help-43

[compact pewter]

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[peak coral]

hello, is this correct?

[odd bolt]

i over 2 in the end, ill recheck

i believe there might be an extra 1/2 in the third step

[peak coral]

but its there to neutralize the inner function -x+12 derivate *(-1) no?

wait never mind!

[odd bolt]

hi! the-

[peak coral]

-1 isnt -1/2

[odd bolt]

oh well you got it right

[peak coral]

the (-) just disappears but thats it

thanks! 😄

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[warped radish]

can somebody explain how to do the translations from the equation to the graph

[compact pewter]

@warped radish Has your question been resolved?

[compact pewter]

@warped radish Has your question been resolved?

[molten coral]

how do you mean?

im confused what the problem is

or is this a solution?

or is the remaining part of the problem just "graph the equation"

[warped radish]

ya its asking me to graph it but i genuinely dony know

[molten coral]

can we assume that you have the asymptote and the period?

for the purposes of graphing

you either are given it or you are able to get that

[warped radish]

ok wait im on a new one rn

i had to solve for the period and asymptote and now its asking me to graph it again on the second ss

[molten coral]

so graphing with this information is pretty straightforward

because theres an asymptote every period

and the graph of tan looks the exact same between each of those asymptotes

you can kind of see this in the graph you posted

so with one asymptote, and the period, you have all the information youneed, since the graph just repeats over and over

[warped radish]

wait im rly lost

what do i do w the equation

[molten coral]

you dont need the equation

the asymptote and period describe the graph

[warped radish]

https://tenor.com/view/long-tears-gif-11087877242190144149

ok wait...........

[warped radish]

can u show me how to find the vertical stretch pls

[neat fossil]

https://tenor.com/view/monk-monkey-monkey-shocked-gif-8841291472583398259

[warped radish]

same

[molten coral]

vertical stretch?

[warped radish]

yes

i think its called compression too idk

[eternal pulsar]

That would be horizontal

which involves the period

[warped radish]

Stop

I GET IT NOW

no I dont

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[quartz yoke]

determine a condition on |x-1| that will assure that |x²-1|<1/2

Can someone help me with this?

[kind viper]

factorize x^2 - 1

[quartz yoke]

yes

[woeful schooner]

What did you get?

[quartz yoke]

|x+1||x-1|<1/2

and also that |x+1||x-1|<=3|x-1|

but what after this

[kind viper]

set up 3|x-1| < 1/2

[quartz yoke]

so the condition is |x-1|< 1/6?

[kind viper]

it's one possible condition yeah

[quartz yoke]

alright, thank you!

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[delicate musk]

Can anyone help me solve this

[delicate musk]

So far my thinking is something like

order of g | order of G

and order of gH | order of G/H

I think i need to prove order of G/H | order of g but im not sure

[kind viper]

let n = ord(g)

then g^n = 1_G

what can you say about (gH)^n

[delicate musk]

whats 1_G

oh identity in G

[kind viper]

yes

[delicate musk]

I think its identity because $(gH)^n \cong (g^nH)=eH=H$ which is the identity in the factor group

is that the proper way to prove it tho?

[boreal girder]

BOSS

[kind viper]

i mean why \cong when you can write =

but yes

(gH)^n = 1_{G/H}

[delicate musk]

did a lot of iso stuff lmao

[kind viper]

so what's that tell you about the order of gH

[delicate musk]

Ok that makes sense

[delicate musk]

its a multiple of or devides n

how do we know its smaller

[kind viper]

it cannot possibly be bigger than n

[compact mortar]

I just wanted to add my too cents, with another method on how to prove this. Without giving too much away. I would first note that left cosets of H partition G. Then you can say that all left cosets have the same size. So ord(gH) | ord(G). Then you should be able to go from there

[delicate musk]

I was originally thinking of ord(gH) | ord(G) = [G:H] | G

but idk if this would go anywhere in terms of a proof

[delicate musk]

I do have to understand/explain why

for the proof to be valid

[kind viper]

given any group G and element x therein, if you know x^n = 1 then ord(x) divides n

[delicate musk]

yeah that makes sense

oh

yeah

my question was more

how do we know order g is not a multiple of order gH

not the other way around

yk

[compact pewter]

@delicate musk Has your question been resolved?

[delicate musk]

which n is this

order of the group?

In the question its the same g as in front of the coset

[kind viper]

n is just n

[compact pewter]

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[granite pendant]

Do I solve the integral to find g(x) somehow?

Even then I have no clue how to solve that 😭

Also how is f(x) even a function

It's not in terms of x

[mild pivot]

you can express n as a function of x, so f(x) can also be expressed in terms of x

[granite pendant]

How is n a function of x?

Only the limits are given..

Oh right n = f(x)²

But that doesn't tell us f(x)

[mild pivot]

you have 1/(n+1) <= x < 1/n.
that is the same as n-1 < 1/x - 1 <= n.
so n = ceil(1/x - 1)

[granite pendant]

Wait how did you get the last step?

Oh right the definition for greatest integer

How do you find the inv of ceil?

To get f(x)

[mild pivot]

you don't need to

f(x) = sqrt(n)

[granite pendant]

Oh..

Ok so I have f(x) in terms of x.. what about g(x)?

[mild pivot]

you have an upper bound you can use at least, I haven't found a solution to the integral so far

[granite pendant]

Getting it till here... how do you solve that lim?

[young raft]

,rccw

[granite pendant]

But what about the other side?

Ok it's 2 I think

If it were a sandwich thm q the ans should be 2 but how do I simplify the left side?

And the answer is 2 but that's beside the point

Small error in the right side, but it's still turning out to be 2

Can I use l'hospital here?

I mean it's in the form of 0×infinity.. I don't think I can do that

[mild pivot]

0 x inf is same format as 0 / 0

[granite pendant]

But to get it into that form I'd be taking the reciprocal of a ceil function.. is that allowed?

That's within a square root

[mild pivot]

sqrt(ceil(1/x - 1)) < sqrt(1/x)

[granite pendant]

Can I do this

[mild pivot]

I don't think so

[granite pendant]

Yes.. so do I substitute sqrt(ceil(1/x)) instead of f(x)?

Is the derivative of f(x) always 0?

[mild pivot]

at some points it is undefined

[granite pendant]

I tried using l'hospital but I'm so confused now

[mild pivot]

Im getting that the answer is strictly smaller than 2 for all x not equal to 0:
f(x)g(x) = sqrt(ceil(1/x - 1)) * g(x) < sqrt(1/x) * g(x) < sqrt(1/x) * 2 * sqrt(x) = 2.

[granite pendant]

But the sandwich theorem would still work if the left side were also 2, wouldn't it?

[mild pivot]

if the left side is 2, then we get 2 < 2, which is a contradiction so no such functions exist.
if the left side approaches 2 from below, then the solution of the limit is 2

[granite pendant]

Ok I'll try using that then

I don't know where I'm going with that

Literally just wrote it out

To get it to tend to 2 from the lower side wouldn't we need one more inequality on the left side?

[compact pewter]

@granite pendant Has your question been resolved?

[granite pendant]

I converted it into a variable tending to infinity, but what now

Sobbing I've spent 2 hours on this question I'll try it another day

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[fleet frost]

Evaluate the integral and then try to put in limx tends to 0

Mb you can do it without evaluating the integral as well, I think it should turn out to be 1

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[quartz yoke]

Can you diverge against convergence

[quartz yoke]

Can you converge against divergence

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[severe pelican]

I just had a question about ranges, is it essentially just rearranging the function until you find an instance where there might be a restriction on y?

[severe pelican]

I'm re-learning calculus and I did not expect ranges to be as hard as I'm finding them now lol

[slate sand]

Yes

[severe pelican]

alrighty, thank you!

[slate sand]

Range isn’t that awfully useful for arbitrary functions in general

It becomes very important when you encounter linear algebra but for general functions, eh

¯_(ツ)_/¯

[severe pelican]

yeah I remember barely using them, so my brain probably erased any knowledge it had regarding ranges

still part of the curriculum though

[slate sand]

Yeah but I’m just saying it’s good to know how to do it but it’s not terribly useful for general functions

It’s sorta just a: hey find this for this particular function

Which boils down to knowing you can’t divide by 0

You can’t log non-positive numbers

You can’t square root negative numbers

And that’s pretty much it

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[shy nacelle]

Is there formal language for what the n and N are called?

Like in the quadratic formula a, b, c are called coefficients

[kind viper]

favorable outcomes and total outcomes i guess?

not really all that formal

but if there's a pair of names for them those are the names

[shy nacelle]

No but that is very helpful for me uhh

I guess trying to structure how to approach these problems

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[knotty basin]

i read that the mean of a binomial distribution being given by np and variance by npq comes from bernoullis (i have no clue about any of that). if i had to conceptualise it in rudimentary terms i get that the mean is the the product of the number of trials and probability of successes to get the expected value somewhat but i kinda just put the thing into words and i dont actually get how that comes about. and i have no clue about the variance

[knotty basin]

oh a bernoulli is just one specific trial

[storm lichen]

nerdddd

[knotty basin]

water beam my goat

oh i get the mean

what about variance tho

OH SO IN A BNOMIAL YOURE EITHER -P OR (1-P) AWAY FROM THE MEAN

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[knotty basin]

thank you all

[near sierra]

Did bro solve his own question

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[rigid thunder]

Is my answer for option c is correct?

Like they are neither symmetric nor skew symmetric right ?

[kind viper]

yes your answer is correct

[rigid thunder]

Thanks 🫂

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[sand dune]

stuck on (1) i want to prove that the opp of the quad =180 by using 'let one of the angles=x ',but i am not sure which angle to use. PLEASE HELP MEEEE

[digital delta]

Okay so I think theres too many colors going on in this picture

this can be done just by angle chasing alone

taking <PAB and <BRP its sufficient to show their sum is 180

[sand dune]

what do you mean?

[sand dune]

yeahh im trying to do but can't figure it out how

[digital delta]

I mean that its enough to just figure out some equal angles

[digital delta]

okay ill give you a hint

notice that <PSA = green angle, do you see why?

[sand dune]

yh tan chord

let q= x??

[digital delta]

could you elaborate?

[sand dune]

since CBR is a tangent, then we can use the tan chord theorem so Q=R bc the angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment

[digital delta]

okay yea, but that doesnt explain why <PSA is green does it?

[sand dune]

wait, which green is it equal to the Q or R??

[digital delta]

well from here <CRP = <RQP so they are the same

[sand dune]

so <PSA is also equal to both of the Green <s right and there's a reason for that

wait me see if i can figure that out

[digital delta]

sure

[sand dune]

OHHHHH S1=Q /ext <

[sand dune]

so am right by saying let Q=x?

[digital delta]

hm Im not sure what you mean by this

[sand dune]

It's a method my teacher taught me for proving angles. For example, if angle A is equal to 180 - x and angle R is equal to x, we can see a connection between A and R, as both add up to 180 degrees. not sure if this makes sense to you. but if i let Q= x then R will be x as well, which means i gotta find a way to get A to be 180-x.

[digital delta]

oh yeah I see what you mean

hm its hard I think this way, because <PAB doesnt equal anything nice itself

[sand dune]

do you have a better way of doing this?that would be very helpful ??

[digital delta]

well we can let <R_2 (Green) equal x

let <A_1 = y

and <A_2 = z

we want x+y+z = 180 right?

[sand dune]

um yeah

[digital delta]

ok what is angle y (A_1) equal to

any other angle?

[sand dune]

P1?

[digital delta]

yes exactly

okay we are pretty much done

look all the angles x,y,z are angles of triangle PAS

[sand dune]

oh thats all really

[sand dune]

oh

[sand dune]

thank you anyway

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[dusk elm]

how does this problem work

[dusky nymph]

do you know how to compute the average cost?

[dusk elm]

no i do not

[dusky nymph]

probably $$\overline{C}(x) = \frac{1}{x}\int_0^x C(y),dy$$

[boreal girder]

Bungo

[dusky nymph]

does that look familiar?

[dusk elm]

no it should not be a integral

[dusky nymph]

ok

what should it be then?

[dusk elm]

derivative i believe

[dusky nymph]

to compute the average cost?

derivative of what?

[dusk elm]

that im not sure maybe the equation?

ah nvm its quotient rule

[dusky nymph]

well the first step should to be to find out the definition of average cost, you can't really proceed without that

what does your text say

[dusk elm]

waht i sent

i figured it out tho

[compact pewter]

@dusk elm Has your question been resolved?

[eternal pulsar]

The derivative of $\overline{C}(x)$ which Bungo provided.

[boreal girder]

;(

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[heady osprey]

is this a fine final form?

[short silo]

What's the question?

[heady osprey]

here

using chain rule on it

[short silo]

Oh

So u want s'(t)

[heady osprey]

yeah

[short silo]

Thats wrong i believe

How did you differentiate inside the bracket?

[heady osprey]

quotient rule and pythagorean theorem shows up which is why there's a 1

[short silo]

Ah right

Then yh sure looks right actually

Sorry i didnf actually attempt it or anything

[heady osprey]

(1+cost)cost-(-sint)(1+sint)

[short silo]

Looks right then

Why do u doubt yourself

[heady osprey]

yeah i'm mainly asking if that is a simplified enough form to not get marked down on

[short silo]

I don't see how u can make it simpler

[heady osprey]

fair

alright thanks

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[delicate musk]

this is from a michel pham video, just to make sure im not loosing it $(1+\sqrt5)(a+b\sqrt5)=(a+5b)+(a+b)\sqrt5$ where did the $1+a$ come from?

[boreal girder]

BOSS

[eternal pulsar]

You’re right lol

Wait is this statement for his entire proof 💀

[delicate musk]

alr good to know

lmfao

[eternal pulsar]

Ohhh my how Micheal Penn has fallen off

[delicate musk]

https://youtu.be/BTZXfw6Z3ok?si=BPtWOB5VsRqvDDTv&t=533

included time stamp

tbh im just learning this rn so idk what hes doing wrong

wait so

how would i do it properly

[eternal pulsar]

I don’t know this

[delicate musk]

ah

rip

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[lavish moat]

pls

[lavish moat]

515 nm to m is what

Im getting mixed answers

Or isnt that the same

[rigid perch]

,, \qty{515}{\nm} = \qty{515e-9}{\m} = \qty{5.15e-7}{\m}

[boreal girder]

cloud

[rigid perch]

the only difference is in using engineering vs scientific notation

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[river shuttle]

helo

[pine osprey]

you can replace x by its value and see what happen first for a_0 and then in the second time for q

[compact pewter]

@river shuttle Has your question been resolved?

[pine osprey]

first of all if x = 0 this means that p is 0 and q is 1 and so for P(x) = 0 => P(0) = 0 and so a_0 = 0 and indeed 0 divide 0 (in the sense of the euclidian division in Z) and also that q = 1 which divides any numbers so the case x = 0 is true

now for x != 0 which means p != 0

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[winged lion]

just post it

[kind viper]

!noping

[compact pewter]

Please do not ping individual helpers unprompted.

[kind viper]

but also im not qualified for stochastic calculus most likely

[compact pewter]

@junior iris Has your question been resolved?

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[flint wing]

where do the equations for x =...... and y =......... come from?

[bronze narwhal]

from the RREF matrix in the end?

[knotty basin]

hello beam water

[flint wing]

but we have x + 4s - t = -2

if we use the rref matrix

[bronze narwhal]

add and subtract 4s and t

[flint wing]

oh

🤣

true

.solved

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[prisma karma]

Hi, I just created a simple game, but I’m wondering if it is fair enough. I feel that it is an asymmetric game, but I’m not sure how to prove it. I think there might be a way to predict which player will win based on the rules of the game. if u want, u can play the game at here https://gkhang17.github.io/A23-game/

[hushed magnet]

given that the grid is random one player will have an advantage probably

[prisma karma]

that's right

[hushed magnet]

I think there was a result in game theory that necessarily either one player can force a win or a tie

but just because such a strategy exist doesnt mean that anyone knows what it is

[long vigil]

This seems like a graph problem

[prisma karma]

I think if the grid increases in columns or rows, there is still a possibility of a strategy to beat the other player, however assuming that both players are equally skilled, there is still a possibility of bias.

[hushed magnet]

there is always a strategy to win finite games

[prisma karma]

Sometimes random map generation can result in one player side being biased, for example this map shows that there is no chance for yellow to beat blue by attacking the "home" position, so yellow can only attack blue's position directly

[glass nest]

what denascite is saying is that, given a particular board configuration, one player has a way to force the other to lose

but it's not necessarily always the first or the second player

[prisma karma]

i see

[glass nest]

put differently, who wins is predetermined by the board configuration if the players play perfectly

[slate sand]

can't one player block off all routes to their home?

reaching the other player's home isn't the only way to win though

[glass nest]

they can still lose by collision or running out of moves

[prisma karma]

that's right, You can still win if your next position is the position the other player is standing in or the other player falls into a position where they have no way to go.

[prisma karma]

there is a link that you can play on the website

[glass nest]

I think I can contrive boards where player 1 can force a win and where player 2 can force a win, so that's about all you can say

[prisma karma]

Technically, we could probably use probability theory to calculate the combination of all paths that a player is bound to win on the minimap. But as the rows and columns increase, the number of elements in the set will also increase greatly.

[glass nest]

then you need to get into constructing an optimal strategy and refining what you mean by the lines are generated "randomly"

[prisma karma]

ok, I will try it

[compact pewter]

@prisma karma Has your question been resolved?

[prisma karma]

.close

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[waxen scaffold]

Did I calculate this correctly?
17.9 - 10.496 =~7.4 = x?

[olive field]

Yes

[waxen scaffold]

Ok what about this, I put ? = 4.9 but i'm not entirely sure that right angle is a right angle

If it is how do I tell that it is without there being the red square thing

[olive field]

I think you can assume that the 7.7 side is tangent to the circle

which implies it's a right angle because otherwise you don't really have a way of doing the problem

so yes it's correct

[waxen scaffold]

Ok good, one more thing, when finding an arc length of a circle, its the outer section of the circumference that i'm finding right

might be a dum question

Like a 120 degree arc would be 1/3 the length of the total circumference of the circle

[olive field]

Yes

Assuming the 120 degree arc is formed by two radii

i.e it's a central angle

[waxen scaffold]

Yes

Ok great thanks so much man

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[tiny hamlet]

just wanted to check if this is how u solve this

[long vigil]

what's the subscript notation?

[keen granite]

(f_u=\frac{\partial f}{\partial u})

[boreal girder]

PajamaMamaLlama

[long vigil]

oh

okay cool

not sure how to work with those so uh

[tiny hamlet]

well 😭 to whoever knows

[olive field]

Yes this looks good

[tiny hamlet]

alright thanks, first time solving so wasnt sure

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[tiny hamlet]

idk what to do

[devout latch]

the gradient has a cool property when doing optimization.

[tiny hamlet]

mmmmm

[devout latch]

it points in the direction of …

[tiny hamlet]

max increase

soo it should poing torwards like Q?

[devout latch]

yep also at what angle is it with respect to the level curves

[devout latch]

if you start with Q it will be anchored at Q and with the two properties i just eluded to

i mean you can visualize it anchored at Q

[tiny hamlet]

do you mean it will start at 0 and be anchored at Q right

[tiny hamlet]

parallels maybe..

[devout latch]

if it were parallel it would mean the direction of greatest increase point in the direction where the function keeps the same values (on a level curve f(x,y) = k)

[devout latch]

for P it’s more like this

[tiny hamlet]

ohh we have to check for each case

alright gradient direction i understand

and it will be perpendicular

[devout latch]

yep the gradient is perpendicular to the level curve

[tiny hamlet]

what about fx, fy signs

[devout latch]

what’s the definition of the gradient?

[tiny hamlet]

its <fx, fy>

soo the value of the x would determine fx value?

[devout latch]

then can you put the arrows f_x and f_y on this picture

[devout latch]

the gradient is dependent on the (x,y) value

[tiny hamlet]

i cant draw on the picture but it would basically be like right arrow and down arrow for P so Fx+ and Fy- ?

[devout latch]

i like to think of it like adding an row to each point of R^2 depending on the way f(x,y) varies

[tiny hamlet]

okayy that helps a lot

so does fv depend on the angle

or well im not sure fv isnt about fx fy

[devout latch]

it depends on the sign of f_x and f_y

there is a way if i remember to compute the directional derivative with a dot product

[tiny hamlet]

it says just i+j tho, doesnt that mean of the point and not the gradient

[tiny hamlet]

delta F . u ?

[devout latch]

yeah but since v = i + j, then using gradient dot v compute the directional derivative

we don’t have numbers but analyzing the sign is still possible imo

[tiny hamlet]

i dont super get it lol..

[devout latch]

we want the rate of change of f in the direction v = i + j

or at least because there is no number to have an idea of it’s sign

here I have to admit it’s a bit ambiguous because it’s close to 0 or even 0

[tiny hamlet]

yeah that part confused me a bit

but i think i got it

[devout latch]

for the other points it should be clearer hopefully

[tiny hamlet]

alrightt thank you soo much

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[copper lion]

An object with a weight of 60 N is suspended by two lengths of rope from the ceiling. The
angles that both lengths make with the ceiling are the same. The tension in each length is
40 N. Determine the angle that the lengths of ropes make with the ceiling

[forest token]

did u draw a diagram

if u did send a pic

[potent berry]

show us the FBD!

[forest token]

atb for bitsat bro

[potent berry]

thanks

[copper lion]

i didnt know

no

[forest token]

didnt know what

[copper lion]

man i dont know physics

can you just explain it like that

[forest token]

dude 💀 we can try and guide you through the problem we cant explain whole concepts

do you know what a force diagram is?

[copper lion]

no

[kind crane]

https://en.khanacademy.org/science/up-class-11-physics/x3a9a44f124d01cf7:laws-of-motion/x3a9a44f124d01cf7:common-forces/v/free-body-diagram-with-angled-forces-worked-example

[copper lion]

okay thanks

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[frigid cape]

okay, so I'm currently going over differential equations, when solving second-order non-homogeneous differential equations and you're finding a particular integral, I'm told not to make one that has terms that are in the complementary function

[short silo]

Yws

Yes true

[frigid cape]

why does it matter that when you find a general solution in the end and add the C.F and P.I. together they need to have no similar terms

[short silo]

Because if it hss the same term

It will equate to 0

When you substitute

[frigid cape]

ooooohhhh I see

[short silo]

An example q would help you

But essentially if your solutions are Ae^mx+Be^3mx

[frigid cape]

honestly I can't find one in my a-level topical questions anywhere 🥲 otherwise I'd give it a try, I just know it's something that can show up on a paper

[short silo]

And ur rhs bit looks the exact same thing

When you substitute the complimentary function you will get 0

So clearly that is not the particular integral

[frigid cape]

okay, that makes sense. thank you. why does multiplying by, for example, x allow it to work?

[short silo]

Well it just works tbh, there was a formal proof on how u can derive the solutions

Tlmaths shows it if you want

[frigid cape]

oh as if lmfao, they actually have all the further maths stuff which is what I needed anyways

that's useful

thanks for that

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[twin snow]

I'm stuck at integrating $\int_{-3}^0(\sqrt{9-x^2})dx$. If the practice test allowed me to use the calculator... but it doesn't.

[twin snow]

Idk if I need to use something from those first questions

[boreal girder]

WAHAMBA LA GAMBA

[pseudo rapids]

x = 3sin(u)

[twin snow]

dude... as someone who is taking AP calc, idk if i have to know trig substitution

[pseudo rapids]

Well I really don't see any other way :(

[twin snow]

i think i'm not expected to do trig substitution in ap

idk

lemme ask chat

[pseudo rapids]

Was about to say lmao

Also people lose their shit if you say chat here

[twin snow]

💀

[pseudo rapids]

so be careful 🙃

omfg

we are so dumb

[twin snow]

what

gahahahahahaha

[pseudo rapids]

9-x^2 = y^2 is just fking circle

[twin snow]

this is so dumb lmaoooo

[pseudo rapids]

well there u go I guess

[twin snow]

lmao

[pseudo rapids]

no calculator but chat is allowed

crazy take

[twin snow]

practice test anyways

[pseudo rapids]

fair

[twin snow]

welp

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[twin snow]

boop

.close

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[fluid stag]

!status

[compact pewter]

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

[fluid stag]

How did you go from (2/3)^(1/x) < (2/3)^(-1) to 1/x < -1?

||(2/3)^t is not an increasing function of t||

[slim swan]

2

[molten coral]

the goal is to just get a restriction on x?

[slim swan]

oh it’s strictly

thx

[slim swan]

yes

[molten coral]

I think you can use the shape of the function here

if you wanted

by that i mean, just solve (2/3)^(1/x) = 3/2

and then work from the properties of the function

[molten coral]

this can be solved by inspection, too

[eternal pulsar]

After solving you can plug in, say, x=1 into the original equation and then determine which way the inequality should go

[molten coral]

well you have a couple of regions of different behavior

[slim swan]

thx

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[stable kite]

It asks to find values of a and b for which the function is continuous

[stable kite]

I'm kind of stuck with the evaluation of the limit of the first function as x goes to -1

[ripe ether]

Lhospital should work unless b=0. also you can use the fact that x<-1 to simplify |x+1|.

[compact pewter]

@stable kite Has your question been resolved?

[stable kite]

ywah i did that

i could show you my work if you want

[ripe ether]

sure

[stable kite]

ok i need to write it in a better way but

i analyze how the limit of the first function going to -1 changes while b changes and found the cases of b=1 and b>1 to be possible while having the function continuous

but both ways lead me to values of b that arent in the range i defined before

so i just assumed it's impossible

unless maybe 0^0 != 1

[ripe ether]

I think Id need to see the detailed solution to judge

[stable kite]

what solution do you obtain?

[ripe ether]

I havent done it in detail yet

[stable kite]

wait i have a question

[ripe ether]

yeah?

[compact pewter]

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[ripe ether]

.reopen

[compact pewter]

✅

[stable kite]

would you say this limit is equal to 1 for b^2 =1?

cuz if it is

i dont think there are solution

if for b^2 =1 or greater than 1 lim =0 then i have a solution for a and b

[ripe ether]

you know the value of $ln(u+1)^{b^2-1}$, for all $u\neq -1$ in the domain, so you can compute the limit

[boreal girder]

qwertytrewq

[ripe ether]

this has nothing to do with the convention of $0^0$

[boreal girder]

qwertytrewq

[stable kite]

yeah for u =0 then ln = 0 right?

since it's ln1

but then the exponent

if its less than 0

then isnt it 1/0

so infinity

if its equal to 0

i have 0^0

[ripe ether]

plugging in 0 does not work if your function is not nice

[stable kite]

wait wdym bu that

this limit is equivalent to the limit of the original function with x that goes to -1

[ripe ether]

I mean it is possible, for a function $f$, that $\lim_{x\to a} f(x)\neq f(a)$

[boreal girder]

qwertytrewq

[stable kite]

Yeah i know that

but then isnt it impossible to determine b

[ripe ether]

so in this case you should be cautious about plugging in 0

the limit is well defined

and it does not depend on how you would define 0^0

[stable kite]

mhm

then i have no idea how to hanfle the problem

[ripe ether]

compute the value of your function at $u\neq -1$

[boreal girder]

qwertytrewq

[ripe ether]

limit is asking how the function behaves around $-1$, the behaviour at $-1$ is irrelevant

[boreal girder]

qwertytrewq

[ripe ether]

so ud need to think abt how it behaves around $-1$

[boreal girder]

qwertytrewq

[ripe ether]

at the very least, this is computable, and not ambiguous

[stable kite]

mhm and that helps me determine valid values of b?

[ripe ether]

i mean u r working with the assumption that b^2=1 already

[stable kite]

I am not?

[ripe ether]

as far as i understand what u r asking, i thought u want to find the limit when b^2=1

[stable kite]

nope

[ripe ether]

that is a crucial bit of the whole problem

[stable kite]

i need to find values of b and a for which the function is continous

[ripe ether]

im aware

[stable kite]

putting the limit at the point where the function changes

[ripe ether]

but im answering the subquestion u asked

[stable kite]

i found b>=1 to be acceptable

then with that i distingued the cases

b=1 and b>1

cuz the first limit evalueated to 2 different values

[ripe ether]

yes, great. can you show your work for the limits?

[stable kite]

yeah

i need to rewrite tha tin order tho

cuz it's pretty chaotic

rn srry

[ripe ether]

because it seems like you have struggles with the b=1 case

[stable kite]

yeah i guess so

lemme do this

i dont think it will take more then 15mins

wait if you can thx

i'd be grateful

[ripe ether]

ok nw

[stable kite]

idk how to handle that limit

[ripe ether]

b^2=0 should be considered separately, since in that case lhospital don't apply

[stable kite]

oh okok

ill do that now

so for b^2 =0 as an exponent you do have 0^0 =1

?

btw imma go to sleep now thx for the help anyways

.close

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[ripe ether]

no we are not assuming the value of $0^0$, but rather the limit as $x\to -1$ of $\frac{-\ln(|x+1|+1)^{b^2}}{x+1}$ is $1$

[boreal girder]

qwertytrewq

[ripe ether]

again, the limit is unambiguous, while the value of $0^0$ may be.

[boreal girder]

qwertytrewq

[ripe ether]

it's better not to think of it like "oh in this case 0^0 is 1 and in this other case 0^0 is 0"

[zenith berry]

got it

ill check it with him again tmr

[ripe ether]

same class?

[zenith berry]

kinda

i suggested this problem to him cuz my teacher couldnt do it

ty btw

[ripe ether]

np, u can dm me if you want further guidance from here cuz i might not be online

[zenith berry]

alr, i'll tmr if i can do it better when im not sleepy

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[sudden carbon]

did i do this quesiton wrong

are we not supposed to add the displacements

[compact pewter]

@sudden carbon Has your question been resolved?

[rigid perch]

we need to add the displacements as vectors

[sudden carbon]

what

how

[rigid perch]

so we need to find the magnitude (and possibly direction) of the vector sum A + B + C + D

one way you could co about this would be to add up (A + B) (find the magnitude and direction of that vector sum), and then add up (C+D) in the same way, then find the vector sum of (A + B) + (C + D)

although any order of adding up would work

[sudden carbon]

but

The vector itself has a bearing

It’s not like

NSEW so you can’t assign positive/ negative signs to it right

[sudden carbon]

OH WAIT

i draw more triangles? 😭

[rigid perch]

well if you do it in that order you would end up drawing 3 triangles in total (one for summing A+B, one for summing C+D, and then one for summing the results)

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[sudden carbon]

ty

[sudden carbon]

.close

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[tribal blade]

I don't know how to format this better so I wrote it out as best as I can:

I'm trying to work through solving for a different variable in a formula.

A = 1/2ah + 1/2bh  <-  Solving for b

Multiply both sides by 2:
2A = ah + bh

Factor h from right side:
2A = h(a+b)

divide both sides by h:
2A/h = a+b

Subtract a from both sides:
b = 2A/h - a

I'm confused why the above doesn't work, seems to come down to the factoring or something I'm failing to do after subtracting a.

I understand the correct solution to be:
b = (2A-ah)/h

which I do understand how it's arrived at, but the process I detailed first seems like it should work.

[tribal blade]

Where am I going wrong in my understanding?

[rigid perch]

those are equivalent

[thorny urchin]

nowhere, your work is fine

[tribal blade]

Ahh

[thorny urchin]

unless they requested the answer in a specific form, your result is fine

[tribal blade]

They didn't, it just had me confused as the answer I have available in a key showed the second form.

I'm just teaching myself.

[thorny urchin]

from there combining the fractions you have will get you their result

to efficiently reach their result,
from
2A = ah + bh
instead of factoring out h,
instead subtract ah from both sides

[tribal blade]

Yeah, I see how they reached their result.

[thorny urchin]

as it's easier to reach b when you have bh rather than h(a+b)

but if the algebra is valid, you'll get there in the end

[tribal blade]

When combining the fractions b = 2A/h - a it's effectively because I have to do it as follows: 2A/h - ah/h` = (2A-ah)/h

I couldn't see the equivalence before I guess.

Thank you!

[thorny urchin]

np

[tribal blade]

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[compact pewter]

@compact nimbus Has your question been resolved?

[compact nimbus]

[unkempt sail]

Yeah the sketch is accurate

Well you can find coordinates only if one of them is given

[compact nimbus]

ok sweet, ty man

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[unkempt sail]

Wait what

That’s you doubt!

?

[compact nimbus]

?

[unkempt sail]

Is that all you wanted to ask

[compact nimbus]

yea

[unkempt sail]

Oh ok

[compact nimbus]

ty

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[forest token]

never mind i got it

.close

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[fallen frost]

Is trigonometry needed in order to do well in a geometry course?

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[echo elm]

hey ! I have a question regarding this problem

So to prove if its diff. at x = 0, we have to take the lim as x -> 0 and prove it exists. We prove that by taking the limits from both sides, however when I take the lim as x -> 0+, I get f(x) / x which is (2 tan(x/2)) / x, but is that not equal to 0 ? The answer guide says it is 1, but im not sure if im missing something? thanks!

[echo elm]

and i see that one can use L'Hospitals rule, but why can we just not plug in 0 directly and solve from there ?

[kind crane]

Can you just show your work

[echo elm]

sure

[dense dagger]

"So to prove if its diff. at x = 0, we have to take the lim as x -> 0 and prove it exists.". You mean the lim of the derivative, or the function itself?

Cause if its the function itself, I don't think that's valid- continuity doesnt imply differentiability

[echo elm]

theres a thm in the book im using that says

f(x) is diff. at x = a if the following limit exists:

lim x -> a ((f(x) - f(a)) / x - a)

[eternal pulsar]

Yes, use that, also strange why they wrote $0\le x\le 1$ first...

[boreal girder]

;(

[dense dagger]

okay, so if the derivative exits, yeah

[eternal pulsar]

Automod

In reference to the .heic

[echo elm]

ah , wack

[eternal pulsar]

Try to screenshot it

[dense dagger]

It stops it now? Damn

[echo elm]

the answer key says that the lim exists and = 1, but im confused on how he got that lol

[eternal pulsar]

Your limit is wrong there

Notice that $\tan(\frac{x}2)=\frac{\sin(\frac{x}2)}{\cos(\frac{x}2)}$

[boreal girder]

;(

[eternal pulsar]

Then transform your limit and do 2 things:\
\begin{enumerate}
\item Note that $\lim_{x\to 0}\frac{\sin(x)}{x}=1$.
\item Let $u=x/2$ in your limit and proceed.
\end{enumerate}

[boreal girder]

;(

[eternal pulsar]

Too much info?

[echo elm]

no , that makes sense , but if this is true, why cant i just plug in 0 into x and itll be = 0

[eternal pulsar]

Because you get a 0/0 (indeterminate) form.

tan(0)/0=0/0

[echo elm]

so since id get 0 / 0 , im forced to evaluate the lim another way

ah , thats the step i was missing

idk why i completely missed that

tysm 🙏

[eternal pulsar]

[echo elm]

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[warm sigil]

$3x^x=x^4$

[boreal girder]

GoldenPhoenix

[warm sigil]

I've been able to reduce this a bit, but I feel like I need a transformation to find the second solution

$x^{x-4}=\frac{1}{3}$

[boreal girder]

GoldenPhoenix

[warm sigil]

$\ln(x)e^{ln(x-4)}=-\ln(3)$

[boreal girder]

GoldenPhoenix

[warm sigil]

if I can find some transformation x=t+k I would be able to use the lambert W function, but the problem is I'd have to figure out the value k s.t. $(t+k)^{t+k-4}=t^t$

[boreal girder]

GoldenPhoenix

[warm sigil]

and I think such a value is technically dependent on what the value x is

I know through inspection that an x value of 3 is valid, and there is another solution I can find by graphing of about 1.57, which I would guess to be transcendental (since such numbers are more numerous than algebraic numbers), but I was wondering if there was a way to extract this solution in a more closed form than simply a numerical method of approximation.

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@warm sigil Has your question been resolved?

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[iron lantern]

A particle moves along the x axis. The velocity of the particle at time t is given by v(t)= 4/(t^3)+1))
if the position of the particle is x=1 when t=2, what is the position of the particle when t=4?

I don't know how to start because I don't think u sub works correct me if im wrong

[carmine garden]

okay,so what you know is $\frac{dx}{dt} = \frac{4}{t^3}+1$, yes?

[iron lantern]

uhhh

no not really

[boreal girder]

What a wonderful world !

[iron lantern]

oh no its

[carmine garden]

oh

that's rough

okay, the best you can do is probably partial fractions

[molten coral]

,w Apart[1/(t^3+1)]

[boreal girder]

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Requested by janniku

[iron lantern]

im not in bc but they allowed the grapher

so i got it

im sure id get this if i was

thanks though

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[zenith ore]

I was trying to find the domain of this function, so i made the denominater = 0, but when finding the natural log to move down the cos x, i made the right side undefined

[zenith ore]

im not sure how to approach this question

[kind crane]

e^thing is never zero

[zenith ore]

its to find where the function is undefined

[kind crane]

Yes

[zenith ore]

what is the procedure?

[kind viper]

what is this (0)ln

what does that mean

[zenith ore]

natural log of 0

[kind viper]

ok but why is the ln on the right

what's it doing on the right 😭

that would've been ln(0)

point stands tho, e^cos(x) is never zero

[zenith ore]

i kinda tried to apply it to both sides visually but it doesnt look good for other people

[kind viper]

because e^(whatever) is never zero

[kind viper]

well you should have written ln(e^cos(x)) = ln(0)

[zenith ore]

yeah i was following something i found online for finding the domain of exponential functions

[zenith ore]

ill do that in the future

but its kinda just how i think

anyways

how should I approach this question?

[kind viper]

e^thing is never zero

point stands tho, e^cos(x) is never zero

[zenith ore]

ooh wait

[gilded kayak]

domain can just be all real numbers

[zenith ore]

oh

[gilded kayak]

doesnt need a limit/hole/tiny range

[zenith ore]

i mean if it is never 0 is that not in the domain?

[broken trail]

thats a problem for the range, cus if you cross multiply, y >0, which affects range

[zenith ore]

im just thinking abt it wrong

ok bet

thank you guys

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[gilded kayak]

x=0 is possible so its in the domain, denominator tho its never 0

o

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[iron lantern]

can someone explain how to do this problem? I just guessed because didn't know what to base my answer off of

[iron lantern]

this is all i know so far

[slate sand]

But look at the secant line from 3 to 5

[iron lantern]

like that?

[slate sand]

Yeah your drawing is good

So what is the lowest line at 4.2

[iron lantern]

s

ah i see

[slate sand]

Yeah

[iron lantern]

ok yea i get it

thanks

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[gilded kayak]

i have all these relations now, how do i find p(5)?

[kind viper]

the p(n) form an arithmetic progression do they not

so p(n)=p(0)-nd where d is the common difference*

and the sum of all p(n) is 1 bc they are probabilities of an exhaustive and mutually exclusive set of events

[compact pewter]

@gilded kayak Has your question been resolved?

[gilded kayak]

yes i did write that down whats next tho

[kind viper]

call p(0) = a for convenience

a+(a-d)=0.4

a+(a-d)+(a-2d)+...+(a-5d)=1

2 linear eqs in 2 variables

[gilded kayak]

i try

btw p5 isnt in the sequence

its just p0 to p4

[kind viper]

yes it is lol

the q mentions p(5)

[gilded kayak]

2nd observation doesnt have p5

[kind viper]

it involves p(n+1) for, among others, n=4

that means p(4)-p(5)=d still

[gilded kayak]

oh

damn

done thx

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[solid socket]

Can someone please help in help-25 | jesus20

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[solid socket]

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[summer cedar]

how do u do this

[bleak dock]

the important part is that the seabird's nest is 20m above sea level

it's telling you about the y-intercept being 20

[summer cedar]

i thought just plug 20 into the equation and solve for A

[bleak dock]

you want to find H when t = 20? really?

[summer cedar]

oh no i mean

H = 20

when h = 20

[bleak dock]

yeah that's why that's incorrect

[bleak dock]

yes so you don't sub in t = 20 but t = ... ?

[summer cedar]

oh i saw the mark scheme said t = 0

I think i get it now

[bleak dock]

yeah, when the bird is still in the nest, that's 0 seconds after the bird leaves the nest

[bleak dock]

cool

[summer cedar]

uhh

part b i also didnt get 😅

[bleak dock]

alright, show part b pls

[summer cedar]

i thought youd use addition formula

but its only 2 marks

so that seems way too complex

[bleak dock]

ah it's the R cos(x - alpha) formula

[summer cedar]

cos?

[bleak dock]

https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf

[bleak dock]

you have to adapt it a bit for sin

[summer cedar]

idk cuz in the mark scheme

they just instantly got this

[bleak dock]

check your syllabus; it should be there

which paper is this for Edexcel?

[summer cedar]

mock set pure 2 set 2

[bleak dock]

ah okay, pure 2

[summer cedar]

nah this is all they put

[bleak dock]

if you try giving this a read, it will explain

it is based on the angle addition formulas yes

it's a big shortcut

[summer cedar]

OMGGGGGG

its matching coefficients?

[bleak dock]

yes!!

[summer cedar]

sweet

[bleak dock]

also don't worry about the exact order, whether it's 28/45 or 45/28

you have your graphing calculator so if you want to check, you just graph it and see if it matches

[summer cedar]

not sure how to do this

[bleak dock]

first of all you graph 45 sin 2x - 28 cos 2x

then you say graph 53 sin(2x - arctan(28/45))

check if the roots match

[summer cedar]

ah its not working for some reason

I think ill try finish off this question for now

[bleak dock]

if the roots match then the entire function will match for this kind of question

[summer cedar]

what dyu plug in for min height here

i thoguht its t = 0

oh wait

[bleak dock]

ah, so because the trig function is in the denominator

[summer cedar]

it would be as big as possible

[bleak dock]

to minimise H, you maximise the denominator