help-17

[restive crag]

ohhhh

[dull bear]

So c is the absolute maximum, which is useful, buuutttttttttttt!

We have no idea where c is we know that it exists at least in between, so it's something like we go up from x = a, then reach our maximum, then go back down to x = b (where h(a) = h(b) of course)

So what could we say about h from those statements?

[also the difference between here and the last question is that in this one, we introduced that point c, whereas in the last one, we were given it straight away, so the latter we knew where it was, this one we don't ]

[restive crag]

it means, that h is

i get what you're saying about the last problem and this, but am still just confused on what you're asking exactly when talking about h

[dull bear]

Well I'm probably trying to ask you where the minimum of h would be

[restive crag]

OH the minimum would be h(a)?

wait wait

no

OH

WAIT YES

WAIT... which one

[dull bear]

Yep the minimum is at x = a (and also at x = b, cause those give you the same output!)

[restive crag]

OHHH

OMGOSH

THAT MAKES SO MUCH SENSE

[dull bear]

h(a) [ = h(b)] must be the smallest value the function can output for you

[restive crag]

gotchaaaaa

[dull bear]

An example of this is if we take e.g. h(x) = -x^2 on the interval, say [-2, 2]: we have h(-2) = h(2), and of course those two are the smallest you can get

,w plot y = -x^2

Bruh

[twin meteor]

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Requested by chartbit

[dull bear]

Anyways, you get what I mean

[restive crag]

i get what you mean! tysm

[dull bear]

So of course here, now we're at:
The minimum must occur at x = a and x = b: the fact that we have h(a) = h(b) and h''(x) < 0 tells us that h is differentiable, that the derivative h' is always strictly decreasing, and that there is some point c in (a,b) such that h'(c) = 0: the decreasing-ness forces that to be the only such point. As also h''(c) < 0, this makes x = c an absolute maximum point, so the minimum must now occur at the points x = a and x = b
We do not know where the maximum is, as the previous para tells us nothing about the location of the maximum point

How'd ya find that?

[restive crag]

wait, so is c the absolute max or not? i thought we agreed it was when it equals 0

[dull bear]

It is, but we have no idea where it is!

[restive crag]

OHHHH

that makese sense

[dull bear]

It's like the difference between me telling you "at that junction, there are traffic lights" as opposed to me saying "there must be a junction in this city that has traffic lights"

But yea, happy with that?

[restive crag]

yes i am! tysm

[dull bear]

Great so, that last one?

[restive crag]

we have ONE more

but idk if i have it in me

cuz i have so many other more complex problems i need to do

[dull bear]

Awwwww well up to you! If you feel like it or if you'd rather do something else, whichever you feel is best

[restive crag]

tysm for your help! you've been a blessing

[dull bear]

Awwww it's been a pleasure working with you

[restive crag]

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[upper zodiac]

just lost

or how do you find the domian

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[fading seal]

need help understanding how to find if its surjective or not

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[open nacelle]

How do I solve the last problem

I do not know

[vast shale]

There is some story too if you give that then someone one might be able to help

[open nacelle]

hmmmm

i have mainly an image

of like just the shapes for visual

but off of like context

this is it

ITS HARD THO

I DONT GET IT

[vast shale]

Send that

Whatever you have been given

[open nacelle]

This is what I have

[vast shale]

It's asking distance from decision point

[open nacelle]

hm

so just how far it is

ok

now i just gotta

figure out

[vast shale]

That is your work

[open nacelle]

the way to solve it

[vast shale]

Yess

[open nacelle]

now idk what to do for that

[vast shale]

To find distance between decision point and airport just use Pythagoras

[open nacelle]

oh, if it's the airport then, how does that help, since that's for finding missing sides

[vast shale]

Find distance between ground and decision point and distance between airport and point below decision point then use Pythagoras you would have calculated possibly in previous question?

The Hudson river is landing spot then compare the distance between them you will have amswer

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[wind orchid]

I'm confused, I got quartile 1 as 25 but the lowest grade is 51-55th

[wind orchid]

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[snow cove]

1+1=2

[snow cove]

omg

[hushed pewter]

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[hushed pewter]

@snow cove stick to #discussion and not help channels.

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[hoary nebula]

Help please. i am stuck on inductive n = k+1 step

[steep finch]

what have u tried so fa r

[hoary nebula]

hold on lemme send

after the LEFT HAND SIDE, the denominator is 2^k(2-i). i am just too lazy to rewrite

[steep finch]

the (i/2)^k+1 term in LHS

doesnt need to be multiplied by 2^K+1

when combining the fractions

[hoary nebula]

it would make it harder?

[steep finch]

it already has 2^k+1 in denom

[hoary nebula]

ohh

[steep finch]

no i mean it's incorrect

[hoary nebula]

uhmm. can you just not multiply and cross like how i did?

[steep finch]

yes you can but that term is a mistake

id say do it again and ull probably see the mistake im pointing out

[hoary nebula]

let me retake

it's the same 😂

you would still get i^(k+1) (2-i), shown in my second working out in LHS @steep finch

[steep finch]

Hmm I see

Yeh you're right it just looks weird

Ok

[hoary nebula]

i just need to factor out. but i dont know what

[steep finch]

I'm gonna see what I can do next

[hoary nebula]

okay

[steep finch]

Yeh keep looking for something to factor

I'll help too

[hoary nebula]

cheers

[steep finch]

Those first 2 terms in third line should cancel no?

2nd and 3rd term*

[hoary nebula]

yes i just saw that

you're right

i'm getting nothing bruh

hopefully i didnt make a mistake making same denom

taking too long 😭

this question hard for no reason

<@&286206848099549185>

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[hoary nebula]

f it

.closr

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[urban summit]

An individual has taken a loan of tk. 50000 at the rate of monthly 11% compound
interest from a Credit organization. If that person pays his loan tk. 12000 monthly
then,
a ) How much loan will remain after 1 month?
b ) How much loan will remain after 2 month?
c) How much loan will remain after 3 month?

[urban summit]

im having trouble solving this question

havent done these in a long time

[vast shale]

good start is to calculate the interest for the first month

[urban summit]

could you

explain it to me like im a 7 year old

💀

[vast shale]

you have a percentage

you have a number

[urban summit]

yeah

[vast shale]

how do u find how much of the number there is based on the percentage

[urban summit]

dividing by 100?

[vast shale]

your favourite iphone costs $10000 and ur favourite store nearby has a 70% discount on it

what is the price of it

[urban summit]

uh

[vast shale]

no its 3000, but 7000 is how much the price decreased

[urban summit]

mb

ohh

oh

mb mb

[vast shale]

anyways like

try calculating the interest for the first month by multiplying the percentage by the base value u havw (50000)

[urban summit]

okay

then

[vast shale]

then?

[urban summit]

yk what ill try to solve this again

ill come back if i cant

thanks a lot tho

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[vast shale]

ok

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[sleek moss]

The point P lies in a square ABCD at a distance 1 from the vertices A and B and
equal to a distance 1 from the side [𝐶𝐷]. How much is the area of ​​the
triangle-ABP?

[sleek moss]

So I tried calculating ut

it

but then I get sqrt2/2

but the answer is something totally else😭

AP=1

AB=1

So I thought

angle A=45 degrees

B=45 degrees

and P=90 degrees

So AB=sqrt2

but it's totally wrong...:(

[tulip cedar]

Why is AB sqrt2?

[sleek moss]

I thought so because P must be 90 degrees right?

So with pythagoras

sqrt(1+1)

[tulip cedar]

I hate geometry, but if P is a distance 1 from CD then shouldn’t that tell you that CD is 2 from AB?

meaning AC = BD = 2

[sleek moss]

tried that but that also gives a wrong answer

The side is then 1+x

[wicked shard]

Notice that the subdivisions of the squares arent squares

[sleek moss]

The drawing is from the solution😅 normally we only get this so I don't really understand what that means

I think

I have to substitute soo

AB=X

BP=x-1

so Ab=x/2

yup got it thx

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[vast shale]

I am searching for videos related to a particular topic. However, I would prefer more didactic learner sheets instead. "The concurrency of arbitrary triangles BKB KBK BBB and right-angled triangles."

[vast shale]

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[sweet ingot]

Find x is the question. I dont understand where to start

[sweet ingot]

,rcw

[sweet ingot]

Find x is the question. I dont know where to start or ehat to do

Wait nvm its 35 i overthunk it lol

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[sweet ingot]

.reopen

[vocal sleet]

✅

[sweet ingot]

Find BAC is the question

Tried a few things didnt work need help

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@sweet ingot Has your question been resolved?

[sweet ingot]

<@&286206848099549185>

[silent arrow]

What is AD value

?

And DE?

[sweet ingot]

The value isnt given

[sweet ingot]

Ab=Ad=Ce=Cd and Bd=De

[silent arrow]

Value THAT is I need to solve

Number please

Ooh

Value is not given

Mind me

Wait

[sweet ingot]

Yeh np find bac is the question

[silent arrow]

Use SOH CAH TOA

[sweet ingot]

Whats that

;SOH

[silent arrow]

Sine cos tan

Wait

[sweet ingot]

Nvm

[silent arrow]

I'll send you a book on geometry

You will find such problems

It is downloading

[sweet ingot]

Ight im downloading it too

Btw how do you use sin cos tan on this

[silent arrow]

My friend that is your quest I am just guessing

A way

[sweet ingot]

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[sweet ingot]

Find bac is the question still couldnt find it

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[sweet ingot]

N

[sweet ingot]

T

[sweet ingot]

So

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[split dust]

why did they do that

[pale perch]

to show values above and below the calculated one have higher values presumably

its not really significant to the question

more a check of sorts

[split dust]

!close

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[vocal sleet]

Please don't occupy multiple help channels.

[rose scarab]

im stuck

can anyone help me?

i can only use transformations

[wraith python]

What is the equivalent of sem() in english?

[rose scarab]

what?

its sine

sorry for my handwriting

[wraith python]

sem(2x) = 2sem(x)cos(x)

[rose scarab]

ok

can you solve on a notepad and send the solution?

[wraith python]

Try splitting it into three different fractions.

[rose scarab]

ok

it doesnt work

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[storm stone]

i know this is wrong but tell me when i make the mistake please

i mean sin squared x + cos x oops

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[buoyant notch]

you cant cancel

at that stage

like if u have (x^2 +y)/xy
you cant cancel that into x

so you made the mistake at that second last step

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[pastel sorrel]

to find the range of a function, I can isolate the x and whichever restrictions my new expression has are the restrictions of my range. Does this method always work?

[pastel sorrel]

keep in mind that Im(f) is the same as the range of the function

<@&286206848099549185>

[viral terrace]

The first time

Ent one no how do you spell know do you know how to spell no

[pastel sorrel]

um

<@&286206848099549185>

[waxen harness]

you cant always just "isolate" the x....ex f(x)=xe^x

[pastel sorrel]

logarithm?

or is it impossible

I need a reliable method to find the range of a function

I cant find anything online

Ill ask my teacher tomorrow

but any info is appreciated

[waxen harness]

for the particular one f(x)=xe^x there is no function g(x) in terms of standard mathematical notation such that g(f(x))=x

[pastel sorrel]

Im at the first year of college

I wont be seeing any of those functions

so just the basic ones will do

[waxen harness]

dont you think you might see xe^x? or x^x?

[pastel sorrel]

.

[pastel sorrel]

for now, no

we are just getting started

I need to be able to solve basic functions

then Ill focus on others

so

what methods can I use

[waxen harness]

do you know derivatives?

[pastel sorrel]

hahah no

we barely saw limits

we really are just getting started

but we need to know range

[waxen harness]

so the most common method to find ranges is with derivatives

your method of isolating will work in linear functions such as ax+b and some others

but you ll pretty much stop using it after you reach derivatives

[pastel sorrel]

okay

thank you

[waxen harness]

until then yeah isolate

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[vast shale]

idk how to do this without using if statements

[vast shale]

please help

i need help please

[zinc quail]

what do you define as conditional statements?

<=>, => and <=?

[vast shale]

the question says make a statement similar to that

without using conditionals

in other words, i'm only allowed to use and, or, not

[zinc quail]

kk how would you translate P <=> Q

notice that this is only true

if P is 1 and Q is 1

or P is 0 and Q is 0

meaning P <=> Q is equivalent to (P and Q) or (notP and notQ)

[vast shale]

no because if you put p and q to be true, the statement will be true
and in the conditional it's false

[zinc quail]

?

no it's true in both

I meant if you only regard P <=> Q

instead of not(P <=> Q)

[vast shale]

its asking for the equivalent of not(p <->Q)

[zinc quail]

and above I explained how to insert

"P <=> Q is equivalent to (P and Q) or (notP and notQ)"

therefore

not(P <=> Q) is equivalent to not((P and Q) or (notP and notQ))

[vast shale]

ok i understand now

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[halcyon girder]

need brief explaination on how the 1/4^2 is aquired and the -3/4

where does the 1 come from ? and do you just remove the n in 4^n+ 2 ? to get 1/4^2 ?

[edgy gulch]

exponent laws

[halcyon girder]

so the n cancels out ? and your left with just the 2 ?

[edgy gulch]

dont know if "cancels out" is accurate term but

its basically breaking 4^(n+2) to 4^n * 4^2

and then you also have a power of n on the top

so you can combine them together

[halcyon girder]

ok

one last thing

is the 2 the first power to start because n = 2 ?

[edgy gulch]

yes

[halcyon girder]

so if it was n = 3 it'd start with ^3 right ?

[edgy gulch]

yes

[halcyon girder]

ok

thanks

[edgy gulch]

and the 1 is there because you're basically factoring 1/4^2

[halcyon girder]

for the -3/4 you get that because you cross out the n from top and bottom and the 2 ?

from n = 2 * ?

[edgy gulch]

well on the top you have (-3)^n

[halcyon girder]

right

[edgy gulch]

on the bottom you have 4^n * 4^2

you have two power of n

so it can be combined to be (-3/4)^n

but you also still have that 4^2 on the bottom

so you have something like (-3/4)^n * 1/4^2

[halcyon girder]

do you know by chance how to get the n variable on a ti84 ?

still been trying to figure it out

[edgy gulch]

nope never used a ti84 before

[halcyon girder]

ok

can i tag you again in abit, im reviewing for an exam rn skimming through stuff I still have questions about

[edgy gulch]

if i know it sure

[halcyon girder]

ok cool appreciate it

[vocal sleet]

@halcyon girder Has your question been resolved?

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[halcyon girder]

.reopen

[vocal sleet]

✅

[halcyon girder]

ok got a question again @edgy gulch

for this

how do you get -1/8 here ?

the two values are -1/8 & (-8/9)^n

I know how to get the (-8/9)^n but not the -1/8

[edgy gulch]

exponent rules again

(-8)^(n-1) = (-8)^n * (-8)^-1

(-8)^-1 is 1/-8

[halcyon girder]

I still dont get it can't visualize it

[edgy gulch]

you do know the exponent rules right?

[halcyon girder]

after seeing this probably not fully ?

[edgy gulch]

so (-8)^(n-1)

you can break it into (-8)^n * (-8)^-1

and (-8)^-1 is 1/-8

[halcyon girder]

so you are only working with the top value, you dont introduce the 9^n ?

[edgy gulch]

yes

the 9^n can be combined later with the (-8)^n

so you have (-8/9)^n

but you asked why the -1/8 is there and there you go

[halcyon girder]

so can I just like remove the n from (-8)^n-1 get (-8)^-1 and move the 1 to the top to get the -1/8?

would that work just to make it easier for each case ?

[edgy gulch]

if it makes you understand then yeah

but remember you still have (-8)^n remaining

it doesnt just go nowhere

[halcyon girder]

yeah so then it works out as 8n/9n which turns to - 8/9

[edgy gulch]

yes

[halcyon girder]

ok now i get it

im slow lol

if you don't mind so that I understand this one now 100% cause now I mostly get the concept

back to the first question

so I now understand you have the remaining -3n / 4n to get -3/4

but where does the 1 come from in terms of the way I explained it?

for 1/4^2

[edgy gulch]

well you combined the top with 4^n and get (-3/4)^n

so whats left of the top?

[halcyon girder]

n ?

[edgy gulch]

nope

[halcyon girder]

nothing ?

[edgy gulch]

(-3)^n is the same as 1 * (-3)^n

so when you combine the (-3)^n with the 4^n

you only have 1 left

[halcyon girder]

ohhhh

so if it were like (-2)^n / 4^n

it would be like 2/4 or 1/2 ?

[edgy gulch]

yeah

[halcyon girder]

ok now i get it

for this telescoping problem

the 1/4 is from -2 - 2 ?

@edgy gulch also another thing for this new problem

where does he start and get (1/3 - 1/7) ? I notice it goes up by 1 each time

[edgy gulch]

hmm lemme see

[halcyon girder]

ok I was working on figuring it out

ok the first part yeah check that

ill ask a question about the second part in sec

[edgy gulch]

its just partial fraction decomposition

which you should be able to do?

[halcyon girder]

so it isnt -2 - 2 ?

[edgy gulch]

nope

[halcyon girder]

so like... how would u do that xD

before we do that one thing

I just wanna make sure this question I figured out

is this correct ? red is what I believe is how its being done

[edgy gulch]

yeah

[halcyon girder]

and then the other side like 1/7 is just like 2 + 5 ...

[edgy gulch]

yes

[halcyon girder]

ok cool

so back to the 1/4

so like

[edgy gulch]

well i dont think you need to do partial fraction decomposition

they already gave you the formula for it

the use thing

[halcyon girder]

the formula doesnt show 1/4 though

wait so was I right ?

the b - a ?

[edgy gulch]

your b is 2

a is -2

2-(-2) is 4

[halcyon girder]

ok bingo

one last thing for this problem

how does it go from 1/3 to 20/60 ?

[edgy gulch]

you're adding fractions

so you need to find the greatest common factor of 3 4 5 and 6

to make their denominators equal

[halcyon girder]

the gcf wouldnt be 1 ?

[edgy gulch]

well actually idk the term for it

but you need to make the denominators equal is the idea

[halcyon girder]

but my question is how did the values increase so much ? is it multiplied ?

[edgy gulch]

yes

you know how to add fractions?

[halcyon girder]

yeah

[edgy gulch]

then just add all those together

[halcyon girder]

you know what yeah I can just do it in the calculator

I got my answer

ok one last thing for this problem @edgy gulch

does he go up in n = values until get gets 4 left that arent terms cancelled out because of the 1/4 ?

[edgy gulch]

nope

nothing to do with the 1/4

its just how the series cancels out

[halcyon girder]

but how does he know how many n = 's to go up to ?

I know its infinite but

[edgy gulch]

i suppose you can go until you see a pattern

[halcyon girder]

got it

[edgy gulch]

but if you go like 50 terms and still cant see a pattern then idk you need to look closer lol

[halcyon girder]

yeah lol

[vocal sleet]

@halcyon girder Has your question been resolved?

[halcyon girder]

For this question I try doing the last portion on the calculator and it gives me 1 each time not 0 ?

@edgy gulch not sure if you know how this one works different style problem

[edgy gulch]

the (3/6)^n as n goes to infinity?

[halcyon girder]

ahh wdym ?

sry

[edgy gulch]

the last part

(3/6)^n

[halcyon girder]

yeah how does that equate to 0 though ?

I don't understand the rule maybe ?

[edgy gulch]

well its like you keep multiplying 3/6

infinite times

3/6 is less than 1, when you keep multiplying it by 3/6

you get closer and closer to 0

thats why its 0

[halcyon girder]

ohh ok so just summary the 3^n and 6^n are the dominant largest ones to choose from. With them you get (3/6)^n. Because its a fraction and not greater than 1 its always equal to 0 ?

So in any case that its a fraction less than 1 it'll always = 0 and convergent ?

[edgy gulch]

its not greater than 1

so as n goes to infinity it always goes to 0

the rest im not sure

[halcyon girder]

right. So because its not greater than 1 and a fraction that means it'll = 0. But you aren't sure if thats the case everytime you get a similar result is what you are saying ?

[edgy gulch]

yeah

[halcyon girder]

got it

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@halcyon girder Has your question been resolved?

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@halcyon girder Has your question been resolved?

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[toxic obsidian]

how do i solve this it don't make sense

[bronze osprey]

Well the answer to a 'logic problem' like this can be anything

You can make up any pattern

[toxic obsidian]

bruh

thanks

[bronze osprey]

But it seems that they're doing 8 + 6 = 14

6 + 8 = 14

So 12 + 14 = ...

No worries

[toxic obsidian]

26

yeh

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[scenic leaf]

What do u do after this point

[nova yoke]

Use the division rule

[scenic ravine]

ln(a)-ln(b)=ln(a/b)

[scenic leaf]

.close

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[marble tinsel]

im confused

[waxen harness]

do you know when ax^2+bx+c > 0?

and when it s <0

[marble tinsel]

nop

[waxen harness]

well you ve probably been taught...you write the polynomial as a(x-r1)(x-r2) (if possible) an then you know that it will be positive when x>r2 and x<r1

and negative when r1<x<r2

(assume r1<r2)

[waxen harness]

you just split the double inequality into 2 singles and do that

[marble tinsel]

2 singles?

u mean factor?

i.e (x-r1)(x-r2)?

[waxen harness]

i mean that the 2 inequalities will be x^2-2x-8>0 and x^2-2x-8<27

[marble tinsel]

ah

and then solve those

[waxen harness]

yeah

[marble tinsel]

i forget how to do 0<(x-4)(x+2)

do i set the < to a = and solve?

[waxen harness]

you ve already solved it

it s just theory...you should probably read it from your text book

[marble tinsel]

the reason im here is because the textbook doesnt do a great job at explaining it

[waxen harness]

in order for (x-4)(x+2) to be positive you need both x-4 and x+2 to be positive or both be negative

for both of them to be positive you need x-4>0 and x+2>0 which gives us x>4

and i ll let you do the negative

[marble tinsel]

which is x>-2

[waxen harness]

not really

maybe do it analytically like i did

oh wait mb you are correct

getting tired 😦

wait x<-2*

but yeah

[marble tinsel]

wait y does the < flip

i forgot why exactly but ik it has to do with negative numbers in general

[waxen harness]

yeah -5<-4 but 5>4

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@marble tinsel Has your question been resolved?

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[rancid kestrel]

Hii

[rancid kestrel]

I’m here rn

I think for trip back time I need to put x + 1 because the trip there took one hour less than the trip back meaning the trip back is + 1 hour

When I did it I got the wrong answer tho

[dull bear]

How did you do it?

[rancid kestrel]

Hold on I think it’s correct

[dull bear]

@brisk moss you’re back

[rancid kestrel]

How does this look

[dull bear]

Looks good to me

[rancid kestrel]

Tysm!

.close

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[karmic abyss]

how do i solve this

[fleet quiver]

its the same as solving for tan(theta)-1 =0 if that provides any insight

[karmic abyss]

okay i can see why tan(theta) - 1 = 0

so stilll tan = 1?

[fleet quiver]

well first, consider than cot(x) is has period 2pi

which means that for any x, cot(x)=cot(x+2pi)

[karmic abyss]

okay can u help me how they got pi/4

[karmic abyss]

i dont understand this atm

sqrt2/2?

sin pi/4?

[fleet quiver]

yes

what is cos(pi/4)

[karmic abyss]

same

[fleet quiver]

so then cot(pi/4)=cos/sin = (sqrt2/2)/(sqrt2/2)

=1

[karmic abyss]

how does knowing what pi/4 help with getting it from the cot=1

[fleet quiver]

well we are verifing that pi/4 is a solution right?

[karmic abyss]

if i was looking at this prob. i woudlnt know why you woudl get pi/4 from cot =1

[karmic abyss]

yes but i wanna know how to get the pi/4 initially

from cot=1 if that makes sense

[fleet quiver]

you can look at the problem geometrically

tan is oposite over adjacent

if angle=0 is the positive x axis, then tan can be thought of as slope

because rise = opposite and run = adjacent

so if you are trying to find tan = 1

you are looking for what lines through the origin make a slope of 1

and what angles they correspond to

[karmic abyss]

same thing as saying tan = 1/1 right?

and both x and y is 1

[fleet quiver]

yes

[karmic abyss]

i mean y over x

[fleet quiver]

so there are only two rays that solve this

[karmic abyss]

so since the pi/4 has 1 for both x and y

its pi/4

[fleet quiver]

one pointing in the positive x and positive y direction

and one pointing in the negative x and negative y

[karmic abyss]

ok

so i understand why pi/4 is there now

now i wanna understand how to get all the other values

[fleet quiver]

this

[karmic abyss]

so just add 2pi

also subtract? since its -4pi?

[fleet quiver]

add or subtract until you cant anymore

[karmic abyss]

okayyy then

[fleet quiver]

but remember this

there are two solutions from 0 to 2pi

[karmic abyss]

what is this for?

[karmic abyss]

i also saw one other problem with 4 solutions

[fleet quiver]

well consider a line pointing from 0,0 to 1,1

corresponds to one angle

in 0,2pi

theres another line pointing from 0,0 to -1,-1 that corresponds to another angle in 0,2pi

[karmic abyss]

yes

[fleet quiver]

another way to think about this is that the line 180 degrees away from ur current line is always parallel

[karmic abyss]

how about for the problems with 4 solutions

[karmic abyss]

for the 0, 2pi i just find 1 angle then see its opposite angle is that fair

[fleet quiver]

for specifically tan

since the "opposite angle" is always parallel

[karmic abyss]

ok i have another specific question

for graphing tan

to get its asymptotes its supposed to be

starts ( ) = - pi/2
end ( ) = pi / 2
?

how come this one is 3pi/2

[fleet quiver]

consider where you have a slope of 1/0

so whenever cos=0

you have an asymptote

[karmic abyss]

i dont get it

[fleet quiver]

so think of tan as a slope

sin/cos

you get an indeterminent slope whenever the denominator is zero

denominator is cos

[karmic abyss]

im referring to this part over here of the solution

[fleet quiver]

you can also do that

[karmic abyss]

yes but how come the other solution on the right is 3pi/2

i mean its the same as -pi/2

but why on this specific one its 3pi/2

[fleet quiver]

do you understand why there is an asymptote at -pi/2

[karmic abyss]

no

[fleet quiver]

what is cos(-pi/2)

or just think of it geomterically

-pi/2 is a vertical line pointing down

vertical lines have indeterminant slope

[karmic abyss]

what does interdeterminant mean

[fleet quiver]

unable to determine

basically does not exist

[karmic abyss]

okay

appreciate your help

i got an exam later so

ill keep it in mind

[fleet quiver]

gl

[karmic abyss]

.close

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[cunning plaza]

Pretty straightforward, make two sequences, find when they match.

Your first sequence would be 2+8n. Try to get the second yourself.

Yes

Your first sequence expands to 2, 10, 18, 26…

Do the same with the other one and see when they match

Hm

It’s tough when neither the remainders nor the difference of the divisor and remainder don’t match

27-3, for example

The solution probably involves the CRT

Yeah, it’s CRT

Chinese Remainder Theorem

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[vast shale]

.5 cookie

half of a zero cookie...̄?

[low pewter]

what is your question?

[jagged cargo]

half a one

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@vast shale Has your question been resolved?

[vast shale]

.close

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[river kettle]

i found the gradient k to be 300 but im not sure what to do after that

[heavy yoke]

use conservation of energy

[river kettle]

how tho

[viral copper]

the potential energy of the spring goes into gravitational potential energy and kinetic energy

[river kettle]

uh

[somber portal]

you have x. Thus you can calculate the energy on the spring

at launch, the spring transfered all that energy to the projectile, in the form of kinetic energy

if you have the total kinetic energy, and you have the mass, can you compute the velocity?

[viral copper]

okay

[viral copper]

and gravitational potential

the spring is tilted

[river kettle]

do we use potential?

[twin meteor]

uoǝu

[viral copper]

i don't know what you mean by "use potential"

[river kettle]

potential energy

u = mgh

[viral copper]

yeah

you know compression of the spring

so you'd use that for the change in gravitational potential as well

[fading plume]

are we assuming gravity exists here? there's no angle to compute delta height

[viral copper]

probably being shot straight up then

[fading plume]

if it's straight up that wouldn't be much of a projectile launcher

[viral copper]

i mean

[river kettle]

the image shows its being shot at an angle

[viral copper]

a projectile is just a launched object

[viral copper]

with no angle as b said though

it would be hard

[fading plume]

🤷‍♀️

[viral copper]

actually impossible

[fading plume]

you could just as easily assume it's sideways

[viral copper]

😭

[viral copper]

boring

[fading plume]

i think the point of this exercise is just spring energy anyways lmao

[river kettle]

so

[river kettle]

potential + kinetic equals to potential energy of the spring is what u are saying here?

[viral copper]

you should probably follow what b is saying

disregard mgh

put all the spring potential energy into kinetic energy

[river kettle]

huh

im not sure i understand

[fading plume]

ok let's back up

the formula $E = \frac 1 2 kx^2$ should be familiar right

[twin meteor]

b

[vocal sleet]

@river kettle Has your question been resolved?

[river kettle]

yeah

[fading plume]

ok, so you can just plug in x

[river kettle]

do we know what x is?

[fading plume]

to find k, you need to read the graph and use F = kx

[river kettle]

i found k = 300

[fading plume]

oh lol i just saw you did a while ago

that's right

[river kettle]

what is x though?

[fading plume]

displacement is given in the question

[river kettle]

oh so its 0.08

[fading plume]

👍

[river kettle]

so 1/2(0.08)^2(300) = E

what do we do with this information?

[fading plume]

$E = \frac 1 2 m v ^2$

[twin meteor]

b

[river kettle]

uh yeah thats the formula

[fading plume]

you have E and m, and they want v

[river kettle]

o

24/25 = 1/50v^2 so

50(24/25) = v^2

v^2 = 48

v = 4sqrt(3)

6.92

m/s

[fading plume]

yup

nice

[river kettle]

so to recap

[river kettle]

we found the potential energy from the spring and equated it to the formula of the pontential energy?

[fading plume]

yes

that's conservation of energy

when the spring decompresses it gives that stored energy to the ball

[river kettle]

oh i see

so the Potential Energy of Spring = Potential Energy of Ball

[fading plume]

basically

[river kettle]

thank you

.close

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[cerulean talon]

How to find coordinates of point A

[cerulean talon]

There is no other further information

[fading plume]

oh that's an interesting question

the first step is to find the tangent line of O

then you find the slope of O's normal

[cerulean talon]

I'm not sure line p has pass through origin or not

If I guess it has then equation of p should be y=4x

[fading plume]

O means origin, so y = 4x is right

[cerulean talon]

Oh I get it

Thanks~

.close

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[charred dune]

i don’t understand how to solve this
if X+Y=43 and
X*Y=280??

[hidden kelp]

One way to do this is isolate a variable in one equation then plug it in the other

[charred dune]

okay but how-

[hidden kelp]

Using operations? Like, in x+y = 43, you could subtract x on both sides to isolate y

[charred dune]

ohh so y=43-x?

[hidden kelp]

Yeah

[charred dune]

ooh-

[hidden kelp]

Then you can substitute it in the other equation and get one with only X

[charred dune]

then it eould be x* 43-x =280

[hidden kelp]

Careful, you should put parentheses

[charred dune]

like this x*(43-x)=280?

[hidden kelp]

Yes

[charred dune]

now im really stuck-

[hidden kelp]

You should multiply x inside the parentheses to get rid of them

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@charred dune Has your question been resolved?

[vocal sleet]

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• 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.

[viral copper]

Do you know the base change theorem for logarithms?

[crude goblet]

what is that?

please teach me

[crude goblet]

.

ik only 10 rule of logs

[twin meteor]

uoǝu

[viral copper]

standard restrictions on a, b, and c

[viral copper]

is this not familiar?

[crude goblet]

we wrote it like loga/logb

[viral copper]

,tex .log rules

[twin meteor]

uoǝu

[viral copper]

yeah same

[crude goblet]

ik this rules

[viral copper]

thats what this is

you just didnt specify a base

[crude goblet]

okay then

[viral copper]

(you dont need to)

[crude goblet]

okay

i got logx/log5 = log 5 base x

then?

i can rewrite it as

log x base 5 = log 5 base x

@viral copper

[viral copper]

yeah

or you could base change the right as well

[crude goblet]

how could we change base?

there is no rule for it

[viral copper]

(log 5)/(log x)

[crude goblet]

oh

(log x )^2 = (log 5)^2

[viral copper]

yeah

[crude goblet]

then?

im stuck

[viral copper]

square root

[crude goblet]

then log x = log 5

[viral copper]

and log x = - log 5

[crude goblet]

+or-?

+-log 5?

[viral copper]

both

[crude goblet]

then?

[viral copper]

solve for x

😭

[crude goblet]

how

log x = +- log 5

[twin meteor]

uoǝu

[crude goblet]

oh

then

[viral copper]

...

[crude goblet]

1/5?

[viral copper]

and 5

x = 1/5 or 5

[crude goblet]

how ?

[viral copper]

because log x is log 5 or - log 5

[crude goblet]

then it means x = +5 - 5

[viral copper]

no

[crude goblet]

then ?

[viral copper]

first of all x cant be negative

[crude goblet]

okay

[viral copper]

I told you, 5 or 1/5

use the properties

[crude goblet]

1/5 how came?

[viral copper]

third one

[crude goblet]

then i get log x/5

[twin meteor]

uoǝu

[crude goblet]

@viral copper you are genius

[viral copper]

Can't tell if that was sarcastic or not

If it wasn't thank you

[crude goblet]

can i get help from you whenever i want please cuz i want someone like you to teach me

[crude goblet]

?

thanks a lot

[viral copper]

Open a channel

If I'm around I'll anwer

There are others here

[crude goblet]

i can msg you dm?

[viral copper]

Much much better

[crude goblet]

oh

[viral copper]

Please don't 🫠

[crude goblet]

okay

as you wish

[viral copper]

You'll probably get help faster if you open a channel anyways

[crude goblet]

hmm yeah

.colse

.close

[vocal sleet]

Channel closed

Closed by @crude goblet

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[steep merlin]

Have i messed up on a step

[steep merlin]

The answer should be -4/3

Oh wait lmao

.close

[vocal sleet]

Channel closed

Closed by @steep merlin

Use .reopen if this was a mistake.

[strong grove]

this is wrong btw but u still get the same answer

shouldve used power rule and chain rule to get
-8 sec^2x secx tanx

[steep merlin]

Yeah I know

I realized i mistake

I just fixed it up

[strong grove]

also to get ur desired answer just plug in the limit

[steep merlin]

ya

sec^2(0) = 1

cos(0)=1

all left is -8/6 to -4/3

thank you tho

[vocal sleet]

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>.
• 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.

[noble whale]

\frac{x}{x - 2} + \frac{1}{x - 4} = \frac{2}{x^{2} - 6x + 8}

[daring needle]

$\frac{x}{x - 2} + \frac{1}{x - 4} = \frac{2}{x^{2} - 6x + 8}$

[twin meteor]

Owen

[noble whale]

i dont know what the answer is

[strong olive]

you want the value of x?

[noble whale]

i put in a answer and i got it wrong and i have one more try

it just says solve

[strong olive]

!status

[vocal sleet]

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

[noble whale]

3

[strong olive]

show it

did you just try to add those two fractions

[noble whale]

ye

[nova yoke]

Multiple it all by the LCM

I think I used the right acronym

[strong olive]

what did you got