune]

oh

this is what i have rn

its loading

ignore the stuff on the top\

that was from the previous question

[minor needle]

oo I have cool form

[real rune]

wdym

oh btw i dont know hwy but i alternate between theta and x

bad habit of mine

[minor needle]

I've got

equivalent form

as

[ocean seal]

Modus

[real rune]

wait where does this occur?

im blind

[minor needle]

some trig identities and that's it

and I see it works

[real rune]

what exactly is this identity

like the inside is cos4x right

wait what

[minor needle]

aa

but still not

[real rune]

hm

[minor needle]

it's not cos(4x)

[real rune]

oh wait its not squareed

square

squared

mb

[minor needle]

cos(4x) would be 1 - 2sin^(2x)

[real rune]

ye

i didnt see there was no square mbmb

[minor needle]

np, now

lemme write my work out

and from here it's easy to solve

[real rune]

wait why do u multiply by sin x on both sides

[minor needle]

I can do it since cot(x) provides that sin(x) isn't 0

[real rune]

wait what

[minor needle]

and main problem is sin(x) in the denominator, we want to have no fractions there

so this is why

[real rune]

also i got the answer after

bt im still conbfused in your process

the answer is pi/24 and 5pi/24

but

like going from

[minor needle]

and pi/2 also

[real rune]

step 1 to 2 after the multiply both sides by sinx

no pi/2 wouldnt be one

oh i forgot

to mention the range

i mean domain

its 0 <= x <= pi/4

but from ur work idk how u go from step 1 to 2 after the sinx multiplicatin

[real rune]

oh wait the fk?

i wrote pi/4 on my sheet

MB

MB

[minor needle]

ye

[real rune]

LMAO

but anyways

[minor needle]

I've just distributed terms in the bracket

[real rune]

where does the -sinx come from

oh wait

did u use the

identity

[minor needle]

I've multiplied by sin(x) and that's it

[real rune]

wait no

[minor needle]

• was there

[real rune]

wait what

OHHH

[minor needle]

in the last few lines I've used sin(2x) twice

[real rune]

wait but then

how does this portion work

like when u multiply

cos2x into 8cos x

and 1/sinx

how do u get cos2x + cos 2x

[minor needle]

sin(x) * 1/sin(x) = 1

so from sin(x) * cos(2x) * 1/sin(x) you have cos(2x)

and 8 * sin(x) * cos(x) * cos(2x)

I've used sin(2x) = 2sin(x)cos(x)

and fact that 8 = 4 * 2

[real rune]

also

when u multiply sinx into 8cosx

how does that become 8sinx

by itself

idk this work blows my mind so i dont really know what im looking for tbh

[minor needle]

Multiplication is commutative.

[real rune]

more of a brain gap tbh

im not smart enough to comprehend it all

[minor needle]

then if you have sin(x) * 8cos(x)

u can rewrite this as 8 * sin(x) * cos(x)

it's same thing

like a * b * c = b * a * c

[real rune]

oh

i see

thats where the other cos x came from

ok

so now

how does the cos2x * 8cosx - 1/sinx work again

u did like

cos2x * 4cosx * 2 cosx right

or something

[minor needle]

? We've just discussed it, haven't we

[minor needle]

ye

look at this

[ocean seal]

Modus

[real rune]

ohhhhh

i see

just gonna send this again so i dont have to scroll up

one last thing

how does cos2x * 1/sinx work again?

o-o

i understand EVERYTHIGN else tho

B)

[minor needle]

it's sin(x) * cos(2x) * 1/sin(x)

not cos(2x) * 1/sin(x)

[real rune]

oh

so then its just 1 x cos2x

[minor needle]

ye

[real rune]

OH MY GAH

ok so uh

i guess that odes it for that question

just quick question how much time do u have

cus i dont wanna keep u here too long if ur busy

nvm i think im good

thanks so much for ur time and patience with me B)

[minor needle]

it's fine

[real rune]

.close

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[lyric fable]

is this correct?

a simple yes or no would do

[alpine sable]

Yes

[raven rover]

Yes

Good to go 👍

[median oar]

i wonder if the 3rd side being 4 counts as a triangle

[alpine sable]

Do this for me please

[median oar]

it will still have 3 sides and have all the angle properties of a triangle

[waxen osprey]

Consider the triangle inequality

[lyric fable]

thanks guys

[median oar]

yeah but is a triangle that's stretched to a line not a triangle

$z\leq x+y$

[ocean seal]

Frosst

[median oar]

wikipedia writes this

[sage jolt]

help me describe the solution

[waxen osprey]

z=x+y is not a triangle it is a line

[median oar]

a degenerate case for the triangle is still a triangle is it not?

[median oar]

no a line only has 1 line

[waxen osprey]

if is a line

it*

[median oar]

the degenerate case of a triangle still does have 3 sides

they just happen to lie on the longest side of the triangle as well

[waxen osprey]

well can you have a triangle with no area

[median oar]

well yeah why not

any 3 points in 2D space can make a triangle

even if 2 or 3 of the points are on top of each other

[waxen osprey]

In Euclidean geometry, any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane

[median oar]

or are collinear

[waxen osprey]

from wiki

[median oar]

that doesn't say it's not a triangle

that just says it's not a unique triangle and does not simultaneously determine a unique plane

[waxen osprey]

I see your point but it wouldnt get the person asking the marks

its just to do with definition

[median oar]

yeah i wasn't telling him to say 4

i was just curious to see if a degenerate triangle is a triangle

and it seems so

[lone heart]

@lyric fable Has your question been resolved?

[waxen osprey]

Wait does that mean a triangle is a degenerate quadrilateral ?

if you let one of the points be on the side of the triangle?

[lone heart]

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[nimble estuary]

Angle A is trisected. Find the length of the shortest side in triangle ABC. I’m really stuck on this, stuff in this unit is generally law of cosines but I can’t seem to figure out if that’s possible.

[pliant cedar]

,rotate

[pliant cedar]

maybe the law of sines can help?

[nimble estuary]

wouldn't I need at least an angle?

what about angle bisector theorem multiple times?

[lone heart]

@nimble estuary Has your question been resolved?

[nimble estuary]

<@&286206848099549185>

[alpine sable]

Sup

[nimble estuary]

hello

[alpine sable]

Tried triangular inequality?

[nimble estuary]

how would that work with only one side?

[tepid jungle]

Ya got a ratio

[nimble estuary]

i did bisector angle then attempted stewarts

[alpine sable]

Take AB as x

Lemme try

[nimble estuary]

I put point D in between 2 and 3 and point E between 3 and 6 btw

I think if I do stewarts and make a system it could work but idk how i would make the system

[tepid jungle]

This is the idea you want

Oh you’re wanting the short side, but still same idea

Oh my math bad, gimme sec

[nimble estuary]

what i l1 here?

wouldnt law of sines be sin(angle1)/a

[pliant oak]

Set B-D-E-C colinear
AD is the line bisects ∠BAE
AE is the line bisects ∠DAC
Then
AB:AE=BD:DE=2:3
AD:AC=DE:EC=1:2
Set AB=2x,AE=3x,AD=y,EC=2y
∠BAD=∠DAE=∠EAC=z
Then by law of sine
2/sinz=y/(sin∠B)
5/sin(2z)=3x/(sin∠B)
11/sin(3z)=2y/(sin∠B)
3/sinz=3x/sin(∠B+z)
4 equations,4unkowns
But it is too complicated to solve

[nimble estuary]

so just no answer?...

[pliant oak]

Maybe use law of cosine in ∠B or z can list equations more easily to solve

Solve these equations to reach the answer but it‘s hard lol😩

[carmine reef]

Ah I've seen this problem before

Exact same one with the exact same side lengths

Twice before

[nimble estuary]

really

[carmine reef]

Kinda forget tho

[nimble estuary]

😭

[carmine reef]

Once on a high school math club comp and once in an aops book

You have to set one side length to x I think

If you choose AB then AE is 3x/2

I suppose you could say AD is y and AB is 2y and try stuarts

Maybe Stuarts with triangle ABC and both the AD and AE cevians

iirc the algebra somehow is actually ok

[nimble estuary]

how would you do stewart's with both cevians?

[carmine reef]

One at a time

[nimble estuary]

ah

[carmine reef]

system of equations 😭

[nimble estuary]

using AB and AC as the outer lines or AB and AE to find AD

[carmine reef]

but its ok I think

[carmine reef]

Second one is a special case: angle bisector

If you know that one you can do it that way instead

But in essence we are finding every single cevian of every single triangle and performing Stuarts on it lol

[nimble estuary]

i didnt even know stewarts til about 30 minutes ago lol

[carmine reef]

It be like that

a man and his dad put a bomb in the sink

man + dad = bmb + cnc

lol

[nimble estuary]

so for my first case here

i would add DE and EC?

[carmine reef]

Are you doing Stuarts on CDE?

Or wait no lol

ACD

[nimble estuary]

I'm doing it on cevian AD

i thought u said use triangle ABC 💀

[carmine reef]

oh

No that's right ok

[nimble estuary]

i get like

(2x^2+48y^2)/11-16

[carmine reef]

Yes CD = CE + ED

[nimble estuary]

good lord theres no way im doing this right

[carmine reef]

m = 2, n = 9, right

a = 11

[nimble estuary]

i did math wrong

[carmine reef]

b = x, c = 2y

[nimble estuary]

2x^2+36y^2

yep

[carmine reef]

I think unless I've swapped m and n

[nimble estuary]

BD=m=2

DC=n=9

i got (2x^2+36y^2)/11-16

=y^2

and (2x^2+36y^2)/11-30=(9x^2/4)

this seems really bad

[carmine reef]

198 + 11y^2 = 2x^2 + 9y^2 is what I get

,w Stuarts theorem

[nimble estuary]

bruh

[carmine reef]

Oh c and b are swapped

Or m and n are

[nimble estuary]

ohhh

b and c are

waitno

[carmine reef]

Well making either swap makes it work

symmetry

[nimble estuary]

ohhh

i see

[carmine reef]

b = 2y, c = x

man + dad = bmb + cnc

198 + 11y^2 = 8y^2 + 9x^2

I think

[nimble estuary]

whats d in this case

[carmine reef]

If you want you can even treat this as a linear equation in x^2 and y^2 and then solve for x^2 and y^2 once you do the other one

[carmine reef]

Length of the cevian, I called it y

[nimble estuary]

the cevian?

ah

yeah i got what you got

[carmine reef]

nice

[nimble estuary]

330+(99x^2/4)=20y^2+6x^2?

[carmine reef]

That's what I get too

[nimble estuary]

great now

system...

uh

do i multiply smth

do cancel out a variable

[carmine reef]

Probably add the systems together to cancel out the ys

[nimble estuary]

yeesh

that math is nasty

multiply by 4 and 2

[carmine reef]

Well first you cancel like terms

Or you can do substitution if you want

I recommend going about it in a way that you'll solve for x

Since the shortest side is definitely either x or 11

[nimble estuary]

ok

198=-3y^2+9x^2

and 330=20y^2-(75/4x^2)

[carmine reef]

Divide all terms of 1st equation by 3

66 = -y^2 + 3x^2

[nimble estuary]

you just saved me a lot of time

[carmine reef]

:D

[nimble estuary]

1650=165/4x^2

x^2=40

[carmine reef]

330 = 60x^2 - 1320 - 75x^2/4

1650 = 165x^2/4

so yr

ye

[nimble estuary]

i got an imaginary num for y lmao

[carmine reef]

y^2 = 3x^2-66

120-66

[nimble estuary]

oh

lmao i did it weird

[carmine reef]

hopefully y = sqrt(54)

[nimble estuary]

yup

oml that took me an hour

i have a problem solving test tmr

and we get 100 minutes

for 4 questions like this

[carmine reef]

Oof

Competition math?

[nimble estuary]

nope

just a plan ol

honors precalc bc

final

theres no set curriculum cause its not AP

so he just does whatever he wants

[carmine reef]

oof

gl

[nimble estuary]

my math teacher is lowk psycho lmao

giving us induction proof and stuff within a week of starting

no idea how people like math

no offense lol

[carmine reef]

lol dw

that's rough

Proof writing is a pain to get into for sure

[nimble estuary]

i like proofs but not computing

[carmine reef]

oh yeah thats pain too

For a problem like this it's rough because unless I get into the computational mess to some degree it's not clear that the method I'm thinking of will work

[carmine reef]

Once I see something like this I would know it works anyways

[nimble estuary]

dyk geometric series

91=7(1-r^6)/1-r

[carmine reef]

That seems like a degree 6 polynomial equation

Geometric series or not

I severely doubt it has a rational solution

You can factor a 7 out idk what else to do

[nimble estuary]

is there a diff way to go abt it
given
a+ar=7
a+ar+...+ar^5=91

thats the original problem

and the question asks to find the sum of the first 4 terms

[carmine reef]

yeah

Original equation isn't right I don't think

Then

[nimble estuary]

yeah its not

[carmine reef]

(a+ar)(1+r^2+r^4)

[nimble estuary]

i got r^4+r^2+1=13

[carmine reef]

(idk how to call it but this is a factorization you'll sometimes run into, it works for geometric series of any composite length)

[carmine reef]

Ye

[nimble estuary]

now what tho

[carmine reef]

There's a useful substitution

You see these a lot for solving high degree polynomials that would otherwise seem nearly impossible

[nimble estuary]

just substitute x for r^2?

[carmine reef]

ye

exactly

[carmine reef]

Example: is 1111111111 prime

[nimble estuary]

x=-3,4

[scenic crescent]

pls help w 2 i've done 1

[nimble estuary]

.

[carmine reef]

#❓how-to-get-help

[nimble estuary]

is it even possible for -4=x

wouldnt that be imaginary

[carmine reef]

Yeah

it would be imaginary

[nimble estuary]

so r=sqrt3?

[carmine reef]

So we have a case where a geometric series with imaginary components ends up having certain real partial sums

But presumably we're not talking about imaginary sequences so we can ignore the solution

r = sqrt(3) should work

[nimble estuary]

i got a=14sqrt3/3

thats some ugly numbers

[carmine reef]

I don't think that's right

What did you do?

[nimble estuary]

a+ar=7

substituted r

divided 7 by r

and divided that by 2

[carmine reef]

a+asqrt(3) = 7

[nimble estuary]

yes

so 2a=7/sqrt3

rationalized

[carmine reef]

no

[nimble estuary]

no?

[carmine reef]

a + asqrt(3)

What would happen if we divided that side by r

[nimble estuary]

it would end up a/sqrt3?

factor out a?

[carmine reef]

(a+b)/c is not a + (b/c)

[nimble estuary]

a=7/1+sqrt3

?

[carmine reef]

with parentheses yes

[nimble estuary]

yeah

[carmine reef]

could rationalize if u want not too important

[nimble estuary]

-(7-7sqrt3)/2

even worse than before

[carmine reef]

yeah

hopefully this one should work tho

[nimble estuary]

thats weird

i got 112+112sqrt3

wait no

its 28

[carmine reef]

for what?

first 6 terms sum?

[nimble estuary]

4 terms sum

i actually coulda just

found r

then subtracted a+ar from 6 term series

then divide by r^2

to get sum

[carmine reef]

(a+ar)(1+r^2) too yeah

7*4

[lone heart]

@nimble estuary Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[spare fern]

hey guys

[spare fern]

could someone help me out with this question

8m/ square root of m

[vague mirage]

$8m^{1-1/2}$

[ocean seal]

Arnab Pal

[spare fern]

its a fractional powers question and I'm not sure how to write what it is equivalent to

[vague mirage]

It is

[vague mirage]

Yeah

$8m^{1-1/2}$

[ocean seal]

Arnab Pal

[vague mirage]

What is 1-1/2?

[spare fern]

1/2

[vague mirage]

Yeah so 8√m

Is the answer

Cause ^1/2=√

You know what I mean

Laws of exponents

[spare fern]

oh I see

thanks

[vague mirage]

Yes

Wlcm

[spare fern]

are you really taught this at 14?

.close

[lone heart]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[alpine sable]

do you know how to calculate 8 - (-15) ?

[vague mirage]

It is that easy

Bope

[alpine sable]

no

[vague mirage]

Nope*

8-(-15)

What's -1×-1

$-1×-1$

[alpine sable]

because subtraction is not commutative

[ocean seal]

Arnab Pal

[alpine sable]

order matters

[vague mirage]

Yupp

[alpine sable]

or at least you would have to write it as -(-15)+8

[vague mirage]

Nooo

It wasn't right

What is 8-(-15)

Actually tell me what's 8+15

Yeah that's it it's the range

Yess

[alpine sable]

also it's important to note that range cannot be a negative number

because you are subtracting the lowest number from the highest number right

therefore it's always a positive number

both B and D are negative numbers

therefore you can immediately know that they are wrong

[lone heart]

@dusty spire Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
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[warped berry]

Hello guys, I have a question regarding to potence series and the radius of convergence. Maybe somebody can help me out with the following exercise where you have to determine the radius of convergence

[warped berry]

I'm just unsure how to solve this problem. I know that this is the harmonic series

[rose sigil]

it's def at most 1 bc it diverges when x=1

i haven't done anything yet but i'd guess it converges when 0 < x < 1

maybe you can compare it to smth

like a geometric series

[keen plinth]

ig ratio test?

[rose sigil]

oh yea i think that'll do it for when 0 < x < 1

[rose sigil]

,rotate

[lone heart]

@warped berry Has your question been resolved?

[lone heart]

Channel closed

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[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[halcyon delta]

Hey, thanks in advance for the time people spend on solving this problem, I really appreciate it!

The gps system requires 4 satellites to determine the position of a point, however, a satellite must orbit the planet and the planet must not be between the satellite and the point in order to be able to reach the planet. The question is: given a radius of the planet and the radius of circle the satellites are orbitting in, how many satellites are needed such that every point on the planet can be seen by at least 4 satellites at all times?

I’ve solved the problem already in 2 dimensions, where 3 satellites are needed instead of 4; see picture

[limpid turret]

assuming general relativity is not considered

[halcyon delta]

yeah

[green folio]

The legendary help 0

[limpid turret]

"a satellite must orbit the planet and the planet must not be between the satellite and the point in order to be able to reach the planet"

[halcyon delta]

I’ve called the distance where a satellite can be in ‘o’. and the radius of the satellite, r2, I’ve set to 1. o = pi - sin^-1(r1/1)

[limpid turret]

What does this mean

[halcyon delta]

https://cdn.discordapp.com/attachments/903430334681608232/1055159057998155937/image.png

[limpid turret]

gotcha

So for any point in 3d, you need 4 satellites in that valid range, yeah?

[halcyon delta]

yes

and in 3D, the part where the satellite must be in to see a certain point is a cut sphere

[limpid turret]

and I'm guessing the number of satellites is variable based on the two radii

[halcyon delta]

yeah

[limpid turret]

Interesting problem

[halcyon delta]

wel technically the ratio between the two radii

I agree

[limpid turret]

I think it would depend on ratio

Ratio of the radii should create an invariant

[halcyon delta]

err yeah ratio*

[tribal haven]

this does not feel at all easy and my brain keeps going to infinitely far away satellites in which case you only need 8

but i guess you can try to cover the sphere using several circular patches where you stack 4 satellites on the same point at a time

[limpid turret]

My intuition is thinking:

Total solid angle of a sphere is 4pi

Figure out how much solid angle a single satellite covers

Then you need 4 satellites in that solid angle

[halcyon delta]

by the way here is some clarification on the 2D version of the problem

[limpid turret]

It's 3am here so I cannot offer much more discussion

But I will probably ruminate on this in my sleep

Feels free to DM me as I find this problem interesting.

[halcyon delta]

thanks! you should go to bed xD

[green folio]

Tf?

[halcyon delta]

will do

[alpine sable]

what does

ruminate mean

[halcyon delta]

you should ask your question in one of the listed available help channels

[halcyon delta]

I assumed it means thinking about it, I’m not a native enghartig speaker

[alpine sable]

which chanel should i use?

[green folio]

Yep

To think extensively abt it

[halcyon delta]

#help-15

[limpid turret]

lit. definition: to think deeply about something

[keen plinth]

some kind of sphere covering problem

getting upper and lower bound is easy

but theyre pretty far apart

[halcyon delta]

the radius of the circle that the satellite have to be in is pi-2*sin^-1(r1/1) if you take the distance as having to walk over the big sphere and r2=1, is that what you mean?

[halcyon delta]

yeah

[keen plinth]

i mean you can work out a rough lower bound and upper bound on how many satellites you'd need

but they can be quite far apart meaning theyre basically useless

[halcyon delta]

true, what makes it difficult is the many ways you can arrange the orbits, even if you assume the orbits are in the shape of a circle

[lone heart]

@halcyon delta Has your question been resolved?

[lone heart]

@halcyon delta Has your question been resolved?

[lone heart]

@halcyon delta Has your question been resolved?

[lone heart]

@halcyon delta Has your question been resolved?

[opal sphinx]

H

[lone heart]

@halcyon delta Has your question been resolved?

[lunar bluff]

Value x equation 4(6x-9.5)=46

[halcyon delta]

please ask your question in an available help channek

if you do, im willing to help you with it. Feel free to ping me

[halcyon delta]

<@&286206848099549185> any chance you could have a look at this problem?

[last ether]

Why not make a square around the earth, place four satellites on the corners, then rotate that square 45° and then place 4 more

Oh wait nvm that won't work

Big brain moment

[raven haven]

,w gps satellite coverage

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by DerpZ#9141

[raven haven]

welp

[lone heart]

@halcyon delta Has your question been resolved?

[left crypt]

@halcyon delta are you still looking for some help here?

[bold gate]

i need help

badly

i feel so stupid

i dont belong here

i suck at math

[left crypt]

@bold gate open a help channel.

[bold gate]

idk why i try anymore i just wantt o drop out of life

how do i do that

[left crypt]

Follow the instructions in #❓how-to-get-help

Don't hijack open help threads.

[bold gate]

what does hjiack mean

im new here

[left crypt]

maybe read the instructions. Like #rules

[bold gate]

thats the question i need help with

[left crypt]

Stop posting here.

[bold gate]

i dont get it

[left crypt]

This is someone else's help thread.

[bold gate]

ok how do i delet eit

[left crypt]

Follow the instructions in #rules and open your own channel.

Anyway @halcyon delta :

I have some pointers to help you frame the problem a bit better if you are still looking for help

-you don't really need to consider the orbits of the satellites for the coverage problem; a static arrangement is sufficient.

-any particular satellite cannot see past the great circle arc which forms the plane orthagonal to the line between the satellite and the center of the globe (i.e. coverage is a hemisphere assuming sufficiently wide field of view, precision, and distance from the earth)

-it's relatively simple geometrically to construct a pattern which clearly provides quadruple coverage to the whole planet; the tricky part is showing when the removal of any additional satellite makes complete surface coverage impossible--ideally, you would like to construct a minimal coverage pattern that has exactly 4x coverage along each point except possibly some great circle arcs, such that clear removal puts you below the lower bound of coverage (namely: total coverage < 4x surface area of the earth, or in other words an arrangement that does it with exactly 8 satellites.

The problem becomes more complex if you limit ranges and fields of view, but if you construct the mathematical relationships for the minimal cover it should become apparent how to start attacking the more restrictive variant...

[halcyon delta]

thanks for your message! Are you sure I don’t need to consider the orbits? since the middle point of all orbits have to be the middle point of the planet, I don’t see why you don’t have to consider them

[left crypt]

so, if you read through, I was trying to lead you through a process to illuminate the "more restrictive problem".

if you think about the symmetries of a sphere and an orbit, the same notions that take you from an unrestricted field of view to a restricted one are the same notions that solve the orbits problem. In the words of Polya: solve a simpler problem first.

(Think in radians about how a narrowed field of view changes the math on the static problem and then consider the symmetries of an orbit and a sphere and how the angular sweeps of those symmetries interact)

[lavish oyster]

,w a_1+4a_2=6, a_2+4b_1=-3, a_1+b_1=1

[ocean seal]

Results provided by WolframAlpha

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

Requested by ProZeLL#9557

[lone heart]

@halcyon delta Has your question been resolved?

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[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[strong flame]

Hey

[strong flame]

can i get some help?

how do i calculate the maximum and minimum value of this function:

5sinx - 12cosx

my idea was to make:

5*sqrt(1 - cosx^2) - 12cosx

so its only cos

but what now?

[worn fox]

Perhaps a more helpful method might be to write it in the form Rsin(x+α)

[long axle]

Bruh

[worn fox]

Please put a disclaimer if you're going to use chatgpt

[strong flame]

is it possible without derivatives?

[worn fox]

Yes by doing this

You know the max an min values of just a sin function

[strong flame]

-1; 1

[worn fox]

So you need to find R

Wow so helpful

[strong flame]

what do u mean by that

how do i find 'R'

[worn fox]

Expand out Rsin(x+α), and set equal to your original expression

Using angle sum formula^ that'll let you find R

This stuff is called "harmonic form"

Okay you should probably go elsewhere

<@&268886789983436800> 🤷‍♂️

[strong flame]

i know the sum formula