help-33

[brittle vessel]

change everything back

preferably

[queen prairie]

what

wdym

wdyn change back

[brittle vessel]

bring back the sqrt(3) forms

[queen prairie]

ah

OH

-2sqr3 + 4 - 3sqr3 + 6

[brittle vessel]

and now you can simplify

[queen prairie]

so we get -5sqr3 + 10 ?

wait i feel like i've fucked it up

-2 -3 = -5 ?

[brittle vessel]

xd

[queen prairie]

i mean it is

ig

[brittle vessel]

yeah

[queen prairie]

so it's alr

show that a > b

this will be easier

i think..

nvm it feels like hell tf

so we write it first tf

oh

i think ik

2sqr 3 - 4 > 3 sqr 3 - 6

i feel really nice so i'll sqruare it all

2sqr 3 - 4 > 3 sqr 3 - 6 | ^2

we get uhm

[brittle vessel]

let me stop you now xd

[queen prairie]

4*3 - 16 > 27 - 36

no?>?

DAMN

alright

[brittle vessel]

oh

i did a good call

so

you cant square them like that

[queen prairie]

why not

[brittle vessel]

these are not multiplied

[queen prairie]

wdym

[brittle vessel]

(2-1)^2 is not 4-1

[queen prairie]

so what must we do

[brittle vessel]

you can square them correctly

if you want to practice

but i dont feel like its gonna get any more nice looking for us

[queen prairie]

idrm

if it'll help me learn faster

i need it

be it drugs or selling my soul idrc

[brittle vessel]

okay, do it, but after your test xd

now we need to know which one is bigger, right?

[queen prairie]

ye

my g my g

i got an idea !!

[brittle vessel]

so if we change them by the same amount thats not gonna change it

[queen prairie]

we add 6 to both of them??

[queen prairie]

ye

[brittle vessel]

good idea!

[queen prairie]

2sqr 3 - 4 > 3 sqr 3 - 6 \ + 6 is

2sqr 3 + 2 > 3 sqr 3

THEN

we can easily square it all

[brittle vessel]

but do something with the other part too pls

[queen prairie]

other part ?

[brittle vessel]

the roots

[queen prairie]

we'll square it all

wont it work?

[brittle vessel]

xddddd

[queen prairie]

oh

..

SHIT

or

[brittle vessel]

what is with you and the squaring?

[queen prairie]

no

wdym

i did the adition correctly

[brittle vessel]

you like squaring things?

[queen prairie]

i find it easier

and faster

but wont it work?|

[queen prairie]

i actually hate it but it's the easiest one ik

and hope and pray for it to work

[brittle vessel]

oh, then please DO square it, and then youll see its not always easier

[queen prairie]

2sqr 3 + 2 > 3 sqr 3 \ ^2

[brittle vessel]

just kidding, dont do it

[queen prairie]

we get uhm

damn

Why nott

we get 12 + 4 > 27 ..

wait wtf

[brittle vessel]

lets call sqrt(3) an apple, so we have 2apples+2 and 3 apples

[queen prairie]

ye

[brittle vessel]

how would you solve this

i got rid of the confusing root signs

[queen prairie]

idk..

another hint|?

[brittle vessel]

the next step is the solution

so no hint available anymore

[queen prairie]

ahah

[brittle vessel]

take away 2 apples

[queen prairie]

damn

[brittle vessel]

from both side

[queen prairie]

but we end up with 2 > sqr 3

dont we ?

[brittle vessel]

we dont know the relation

thats the question

[queen prairie]

show that a > b

so we know for sure that it is higher than b ?

or it is not sure

[brittle vessel]

ah, sorry, so they told us this

[queen prairie]

yeah..

[brittle vessel]

anyway we could figure it out

[queen prairie]

so we get 2 > sqr 3

[brittle vessel]

yes

[queen prairie]

do we square it up now ??

[brittle vessel]

if you need to

[queen prairie]

FINALLY !!! FINALLY

[brittle vessel]

i mean

i know you WANT TO

[queen prairie]

i mean it is fucking evident it is higher BUT

4 > 3

GG

squared it up

no

w

where tf were we

at C

shit

tf was that sign

that bracket i mean

[brittle vessel]

i think its the whole parts sign

[queen prairie]

it is a squared bracket

so it is different

[brittle vessel]

different from?

[queen prairie]

from the | |

but i think

we just put the first number in the || ?

[brittle vessel]

who said anything about ||

[queen prairie]

idk

what is the whole parts sign

[brittle vessel]

[ ]

you asked this

[queen prairie]

nvm

i remember it now

[brittle vessel]

what is this? xd

[queen prairie]

idfk

so how we doing

this is primary school shit

[brittle vessel]

what is your question?

[queen prairie]

show that

[brittle vessel]

what you got so far?

[queen prairie]

4 > 3
?

wdym by what i got so far..

i got |a| and |b|

Oh

do we just

take that

and uhm idk what with the original number?

a = |2sqr3 -4 |

b = |3sqr3 -6|

[brittle vessel]

a is not that

neither is b

[queen prairie]

what

ik that's not the []

but isbt the ||a part of the [] ?

like || - the normal number

or

[brittle vessel]

those 2 numbers are obviously cant be equal

we just did b)

[queen prairie]

ye..

[brittle vessel]

now they ask you to calculate their whole parts

[queen prairie]

idk how to calculate that tho

[brittle vessel]

why not

[queen prairie]

i dont remember

or

[brittle vessel]

.

[queen prairie]

nah i dont

[brittle vessel]

.

[queen prairie]

oh

[brittle vessel]

what integers are next to it, basically thats the Q

[queen prairie]

WAIT HAVE I DONE THAT SHIT WITHOUT KNOWING WHAT I AM DOING

so can we redo this one

for the third time ik

but

3 - sqr 6

we will

take the closest integers that are sqrt

4 and 9

which are 2 and 3

and wince sqr 6 OBVIOUSLY isn't sqr 4 or sqr 9

it'll be < anbd <

so

now our problem

2sqr3 - 4

and 3 sqr 3 - 6

we put them under the squareroot?

sqr12 - 4

and sqr 27 - 6

[brittle vessel]

we can forget the 4 and 6

[queen prairie]

then we work with the square roots ?

[queen prairie]

for the moment

[brittle vessel]

they wont help us

[queen prairie]

ye

[brittle vessel]

yeah

[queen prairie]

so

sqr 12 and sqr 27

3 and 4

sqr 9 and sqr 16

sqr 9 < sqr 12 <sqr 16

right?

[brittle vessel]

yes

[queen prairie]

so out first is 3 < x < 4

• 4 tho

-4

since X is smaller than 4

and HIGHER than 3

higher than 3 so it cant be 1

it must be at least 1,...

[brittle vessel]

uhm

[queen prairie]

wait

wtf im doing

[brittle vessel]

so the whole part is 3

end

[queen prairie]

.

[brittle vessel]

and now do b

[queen prairie]

so basically

|x| = 3

3 < x < 4 ?

well forget about it

the first number is the whole part

i got it

sqr 27

[brittle vessel]

minus -4 is there too

[queen prairie]

so we get -1

[brittle vessel]

so [a]=3-4

[queen prairie]

so it is -1

so we must demonstrate b and get -1 as well

sqr 27

ik that sqt 25 is 5

[brittle vessel]

yes, and after this i have to go

[queen prairie]

and sqr 36 is 6

so it is between 5 and 6?

[queen prairie]

alr

[queen prairie]

right?

so we get 5 < x < 6

so the whole part is 5.. ?

WELL IT WILL GIVE ME THE RIGHT ANSWER SO IDFC

5 - 6 = -1

GOD BLESS IT

FINALLY

-1 = -1

[brittle vessel]

indeed, good job

[queen prairie]

thank you !

since u have to go

i'll do these on my own

or try to

or . find someone else

but thank you spreeter

[brittle vessel]

okay

[queen prairie]

for wasting 1 hour of your time educating me

[brittle vessel]

(ideally you try them before coming here)

[brittle vessel]

youre welcome, good luck tomorrow

[queen prairie]

thankyou

.close

[marsh citrus]

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[marsh citrus]

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[weary night]

Assume you have an array A with n elements. Prove that there can be at most 1 majority element i.e. an element in A that appears more than n//2 times

[weary night]

Just smthn I was wondering about and if it could be proven mathematically

so i could also find an upper bound for the number of majority elements that appears n//3 times

[idle pumice]

suppose there are (at least) two majority elements, then we get a contradiction

[marsh citrus]

@weary night Has your question been resolved?

[weary night]

ohhhh wow didnt think to prove with a contradiction

thanks

.close

[marsh citrus]

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[marsh citrus]

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[gentle fable]

How we get -2sqr3/3 from -1/sqr3/2

[final bolt]

are you meaning (sqr3)/3 or sqr(3/3)

[gentle fable]

this

[marsh citrus]

@gentle fable Has your question been resolved?

[stark trail]

that is because you are dividing fractions

keep change flip

you have

(1/1)/(sqrt(3)/-2))

and then you just keep change flip

[gentle fable]

i see, but from where that additional 1 came from

[marsh citrus]

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[stark trail]

put g(x) into f(x)

[vapid bough]

ohhh

that makes sense lol

thanks

[stark trail]

np

[vapid bough]

would i just plug the whole equation with x into f(x)

like this

1/ (5x+8) - 5

[stark trail]

yes

[vapid bough]

I see got thank you

.close

[marsh citrus]

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[trail cove]

Can someone confirm if the proof is ok?

[trail cove]

This is the claim

$f(X)$ is defined as $X\cap 2\mathbb{Z}$

[elfin berry]

42

[sand kindle]

I don't really get this, but the third row doesn't follow from the second

[trail cove]

how so?

im just subbing in the def of f(X)

[trail cove]

since f(X) is defined as this

[marsh citrus]

@trail cove Has your question been resolved?

[marsh citrus]

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[orchid oxide]

hi

[orchid oxide]

this has to be done on an excel spreadsheet

and im lost

on how to do it

with a setup somewhat similar to this

im a little bricked in the head i just reread the question and realized the tpc is the payment

i got this

.cloes

.close

[marsh citrus]

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[still temple]

why does the cross products of the gradients give a vector parallel to the tangent line

I know gradient "points" toward the direction of steepest ascent

so the cross product would be a vector that is orthogonal to both gradients

how is this parallel to the line

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[young owl]

the tangent line itself is orthogonal to the curve by definition of tangent line, so if you find any orthogonal vector, it's parallel to your line

and the gradient gives you the tendency of the curve along all directions so there's no dimension problem since the cross product will be orthogonal to it for all directions

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

Tendency of the curve for all directions?

[young owl]

the partial derivative relative to x gives how the curve evolves from a point in x direction
and the gradient has all partial derivatives

it is some sort of summary of directional derivatives

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

So the cross product of a gradient will always be orthogonal to a tangent line

[still temple]

.close

[marsh citrus]

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[hallow marten]

Could someone help mne with trig?

[hallow marten]

My Personal Numbers:

Read and follow the directions for the following activity. You must show work for full credit.

For this activity you will use two numbers specific to you.

1st number is your shoe size to the nearest whole number

2nd number is your age.

Instructions:

A. Subject Line: Use your two numbers for the Subject Line of your Post. (Ex. SHOE SIZE-AGE) (Ex. 11 - 21)

B. Body: For the body of your post, answer the following questions, showing all work.

For this activity, use your shoe size as VR and your age as E in an RC Circuit. Using those values, find VC and θ.

A =Followed all directions, including correct subject line. All work was shown. Post was neat and organized. All answers correct.

B =Followed most directions. Most answers correct/work shown. May have lacked organization or not shown sufficient work.

C =Posted a response demonstrating minimal proficiency and understanding. May or may not have followed all directions. Errors were present in the mathematics or included answers with no work shown.

D = Posted a response but without demonstration minimal proficiency or understanding.

F = Did not post a response.

cant barely think have been working all day and would really appreciate some guidance.

[marsh citrus]

@hallow marten Has your question been resolved?

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[boreal jetty]

how did they go from the circled step to the next step? namely, the 1

[tropic forge]

u separate it into h/h - the right part

h/h = 1

for example: (2+4)/10 = (2/10) + (4/10)

[boreal jetty]

ohh i see, thank you

what about this step?

oh wait nevermind

that's all, thank you!

.close

[marsh citrus]

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[supple jolt]

I don’t really remember how to do this, can someone help me

[elfin berry]

tatpoj

[trim quest]

"signed" area, meaning that we count it as positive if it's above the x-axis, and negative if it's below the x-axis

@supple jolt

[marsh citrus]

@supple jolt Has your question been resolved?

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[cosmic veldt]

how do you simplify

[cosmic veldt]

(x^-3)(6x^2)(x^3 - 4) + (-3x)^-4(x^3-4)^2

aka derivative of

[stoic saddle]

were you forced to do it in that particular way

[cosmic veldt]

nah

[stoic saddle]

cause i know of a much less painful method

[cosmic veldt]

sure

induldge me

[stoic saddle]

$\frac{(x^3-4)^2}{x^3} = \frac{x^6 - 8x^3 + 16}{x^3} = x^3 - 8 + 16x^{-3}$

[elfin berry]

Ann

[cosmic veldt]

f you type fast

yooo

that aint half bad

will do

[cosmic veldt]

but how do i smplfy this

for math knowledge skills

since i am unable to

[fathom ridge]

expand all parenthesis

[cosmic veldt]

can you rly with like -3x^-4

[fathom ridge]

sure

[fathom ridge]

You'll end up with what you'd get from differentiating this

[cosmic veldt]

howde you expand this then

[cosmic veldt]

ye

[hard gull]

shouldnt there be a 2 there?
(x^-3)(6x^2)2(x^3 - 4) + (-3x)^-4(x^3-4)^2

[fathom ridge]

(a - b)^2

[cosmic veldt]

where is there

[hard gull]

where i put it

[cosmic veldt]

hmm

[hard gull]

cuz power rule

[fathom ridge]

(bold it)

[cosmic veldt]

nah

cuz 3x^2*2

=6x

i already did it

[hard gull]

d/dx (x^3 - 4)^2 = 2(x^3 - 4) 6x^2

[cosmic veldt]

ehh

well idk

i dont think so

[cosmic veldt]

why 6x^2 on the end

u get that from derivative of x^3

to 3x^2

then * 2 (the 2 youa dadded)= 6x^2

[hard gull]

oh right right

my brain need time to function

[cosmic veldt]

bro ive been doing math for like 4h and achieved basilcy nothing

this is why i dont study smh

[hard gull]

$$\frac{(6x^2)(x^3 - 4)}{x^{3}} + (-3x)^{2}\times-4(x^{3}-4)^{2}$$

[elfin berry]

JustToPro

[cosmic veldt]

u get (x^3-4)^2/(-3x)^4

dont think thats expandable

unless u want to do -3^4

which is.. scuffed

[cosmic veldt]

so what to do

the main question..

[hard gull]

multiply x^3 first

[cosmic veldt]

$$\frac{(6x^2)(x^3 - 4)}{x^{3}} + (-3x)^{-4}\times(x^{3}-4)^{2}$$

[elfin berry]

™Vlad The Lad

[hard gull]

and then open all brackets and simplify

[cosmic veldt]

you missed ^-4

i may have typed it worse originally so

[hard gull]

$$\frac{(6x^2)(x^3 - 4)}{x^{3}} + \frac{(x^{3}-4)^{2}}{(-3x)^{4}}$$

[elfin berry]

JustToPro

[cosmic veldt]

ye i get that

same thing

[hard gull]

the ^4 would cancel the negative

so its 81x^4 there

multiply the left fraction with 81x so u can combine both and add and stuff

[cosmic veldt]

43,046,721

[fathom ridge]

you can comfortably do 3^6 in your head lol

[cosmic veldt]

i... cannot

[fathom ridge]

[cosmic veldt]

to 729?

nahhh

anyways if you expand thats like... not good

u get 486x^6 - 1944x^3 ......

cant do jack from that

for

[fathom ridge]

In still kind of confused as to what you're expanding

[cosmic veldt]

i found the deriviative of the 2nd image i posted

and that equals that long thing i wanna simplify

since uv' + u'v

[fathom ridge]

Where does this come from then

[cosmic veldt]

expanding... as you said

[fathom ridge]

wait they deleted their message

[hard gull]

that is wrong , no?

[cosmic veldt]

$\frac{486x^6 - 1944x^3 + x^2 - 8x^3 + 16}{81x^4}$

[elfin berry]

™Vlad The Lad

[hard gull]

how do u get those big numbers?

486 , 1944

there is hardly any exponents on the constant / coefficients

[cosmic veldt]

idk whatevs

.close

[marsh citrus]

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[delicate anvil]

Hey, I need help with this problem. Please note that english is not my main language.

From an external point P to a circle with center O and a radius of 1 cm, tangents are drawn to the circle, intersecting it at points A and B. The area of the polygon PAOB is √3 cm² if the distance from P to O is one of the following:
A. 3 cm
B. √3/2 cm
C. 4 cm
D. 2 cm
E. 3/2 cm

I have tried considering the triangle OAP and trying to figure out the length of OP by knowing the area of said triangle and the length of AO but I can't seem to get anywhere. I also tried considering the rhombus AOBP and trying to find the answer that way but I had no luck. Maybe I'm missing some theorem I don't know about?

[brittle vessel]

did you make a diagram?

[gentle sorrel]

well

polygon PAOB

can be splited into two triangles

[delicate anvil]

yeah

but idk what to do with it

only think i know is that the area of each triangle is √3/2

[gentle sorrel]

use Pythagorean theorem

[delicate anvil]

and that AO is 1

but idk AP

[gentle sorrel]

my national language keyboard be like:

💀

but anyway

triangle

AOP

's area is

wait.. should start from here first

what is ∠PAO?

[delicate anvil]

90

[gentle sorrel]

great

[delicate anvil]

oh

bruh

[gentle sorrel]

if you solved this problem, close the ticket with .close

gets out of here

[delicate anvil]

thought i had it

still have no clue of what im doing

[gentle sorrel]

uh

i had to go somewhere but sorry

you can get the length of AP at that point

using the area of triangle

[delicate anvil]

√3/2=(OP * OA * sin90°)/2?

[gentle sorrel]

uh?

sin90 is 1 (degrees)

wait

you should multiply like $\frac{1}{2}$absinθ

[elfin berry]

A_Note

[gentle sorrel]

where θ is an Included angle

[delicate anvil]

i took a as OP and b as OA

oh i see

but idk the angle between them

i should take AP then and not Op

[gentle sorrel]

should took a as AP and b as AO (the order of a,b doesn't matter)

[delicate anvil]

so AP is √3

[gentle sorrel]

yes

[delicate anvil]

and now i just go on with √(AO² * AP²)

and find OP

[gentle sorrel]

yes

wait

it isn't *

[delicate anvil]

plus

[gentle sorrel]

it's +

[delicate anvil]

yes

[gentle sorrel]

yes

[delicate anvil]

my bad

ok the result is right

but how

is AP the same as the area of the whole shape

I don't get it

[gentle sorrel]

it isn't exactly same

AP is a length

but area is area

their units are technically different

[delicate anvil]

true

ohhh i never write the units down

forgot that cm is a thing and cm² is another

right

[gentle sorrel]

you don't need to write untis down when the problem doesn't give unit

wait

there's cm

in the problem

so maybe you should

[delicate anvil]

ye

but anyway

thanks a lot

saved me hours of agony

[gentle sorrel]

lol

gets out of here

by the way

[delicate anvil]

.close

[marsh citrus]

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[cold sundial]

What does it mean by "selecting the θ that won't change the trigonometric function"
ex: they took (θ-π) for tan

[marsh citrus]

@cold sundial Has your question been resolved?

[cold sundial]

<@&286206848099549185>

[marsh citrus]

@cold sundial Has your question been resolved?

[cold sundial]

<@&286206848099549185>

[pastel granite]

You use the angles along axis like 2pi, pi/2, pi, 3pi/2 to place the unknown angle into different quadrants.

for example, if you wanna place an angle in the 2nd quadrant, it either has to be
more than pi/2 = (pi/2 + θ)
less than pi = (pi - θ)

the general rule of thumb is, when you use the fractional axis angles like pi/2 or 3pi/2,

the trig function changes to it's cofunction.

for example

Sin(pi/2 + θ) = cosθ

But, if the axis angle isn't a fraction, it remains the same.

Sin(pi - θ) = sinθ

[pastel granite]

here, you used (pi - θ) instead of (pi/2 + θ) to put the angle in 2nd quadrant cause (pi - θ) won't force you to change your tan to it's cofunction cot

does that make sense?

[pastel granite]

refer to this if you don't know how axis angles and positive and negatives work for trig function values

[marsh citrus]

@cold sundial Has your question been resolved?

[cold sundial]

It somewhat made sense...
like... isn't the angle still the same? How did it not change the function just by interpreting the angle in a different way...

Also, why did we take θ-π instead of π-θ in that formula derivation I put pic of in original question?

[pastel granite]

1. yes the angle is the same. As for why they're converted to cofunctions, I can't think of a concise way of explaining rn, I could explain it if I had the chance to draw in front of you, but not through typing I can't. Keep searching, you'll find the answer somewhere.
   you could also figure it out on your own if you know basic triangle trig and can apply it to the unit circle.

2. if you notice the screenshot I sent above, tan is not positive in 2nd quadrant.
   and, -tanX = tan(-X)

so, -(pi - θ) = θ - pi

[cold sundial]

1. Hbt a drawing app
2. Oh. Ty, that explains it.

[cold sundial]

Like, I'm not asking why they are converting to cofunctions. I'm asking why they aren't...

[pastel granite]

you're asking a bit too much from me,
im happy to help but Im busy rn

[cold sundial]

Sorry.

[pastel granite]

no worries

feel free to dm me some other time

[cold sundial]

I'm well acquainted with why they are converting.
That's why, I have problem understanding why they aren't.

[cold sundial]

Any particular time/date

[pastel granite]

Im gonna be honest, I can't comprehend any of that

[pastel granite]

leave me a dm, I'll reach out to you

[cold sundial]

kool

[pastel granite]

and close the channel if you're done for now, thanks

[cold sundial]

Thank you.

.close

[marsh citrus]

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[whole linden]

https://cdn.discordapp.com/attachments/975379968714477631/1161346208938336377/766268632703434794.png?ex=6537f6f0&is=652581f0&hm=d47ba58ef54fd99c103b979baa16c06325f7543a6773e5271cf021f3ee9e508d&

[whole linden]

can this be differentiated w.r.t x, if I don't know the relation b/w x & y?

[static quarry]

why is x both an integration limit and the dummy variable for integration

and does y depend on x or what

[whole linden]

$\int_{0}^{t} y ,dx$

[elfin berry]

kinglacto

[whole linden]

ig I dont know

[static quarry]

well if y does not depend on x, then it's a constant as far as the integral is concerned and so the integral is just yt, which of course can be differentiated

otherwise you need the fundamental theorem of calculus

with appropriate hypotheses for y

[whole linden]

what if it was dependent tho? then it wouldn't be treated as a constant, right?

[static quarry]

correct

[whole linden]

hmm then it cant be solved, nvm im dumb

[static quarry]

formally you would have something like
$$\int_0^t y(x),dx = Y(t) - Y(0)$$
where $Y$ is an antiderivative for $y$, then you would differentiate that and get back $y(t)$, if the hypotheses for the FTC hold

[elfin berry]

Bungo

[whole linden]

hm yea makes sense, thank you

[static quarry]

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[queen prairie]

So my comp science teacher is like " you know geometric and algebric progressions ?" and we were like no. then he said that we'll learn it for the next class so NOW i have to learn stuff i'll do in 4 years till tmrw..

[queen prairie]

so..

wtf is that

[fathom ridge]

You're better off looking on YouTube and khan academy

[tall umbra]

Agreed.

[static quarry]

what do you mean "in 4 years", what level are you at, high school?

[fathom ridge]

These channels are meant generally for helping with single problems

[queen prairie]

frewshmen at highschool

[static quarry]

i mean it's not like these won't be useful to you throughout all math courses

[queen prairie]

and it's what we'll have at the exam in 4 years

[static quarry]

ah i see

yea try khan academy, they're usually decent for what they cover

[queen prairie]

alright , thank you..

i'll try to leanr that

[static quarry]

gl!

[queen prairie]

.cancel

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[marsh citrus]

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[uneven pier]

I'm supposed to define this set, but I'm getting thrown off by the x = 1/x.

[uneven pier]

I just need an explanation as to what the x = 1/x means in this scenario.

[runic temple]

Its the set of real numbers that satisfy that equation

[uneven pier]

So, the answer is just every real number except for 0?

[runic temple]

does 2=1/2?

[uneven pier]

Oh wait.

No, I get it now.

I'm forgetting basic math over here.

[runic temple]

It happens

[uneven pier]

Thanks.

[runic temple]

Youre welcome!

[uneven pier]

.close

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[abstract kernel]

need help with this

[marsh citrus]

@abstract kernel Has your question been resolved?

[abstract kernel]

<@&286206848099549185>

[marsh citrus]

@abstract kernel Has your question been resolved?

[marsh citrus]

@abstract kernel Has your question been resolved?

[abstract kernel]

<@&286206848099549185>

[tacit vale]

which q?

[abstract kernel]

all honestly 😟

[marsh citrus]

@abstract kernel Has your question been resolved?

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[solar swan]

Hello, I have a question on my volume by slicing calculus homework. I have determined on this problem that since each cross section is perpendicular, I am integrating on the x-axis. Given this I have determined that the base of each square to be the function I wrote down, squared. My understanding is now that I need to integrate across this value, however finding the anti-derivative of this function seems quite unreasonable to do by hand. Could use some help regarding where to go from here or any mistakes I've made until this point.

[marsh citrus]

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@solar swan Has your question been resolved?

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[pine zealot]

In a triangle rectangle ABC, a point D is chosen on the hypotenuse AC. We draw perpendiculars DE and DF to the sides AB and CB respectively. Determine the point D for which EF has minimum length.

[marsh citrus]

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[echo folio]

Here, why is f_x and f_y not continuous at (0,0)?

I don't get how they are not continuous here

[main idol]

show what you get for the partial x

[echo folio]

Uhh sorry

I'll rephrase

I made a mistake

didn't include at (0,0)

Got this for partial x tho

@main idol Are you there

[main idol]

take a path y=kx and show it doesn't have the same value for different k

[echo folio]

So basically using the limit definition?

To find continuity?

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[tawdry stone]

How do i solve for x?

[random palm]

equate them

[marsh peak]

"Q is the midpoint of PR", what does that tell you about the distance between Q and P and between Q and R?

[tawdry stone]

They’re equal?

[marsh peak]

Yes, and you are told that the distance between Q and P is 7x - 16 and between Q and R is 4x + 2

Hence 7x - 16 = 4x + 2

Can you solve for x here?

[tawdry stone]

Got it

I should be able to solve

[tawdry stone]

-9 right?

For x

[marsh peak]

Could you show your work?

[tawdry stone]

Very bad handwriting

Also brb im getting food

[marsh peak]

Should be 7x instead of 2x

[marsh citrus]

@tawdry stone Has your question been resolved?

[tawdry stone]

How do i get that?

[marsh citrus]

@tawdry stone Has your question been resolved?

[marsh peak]

I mean it's given to you as 7x - 16 instead of 2x - 16, which is what you wrote

\begin{align*}
7x - 16 &= 4x + 2 \
3x &= 18 \
x &= 6 \
\end{align*}

[elfin berry]

A Lonely Bean

[marsh citrus]

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[broken sorrel]

this cos is really throwing me off, i tried converting to polar, i tried 2 paths for x=0 and y=0 and y=x i am unsure of how to solve this otherwise

[marsh citrus]

@broken sorrel Has your question been resolved?

[tawdry stone]

Oh.

[marsh peak]

I would try x = y^2

[broken sorrel]

this gives 0/2 no?

y^4*cos(y)/2y^4

[marsh peak]

Why?

Right

So cos(y)/2

[broken sorrel]

yes

[marsh peak]

What does that approach?

[broken sorrel]

taking the limit gives 0

[marsh peak]

So you are saying cos(0) = 0?

[broken sorrel]

omg

i was thinking sin

mbmbm

ok big ty

.close

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[plucky trench]

I have a question on a math problem. I know the answer to the problem but I am confused on how to get there. The question is what option is perpendicular to y=3x+5. Can anyone help me?

[main idol]

https://www.cuemath.com/perpendicular-lines-formula/

[plucky trench]

I know the formula m1xm2=-1

but the answer to the problem is confusing(I had a circuit and the only leftover answer on the answer key applies to this problem)

A:3y+x=15

B:2y=3x+6

[main idol]

put your choices in that form

[plucky trench]

C:2y=-3x+6

[main idol]

solve for y in all your choices

[plucky trench]

the answer is C and it is y=-3x+3

I am just confused on how to get there

since if you follow the perpendicular slope formula it should be -1/3x not -3/1

[main idol]

can you screenshot the options and exact problem

[plucky trench]

sure

one second

none of the answers make sense to me. I thought the slope is supposed to be -1/3 or is it just a program mistake?

[main idol]

did you solve for y in option A?

[plucky trench]

3y=15-x

and divide by 3

oh 0_0

[main idol]

yea so it isn't C

[plucky trench]

thank you for your help xd

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[pulsar bluff]

hey guys, i solved the problem just fine but I feel like there's something fundamental i'm missing and it should've been a lot easier

[pulsar bluff]

so here's the problem

and here's my long ass solution

to find k, i basically had to solve P(t)

and i used bernouli

but the way webwork words it, it feels like i could find k on its own and then solve p(t) with it?

thing is that's a weird form of the logistics equation that the professor nor the textbook uses

like i'm used to this, and there's no k in it. there's K which is different

soo was the way i did it the only way or am i missing something a lot simpler?

[sacred kestrel]

the way my book writes it

[pulsar bluff]

interesting. i tried to get to that by factoring the r out from my textbook to get dP/dt = rp(1-P/K) and maybe my book just replaced k with r but that's not right either. because my book says r is the initial condition. which is not 1.23

[pulsar bluff]

what book is this? i'll see if i can find a copy and read more

[sacred kestrel]

Stewart's Calculus, Single Variable

the solution

[pulsar bluff]

damn this book goes in a lot more detail than mine

ok thank you very much

.close

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[still temple]

Explain each step

i'd like to see your explainations

[marsh citrus]

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

[still temple]

2

<@&286206848099549185>

im super stuck please

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[runic dust]

whats a common base there

for 27 and 9

I mean without even doing that

just do it straight up

[marsh citrus]

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[short sparrow]

what do i do after ive found the whats between (ping me if you have a answer)

[desert dirge]

,rccw

[desert dirge]

it is not between 100 and 121

[thin bison]

i think you didn't see that i has a sq root

[short sparrow]

forgot to add the square root

sign

100sq and 121sq

what do i do after i find whats between?

[wary condor]

Is 101 closer to 100 or 121?

[short sparrow]

100

[wary condor]

so sqrt(101) is close to what number?

[short sparrow]

uh

10?

[wary condor]

yeah

[short sparrow]

what about

sr97 is between 81sr & 100sr therefore it is =~

[short sparrow]

recreating this question

[wary condor]

Mhmm, did your teacher cover how you're supposed to be doing these approximations?

[short sparrow]

yeah but i didnt quite understand it

[wary condor]

There are many roughly okay ways of doing this approximation, and I can't be sure how you're expected to carry it out

sqrt(97) will be close to sqrt(100), but it depends on how close 97 is to 100, for instance

[short sparrow]

,rccw

[short sparrow]

mhm

[wary condor]

Did you do some sort of interpolation?

[short sparrow]

interpolation?

[wary condor]

If you haven't heard of it, don't worry

What course is this?

[short sparrow]

uh

yeah mb

i dont really know

nvm i got it

.close

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[sudden mauve]

The answer is -7-12x but I got -9-12x. I was wondering on what line I messed up on

[still light]

(-1)^4=1 not -1

I think that should fix your answer.

[sudden mauve]

.close

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[still temple]

how to do this

[tacit vale]

evaluate the bracket and the power seperately

[still temple]

I was thinking about: (x-2+2)/(x-1)ln (x^2+1)/(x^2-x)

,w differentiate (x-2+2)/(x-1)ln (x^2+1)/(x^2-x)

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by sherby11999

[woeful cipher]

the answer, doctor hugo

is e!

[still temple]

lint

expand

how

[woeful cipher]

then

lets start over.

[still temple]

ok

restart

[elfin berry]

johnny123

johnny123

johnny123

[woeful cipher]

$(x-2+\frac{2}{x-1}) \cdot \ln\left(\frac{x^2 + 1}{x^2 - x}\right)$

[still temple]

what about this?

please?

[woeful cipher]

dude

idk why its not working

[still temple]

WHY IS IT NOT WORKING

ASNER

ANSWER

what is not working

[elfin berry]

johnny123

[woeful cipher]

ur such a hugo

ok

now lim x to inf of this

Left Side races to inf

while Right side races to 0

we need 0/0 or inf/inf to use Lhopital

SO

replace right side with divide 1/rightside

$(x-2+\frac{2}{x-1}) \div \frac{1}{\ln\left(\frac{x^2 + 1}{x^2 - x}\right)}$

[elfin berry]

johnny123

[woeful cipher]

$\frac{(x-2+\frac{2}{x-1})}{\frac{1}{\ln\left(\frac{x^2 + 1}{x^2 - x}\right)}}$

[elfin berry]

johnny123

[woeful cipher]

now differentiate top and bottom

byebye

,w d/dx (x-2 + 2/(x-1))

[elfin berry]

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Requested by johnny1230851

[woeful cipher]

,w d/dx 1/ln((x^2 + 1)/(x^2 - x))

[elfin berry]

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Requested by johnny1230851

[woeful cipher]

,w lim x to inf of (1-2/(x-1)^2)/((x^2 + 2x - 1)/((x-1)x(x^2 + 1)ln^2((x^2 + 1)/((x-1)x)))

[elfin berry]

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Requested by johnny1230851

[woeful cipher]

trivial

then e both sides

e^1 = e

byebye

exp both sides

[woeful cipher]

cuz this

@still temple

wolf

[still temple]

.close

[marsh citrus]

Channel closed

Closed by @rough stag

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[fickle sphinx]

This is more of a physics question

But how do I do this basically

Would I just do

12x-7x=15

Then multiple what I get for x with jakes speed?

[sacred kestrel]

idk if its right, its been a while for this type of question, but i found the difference in their velocity, divided the distance between them by that difference to get the time it takes for Tylor to close the distance. then we just multiply that time by Jake's velocity to get the distance he travelled.

[fickle sphinx]

Wait

Oh

Ya ok

[fickle sphinx]

So yeah this equation

Or start w it

[sacred kestrel]

where'd you get 7 from

[fickle sphinx]

Oh mb

I meant 12 and 5

[sacred kestrel]

have you tried drawing it out

[fickle sphinx]

Ya

But I don’t rly get anytbjng from it 😭

[fickle sphinx]

I mean when I used this equation it worked for a problem

But idk if it would work for like

All problems

Like that

[sacred kestrel]

T for tylor who's behind J for Jake

15m between them, T moves at 12 m/s and J moves at 5 m/s, both in the same direction.

the difference in their speed is 12-5 = 7 m/s

that means that distance between them will reduce by 7 m/s since we have accounted for both of their velocities.

[fickle sphinx]

Ahh ok

Yes

[fickle sphinx]

I get this part

[sacred kestrel]

so 7x = 15

x is how many seconds it takes to be 0

[fickle sphinx]

Yes

Ok

[sacred kestrel]

we get x = 15/7 ~ 2.1428571...

then we multiply that time by Jake's velocity to get how far he went

[fickle sphinx]

Ok

Yeah I get t

It

Thank you thank you :))

.close

[marsh citrus]

Channel closed

Closed by @fickle sphinx

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[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[pliant island]

I need help figuring out the inflection point of problem 3 and 4

[sullen ice]

are you using derivatives?

[pliant island]

No, my teacher didn’t mentioned anything about derivatives

[sullen ice]

would be much easier with them lol

im not entirely sure how else you would solve this tho

[pliant island]

Please teach 🤧 I gotta submit this at 11:59 and it’s 10:12 for me rn

[sullen ice]

it seems to me you can fairly estimate where it is

can you take a more up close picture of the graph?

[sullen ice]

so that i can see the lines

yeah see

its right on the intersection of a box

oh nvm

wait

[pliant island]

Wait, not inflection point 😭😭 like where would it land on the x-axis

[sullen ice]

hm?

[pliant island]

So I can do concave

[sullen ice]

x intercept?

[pliant island]

Yah

[sullen ice]

you're not talking abt this point?

i really have no clue how you would find that point without using calc

<@&286206848099549185>

you can estimate it

[pliant island]

It’s definitely above 1.5

[sullen ice]

do you have any examples from your teacher?

[sterile yarrow]

this is calc??

[pliant island]

Advanced algebra

[sullen ice]

they say they arent using derrivatives

[sterile yarrow]

huuuuh

[pliant island]

😭😭😭

[sullen ice]

do you not have any examples?

[pliant island]

1-3

Concave

But those ones aren’t decimal so they were easy to do

[sullen ice]

but its obvious where the inflection point is on there

[pliant island]

This new homework does have decimal so how am I suppose to figure out where would it land on the x-axis 😢

[sullen ice]

we could do some estimation

[pliant island]

I messed up with the question 😭😭 I know what the inflection point is just not where would it land on x-axis yk

[pliant island]

Yah

[sullen ice]

so see its a little bit above 2.5