help-13

[lyric narwhal]

you can factor out an x from the first term and a 2 from the second term

[tropic oxide]

what's the goal?

[opal basin]

it is correct yes

[marsh pond]

all this work just for the numerator lmao

[tropic oxide]

why in blazes would you expand

you can factor a lot of shit out

as is

without busting your ass expanding only to then get stuck with a hard factorization

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

(2x-4)^2 = 4(x-2)^2

[marsh pond]

but the power 3

i'm allowed to do that?

[tropic oxide]

well i don't know, are you allowed to rewrite (x^2+x)^3 as (x(x+1))^3?

[marsh pond]

oh you wrote that differently

alright

[lyric narwhal]

$(a\cdot b)^c = a^c \cdot b^c$

[wraith dagger]

kheerii

[lyric narwhal]

use laws of exponents

[tropic oxide]

on the top you have 4 x^3 (x+1)^3 (x-2)^2, from which you care only that it behaves as 4x^8

on the bottom you have x^7 * sqrt(4x^2 + 2), and squinting at sqrt(4x^2 + 2) turns it into sqrt(4x^2), or -2x (because x is approaching negative infinity)

[lyric narwhal]

good catch

i would not have realised that

[marsh pond]

this would help me?

everything is in it's own jail cell, so to speak, with the brackets

hard to cancel anything out

https://tenor.com/view/escape-stuck-trapped-endless-doors-gif-16956749

I need to get some of these terms out of the brackets in order to cancel them, right?

or is that not the goal here, just start to evaluate where x is -Inf?

[lyric narwhal]

you can figure out how the function behaves near the limiting case

i.e what the apparent degree of the numerator and denominator is

[marsh pond]

so the odd powers will keep it -Inf, the even powers will make it +Inf

hmmm, so with the denominator I believe it would be power 8

the numerator power 5

oh it's 6 * 2?

[lyric narwhal]

dont get what you mean here

[marsh pond]

numerator is power 12?

[lyric narwhal]

why *

[marsh pond]

x^2 to the power of 3

and x to the power of 2

in numerator

[lyric narwhal]

right

[marsh pond]

oops

that would be 6 + 2

not 6 * 2

for exponents

so we have powers 8/8?

[lyric narwhal]

in most cases, the degree of the numerator and denominator should be the same

[lyric narwhal]

right

[marsh pond]

interesting

so this tells me that it will have to be POSITIVE

even powers

it's like saying +Inf/+Inf?

[lyric narwhal]

nope

[marsh pond]

oh

[lyric narwhal]

well

yes

[marsh pond]

lol

1?

the answer is 1?

[lyric narwhal]

this is wrong

[marsh pond]

how so?

[lyric narwhal]

what would be the coefficient of x^8 in the numerator

[marsh pond]

2, I think?

.. not sure on that

[lyric narwhal]

nope

[marsh pond]

1?

its the coefficient of the highest power term, ya?

if it's not 1 or 2, then no clue

[lyric narwhal]

how did you get 1

[marsh pond]

this

[lyric narwhal]

yeah

but

why would it be 1

[marsh pond]

x^2 to the power of 3

is the highest power term in the numerator

[lyric narwhal]

but you

[marsh pond]

x has a coefficient of 1

[lyric narwhal]

have to multiply it with the terms of the other multiplicant

[marsh pond]

but i already said 2, and that's wrong?

[lyric narwhal]

(2x-4)^2

dont forget the square

[marsh pond]

oh

so it's 4

[lyric narwhal]

right

and what about the bottom

[lyric narwhal]

remember this

[marsh pond]

2

[lyric narwhal]

no.

read what Ann wrote

[tropic oxide]

imagine reading what i wrote

[marsh pond]

sqrt(4)

[tropic oxide]

couldnt be me

[marsh pond]

i don't understand this

[tropic oxide]

and you didn't think to ask until now?

[marsh pond]

there is a lot going on lol

sorry

[lyric narwhal]

square roots are defined to be strictly positive

[marsh pond]

late to the party

oh because it's already included?

[lyric narwhal]

since x is tending to a negative value, sqrt(4x^2) = -2x

since x is negative

[marsh pond]

if i add my own sqrt i would have to use +-

but since it's already included it's just implied as +

[lyric narwhal]

what?

[marsh pond]

ONLY when they are written for you

not when you write them yourself

if you add a sqrt out of nowhere

like sqrt(4)

the answer is not just +2

it's +- 2

[tropic oxide]

like sqrt(4)
the answer is not just +2
it's +- 2

no

[marsh pond]

???

[lyric narwhal]

sqrt(4) is always equal to +2

that's what I said

square roots are strictly positive

[tropic oxide]

"evaluate sqrt(4)" is a different problem than "solve the equation t^2 = 4 for t"

[marsh pond]

(-2)(-2) = 4

(2)(2) = 4

[tropic oxide]

sqrt() always returns the positive solution, and it is your job to account for the negative root too IF CONTEXT CALLS FOR IT

which here it does not.

[marsh pond]

i'm saying if YOU add your own sqrt onto the paper

just to clarify

[lyric narwhal]

what do you mean by that

[marsh pond]

where it's not already included in the question

in this case it's already included

so it's just +

but if YOU add your own sqrt

[lyric narwhal]

no

[marsh pond]

you gotta account for +-

[marsh pond]

???

[lyric narwhal]

no matter how you introduce a square root, whether it's already there or if you add it

[tropic oxide]

misconception time huh.

[lyric narwhal]

sqrt(x) ALWAYS represents the positive value

[marsh pond]

this is exactly how professor leonard taught it

[tropic oxide]

video+timestamp

of all instances where he says this

[lyric narwhal]

$t = \sqrt{4} \implies t = 2
t^2 = 4 \implies t = \pm \sqrt{4} = \pm 2$

whenever you have an even power, that's where the plus minus shows up

[wraith dagger]

kheerii

[marsh pond]

https://youtu.be/mXAd6rkNSK0?t=1033

[lyric narwhal]

again

that's the same as what i JUST highlighted

[tropic oxide]

oh that wording is yuck for PRECISELY the reason that we're running up against here.

[lyric narwhal]

agreed

[tropic oxide]

this "always put a \pm in front of your sqrts" is a crutch that eventually breaks

[marsh pond]

https://tenor.com/view/unimpressed-not-impressed-amused-gif-13997644

[tropic oxide]

what doesnt break is the knowledge that $\sqrt{x^2} = |x|$

[wraith dagger]

Ann

[marsh pond]

i've been going with this rule every time i add a sqrt

it's wrong?

[lyric narwhal]

did you even bother to read what I wrote?

[marsh pond]

oh well yeah

sqrt(x^2) = |x|

this is the one exception

but for everything else, i think

[lyric narwhal]

there is no exception

.

[marsh pond]

even power roots, if YOU add it to the paper yourself, you gotta add +- in front of it?

just like professor leonard stated

[lyric narwhal]

the plus minus is completely disconnected to the square root sign

the square root is still returning a positive value

[marsh pond]

so why do we see it all the time?

why does the QF include +- if it's not needed?

[lyric narwhal]

$(...)^2 = a^2 \implies (...) = \pm a$

[wraith dagger]

kheerii

[marsh pond]

right i understand that, i'm saying IN FRONT of the sqrt, add +-

like how professor leonard shows it in the timestamp

[lyric narwhal]

you can justify this by saying

$(...)^2 = a^2 \implies (...)^2 - a^2 = 0 \implies ((...)+a)((...)-a) = 0 \implies (...) = a, (...) = -a$

[wraith dagger]

kheerii

[tropic oxide]

kheerii im so sorry but i have to delegate this entirely to you for the sake of my own sanity

[lyric narwhal]

it's alright lol

[marsh pond]

you don't wanna put +- inside a sqrt, otherwise you would be getting complex solutions

[lyric narwhal]

how did we even get to the topic of putting a plus minus INSIDE the square root

[marsh pond]

I don't know if that's what you were implying with this

here you have +-

so we agree

[lyric narwhal]

$(...)^2 = a^2 \implies (...) = \pm \sqrt{a^2} = \pm |a|$

[wraith dagger]

kheerii

[lyric narwhal]

the modulus is unnecessary here, I just put that for your clarity

[marsh pond]

a is not a variable? a is a constant?

[lyric narwhal]

both ... and a can be absolutely anything

an algebraic expression

a

number

anything

[marsh pond]

why the pipes?

[lyric narwhal]

.

[marsh pond]

you said unnecessary, just wondering why they were added

[granite knoll]

...pipes?

[marsh pond]

modulus

absolute

[lyric narwhal]

i added them because technically sqrt(a^2) = |a|

but the +- negates the modulus anyway

[marsh pond]

so it's not just sqrt(x^2) = |x|

it's sqrt(4) = +-|2|?

[lyric narwhal]

no...

okay.

let's go back

[marsh pond]

but you stated this

[lyric narwhal]

sqrt(a^2) = |a|

[marsh pond]

sqrt(4) = |2| ?

[lyric narwhal]

for example, sqrt(4) = 2

yes

sqrt(9) = 3

[marsh pond]

ummmm

[lyric narwhal]

sqrt(anything) >= 0

[marsh pond]

so where is the +-?

[lyric narwhal]

im getting to that.

[marsh pond]

ok

[granite knoll]

@marsh pond stop rapid-fire replying... read and think about it

[lyric narwhal]

make sure you are really comfortable with this

[marsh pond]

yes, the domain of even roots must be [0, Inf)

[lyric narwhal]

correct

now

about the plus minus

whenever you have an even power in a polynomial, that's when the plus minus shows up

the simplest example is x^2 = a^2

this is where your point of (2)(2) = (-2)(-2) = 4 comes into play.

$x^2 = a^2 \implies x = \pm\sqrt{a^2} = x = \pm |a| = \pm a$

does this make sense to you?

the plus-minus sign is completely disconnected to the square root

wait, let me make it more clear

[wraith dagger]

kheerii

[marsh pond]

you are taking sqrt(a^2) = +-a

[lyric narwhal]

No, I am not

sqrt(a^2) = |a|

[marsh pond]

oh right

[lyric narwhal]

the plus minus only shows up because of the even power of x

[marsh pond]

+- sqrt(a^2) = +-a

this is a better way to write it?

[lyric narwhal]

to write what

[marsh pond]

instead of this sqrt(a^2) = +-a

this +- sqrt(a^2) = +-a

[lyric narwhal]

this is just factually incorrect

[lyric narwhal]

this is right

[marsh pond]

isn't that what professor leonard is stating in the timestamp?

i don't understand where his wording is incorrect

[lyric narwhal]

whenever you have squares on both sides, THAT'S when the plus-minus shows up

the square root does not inherently have anything to do with plus-minus

and I have repeated this like 5 times atp

[marsh pond]

professor leonard states "whenever YOU add an even root to the paper, you must include +- in front of it"
but you are saying this fails for x^2
sqrt(x^2) = |x|

[lyric narwhal]

yes, because the way he says it is VERY bad wording

[marsh pond]

would this be the only scenario where +- is not needed in front of even root?

[lyric narwhal]

and that's why you are so confused

[marsh pond]

yes

but it's still good to know

when dealing with even roots

you gotta account for +-

unless it's just a variable? i guess

such as x^2?

what about sqrt(x^2 + 3)

[lyric narwhal]

always positive

the simplest way to remember it is the same two equations I have been repeating for the past 20 minutes.

$sqrt{a^2} = |a|$

[wraith dagger]

kheerii

[marsh pond]

what about sqrt(x^2 - 3)?

[lyric narwhal]

and, $(...)^2 = (a)^2 \implies (...) = \pm a$

[wraith dagger]

kheerii

[lyric narwhal]

it's always positive.

[marsh pond]

but what if x is 1

[lyric narwhal]

then the square root is not defined in the reals anyway

and if you're including complex numbers, all the rules go out the window

[marsh pond]

but it's still positive?

[lyric narwhal]

the square root is not defined.

[marsh pond]

oh

we go into third dimension

z axis

complex realm

anyways

i haven't learned about that so i won't bother asking yet

[lyric narwhal]

what you just said was very wrong but

let's just move on

i HOPE you've understood what i am trying to tell you now.

[marsh pond]

tbh, it probably adds more confusion (sorry)
I thought +- needed to be added in front of even roots, if I write them on paper
I thought just + is implied, if even root is already included on paper
and i thought sqrt(x^2) = |x|

[dire geode]

they did tell you that

[lyric narwhal]

I have to go eat dinner now

riemann can you take over

[marsh pond]

i never see it written like this, except for x^2

[dire geode]

oi

[marsh pond]

ty @lyric narwhal!

sqrt(4) = |2|

this is a fact

do I have that correct?

[dire geode]

yes

[marsh pond]

and now 2 is treated like a piecewise function?

|2| =
2 if 2 >= 0
-2 if 2 < 0

lol

I do realize this doesn't make sense, but just trying to understand the next piece of logic after sqrt(4) = |2|

for sqrt(x^2) I can certainly see this as a piecewise, defined above

as x is a variable and can change signs

-2 if 2 < 0 but this is where the logic fails for constants

https://math.stackexchange.com/questions/225388/why-sqrt4-isnt-equall-to-2 taken from this page

[marsh pond]

I still don't see the issue with what Professor Leonard is saying here

"plus or minus in front of the square root of your CONSTANT"

this is correct

he didn't say "plus or minus in front of the square root of your variable"

[marsh pond]

@tropic oxide fyi

[dire geode]

$\sqrt{4} = 2$ is just convention in most cases

[vagrant elbow]

|2| is just two

[dire geode]

is reserved for the positive square root

[vagrant elbow]

It's not a function

There's no point of doing that whole piecewise thing

[wraith dagger]

riemann

[vagrant elbow]

It's just in place to tell you that the square root of a positive real number is always positive

[dire geode]

this piecewise only makes sense because x can be negative a priori. you've already ruled that case out by setting x=2

[dire geode]

i don't even understand why you're arguing over this semantics. you're supposed to do calculus, not basic high school algebra.

[cedar kiln]

@marsh pond Has your question been resolved?

[marsh pond]

.close

[cedar kiln]

Channel closed

Closed by @marsh pond

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wraith fog]

these questions give me migraines

i think i got the answer but i need a confirmation

a=0,b=-2?

<@&286206848099549185>

[granite knoll]

how'd you get the answer?

[crimson obsidian]

Btw can you multiply X^(ab) both sides? Idk coz x^ab may be negative as well but you know- exponential functions a^x>0

[granite knoll]

and your answer is wrong FYI

[dire geode]

,tex .exp rules

[wraith dagger]

riemann

[teal mountain]

X^a^2 / X^(ab) can be written as X^(a²-ab).

[wraith fog]

.close

[cedar kiln]

Channel closed

Closed by @wraith fog

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ashen rain]

I miht be interpretting this wrong, but im confused on how they got the new height of triangle A’B’C’ to be 10 instead of 12 - root(3)

[cedar kiln]

@ashen rain Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson sedge]

dont judge i beg

[crimson sedge]

@lusty birch

MB

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

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

[crimson sedge]

3

[cedar kiln]

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

[crimson sedge]

i didnt write my work

[obsidian coral]

Then write it out

It's hard for people helping to understand what you did wrong

[crimson sedge]

true

[obsidian coral]

What's the x = -1 mean?

[crimson sedge]

idk

nvm forget it its fine

[obsidian coral]

Do you know what m and c means?

[cedar kiln]

@sour berry Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[south oracle]

Hi, I'm having trouble solving a problem, In this task I have to calculate the ratio a:b if this calculation = 6dB.

[upper abyss]

10lg(a/b) = 6

lg(a/b) = 0.6

That's just dividing both sides by 10

Then raise both sides by 10^ to get rid of the log

[south oracle]

In the solution book it says 3,98.

I was translating the problem because the original problem is in German and I'm not sure if ratio is the correct translation (in German it says "Verhältniss")

I first tought the solution to the problem would be something in the way of 2:5 or 6:1

This problem has more smaller parts to it where the change is the number coming out of the calculation[calc...calculation] (6 was for sub problem b) ). Others are for calc. =1 solution 1 ; calc. = 9 solution 7,94 (or 8 if you round what seems tho be wanted guessing from how the solutions do it as well).

@upper abyss do you happen to have another idea how this could work?

[cedar kiln]

@south oracle Has your question been resolved?

[south oracle]

<@&286206848099549185>

[south oracle]

I found out how it works, not really sure why it works like that. I`ll ask my teacher if he could explain it to me.

.close

[cedar kiln]

Channel closed

Closed by @south oracle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[karmic palm]

I need help simplifying this horrible expression, anyone got any tips?

d,q,l,i,t,E ∈ ℝ+

[karmic palm]

Idealy I would like to get rid of the square root if possible

[cedar kiln]

@karmic palm Has your question been resolved?

[soft cloak]

is "lit" a term on its own, or are l, i and t separate ?

[karmic palm]

l,i,t are separate

[soft cloak]

oh right, i can't read

i doubt this expression can be simplified

[cedar kiln]

@karmic palm Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson sedge]

I just have a question

I will try to translate

So we have a<b and delta is a partition of [a,b] interval

P(delta) is the set of all sample points for the partition delta

And f:[a,b]->R is a function

We say sigma(f,delta,c) is the Riemann sum for function f, partition delta and the sample points c.

At a) it says that: if f has Darboux Propriety then we want to show that I(delta)={sigma(f,delta,c) | c in P(delta)} is a interval

Can someone explain me what that I(delta) exactly means?

Isnt the set of all riemann sums for function f, for a fixed partition delta and all sets of sample points for the partition delta?

I dont understand how a set of riemann sums can be an interval?

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dreamy zenith]

$2(10)^{x+2}=35$

[wraith dagger]

putridplanet

[dreamy zenith]

solve the exponential equation

idk where to start

[grand forge]

Id divide by 2 coz fuck it

[dreamy zenith]

can you make the left 20

then divide by 5

[upper abyss]

You could pull the +2 out by multiplying the left by 100

200(10^x) = 35

[dreamy zenith]

answer sheet says log 35/2 -2

[valid tendon]

yeah

divide by 2

take log

[earnest fiber]

Yeah, I forgor to divide by 2

[valid tendon]

minus 2

[cedar kiln]

@dreamy zenith Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson hare]

was wondering how to solve this equation

[crimson hare]

ive forgotten the process of how to make sure theyre to the same log

[carmine bronze]

$\log_{5}{x} = \frac{\log{x}}{\log{5}}$

[wraith dagger]

Kookiemon

[crimson hare]

i understand that

but what do you do after that point

[storm mural]

do that with both of the logs

[crimson hare]

i think one time someone said something like sub in a random value

like on both sides?

[storm mural]

mhm

[crimson hare]

so like this or?

2nd line

[storm mural]

yes

[crimson hare]

but at this point what do i do

divide 16 onto the other side but then

its just on the other side

[storm mural]

this is something = 16/something

[crimson hare]

im a bit confused honestly

and its because of the 16, the after that whats left

[storm mural]

can you solve for something
then you know what logx/log5 is

[crimson hare]

AHHH

im not quite getting at what you mean

[storm mural]

log(5)/log(x) = 1/(log(x)/log(5))

$16\frac{log5}{logx} = \frac{16}{\frac{logx}{log5}}$

[wraith dagger]

zfnQRZJT

[crimson hare]

ahhh i have got another method to trying this

just then

i found its a bit easier however

when i get to the last stage

i’m wondering would i just use guess and check

5=x^(1/4)

[storm mural]

or 5 = x^(-1/4)

?

[crimson hare]

ye

is there an easy way to solve this or just use guess and check

[storm mural]

there is an easy way

take both sides to the power of 4

or -4

that isolates x

[crimson hare]

i feel a little stupid

WHY DIDNT I GET THAT SOONER AGHHH

thank you though

i really appreciate the help

.close

[cedar kiln]

Channel closed

Closed by @crimson hare

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ionic nimbus]

How would i go about finding "y" so it satisfies both scenarios? Or how would i find a pair of x and y that satisfies both?
x = 20
y = ?

0.068 = (y(dx997)) /((x1000)+(dx997))
14.8 = (x(dy997)) / ((y1000) + (dy997))

If i replace for the above i get: 1.4 (x = 20, dx = 1)
For the bellow i get: 0.35 (x = 20, dy = 1)

Id like to find a balance between x and y for them to satisfy both outcomes, how would i calculate this?

[cedar kiln]

@ionic nimbus Has your question been resolved?

[ionic nimbus]

Some examples of it working:
x = 691
y = 46

x = 2593
y = 173

How could i write an equation to get to these?

[cedar kiln]

@ionic nimbus Has your question been resolved?

[slow thicket]

do \ * instead of *

cant read that properly

[ionic nimbus]

Sorry, i use wolfram to like calculate. And i usually just write it like that

[ionic nimbus]

.close

[cedar kiln]

Channel closed

Closed by @ionic nimbus

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bronze scroll]

why does this work?

like I do not understand the mark scheme.

but i get confused

[cedar kiln]

@bronze scroll Has your question been resolved?

[bronze scroll]

<@&286206848099549185>

[gritty viper]

did you multiply those all thogether?

they aren't 3 separate polynomials

[gritty viper]

.

[bronze scroll]

sorry I got

3 roots

so

(x-2+i) and (x-2-i) and (x+1)(x-2) four I mean

but I do not know where the (x-2)^2 comes from

i mean i know i am left 1 because n^5 but still why is (x-2) the repeated one?

[gritty viper]

because of tangency

[bronze scroll]

??

[gritty viper]

,w plot (x+1)(x-2)^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by RedstonePlayz09#0205

[gritty viper]

when you have a double (or more) root it is tangent to the x axis

like here, the graph is tangent to the axis at x=2

[bronze scroll]

dy.dx = 0

[gritty viper]

yeah

[bronze scroll]

but why like how do you know it is repeated life if i see a min on that point in x

do i just know that it is (x-2)^2?

[gritty viper]

if its on the x axis

[bronze scroll]

is there any video on this topic?

[gritty viper]

and its a polynomial

[bronze scroll]

cuz i do not really understand why

like i get that its a tangent

[gritty viper]

https://www.youtube.com/watch?v=jrFLb9ZoZH0&pp=ygUdbXVsdGlwbGUgcm9vdHMgb2YgcG9seW5vbWlhbHM%3D

[bronze scroll]

ok thanks

[gritty viper]

np

[bronze scroll]

.close

[cedar kiln]

Channel closed

Closed by @bronze scroll

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tropic oxide]

what's the goal here

[young urchin]

This is the question i have to simplify it

How to solve inverse trigonometric functions

[tropic oxide]

uh ok so like. you should always post the whole problem lmao

arccos(sqrt(2/3)) + arccos(sqrt(3/8)) could probably be rewritten as the arcsin or arccos of one angle

arccos(sqrt(2/3)) + arccos(sqrt(3/8)) can prob be expressed as arcsin or arccos of one number

[young urchin]

<@&286206848099549185>

.close

[cedar kiln]

Channel closed

Closed by @young urchin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[acoustic shoal]

working on a problem about dilations and enlargements and i'm confused because i have no other idea of what it could be. because if the preimage is scaled by 3 and the area of the preimage was 4 shouldn't it be 12?

[tropic oxide]

when you scale something up by 3 its area does not multiply by 3 but by 3^2

[crimson fractal]

.close

[cedar kiln]

Channel closed

Closed by @crimson fractal

Use .reopen if this was a mistake.

[tropic oxide]

.reopen

[cedar kiln]

✅

[acoustic shoal]

thank you

[tropic oxide]

@crimson fractal

[crimson fractal]

sorry

mistake

[acoustic shoal]

dw it's fine

we're in a call

thanks a lot @tropic oxide

.close

[cedar kiln]

Channel closed

Closed by @acoustic shoal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic eagle]

this is the original problem. i pretty much figured it out. but i ran into something weird when i was optimizing. i arrived at a distance formula of d = x^2 - x - 2 and calculating the first derivative of that distance as d' = 2x - 1, but this was wrong. putting it into desmos, desmos calculates a derivative of d' = 1 - 2x. how did i mess that up?

[civic eagle]

this essentially inverted my derivative, showing me an absolute min at x = 1/2 rather than the absolute max that it is.

[gentle flower]

the derivative is 2x - 1

desmos calculates 2x - 1 as well

[civic eagle]

like this it doesn't look right though. so i added the absolute value to the distance since a distance can't be negative

giving this

which is the proper output i would expect. how does the absolute value affect the derivative in such a way?

[gentle flower]

derivative of $x^{2} - x - 2 \neq |x^{2} - x - 2|$

[wraith dagger]

! saad

[heavy mulch]

2x-1

[gentle flower]

$\frac {dy}{dx} |x| = \frac {x}{|x|}$

[wraith dagger]

! saad

[civic eagle]

so in effect, how should i have gone about this?

[gentle flower]

formula is $\frac {dy}{dx} |f(x)| = \frac {f(x)}{|f(x)|} \cdot f'(x)$

[wraith dagger]

! saad

[civic eagle]

because using the first derivative test on 2x - 1 gives an abs min rather than a max

that definitely works

was it the right choice to think of using the absolute value since this is a distance? the graph looks way better that way, and doesn't give an absolute min by the first deriv test

[heavy mulch]

You got the right answer so everyone should assume so

If it ain't broke don't fix it

[gentle flower]

absolute value is correct, find the maximum for the absolute value function

[civic eagle]

just threw me for a loop thats all. thought i had it right and then when i graphed it in desmos it looked messed up until i used abs value. i guess i gotta memorize that derivative

thanks guys

.close

[cedar kiln]

Channel closed

Closed by @civic eagle

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[nova snow]

How would I draw this?

[nova snow]

would it look like a rhombus or something or do i just wing it

[storm mural]

do you have to draw to scale

[nova snow]

no

[storm mural]

I would say draw a generic quadrilateral

[nova snow]

but all the sides are unequal

[storm mural]

don't even bother making the sides the correct ratios

unless it matters, like, which side of a line a point is on

totally wing it

[nova snow]

so i can just draw a rectangle

[storm mural]

pretty much

[nova snow]

im so confused

[tropic oxide]

this is a rectangle

[nova snow]

yes

[tropic oxide]

or rather looks like one

you might be better off drawing it to not look like a rectangle

[nova snow]

what should it look like then

[tropic oxide]

just a generic quadrilateral

[nova snow]

[tropic oxide]

like this maybe

[nova snow]

alright

[storm mural]

find them in the order they ask

[nova snow]

okay

i can use cosine rule for AC

[tropic oxide]

you sure can.

[nova snow]

AC = 11.94cm

[tropic oxide]

sounds approximately right

[nova snow]

i also found angle A from DAC to be 53.84 degrees

[storm mural]

hmm

[nova snow]

o

im not sure what i did wrong

[nova snow]

So I’m incorrect right?

@tropic oxide did I miscalculate an angle

[tropic oxide]

nyeh?

[tropic oxide]

you know sqrt exists and is understood by desmos right

[tropic oxide]

show work

if only to ensure no stupid arithmetic mistakes happen

[nova snow]

ignore red

[tropic oxide]

you found angle ACD not angle DAC

[nova snow]

huh

[tropic oxide]

the sides in the denominator, and the sides whose squares are added in the numerator, are those between which the angle is.

the third side's square is subtracted in the numerator and absent from the denominator.

so it's a case of misidentifying which side goes where and as a result calculating the wrong angle.

[nova snow]

so i subtracted and added the wrong sides?

[tropic oxide]

yes that's what i said

[nova snow]

actually i should label A as C

@tropic oxide Did I add and subtract the correct sides?

[tropic oxide]

no

denom is correct num isn't

[nova snow]

a^2+b^2-c^2 no?

[tropic oxide]

... okay hold on you swapped the letters

[nova snow]

yeah

[tropic oxide]

now num is correct and denom isn't

denom should be 2 * 11.94 * 9.7

[nova snow]

ohh

yeah

cos^-1(0.72)

so angle C is 44 degrees

[tropic oxide]

so it appears.

[nova snow]

great

now i use cosine rule again to find the last angle?

[tropic oxide]

could do that, sure.

[nova snow]

cos^-1(0.14) and then

82 degrees

[tropic oxide]

it's a little bit unkosher to shuffle around point names like you've done, tbh.

not every triangle you come across will be named ABC.

[nova snow]

well i guess i could always change up the letters

it just makes it easier for me to understand

[tropic oxide]

no the thing is

you'll be using the point names in ways that conflict with how the original problem uses them

your B and their B won't be the same point.

i would recommend that you not risk such confusion.

[nova snow]

its just using the point names that the problems uses ex ABD kinda confuses the formula up for me and swapping the values around

[tropic oxide]

and this is why i suggested you understand the formula in a name-free manner.

an angle has two adjacent sides and one opposite side. to find the cosine of your angle, you take:

the sum of the squares of adjacent sides minus the square of the opposite,

all divided by 2 times the product of the adjacent sides

[nova snow]

by two adjacent sides ur talking about the hypotenuse as well right?

[tropic oxide]

what hypotenuse

we are not talking about a right triangle

[nova snow]

o

right

[tropic oxide]

i am talking about the two sides that border your angle

[nova snow]

okay

[cedar kiln]

@nova snow Has your question been resolved?

[nova snow]

why is the final answer in units squared

[tropic oxide]

it's an area

[nova snow]

oh

right

[nova snow]

is length OA also 4cm

it looks like it would be equal

is the circle just there to be confusing

[cedar kiln]

@nova snow Has your question been resolved?

Channel closed

Closed by @nova snow

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[digital mantle]

In a right triangular prism and in a orthogonal frame of reference as well as Oxyz monometer, it's known that:
Point A (4 ; 4 ; 0)
Point B ( 2; 2; 4)
the vector OC is (-2 ; 2; 0)

» Find the angle formed by the vectors BC and BA

» Using the dot product, find an equation that can define the spherical surface of diameter [BD]

» Determine a general equation of the plane perpendicular to the line OC and containing the point B.

[cedar kiln]

@digital mantle Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[digital mantle]

.reopen

[cedar kiln]

✅

[digital mantle]

@tropic oxide are you able to help me out when you have some time?

[cedar kiln]

@digital mantle Has your question been resolved?

[cedar kiln]

@digital mantle Has your question been resolved?

Channel closed

Closed by @digital mantle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dreamy shell]

Can some one solve it for me?
Thank you

[nova snow]

!show

[cedar kiln]

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

[buoyant latch]

have you tried anything

[steel heart]

no one solves it for you, someone can only guide you

[dreamy shell]

yeah help me

very hard question

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[eternal ingot]

it says the first step is putting AD in terms of h. have you made any progress on that?

[cedar kiln]

@dreamy shell Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silk beacon]

,tex \left. \begin{cases} { A = \left(\begin{matrix} 2 & -1 \-1 & 2 \end{matrix}\right) } \ { { A }^{ 2 } -4A-n \left(\begin{matrix} 1 & 0 \0 & 1 \end{matrix}\right) } \end{cases} \right.

[wraith dagger]

sarang
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[silk beacon]

can anyone help me with this

whenever i try thos

i do not get a diagonal matrix after
A^2 - 4A

[cedar kiln]

@silk beacon Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[silk beacon]

.reopen

[cedar kiln]

✅

[sacred stone]

what's your result?

[dreamy shell]

.reopen

[eternal ingot]

the channel timed out! you might need to get a new one.

[cedar kiln]

@silk beacon Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[waxen prawn]

need help on 4

[waxen prawn]

not sure what my common ratio is

[south tundra]

Note that $\sin\theta - \frac13\sin2\theta + \frac19\sin3\theta - \frac1{27}\sin4\theta + \dots = \ = \Im((\cos\theta + i\sin\theta) - \frac13(\cos2\theta + i\sin2\theta) + \frac19(\cos3\theta + i\sin3\theta) - \frac1{27}(\cos4\theta + i\sin4\theta) + \dots)$

[wraith dagger]

A Lonely Bean

[south tundra]

Can you apply the Euler's formula here?

[waxen prawn]

im e^iθ and so on

so its just that

ive been really over complicating this 😭

[south tundra]

Yeah, and you will notice that it becomes a geometric series

[waxen prawn]

yeah thank u

[cedar kiln]

@waxen prawn Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[keen anvil]

could someone help me with #11

[inland ocean]

Do you know what a mole is?

[keen anvil]

isn’t it something to do with 6.02 * 10^23

@inland ocean

do i divide 7.5 x 10^23 by 6.02 * 10^23?

cause then i got 1.25 (rounded)

but then what am i supposed to do

[inland ocean]

Yeah it seems correct

I was confused what they meant by formula units

Ig it's just the no. Of molecules

[keen anvil]

so is it 1.2?

[inland ocean]

,w 7.5/6.022

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ColdTee#6459

[inland ocean]

Yeah

[keen anvil]

aaa alr

thank you

.close

[cedar kiln]

Channel closed

Closed by @keen anvil

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[coral plover]

Can somebody help me with my understanding of the chain rule?

[coral plover]

the formula i was given seems to be both the chain rule and the power rule simultaneously

[upbeat frigate]

this video is excellent for it https://youtu.be/YG15m2VwSjA

[buoyant latch]

what's the formula you're given?

[storm mural]

df/dx = df/dg * dg/dx
is easy to remember

[coral plover]

I will write it out real quick

i'm familiar with this and am able to use it

but is this not the power rule and the chain rule simultaneously?

[storm mural]

it is both

[coral plover]

okay

because sometimes i'm given questions where

[storm mural]

you can also get that with product rule only and induction but yeah

[coral plover]

well, i have to use other forms of derivation and then multiply by derivative of inner function

and it just seemed strange to me that two completely different looking derivatives were both being described as "chain rule"

[storm mural]

they are though
3x - 1 is a function of x

so its basically the same

[coral plover]

my confusion basically lies in the fact that I was taught the chain rule as a formula where it was only being used in addition to the power rule

[storm mural]

ah

[coral plover]

i was simply told, chain rule is this:

and sort of found out myself that i can do it to non-exponent parts of an equation

[storm mural]

they didnt teach it as f(g(x)) ‘ = f’(g(x))g’(x)?

[rugged palm]

This is a very specific version of the chain rule
In this case the "outer" function is x^n, and the "inner" function is f(x)

[coral plover]

nope, just the specific version of where it applied to the power rule

[storm mural]

weirdddd

[coral plover]

so basically i've just sort of adopted the strategy of

derivate the outer function as if it just said X and then retain all the original inside parts of the outer function

then multiply by derivative of inner function

[coral plover]

which i sort of see as the same thing as this now

well, this was helpful, thanks guys

.close

[cedar kiln]

Channel closed

Closed by @coral plover

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[river frigate]

hi, can someone help me with this double integral question pls. Tried and didnt get the same ans as the calculator

[river frigate]

nvm found it

.close

[cedar kiln]

Channel closed

Closed by @river frigate

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lucid turtle]

ideas on where to start?

[buoyant latch]

maybe split them into their fractions

[vagrant elbow]

Write everything in terms of sin and cos

[buoyant latch]

^^

[lucid turtle]

(1+cosx)/sin(x)

————————

(Sinx/cosx) + sinx

[buoyant latch]

try simplify that

[lucid turtle]

(1+cosx)/sinx * (cosx +cosxsinx)/sinx ?

[buoyant latch]

keep going and see if u get the answer

[lucid turtle]

Uhh I’m at (cosx(2x))/(sin(2x)), how would I use the double angle identity for cosx(2x)

[buoyant latch]

how did you get cosx(2x)

[lucid turtle]

Multiplied this out

Then factored a cosx

[cedar kiln]

@lucid turtle Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thin skiff]

hey

[thin skiff]

.reopen

[dusk finch]

.close

[cedar kiln]

Channel closed

Closed by @dusk finch

Use .reopen if this was a mistake.

[dusk finch]

you're in #help-25 and aparrently have no question

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[plain badge]

I'm stuck on 2

so if A is independent the scalars must all add up to 0

aka the scalars are all 0

c1,c2,c3,c4 in F

now i'm not quite sure how to go about this now

[dim stump]

Well occupy a different channel , since i can't now

[plain badge]

oh lmfao

[cedar kiln]

@dim stump Has your question been resolved?

[dim stump]

<@&286206848099549185>

[cedar kiln]

@dim stump Has your question been resolved?

[cedar kiln]

@dim stump Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[static fern]

Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0,$ and let $s = \sqrt{a} + \sqrt{b} + \sqrt{c}.$ Find $s^4 - 18s^2 - 9s.$ (Source: HMMT)

[wraith dagger]

FireBlazer

[static fern]

im pretty sure Vieta is needed but not sure how it comes in

[cedar kiln]

@static fern Has your question been resolved?

[static fern]

@heady spindle this channel is occupied.

[heady spindle]

sorry sorry I don't see

[static fern]

its ok

[odd snow]

find s²

first

@static fern

$s^2=a+b+c+2(√ab+√bc+√ac)$

with Vieta's formulas we can say that $a+b+c=9$

[wraith dagger]

Ozgr

Ozgr

[odd snow]

to find $2(√ab+√bc+√ac)$ we can square it which makes $ab+bc+ac+2√abc(√a+√b+√c)$, and we can find $ab+bc+ac=11$ and $2√abc(√a+√b+√c) = 21s$

[wraith dagger]

Ozgr

[odd snow]

so $(√ab+√bc+√ac)^2=2s+11$

[wraith dagger]

Ozgr

[odd snow]

which makes $s^2=9+2√(2s+11)$

[wraith dagger]

Ozgr

[odd snow]

$s^2-9=2√(2s+11)$

[wraith dagger]

Ozgr

[odd snow]

by squaring both sides $s^4-18s^2+81=8s+44$

[wraith dagger]

Ozgr

[odd snow]

$s^4-18s^2-8s=-37$

[wraith dagger]

Ozgr

[cedar kiln]

@static fern Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[elder elm]

I may be blanking but I'm having problems evaluating this..

[brave wedge]

Why is change in x^2 + change in y^2 sqrt equal to speed

[elder elm]

I have this so far:

[brave wedge]

Lol

[elder elm]

-k/4-k/1..

[hasty fulcrum]

mistake on the last line

it should be a +, no?

$-\frac k4 + \frac k1$

[wraith dagger]

blanket

[elder elm]

Yeah, I have two negatives though.

[hasty fulcrum]

right

[elder elm]

I mean subtraction signs

Sorry

[hasty fulcrum]

subtracting a negative is the same thing as adding it

[elder elm]

Yeah, a +

[elder elm]

Ik.

[hasty fulcrum]

then you just simplify what you have?

there's nothing else to it if k is just a constant

[elder elm]

The answer is supposed to be 3/4k but like..I’m not getting that?

[hasty fulcrum]

$-\frac k4 + \frac k1 = -\frac k4 + \frac{4k}{4}$

[wraith dagger]

blanket

[elder elm]

I didn't turn k/1 into 4k/4..

😦

is this intuition

[hasty fulcrum]

not necessarily intuition

to add fractions with no common denominator, you just multiply something on the top and bottom :)

[elder elm]

Thank you. May I ask a question on integration? It's related to this problem..

[hasty fulcrum]

yeah sure

[elder elm]

I'm still a little confused on the steps of integration for a function like k/x^2..

so they pull a constant k out

[hasty fulcrum]

right

[elder elm]

we get k* integral of 1/x^2

but I thought that should turn to x^-2...

I must be confusing it with something else

[hasty fulcrum]

thats taking the derivative

[elder elm]

oh

[hasty fulcrum]

yep

you're taking the anti derivative of k/x^2

actually, a revision to my old statement, you'r enot taking the derivative if k/x^2 turns into 1/x^2

you just magically removed the constant

the derivative of that would be -2k/x^3 i think

im just not sure what happened there

[elder elm]

Yeah, in that case it becomes -1/x, but what's the step using the power rule for this? Don't we add 1 so it becomes 1/x^3 and then divide by 3 so its 1/x^3/3?

[hasty fulcrum]

thats a negative exponent, remember?

$x^{-2 + 1} = x^{-1}$

[wraith dagger]

blanket

[elder elm]

Oh, so because its in denominator we remember to treat it as 1/x^-2....and then apply power rule?

[hasty fulcrum]

no

$\frac{1}{x^2} = x^{-2}$

[wraith dagger]

blanket

[hasty fulcrum]

so $\int_1^4\frac{k}{x^2}dx = \int_1^4 kx^{-2}dx$

[wraith dagger]

blanket

[hasty fulcrum]

also unfortunately i do have to go now

<@&286206848099549185> if they still need help

[elder elm]

ok

[hasty fulcrum]

[storm mural]

no dx or dk

answer is infinity for k =/= 0 gg ez

[elder elm]

May I ask another question?

unrelated

[storm mural]

sure

[elder elm]

So this is supposed to be straight forward but the functions my professor is giving is seriously difficult to work with

for context...

Our p(x) was given as k/x^2

and our k is 4/3

[storm mural]

uh huh

[elder elm]

but like, I'm stuck on trying to evaluate x(4/3/x^2)dx...

[storm mural]

the x's cancel out

4/3/x

[elder elm]

Got it, not sure what I was thinking..We now have integral from 1 to 4 of 4/3x, we take the 4/3 out and get 4/3* integral 1 to 4 of 1/x, and then we take the integral of 1/x and get 4/3* integral 1 to 4 of ln|x|?

[storm mural]

4/3 * ln(4) - ln(1)

[elder elm]

My final question is on this...does it seem like I'm on the right track:

im ending up with 4/3*-1/x, where I will be plugging in m and 1

[storm mural]

yes

[cedar kiln]

@elder elm Has your question been resolved?

[elder elm]

Interesting, but I'm not getting that..

check this:

[elder elm]

I’m getting -8/11..

<@&286206848099549185> Can anyone help me understand what I did wrong?

[cedar kiln]

@elder elm Has your question been resolved?

[storm mural]

remember it's F(m) - F(1)

(had to go for an hour sorry ;-;)

[elder elm]

Oh wait so its supposed to be addition because we should have a minus symbol and the value is negative?

[storm mural]

ye

[elder elm]

I'm still ending up with incorrect value...I'm getting 30/15...

[storm mural]

1/2 = -4/(3m) + 4/3
-4/(3m) = 3/6 - 8/6 = -5/6

[elder elm]

Wrong picture

I’m now getting 30/15..

[storm mural]

4*6 = 24

[elder elm]

oh snap

[storm mural]

it's always the arithmetic that gets you (not you specifically ppl in general)

never the hard math

/joke

[elder elm]

awesome!

[elder elm]

yeah seems like somethings going wrong with my arithmetic..

[elder elm]

Could you help me understand this?

I do not understand how they got e^5k=305/330=61/66..
hey lost me at this part. But I understand everything up to y(t)=70+33oe^kt

[royal loom]

given the initial condition y(5)=375, they substitute this into the equation to solve for "k".

replace y(t) with 375, and replace t with 5

and then solve for k, as they did

[cedar kiln]

@elder elm Has your question been resolved?

Channel closed

Closed by @elder elm

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[buoyant latch]

Is $|\lim_{x\to x_0} f(x)| = \lim_{x\to x_0} |f(x)|$?

[wraith dagger]

Frosst

[modern compass]

nope

f(x) = x/|x|
as x -> 0:
|lim f(x)| does not exist
lim |f(x)| = 1

[buoyant latch]

What if the first one exists

The question is show if $\lim_{x\to x_0} f(x)=A$, then $\lim_{x\to x_0} |f(x)| = |A|$

[wraith dagger]

Frosst

[buoyant latch]

I’m pretty sure I’m supposed to use limit definitions but I want to see if this other way works as wel

[modern compass]

if f is continuous at the point, then yes it's true

[buoyant latch]

Does it work at all if it’s not continuous? Like via the limit def

[modern compass]

if it's continuous at the point x0 it should work. because you basically have a neighborhood of continuity around x0 in that case.
if it's not continuous at x0 then you just have a bunch of 'does not exist' statements, where equality doesn't make sense anyway

but you should be able to get this without the limit definition

that seems like overkill, imo

[buoyant latch]

That’s what my teacher implied is used for the solution pdf

But I suppose since it doesn’t explicit say to use the limit def this should do

Now that I think about it it seems just like a composition of limits

And limits of composition

Aight thanks

.close

[cedar kiln]

Channel closed

Closed by @buoyant latch

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.