help-43

[kind viper]

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[true coral]

I was supposed to write for what parameter k x are real numbers

[kind viper]

sorry but you aren't expressing yourself clearly rn

are you able to post a PICTURE?

[true coral]

This picture above is only one i have rn

[gilded kayak]

in f(x) = sthsthsth, for x to only be real numbers, what is the range of k?

is that what it is

is this just a thought experiment how is there no original question

[kind viper]

ok then what, you had the problem stmt given to you orally and you did not write it down?

[true coral]

I did write it down

[kind viper]

then show what you wrote down

aside from the function

[gilded kayak]

i think thats all he wrote down

[kind viper]

the function alone is insufficient

[true coral]

Dont have it rn

[gilded kayak]

does f(x) = 0

[kind viper]

then come back when you do

[true coral]

Sure

[gilded kayak]

if f(x) is another variable theres probably no answer

[true coral]

Wiat

Wait*

For what values of the parameter k the domain of the function f is the set of all real numbers?

I think this is it

@compact pewter

[gilded kayak]

alr

what makes a rational function undefined

[true coral]

Domain???

[gilded kayak]

what makes a function that takes the form of a fraction undefined

[true coral]

No clue

I failed it

[gilded kayak]

its when the denominator = 0

...

[true coral]

Ok

[gilded kayak]

does this make sense

[true coral]

Yea

[gilded kayak]

yeah so you can now focus on making the denominator NOT 0

[true coral]

So x^2 - kx + k =/= 0

[gilded kayak]

yeah

do you know what happens if denominator = 0

[true coral]

It cant be 0 bc i cant divide by 0

[gilded kayak]

yeah if you do its either inf or -inf

[true coral]

Yea

[gilded kayak]

on a graph the curve would peak or dip hard at that x value

just giving you more facts

ok so for this to not = 0, it means the solutions to thatthing = 0 are imaginary

[neon dragon]

wsp

[gilded kayak]

hello

[true coral]

Imaginary?

[gilded kayak]

imaginary

in math it means it contains sqrt(-1) somewhere

[true coral]

Wait so if x^2 - kx + k =/= 0 then ∆ = k^2 - 4k???

[neon dragon]

what grade is this again?

[gilded kayak]

doesnt matter if thatthing = 0 or not, delta will always = k^2 - 4k

but does k^2 - 4k >0, =0 or <0?

[gilded kayak]

10th where im from

[true coral]

for <0 theres no solutions for (f)x i think

Ohh i think i get it now

[neon dragon]

...

[neon dragon]

I am in 8th grade

[gilded kayak]

are you learning this

[neon dragon]

nope

[gilded kayak]

then bye

[neon dragon]

hm

k

[true coral]

I got the solution for this on the other server. I speak in different language, so i couldnt get to the point sorry for misunderstanding.

[gilded kayak]

cool

try ur best to be clear regardless

[true coral]

I will thanks for felp

[compact pewter]

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[quartz yoke]

how would I find the points this corresponds to on the unit circle? (without the unit circle)

[quartz yoke]

and even questions like these which arent even on the unit circle

[kind viper]

that one is tho

it is just in quadrant 3

[kind viper]

pi is a half turn

[quartz yoke]

how am I to know this? and does it help when it comes to solving it?

[kind viper]

you should be able to visualize a unit circle and where you end up if you start at (1,0) and go halfway around

[kind viper]

the third quadrant begins at pi and ends at 3pi/2

[quartz yoke]

oh so its between those values

[potent berry]

that totally is on the unit circle lol

[kind viper]

and knowing 7pi/6 is in Q3 will inform you as to what symmetry you can use to pass to an acute angle

or to find the reference angle, if that's a thing you were taught in class

[plain quail]

If you can tell it's in quadrant 3 then you can easily tell that:
$\tan{\frac{7\pi}{6}}=\tan{\frac{\pi}{6}}$

[boreal girder]

denzio321

[quartz yoke]

i dont understand what you mean

[potent berry]

[quartz yoke]

i get now that its in the 3rd quadrant

but where does the 7 go

[potent berry]

$\tan \frac{7\pi}{6}$ is what now?

[kind viper]

tan(x) = tan(x-pi) @quartz yoke

denzio subtracted pi from it to exploit the fact that tan is pi-periodic

[plain quail]

I mean it is though

[quartz yoke]

holdon what is that asking me

[potent berry]

yes it but lol

[mortal bay]

perhaps you should brush up on periods of trig ratios

[quartz yoke]

so 7pi/6 - pi?

[potent berry]

anyway nvm

[mortal bay]

and cyclicity stuff

[plain quail]

Yea mb bruh

[quartz yoke]

can you suggest me a resource

[quartz yoke]

and again what are you asking me here lol

i dont even understand what to answer because i dont even know what the question is

[mortal bay]

reduce it to a smaller domain like [-pi,pi]

i.e. a smaller value

[quartz yoke]

by subtracting by pi?

[mortal bay]

yes

[potent berry]

wasnt for you

[potent berry]

no i get the thing i just didnt thing straight up stating it was particularly pedagogical

i just woke up so i took the sarcastic route

sorry about that lmao

[mortal bay]

huh

[quartz yoke]

and thus its pi/6, but what is the best way to go from that to a point without having a unit circle infront of me? ive memorized the first section of the unit circle and can replicate it (but is there an easier way)

[mortal bay]

"to a point" huh?

i dont quite follow

[quartz yoke]

to coordinates

[mortal bay]

yeah you just have to know the standard angle and their trigonometric ratios

no other way out of this 🤷

[quartz yoke]

ok, and one more thing.. does that -pi thing work for things other than tan as well or not

[mortal bay]

https://en.m.wikipedia.org/wiki/Trigonometric_functions

go to the periodicity section

you'll get what you want

anyways, as you probably read there, no you cant do it to other functions

only for the tangent

[quartz yoke]

it looks like its +2pi for cos and sin

[quartz yoke]

and in this case it becomes sin2pi

is that a valid conclusion?

[mortal bay]

uhm what?

how?

$\sin{\left(\pi + \frac{\pi}{3}\right)}$ notice this

[boreal girder]

parabolicinsanity

[quartz yoke]

yeah

that becomes 2pi/3

[mortal bay]

if you read a bit cautiously, you'd see that $\sin{(x+\pi)} = -\sin{(x)}$

[boreal girder]

parabolicinsanity

[mortal bay]

then you realise that the answer should be $-\sin{\left(\frac{\pi}{3}\right)}$

[boreal girder]

parabolicinsanity

[compact pewter]

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[elfin surge]

Hii not sure how to do b
For a it's 4sin(2x)+4

[plain quail]

Try imagining a few lines that pass through the origin

When does a line have 3 intercepts

Use the diagram of the graph that has been given to you

[plain quail]

If u use the graph you can notice that all lines between these two lines have three intercepts

[charred finch]

the lower one should be a bit more down

but it's fine

[charred finch]

I have a doubt, i don't see an option to do this on a computer, is there a way to?

[plain quail]

Well you can't find the points each line passes through by noting where the maximum of the function is

And for the second line just plugging in 2π

[charred finch]

no, i mean remixing the image

[plain quail]

Oh ion

Maybe just your device I guess

[plain quail]

@elfin surge

[elfin surge]

wait 1 sec!!

[elfin surge]

So these are the angles right

[elfin surge]

But technically speaking if there's 2 possible angles in the 1st quad shldnt there be 2 more In the 2nd quadrant then we end i with 4 soln?

Since sin(180-theta)=sin(theta)

[plain quail]

No no

[elfin surge]

Oh

[plain quail]

The uh angles

Don't matter here

I gtg eat dinner so

[elfin surge]

😭😭

[charred finch]

I'll take over

[plain quail]

Ye

[elfin surge]

Erm sure I don't rlly understand what's gg on honestly

[charred finch]

but i don't think u can get the line on top without differentiation

do u know differentiation?

[elfin surge]

Yep

[charred finch]

okay good

[elfin surge]

Hmm. how do we even proceed with it tho like after differenating what do we even set the differential to

[charred finch]

just a sec

[elfin surge]

Np

[charred finch]

u can see that in the region in beween the black lines, the graphs have 3 intersection points

(and the values of k, i.e. 2.08 > k > 0.63, are an approximate)

hence you can find the slopes of the 2 black lines and k is in between them

[elfin surge]

I genuinely don't understand anything 😭 r u saying like the line and the curve intersect at the same degree (In rad) so like same x coordinate erm what 😭

[charred finch]

no

[plain quail]

Hallo am back

[charred finch]

hi!

[elfin surge]

Man my math 😭

[charred finch]

so observe that as i am changing the value of k, the number of points of intersection of the graph and sin x is changing

i have marked 2 lines in black

if we take k values such that the line y = kx is in between them,

we can observe that it intersections the graph at exactly 3 points

also try taking lines which are outside that region, you get only 1 intersection point

[elfin surge]

Ohh so we want a line where the intersection between the curve and the line has 3 possible coordinates?

[charred finch]

yes

[elfin surge]

How does differenating the curve get it tho erm

I only know we can get the graident of the curve

[charred finch]

the top line is a tangent to the sin x graph

observe properly

[elfin surge]

Hmm u mean y=8

[charred finch]

no, its not 8 exactly

[elfin surge]

Oh

[charred finch]

take a general tangent equation of the graph and substitute in (0, 0) as tangent should pass though the origin

then take the tangent equation which makes sense there

[elfin surge]

U mean f(x)=4sin(2x)+4 find f'(x) sub f'(x)= and x=0
Which gets me 8

[charred finch]

no no

we get f'(x) = 8 cos(2x) right

[elfin surge]

Hmm yes

But sub x=0? Rifbt

[charred finch]

take a general value for x, for example, t

[elfin surge]

Hmm ok

[charred finch]

so f'(t) = 8 cos(2t)

let this be the slope of a line

y = 8 cos(2t) x + c

and since we want the tangent to pass through the origin, substitute (0, 0) to get c = 0

we get y = 8 cos(2t) x

and since this point is also a tangent at (t, 4 sin(2t) + 4), substiute it in

??

It's not exactly 8, is it?

[elfin surge]

No they want the range 😭

[elfin surge]

Hmm so ur saying we treate the differenatal as y=kx?

[plain quail]

No nvm

[charred finch]

yes

[plain quail]

It is exactly 8

So you can skip the derivatives

Oh nvm

[charred finch]

we need this

[plain quail]

Yea that's fair

[elfin surge]

Genuinely I can't even read the ans keu

[charred finch]

,w 4 sin(2t) + 4 = 8t cos(2t)

[elfin surge]

So if 8cos(2t) is the slope the line will always pass through a point of the curve?

[boreal girder]

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[charred finch]

,w 4 sin(2t) + 4 = 8t cos(2t) solve for t

[boreal girder]

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[charred finch]

,w 4 sin(2t) + 4 = 8t cos(2t) give a general solution for t

[boreal girder]

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[charred finch]

,w 4 sin(2t) + 4 = 8t cos(2t) general solution of t

[boreal girder]

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[charred finch]

,w 4 sin(2t) + 4 = 8t cos(2t) solve for positive t

[boreal girder]

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[charred finch]

damn

it gives t < 0 but clearly there exist a solution for t > 0

[elfin surge]

😭

[charred finch]

oh, wait

we can solve it ourselves

[plain quail]

Yea probably the better idea

[charred finch]

since t is n pi (n being not necessarily an integer)

and have a coefficient of pi does not help in any way, we can try putting cos(2t)=0

2n pi = 0

when 2n is (2k+1)pi/2 where k is an integer

hence n = (2k+1)pi/4

now, substitute this in the earlier equation,

4 sin(2t) = -4

sin(2t) = -1

sin( (2k+1) pi/2) ) = -1

when (2k+1)pi/2 is 3pi/2, 7pi/2, etc.

hence n = 3pi/4, 7pi/4, etc.

in our range, n = 3pi/4

wait...

but doesn't it give slope = 0 too

[elfin surge]

Ok u guys erm I have no idea where u guys are gg I know how to get it from the visualization method but idk how to do it mathematically

😭💔

[charred finch]

lol

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[thin basalt]

Both (-1/(x+1) )+ C and (x/(x+1)) +C differentiate to give 1/(x+1)^2. Both integrals have different expressions but seem to valid. How is that possible? Does this have anything to do with the constant C?

[quartz yoke]

Because x/(x+1) = 1 - 1/(x+1)

The 1 gets "absorbed" in C

[thin basalt]

Oh

I see

[quartz yoke]

C just represents a constant

C+1 = C' also represents a constant

3C = C'' also represents a constant

So to get rid of notation complications, we just "absorb" any numeric value into C

Final goal is to say that after indefinite integration, we also have some added constant to it. C just represents it

[thin basalt]

I understand

Thanks

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[timber bison]

Hi, could anyone please help with this. I tried looking at the solution they gave, but it literally says "is in notes" and it's not even in the notes. Thanks!

[timber bison]

This is what I have so far but I'm not too sure if it's right

<@&286206848099549185>

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[low garden]

I need my rad-deg conversions checked

[knotty basin]

,rccw

[knotty basin]

can you specify for 347, -820, and -250 degrees since you specified a recurring decimal in other places

rest looks perfect

[remote knot]

Just checked the rest and I agree

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[hard quartz]

x+2y=z^2 why is cylinder

[young raft]

? it's not a cylinder

[trail cave]

make ur graph bigger

[young raft]

it's 100% not a cylinder

[trail cave]

who knows maybe it's an infinitely wide cylinder

[rigid perch]

the plane of the cylinder might not be a coordinate plane

[hard quartz]

the answer c, the surface x+2y=z^2 is a cylinder my teacher told me its correct

[rigid perch]

you can see it from this perspective

[hard quartz]

by in the exam i cant use desmos

what should i do now

only a calculator and a pen

[kind crane]

re-learn how to plot 2d conics

[hard quartz]

why

3d is different

[trail cave]

ask your teacher why it's a cylinder? it's definitely not

[hard quartz]

the lecture happened this afternoon i missed the chance

[rigid perch]

a cylinder is a surface consisting of parallel lines all passing through a given plane curve

[hard quartz]

how to draw by hand

[young raft]

oh yeah

i forgot about that

i love math terminology

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[sacred island]

Can someone explain why the bits of v1 are being non deterministically guessed

Instead of

The nodes

As a whole

Shouldnt the bits be

Known

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[sacred island]

#theoretical-cs

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[stray flint]

hers f(x) and f'(x)

how come theres only one critical point for f(x) when you f'(x) = 0 gives you two solutions

[pine osprey]

Only one ?

Where

[stray flint]

x = 0, the f(x) is flat

[pine osprey]

Same for x = 1

[stray flint]

when x = 1 , you can see the f(x) graph increase

shouldnt it be flat as well?

[pine osprey]

Its flat as well

[stray flint]

right

gotta check my eyes

I was focusing on the intersection between f(x) and f'(x)

thanks for the help

[pine osprey]

Yw

[stray flint]

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[wary helm]

Whats the largest integer value of x^2 where -4<x<=1

[wary helm]

im pretty sure its 9 my answer key says 15

idk where it pull it from

[kind viper]

at $x = -\sqrt{15}$

[boreal girder]

Ann

[kind viper]

(nobody said x itself had to be an integer, did they?)

[wary helm]

oh ys

ya

but wait

why not 16 then?

[kind viper]

at what values of x would x^2 be equal to 16?

[wary helm]

wait im dumb

mb

[wary helm]

sqrt 16 but they told less than 16 mb

thnx

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[wary helm]

where did my shoelace go wrong?

[floral fjord]

shoelace?

[stoic dove]

maybe they mean that weird determinant

[floral fjord]

but how

Is determinant possible?

I think they are supposed to take 1 1 1

as last row

@wary helm

[stoic dove]

yeah but it could be a local trick to compute the area

[floral fjord]

You can take determinant for square matrices

[wary helm]

ya?

[floral fjord]

Your matrix is not a square matrix

You are supposed to use 1 1 1 as last row

[wary helm]

why

[floral fjord]

Because

That is not a square matrix

That's a 2x3

Determinant is not possible for non square matrices

[stoic dove]

wait do u k wt matrices and determinants are?

[wary helm]

nope

[floral fjord]

.

[wary helm]

all i know is this method to find area of a polynomial

[floral fjord]

Oh that's why

So what you are doing "shoelace" is called determinant

And that is possible only for square matrices meaning they have the same number of rows and columns

[wary helm]

bruh

im pretty sure im abel to use it even without same number of colums and rows

ive used it mutliple times

[floral fjord]

..

[stoic dove]

yeah i g it doesn't change the result

can u show me exactly wt the method is given to you

[floral fjord]

ur determinant

[wary helm]

lemme send it

the formula or whatever is something like this

Wait nvm im washed

i forgot to close the lace💀💀💀

sry

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[copper lion]

does the second A have anything to do with the first one?

[copper lion]

before the and

because if not, then it should've chosen a different letter..?

i guess i could think of x is in A and in B as a generic P(x) function, the set name doesn't matter

ok thnks

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[reef bronze]

the hint says to assume a >= b >= c and show a | bc and then put bounds on lcm(a,b,c) to show lcm(a,b,c) = bc, which i think i understand,
and then it says "if p | (b+c) then p | bc, what does this give?" which i don't understand
how does that help?

[winged lion]

what's p? a prime number?

[reef bronze]

it doesn't say, but i think so

[winged lion]

is this even true

3 | (2+1) but 3 -|- 2.1

[reef bronze]

yeah i think it only works for primes other than 3

because you have 3bc = a(b+c)

so if p | (b+c) you get p | 3bc and then if p is not 3 you get p | bc by euclid's lemma

[azure vault]

5|(4+1) then 5|4•1?

[reef bronze]

well it only works in the context of the problem

[azure vault]

??

Then what's the missing context then

[reef bronze]

because you know 3bc = a(b+c)

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@reef bronze Has your question been resolved?

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@reef bronze Has your question been resolved?

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@reef bronze Has your question been resolved?

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@reef bronze Has your question been resolved?

[winged lion]

you need to post a wrong solution like Cleo so you attract the klugscheißer

[reef bronze]

actually i found the answer with a different method but i want to understand the suggested method

the only solutions are permutations of (2,2,1)

[compact pewter]

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[reef bronze]

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[idle shard]

$$ \int_{1}^{3} \frac{g(x)}{g(x)+h(x)} dx=4 $$

[boreal girder]

꧅

[idle shard]

ok the above is a statement that is given to be true

so then i have to find this following integral

$$ \int_{3}^{1} \frac{h(x)}{g(x)+h(x)} dx $$

[boreal girder]

꧅

[idle shard]

basically i hvae no idea where to start

fyi the answer is 2

[olive field]

Notice that the integrand in the second integral is 1 minus the first integrand

[idle shard]

damn ok

ty

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[urban glen]

i dont quite get what part c is trying to do... is it just showing that f and f' converges by the same |x-1|<R?

[compact pewter]

@urban glen Has your question been resolved?

[native shard]

hello

where did you get up to?

did you find the function for f?

[urban glen]

hello

i solved it but i dont quite get the problem

[native shard]

what part is confusing

[urban glen]

like finding the range that f' converge and trying to force that to f

[native shard]

well let’s start with finding f

what did you get for R btw

[urban glen]

R=1/2 from part a

[native shard]

mhm

so

did you find f?

it’s geometric

[urban glen]

yea i did a/1-r

[native shard]

$f(x) = \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{2^n}{n} (x-1)^n$

yes?

[urban glen]

yea

[native shard]

$f(x) = \frac{2x-2}{2x - 1}$

oh wait

f’

my fault

[boreal girder]

knief

[native shard]

so we find f’ by differentiating inside the sum

taking the derivative wrt x gives

$f’(x) = \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{2^n}{n} \cdot n (x-1)^{n-1}$

[boreal girder]

knief

[native shard]

n cancels

[urban glen]

wait why should f and f' converge by the same |x-1|<R?

[native shard]

well the proof in the general case is a bit hairy and technical because youd actually have to use the root test i believe so you essentially just have to take this for granted, but btw this doesn’t mean the interval of convergence is the same

[urban glen]

so just the radius is same?

[native shard]

and btw the statements of ratio/root test are actually much more involved than what you’re told they are

[urban glen]

i js know some a_n+1 / a_n type thing

thanks for helping 😀

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[quartz oasis]

!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

[static fable]

1

[snow finch]

draw a picture

try some examples for M

make a chart

find the equation!

[static fable]

but how to do it in an analytical way

i can't do examples in the demonstration

[static fable]

<@&286206848099549185>

[kind crane]

the analytical way is to draw a picture and come up with the algebra

[kind crane]

what does this mean

[static fable]

i figured it out tnx

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[warped oar]

Hi.

[rancid light]

Hi.

[warped oar]

The Erdős squarefree conjecture that central binomial coefficients C(2n, n) are never squarefree for n > 4 was proved in 1996. - Wikipedia

[rancid light]

thats good to know

[warped oar]

Okay.

Given any positive integer n, we can uniquely express n as a product of a non-negative power of 2 and an odd number.

Prove this.

Let n = the number.

Then either n = odd or n = even.
If n = odd, then n = (2k + 1) 2^0 and we're done.

If n = even, n = 2^1 p.

Now if p = odd, we're done.
If p = even, p = 2q.

n = 2^1 p = 2^2 q.
If q = odd, we're done.
At some point in this process, we must get n = 2^k (some odd number), because if this is not the case, then n = infinity.

Is this proof correct?

<@&286206848099549185>

What is even to prove?

For uniqueness, consider the following two cases:

1. When n is odd.
   When n = odd, n = 2^0 (2k + 1) for some non-negative integer n.
   Let there exist k_1 such that (2k_1 + 1) 2^0 = n
   So (2k_1 + 1) 2^0 = 2^0 (2k + 1).
   k = k_1
   So all assumed values must coincide to a single solution.

2. When n is even.
   When n = even, n = 2^k (odd) = 2^k (2m + 1)
   Let another a possible factorization of n be that n = 2^l (2n + 1)
   So n = 2^l (2n + 1) = 2^k (2m + 1)
   For the sake of convenience, let the variable k denote the higher power of 2. That is, let k > l.
   Now 2^k/2^l = 2^(k - l) = a multiple of 2 = integer = (2n + 1)/(2m + 1)
   So 2n + 1 = (a multiple of 2)(2m + 1).
   Since the LHS is odd and the RHS is even, it must be that k is not greater than l. k cannot be less than l because then the roles of k and l would be changed. So k = l.
   So there is a unique factorization of n into a non-negative power of 2 and an odd positive integer.

Q.2 Is this true?
min(a, b) + max(a, b) = a + b

Q.3 Is this true?
(a1 + a2 + a3 + ........... + a_n)! is a multiple of (a1)! (a2)! (a3)! ......... (a_n)!?

Helpers are sleeping, no?

Useless server, isn't it?
||JK||.

<@&286206848099549185>

[azure vault]

what have you tried for those two questions Q2 and Q3?

[warped oar]

I was just asking.
Anyway:
Ans. 2:
If a > b, then min(a, b) = b and max(a, b) = a.
Proved.
Since min(a,b), max(a,b) and (a+b) are symmetric, it is also valid for b > a (a < b).
If a = b, then min(a, b) = ? and max(a, b) = ?
I am stuck on this step.

Ans. 3.
From the fact that the product of r consecutive integers is divisible by r!, we can get that (a! b!) | (a+b)!

Consider x = a and y = b + c.
(x! y!) | (x + y)!
(a! (b+c)!) | (a + b + c)!
Since (b! c!) divides (b + c)!, replacing it with a factor of it (which is b! c!) will not give a non-integer.
So (a! b! c!) | (a + b + c)!

Consider x = a and y = b + c + d
(x! y!) | (x+y)!
(a! (b + c + d)!) | (a + b + c + d)!
Since (b! c! d!) | (b + c + d)! (proved above), (a! b! c! d!) | (a + b + c + d)!.
By induction, this fact can be proved.

[warped oar]

What about the first question?

[azure vault]

This one?

[warped oar]

Yes.

[azure vault]

Well seems like you just proved everything

So

[warped oar]

This is the solution given in the book.

Is it any better than mine?

[azure vault]

It's the same idea

so equivalent

[kind viper]

tfw they didnt have a 0 superscript and used the degree symbol

[warped oar]

Okay. Thanks.

I am closing this. Just a minute.

[kind viper]

don't do this troll shit btw

the number of people here who find this sort of misdirection funny is \leq 1.

[warped oar]

,, \leq 1.

[boreal girder]

Mathematician

[warped oar]

Okay.

.clsoe

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[coarse plume]

Hi

[coarse plume]

How do I split fractions

[glacial prairie]

i.e. which fraction do you want to split?

[coarse plume]

In general I mean

[hard trellis]

Give me a fraction

Make one up that you want to split

[coarse plume]

I don’t know dude cna you give me some because I don’t know when the cases I can split and not split

Thsi is my problem

[glacial prairie]

you want an answer to a 1/((x+1)*(x-2)=A/(x+1)+B/(x-2) type of problem?

[coarse plume]

Yes sure

[glacial prairie]

so this is what you want, right?

[coarse plume]

Yes but I just want to know the rules of splitting fraction

[glacial prairie]

i.e. this one (?)

[coarse plume]

Yep

[glacial prairie]

multiply everything by (x+1)* (x-2) then do the products and then you have 1=Ax-2A+Bx+B => A=-B, -2A+B=1 => B=1/3,A=-1/3, so 1/((x+1)*(x-2))=-1/(3(x+1))+1/(3(x-2))

maybe this is what you mean by splitting fractions

[coarse plume]

Thsi is too much advanced for me sorry nvm I meant like split n+x/2 but I don’t know how

[glacial prairie]

n+x is the numerator here

if n+x is the numerator all you have to do is just n/2+x/2 you just dont mess with the denominator

[coarse plume]

@crisp cedar

Why are we laughing

[coarse plume]

Ohhhhhhhhh Ok makes sense thx

But what if it was weird thing like n+x/2+x

[glacial prairie]

if you have a denominator is the form of (x+y) you cannot just split it you dont touch it at all, only the numerator can be made from (x+y) into separate x and y with the same denominator

[coarse plume]

Thx dude

[glacial prairie]

try it yourself using this

[coarse plume]

Hmmm so if I have 8+x/2 it will be 8/2 + x/2 because denominator don’t have anything beside it

[glacial prairie]

ye

[coarse plume]

Ohhhh Ok then but if I have 8+x/2+x this cannot work

So there is No way to simpifu

Simplify

[glacial prairie]

you can only do 8/(2+x)+x/(2+x) here

[coarse plume]

Okkkkkkkkkkk I think I get it
Now

But that’s all ?

There is anything else I should watch for

[compact pewter]

@coarse plume Has your question been resolved?

[compact pewter]

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[sullen canopy]

find the value of ((3+h)^2 -9)/h as h goes to 0
create a function f(h) = ((3+h)^2 -9)/h and you notice plugging in 0 gives you a division by zero which is undefined but if you simplify the (3+h)^2 by expanding the binomial and you find that it is equal to 6+h which now you can plug 0 into as there is no more division error so the limit as f(h) as h goes to 0 is just 6

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[wet gale]

if u have to do this with respect to y then how come the area is split into two parts?

[keen granite]

it's easier to compute the blue area as an integral than doing (blue area + rectangle (pink area)) as an integral

area of a rectangle is trivial

and we work out the integral

sum these to get the total area

[wet gale]

but we cant do the whole thing as one integral with respect to y ?

[keen granite]

well that's kinda difficult because how do we you bound it with the rectangle included?

[wet gale]

i know that the area of a rectangle is trivial but then i need to find the line where the curve crosses the y axis for the upper bound

[keen granite]

you certainly can but it's not nearly as easy to do it than just split into rectnagle and integral

[keen granite]

indeed, because the rectangle touches the y-axis

so the blue area starts when the rectangle stops

which is when the function 4ln(3-x) touches the y-axis

[wet gale]

why arent you able to just do the integral from 0 to 6 of 2-(3-e^(y/4)) ?

[keen granite]

bcz that'd give this guy

[wet gale]

ohhhh

i get it now thank you

[keen granite]

[wet gale]

have a good night

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[keen granite]

you too

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[cursive granite]

How do you find where the local maximum/minimums are from the first derivative?

[cursive granite]

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[brazen bramble]

Idk where to start from

[compact pewter]

@brazen bramble Has your question been resolved?

[brazen bramble]

Nvm i got it

Close

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[rough wagon]

got to deriving it, not sure what to do after

[rough wagon]

i see that we have to use the IVT but idk how that applies here

[azure vault]

Yeah it's the IVT corollary more specifically, the one that involves increasing/decreasing behavior

[rough wagon]

the wat

corollary

[azure vault]

Like

IVT says that for every y between f(a) and f(b) you can find y = f(c) with a <= c <= b

[rough wagon]

oh i see

wait

[azure vault]

The corollary

[rough wagon]

so what would be my next step

[azure vault]

Adds that it's unique

Under the condition that f is increasing or decreasing on that interval

!occupied

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[rough wagon]

blud is stealing my spotlight 😔

[rough wagon]

hmmm

ohhh

i see okay

[azure vault]

So

Finding out the derivative is not bad

It gives you insight on if f could be decreasing everywhere

Or increasing everywhere

Or a mix of both

[rough wagon]

icic

so then i should work with the original function instead right

[azure vault]

Well we're working with f(x) = 3x + 2cos(x) + 5 right?

[rough wagon]

yeye

[azure vault]

What f' did you find

[rough wagon]

3 - 2sinx

[azure vault]

Alright

Can we find the sign of that?

[rough wagon]

errrr

do we use sin's domain for the like x?

[azure vault]

Domain?

You mean range?

[rough wagon]

or yea

range

[azure vault]

Domain is all x you can input in sin

which is all real numbers

Range is all the possible outputs of sin

[rough wagon]

yeye

so then -1,1

[azure vault]

[-1,1] yes

[rough wagon]

okok

just a conceptual question bc i forgot exactly why, but why is it [-1,1] and not [1,-1] again

[azure vault]

By convention real intervals are written from lower bound to upper bound

From 'left to right' on the real line

[rough wagon]

right right

okay makes snese

sense

i got [1,5] btw

or ig 1 and 5

[azure vault]

So that's the range of f' right?

[rough wagon]

yea

[azure vault]

[1,5] is the range of f' yeah

f' can take any values between 1 and 5

But not outside

So does that help us find the sign of f'?

[rough wagon]

errrr

it means its positive right cus its from [1,5] n it doesnt have any negatives

in the derivative

[azure vault]

Yeah

f' >= 1

Which is positive

[rough wagon]

yeyee

[azure vault]

So in fact we didn't need to exactly find the range of f', the lower bound was all that mattered

[rough wagon]

icic

[azure vault]

So f' is positive, which means f is...?

[rough wagon]

should be positive aswell

[azure vault]

Uhhh

[rough wagon]

errrr

[azure vault]

Not necessarily

I can give examples of f' positive but f negative everywhere if you wanna find out

[rough wagon]

mmmm maybe another time

[azure vault]

the sign of f' will not relate to the sign of f, but to the ... of f

Remember, what does the derivative represent

It's a slope

[rough wagon]

oh right

[azure vault]

So by having f' > 0

We know f has a positive slope everywhere

Which means that f is what type of function?

(We mentioned it at the beginning)

[rough wagon]

increasing?

errr

that doesnt sound right

hol up

[azure vault]

No it was right

[rough wagon]

oh dang ok

oh wait nvm it makes sense

[azure vault]

Positive slope -> the function shoots up from every point

So increasing

[rough wagon]

mhmhm

[azure vault]

Alright

[rough wagon]

hmmm

[azure vault]

Now that we know that f is increasing

Say I had some a such that f(a) < 0

Is there a chance that I could find f(x) = 0 with x < a?

[rough wagon]

wait decreasing?

[azure vault]

Increasing****

[rough wagon]

ohh

okok

err

[azure vault]

So at a, f(a) is below 0

Is there a chance that we reached 0 prior to a?

[rough wagon]

no cus its positive right

[azure vault]

Cause it's increasing*

[rough wagon]

or yea increasing

[azure vault]

If I had reached 0 before a, I would have needed to go back down to f(a) afterwards

[rough wagon]

icic

[azure vault]

Which isn't possible

And similarly

If I have some b such that f(b) > 0

Can I have f(x) = 0 for x > b?

[rough wagon]

yes

[azure vault]

Uh

[rough wagon]

cus its increasing no? n like if the function is positive we can have some positive that could be bigger than be b right

orrrr

[azure vault]

Uh

if f(b) > 0

Is there a chance to reach f(x) = 0 after b?

[rough wagon]

oh

no

[azure vault]

No, indeed

Because once again

That would require us to go from f(b) down to 0

[rough wagon]

yup yup yup

[azure vault]

So

If I have a and b such that f(a) < 0 and f(b) > 0

Where do I look if I want f(x) = 0?

[rough wagon]

errr

f(b) > 0 > f(a) so you'd look between f(b) and 0 ?

or no

wait

mmm

[azure vault]

I'm looking for the value(s) of x that would give f(x) = 0

Where could it/they be

[azure vault]

Can they be here?

[rough wagon]

nop

[azure vault]

Can they be here?

[rough wagon]

yes

[azure vault]

🤨

[rough wagon]

aaaaa

is it the opposite

oh

wait

uuuurgg

cus u cant do it after b and you cant do it prior to a

right

[azure vault]

Yeah

So

If it's not before a or after b

It's between a and b

[rough wagon]

okokok

[azure vault]

But now say we did have those a and b

[rough wagon]

mhmm

[azure vault]

Remember what IVT says

[azure vault]

So this

As well as the corollary

[rough wagon]

yeye

[azure vault]

So

There is a unique c between a and b such that f(c) = 0

This is possible because f(a) < 0 < f(b)

[rough wagon]

mhmm

[azure vault]

Since no one outside of [a,b] is gonna have f(x) = 0

c is the only input on the entirety of real numbers

That gives f(c) = 0

[rough wagon]

yeyeye

[azure vault]

So

[azure vault]

All that's left is to show

That this is not just an assumption

And we can actually find such a and b

[rough wagon]

okok

so we use the interval we found before?

[azure vault]

Uh

[1,5] was the range of f'

Not sure it has a lot to do with the domain of f

[rough wagon]

oh i see

[azure vault]

Just find some a

Such that f(a) < 0

[rough wagon]

oh so like idk -1?

[azure vault]

If you pick a small enough you'll find it

[rough wagon]

ok i see

[azure vault]

Try it

If it doesn't work aim lower

And find b such that f(b) > 0

[rough wagon]

ahh i got 3.99

ill go -2

nvm ok i got -2.003 at x=-3

and for b i did 3 n got 15.997

[azure vault]

Well -1 could serve as b then

[azure vault]

Or that sure

[rough wagon]

bet

[azure vault]

So

We found that there is indeed a unique solution to f(x) = 0

And it's between a and b

Also, side note if you're interested

[rough wagon]

yesyes

[azure vault]

This is also how you can find with more and more precision the zeroes of some hard function like f here

It's called dichotomy and the way we proceed from here:

Our search interval right now for finding the solution to f(x) = 0 is [a,b]

Take the middle of the interval, d = (a+b)/2

If f(d) < 0

Then we haven't reached the 0 yet

So [d,b] becomes the new [a,b]

And if f(d) >0, then the zero was somewhere before

And [a,d] becomes the new [a,b]

Repeat

[rough wagon]

daannngg

[azure vault]

Since you're dividing the length of the search interval by 2 everytime

You're bound to get any precision you can hope for

[rough wagon]

oh its like that computer science thing right idr what its called

[azure vault]

By doing this enough times

This is an "algorithm" to be more precise

[rough wagon]

yeaaa

[azure vault]

Not necessarily linked with computer stuff

[rough wagon]

icic

[azure vault]

One can do this by hand

[rough wagon]

dam thats rlly cool ngl

what lvl math is that

[azure vault]

Not that far advanced

Since you're already using IVT, it can be understood from your current level I think

[rough wagon]

oooo icic

we're done right

ty for help

.close

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[steep hill]

Guys I know this is a math help server but I reallt need to figure this out. Why is the force of gravity that adds onto the centripetal acceleration mgcos(theta). It doesnt make sense.

[rigid perch]

you can split the gravity force (as with any other vector) into components tangential and normal to the ball's trajectory

[steep hill]

I dont get how the weight mg becomes the hypothenuse

wait sorry

Why does the vertical vector become tangential velocity?

[rigid perch]

the vertical vector mg is split into two components, a component normal to the circle mg cos(theta), and a component tangent to the curve mg sin(theta)

[steep hill]

Ah ok thanks

close.

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[compact pewter]

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[quartz yoke]

Can someone help me do 6 amd 7

[trail cave]

6
ABD is isosceles so AOB is also isosceles
find angles OBA and OAB?
find angles BAD and ABD?

[quartz yoke]

Ok wait

[floral fjord]

I thought of just drawing OD

[trail cave]

ur so right

[floral fjord]

?

[quartz yoke]

That means?

[floral fjord]

ten who

[trail cave]

u

[floral fjord]

Draw line OD

[trail cave]

that still needs to assume AOB is isosceles tho

[floral fjord]

It is

[trail cave]

yeah i know

[floral fjord]

OA=OB radius?

[strange latch]

n is 40

[quartz yoke]

So i cut it into two parts?

[quartz yoke]

How 🥲

[floral fjord]

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[floral fjord]

DRAW OD

[winged dock]

I don't think there is a need for that

[floral fjord]

Did you draw?

[strange latch]

herons rule and jensons inequality

[floral fjord]

Random stuff

[strange latch]

no

[quartz yoke]

Idk rhat

[floral fjord]

I will help

[quartz yoke]

Havent learned it

[strange latch]

how else is he meant to draw OD without herons

[floral fjord]

Draw OD

with a pencil/pen

[quartz yoke]

Like a lime through it?

[floral fjord]

or mouse

[quartz yoke]

Gimme a min

[floral fjord]

See the angles at O

[quartz yoke]

🥲like this?

[floral fjord]

yes

[strange latch]

How do u know OD bisects the 40 degree

[winged dock]

Congruence

[floral fjord]

triangles are congruent

..

[quartz yoke]

So i divide 40 by 2 right

[floral fjord]

ok

now what is <AOD?

Did you get <ADO=20?

[quartz yoke]

20+20+x=180?

[floral fjord]

what

what triangle are you talking about

Look in triangle AOD

[quartz yoke]

Ok

Yea Ado i got 20

[floral fjord]

ok now <AOD?

[quartz yoke]

How do i get that?

[floral fjord]

Uh so basically

Both the triangles AOD and BOD are congruent

[quartz yoke]

Ok

[floral fjord]

So <AOD=<BOD

and <AOB+<AOD+<BOD=360

[quartz yoke]

Ok

But i only mknow AOB

[floral fjord]

uh

[floral fjord]

@quartz yoke

[quartz yoke]

Ok

[compact pewter]

@quartz yoke Has your question been resolved?

[quartz yoke]

Ok i got it

.close

[compact pewter]

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[compact pewter]

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[hallow spade]

does anyone know what is the command to solve this on rstudio

[compact pewter]

@hallow spade Has your question been resolved?

[compact pewter]

@hallow spade Has your question been resolved?

[compact pewter]

@hallow spade Has your question been resolved?

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@hallow spade Has your question been resolved?

[honest kernel]

hello

[brazen quiver]

What is f? What is S1 and S2?

[compact pewter]

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[tropic agate]

Silly question perhaps, but consider a diagonalizable linear operator $\mathsf T$ on a finite-dimensional vector space $\mathsf V$, with distinct eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_k$. Why is it that the characteristic polynomial of the restriction of $\mathsf T$ to $\mathsf E_{\lambda_i}$ is $(\lambda_i-t)^{m_i}$ where $m_i$ is the dimension of $\mathsf E_{\lambda_i}$?

[boreal girder]

psie

[tropic agate]

.close

[compact pewter]

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[devout topaz]

apparently these bounds indicate this shape over here, but I'm sort of confused. Why isn't there a z=y line for the bottom y bound? What am I missing here?

[topaz cargo]

It is there, no?

second integral from z to 1

[devout topaz]

yeah, I think I'm having trouble visualising it, do you mind pointing out which line it is?

[topaz cargo]

So basically you need to imagine the projection of this shape on yOz plane. it will be triangle with y, z coordinates (0, 0), (1, 0), (1, 1)

so for given z valid y lies in interval [z, 1], so point is in projection

[devout topaz]

fair

thanks

.close

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[lime kestrel]

i ve got ln^2 in the denom and - in between in the num unlike what is suggested in the answer

[acoustic yarrow]

,w differentiate ln(x^2-x)/ln(x)

[boreal girder]

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Requested by _mia_34

[acoustic yarrow]

is that what u got?

[lime kestrel]

yup

[acoustic yarrow]

u r correct then

[lime kestrel]

thankss