help-13

[cedar kiln]

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[fallow gorge]

System of differential equations

Is what I am doing correct?

[cedar kiln]

@fallow gorge Has your question been resolved?

[cedar kiln]

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[ember warren]

Trying to simplify this, got 16^x.

[ember warren]

Teach said the right answer is 2^8x-4/x

[jade sonnet]

Do you have your Working

[ember warren]

Ah yes, 2^(4x)*4^(2x)/16^x = 16^x times 16^x/16x

And then I cancelled the denominator and numerator

(bottom number)

[gentle flower]

are you familiar with exponent rules

[ember warren]

yes

[gentle flower]

can you find a way to apply them here?

[ember warren]

I thought of doing the product rule but apparently thats wrong, then I went to power of a power rule to get the answer I got

[gentle flower]

can you convert everything to base 2 in your problem?

[ember warren]

yes, the 4?

[gentle flower]

even the 16

[ember warren]

technically yes, thats true as well, thought you meant the base question

So I just have to convert everything to two and divide 256^x by 16^x?

[gentle flower]

$\frac {2^{4x} \cdot (2^{2})^{2x}} {(2^4)^{x}}$

[wraith dagger]

! saad

[gentle flower]

like this

[civic eagle]

if this assignment is specifically for practicing changing of base i can see that being the answer. but your answer 16x and the teacher's answer is actually equivalent.

[ember warren]

I just need to simplify it as much as possible

Thats my assignment

Currently

[civic eagle]

16^x imo is the most simplified, the teacher's answer keeps the base of 2 instead.

ohhhh 16x. you didn't get it right.

[ember warren]

I wrote wrong

Forgot the ^

[civic eagle]

then yes your answer is equivalent.

[ember warren]

So is this the finished product as well?

[civic eagle]

but to see how your teacher came up with the answer, what @gentle flower is showing you is the correct way to get there.

16^x == 2^[8x-4x]

[ember warren]

Ah okay, he just moved the bottom exponent up and plussed the numerators

Interesting way of doing it

[civic eagle]

the key part is that the denominator is the same as multiplying by 2^-4x.
2^-4x == 1/(2^4x)

imo its not the most simplified, 16^x would be. it must be some goal of the topic youre on to practice this.

[ember warren]

I see, well, ill store it in my head for the future just in case

Thanks guys!

.close

[cedar kiln]

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[forest swan]

if arg(b) = 3pi/4 and |OB| = |OA| = 2
how became arg(o) also 3pi/4 ?

[cedar kiln]

@forest swan Has your question been resolved?

[forest swan]

<@&286206848099549185>

[sage forge]

What is meant with arg(b)?

[forest swan]

$b = 2e^{i 3\pi/4}$

[wraith dagger]

Amer

[stable edge]

the angle needed to point at b going anticlockwise from the positive x axis

[forest swan]

ooooh

i think my problem is only not knowing the definition 🤦‍♂️

ty!

.close

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[crimson sedge]

hello

[crimson sedge]

how do i integrate a thing like this to get the area of almost

0 to 1.5

[bright bridge]

uh

[dull oxide]

Are you trying to get the area inside the loop?

Immediately, it looks like doing it in polar coordinates would be easier

@crimson sedge

[crimson sedge]

yep

[crimson sedge]

idk wat that please explain

[dull oxide]

First, why are you trying to find this area?

[lethal jackal]

might be cool to know the area of flower petals

[inland fossil]

this could be xyproblem lol

http://xyproblem.info

[dull oxide]

Yeah I highly suspect

What is your purpose of doing this, @crimson sedge

[crimson sedge]

because exam

[bright bridge]

oh

[crimson sedge]

and just started studying

[dull oxide]

Wdym by this? It's in an exam? You think it will be? Who gave you this problem and why?

[crimson sedge]

i think things like this may appear

[dull oxide]

Have you had problems similar to this? I can't think of a better way to get the area than to use polar form. But if you have no idea what that is, I have no idea why you think this will be on an exam

[crimson sedge]

i have no idea because of no attention

[dull oxide]

Wdym no attention?

I hope that you can respect my position where you are giving me very few resources to offer help to you

I can help you, but you need to put in more effort. Your explanations are very short and giving me very little

[crimson sedge]

the thing is that i dont attend to some classes

and a friend told me

[dull oxide]

Well yeah. You need to find the area using polar coordinates then

If you don't know how to do that, then you need to study it

[crimson sedge]

thats the thing

what do you recommend for studying that

[crimson sedge]

.

.close

[cedar kiln]

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[thin hedge]

Can someone please explain this problem to me

[runic vector]

Do you know how to calculate the slope of a line?

Consider the formula for it

y2 - y1/
x2 - x1

Or
0 - a/
B - 0

-a / b

[gentle flower]

$\frac {y_2 - y_1}{x_2 - x_1}$

[wraith dagger]

! saad

[runic vector]

$\frac {0 - a}{b - 0}$

[wraith dagger]

Luke.

[cedar kiln]

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.

[runic vector]

Mb

[long idol]

The positive difference of the cubes of two consecutive positive integers is 111
less than five times the product of the two consecutive integers. What is the
sum of the two consecutive integers?

[bright bridge]

huh

[long idol]

i need help

[bright bridge]

!help

[cedar kiln]

Please read #❓how-to-get-help

[bright bridge]

you get help

[nova snow]

You’re the biggest troll ever

[bright bridge]

:/

[nova snow]

Don’t you :/ me I see you go into channels and troll with you selecting every single role

why do you do this

[cedar kiln]

@thin hedge Has your question been resolved?

[cedar kiln]

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[sharp sinew]

for the first boundary value

0 = wC2

right?

[cedar kiln]

@sharp sinew Has your question been resolved?

[cedar kiln]

@sharp sinew Has your question been resolved?

[cedar kiln]

@sharp sinew Has your question been resolved?

[cedar kiln]

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[boreal epoch]

@proud geyser like you have 837373$ dollars and you want to know the "different ways" to change that with those bills

[tropic oxide]

ever done shit with generating functions before?

[boreal epoch]

google recurrence relation, this is definitely something you solve with programming rather a simple math equation

[cedar kiln]

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[boreal epoch]

breh deleted them

well @tropic oxide is there actually a mathematical way to represent what she asked?

[cedar kiln]

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[gray blade]

This is a telescopic series

what have you tried tho?

[long walrus]

I used partial fractions

but i can’t see the pattern when I write some terms out

[grave gazelle]

Bottom expands to n^2 + 4n

n^2 has a higher order so can ignore 4n

When n tends to infinity, what does 1/n^2 become

[long walrus]

that doesn’t give us the value for the series

[grave gazelle]

Oh I get what you mean

[cedar kiln]

@long walrus Has your question been resolved?

[long walrus]

<@&286206848099549185>

[cedar kiln]

@long walrus Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Henlo

[crimson sedge]

Sorry

It's taking so long for the image to upload

But I'll just write it out

After doing limits by conjugation or rationalization

I ended with:

3/(5+sqrt(3x+28))

limit is approaching -1

[long swan]

Ok now just rationalize the denom

[crimson sedge]

Now do I use the limit property of quotient

wait wot

That would reverse the operation I did

Original was 5- sqrt(3x+28)/(x+1)

Plugging in -1 in to the original gave 0/0

[granite eagle]

it seems like you made a error

those two dont check out

[crimson sedge]

Hm

[long swan]

$$\frac{5-\sqrt{3x+28}}{x+1} × \frac{5+\sqrt{3x+28}}{5+\sqrt{3x+28}} = ???$$

[wraith dagger]

Mr. Gamer

[crimson sedge]

Nvm that's right

That's what I did

[long swan]

Check your multiplication

[crimson sedge]

Oh

Well

I get 25-sqrt(3x+28) for the top

Well

[crimson sedge]

The answer I got was right

But I shouldn't use quotient property of limits Forgive me I thought the limit of a constant was 0 but it's the constant itself

So then I would -3/10

.close

[cedar kiln]

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[lofty sky]

can someone pls explain this question

[wind nexus]

with any questions like this, consider what properties of the object you're given you can figure out with the given info

and then see if one of those properties can somehow be interpreted as/related to what the question is asking you to find

[cedar kiln]

@lofty sky Has your question been resolved?

[lofty sky]

<@&286206848099549185>

oh wait

i understand

nvm

[crimson sedge]

Lol

It happened with me too

[lofty sky]

lol

.close

[cedar kiln]

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[cedar kiln]

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[honest bobcat]

is there a general method for questions like these

[honest bobcat]

normal(inverse(x)) = x i think

inverse(normal(x)) = x if x is within the bounds of the inverse

but then for g and h i have no idea

[south tundra]

1. Memorise the values of sрecial angles’ trig values
2. Recall what inverse function means (e.g. cos^-1(cos(-рi/3) is equal to -рi/3
3. For stuff like cos(sin^-1(3/5)) you can draw a right triangle where one of the angles has a sin value of 3/5 and calculate the cos value using рythagorean theorem

[runic garnet]

For something like g, sec^-1 (x) = cos^-1 (1/x)

[south tundra]

You can have a right triangle where the opposite side is 3 and the hypotenuse is 5; or you can have a right triangle where those are 3/5 and 1 correspondingly

Anything, as long as the sin value is 3/5, will work

[cedar kiln]

@honest bobcat Has your question been resolved?

[cedar kiln]

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[woven moon]

Checking my answer on a math question that doesn't have a mark scheme to it. I need someone to help me by verifying if I am right.

Q:
The sun is 1.507 * 10^9 km away from the Earth.
Mercury is 9.37 * 10^8 km away from the Earth. What is the distance between the sun and Mercury assuming that they are colinear?

My answer:
1.507 * 10^9 + 0.937 * 10^9 = 2.444 * 10^9 km

[mystic bluff]

Mercury lies between the sun and the Earth.

So it should be minus.

@woven moon

[woven moon]

Since this is a math question I really doubt knowledge of this information would be tested

Also this question was made by my own math teacher, so... I can understand any scientific errors

[fair geyser]

well it's clearly being tested

even without this fact you gave one answer out of two possible

often marked down in math

[woven moon]

.close

[cedar kiln]

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[desert zephyr]

is this proof correct? I want to prove RoS is symmetrical if and only if RoS = SoR

[radiant topaz]

It's really hard to see what you did

[desert zephyr]

in what way

reading or understanding lol

[radiant topaz]

Reading

[desert zephyr]

ok ill send another pic

[radiant topaz]

Idk because the lines make it hard

[desert zephyr]

ik the blue is very intense

but this notebook is almost full

is this better?

[radiant topaz]

Ok wait

I'm having trouble seeing the part after RoS = SoR

That wasn't necessary to write

[desert zephyr]

yeah now tht i think abt it wasnt ur right

[radiant topaz]

Let me just write the things here

[wraith dagger]

fäf

fäf

[desert zephyr]

yes

[wraith dagger]

fäf

fäf

[desert zephyr]

wouldnt the inverse of this be true aswell?

[radiant topaz]

Yes but you have to show it

As of by just doing this you can't be sure they are equal

[desert zephyr]

but by proving this it would be

[radiant topaz]

Well looking at sets it's pretty obvious, but let's not keep any unattended doubts

[desert zephyr]

yes ur right

could i say that because SoR is the inverse of RoS, they have the same domain and are therefore equal?

@radiant topaz

[radiant topaz]

Nah just write doing it the same way for RoS

[desert zephyr]

okok

@radiant topaz

[radiant topaz]

Alright

[desert zephyr]

K thanks m8!

.close

[cedar kiln]

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[turbid spoke]

how i can solve this

[radiant topaz]

Write terms for first few values of k

[turbid spoke]

i dont undertand

[radiant topaz]

What is the first term of the sum?

[turbid spoke]

I never did summation

k

[wraith dagger]

fäf

[foggy merlin]

you wrote a_n instead of a_k

[turbid spoke]

i din t understand}

[foggy merlin]

basically :
$\sum_{k=1}^n (c_k - c_{k+1}) = c_1 - c_2 + c_2 - c_3 + c_3 - c_4 + ... + c_n - c_{n+1}$

jesus

[wraith dagger]

Ｈｅｒｅｌｓ

[foggy merlin]

the term in the middle cancels

and you are left with c1 - c_{n+1}

this thing is called telescoping sum

[turbid spoke]

let's go from the beginning, which is n

[foggy merlin]

?

[turbid spoke]

what doing n

[foggy merlin]

???

[turbid spoke]

n in the sigma

[foggy merlin]

its just a number

natural number if we want to be rigorous

[turbid spoke]

why you determinate is one

[foggy merlin]

????

[turbid spoke]

what number is n

[foggy merlin]

anything

except numbers beyond natural numbers

[turbid spoke]

infinity?

like a general solution as trigometric function

[foggy merlin]

there is no infinity here

[turbid spoke]

this expresion

[foggy merlin]

yea whats wrong with it ?

[turbid spoke]

oyou can show me the stepes

[foggy merlin]

I... just did here

[turbid spoke]

oh

[foggy merlin]

$\sum_{k=1}^n (c_k - c_{k+1}) = (c_1 - c_2) + (c_2 - c_3) + (c_3 - c_4) + ... + (c_n - c_{n+1})$

[turbid spoke]

how i can solve this

[wraith dagger]

Ｈｅｒｅｌｓ

[radiant topaz]

Sorry

[foggy merlin]

applying the result of this

[turbid spoke]

oka thx guys

.close

[cedar kiln]

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[cedar kiln]

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[coral valley]

can anyone help me solve this?

[tropic oxide]

any progress thus far?

[coral valley]

wait taking new pic

,rotate ccw

[coral valley]

using l'hopital's rule is very messy and doesn't seem to work in this case

I've tried before

[cedar hull]

why hopital, can't you just divide by x ?

[runic garnet]

It’d be -(4x^2 + x) in the numerator after the a^2 - b^2

U forgot to distribute the -

[coral valley]

divide by x?

[cedar hull]

wait let's confirm what stephen says first

[crimson sedge]

multiply by the conjugate, write everything over x and divide by highest denom power, then use lhopitals

[fringe birch]

Raise e to the power of the inside then take the natural log from the outside. Now use tricks you already know

[cedar hull]

I assumed you did the multiplication correctly

[coral valley]

the numerator should be -x yes

[cedar hull]

then if you divide by x in both numerator and denominator, what will you get ?

[fringe birch]

$\ln\left(\lim_{x\to \infty}\frac{e^{2x}}{e^{\sqrt{4x^{2}+x}}}\right)$

[wraith dagger]

Chixen

[coral valley]

Like this?

[cedar hull]

yes then get the x inside the root

[coral valley]

It'll be in the form of infinite/infinite

ITs indeterminate still

[cedar hull]

it becomes sqrt of 4 + 1/x

[coral valley]

What

[runic garnet]

I’ll látex it

[cedar hull]

yes please

[fringe birch]

Why do you write LaTeX like that?

[coral valley]

you cannot divide like this

[cedar hull]

why not ?

[fringe birch]

because that’s not how radicals work

[cedar hull]

just get the division inside the root

[coral valley]

this is what I know aswell

[cedar hull]

then how do they work ?

[fringe birch]

Well you would end up with $\sqrt{4+\frac{1}{x}}$

[wraith dagger]

Chixen

[cedar hull]

yes exactly that

[fringe birch]

How would that help

[runic garnet]

$\frac {\sqrt {4x^2 +x}}{x} = \frac {\sqrt {4x^2 +x}}{\sqrt {x^2}} = \sqrt {\frac {4x^2 + x}{x^2}}$

[wraith dagger]

Stephen

[cedar hull]

just take the limit now

1/x vanishes

and you get your limit

[fringe birch]

I still like this way.

[coral valley]

how do you even end up with this lol

[cedar hull]

it becomes -1/4

[runic garnet]

Look at this rx8

[fringe birch]

$\ln\left(e^{x}\right)=x$

[wraith dagger]

Chixen

[cedar hull]

I feel the exponential is an overcomplication

[coral valley]

so the answer would be -1/1.75?

[fringe birch]

It allows that derivative thing I can’t spell

[cedar hull]

hopital yes but it's an over complication

[coral valley]

confirm?

[cedar hull]

you don't even need to take derivatives

I upvote multiplying by conjugate

[coral valley]

this is the conclusion?

because it's wrong

[fringe birch]

Eh. It’s how I would solve it but you still solved it so congrats on that part. You don’t need to do my way.

[runic garnet]

How’d u get that

[cedar hull]

I'm not insulting your answer xD

[runic garnet]

How’d u get -1/1.75

[runic garnet]

That’s not what the lim to infinity becomes from this

[coral valley]

oh I see

[coral valley]

also, what is this property called?

never seen it before

[cedar hull]

that's what we meant

[coral valley]

im particularly confused about this part

where the x becomes x^2

and gets inside the root

how do I google this?

[cedar hull]

look

do you agree with this (x*y)^z = x^z * y^z

[coral valley]

nono, can you link me to a table of properties containing that?

[runic garnet]

$\frac {\sqrt a}{\sqrt b} = \sqrt {\frac ab}$

[wraith dagger]

Stephen

[coral valley]

I'm not too good at math lol

[cedar hull]

but you gotta learn though

[coral valley]

the problem is the denominator is not a root

[cedar hull]

yes he turned it into a root

square root is the inverse of square

hence sqrt (x^2) = x

[runic garnet]

We had an ‘x’ in the denominator originally. We changed it to the square root of x^2 (which is just x)

[coral valley]

this is what I seee

[runic garnet]

Look at this

[coral valley]

oh shit

i get it now

that's an interesting way to solve this.

thank you so much you both

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[cedar hull]

I'm Glad we were able to help ^ ^

[runic garnet]

Np

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[lusty abyss]

THE EXERCISE:
Two sharpshooters, one of whom (A) hits a target 75% of the time and the other (B) only 25% of the time, aim at a target. Exactly one bullet hits the target. What is the probability that this bullet came from A?

To calculate the probability that exactly one bullet hits the target, I have two ways in mind, but they differ in solution. Which one would be the correct one?
P(A xor B) = P(A) + P(B) - P(A ∧ B) = 0.6923 (appr.)
Or
0.25 * (1 - 0.75) = 0.0625
0.75 * (1 - 0.25) = 0.625
P(A xor B) = 0.0625 + 0.5625 = 0.625

How do these two calculations differ from eachother?

[cedar kiln]

@lusty abyss Has your question been resolved?

[fair geyser]

the first one just gives you A or B

P(A) + P(B) - P(A ∧ B) - P(A ∧ B) should be right

[cedar kiln]

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[chrome vigil]

I can’t figure out any of the coefficients after the first one, I’m not sure how to proceed

[chrome vigil]

This is what I got as the power series for the function

[tropic oxide]

yes, so now you need to return the coefficients on x^0, x^1, x^2, x^3 and x^4 for this

[chrome vigil]

So for x^1 the coefficient should be -100?

Cuz I typed that in and it said incorrect

[tropic oxide]

no

is there an x^1 term in your sum?

[chrome vigil]

I don’t think so since x is raised to the 2n

[tropic oxide]

yes exactly

so then what's the coefficient on x^1?

[chrome vigil]

Oh it’s 0 lol 💀

Aight I got it thx for help two days in a row

.closd

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[teal root]

What's wrong with this? Because it makes no sense the (-108/b)^2

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@teal root Has your question been resolved?

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[sharp helm]

I need some help with problem 5

[crimson sedge]

which is ?

[sharp helm]

Sorry the image isnt loading

Bad connection

[wanton summit]

that’s ok

[sharp helm]

This is painful

So i set up my sigma as n goes to infinity and applied sum of geometric series

Let me send a pic of that as well

So i havent been able to evaluate the limit, and he does reference using the taylor approximation for e^x but i still dont see how it helps, im getting 0 as my limit

[cedar kiln]

@sharp helm Has your question been resolved?

[sharp helm]

<@&286206848099549185>

[paper raft]

which question?

[cedar kiln]

@sharp helm Has your question been resolved?

[pulsar pier]

Try to re-do the sum

The upper limit of the sum is n not infinity

[pulsar pier]

n - 1 not n

[sharp helm]

Ok ill double check in a min ty

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[half quiver]

Hi hello

[half quiver]

So

for question (b) Could anyone explain how u could calculate C^2

[crimson delta]

well standard matrix multiplication

C^2 = C*C

[half quiver]

and how would you find the eigenvalues of C^2 when uve already found the eigenvalues of C

[half quiver]

makes- sense thanku😭

[crimson delta]

well as you see in the second picture, if lambda is an eigenvalue of C, then lambda^2 is an eigenvalue of C^2

[half quiver]

but like the eigenvalues of C^2 are + and - 1 but for C its +-1 and +-i

so why cant +-i be eigenvalues of C^2

[crimson delta]

they could in theory be. but they just arent

[half quiver]

huh?..

[half quiver]

idont really understand:(

[crimson delta]

well they aren't eigenvalues

end of story

[half quiver]

i

Why tho😭

[crimson delta]

what do you want me to tell you. calculate the eigenvalues of C^2 by hand and see that +-i are not among them

[half quiver]

But were not supposed to find them by hand, i just didnt really understand how we could know +-i wouldnt be eigenvalues of C^2 only of C

[crimson delta]

for C you have 4 linearly independent eigenvectors, one for each of +-i, +-1

two of each of these fall together to the same eigenvalues for C^2

so now you have 4 linearly independent eigenvectors for C^2, 2 for the eigenvalue -1 and 2 for the eigenvalue 1

there isnt any space left for any other eigenvalues

[cedar kiln]

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[crimson sedge]

.reopen

[crimson sedge]

Does this mean that sin(180-theta)=sin(theta) and why is that

[crimson sedge]

<@&286206848099549185>

[plain breach]

Yes

Because sin is positive in second quadrant

[crimson sedge]

but im confused how sin, cos, and tan are defined when theta is greater than 90

im new to trig

[plain breach]

It's easier in radians

Have you learnt radians

[crimson sedge]

no

[plain breach]

Quadrant 1 all three are positive

Q2 only sin is positive

Q3 tan is positive

Q4 cos is positive

[crimson sedge]

i see

[plain breach]

so if you have an angle greater than 90, you find the angle which it makes with the horizontal

for example

because the angle is in Q3 and sin is negtive in Q3 you make the whole thing negative

[crimson sedge]

i see

yhat makes sense

thank you

i got it

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[dreamy zenith]

$f(x)=x^4 + 100$

[wraith dagger]

putridplanet

[dreamy zenith]

show that f is not a one to one function by finding two inputs that result in the same output

idk where to start

[civic nymph]

f(-2) = f(2)

but more generally even f(-x) = f(x) for all x

[dreamy zenith]

i have the answer sheet thats wrong

[civic nymph]

xD

but its not wrong

[dreamy zenith]

-2^4 = -16

[civic nymph]

wrong

are you a math student or something math related?

[dreamy zenith]

so f(-1)=f(1) would also work

[civic nymph]

ofc

as i said

but more generally even f(-x) = f(x) for all x

for ALL x

if you are math student or smth math related pls stop and change subject..

-2^4 = -16 if that was meant serious pls change subject

[lusty abyss]

@dreamy zenith please note that f(-2) should be calculated as (-2)^4 = 16

Else it is calculated as -1*2^4

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[storm pebble]

hello. i know you can find local minimum or local maximum points of a function by solving for the first derivative, however if said function had for example a local minimum and a local maximum how would you be able to select which one you solve for?

[carmine bronze]

Use the second derivative. If it is negative, then it is a maximum. If it is positive, then it is a minimum. If it is zero, then it is an inflection point.

[dreamy zenith]

bro really has a jah pfp telling people they are are stupid LMAO
I know a negative times a negative is positive

[carmine bronze]

You need to open a new channel.

[civic nymph]

why you trippin get your mood right

[dreamy zenith]

you are illiterate

[civic nymph]

okay

[carmine bronze]

This is not the place for that discussion. Take it to #discussion

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[nimble goblet]

hello can someone help me with this!

[nimble goblet]

hm?

[dull oxide]

what have you tried

[wraith dagger]

hannibal

[nimble goblet]

i’ve done a few scribbles but none that produces the right answer

do i use that arctan formula

oh wait i cant

am i using the theta equals arctan 7/x?

right

[wraith dagger]

hannibal

[nimble goblet]

the derivative of an arctanx is 1/1+x^2 right?

i got to here but i forgot the chain rule step i need to use

is it the derivative of 7/x?

that i multiply by

oh is it dtheta / dtime

[wraith dagger]

hannibal

[nimble goblet]

so 10tantheta = 7 or am i leaving it as x

nvm i think i skipped steps

oooh i see

thank you so much

i can take it from here

[cedar kiln]

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[molten kernel]

,rotate

[molten kernel]

i dont understand where im going wrong

heres the question:

<@&286206848099549185>

[cedar kiln]

@molten kernel Has your question been resolved?

[molten kernel]

<@&286206848099549185>

[lucid plover]

@molten kernel

[lucid plover]

You flipped your y and x

[molten kernel]

can you specify please idk where i flipped it

[molten kernel]

@lucid plover

[lucid plover]

@molten kernel As in you set x = 1 and y= 2. The prompt stated x=2 and y =1.

[cedar kiln]

@molten kernel Has your question been resolved?

[molten kernel]

I have corrected it but keep not getting the right answer

[lucid plover]

I made a mistake with my z(y) similar to you.

z(y) should be 10

My bad

[molten kernel]

,rotate

[lucid plover]

That should be right

[molten kernel]

its wrong

[lucid plover]

You need to subtract 2 from x, not 1

[molten kernel]

can you please explain why?

[lucid plover]

You will be shifting the plane by a factor of 2 since the initial x was at 2

[molten kernel]

oh thank you for that

[lucid plover]

No problem. Good luck with calc 3.

[molten kernel]

thank you i matched with the answer shheet!

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[acoustic plinth]

any help on how to solve this?

[acoustic plinth]

i know we should use the FTC but which rule exactly?

[tropic oxide]

are you sure you didn't miss anything?

like maybe that you were in fact not asked to calculate this integral in particular, but only its derivative?

also missing dt

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[merry tinsel]

is this correct

@cerulean sail im sorry if i pinging but tom my exam

and i really need the help u come in clutch 💀 in these times

<@&286206848099549185>

[gritty galleon]

Ig it's right..wait doesn't this become a double integral

[merry tinsel]

i did it easier way

now lol

[cedar kiln]

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[merry tinsel]

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[coral valley]

can anyone help me solve this?

I've tried lhopital's rule but it won't work

it seems like it needs somehting else

,rotate ccw

[coral valley]

I did distribute the denomimator

Just try it. It wont work

[crimson sedge]

write x/(x-6) as 1 + 6/(x-6)

[coral valley]

how does that work?

[crimson sedge]

if x tends to 0 then (1+x)^n = 1 + nx
and if you want to see the whole thing then use binomial expansion for
(1+x)^n and the higher terms will tend to zero

[coral valley]

it's x going to infinity

[crimson sedge]

so (1+ 6/(x-6) )^6x = 1 + {6/(x-6)} 6x +

this wont work tho because every term will have equal powers of x in the numerator and denominator

[coral valley]

can you just write it on papar lol

[crimson sedge]

sorry

[crimson sedge]

yes but because x goes to infinity 6/(x-6) goes to 0

[coral valley]

can you send a pic of ur work?

[crimson sedge]

im going to tell a different approach because this one is unnecessarily complicated

[eager sparrow]

I got e^36

lemme show u my work first

[crimson sedge]

yep

[coral valley]

this is what I got aswell

and it's wrong

this is the true result on desmos

should be e^0 but I have no clue how to come to that

[eager sparrow]

Wait it does

e^36

It's e^36

[crimson sedge]

no that's for x tends to 0

[coral valley]

oh shit u right

lmao

[eager sparrow]

Ur x tends to infinity

[coral valley]

i have brain fart

sorry

[eager sparrow]

lol

[coral valley]

yea i think it's e^36

I can't scroll up too far either desmos will bug out

thanks for the help u both

[eager sparrow]

Heres my work btw

u can use it if u want, but u might be better using other websites beside desmos

Desmos is just mainly for graphing\

[coral valley]

yea, it's good to see the result quickly for x going to 0 but I forgot this one to infinity

thx

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[rain folio]

How can I find the ellipsical (I know it's gonna be ellipsical, not parabolic or whatever) cross section of a cone? I'm overcomplicating a question meant for 14 year olds, but here's the question:

Ms Ng (random name) ordered half a glass of non-alcholic cocktail from restaurant X (random). The bartender filled the conical wine glass (radius 5 height/depth 12) to 3/4 of it's depth. Is this indeed half a glass of cocktails (plural, yes)? Justify your answer.

So, like what if, I tilt the glass as much as I am in the head
my plan for now:
find cross section of cone that's 3/4 the height of the tilted cone (I'm sure I can figure a height using pythagoreas' theorem)
find volume of that obliesk ellipsical cone (above * 1/3 * height)
make an excel spreadsheet of the ratio between that and the full volume

I and stuck on finding the cross section, and I don't understand anything online, everything I got was
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 1
A != 0, C != 0, A != C
where do I get all the numbers, and if this is an overly complicated way of getting the answer, how can I get it faster?

[cedar kiln]

@rain folio Has your question been resolved?

[rain folio]

<@&286206848099549185>

should I close this and post on the help-forum thingamajing

[crimson sedge]

wait the question is so long can you ask precisely what u want

?

[rain folio]

How can I find the ellipsical (I know it's gonna be ellipsical, not parabolic or whatever) cross section of a cone?

[crimson sedge]

you can use pythagoras to find semi major axis and semi minor axis

once you get a and b use pi.a.b to find cross sectional area

[rain folio]

how

[dull oxide]

Why do you need the ellipse

You just need to determine if the glass is half full, do you not?

[rain folio]

read my question kekw main parts is 3rd paragraph

[rain folio]

well, yeah, but I'm 1. really bored 2. really interested 3. really confused

[dull oxide]

Does height/depth 12 describe their ratio?

[rain folio]

no, just didn't know which word to use, the height of the cone perpedicular to the base is 12 cm

[dull oxide]

Oh. Height and depth mean same thing. I see

So what is the volume of the cone

[rain folio]

um, 25pi * 1/3 which is around 30

[dull oxide]

You forgot height

[rain folio]

right

or 100 pi

[dull oxide]

100pi

[rain folio]

idk lost my work-answer sheet

[crimson sedge]

when you tilt the drink, it will have an upper height and a lower height from the height it was before. take radius at the new upper height as base and distance from the original centre to the new upper height as the perpendicular and then find the hypotenuse which will be the semi major axis

[dull oxide]

What are radius and height at 3/4 full

[autumn cradle]

you could use the equation of the cone and find the equation of the intersection of a plane with the cone i think

you'd get a 3d ellipse

you'd actually get different conics depending on the angle of the plane

[dull oxide]

Or you could transform the cone first and intersect it with the xy plane

[rain folio]

well this is confusing, but I'll try and understand it

so Soviet onion is giving a formula to somehow find the equation for the semi major and semi minor axis of the ellipse for the intersection of the cone, basically what pl4y3r is trying to find also

[rain folio]

then this is just some mad scientist esque shit that I never thought of

[dull oxide]

Matrix transformations are nice if you know them

[rain folio]

that's awesome, I'm 14, don't know matrix transformations, time to learn then,

[dull oxide]

Good luck. It's not trivial

[rain folio]

perhaps I'll try these one by one because they happen to be in increasing complexity

[dull oxide]

Just do the problem the normal way for now

Then mess around

[rain folio]

well then I'd be done in less than 20 minutes of googling, I have wayy too much time

actually ur prob right, that first then this thing...?

[dull oxide]

Then don't Google it. :p

[dull oxide]

[rain folio]

ty guys <3 I die then I'll ask again hopefully I don't die with the matrix transformations

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[livid ravine]

Any ideas on how to solve this ode? I know that x = 1, so this gives
$$\begin{cases}
\dot y = z + y^2z\
\dot z = -y - y^3
\end{cases}$$

[wraith dagger]

FrankF

[cedar kiln]

@livid ravine Has your question been resolved?

[livid ravine]

<@&286206848099549185>

[storm mural]

it may be helpful to find ydot/zdot = dy/dz

[livid ravine]

hmm, it is not clear to me how to work that out.

[storm mural]

it simplifies

[livid ravine]

It does not from what I can see.

[livid ravine]

You have -y-y^3 in the denominator and it does not go away

[storm mural]

factor

[livid ravine]

I don't see how this makes it easier to solve $$-(\frac{z}{y(1+y^2)} + \frac{yz}{1+y^2})$$

[wraith dagger]

FrankF

[storm mural]

um

factor both

[radiant topaz]

You get $\dot{y} = -z \times \frac{\dot{z}}{y}$ from these two equations and then I think you can solve further

[wraith dagger]

fäf

[storm mural]

equation of the field line is not very hard from that but it says find a parameterization so does that imply finding a function of time that gives the point

[livid ravine]

For me it's hard though because both z and y are in one equation, what technique are you supposed to apply here?

[storm mural]

ydot/zdot = dy/dz = -z/y
separable

[radiant topaz]

Yep

[livid ravine]

oh

separation of variables

[radiant topaz]

I wrote it here

[storm mural]

you wrote it in the most cursed format of all time

[livid ravine]

Ok thanks, I'll come back with an answer

[radiant topaz]

What?

You need to see some solved examples then

[storm mural]

this differential equation also has a nice geometric interpretation

[storm mural]

yeah I do I haven’t taken like a class on this

[radiant topaz]

Then why even say that

[storm mural]

this is more intuitive to me

[radiant topaz]

That's the same thing, not really hard to figure out

[storm mural]

well yeah

[radiant topaz]

Ah well I should've known

[livid ravine]

I get that $$y=\frac{C}{z}$$, so $C = 0$. Then filling it in $\dot z = -y - y^3$ gives $z = 1$. So my answer is $$\begin{cases}
x = 1\
y = 0\
z=1
\end{cases}$$

[wraith dagger]

FrankF

[storm mural]

y is not C/z

[radiant topaz]

I think you did the integration wrong

[livid ravine]

But it's just this right?
\begin{align*}
\int \frac{1}{z}dz &= -\int \frac{1}{y}dy\
ln(z) &= -(ln(y) + C)\
z &= \frac{e^{-C}}{y}\
&= \frac{D}{y}
\end{align*}

[radiant topaz]

Nah

Don't use $

If you are using align

[storm mural]

also ^

[storm mural]

cross multiply

[wraith dagger]

FrankF

[livid ravine]

Ok, it should be $$y = \sqrt{C-z^2}$$ I am lost again on what to do

[wraith dagger]

FrankF

[storm mural]

y^2 = C - z^2
y^2 + z^2 = C

[livid ravine]

hmm, I see. So $C=1$. However, how should I write it as a field line? Would it be valid to say that the general solution would be $\begin{cases}
x(t) = A\
y(t) = \sqrt{C - z(t)^2}\
z(t) = \sqrt{C - y(t)^2}
\end{cases}$

[wraith dagger]

FrankF

[livid ravine]

Btw, so the parametrization is just $y^2 + z^2 = 1 = x$

[wraith dagger]

FrankF

[storm mural]

this looks a lot better and also allows y and z to be negative

they might be looking for smth like (1, sin(u), cos(u))

[livid ravine]

Ah yes, I see

[livid ravine]

Thanks for the help

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[ebon drift]

I don't understand how this is done at all

Well I understand the calculations

But not the concept

Like how do I know how to make the variation table depending on if a is greater than 0

Or less than 0

[surreal cave]

Are you talking about how to determine if the graph is increase or decreasing?

[ebon drift]

Well I know it's decreasing if it's negative

I think

And increasing when positive

[surreal cave]

Yes, so what your asking is for the basic concepts of a derivative then?

[ebon drift]

Well how do you make the variation table?

[ebon drift]

I understand it somewhat

Just confused with the variation table

[spiral fog]

1. Determine the function for which you want to create the variation table.

2. Find the derivative of the function using differentiation rules.

3. Identify the critical points of the function by setting the derivative equal to zero and solving for x.

4. Determine the sign of the derivative on either side of each critical point by plugging in test values into the derivative. If the test value produces a positive result, the function is increasing at that point, and if the test value produces a negative result, the function is decreasing at that point.

5. Record the information in a table, including the critical points, the sign of the derivative on either side of each critical point, and whether the function is increasing or decreasing on each interval.

[surreal cave]

ignore this guy lol

[ebon drift]

Ok

[surreal cave]

But the table describes the relation of a function and its derivative

[ebon drift]

Right

[spiral fog]

I have just listed every single point on how to use variating tables using derivatives and you tell them to "ignore me"?

[surreal cave]

so then the derivative describes the slope of the function, yes?

[ebon drift]

Well I'm just confused with the a<0

[ebon drift]

Yeah

[surreal cave]

oh well that's just because if the leading term is negative then the shape of the graph changes

[ebon drift]

So if it's negative it goes down right?

[spiral fog]

no, that would be the opposite

[surreal cave]

Not always: (https://www.desmos.com/calculator/kdvzibordk)

[ebon drift]

How are you able to tell then

Without using desmos

[surreal cave]

The leading coefficient

On an odd leading power polynomial then there are two cases:

1. Leading coeff. is positive, then it is down-up.
2. Leading coeff. is negative, then it is up-down.

If it is even power on the leading term:

1. Leading coeff. is positive, then it is up-up.
2. Leading coeff. is negative, then it is down-down.

[spiral fog]

the number of critical points on a function's variation table is directly proportional to the complexity of the function's behavior. In other words, the more critical points a function has, the more intricate its pattern of behavior becomes.

[ebon drift]

What about something like this

[ebon drift]

Oh ok

[surreal cave]

Well $f(x)$ is a cubic and it's higher power is 3. So we look for it's coefficient, which is $1$ which is strictly positive so we know that the function goes $down-up$ then like that guy was saying, we take the derivative and set equal to zero to find the "critical points" on which the function turns to really get a good understands of what it's going to look like.

[wraith dagger]

XxMrFancyu2xX

[surreal cave]

If that answers your question

The reason is turns on the critical point is because the slope or derivative is zero

[ebon drift]

Ohhhhhhhhhh

Ok

Right I get it now

Thank you!

[surreal cave]

Also one more thinmg

when the $a$ happens to be in the coefficient in front of the term with the highest power and when $a<0$ then we know $a$ is strictly negative, and so it either $up-down$ or $down-down$!

[wraith dagger]

XxMrFancyu2xX

[surreal cave]

(depending on the highest power)

[ebon drift]

Right ok

Thanks for your help

[spiral fog]

👌

[ebon drift]

Thank you!!

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[quasi bear]

how do i start?

[cedar kiln]

@quasi bear Has your question been resolved?

[gritty viper]

probably find a closed form for each function and multiply those

or maybe you can do it the polynomial multiplication way

(3+3x+3x^2+3x^3+...)(x+2x^2+3x^3+...)

= 3x + 9x^2 + 18x^3 + ..., the pattern followed is findable

[cedar kiln]

@quasi bear Has your question been resolved?

[radiant topaz]

How does the other sum starts from n=1

n=0 gives 0

[gritty viper]

?

[cedar kiln]

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[stone dagger]

Can someone help me to solve this linear ODE equation? I'm having a trouble rearranging the equation.

[sullen saffron]

First write it as

$x \dv{y}{x} + (x + 1)y = 1$.

[wraith dagger]

stabulo

[stone dagger]

Oh I see now.

Thanks

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[tawny sand]

It‘s about the question c.). It‘s in german so here a translation of c.): The card is positioned so that it is vertical (see figure). The corner point K creates a shadow point K' on the top side of the concrete body. Find the coordinates of K'.

I dont know how I can find K.)

[tawny sand]

What I have for now:

<@&286206848099549185>

[cedar kiln]

@tawny sand Has your question been resolved?

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@tawny sand Has your question been resolved?

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[modest crystal]

Is it possible to get an R^2 value for the purple line that I made?

[dull oxide]

R² value?

wdym

[red pumice]

is that 2 regression lines?

stitched together

6 degrees of freedom for 11 datapoints?

youre gonna need to calculate the adjusted R^2 lol

[modest crystal]

The question said "Attempt to find two or more functions that can model different parts of the data to create a better overall model. State the equation for each function and its domain."

so I was just wondering if I could get a value for R^2 for my line to see if it was better than the quadratic regression thing

idk why we're being condescending in here

[red pumice]

oh yeah you can just calculate R^2 by hand

[modest crystal]

okay

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[red pumice]

^not trying to be condescending, just trying to say that using 6 degrees of freedom to fit 11 data points would be overfitting

[modest crystal]

I dont know what that means

[red pumice]

youre estimating 6 different parameters to best fit 11 data points

the more parameters you allow yourself to estimate, the better the fit will be with optimal estimates for the parameters

for example if you have 2 data points

and you fit a standard regression model of Y = alpha + beta * X + error on it

you would get a perfect fit

but your model could be very wrong

[modest crystal]

I'm just trying to answer the question. My professor wants me to put together 2 or more expressions to model the data from what I understand

[red pumice]

alright well fair enough

[modest crystal]

its not asking me to predict anything

[cedar kiln]

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[gritty viper]

try to find the height and the width

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[red pumice]

thats not one of the steps hahah

[fair geyser]

the three centers make a triangle

[red pumice]

you can use pythagoras

yes

[fair geyser]

find the height of the triangle, easy with trigonometry

if you can use pythagoras that's smarter probably, i don't know how

oh i see

[sweet pelican]

also add the 2r

[fair geyser]

nice

very true

[cedar kiln]

@crimson sedge Has your question been resolved?

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[latent fox]

how do i solve this

for M

[mental trail]

Get rid of the denominator first

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✅

[civic eagle]

0.02 PLN/1 LVL
40-22 = 18 LVL needed
18 * 0.02 = 0.36 PLN for 18 LVLs

idk how the xp factors into this because you kinda just tacked that onto the end.

[cedar kiln]

@crimson sedge Has your question been resolved?

[civic eagle]

these are the values that work given what you posted.

every 2 PLN is 100 lvl

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[silent tide]

2 pln = 100xp
1 pln = 50xp
(1/50) pln = 1 xp

You are at level 22 and you need to reach level 40, so you need to increase your level by 18 levels. Each level is 300 xp so you need:
18 levels
18 (300 xp)
18 (300 ((1/50) PLN))
18 (300/50 PLN)
18 (6 PLN)
108 PLN

[cedar kiln]

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[still raft]

Are these two homogeneous?

[tropic oxide]

what's your definition of a homogeneous function?

[still raft]

This one?

Also i tried answering it

Idk if its righ

[tropic oxide]

you don't have any y.

[still raft]

Oh so its homogeneous?

[tropic oxide]

no, i didn't say that.

i didn't say anything about homogeneity.

i said that you didn't apply the definition properly.

[still raft]

Ok so what do i do?

[tropic oxide]

can't fault you for it since FORMALLY it was only given for a function of two variables, not one as you have,

for one variable, homogeneity means f(λx) = λ^n f(x)

[still raft]

Oh ok

[tropic oxide]

it should be clear that e^(λx) is NOT a multiple of e^x, and tan(λx) is NOT a multiple of tan(x).

[still raft]

What does that mean? Im sorry i dont underatand

[tropic oxide]

e^(λx) is not equal to any constant times e^x

[still raft]

So what do I do?

[tropic oxide]

definition of homogeneity is not satisfied
"so how do i tell if it's homogeneous or not?"

[still raft]

Oh ok so its not homogeneous?

[tropic oxide]

yes!!! if a function doesn't satisfy the definition of homogeneous then it is not homogeneous!

[still raft]

Ol thank you

Ok

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[quasi plover]

5 mark question

[cedar kiln]

@quasi plover Has your question been resolved?

[quasi plover]

do i suppose to ping someone?

[livid hound]

you should consider the ambiguous case when applying sine rule here