help-23

[obtuse badger]

idk

[woven nova]

wdym

is it a question

or

[obtuse badger]

o shit

ok

but why the ansvere is 4^2-6

[hard crest]

dog you're gonna need to provide much more context

[obtuse badger]

?

[summer jackal]

we can't respond without the otherside of the equation

[obtuse badger]

no ther is no oter side

[hard crest]

is there more to the question?

can u take a picture of it

[obtuse badger]

i misunderstod

no

hold up

a

a)

[brave wolf]

are you finding the roots?

[summer jackal]

can you traduce please

[obtuse badger]

traduce

[summer jackal]

translate*

[obtuse badger]

?

[brave wolf]

it says "write in the simplest form"

right

[obtuse badger]

type as easy as posibele

[obtuse badger]

yes

hello?

[hard crest]

do you mean 4x^2 - 6

[obtuse badger]

yes

[hard crest]

those are different things

[obtuse badger]

thats the answere

[hard crest]

4x^2 - 6 is not the same thing as 4^2 - 6

anyway

you have already drawn the lines for the distributive property

[obtuse badger]

i got 4x -6

[hard crest]

can you show your work

[hard crest]

x * 4x is 4x^2

[obtuse badger]

but what if i turn x to 2x what do i get then

[hard crest]

?

2x + x(4x - 2) - 6
2x + x(4x) - x(2) - 6
2x + 4x² - 2x - 6
4x² - 6

[obtuse badger]

ye but 2x + 2x(4x -2) -6

what then

[hard crest]

ok....

2x + 2x(4x) - 2x(2) - 6

2x + 8x² - 4x - 6

[obtuse badger]

?

ok

tnk

.close

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[lean patio]

is this correct im getting 0/1 whilst the book got 1/1 :

[rough storm]

!showwork

[safe radish]

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

[obsidian oracle]

is this the book correction? looks pretty bad

[lean patio]

this is a book with problems and their solutions, the assistant said there may be mistakes

[obsidian oracle]

the limit is indeed 0 i think

[lean patio]

A bit messy

[obsidian oracle]

the leading term of the numerator is n^(4/3) while the leading term of the denominator is n^(6/4) = n^(3/2)

since 4/3 < 3/2

the denominator wins

limit = 0

[lean patio]

isnt it 3/4?

[lone arch]

,w ((n^3 - 2n^2 + 1)^(1/4) + (n^4 + 1)^(1/3))/((n^6 + 6n^5 + 2)^(1/4) - (n^7 + 3n^3 + 1)^(1/5)) as n to infinity

[obsidian oracle]

where?

the numerator?

[lean patio]

the numerator

yea

[obsidian oracle]

n^(3/4) is the first term

which is not the leading term

[lean patio]

ah you meant the second sqr root

[obsidian oracle]

the leading term is the 2nd term

n^(4/3)

so :

[lean patio]

so the book is wrong basically

[obsidian oracle]

yep

[lean patio]

thanks a lot

[obsidian oracle]

• either divide both numerator and denominator by n^4/3, and you'll get something like 1/(n^...)

or

divide both numerator and denominator by n^3/2, and you'll get something like (n^(-...))/1

[lean patio]

i divided by n^3/2 since its the biggest

[obsidian oracle]

yep that's the second way

[lean patio]

good to know it works both ways

thanks again 🙂

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[fluid wren]

hello

[fluid wren]

i am having troubl

[white schooner]

ooh no why

[fluid wren]

i have this problem and i soled the domain and range

but idk how to solve the other question "how many solutions" : (

[white schooner]

Oh

wait lemme read 9t

it

[fluid wren]

i found this by plugging into desmos, but dont know how to explain alegbraically

[white schooner]

what do u dont understand/

[fluid wren]

bc i have equation 2 = sin (3x-1) and dont know where to go next

how to isolate x

[white schooner]

Hm

[fluid wren]

Bc teacher say that arcsin of sin != x

[white schooner]

lemme caculate it i dont want to give ur the wrong answer x

the teacher is right

[fluid wren]

ya so hwo to isolate x then?

how*

[white schooner]

um so

[fluid wren]

Teacher said one time about pi - or something symmetry solution but i forgot how to apply here

[white schooner]

lemme send ur the screenshot

2
sin
3
x

1

or
sin
3
x

1
2

or
3
x

sin
−
1
(
1
2
)

or
3
x

30

or
x

30
3

or
x

10

[fluid wren]

Im not understand how 2sin3x = 1

[white schooner]

here check it

[fluid wren]

Wait but question say 2sin(3x-1) = 4

Not equal 1

[white schooner]

Yh

do u understand it

if u dont understand it read the line right of it they will explain it

[fluid wren]

I do not understand it because that is different question I think

[white schooner]

Oh

i think its the same

[safe radish]

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[lean otter]

I need help with finding the distance from a point to a line using vector operations.
I understand cross products and scalar products well, but I couldn't do it.
When I looked it up on YouTube I found the formula:

d = |a x b| / |a| where a is a vector where (x, y, z) are the coefficients of the x, y, z values of the line, and b is the vector from a point on the line to the point you want to find the shortest distance to.

But I have no idea how they came up with this formula or why it works, any help?

So for instance, if we have the line (x,y,z) = (1,3,4) + t(1, -2, -3) and the point (4, 1, -2)
a would be (1, -2, -3)
and b would be (4-1, 1-3, -2-4) = (3, -2, -6)

[obsidian oracle]

b is a distance?

[lean otter]

both are vectors. I can draw a picture if it helps.

[obsidian oracle]

I was just reflecting on what you wrote about b here

[lean otter]

Oh you're right, typo.

[obsidian oracle]

Well you can always compute orthogonal projection

[lean otter]

Is that by using cross products?

[obsidian oracle]

Scalar products

But cross products might intervene

Given P the point, line (x,y,z) = Q + tv

Then to find its orthogonal orojection onto the line

You subtract Q to both :

u = P-Q = vector QP

[lean otter]

t is a parameter and v is a vector with the coefficients of (x, y, z)?

[obsidian oracle]

t is the parameter and v is vector

So

[obsidian oracle]

Coming back to this

New line is (x,y,z) = tv

It's a line that goes through the origin

We can compute orthogonal projection :
proj(u) = (u•v)/(v•v) v

Then the distance between P and our original line is exactly ||proj(u)||

Which is |u•v|/||v||

[lean otter]

what does double | mean?

[obsidian oracle]

|.| is for absolute value, which we make sure to not mistake with the norm of a vector ||.||

[lean otter]

Yeah, |.| I know is the absolute value, what is the norm of a vector?

@obsidian oracle I found a great visual explanation. |a x b| is the area of the parallelogram, so the height which would be the shortest distance is just area / width and the width is the a vector I mentioned earlier.

Thanks for your help!

.done

/done

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[marsh kayak]

would this be considered a positive solution? because i have the answer key for this and there are 2 answers, 1/16 and the one above, how do i know which 1 to put and would both be necessary?

[median vigil]

64^(-2/3) = 1/16, it is the same answer but simplified

[marsh kayak]

ok thanks also where would i go from here to solve for x

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[wind echo]

regarding 2nd order linear equations (application of)

[river field]

anticipates the most sophisticated question in the history of mathematics

[wind echo]

Suppose a spring has a mass m and spring constant k and let w=sqrt(k/m). Suppose that the damping constant is so small that the damping force is negligible. If an external force F(t)=[Fnaught]cos(wnaught)t is applied, where wnaught =/= w, use the method of undetermined coefficients to show that the motion of the mass is described by x=(csub1)coswt+(csub2)coswt+([Fnaught]/[m(w^2-wnaught^2)]coswnaughtt

[wind echo]

well it's confusing to me, so maybe it is

i've gotten the back half of the equation ([Fnaught]/[m(w^2-wnaught^2)]coswnaughtt

by solving for a particular solution

but when i try to solve for the front half (csub1)coswt+(csub2)coswt

by solving for the complementary solution, I end up with exponentials instead of cos/sin

[river field]

What in the name of all that's holy is a "coswnaughtt"?

[wind echo]

lol

[remote verge]

$\cos \omega_0t$

[wind echo]

cos([wnaught]t)

yea that

[remote verge]

I'm guessing

wait

[flat frigate]

otheol

[wind echo]

any insights?

[river field]

Hold on, I gotta decipher your question

[wind echo]

sure

[remote verge]

Can i put this into TeX

[river field]

"a spring has a mass m", does the spring have a mass or is there a mass attached to the spring?

[median vigil]

remember that euler's formula says the exponential of a complex number results in a combination of sines & cosines

[wind echo]

the spring doesn't have mass

its attatched

[wind echo]

o wait u right

lemme check that

[remote verge]

Suppose a spring has a mass m and spring constant $k$ and let $\omega=\sqrt{\frac{k}{m}}$. Suppose that the damping constant is so small that the damping force is negligible. If an external force $F(t)=F_0\cos(\omega_0t)$ is applied, where $\omega_0\neq\omega$, use the method of undetermined coefficients to show that the motion of the mass is described by $x=C_1\cos\omega t+C_2\cos\omega t+\frac{F_0}{m(\omega^2-\omega_0^2)}\cos\omega_0t$

Dang it

[flat frigate]

otheol

[wind echo]

yea

Here’s my work for the particular solution which gave me a good part

[river field]

break neck Ah, I see

[wind echo]

Here’s my attempt at the complementary which still isn’t what I’m loooking for

[wind echo]

Lol mb

[remote verge]

,rotate

[remote verge]

Oh dang

How do i get it to rotate any image

[wind echo]

thats dope woah

[median vigil]

in general if you have a solution in the form [ x(t) = e^{(\lambda \pm i \mu) t} = {e^{\lambda t}(\cos(\mu t) + i\sin(\mu t)), e^{\lambda t}(\cos(\mu t) - i\sin(\mu t))} ] then by taking linear combinations of those solutions we get the real-valued fundamental solution set [ x(t) = { e^{\lambda t}\cos(\mu t), e^{\lambda t}\sin(\mu t)} ]

[flat frigate]

cloud

[wind echo]

well that looks pretty damn close to what i need

but why am i missing a t in the complementary

[median vigil]

your complimentary solution assumed the form $e^{rt}$ but you only wrote $e^{r}$

[wind echo]

o im stupid

[flat frigate]

cloud

[wind echo]

yea got it

WOOOOO IT WORKS

merci beaucoup everyone

uh how do i close this

[median vigil]

using .close or .solved

[wind echo]

.solved

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[safe radish]

@sudden hare Has your question been resolved?

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[clever ridge]

i think its a b d c but my explanations need bit more understanding

[stray axle]

is this transformations?

[clever ridge]

no dont think so

[stray axle]

oh then idk

[clever ridge]

just derivatives

[stray axle]

gl

[severe pond]

so you’re saying a is f?

[clever ridge]

yes

[stray axle]

lmao the amnt of questions related to derivates makes me scared to learn them

[lean otter]

All are graphs of polynomials

[versed wave]

you might want to identify what types of function f is

yeah, polynomial

and use the fact that higher order derivative of polynomial result in degree reduction

[clever ridge]

a is f b is f' d is f'' c is f'''

[versed wave]

check again

[severe pond]

no

[clever ridge]

oops

sry

d then

c

[lean otter]

Take two cases
i.f has a odd degree
ii.f. has a even degree

[clever ridge]

but how to explain y

[versed wave]

what degree is function a?

[clever ridge]

2

but i dint have the equations of the lines

[remote verge]

You could also consider maximums and zeroes

[versed wave]

a is indeed 2 degree

[clever ridge]

its just identifying by looking at the graph

yea idk y he gave X

my prof said sonething abt line being pos

or neg

relating to other lines

is there like a theorem

idk

nm

thanks

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[remote verge]

A function has a horizontal slope (which tends to be a local maximum or minimum) when its derivative = 0 (when it crosses the x-axis)

[safe radish]

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[smoky goblet]

i am a bit confused on how i should explain this

[safe radish]

@smoky goblet Has your question been resolved?

[amber trellis]

I don't have a great answer but what's the maximum distance the parallel planes can have between them?

[safe radish]

@smoky goblet Has your question been resolved?

[smoky goblet]

i guess infinite?

it doesnt really specify

so i guess it just depends on the two parallel cuts

[amber trellis]

If a sphere has a radius r and I cut at r+1 am I cutting the sphere?

[smoky goblet]

i think so?

[amber trellis]

The cut will miss. What I'm trying to say is that in order to cut the sphere the parallel planes need to intersect the sphere. So the distance between the planes will be limited by the radius of the sphere.

In 2 dimensions if we have a circle centered at the origin with radius r then a verticle line at x = r + 1 won't intersect the circle, neither will a line at x = r + .0000001. So what's the maximum value of x we can choose for a circle of radius r so the line touches the sphere?

[smoky goblet]

just r?

[amber trellis]

Yeah, so that's just one cut if we make another cut what's the furthest distance from the first cut we can make?

and it has to be parallel to the first

[smoky goblet]

0?

[amber trellis]

its easier with a visual. This is a circle with radius 1. The blue line is x = r = 1 so what's the furthest I can place a second line from the first while still touching the circle and being parallel?

[smoky goblet]

oh at -1

[amber trellis]

Another way to say that is -r so what's the distance between r and -r?

[smoky goblet]

2

[amber trellis]

What is it in terms of r?

[smoky goblet]

2r?

[amber trellis]

Yep, so what we've shown is that the maximum distance between 2 parallel cuts is 2r. So any 2 points bound by a circle of radius r will have a distance less than or equal to 2r right? And if we plug 2r in for D in your original question does it look familiar?

[smoky goblet]

would that be the surface area of a sphere? if we plugged 2r in for D

[amber trellis]

Yeah the only problem is I don't know if that proves that 2(pi)rD is the surface area for anything less than 2r. Just that at the maximum distance of 2r we get the formula for the surface area of a sphere.

[smoky goblet]

wdym less than 2r

like if it was one line at r = 1 and one at r = 0

?

[amber trellis]

Yeah. I can't think of how to show that for any arbitrarily chosen points 2(pi)rD will give accurate surface area

[smoky goblet]

i understand ur other explanation at least

same though i dont really know if it would work for something if it wasnt at the maximum distance

i thought i would have to prove it using integrals or something but idk

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[deft pivot]

so i first calculate line integral and then compare it to surface integral and they do not match

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[deft pivot]

<@&286206848099549185>

tnx

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[versed wave]

Given a cylinder $(T)$ whose bases $(C)$ and $(C')$ are 2 incircles of 2 opposing sides of a cube. In one of the concave triangles created by $(C)$ and its circumscribing square, there is a rectangle with dimensions $a\times2a$. Calculate the volume of the cylinder in $a$

[eternal carbon]

reading this problem description is interesting

gotta love those concave triangles

[versed wave]

idk how to call the triangle in english

but ygwim

[eternal carbon]

yessir

anyway you can solve this by setting up a system of equations (if it makes the process easier for now you can decide to use a = 1)

pythagorean theorem helps here

try enlarging the diagram you have, drawing lines and labels on it, and seeing if anything comes to mind

[versed wave]

ok so

[versed wave]

perhaps this helps?

[safe radish]

@versed wave Has your question been resolved?

[silent token]

I'd be more interested in this triangle

@versed wave

[versed wave]

hm well

HD = asqrt5

EH = R

DE = Rsqrt2

we dont know the angles, nor the relation between R and a

[silent token]

We know angles

Use a bit of tangent here and there

IDH will have measure arctan1/2

[versed wave]

its tan^-1(1/2)

yeah

but thats pretty ugly init

[silent token]

EDH is pi/4

It's all ugly

Alternatively

Take the circle centered at (r,r) with radius r and the line y=2x

Find the appropriate intersection

Let that be a,2a
Or 2a,a
Draw it out, it'll make more sense
Then solve for a or r

@versed wave

[versed wave]

hmmm alright

thanks for the help

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[spare grotto]

"9. Enter the tangent plane to the surface
S : e^x−3ycos(5y − z) + yz = 6. at point A = (3, 1, 5). Find an approximate value for the height z of the point B = (3.06, 0.95, z) ∈ S."

[spare grotto]

i am not sure how to solve this

[safe radish]

@spare grotto Has your question been resolved?

[fathom jewel]

there is a formula for the tangent plane

and for the approximation we need the total differential

[mortal sandal]

That point is not on that surface

Oh nvm you typed it up without parentheses

[spare grotto]

oh yeah sorry, wait let me try to find the formula

is it one of these?

[mortal sandal]

Looks like it yeah

Which one looks like it should be applicable

[spare grotto]

the second one maybe

[mortal sandal]

Right

[fathom jewel]

$\Delta z \approx f_x(x_0,y_0) \Delta x + f_y(x_0,y_0) \Delta y$

[flat frigate]

𝔸dωn𝓲²s

[mortal sandal]

I'm gonna sleep now gl

[spare grotto]

okay thank you

[mortal sandal]

Probably shouldn't use that one as that's for when you have z = f(x,y)

[fathom jewel]

You basically do

It's implicit the equation

[spare grotto]

is ∂f/∂x = e^(x-3y)*cos(5y-z)

∂f/∂y= -3e^(x-3y)cos(5y-z) -5e^(x-3y)sin(5y-z)+z

∂f/∂z = e^(x-3y)sin(5y-z)+y

[fathom jewel]

minus?

-sine

Oh no nvm

cancels

good work

[spare grotto]

Thanks and now we put in A?

so ∂f/∂x = 1?

[fathom jewel]

yea use the right one

formula

you got this

[spare grotto]

do you mean ∇f(3,1,5) = (1,2,5)?

[fathom jewel]

Where is x,y and z

Your tangent plane has variables

[spare grotto]

∇f(3,1,5) * ((x,y,z) -(3,1,5)) = 0

[fathom jewel]

(x-3,y-1,z-5)

Yeah

[spare grotto]

3(x-3) + 1(y-1) + 5(z-5)

3x+y+5z-32=0

now we put in B?

z = 2,974?

[fathom jewel]

Oh right the tangent plane should also be applicable for broad approximations

[fathom jewel]

forget this then

[spare grotto]

Yes okay

Ohh wait

[spare grotto]

I see what I did wrong

[fathom jewel]

,w e^(3.06-30.95)cos(50.95-z) + 0.95z = 6

yeah very rough approximation

[spare grotto]

I got z = 4,374

[fathom jewel]

it's a tangent plane after all

[flat frigate]

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[fathom jewel]

,w |5.092 - 4.974|

[flat frigate]

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[fathom jewel]

thats the error

so actually not rough approximation lol

[spare grotto]

Wait what

[fathom jewel]

it's a good one

I am saying the approxmation is good

[spare grotto]

Ohh yeah okay

[fathom jewel]

in comparison to the actual value

[fathom jewel]

how

[fathom jewel]

is this wrong

[spare grotto]

3(3.06) + 0.95 -32 +5z = 0

Z = 4,374

Or maybe I am wrong

[fathom jewel]

How did you get there

,w 3(x-3) + 1(y-1) + 5(z-5)=0

[flat frigate]

Results provided by WolframAlpha

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Requested by adoniszgks

[spare grotto]

We have 3(x-3) + 1(y-1) +5(z-5) = 0

[fathom jewel]

Yes

3x + y + 5z + (-9-1-25) = 0

[spare grotto]

If we expand we get 3x - 9 + y -1 + 5z -25 =0

[fathom jewel]

-35

3x + y + 5z + (-9-1-25) = 3x + y + 5z -35 = 0

[spare grotto]

Ohh wait yeah

[fathom jewel]

yeah

[spare grotto]

Then we get 4,974

[fathom jewel]

hmm

[spare grotto]

Not minus

[fathom jewel]

haha

[fathom jewel]

Yeah

See how small the error is

Thats good

[spare grotto]

Yes

Well that’s it then

Thank you

.solved

[safe radish]

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[rain vault]

Hi can someone explain to me how I would go about solving this? I don’t even know where to start!

First number is -2 (in the numerator)

[lean otter]

Write -8 as -1 ×2×2×2

√(abcd)=√a√b√c√d

Do this

[rain vault]

wouldnt square root of -8 be better written as sqrt of -1, sqrt of 4 and sqrt of 2?

if im breaking -8 into all square roots?

? @lean otter

<@&286206848099549185>

[somber parcel]

Yes so what would that be

[rain vault]

hi, so the new equation after breaking it down, wouldnt it be:

-2 + i * 2 * sqrt (2) / 2?

[somber parcel]

Yeah

So you see how there 2 common in both numerator and denominator

So to simply it further you multiply the fraction by 2/2

[rain vault]

dumb question but since square root of 8 is equal to

2 * i * sqrt(2) do i multiple all of this by 2?

since i multiply the fraction by 2

[somber parcel]

Ok my bad hold on

No need to multiply sorry

[rain vault]

i thought u meant multiply the whole thing by 2/2 to get rid of the denominator

[somber parcel]

Just divide by you have in the fraction

[rain vault]

OHH

[somber parcel]

Yeah, but it’s easier to just directly do it

[rain vault]

how do i divide square root of 2 by 2?

[somber parcel]

Well the whole thing is 2i rt(2), so by dividing this by 2, you would have i rt2

[rain vault]

OHH

I see

so the final answer would just be -1 + i *Sqrt2

[somber parcel]

Yeah

[rain vault]

and then the denominator would just be 1, which wouldnt matter

so thats the final answer

[somber parcel]

Yep

Sorry for the confusion in the middle

[rain vault]

sorry im not too good at math

[somber parcel]

Nah it’s ok

[rain vault]

so id just like some clarification on this but

2 i rt(2)

this is the same as

2 * i * rt(2) correct?

[somber parcel]

Yeah

[rain vault]

so how come when you did the step to simplify

divide everything by 2

you only did it to the 2

i thought it would be 1 * 1/2i * 1/2(rt(2)

[somber parcel]

Because since it’s in multiplication, you only have to divide a factor of it,

You can think of it using numbers like 6/2, is the same as 2x3/2, you only divide a part of it

[rain vault]

i see okay thanks!

and last question

the example i just solved, the simplyfying step was easy because it was all 2's

but what happens if its different numbers everywhere?

ill send u a pic of the same problem just different #'s

[somber parcel]

Ok

[rain vault]

so i did the simplying of sqrt 20 already

but now theres an 8, 2 and 24

so im not sure how to get rid of 24?

[somber parcel]

You cannot completely remove the 24

The hcf of 2,8,24 is 2

So you can simply divide the numerator and denominator both by 2

[rain vault]

so would it be -4 + sqrt 5 / 12?

[somber parcel]

Yeah

Cannot simplify it any further

[rain vault]

ok thx! how do i know when a problem is completely simplified, if you werent explaining it to me

id assume i can divide in half again

is it because theres no number in front of the square root

aka there's a 1

[somber parcel]

Yeah

You can just try to take th hcf of all the numbers

If it is anything other than 1, then divide by that number

[rain vault]

i see ok

i typed in (-4 + sqrt(5) ) / 12 into my homework problem

and it gave me a completely different correct answer so im so confused

[somber parcel]

What did it give?

[rain vault]

im not sure if this is an error bc it seems way off?

[somber parcel]

You took rt20 instead of rt-20

[rain vault]

oh..... my...

i see it now

[somber parcel]

It split the fraction into two, so that both the fractions can be fully simplified

So -4 + i rt5 / 12 = -4/12 + i rt5/12

[rain vault]

wait

i see, you just divided everything by 2

aka simplified it

[somber parcel]

I didn’t take the original question, I took it from here

[rain vault]

ur right

ok i get it now, thank you for the help.

i am getting back into math so i always miss little details that throw the whole problem off

like the negative -20 lol

[somber parcel]

Mhm

[rain vault]

anyways i can close this now thanks for ur help sir

[somber parcel]

You’re welcome

[rain vault]

.close

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[restive palm]

A family purchased a new home for $225 000. The appreciation of the home’s value can be
modeled by the equation 𝐻(𝑡) = 225 000 (1.08)𝑡, where 𝐻(𝑡) the value of the home and
𝑡 is the number of years they have owned the home.
a) What is the average rate of change in the home’s value between years 5 and 8?
can someone help me with this

[restive palm]

im not sure how to start

[topaz tree]

The average of an function over a and b

[restive palm]

i still didnt take calculus

this is functions

[topaz tree]

Since this is a linear function

[split fulcrum]

$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

[flat frigate]

water beam

[topaz tree]

(H(8)-H(5))/(8-5)

[restive palm]

oh alright let me do it

[restive palm]

2862.016?

[topaz tree]

lemme check

[restive palm]

alright

[topaz tree]

I think it's wrong

[restive palm]

why

[topaz tree]

I get 243000 per year

[restive palm]

can you show ur calculations

[topaz tree]

225000* 1.08 * t = 243000t

,calc 225* 1.08

[flat frigate]

Result:

243

[restive palm]

416459.2973 - 330598.8173 / 3

this is what i do

[topaz tree]

How did you get it?

[restive palm]

i js plugged in 8 and 6

5

[topaz tree]

in which equation?

[restive palm]

225000(1.08)^t

[topaz tree]

Ah! i see the problem now

[restive palm]

what?

o i forgot the square

[topaz tree]

See equation in the question

[restive palm]

my bad

[restive palm]

if its 225000(1.08)^t, then is my answer right?

28620.16

dollars per year

[topaz tree]

No, ig, because the function isn't linear anymore

[restive palm]

huh

[topaz tree]

Can you differentiate?

[restive palm]

nope

we're using the same formula

you gave me

[restive palm]

this

only that

[topaz tree]

mb you're right, your answer should be right then ig

[restive palm]

alr can I ask another question

[topaz tree]

Yes you can

[restive palm]

how do I find the instantaneous rate of change at 15 mins here

[topaz tree]

$28,620.32 per year is the answer i got

[topaz tree]

Is there any additional information?

[restive palm]

yeah i got around the same answer

[restive palm]

nope just that

[topaz tree]

take any two points through which line passing through p is passing and find their slope

based on the graph the two points i would take is P(15,4) and Q(35,6)

Now just find the slope and you would be fine

[restive palm]

oh ok, calculated it and got 10

looks good?

[topaz tree]

yeah

[restive palm]

aight tyyy, one last question

like the other one, but i wanna make sure im right

[topaz tree]

bring it on

[restive palm]

The population of a town is modeled by 𝑃(𝑡) = 25𝑡^2 + 150𝑡 + 5000, where 𝑃(𝑡) is the
size of the population and t is the number of years since 2000.
a) Calculate the average rate of change in the population between 2008 and 2016

is the answer 750?

[topaz tree]

Let me see

Yes it's right...

[restive palm]

cool, what are the units tho?

not sure

[topaz tree]

750 people per year

🤣 It's interesting if you keep the units

[restive palm]

lmao I see now, im also being asked to tell whether average rate of change in population increase, decrease, or stay the same, how do i find that out

[topaz tree]

if rate of change is positive that population will increase

if it's zero it'll stay the same

if it is negative, it'll decrease

let r be rate of change of population and p be the population

if r>0, p will increase
if r=0, p will be the same
if r<0, p will decrease

[restive palm]

ahh, so in this case it increases?

[topaz tree]

Yes

[restive palm]

alright, a silly question (ik this is very basic), the slope of the secant drawn through thsee two points is 2?

rise 6 and run 3

so 6 / 3 = 2

[topaz tree]

Yeah

You're doing pre-calc it seems

[restive palm]

yep

[restive palm]

thanks

[safe radish]

@restive palm Has your question been resolved?

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[void cedar]

Help idk what to do

[river field]

Yeah so solve the equation

expand the right side

and compare the coefficients

[void cedar]

I can do this

[void cedar]

but Im stuck on this

[river field]

When you expand what do you get?

[void cedar]

x^2 - 2bx + b^2 + c

[river field]

good

now see how x has the coefficient -2b

and on the left side x has coefficient 6

so -2b = 6

[void cedar]

ahh okay

[river field]

similiarly b^2 + c = -2

[void cedar]

tysm

.close

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[daring sparrow]

Find all Natural Numbers for Which : (x+2)^4 -x^4 = y^3 .
I have Found no solutions can someone confirm and maybe present a Solution themself

[river field]

all natural numbers x and y I assume?

Graph says there are none

By expanding the left side we get

uhhh, hell

[solar hazel]

how could you possibly tell that from a graph

[river field]

Only meeting point is here

on the left and right they drift apart

[old chasm]

There is integer solution but not natural number

[solar hazel]

(-1,0) is the only integer solution i think

[solar hazel]

this is not a relevant graph

[river field]

ah nvm

[old chasm]

that checks out

[solar hazel]

[safe radish]

@daring sparrow Has your question been resolved?

[solar hazel]

it would be nicer to factor with difference of squares

[old chasm]

x^2 + 2x + 2 is your ...

[solar hazel]

not sure that’s right

[river field]

ye

[river field]

is it?

[old chasm]

it is

[river field]

doesn't seem correct

[solar hazel]

isn’t it 8(x+1)(x^2 + 2x + 2)

[old chasm]

$8(x+1)(x^2+2x+2) = y^3$

[flat frigate]

print("NAME")

[river field]

ah nvm

[old chasm]

oh... yeah plus sign

[river field]

hold on

ye is good

So we get this then?*

[safe radish]

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[safe parrot]

Determine the partial derivate of the function $f(x,y) = (xy^2+1)^5$

[flat frigate]

Merineth

[safe parrot]

I'd assume that i have to differentiate with wrt x and then wrt y

however how do i apply the chain rule here?

[summer crow]

you apply the chain rule the same way as you usually do for derivative

if you're differentiating wrt x

just treat y as a constant

so imagine you're d/dx (k^2 x + 1)^5

where k is constant

[safe parrot]

Right, that makes sense but

how do i handle the ^5 ?

[summer crow]

what's the derivative of x^5?

[safe parrot]

5x^4

so it's fine to do as such?

$5(xy^2+1)^4 * y^2$

[flat frigate]

Merineth

[safe parrot]

This would be wrt x?

[summer crow]

yes

[safe parrot]

$5(xy^2+1)^4 * 2xy$

[flat frigate]

Merineth

[safe parrot]

And this would be wrt y

and these two become a vector?

[summer crow]

yes

[summer crow]

depend on the context

[safe parrot]

or am i thinking of gradients?

[old chasm]

gradient here is vector, yes

but your question might not need to be answered in the gradient format

[safe parrot]

$arctan(y/x)$

Can i have some help differentiating this wrt y

chain rule would be

[flat frigate]

Merineth

[safe parrot]

$1 / (1 + (x/y)^2) * 1/x$

[flat frigate]

Merineth

[safe parrot]

$\frac{1}{1+(x/y)^2} * \frac{1}{x}$

[desert pasture]

$\left(\frac{1}{\left(1+\frac{y^2}{x^2}\right)}\left(\frac{\left(y-xy'\right)}{y^2}\right)=y'\right)$

[flat frigate]

Merineth

ƒ(Why am. I here)=I don't Know

[desert pasture]

you're differentiating x/y wrong

[safe parrot]

am i?

shit sorry

it should be

y/x

[desert pasture]

ah

[safe parrot]

$arctan(\frac{y}{x})$

[flat frigate]

Merineth

[safe parrot]

This is what i'm differentiating wrt y

$\frac{1}{1+(\frac{y}{x})^2} * \frac{1}{x}$

[flat frigate]

Merineth

[desert pasture]

you still haven't differeniate y/x right imo

[safe parrot]

Hmm

[desert pasture]

wrt y, you're differentating

right?

or partial derivatives

[safe parrot]

$y/x = yx^{-1}$

[flat frigate]

Merineth

[safe parrot]

This is what i'm thinking

which is just $x^{-1}$

[flat frigate]

Merineth

[desert pasture]

that changes the answer a lot , so are you differentating wrt y or the partial derivative ?

[safe parrot]

partial derivative for the given functions

$f(x,y) = arctan(y/x)$

[flat frigate]

Merineth

[desert pasture]

oh

[safe parrot]

I've already managed wrt to x

now i want to do it wrt y

$artcan(x) = \frac{1}{1+x^2}$

[flat frigate]

Merineth

[safe parrot]

This is the rule for differentiating arctan

where my x is (y/x)

then i just need to differentiate inside arctan

[desert pasture]

I can't help if it's partial derivatives, sorry

[safe parrot]

which is y/x

rip

:c

.close

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[solemn anchor]

can someone help me solve this

[quiet juniper]

Write out tan and csc in terms of sin and cos

[solemn anchor]

this is what I have so far

[quiet juniper]

where did the + come from

[solemn anchor]

holy guacamole

you're right

i dont even remember throwing that in there

[quiet juniper]

Also you're thinking of cos(2theta) = cos^2 - sin^2

[solemn anchor]

ohhhh

yeah

[quiet juniper]

And I think you're also thinking of cos^2 + sin^2 = 1 in that last line :p (plus, not minus)

[solemn anchor]

this should be right yes?

[quiet juniper]

Yup

[solemn anchor]

thank you

.close

[quiet juniper]

np

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[hasty marsh]

For (b) I need to use a graphing calculator(ti nspire 84) to find the roots of this function, I tried using nsolve to do this question, but the answer always pops up as zero, I would like to know how to set specific ranges of x when using nsolve

[real fern]

a container divided into three regions: A, B, and C. A gas molecule, M, initially located in region A. The problem requires calculating the probability that the gas molecule M will eventually move from region A to the outside. A particular twist in this problem is that after moving from one region to another, the gas molecule M may return to its original region, so it cannot be treated as a simple probability problem. The author shows great interest in this problem, and the teacher has promised to exempt one week's homework for the first person to solve it.

[safe radish]

@hasty marsh Has your question been resolved?

[safe radish]

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[desert pasture]

stupid question, but
lim of xsin(1/x) at infty
what is it?
not defined
right?

[chrome pond]

yh

thats: infty.sin(0)

[desert pasture]

got ti

thanks

.close

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[thin bridge]

no

Thats incorrect @desert pasture

[buoyant shadow]

it's 1

[lime dust]

.reopen

[safe radish]

✅

[desert pasture]

my thinking was sin(1/x) tends to zero

[thin bridge]

inf * 0 is also indeterminate indicating more work is required, not inherently undefined

[desert pasture]

ah yeah

thanks

got it now

[thin bridge]

can be expressed as
sin(1/x)/(1/x)
then u = 1/x gives you the trig classic limit

[lime dust]

U can yes

What i was about to say

[desert pasture]

thanks

.close

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[hasty marsh]

how can i find the range of this function

[narrow ridge]

Oblique

Asymptote

[hasty marsh]

plz explain

[safe radish]

@hasty marsh Has your question been resolved?

[mortal sandal]

personally I would recommend graphing it

do you know calculus?

[safe radish]

@hasty marsh Has your question been resolved?

[dusk sparrow]

do you know how to find inverse?

[safe radish]

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[gilded scaffold]

why is that not factorizable??

[safe radish]

@gilded scaffold Has your question been resolved?

[mellow cypress]

There is nothing to factorise. You either have 8xy when in A or 0 elsewhere

[gilded scaffold]

ok thx

[safe radish]

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[hasty blaze]

how do you do this

[chrome pond]

replace y with x+1/x and calculate dy/dx

[hasty blaze]

i get that this is quotient rule but the way the answer are arranged is confusing

[river field]

Jika maka lmao

[hasty blaze]

if , then

[river field]

Translate plz

[hasty blaze]

yeah its not in english

[river field]

Oh

[hasty blaze]

that meand derive right so thats

[chrome pond]

yeah its the derivative of y with respect to x

and y is a function of x

[hasty blaze]

1/x^2?

[chrome pond]

yupp

[hasty blaze]

ok

[chrome pond]

and if u fill that in some things will fall away nicely

[hasty blaze]

does quotient rule work here?

[chrome pond]

quotient rule ?

for what ?

[hasty blaze]

for this(?)

to make it end up like that

[chrome pond]

u dont need to take a derivative of a fraction here

[hasty blaze]

or is this chain rule?

[chrome pond]

no its quotient rule but u dont need to aply it in this excercise

[hasty blaze]

oh ok

im kinda lost honestly now that i know its
1/x^2

idk what else ther eis to do

[chrome pond]

1/x^2 is dy/dx

[hasty blaze]

oh ok

[chrome pond]

u can just fill it into the equations that are given and see what happens

[hasty blaze]

ok so

x^2*1/x^2 is just 1

[chrome pond]

yeah

[hasty blaze]

xy= 2x+1?

[chrome pond]

no

what is y

[hasty blaze]

y is x^2+1/x

oh wait

[chrome pond]

not x^2

[hasty blaze]

so xy is just x^2/x

[chrome pond]

?

[hasty blaze]

wait why not?

[chrome pond]

y is x+1/x

[hasty blaze]

oh yes

[chrome pond]

so xy is x(x+1/x)

[hasty blaze]

i wanna make it have the szame denominator

[chrome pond]

dont do that

[hasty blaze]

oh ok

[chrome pond]

its gonna give more trouble than needed

[hasty blaze]

ok now heres the issue most of those result in 0

but theres nothing that can take out the

[hasty blaze]

x^2

in this equation

[chrome pond]

what are u left with as an equation ?

[hasty blaze]

1 +/- x(x+1/x)

if i multiply th eback part it becomes

$x^2 +1 right?$

[flat frigate]

veryhuman

[hasty blaze]

yeah theres the problem theres nothing getting rid of x^2

[chrome pond]

is there another part to the question ?

[hasty blaze]

uhh not that i know of i took a picture of the engtire question

[chrome pond]

because u have this

c and e can never happen so they are already wrong

[hasty blaze]

a seems promising

[chrome pond]

a can also not be right otherwise y is +infty

no because then x=0

and y=x+1/x

thats gonna blow up to +infty

[hasty blaze]

oh alr

[chrome pond]

and u have 2 eqs that are

$-x^2 +2 = 0$

[flat frigate]

spookyspaghetti

[hasty blaze]

so clearly b and d arent right

since yeah tehre cant be copies

hold on

wait wait wait

[chrome pond]

those are the only equations that are solvable

[hasty blaze]

i derived y right early on

what did i say it was

[chrome pond]

1/x^2

has to be -1/x^2

[hasty blaze]

1+1/x^2

right?

oh

wait is that where the msitake is

[chrome pond]

hahah

welcome to math

[hasty blaze]

oh yeah i added the xs together

[chrome pond]

gotta do it all over again

[hasty blaze]

i didnt add them this time bc you said somehwere dont ad the denomenators together

ok wait

so now

[chrome pond]

yeah u dont have to

[hasty blaze]

y= (x^2+1)/x^2

[chrome pond]

,derivative 1/x

[hasty blaze]

OH

bc that is dy/dx

[chrome pond]

,derive 1/x

how tf does this bot work

,help

[flat frigate]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[hasty blaze]

oh

the one that does this $\frac{dy}{dx}$?

[flat frigate]

veryhuman

[chrome pond]

yeah

u can ask for a derivative i think

[hasty blaze]

oh im not sure how to do that

[chrome pond]

,w derive 1/x

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spookyspaghetti_

[chrome pond]

aha

[hasty blaze]

but to get the image lie that jsut put $ around the

[chrome pond]

see

[hasty blaze]

oh the wolfram

wait wait it becomes negative?

[chrome pond]

ur dy/dx is wrong

hahaha

1/x is x^-1

gotta remember that negative sign

[hasty blaze]

OHHHH

right right

-1

[chrome pond]

u forget that once on an exam and then never again

[hasty blaze]

ig -1

im glad the exam im studying this for is multiple choice tho

so hoepfully if i mess up it just wont show upa s an option

[chrome pond]

is the question wich one cant be solved, by chance ?

[hasty blaze]

no

[chrome pond]

then idk how u choose an answer

i got this

[hasty blaze]

im askinga friend about this aswell

[chrome pond]

all of em are solvable except e

[granite idol]

it's asking which one is true

[hasty blaze]

they said the answer is

this one

WAIT

THATS D?

[chrome pond]

huh

what is this question

[granite idol]

yes, it's D

[hasty blaze]

how

[granite idol]

what'd you get for dy/dx?

[chrome pond]

-1/x^2

[hasty blaze]

-1/x^2

[granite idol]

nope

[chrome pond]

?

[granite idol]

it is 1 - 1/x^2

[chrome pond]

omg

[hasty blaze]

SO

(x^2-1)/x^2

OH

THAT SHOW YOU

GET RID OF THE

X^2

[granite idol]

you can just leave it as 1 - 1/x^2

but yeah

[chrome pond]

imma just study to be a trashman ig cant even take a derivative anymore