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[distant nymph]

Then she changed this to y=^3sqrt(5x+25)

So if I do the same, I keep 2/3 for this one?

Nice

[lapis ingot]

there is no square root

it's a cube root

[distant nymph]

ohh, it makes sense

thx

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[unkempt nova]

hi, if I replace coefficients of each term in a series by their complex conjugate, will the radius of convergence remain the same

[wary stream]

Not a vacant channel

[slender wedge]

oops

[vale wigeon]

yes

[unkempt nova]

but whyyy

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@unkempt nova Has your question been resolved?

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[scarlet anvil]

im stuck again
This time theres 3 brackets

[scarlet anvil]

so I know
it will be > 0 if all three brackets is positive

or

2 of the brackets are negative and 1 positive

for example if all brackets are positive
im using and since all conditions must be met together
x -pi > 0 and x + 5 > 0 and x - 3 > 0

but then im stuck

I know how to solve it graphically, but I'm looking for a more algebraic/logical way of doing it. Thanks in advance.

[naive valley]

notice that you can simplify the condition "x-pi > 0 and x + 5 > 0 and x - 3 > 0"

it means: x > pi and x > -5 and x > 3

clearly if x > pi then it's also > the other two

so the condition is the same as just x > pi

for the other case where two are negative and one is positive, i suggest drawing the three points on the x axis, -5, 3 and pi

those points divide the x axis into 4 intervals

check each interval to see if the condition holds in that interval

[scarlet anvil]

I understand how to do it now

thank you

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[calm gale]

Hello

[calm gale]

for this question

Arent v1 and v2 already normalised?

.close

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[warm zephyr]

hello

for this question

#13

doesn't y=x bisect the area into two equal pieces

under the curve and the curve towards the y-axis

so why does the axis of rotation matter?

it produces diff results

shouldnt the volume when rotating across the x-axis equal the volume when rotating around the y?

[sly timber]

It should be equal ?

[warm zephyr]

#3 is rotated across the y-axis

[rigid smelt]

no they should not be equal, the area that you are seeing has nothing to do with the volume, when you rotate that region around the y axis, you would get the solid that is on the outside of a cone and is bounded inside of a cyclinder, however, when rotating around the x-axis, you would get the solid inside of a cone

[warm zephyr]

I can visualize the outside solid and the inside solid

but doesn't the washer method derive from placing an infinite amount of circles on the axis

so the radius should still be equal no matter which side of the axis?

[rigid smelt]

not sure what you are trying to say there

if you want to use washer method to find the volume when rotating around the y-axis, you have to integrate wrt y

meaning the integral then would be pi*y^2dy, integrating from 0 to 2

(the bound 0 to 2 here is according to y btw, not x, it's just that it's also coincidentally the bound in x terms)

anyway, as you can see, the integral in this case is very much different from the integral when using shell method

[warm zephyr]

i think im kinda seeing it

for the diagram on #3 tho

I visually flipped the axis

so the new curve would be bounded to the new y-axis

the new integral was under 4-x^2 because top minus bottom

but im confused on that step

what's the point of the 4

if the area is still the same

[rigid smelt]

ok lemme just reassure something, we are revolving around the y-axis here, that's for sure, but which method are you trying to use?

[warm zephyr]

i used both the shell and washer method

[rigid smelt]

ok but lets just agree on one method here, so we dont get confuse, because the main thing that you don't understand is why the volume when changing the axis of revolution changes

[warm zephyr]

yea

im conceptually understanding that the volume should be the same no matter which axes of rotation

i was thinking of a series of rotations of the lines

[rigid smelt]

no, it does not

[rigid smelt]

here if you tried to revolve this region around the x-axis instead, the volume is not the same

test $\int$

god the bot is still dead

alright, i'll just write it out

[warm zephyr]

I wrote this down

its just that isnt the stole purpose of the 2^2 - y^2

to find the radius or smth like that

but wouldnt the radius always be equal because y=x?

[rigid smelt]

no, the washer method do not care about the radius whatsoever, the disc-washer mehod (like the name suggests) works disks with arbitrary area, which is then being washed along the axis of revolution, creating the solid that it forms

so pi*2^2 actually describe the area of the disk with the radius 2

and pi*y^2 describe the disk bounded below that line x=y and the y-axis with arbitrary radius y

so now when taking the integral of pi*2^2 wrt y, it sweeps out a cyclinder, this is because the radius of the disk is constant throughout, and the height of this cyclinder is 2(based on the bound)

the integral of pi*y^2 on the other hand gives you the volume of a cone, a cone with the lattice line being y=x

so now, the difference between those two volumes gives us the volume that the question asks

so all in all, that's what 2^2-y^2 actually is

[warm zephyr]

ohhhhh

ok

tyvm

i kinda got enlightened XD

.close

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[steady raven]

im in desperate need

for help

please can someone help

[rustic ruin]

Use discriminant formula

[modern topaz]

no real solution🤏

when discriminant is negative

[lone heart]

@steady raven Has your question been resolved?

[steady raven]

o

[steady raven]

i need to find M

[modern topaz]

b^2-4ac

is the discriminant

[steady raven]

o

[steady raven]

is m=-1

[modern topaz]

didnt calculate, but there should be a range for a

cause b and c are given

u have an inequality for a

4-4a>=0

1-a>=0

1>=a

a is m

btw

oh wait, no solution

so just set the thing <= 0

same idea

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@steady raven Has your question been resolved?

[steady raven]

o

[lone heart]

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[lapis ingot]

you can prove it with epsilon delta definition of a limit

[rose sigil]

a_n? are you looking at a sequence or a real valued function?

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[supple solstice]

can someone help me with my act math questions?

[supple solstice]

I have like 6

https://cdn.discordapp.com/attachments/486844829209198613/1033991332903006208/unknown.png

[desert tusk]

Do you know what’s the formula of a circle equation

[lapis ingot]

the equation for a circle centered at (h,k) is (x-h)^2+(y-k)^2=r^2

[desert tusk]

and do you know what each variable means

if yes, then it’s obvious

[supple solstice]

yes

but I'm confused if it's square root or not

[lapis ingot]

^

so

[desert tusk]

the k and h can be anything

for example

I can mark a point’s coordinate as an irrational number

and it’s totally fine

[lapis ingot]

they're not asking for any info about the radius

only the center

[supple solstice]

Okay so, not square root

[lapis ingot]

also

be careful

[supple solstice]

abt?

[lapis ingot]

it's (x**-h)^2+(y-**k)^2=r^2

be careful with the negative signs

[supple solstice]

I'm gonna be honest, I'm in high school and I've never been given a worksheet even closely related to this

so it's x minus negative h?

and y minus neg k?

[lapis ingot]

no

so this is the circle centered at (h,k)

so

[supple solstice]

h and k are just place vars right?

[lapis ingot]

you can rewrite this expression to match with the formula

[supple solstice]

so they could also be x and y

[lapis ingot]

no they're not independent variables

they're constants

$$(x+5)^2+(y-5)^2=5$$

[ocean seal]

Asagao 朝顔

[lapis ingot]

is the expression

so

[supple solstice]

so

[lapis ingot]

you can rewrite it as

[supple solstice]

(x-5)^2+(y--5)^2=r^2

Okay so it would be sqrt of 5 for the answer?

[lapis ingot]

no

[supple solstice]

But that's r^2

[lapis ingot]

nobody asked about the radius

._.

also look

$$(x-(-5))^2+(y-5)^2=5$$

[ocean seal]

Asagao 朝顔

[lapis ingot]

• turns into - -

so the center of the circle is (-5,5)

yes

it is the distance formula

since a circle

is all of the points

that are a certain distance from the center

hence the resemblance

[lone heart]

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Please don't occupy multiple help channels.

[serene junco]

hey

You opened up two channels and didn't ask a question in either one

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[alpine sable]

Yesterday I sold 5 apples for the total price of $20 making it $4 per apple.
Today I sold 10 apples for the total price of $10 making it $1 per apple.

Is the average price per apple is $30/15 = $2 ? Or do I need to weight average it differently?

[vale wigeon]

the average selling price per apple over the course of these two days is $2.00

[alpine sable]

ok thank you.

.close

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[outer fractal]

I dont't quite understand how P(S) and P(G) can be greater than 1. Is there somthing wrong with this question, or am i missing something?

[autumn basalt]

They’re not greater than 1?

[outer fractal]

When summed?

[autumn basalt]

Do they need to?

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[alpine sable]

i dont get this rule that we can go through the asymptote if if it is not at extremities!??

[alpine sable]

isnt asympotote a line shouldnt be passed so the function can exist

[clear trench]

An asymptote is a line that cannot be reached by the graph.

The graph is moving towards that line, but will never touch it, not even in the infinity.

[alpine sable]

yes\

but here you can at finite values

sometimes

[alpine sable]

idfg it, it makes no sense

what difference there is. unless it is a piece wise function where there are conditions, then i get it

[clear trench]

Well sometimes graphs can cross or touch horizontal asymptotes

never for vertical asymptotes though.

[alpine sable]

you have any clue why?

[alpine sable]

ok good to know that atleast tbh

tysmm broski

[clear trench]

Well, it is possible to cross horizontal asymptotes as long as the function turns around and kind of follows that asymptote.

An asymptote is a line that the graph moves towards

vertical asymptotes will never be crossed

but for functions like yours, they can cross the horizontal asymptote as long as the direction eventually changes back.

[alpine sable]

oh

[clear trench]

the graph will always follow the asymptote eventually.

[alpine sable]

i think it is a rational function? where oblique asymptote are included etc..

for it to look like this

we took part of rational but it isnt included in the test

i just remember that the rational function graphs look weird asf so ig this oe is one of these cases

[clear trench]

Indeed.

[alpine sable]

F* it then, ama worry about this in the next test/quiz

🙂

[clear trench]

Here is a nice example.

[alpine sable]

there is change of direction etc..

i think get it

ty mike

[clear trench]

Yea, the left half crosses the asymptote but turns back around to follow it either way.

No problem!

[alpine sable]

ig i can close?

.close

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[sly echo]

anytime i plug in the formula and work it out i always get a very small number

[sly echo]

like 0.0002

[vale wigeon]

what formula are you plugging things into?

show the calculation you are performing.

chances are you are doing it incorrectly.

[sly echo]

m = kh^3

1.2 = k x 29^3

1.2 = k x 24389

divide by 24389 on both sides

[vale wigeon]

ok, sure, that'll get you the value of k.

you're not passing that off as the final answer, are you?

[sly echo]

no no

8.374626166475617 it gives me this

so i round to 1 dp like question asks

[vale wigeon]

it should not be giving you that.

[sly echo]

8.4

[vale wigeon]

hold on.

it's too early to round!

this is an intermediate calculation.

[sly echo]

oh ok

[vale wigeon]

and 1.2/29^3 isn't 8.4 anyway

nor in that ballpark.

,calc 1.2/29^3

[ocean seal]

Result:

4.9202509327976e-5

[vale wigeon]

it's 4.92ish * 10^-5

[sly echo]

ohh, it gave me that and i pressed enter again cos i thought it meant 8.374626166475617

cos i didnt know what e was

[vale wigeon]

look up scientific notation in your spare time

anyway imo it's better to do it in a different way

[sly echo]

convert into grams?

[vale wigeon]

$m \propto h^3$, so $h \propto \sqrt[3]{m}$

[ocean seal]

Ann

[vale wigeon]

no, don't convert any units.

[sly echo]

oh ok

[vale wigeon]

the height of the new doll is equal to that of the old doll times the cube root of the ratio of their massses

,calc 29 * (0.5/1.2)^(1/3)

[ocean seal]

Result:

21.660122941693

[sly echo]

would that be the answer when i round it?

ah ye it was

thank you very much

.close

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[fresh horizon]

i just asked helped for this but still dk how to do it

pls assist

[velvet cliff]

ok so sin(2theta - 30) = -1/2

we got that far before

what's arcsin(-1/2)

[fresh horizon]

idk

[velvet cliff]

30

worth remembering

[fresh horizon]

o ok

[velvet cliff]

it's the same as arcsin 1/2

they just go in opposite direcitons from the origin

[fresh horizon]

i see

is arcsin just sin inverse?

[velvet cliff]

wait sorry I meant -30

[velvet cliff]

yeah

[fresh horizon]

okok

[velvet cliff]

from -30, how would you find the next value on the sin wave

[fresh horizon]

add the period?

[velvet cliff]

hmm, that's a way to find a value

however there is a solution inbetween

[fresh horizon]

am confusion

[velvet cliff]

bare with me

[fresh horizon]

allg

these r the answers btw

[velvet cliff]

you see the value I highlighted in blue

you skipped it

[fresh horizon]

ah

i stil understand nothing

[velvet cliff]

what don't you udnerstand

[fresh horizon]

is there not a way to solve for theta algebraically

[velvet cliff]

yes

I'm just illustrating why there's a point in between

[fresh horizon]

okok

pls hurry

[velvet cliff]

ok

that value inbetween

because if you look at the origin

wait there's an easier way to explain

so 330 degrees s a solution

cus -30 + 360

so the other solutoin, would be 330 degrees mirrored about the minimum

which is at 270 degrees

so what would it be

[fresh horizon]

idk

[velvet cliff]

what's the distance from 3 30 to 270

[fresh horizon]

60

[velvet cliff]

yeah

so since 270 is a minimum

the corresponding value would be symmetrical about it

in other words 270 - 60

so 210 degrees is a solution

[fresh horizon]

it's not tho

[velvet cliff]

it is

this is equal to 2theta - 30

[fresh horizon]

but the answers are

[velvet cliff]

well not in this case since it's out of the bound

you're not listening

[velvet cliff]

.

how woudl you find the next value

[fresh horizon]

oh right

wdym

[velvet cliff]

where does y = -1/2 next intersect sin x

we have =30, 210 and 330

[fresh horizon]

not sure

[velvet cliff]

it's easy now

you just add the period to 210, then 330

then repeat

[fresh horizon]

300?

[velvet cliff]

?

[fresh horizon]

bro what

im so slow

[velvet cliff]

what's the period

[fresh horizon]

180

or pi

[velvet cliff]

no

360

[fresh horizon]

???

how

[velvet cliff]

it repeats every 360 degrees

[fresh horizon]

isn't the period 2pi divided by the absolute value of b

[velvet cliff]

and what's b here

[fresh horizon]

2

[velvet cliff]

how

where'd you get tha from

[fresh horizon]

2theta

[velvet cliff]

no

forget abotu that 2theta + 30

just call it x

[fresh horizon]

ok

[velvet cliff]

and then we equate x to 2theta + 30 after we get enough oslutions

[fresh horizon]

okok

[velvet cliff]

so we have 210 degrees and 330 degrees so far

what's the next one

[fresh horizon]

570

[velvet cliff]

yeah nice

and then?

[fresh horizon]

690

[velvet cliff]

mhm, remember that our upper bound for theta is 540 degrees, so in other words are upper bound for x is 1110, because 2(540) + 30

so we need to keep getting solutions until we reach that bound for x

[velvet cliff]

what would be next

[fresh horizon]

makes sense

930

[velvet cliff]

ye, last one now?

[fresh horizon]

1290

[velvet cliff]

there's one inbetween 930 and 1290

add the period not to 930, but the value before

[fresh horizon]

930

[velvet cliff]

no this is the value before

[fresh horizon]

1050

it's fine

this is just confusing me more

dw ab it ill ask my teacher

[velvet cliff]

oh ok

[fresh horizon]

the way i learnt to do it is just find the middle theta values and add or subtract the period

[velvet cliff]

we're nearly done bro

[fresh horizon]

perhaps that doesn't apply here

[velvet cliff]

we just need to equate the values to 2theta - 30

and then sovle for ethta

[fresh horizon]

can u just explain it and ill listen

like u solve the question and ill see what u did

[velvet cliff]

ok so values 2theta - 30 = 210, 330, 570, 690, 930, 1050

2theta = 240, 360, 600, 720, 960, 1080

theta = 120, 180, 300, 360, 480, 540

120 is not in the bound

therefore theta = 180,300,360,480,540

[fresh horizon]

i see

hm

ight ty

don't understand the way u did it

but coolio

.close

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[calm olive]

help

[gray isle]

um where's this coming from?

[ornate condor]

wot

[gray isle]

this looks like a butchered attempt at applying the quadratic formula

[calm olive]

nowhere really i just found it so i can't solve it

[ornate condor]

maybe

this is the original equation

[calm olive]

can i like solve it

[velvet cliff]

to solve it you'd need another quadratic equation, and 2 cases for the plus or minus

it looks like you just applied the qudratic formula wrong

[calm olive]

so that

question is just total bs

[gray isle]

nowhere really i just found it so i can't solve it
found it where

[calm olive]

in some rnadom server someone sent it

and i wanted to try it out but no idea

[gray isle]

you can manipulate like other equations, it is solvable

[calm olive]

i can't really seem to be getting anywher

[gray isle]

what have you managed to do so far

[calm olive]

let me show

[gray isle]

let me double check

[calm olive]

4x=-2x±√ 2x+2√ x

[gray isle]

can you take a pic

[calm olive]

alright

[rustic ruin]

How can both and plus and minus give the result as 2

[calm olive]

here

[gray isle]

you can't split the root like that

[calm olive]

uh

really

i always thought we could

hi help pls

nvm

i'm so fucking dumb what

alright..

where to go from here?

hi plz

someone

[fringe mural]

quadratic formula?

[calm olive]

uh yeah i think so

[fringe mural]

can u give the equation?

btw in the quadratic formula ur not supposed to take variables

only coefficient

[calm olive]

it's not really that

it's just that

the question is just

different af

idk if i should even consider this as a question or not..

nvm this is just bullshit wtf

.close

[lone heart]

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[red belfry]

$$x_{n+2} = \frac{1}{2} (x_n + x_{n+1}) ; for ;all ;n \in\mathbb{N}$$

[ocean seal]

m_g

[red belfry]

How do I calculcate the limit of this sequence?

[vale wigeon]

NEI

[lone heart]

@red belfry Has your question been resolved?

[red belfry]

<@&286206848099549185>

[gray isle]

is there more info

[red belfry]

$x_1 = 1$ : and : $x_2 = 2$

[ocean seal]

m_g

[red belfry]

this

[lone heart]

@red belfry Has your question been resolved?

[raven kraken]

so this is a recursive

equation

so for something like this you would right out the first few terms and find a pattern here

[lone heart]

@red belfry Has your question been resolved?

[lone heart]

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[grizzled ember]

Reserved

[grizzled ember]

I'm confused by the wording of the question kind of. This is a basic combinatorics question.

Is the union of these sets the number of ways you could pick the cards?

[vague spindle]

yes

total 15 ways

actually 16

there are 13 hearts

[grizzled ember]

Yes but 1 heart is also a queen

So that is the intersection of both sets

So only 15 I believe

[vague spindle]

I know, I'm saying there are 13 total hearts, not 12

[grizzled ember]

Isn't there only 12 of each suit?

[vague spindle]

no, 13

1(A)-10, J, Q, K

[grizzled ember]

Oh yeah woops

You're right

Thank you @vague spindle

.close

[lone heart]

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[grizzled ember]

Reserved

[grizzled ember]

In question a am I only looking for the number of combinations? Then in b I need to look for the permutation of 5 books from the set of 10 right?

If so I answered these wrong.

Upon rereading these I think I misunderstood them.

These combinatorics word problems are kicking my butt. I have a hard time knowing how to model them in my head

[vague spindle]

it's easy. when the order matters, use permutation. otherwise, combination

[grizzled ember]

Right

[vague spindle]

thus, first one is not permutation

[grizzled ember]

Right

That's just the number of combinations of 5 novels you could make from the 10 then correct.

[vague spindle]

yeah so your answer is wrong

[grizzled ember]

Yeah

I will fix thanks

[vague spindle]

but b is permutation

[vague spindle]

correct

[grizzled ember]

Thank you so much for your help sir

I am working through a lab that is worth 9% of my grade

And it's due tonight

I'm working on it while at work lol

So I'm kinda stressing

@vague spindle may I send you my lab after and you can review my work?

It's like 11 questions

[vague spindle]

i'm going to be offline in a few minutes and most of today.

[grizzled ember]

Oh ok

[vague spindle]

i can review tomorrow

[grizzled ember]

No worries yeah I will send you it. If you have time whenever just give it a look tyty

.close

[lone heart]

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[surreal meteor]

Is this essentially just like going for each of the questions and like determining them?

So not done yet but like something like this

[echo socket]

You have to show why 0 < x < 1 implies x^3 being smaller in that case

[surreal meteor]

so just calcualtes values

[echo socket]

But there's just an easier way to disprove the statement: just show a counter example

[surreal meteor]

0.5^3=0.125

[echo socket]

x = 1, -1 or 0 works

[surreal meteor]

0.5^2=0,25

hence x^3 is NOT greater than

[echo socket]

Yeah any counterexample works

[surreal meteor]

x^2

Okay so that's fine

So that's fine? but what for the rest?

[echo socket]

Don't forget to add for all x*

[echo socket]

Ah, wait

It's asking you to find what's wrong with the proof

Well the problem is that (x - 1) could be negative and the signs could have changed

[echo socket]

Or it coud be zero

So x^2 > 0 doesn't imply x^2(x - 1) > 0

[surreal meteor]

how is this true though?

[echo socket]

The case x = 1 makes it clear

[surreal meteor]

so first of all

should i delete this

[echo socket]

It's not true, it's an assumption about x

The definition of it

[surreal meteor]

Well if it's not true, then isn't that what's wrong with the thing?

[echo socket]

Yeah I'd just remove everything except for the counterexample

[surreal meteor]

Okay i have this now

[echo socket]

The statement says "For all x such that x^2 > 0 we have x^2(x - 1) > 0"

[surreal meteor]

so we have an assumption for all x x^2 > 0

Ah yes

[echo socket]

Add "Consider the case x = 0.5 at the beginning

[echo socket]

So you have to show why this is false

[echo socket]

Without the "at the beginning" mb lol

[surreal meteor]

But wouldn't this ALWAYS return zero?

[echo socket]

Yeah, 0(x - 1) = 0

[surreal meteor]

1^2(1-1)

And so if it's one it's 0 > 0

[echo socket]

Yes, which is false

[surreal meteor]

And that's it?

[echo socket]

Yeah

Don't forget to mention that 1^2 > 0

Otherwise it wouldn't be a valid value for x

[surreal meteor]

What

[echo socket]

It says "For all x such that x^2 > 0"

[echo socket]

So we have to consider those values

[surreal meteor]

Ye okay

So do i just test it before

[echo socket]

Yeah just mention it somewhere, perhaps in brackets

"considering x = 1 (1^2 > 0)"

[surreal meteor]

So like this?

[echo socket]

That's like really early though, put it after stating the statement

[surreal meteor]

don't i do the first statement

then move on

so like

[echo socket]

Yes but the entire statement is an implication of those statements

[surreal meteor]

What about this

[echo socket]

Before you prove/disprove a statement you have to present the entire statement

[surreal meteor]

Ye makes sense

So I did present the whole statement before now

[echo socket]

Remove "for all x" in "we have for all x x^2(x - 1) > 0(x - 1)"

[surreal meteor]

And then strirll did the first one

Like this?

[echo socket]

Yeah, just remove the "For all x such that x^2 > 0" (the 3rd line), it's unnecessary

[lone heart]

@surreal meteor Has your question been resolved?

[lone heart]

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[lone heart]

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[lone heart]

Please don't occupy multiple help channels.

@alpine sable Has your question been resolved?

[vague mirage]

@dark marten hey bro make new help channel this is not allowed

[dark marten]

I'm sorry I'm new here

[vague mirage]

#help-46 @dark marten go there

[alpine sable]

can any of u guys help me ?

[vague mirage]

Sorry bro it's out of my league

[lone heart]

@lament mantle Has your question been resolved?

[lone heart]

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[devout creek]

(a poor attempt by me)

[devout creek]

I'm kind of new to this, how do I rearrange and change the subject on this formula?

This is the original formula, I'm trying to find qn

[high rapids]

Multiply both sides by the things given in the denom first

[devout creek]

I didn't quite understand :/

[lone heart]

@devout creek Has your question been resolved?

[devout creek]

!solved

[lone heart]

@devout creek Has your question been resolved?

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[tardy hollow]

I need to find angle of intersection between the level curves of $f(x, y) =\frac{x-y}{xy}$ and $g(x, y) = x^3+y^3$

[ocean seal]

RockLEE

[tardy hollow]

Do i just use random constants for level curve and solve for a variable to find point of intersection?

Kinda confused on how to start this question

[modern topaz]

u set the two things to be equal

[tardy hollow]

So we do $\frac{x-y}{xy}=x^3+y^3$

[ocean seal]

RockLEE

[tardy hollow]

damn thats kinda difficult to calculate

so its $x-y=x^4y + y^4x$

[ocean seal]

RockLEE

[tardy hollow]

its too difficult to solve this lmao

is this the right way??

clearly x=0, y=0 is a solution

and it seems to me that x=y, and y=x is also a solution actually i lied this is not true

[lone heart]

@tardy hollow Has your question been resolved?

[modern topaz]

no

xy cant be 0

denominator

ur solving homogenous equation

u prob never learned this before

x-y=xy(x^3+y^3)

let x=ty

u then get

(t-1)y=t(t^3+1)y^5

divide y by both sides

y^4=t-1/t(t^3+1)

y= quartic root thing t-1/t(t^3+1)

x=+-t* t-1/t(t^3+1)

so first conclusion is that there are infinite numbers of overlapping points

[tardy hollow]

wtf oh okay

[modern topaz]

notice this t-1/t(t^3+1)>0

u have to divide into cases

to solve this inequality

t>1 pr t<1

im just gonna give u the solution to inequality cause im not typing

t is either

-1<t<0

or

t>1

and now ur done

so all y= quartic root thing t-1/t(t^3+1) and x=+-t* t-1/t(t^3+1) satisfying -1<t<0 or t>1

@tardy hollow this technique is called parametric equation

[tardy hollow]

okay but if there are infinite point of overlapping points then isnt there infinite angles of intesection???

[modern topaz]

??? wdym

angle

u can try to plot the graphs out

on desmos

[tardy hollow]

cuz the question asks me to find angle of intersection

hm okay letme see on geogebra

wtf lmao

[lone heart]

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[lone heart]

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[dense plume]

im not sure how to go about this question. I equated the two functions and substituted x from the interval but didn't get the answers below. Can anyone explain this step by step to me?

[lone heart]

@dense plume Has your question been resolved?

[dense plume]

<@&286206848099549185>

[wet mulch]

you get this?

[dense plume]

no i didn't

[wet mulch]

mmm ok

so you equated them and solveds for k?

[dense plume]

yeah

but I didn't get any of those answers

[wet mulch]

$e^14k = 14.5$
$14k = \ln{14.5}$
$k = \frac{\ln{14.5}}{14}$

[ocean seal]

OGPanda

[wet mulch]

ope

pretend those are on different lines lol

[dense plume]

is that suppose to be e^14k?

[wet mulch]

yeah totally

my bad

i suck at latex

[dense plume]

okay just making sure

ahh

i totally see my mistake

[wet mulch]

👌

sweet

[dense plume]

thank you so much

i feel kinda dumb now lol

[wet mulch]

lol np

simple mistakes are the ones that always get me too 💀

[lone heart]

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[hexed hawk]

hi

[hexed hawk]

<@&286206848099549185>

[lone heart]

@hexed hawk Has your question been resolved?

[viral field]

uhh

[lone heart]

@hexed hawk Has your question been resolved?

[hexed hawk]

@ember dagger ?

[ember dagger]

??

[hexed hawk]

.close

[lone heart]

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[alpine sable]

HOnestly have been awol form class, any hints on how to go about this problem even?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[lone heart]

@alpine sable Has your question been resolved?

[near saffron]

@alpine sable what step are you on

[alpine sable]

1. Been working on other work my apologies

[near saffron]

@alpine sable For 1 it wants you to rewrite the function as tan(r(x)) = x, nothing more than that.
For number 2, you could think of r(x) as y, and this is like implicit differentiation. So when you differentiate the left-hand side tan(r(x)), you get sec^2 (r(x)) * r'(x) because of the chain rule.

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

sorry just saw this, i’ll continue the problem w my questions tommorow night when i arrive home

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

.open

[hybrid smelt]

Hi

[alpine sable]

For #1 how would rewriting the function as tan(r(x))=x use th relatinoship of arctan and tangent functions?

Not doubting your validity just wondering

[lone heart]

@alpine sable Has your question been resolved?

[serene quest]

help

I have an F in math 9th grade and im willing to get more knowledge

[neon vector]

help

[tardy coyote]

If f and g are inverses, then f(g(x)) = x and g(f(x)) = x.

It is so that tan and arctan are inverses (on a specific domain), so tan(arctan(x)) = x.

From r(x) = arctan(x), you can apply tan to both sides of the equation to get tan(r(x)) = tan(arctan(x)) = x.

[foggy wagon]

.open

H(x) = 1+ 1/x+1
h=gof , g= 1/x
So what is f(x) ?

help

[neon vector]

.open

[alpine sable]

Hello guys, I wonder if you could help me to solve this problem?

Its about 3g

[humble gazelle]

@alpine sable given you're posting in someone else's help channel, no

#❓how-to-get-help

[alpine sable]

ahhhhh sorry bro

[alpine sable]

Right not sure hwo this ticket is still open but it is quite helpful

I was able to get the first two done

I just don't know how to go about 3-6

[near saffron]

@alpine sable 3 says write r'(x) as a trigonometric function evaluated at r(x). This means we're going to write it as something that looks like sin(r(x)), cot(r(x)), something like that.
From 2 we already have that sec^2 (r(x)) * r'(x) = 1. We solve for r'(x) and then rewrite the expression we get using what we know about trigonometric functions.

[spice hornet]

hello

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

@alpine sable are h ok ?

[lethal olive]

Can anyone explain to me what divisible is??😭

[lone heart]

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[alpine sable]

x/x-1=2/x+1?

[wide raven]

Denominators can't be zero and cross multiply

[rose sigil]

[alpine sable]

number 4 and number 5 😞

[wide raven]

LCM and denominators can't be zero and cross multiply+ solve for 5

[alpine sable]

i don't know how to solve this

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@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

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[crude jungle]

I was taking the first derivative to show its decreasing

[crude jungle]

I’m kinda lost idk where to go from here

That’s the first derivative

nvm

got it

.close

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[hushed raven]

i need help with all of them 😦

im dumb

[gloomy thorn]

whats those squiggly brackets for

[hushed raven]

{

i think

[valid ivy]

Absolute value

thats the distance from 0

ghost try to do 9

start by opening the brackets

[gloomy thorn]

ur not dumb bro everybody needs practice

[valid ivy]

try to do exercise number nine ghost
begin by opening the brackets

[hushed raven]

.close

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[sharp nebula]

Can someone help me solve this equation

I’m stuck

[lone heart]

@sharp nebula Has your question been resolved?

[sharp nebula]

<@&286206848099549185>

[empty ridge]

@sharp nebula What is your question?

[sharp nebula]

Here can you solve for x

I’m stuck

[empty ridge]

@sharp nebula It helps to call the common term A and then execute the same strategy (I think that the last line is not correct, there should be just three nested square roots, not four). I'm trying to keep repeating this same strategy to eliminate the square roots... but i'm not seeing a pretty thing...

[lone heart]

@sharp nebula Has your question been resolved?

[empty ridge]

This problem is infernal... sorry ... I can't help you... There must be a better way to solve this....

[lone heart]

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[lone heart]

@vale blaze Has your question been resolved?

[vale blaze]

someone helps?

[inland plank]

any effort?

[tacit arch]

• Show your work, and if possible, explain where you are stuck.

[vale blaze]

i don't have any work to show bc i don't know where to start

[tacit arch]

put everything over a common denominator

[vale blaze]

ok now i have 2x - x^2 - ax / 2x (x+a)

what's then?

[inland plank]

simplify it and calculate the limit

[tacit arch]

find the value of a for which that limit equals -1/4

[lone heart]

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[vale blaze]

how

[vale blaze]

i know that i have to find a😂 but i don’t know how

[inland plank]

divide denom and numerator by x

and then it should be possible to plug in 0

[tacit arch]

did you try taking the limit at least?

[tacit arch]

review the limit laws:
https://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx

[vale blaze]

thanks for help i solved

[alpine sable]

@vale blaze good job

.close

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[sacred wyvern]

I forgot how to convert kelvin into celsius.
Yup, i know i am stupid as hell.

[inland plank]

you can google easily

[vague coral]

google it

[sacred wyvern]

i forgot about that, alr thanks

.close

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[edgy path]

hi

[lone heart]

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[frigid anvil]

Simple physics question, help me with a contradiction. What is negative/positive? The anode or the cathode?

[minor needle]

anode +, cathode -

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[hushed mica]

Guys any hint on how to start to proof $$a-b|a^n-b^n$$

[ocean seal]

Night Wølf

[hushed mica]

Using Modular arithmetic

[tall topaz]

Start by a-b=0 mod(a-b)

Where = is the congruent symbol

[hushed mica]

Ok next?

@tall topaz

[tall topaz]

a=b mod a-b

[hushed mica]

Why how?

[tall topaz]

That’s just moving b to the other side

[hushed mica]

This can be done in this sort of equations?

[tall topaz]

Yea

[hushed mica]

Ic

Ok so next?

[tall topaz]

Raise both sides to the power of n

[hushed mica]

Done

Then?

[tall topaz]

a^n = b^n mod a-b => a^n -b^n = 0 mod a-b

So a^n -b^n is divisible by a-b

[hushed mica]

Wait what this was ez lol

[tall topaz]

[hushed mica]

Let f be a polynomial with integer coefficients. Show
that a − b | f(a) − f(b) for any integers a, b. This is the same as saying f(a + d) ≡ f(a)
(mod d).

Is there any difference on how to approach this one cause this has funcs?

@tall topaz

[tall topaz]

I gtg sorry

[hushed mica]

Oh np thanks for the help btw 😄

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[mellow sorrel]

Use Long Division to divide (5x^3 + 22x^2 + 7x – 4) by (5x^2 + 2x – 1)

[mellow sorrel]

How would I use long division with such a large equation?

[marsh rapids]

It isn't large. Just do

If you're at the point where you're learning long division and you're scared by a sum of 4 things, any computational practice would be most beneficial

[sullen dome]

Since we have a lower degree at the bottom polynominal division would be possible?

[marsh rapids]

It is always possible anyways. It's just that otherwise the quotient is 0

[sullen dome]

Hmm interesting always thought it had to be the same or a lower degree

[marsh rapids]

You can divide x^2 by x^3

You just get 0*x^3 + x^2

It's just that this would just about never be an exercise, since there's nothing to do

It's like dividing 4 by 7. The euclidian division is doable, it's just that there's nothing to do

[mellow sorrel]

tyty

.close

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[alpine sable]

just checking my answers

[tacit arch]

,w plot 3sin(3x) - cos(3x)

[ocean seal]

Results provided by WolframAlpha

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Requested by riemann#2529

[tacit arch]

,w plot sqrt(10)*cos(3x+1.25)

[ocean seal]

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Requested by riemann#2529

[tacit arch]

looks like you messed up some algebra somewhere

[alpine sable]

i got the same graph

[tacit arch]

notice they're not the same

[alpine sable]

what asnwer did you get

[worn bear]

uhhh

[tacit arch]

not how this works. show your work and someone can point out your mistake

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[alpine sable]

.close

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[desert island]

So ik that the series converges but the sum is incorrect somehow

[tacit arch]

you have to split the series first

sum[ 1/7^n ] + sum[ 3^n / 7^n ]

[desert island]

i see now thanks

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[alpine sable]

Need help with this, similar exercise but not the same, rotation needs to be 45 degrees and i need to do 2

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[alpine sable]

<@&286206848099549185>

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[alpine sable]

<@&286206848099549185>

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[alpine sable]

.close

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[alpine sable]

most likely not getting an answer

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[hot moss]

domain D, and let a be a cluster point of D. Suppose that for all x ∈ D, g(x) ≤ f(x) ≤ h(x)
and that lim x→a
g(x) = lim x→a
h(x) = L. Then lim x→a
f(x) = L.```

[hot moss]

I am not sure how to do this at all!

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[hot moss]

@crisp iron

.close

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[alpine sable]

so that cartesian equation i made is half of it

its the top portion