help-10

[hoary rain]

well that's the question

And i just substituted ln(1+x) with the taylor series representation

and ended up with this

but i honestly forgot what to do from here

@steep flax any hints?

[steep flax]

If this is the question, why not simply substitute 0 for x?

[steep flax]

recall that if there's no funny business going on, $\lim_{x\to c} f(x)$ simply equals $f(c)$.

[warm shale]

waris

[hoary rain]

yeah but the textbook answer is 1/2 hence why i'm so confused

<@&286206848099549185>

[steep flax]

If this is actually the question, there is no way on earth that the limit as x goes to 0 is anything other than 0.

[hoary rain]

Well this is the actual question

[steep flax]

Well yes, the answer to this is indeed 1/2

I need you to be active to guide you through the process, as I will not be giving you the solution outright.

Ping me when you can be consistently available in this chat for 10 minutes or more.

@hoary rain

[obtuse pebble]

@hoary rain Has your question been resolved?

[obtuse pebble]

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[rich tapir]

how did we go from line 1 to 2 (blue arrow)

[obtuse pebble]

@rich tapir Has your question been resolved?

[mellow summit]

Man, I get f"(x_0) instead of 1/2f"(x_0)

[obtuse pebble]

@rich tapir Has your question been resolved?

[rich tapir]

how, i have no idea what they did here

[mellow summit]

I have no idea too. But I got the answer f"(x_0)

[mellow summit]

Maybe I'm wrong

[obtuse pebble]

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[open rivet]

Help! Brute forced it and found a solution of a=-1, b=8, & c=-7, but i kinda think brute force isn't the best way to solve this...

[timid silo]

complete the square

[open rivet]

exactly what would be the first step; I'm kinda lost on where to start TT

[timid silo]

given c=-a-b, find the y coordinate of the vertex and the difference between the two x-intercepts

then you can derivate an expression for the area of 🐕

[obtuse pebble]

@open rivet Has your question been resolved?

[open rivet]

I'm trying to understand the idea yet I'm stuck on how to get the y coordinate of the vertex

after

[timid silo]

.

[open rivet]

$a(x+\frac{b}{2a})^2=-a-c$

[warm shale]

LastStand092

[timid silo]

the constant term after completing the square is the y coordinate of the vertex

[open rivet]

So,
$a(x+\frac{b}{2a})^2+(a+b)$

so the y coordinate is a+B?

[warm shale]

LastStand092

[timid silo]

yup

[open rivet]

so the formula would be

$\frac{\frac{b}{2a}*(a+b)}{2}=27$?

[warm shale]

LastStand092

[open rivet]

wait no

So to find the function's zeros, we equate it to zeor

giving
$x+\frac{b}{2a}=\sqrt{-1-\frac{b}{a}}$

[warm shale]

LastStand092

[open rivet]

<@&286206848099549185>

[open rivet]

.close

[obtuse pebble]

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[mellow summit]

I didn't understand your question

[nocturne minnow]

• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.

[sage geode]

You want to complete the square or just expand?

For?

Can you show how the problem is stated?

[vernal musk]

?what about just multiply?

[sage geode]

Can you type the question exactly as it is here?

[nocturne minnow]

No you did not. You did not phrase a question in a way that we can understand so we can help you

All you did was given an expression

[mellow summit]

Is there any example related to it?

[tame narwhal]

do you know how you could multiply "a(b + c)"?

[mellow summit]

do you know what is 2(1+3)?

you don't know the answer for 2(2+6)?

[jolly ginkgo]

Why is the 2 still outside

[tame narwhal]

maybe this will help you: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-distributive-property/v/the-distributive-property

that's ok, sometimes it can help to refresh your memory

you might want to ask in #precalculus , i haven't taken it in forever

but you'll probably need to review algebra, geom, and trig

[obtuse pebble]

@timid silo Has your question been resolved?

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[jolly ginkgo]

What's the matter?

[cloud saffron]

how is he getting 1/3du?

first time seeing u sub

[tame narwhal]

typo?

[cloud saffron]

1/3dx then?

[tame narwhal]

$x^2 dx = \frac{1}{3} du$ I think

[warm shale]

cwatson

fäf

[cloud saffron]

not seeing xˆ2dx in his work

the steps make sense up to 1/3du

[tame narwhal]

I think he just forgot to write it. it looks right after that

[cloud saffron]

o wait

ok

so before performing the u sub what are the steps?

why are we differentiating u with respect to x if u is simply a substitution?

[tame narwhal]

because you need to "replace" dx, too

[cloud saffron]

ok

i'll try to walk through a problem

integral x sqrt(xˆ3+16)dx u= xˆ3 +16

integral xˆ2

mistake

du = 3xˆ2

dx = 1/3du

i worked it out

i need to go back and review the concept though

i'll go watch a tutorial

thanks guys

.close

[obtuse pebble]

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[thorny wave]

hello

can you specify your question more?

@timid silo

ok

let me have a look

yes so

tell me what exactly you are struggling with

@timid silo

yes that is corrrect!

so this is the initial formula: m(1.04)^2(1.04) + m(1.04)^2 + m(1.04)

we simplify it further: m(1.04)^3 + m(1.04)^2 + m(1.04)

factorize it to m(1.04)(1.04^2 + 1.04 + 1)

and you get m(1.04)(3.124)

let me check

with my calculator

yes so its 3.124m(1.04)

and approx 3.244 rounded to the nearest thousand.

[obtuse pebble]

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[sweet kite]

hey anyone here can help me with this?: |x| < |y| ⇒ x ^2 < y^ 2

i have to either prove or disprove

[marsh geyser]

You need to find a counterexample to disprove it.

[thorny wave]

bro

[marsh geyser]

i gave wrong example

[thorny wave]

am i interupting?

[marsh geyser]

xd

[sweet kite]

and how do i do that specifically?

do i just choose 2 random numbers for x and y and then disprove or is there a different way to solve it?

[sweet kite]

wym haha

[thorny wave]

i wanna ask how do i get points for helping people

for practice

to become a helper

[sweet kite]

aah i see

cannot tell u but maybe u can gain some by helping me :D

[thorny wave]

sure

whats the question

[thorny wave]

?

[sweet kite]

yess

[thorny wave]

what were your problems encountering this?

and what were u struggling with?

[sweet kite]

idk how to solve this, ik i have to find a disprove right?

[thorny wave]

okay

let me solve it myself first

[sweet kite]

sure, thank u!

[thorny wave]

okay

looks like you have to prove the equation

lets say that |x| < |y| is true.

[sweet kite]

oh so the statement is true?

[thorny wave]

let me solve it

i had to deal with something irl lol

[marsh geyser]

the statement is true but he didn't solve yet

he just stated something

[thorny wave]

we have to square both sides since |x| is always a non negative.

inequality is x^2 < y^2.

so yes

the absolute value of x is less than the absolute value of y

|x| < |y| ⇒ x^2 < y^2 is equivalent to the inequality

we proved that |x| < |y| ⇒ x^2 < y^2 is true

[marsh geyser]

you can also use one of the definitions of absolute value to solve it

[sweet kite]

oh dw! sorry for being to fast haha

[marsh geyser]

but this one is enough

[thorny wave]

lol

[sweet kite]

is that the full proof of this statement or do i still have to add something

[thorny wave]

yes

this method is simpler tho

[sweet kite]

oh fr it looks so short hahah

[marsh geyser]

you can do squaring the sqrt(x^2)

[thorny wave]

we have to take the sqrts of the equation on the other one and multiply the inequality by the absolute value of x

[marsh geyser]

and reach to x^2, which is the same as squaring |x|

[thorny wave]

so then x^2 < |xy|.

we take sqrt of both so -> sqrt(x^2) < sqrt(|xy|).

Since sqrt(x^2) is equal to the absolute value of x and sqrt(|xy|) is equal to |y||x|/|x|, we can simplify the inequality to |x| < |y|.

omg i tyoed that so fast frr

lol

[sweet kite]

i try to follow ur statement lul

[thorny wave]

the first method is simpler

[sweet kite]

i understood the first one better haha

[thorny wave]

how do i get the point now

lmao

[sweet kite]

i just looked u have to give the role to urself

under info

theres the description of becoming a helper

btw another question, if u choose a bigger number for x, wouldnt that make the statement untrue?

@marsh geyser @thorny wave

[thorny wave]

in what sense

give me an example @sweet kite

ohh

no

choosing a bigger number for x will not make the statement |x| < |y| ⇒ x^2 < y^2 untrue. In fact, the statement is true for all real numbers x and y.

.close

[obtuse pebble]

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[dense narwhal]

studying for an entrance exam for highschool and im reviewing turning words into algebraic expressions. The question is "A coffeeshop found that the number of orders for croissants was 1/6 the number of muffins. If the total number for the breakfast rolls was 56, how many orders were placed for croissants?" I know this is pretty basic but P.S i live in the philippines.

[royal basin]

how exactly does it matter that you live in the philippines...?

[dense narwhal]

public school

[royal basin]

and how exactly does it matter that you go to a public school...?

[dense narwhal]

public high school

[royal basin]

ok but like what are you trying to do here exactly? lower our expectations for your own mathematical skills or what

let's maybe get to the problem instead, shall we?

[dense narwhal]

i really dont know

[royal basin]

so you do something without knowing why you do it, got it.

have you made any progress with this problem?

[quartz gust]

Just look for quantities in the expression and start thinking about what those mean. In this one you # of orders, # of croissants, number of muffins and number of breakfast rolls. You also know that the # of croissants is 1/6 of the # of muffins

[royal basin]

(if the answer is "no, i have not made any progress" then please say it outright and don't shy away from it.)

[dense narwhal]

tried x+(x+1/6)=56

didnt fit

[royal basin]

what does the letter x mean here?

[quartz gust]

start by assigning a variable to each quantity

[dense narwhal]

lemme try

[royal basin]

nyeh

i think i should maybe let EJHans handle this given how eager they are

[quartz gust]

Less judgmental too 😁

@dense narwhal is that the whole question as well in your post?

[dense narwhal]

yes, im trying to separate it into 2 equations

[quartz gust]

"A coffeeshop found that the number of orders for croissants was 1/6 the number of muffins. If the total number for the breakfast rolls was 56, how many orders were placed for croissants?"

of croissants

of muffins

of breakfast rolls

[dense narwhal]

so let
x=croissants
y= muffins

[quartz gust]

but are breakfast rolls both? Or?

[dense narwhal]

is both

[quartz gust]

so muffins and croissants add to breakfast rolls

[dense narwhal]

yess

i also got a little confused because of that

[quartz gust]

so the first thing is that we know x = 6y OR y = x/6

[dense narwhal]

searched what breakfast rolls are

[quartz gust]

Breakfast rolls to me are something different... but I'm not sure what the question is expecting from you.

If we assume that means the total sold

then you should have x+y=56

and then sub in the relation I made above for y and solve for x

[dense narwhal]

yes, orders placed

[quartz gust]

Did you understand what I suggested above?

[dense narwhal]

yep

oh wait

[quartz gust]

two equations:
x+y=56
y=x/6

?

[dense narwhal]

y=x/6 for muffins right?

[quartz gust]

Sorry, if x is croissants, and y is muffins:
then to have 1/6 the croissants vs muffins
y/6 = x

It may be helpful to even write it like this when you are working on these.

Number of muffins/6 = Number of Croissants

Then sub in variables

And ignore jerks like that other person. It had no relevance where you were from, but it's also irrelevant given what you are asking to comment about it

Good luck with the studies!

[dense narwhal]

okay okay, thank you!!!

.close

[obtuse pebble]

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[shadow lava]

when I evaluate, I just gotta be careful right?

2 is not the final answer

x^7 would cause this to be negative?

[rigid lintel]

yes

square root function is nonnegative

[shadow lava]

this is a tricky question

[shadow lava]

but i would get 2x for the highest power of the sqrt in the denominator, right?

2x * x^7 = 2x^8

even power are positive i thought?

I would have assumed this would be 2, but checking with symbolab it's -2

[rigid lintel]

more like |2x| * x^7

[shadow lava]

so whenever there is a variable in an even root, it's always absolute for the answer?

or too vague of a description

[shadow lava]

oh wow

i think I get it finally

we cannot merge |2x| and x^7 to make it 2x^8 or |2x^8|

[rigid lintel]

exactly

[shadow lava]

modulus prevents that

always |absolute| answer, right? regardless of what is under the even root

[rigid lintel]

2x^8 = |2x^8|

theres no point

,w plot |x| x

[shadow lava]

but I mean for even roots with variables within the argument

[warm shale]

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[rigid lintel]

,w plot x^2

[warm shale]

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[shadow lava]

interesting..

even power and even root

they can lock it in to absolute

no negative output

and this only depends on the highest power leading term of the polynomial?

can ignore all other lower power terms within the polynomial?

[tardy epoch]

No

[shadow lava]

oh

[tardy epoch]

,w plot sqrt(x^2-6x+9)

,w plot sqrt(x^2+6x-9)

[warm shale]

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[obtuse pebble]

@shadow lava Has your question been resolved?

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[hearty pendant]

plz explain first derivative of logx by first principle

[obtuse pebble]

Please don't occupy multiple help channels.

[timid silo]

what is your definition of log?

[hearty pendant]

ru trolling

?

[tardy epoch]

no

it matters

[hearty pendant]

base of log is non zero psitive

[tardy epoch]

that's not a definition of log

[hearty pendant]

log base a where a is non zero positive and range is all real

domain is real positive

[west sierra]

again not the definition

[tardy epoch]

just confirm that it's this one

https://en.wikipedia.org/wiki/Logarithm

[hearty pendant]

ok sir

can differentiate it with first principle now plz

[tardy epoch]

where do you get stuck on

do you know the definition of derivative?

[hearty pendant]

yes

[tardy epoch]

show your attempt then

[hearty pendant]

ok wait a min

sry took some timee

[tardy epoch]

use log rules to bring h to the exponent

,tex .log rules

[warm shale]

riemann

[tardy epoch]

third row

then switch the limit with the logarithm

[tardy epoch]

then after that, you'll need the definition of the exponential function

[hearty pendant]

?

[tardy epoch]

a question mark isn't helpful

[tardy epoch]

do you know this third log rule or not?

[hearty pendant]

yes

exponent ?

[tardy epoch]

in this, y is in the exponent

[hearty pendant]

yes i m seeing

wot to do now

can u explainn frm img briefly

[tardy epoch]

explain what

you need to ask complete questions, i can't read your mind

[hearty pendant]

the above img

like how can i get derivative

where is flaw

[west sierra]

The above image is log(x^y). What about it do you want explained?

[hearty pendant]

bro the img i uploaded

[wanton isle]

this

[west sierra]

Log(a+b)= loga + logb?

[hearty pendant]

this eggman

tell where is flaw

im stuck

[wanton isle]

this

[wanton isle]

And this

[hearty pendant]

ok thanks

im dead

.close

[obtuse pebble]

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[hearty pendant]

intuitively y sinx = tanx when x tends to zero

[haughty coyote]

They differ by cos x which is very close to 1 (as it's very very close to 1 - x^2)

[royal basin]

sin(x)/tan(x) = cos(x), and cos(0) = 1.

[hearty pendant]

so do we put value of x tending 0 in cos x?

[royal basin]

if you want to put it that way

[obtuse pebble]

@hearty pendant Has your question been resolved?

[obtuse pebble]

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[long hornet]

Help

[long hornet]

did I do this right by canceling n's

Ignore what I could have done (like dividing by 2 ) I just am wondering if I CAN get rid of n in this way

And yes n is not 0

[viral blade]

ig so sure

you could have gotten rid of it right from the start in that case (if it's not 0)

[long hornet]

So is this possible too?

[viral blade]

yes

[long hornet]

So then how is n disappearing there

[long hornet]

Since im obviously not dividing both sides

[long hornet]

So what am I doing to make n disappear here

[viral blade]

oh what

no it should be 1/2 on the left, not 2

then it's dividing by n

[long hornet]

Oh yeah ,
so like this?

[viral blade]

yeah

[long hornet]

So the left example and the right example are being divided by n ?

[viral blade]

seems like you're multiplying both sides by n or something

on the left

after inverting both sides

[long hornet]

yup , so both "solutions" are correct just written in a different way?

[long hornet]

It looked at first like left solution and the right solution are completely different

[viral blade]

ye ig so

strange question, for these to count as solutions

[long hornet]

well im doing problem where I have to make an equation look like a certain formula so I'm not actually solving it traditionally

.close

[obtuse pebble]

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[timid silo]

Can someone help me

[sage geode]

Tried switching the order of integration in $\int_2^{10}\int_0^y\frac1{x^2-2x+y}\dd{x}\dd{y}$?

[warm shale]

A Lonely Bean

[sage geode]

Because if you don't then you are gonna have something like arctan(.../sqrt(y - 1))

Which I doubt anyone would want to integrate

[timid silo]

Can you help me I dont know how double integrals work

[sage geode]

Okay actually forget what I said about switching the order of integration

First of all that's the same as $\int_2^{10}\int_0^y\frac1{(x - 1)^2 + \sqrt{y-1}^2}\dd{x}\dd{y}$

[warm shale]

A Lonely Bean

[sage geode]

There is a formula stating that $\int\frac1{t^2+a^2}\dd{t} = \frac1a\arctan{(\frac{x}a)}$

[warm shale]

A Lonely Bean

[timid silo]

Oh ok

I got it

[sage geode]

So this is the same as $\int_2^{10}\frac1{\sqrt{y-1}}\arctan{\frac{x-1}{\sqrt{y-1}}}\eval^{x=y}_{x=0}\dd{y}$ or $\int_2^{10}(\frac1{\sqrt{y-1}}\arctan{\sqrt{y-1}} + \frac1{\sqrt{y-1}}\arctan{\frac1{\sqrt{y-1}}})\dd{y}$

Okay that's not what I wanted to write wait

[warm shale]

A Lonely Bean

[sage geode]

There we go

arctan(1/z) = arccot(z) so this can be rewritten as $\int_2^{10}\frac1{\sqrt{y-1}}\arctan\sqrt{y-1}\dd{y}+\int_2^{10}\frac1{\sqrt{y-1}}\arccot\sqrt{y-1}\dd{y}$

[warm shale]

A Lonely Bean

[sage geode]

Oh actually wait there's a simpler way

Generally arctan + arccot = 1

So this is just $\int_2^{10}\frac1{\sqrt{y-1}}\dd{y}$

[warm shale]

A Lonely Bean

[sage geode]

I think, let me check

Oh my bad it's arctan + arccot = pi/2

So this is gonna be $\frac{\pi}2\int_2^{10}\frac1{\sqrt{y-1}}\dd{y}$

[warm shale]

A Lonely Bean

[sage geode]

Which should be really easy to integrate

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Thank you so much!

[obtuse pebble]

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[ebon finch]

I need help understanding a math proof: So I was told to read a proof for the limit of sin(x) over x, that as it approaches 0 the answer is 1, however I'm not getting the first part.

[void elk]

I can help!

[ebon finch]

It says:
Considering an angle in a deleted neighbourhood it verifies that |sin(x)|≤|x|≤|tan(x)|

But I don't quite get why

[void elk]

That's a great question! Let me try to explain it to you.

The limit of sin(x)/x as x approaches 0 is indeed equal to 1.

To see why, we can use L'Hopital's rule, which says that if we have an indeterminate form (such as 0/0 or ∞/∞), then we can take the derivative of both the numerator and denominator and try again.

So in this case, we have 0/0 as x approaches 0:

sin(x)/x as x approaches 0

We can take the derivative of the numerator and denominator:

cos(x)/1

And now we can evaluate the limit as x approaches 0:

cos(0)/1 = 1/1 = 1

Therefore, the limit of sin(x)/x as x approaches 0 is 1.

I hope this helps! Let me know if you have any other questions.

[ebon finch]

I haven't studied derivatives yet so while the explanation is simple and concise, for the purposes of my class it remains useless haha

[void elk]

I see.

[ebon finch]

I was given a trigonometric proof

but it starts with the assumption that in a deleted neighborhood of x=0, |sin(x)|≤|x|≤|tan(x)|

[woeful cove]

You could use the sandwich theorem

[ebon finch]

yeah, that's what I was trying to do

[woeful cove]

There's a great khan academy video on it

[obtuse pebble]

@ebon finch Has your question been resolved?

[ebon finch]

Not really but it rarely is

Whatevs

[obtuse pebble]

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[fiery wraith]

having the following density function:

[fiery wraith]

how can i find k?

if i remember, i had to integrate every subfunction

[tardy epoch]

use definition of density

one of the properties in the definition gives you a formula involving k

[fiery wraith]

is it $\int^1_0 kx(x-1)dx = 1$?

[warm shale]

Pokerface.exe

[tardy epoch]

right

[fiery wraith]

that gives me 6

is that correct?

or am i doing my calculations wrong?

[tardy epoch]

!show

[obtuse pebble]

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

[fiery wraith]

yea

i was gonna latex it

lol

[tardy epoch]

,w int 0 to 1 of 6x(x-1)

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[fiery wraith]

i forgot a minus signal

actually i got k = -6

another question, if you dont mind

from the same density function, how can i make a cdf out of it?

[warm shale]

Pokerface.exe

[fiery wraith]

in this specific case, obviously

[obtuse pebble]

@fiery wraith Has your question been resolved?

[fiery wraith]

.close

[obtuse pebble]

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[covert cove]

Hello I'm doing homework quiz (not a test!) And this is really confusing me

[nocturne minnow]

,rotate

[covert cove]

It's due pretty soon, so I don't really want to know all the steps, I just need an answer

[nocturne minnow]

Fyi, we're not here to give out answers

[tardy epoch]

then you're in the wrong place

[obtuse pebble]

@covert cove Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

s

[timid silo]

Can someone explain to me why the answer had a 6 instead of a 2?

[obtuse pebble]

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

[wooden marten]

because 6(x + 8)

[timid silo]

Yeah but its also 2(x-9), how come the 6 takes priority?

[wooden marten]

what's the LCD of 1/2 and 1/6

[timid silo]

.............

Im dumb

.close

[obtuse pebble]

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[brisk crystal]

on my sign chart this interval is negative but what does that mean for a f' graph?
I can do the math but I'm confused on which choices I choose depending on if it wants f f' or f'' graph

[wooden marten]

f'' is negative, so f' is ... ?

f'' is the derivative of f'

[brisk crystal]

im confused whether it would be decreasing, concave down or negative

[wooden marten]

the derivative of f' is negative
that means something about f'

[brisk crystal]

ye but can u like tell me how they all work cuz im not sure

like how each graph effect the other

[wooden marten]

when the derivative is negative that means the slope is negative

[wooden marten]

concave down f = decreasing f' = negative f''
concave up f = increasing f' = positive f''

[brisk crystal]

ah ok

ty

.close

[obtuse pebble]

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[round jasper]

alright so i’m supposed to rotate it about the y-axis and then i need to calculate the volume of 3D shape created. i need help with finding the volume.

[round jasper]

i rotated it but don’t know how to make into a 3D shape and find the volume

[obtuse pebble]

@round jasper Has your question been resolved?

[obtuse pebble]

@round jasper Has your question been resolved?

[round jasper]

@obtuse pebble

[obtuse pebble]

@round jasper Has your question been resolved?

[timid silo]

This is proper rotation.

You just mirrored it

Volume of this = cylinder - cone

[obtuse pebble]

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[sleek mural]

How do I approach this problem? Someone plz help

<@&286206848099549185>

I’m so lost

[teal turret]

What class is this for

[viral blade]

!15m

[obtuse pebble]

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.

[sleek mural]

Geometry

It’s 1/5 tho

Not -1/5

Therefore the negative reciprocal is -5

So wouldn’t the answer then be (y-4)=-5(x+2)?

@torpid bloom

Plz respond

Hello?

I rlly need help w/ this

<@&286206848099549185>

@teal turret do u know by any chance

[timid silo]

how can I help you ?

[sleek mural]

Hi, so

For this here question

I’m wondering whether I got it correct

I got (y-4)=-5(x+2)

[timid silo]

ok show me your steps

[sleek mural]

Ok, one sec

[timid silo]

that's correct

[sleek mural]

There isn’t much, but this is it

[sleek mural]

Oh, rlly? Great

One other question if u don’t mind

I’ve already solved it and so I’m just wondering whether my answer’s correct

One sec

[timid silo]

yep I used a different approach and got the same exact answer

[sleek mural]

👍

One other question

One sec

Right here

I got 5 for the radius, but I’m not sure if that’s right

[timid silo]

okay give me a few minutes and if I get to anything I'll let you know

[sleek mural]

Alr tysm

[timid silo]

Never mind glad to be able to help

[timid silo]

when he said GH=8 did he mean the arc or the chord ?

[sleek mural]

I believe that’s referring to the chord

[timid silo]

alright

[sleek mural]

Do u think I’m right?

@timid silo

[timid silo]

I thought that it is the chord myself but I'm not sure

but it make sense if he was referring to the arc he should've wrote arc GH

[sleek mural]

Yeah, I’m sure it’s the chord

I’m just asking abt the radiys

*radius

What’d u get for it?

[timid silo]

still solving the problem just give me some time

[timid silo]

it's 5 yeah

and if GH was the arc the problem would be unsolvable

[sleek mural]

Great

[sleek mural]

Right

Tysm, rlly appreciate it

[timid silo]

Don't mention it I'm glad I was able to help

[obtuse pebble]

@sleek mural Has your question been resolved?

[obtuse pebble]

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[toxic spear]

I saw these questions in another help channel, they arent mine but as I dont have any of my own questions and they're similar to what I am donig in school right now, I would really appreciate some help with learning this

[toxic spear]

if someone replies please ping me, thank you!

[wooden marten]

I guess those are all parallelograms
which satisfy some things like opposite angles are congruent, adjacent angles sum to 180, the diagonals bisect each other

[high lily]

how do you know it's similar without your own questions to compare them to

[toxic spear]

I stared the concept today in school and I would like to get a jump on it

[wooden marten]

youtube is good

[toxic spear]

I never got any homework or anything though

could you recommend me any videos?

[high lily]

anything on properties of parallelograms

[toxic spear]

could you help me work through that '12.' one as an example and I should be good from there

[earnest elk]

Do some angle chasing

[toxic spear]

whats that

[wooden marten]

angle chasing means trying to find the angle

using angle properties

[wooden marten]

such as this

[earnest elk]

Try finding angle BDA

[toxic spear]

so all I need to know to do these questions are the angle properties?

[earnest elk]

No, not all, at some point you are going to have to use properties of a parallelogram

[obtuse pebble]

@toxic spear Has your question been resolved?

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[azure anchor]

!status

[obtuse pebble]

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

[azure anchor]

or more specifically they want a dimension but i have no idea how to arrive at a dimension. i only optimized the configuration (whether the fifth fence line should be vertical or horizontal).

not sure how to make the jump from area to dimensions now.

my work

3x+2y was the correct my bad

but the dimension im still lost

[trail musk]

why not try using a bit of common sense here? shouldn't the middle segment be parallel to the shorter sides?

[azure anchor]

right but how do i figure the 1.5 mil sqft into this

[trail musk]

so let's assume that x <= y, such that xy = 1.5 million. then we wish to minimize 3x + 2y

then just write y in terms of x

[azure anchor]

1.5/x

oh wait, but by perimeter? well perimeter is unknown here, no?

unless there's a way to find total perimeter by total area?

i figured the problem is minimizing perimeter to "cut cost" of the fence

[trail musk]

yeah, so you want to minimize 3x + 3000000/x

just find the critical point now

[azure anchor]

its 1

[trail musk]

remember to keep the scale of the numbers the same

[azure anchor]

1 million feet is the perimeter

?

[trail musk]

what is x?

[azure anchor]

1mil as the crit point

[trail musk]

nope

[azure anchor]

????

[trail musk]

you want to minimize 3x + 3000000/x

[little quartz]

why 3000000?

just 3/x

[little quartz]

.

[trail musk]

no, you need to use 3 million because then the scale of your units doesn't stay the same

[azure anchor]

derivative of 3x + 3/x is 3 - 3/x^2

that crit point is 1 or 1 million is it not

[trail musk]

no

[little quartz]

then should be 3000000x + 3000000/x

[trail musk]

not really, the 3 just refers to the number of segments

the 3 million refers to 2 times 1.5 million

[azure anchor]

hmmmmm

if i do that i get 1000

so i was off by an order of 10^3?

[trail musk]

yes

[azure anchor]

so x = 1000

then P = 3x + 2y => P = 3000 + 2y

A = 3000y = 1,500,000

1500/3?

[trail musk]

stop dropping the zeros

[azure anchor]

errr

1000y = 1,500,000

y = 1500

1000x1500

thats right then

[trail musk]

yep

[azure anchor]

weird. i thought it would've been equivalent just multiplying the end result later, but that wasn't the case at all

i was off by 10^3 but idk how i would've known that without doing what you said

[trail musk]

that's because your units don't align if you ignore the 'million'

[azure anchor]

strange the other config even had a critical point of 1.5

[little quartz]

i see, it's would be 3x + 3/x when xy = 1.5(using the million)

[azure anchor]

thanks @trail musk you are a g

.close

[obtuse pebble]

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[obtuse pebble]

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[obsidian isle]

can you use a calculator

if not then good luck buddy!

[rain meadow]

No calculator

[spice citrus]

try the identities from this site I guess: https://www.geeksforgeeks.org/inverse-trigonometric-identities/

[obtuse pebble]

@rain meadow Has your question been resolved?

[sinful bay]

How do I solve this?

[royal basin]

@sinful bay please open your own channel. #❓how-to-get-help

[rain meadow]

<@&286206848099549185>

Is there anyone?

.close

[obtuse pebble]

Channel closed

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[ivory estuary]

ah

[obtuse pebble]

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[haughty coyote]

!status

[obtuse pebble]

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

[haughty coyote]

Show they verify the Pythagorean equality

[half crown]

so incase here c^2 = a^2 + b^2?

uh I have to do a^2 +b^2 and then obtain the c^2?

[haughty coyote]

Verify, using the definitions, that this equality holds

[half crown]

but how do u know c is the hypotenuse

[haughty coyote]

It's the largest one

(c >= b isn't exactly obvious at your level but it follows from something called the arithmetico-geometric inequality)

[half crown]

oh okay, thanks i was going to ask that

also I have more question wait

[haughty coyote]

Which is very useful btw

[half crown]

i will look after this

!status

[obtuse pebble]

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

[haughty coyote]

fun fact: not only does this always give Pythagorean triples, but every Pythagorean triple can be written in this way for the right choice of n and m

[half crown]

okay i will look into it

[haughty coyote]

There's quite a few proofs, but to my knowledge all of them use some kind of undergraduate math

[half crown]

?

[haughty coyote]

You posted nothing

[haughty coyote]

I mean your answer

[half crown]

oh,

For how do their area vary?
-> If length and breath of rectangle is difference but their sum is 500 their area vary.
For the sape of the rectangle with the largest area?
-> length = breath = 250
For What would be your choice of shape for the pool?
-> circle since it has more area then rectangle

is it right?

again, "If rectangular, what is your preference for dimensions?" what does it mean?

also I have last question

[haughty coyote]

A square

[half crown]

oh okay

!satus

!status

[obtuse pebble]

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

[half crown]

2

[haughty coyote]

!show

[obtuse pebble]

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

[half crown]

for Bochure 26miles approximation I got 23.3 miles which is off. I used the process of taking radius of Earth and using trigonometry.
Then after that I don't know what to do to find the approximation of ship and plane. I tried using similar triangle property but failed miserably

Also,
raidus of Earth = 3960 miles
1 mile = 5280ft
the conversion between ft and miles

[haughty coyote]

26m = sqrt((r+h)^2 - r^2) right ?

[half crown]

correct but it doesn't give 26m

r and h are in different unit so sqrt((r + h/5280)^2 - r^2)
where
r=3960
h=362

[haughty coyote]

I know

Perhaps

Side of triangle != Length of the ground

Which I doubt changes it by 10%, but hey

[half crown]

can you re do it and send it to me I don't think I will understand this problem unless I see the solution?

[haughty coyote]

Look at the arc length

Doesn't change it one bit

As expected

[half crown]

I don't know what are u talking about, i don't know if it's a burden or not but can u do the solution

@haughty coyote ?

[haughty coyote]

I still believe in it being 23.3

[half crown]

oh okay, then what to do after that?

[haughty coyote]

It's the same formula on wikipedia, I simulated it on desmos. They all match

[haughty coyote]

have you tried anything for the plane ?

[half crown]

nope i try to do the ship but i failed and i left as it it's

[haughty coyote]

Try to do the plane. You have every information

[half crown]

I don't know how to do it uhh and also what do for the height of the ship

[haughty coyote]

The point of doing the plane first is to have a formula that you can then invert to get the height of the ship

[haughty coyote]

Make a drawing of what happens at the limit distance, where they can barely see each other

[half crown]

i will leave this question I don't know how to solve it very small brain

[haughty coyote]

they're both looking at the horizon

and it's the same point

so you just sum the distances to the horizon

[half crown]

send wiki link

[haughty coyote]

https://en.wikipedia.org/wiki/Horizon#Objects_above_the_horizon

[half crown]

also accept friend request I will ask question personally if i need help thanks again for your time @haughty coyote

.close

[obtuse pebble]

Channel closed

Closed by @half crown

Use .reopen if this was a mistake.

[haughty coyote]

No. Ask in the server

[half crown]

yesterday i wait for 1 and a half hour no one answer me : (

[obtuse pebble]

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[opaque galleon]

I need help splitting $(-1)$ raised to something from $(-3)^{2n+1}$
I thought it would be $(-1)^{2n+1} (3)^{2n+1}$ but i think im wrong

[warm shale]

steamhahasteamhaha

[ivory estuary]

that looks right, can you give the rest of the question?

[opaque galleon]

im checking the end points of the IoC

[rugged kite]

What's the radius of convergence ?

[opaque galleon]

haven't checked yet

im still at the endpoints

[rugged kite]

i would do it backwards

[opaque galleon]

my initial endpoints are (-1,5)

what?

[rugged kite]

once you have the radius of convergence, finding the domain is a matter of checking whether it is defined at those endpoints or not

[opaque galleon]

uhh yeah that's what im doing

[rugged kite]

oh, didn't see the 2n+1 power

sec

yeah your endpoints are right, my bad

[opaque galleon]

no problem bro

also I got a question

i saw someone on youtube said that if i have |x-a| < Number

Number is my Radius

is okay to use that

[rugged kite]

Depending on how you see it, that's the definition of a radius, so yes

anyway, what are you stuck on ?

[opaque galleon]

$\sum_{n=1}^{\infty} \frac{(-3)^{2n+1}}{9^{n} n^{9}}$

[rugged kite]

you forgot closing braces

$\sum_{n=1}^{\infty} \frac{(-3)^{2n+1}}{9^{n} n^{9}}$

[warm shale]

Syst3ms

steamhahasteamhaha

[opaque galleon]

there

goddamn lmao

hahaha

[rugged kite]

anyway just try to type it without latex if you're not comfortable enough with it

[opaque galleon]

i just tried ratio test i got limit = 1

so i can't use ratio test on this

idk what series test to use

i thought maybe AST

[rugged kite]

try to simplify the expression further

[opaque galleon]

but idk how to split a (-1)^n+1 or (-1)^n

OH SHIIT

[rugged kite]

once you do it will be plain to see

[opaque galleon]

IM DUMB

HAHAHAHHAHA

[rugged kite]

you're not dumb it happens

[opaque galleon]

$\sum_{n=1}^{\infty} \frac{3}{ n^{9}}$

[opaque galleon]

is this right bro

[rugged kite]

didn't cancel some things right

that 9^n shouldn't be there anymore

[warm shale]

steamhahasteamhaha

[opaque galleon]

woops

convergent

[rugged kite]

yup

[opaque galleon]

IoC [-1,5]

[obtuse pebble]

@opaque galleon Has your question been resolved?

Channel closed

Closed by @opaque galleon

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.

[sage dagger]

Im not sure how to find length AE

[sage dagger]

I know we can do 14.3 x 10.5 to find the area of the rectangle

but I am not sure how to approach the triangle

[little quartz]

u can use the cosine theorem

[sage dagger]

actually

[sage dagger]

but I still need a length of a side dont I

[little quartz]

there are AB = AC

and have BE = CD = 10.5

[sage dagger]

AB = AC?

[little quartz]

no

AE =Ab

[sage dagger]

you said AB = AC?

hm

[little quartz]

[prisma hatch]

EB = 10.5

[sage dagger]

oh true

[prisma hatch]

then trigonometry

[sage dagger]

cosine rule for triangle AEB?

[little quartz]

yes

[sage dagger]

ok

still need side b so is there an extra step

before using cosine rule

[little quartz]

..

b = c

[sage dagger]

so do we get rid of b?

[little quartz]

sure , then solve the c

[sage dagger]

okay

[little quartz]

where is b?

[sage dagger]

b would also equal 9.8 right

[little quartz]

i mean in the 3th line,b disappeared

[sage dagger]

oh yeah i forgor to aadd that

[little quartz]

..

[sage dagger]

but on the diagram i have it written

[little quartz]

[sage dagger]

so we use 1/2absinC on AEB and then add that with 150.15m^2 right

[little quartz]

what's 150.15?

[sage dagger]

the area of the rectangle

[little quartz]

i see

[sage dagger]

am i correct?

[little quartz]

i not sure c that you solve

[sage dagger]

huh

[sage dagger]

so is my working here wrong?

[little quartz]

yes,unfortunately

[sage dagger]

i thought you only made C the subject when finding an unknown angle

[little quartz]

wait

48 + 48 + 89 = 185

crazy

[sage dagger]

its 48+48+84 actually

idk where you got the 89

[little quartz]

my eyes wrong

[sage dagger]

so am i right

[little quartz]

[sage dagger]

because im pretty sure what you did was find the angle using that formula but the angle still wrong anyways

[little quartz]

i am right

[sage dagger]

why are you finding angle?

[little quartz]

no ,i use the 48

[sage dagger]

but when you make C the subject of the formula in cosine rule it means you are trying to find the angle?

[little quartz]

but i konw c = 48

i just solve the length of c

[sage dagger]

with this are you solving little c

[little quartz]

yes

[sage dagger]

why is this incorrect can you let me know

[little quartz]

u miss c

where c go?

b

[sage dagger]

b or c

[little quartz]

b

[sage dagger]

oh i just removed it

[little quartz]

sorry ,my english is average

[little quartz]

how do u do this

u can't directly remove b

[sage dagger]

oh

[little quartz]

u should use b=c

[sage dagger]

how did you remove b?

[little quartz]

because i let b = c

[sage dagger]

but i thought i did that here too

[little quartz]

so,where are your c that represent b

[sage dagger]

ohhh

i think i get what you mean now

[little quartz]

haha

it's a little hard for me to use english

[sage dagger]

its fine youre doing good

[little quartz]

can we add friend, i want make friend in discord

u can ask me math question anytime

[sage dagger]

sure

[little quartz]

what's time on your there

now

[sage dagger]

9.38pm

[little quartz]

mine is 7:38

[sage dagger]

oh nice

[little quartz]

p.m too

[sage dagger]

oh thats cool

[little quartz]

yes, where are u from

[sage dagger]

aus

[little quartz]

china

[sage dagger]

oh yeah where did the c^2 go

in this step

[little quartz]

the left have c^2 too

[sage dagger]

it just has 2ac * cosC

[little quartz]

i mean the first line

[sage dagger]

yes but what happened to the c^2 in the second line

[little quartz]

i dont wirte it down

[sage dagger]

but shouldnt it be in second line

why did it dissapear

[little quartz]

ee,this my habit

[sage dagger]

did you mean to write 2ac*cosC = a^2+c^2

[little quartz]

no,

this one

[sage dagger]

,rcw

[sage dagger]

how did you get rid of squared term here

[little quartz]

i simultaneously divide by a

left and right

[sage dagger]

But that doesn’t remove the squared term?

Don’t you need to square root a^2 to remove if

[sage dagger]

hmm ok

ty for the help

.close

[obtuse pebble]

Channel closed

Closed by @sage dagger

Use .reopen if this was a mistake.

[little quartz]

you're welcome

[obtuse pebble]

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.

[west sierra]

Is |adjA|=0 if |A|=0? coz |adjA|=|A|^(n-1)?

[west sierra]

I had this question because in the proof of |adjA|=|A|^(n-1) we had taken det on both sides of A * adjA=|A|*I and It became |A| * |adjA|=|A|^n, here if |A|=0 then I can't divide |A| on both sides so the equation |adjA|=|A|^(n-1) seems to be invalid if |A|=0 but if I put |A|=0 directly into this formula, |adjA| comes out to be 0

.close

[obtuse pebble]

Channel closed

Closed by @west sierra

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.

[fleet geyser]

How can I drop the ln

[prisma hatch]

...?

wdym

[fleet geyser]

Pic is sending

What law allows me to drop it?

[warm shale]

Pro_Hecker

[fleet geyser]

So how would I go about knowing how to sketch this

The memo indicates that it’s a circle

[native inlet]

$e^{\ln(4-x^2-y^2)}=e^{0}$

[prisma hatch]

{

[warm shale]

XxMrFancyu2xX

[eternal thistle]

That equation has no solution, no?
Edit: no

[warm shale]

Pro_Hecker

[prisma hatch]

infinite solutions

[fleet geyser]

I am lost

This is the answer

[prisma hatch]

sorry

[timid silo]

no law just convert back into exponential form i believe

[fleet geyser]

Level curve at c = 0

So how would I go about doing it

Is kinda what I’m asking

[teal turret]

Send the original question

[fleet geyser]

C

[prisma hatch]

https://xyproblem.info

[teal turret]

Alr so what are u confused about

[fleet geyser]

I’ve got the answer memo in front of me

[fleet geyser]

So how would I get the ln away

So that I could then

Mess with the equation to sketch