help-33

[bronze niche]

.close

[marsh citrus]

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[still temple]

help pls

[still temple]

i went wrong somewhere- idk where

the expected final ans is 48.41

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[whole jetty]

A box contains 40 coins, which are 50 cents, €1 and €2. We know that there are twice as many 50 cent coins as there are €2 coins and that the total value in euros is €40.
Can we know how many coins there are of each type?

[stoic saddle]

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stoic saddle]

@whole jetty

[whole jetty]

2

x=Number of 50 cent coins, y=Number of 1€ coins, z=Number of 2€ coins
so x+y+z=40

and x=2z

[stoic saddle]

1€ cent coins

are you sure about this

[whole jetty]

sorry

write it so fast

[stoic saddle]

x + y + z = 40
x = 2z
these are correct

now you also need to write down that their total monetary value is 40 euro

[spark otter]

you didn't find the 3rd equation?

I asked you to express the total amount of money in terms of x,y and z

[stoic saddle]

oh is this a !noclopen situation??

[whole jetty]

he has disconnected on the other channel and I decided to open another

[stoic saddle]

!noclopen

[marsh citrus]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[stoic saddle]

for the future

anyway

the question stands

[whole jetty]

how?

[stoic saddle]

well

there are x fifty-cent coins

what is their total value in euros?

[whole jetty]

0,5?

[spark otter]

that's for one coin

[stoic saddle]

no, 0.5 is the value of ONE such coin

but you have x such coins.

[whole jetty]

0.5x

[stoic saddle]

right

can you now write down the total value of all the coins

incl the single-euros and the two-euros in the same fashion

[whole jetty]

y and 2z

[spark otter]

yes

and so the total amount of money is?

[whole jetty]

so 0.5x+y+2y=40

[stoic saddle]

typo

but yes 0.5x + y + 2z = 40

so now you have three equations in three unknowns: $\begin{cases} x+y+z = 40 \ x = 2z \ 0.5x + y + 2z = 40 \end{cases}$

[elfin berry]

Ann

[whole jetty]

ok, i will solve it now

[stoic saddle]

i'm gonna put a preemptive in the air

[whole jetty]

so z=λ, y=40-3λ and x=2λ

λ=any value

[stoic saddle]

not quite any value

we need the number of coins in each denomination to be ≥0 and an integer

but you are right in that there is not a unique solution.

[whole jetty]

so λ≥0?

anyway tysm

.close

[marsh citrus]

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[wary bluff]

not sure what to do here

[wary bluff]

could I do something like this

[devout mauve]

not parallel to e_1, e_2

[wary bluff]

what is e_1, e_2 here doe

[devout mauve]

unit vectors

[wary bluff]

shit im lost

[bitter solstice]

Can you think of two vectors v1 v2 that aren't parallel to e1 e2?

don't overthink it, chances are random vectors you think of will work 🙂

[wary bluff]

could you give me an example of what e1 and e2 are

i still dont get that

[bitter solstice]

e1 = (1,0), e2 = (0,1)

The basis vectors, standard unit vectors or whatever you want to call them!

[wary bluff]

is it always going to be that?

(1, 0) and (0, 1) ?

[bitter solstice]

for R2 yes

for R3 you get (1,0,0), (0,1,0), (0,0,1)

[wary bluff]

(1, 2) and (2, 4)

how abt that

wait

im dumb

it asked for non parallel vectors my bad one sec

(0, 1) and (1, 1) ?

none of them are parallel to e1, e2

[bitter solstice]

(0,1) is e2

so its definitely parallel to e2

[devout mauve]

can you tell me any 4 random numbers

[wary bluff]

can I pick 2 same vectors

like

(1, 1) and (1, 1)

[bitter solstice]

no

[devout mauve]

no then they are parallel

[wary bluff]

(1, 2) (3, 4)

[devout mauve]

that works

[mossy parcel]

Yo

[wary bluff]

this works?

[devout mauve]

try it

what is that matrix ^3

[wary bluff]

should I first

change it to rref

.close

[marsh citrus]

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[crisp kraken]

integral of 1/(9+x^2)

the exercise is to practice using tangent substitution. But i've noticed that it's similar to the derivative of arctan.
So i made x= x/3 dx=1/3 and i got this form
1/(1+(x/3)^2)
now i know the antideriative is arctan(x)

[crisp kraken]

so in the end i have 1/3 arctan(x/3) + c

is this a valid method? or am i doing something wrong?

i'm not 100 % comfortable using the same x for the variable when i look at it.

[rotund pike]

To check whether it's valid, take the derivative and see if you get what you started with.

[crisp kraken]

but to take the derivvative of arctan is NOT easy right?

[rotund pike]

Well, you just use the chain rule with the normal derivative.

[crisp kraken]

normal?

[rotund pike]

Yeah, the normal derivative, as in arctan(x).

You have x/3 inside, which is a bit different than x, so you use the chain rule.

[crisp kraken]

ok but i still use the fact that i already know the derivative of arctan right?

[rotund pike]

Yes.

[crisp kraken]

works ok

how much should i care about the interval of the trig functions in general?

because usually everything works fine in the exercises

also on kahn academy he double check it but it's good all the time.

well that would be another question.

[crisp kraken]

thank you

[rotund pike]

No problem. That isn't the tangent substitution method, though.

[crisp kraken]

yes i know, that's why i wanted to check this other method

[rotund pike]

Oh, I see.

[crisp kraken]

.close

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[coral stream]

.close

[marsh citrus]

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[rose idol]

What is the total number of possible knight moves (as a function of n) on a n×n board

[rose idol]

im stuck tryi

array([[2., 3., 4., 4., 4., 4., 3., 2.],
[3., 4., 6., 6., 6., 6., 4., 3.],
[4., 6., 8., 8., 8., 8., 6., 4.],
[4., 6., 8., 8., 8., 8., 6., 4.],
[4., 6., 8., 8., 8., 8., 6., 4.],
[4., 6., 8., 8., 8., 8., 6., 4.],
[3., 4., 6., 6., 6., 6., 4., 3.],
[2., 3., 4., 4., 4., 4., 3., 2.]])

i have code which can produce arrays like this so i can see how much it is for values ie this is for 8x8 board

so square (a,1) gives 2 knight moves as expected

so for n=8 total no of knight moves =336

[marsh citrus]

@rose idol Has your question been resolved?

[marsh citrus]

@rose idol Has your question been resolved?

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[kind coral]

Could you check my work? I'm not entirely sure if my proof of continuity and differentiability are correct. Please help

[marsh citrus]

@kind coral Has your question been resolved?

[kind coral]

.close

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[flat gust]

hello, im trying to figure out why (-2)^0.4 has a complex answer on google, wolfram alpha and geogebra, but has a normal solution on desmos and my casio fx991ex. isn't this simply (-2)^(2/5) ?

[flat gust]

shouldn't it simply be equal to this?

[vernal forge]

but it's -2

[flat gust]

it doesn't matter

[vernal forge]

desmos doesn't support complex numbers btw

so be aware

,w (-2)^0.4

[elfin berry]

Results provided by WolframAlpha

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

Requested by arte_m_etra

[vernal forge]

here you go

desmos is showing the real root

[flat gust]

yeah but why

[vernal forge]

but there are other roots

[vernal forge]

$\sqrt[5]{x}$ has 5 solutions in complex numbers

[elfin berry]

artemetra

[flat gust]

okay, but why would google decide to show the complex answer if there is a real root?

[proper zodiac]

Why does Google decide to do anything

[dire spire]

try entering it as $\sqrt[5]{x}$ on google

[elfin berry]

saad

[flat gust]

im not sure how i can enter a specific degree root on google

[flat gust]

either i am dumb or i dont see that here

[dire spire]

click the "Inv" button

[flat gust]

oh okay thanks

let me try, just a second

it doesn't even let me do that

like the button to do that ignores the negative sign

[dire spire]

you have to enter the number inside the root before clicking on the root operation

yeah it wont let you do that

[flat gust]

i see, but that shifts the negative sign to the left

[dire spire]

because it's an illegal operation

negative inside root

[flat gust]

i think this is just a case of google's calculator playing dumb, thanks y'all

[marsh citrus]

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[crisp knoll]

I need to create a mclaurinseries

the green is the end result

I don't understand how you can find the end result

This is what i got so far

.close

[marsh citrus]

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[proven jacinth]

hello

[proven jacinth]

i am stuck on a multivariable calc question

I have progress up to this

$\frac{\partial f}{\partial x} = 2x\frac{\partial g}{\partial u} + 2y\frac{\partial g}{\partial v}$

[elfin berry]

george clooney real account

[proven jacinth]

so im stuck on finding the second derivative

i just need to take the derivative of that wrt x

but i dont know how to take the derivative of dg/du wrt x

[marsh citrus]

@proven jacinth Has your question been resolved?

[proven jacinth]

hello :)

[twilit geyser]

the same way you did the derivative of g wrt x

rewrite h(u,v)=dg/du if that helps

maybe use copious amounts of parenthesis

[proven jacinth]

why is it h(u,v) and not just h(u)?

where is v in dg/du

[twilit geyser]

g(u,v) depends on u,v

[proven jacinth]

so dg/du depends on both as well

ok got it

[twilit geyser]

yup yup

[proven jacinth]

let me try that and get back to u

[twilit geyser]

perfect

[proven jacinth]

oh my god

works like a charm

tysm @twilit geyser

idk why i didnt think of that

[twilit geyser]

cool, yeah you're welcome

[proven jacinth]

.close

[marsh citrus]

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[twilit geyser]

all good haha

[marsh citrus]

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[humble river]

hello

[humble river]

I just started the log rules lesson

Why did that get converted

To that

like where does 6/15 come from

[modest adder]

logx (a) - logx (b) = logx (a/b)

this is a formula

x here is base

[humble river]

Ohhh

So its a bunch of rules i need to memorize basicall

ohk

thnx verymuch

.close

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[lost stratus]

Can someone verify real quick that the notation for these 3 scenarios is correct ?

[main idol]

What does ~ symbol mean

The other two are right

[marsh citrus]

@lost stratus Has your question been resolved?

[lost stratus]

approx

Its the same matrix except I removed row 1 from one of them

[main idol]

Approx in what way?

You mean you did 1 row operation?

[lost stratus]

yeah

[main idol]

Then yes that's a valid row operation and correct use of ~

[lost stratus]

I explain above that I did r1 <- r1-r2

then use that

[main idol]

Typically people write something like R1 = R1 - R2

[main idol]

Where

[lost stratus]

Its not in this pic

I just say above: "We now execute the row operation R1 <- R1 - R2"

then this pic

[marsh citrus]

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[still temple]

how do

what do you know about the ratio of the geometric sequence

you divide a value by the value before it to get the common ratio, r

@still temple

in other words, $3^{3-2x}/3^x = r = 9^{x-1}/3^{3-2x}$

[elfin berry]

Pride

[still temple]

YES

caps

yes

but

its not

l.s does not equal r.s

see, they dont equal the same

@still temple

don't ping me. I gave you the method to calculate the solution.

[marsh citrus]

@still temple Has your question been resolved?

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[still temple]

@still temple thanks

[marsh citrus]

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[keen ocean]

help

[keen ocean]

understanding sufficiency statstics

Exercises
9.37 Let X 1 , X 2 , . . . , X n denote n independent and identically distributed Bernoulli random vari- ables such that
P(Xi =1)=p and P(Xi =0)=1−p,
for each i = 1, 2, . . . , n. Show that n X i=1
criterion given in Theorem 9.4.

questions like this for example

[main idol]

are we supposed to know what theorem 9.4 is

[keen ocean]

here it is

i am not looking to solve that problem specifically but i dont really understand this chapter of the book

[main idol]

wot

then why'd you post the problem

"under this chapter of the book" isn't really specific enough to help you

[keen ocean]

just as an example of the questions that are asked

[main idol]

what about that paragraph don't you understand?

[keen ocean]

could u explain what an estimator is

[main idol]

just a random variable for some pre-defined statistic like mean

https://stats.stackexchange.com/questions/317541/why-is-an-estimator-considered-a-random-variable

[keen ocean]

so its a function that helps us determine a RV

but then its also a RV itself

[main idol]

no

the mean is not a random variable, it's a number

if you had perfect information, you wouldn't need estimators for the mean

the estimator approximates the mean because you don't have perfect information

[keen ocean]

okay got it

and for that estimator to be sufficient

we have to include all information available to us?

[main idol]

just read the example they give

with binomial trials

[keen ocean]

i looked at this example

[main idol]

uh huh

[keen ocean]

im not usre

sure

cant phrase the question

.close

[marsh citrus]

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[quaint sky]

how do i get from the second line to the third line

[main idol]

presumably you were given P_i, P_f and gamma

[quaint sky]

well i know the first one

it is pv=nrt

so p1v1 =t1

hence tf/t1=vf/v1

second lime i assume volume is constant

wait give me a second

.close ill come back

[marsh citrus]

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[azure bone]

My answer isn't close to any of them idk wat to do

[lone heart]

How did you get that height?

How did you get 11 and 6?

[azure bone]

Oh I meant like

The side

The side side

Like on the sude

[azure bone]

From point C to point B

[lone heart]

I know what it is, I'm asking you how you found those values

[azure bone]

Oh it's 10 not 11

[lone heart]

Yes

[azure bone]

Uhh wdym

Idk how to say it

The Pythagorean theorem

[lone heart]

You did some math to get 11 and 6, how did you get those numbers?

Did you just randomly pull them out of a hat?

[azure bone]

I got like 34.5 I think I messed up

[azure bone]

I MEANT 10

[azure bone]

IDK HOW TO SAY IT

Points B and C

Make like a triangle

Right triangle

Thing

[lone heart]

I told you before, showing your work means show everything, the entire process

How are people suppose to know where 11 and 6 came from

[azure bone]

Idk looking at it😭

[lone heart]

You showed a rectangle with coordinate points and there's not even an 11 in it, so how did you get it

What was your process in finding it

[azure bone]

I got like 34.8 fr

Its not an option😭 😭

[lone heart]

,w 2sqrt(34) + 2sqrt(136)

[elfin berry]

Results provided by WolframAlpha

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

Requested by captainnova22

[lone heart]

It rounds to one of those options

[azure bone]

Oh

So 35

C

[lone heart]

Yes

[azure bone]

Ty

[azure bone]

This is outside the circle right

[lone heart]

How did you get that

[lone heart]

Yes looks good

[azure bone]

Yay

Ty

.close

[marsh citrus]

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[marsh citrus]

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[lone heart]

You were close

Recall the cases for discriminant

This

So you want to set b^2 - 4ac = 0 and then solve for k

No sqrt needed

When you find the value of k, if you wanted to, you can find the roots to that quadratic, but that's not necessary

You just wanted to find the value of k for 1 solution with a multiplicity of 2

[marsh citrus]

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[compact crescent]

I’m trying to prove that (1) is true

[compact crescent]

That is the geometry approach of The Polar Form of Complex Numbers

[compact crescent]

By the relation between these line segments, r is equivalent to sqrt(x^2 + y^2) without a doubt.

Yet, I disagree that r = |Z| = |x+yi|

<@&286206848099549185>

[marsh citrus]

@compact crescent Has your question been resolved?

[bitter solstice]

You have |z| = |x + yi| but the norm of a complex number is not given by √(x+yi)^2

It's actually given by $\sqrt{(x+yi)(x-yi)}$

[elfin berry]

OssihLikesBlue

[bitter solstice]

Cause $|z| = \sqrt{zz^*}$ or $\sqrt{z\overline{z}}$ whichever you use to denote the complex conjugate

[elfin berry]

OssihLikesBlue

[marsh citrus]

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[orchid oracle]

$$ \lim_{x\to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} $$

[elfin berry]

rainy

[orchid oracle]

How would you calculate this without L'hopital rule

🧐

[violet glacier]

multiply (3+sqrt(5+x))(1+sqrt(5-x)) to the denominator and the numerator

[orchid oracle]

that doesnt make sense?

[lone heart]

They mean, to multiply the conjugate of the numerator to both the numerator and denominator then multiply the conjugate of the denominator to both the numerator and denominator

[marsh citrus]

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[deft apex]

👋 Was just poking some fun at representing fourier series with spinny spinny circles and I have it working as expected for series that represent either odd or even functions, but when I want to represent something more interesting that contains both sines and cosines i'm not sure where to go with it, https://pastebin.com/3aLW3vUs is the rather poor first implementation ignoring all optimizations and focusing just on the logic, around line 62 is where its drawn, and rudimentarily plonked the series args around line 143, if I change the starting angle for terms by 90 degrees, that is what dictates if it represents a sine or cosine series. And probably the source of my troubles in trying to draw series containing both. I'm also fairly out of my depth with knowledge on this topic and wanted to use it to gain intuiton, which I guess fairly worked for the simple cases, but if anyone had resources that would be useful for me to understand and solve it then that would also be great :D

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[echo fog]

Hey I get the x=1 because tan(pi) =1, but what does it mean by multiply by 4?

[bleak dome]

Since arctan(1) is pi over 4

we multiply with 4 to get pi

[echo fog]

Ohh I feel dumb lol I thought it equaled pi for some reason. Thank you

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[bleak dome]

yes

[bleak dome]

happens to everyone

[marsh citrus]

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[harsh zinc]

Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8}, B = {0, 1, 2}, + - the addition operation by modulo 9, * is the addition operation by modulo 3. Establish a surjective homomorphism between (A, +) and (B, *).

[harsh zinc]

I don't know where should i start 😦

[glass silo]

* for addition

You're familiar with the definition of a homomorphism, right?

[harsh zinc]

Yes, i think this homomorphism is f(x) = x mod 3, but i'm not sure

[vernal forge]

that's correct

it's surjective and satisfies all conditions for a homomorphism

[glass silo]

@proud ice

[harsh zinc]

I'm not sure how to prove that f((a+b) mod 9) = (f(a)+f(b)) mod 3, except from looking at every case

[vernal forge]

{0, 1, 2, 3, 4, 5, 6, 7, 8}
after mod 3
{0, 1, 2, 0, 1, 2, 0, 1, 2}

the property of mod is that this sequence repeats

and 9 is a multiple of 3

[marsh citrus]

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[undone veldt]

Just need help on section e

can provide answer for other sections if needed

[spark otter]

You found that x=1 is a solution to 2x^3 - 7x^2 + 3x + 2 = 0

[undone veldt]

yes

[spark otter]

Since this is about finding the roots of a polynomial...

and you found one of them

[undone veldt]

and x also equals 2.85 and -0.351

[spark otter]

ok so you finished section e

[undone veldt]

then what do you do for d?

i thought that was the answer for d?

[spark otter]

No

the answer to d is "y = y"

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

if the line and curve intersect

[undone veldt]

so then do you solve for x? or... @spark otter

<@&286206848099549185>

[spark otter]

You solved for x already as you told me

In part e

[undone veldt]

so where do you take this?

[spark otter]

The equation of the line is y = x+2

[undone veldt]

i get that part

[spark otter]

The equation of the curve is y = (2x+1)(x-2)²

[undone veldt]

but then what do you do

[spark otter]

Like how do you get to this or what do you do afterwards?

[undone veldt]

what do you do afterwards

[spark otter]

Expand

Bring to the LHS

[undone veldt]

that will be 2x^3 - 7x^2 -13x -6 = 0

[spark otter]

No, try again

Don't forget the signs

[undone veldt]

*+2?

that will be 2x^3 - 7x^2 -13x +2= 0

[spark otter]

Your x coefficient is wrong too

[undone veldt]

+3x

[spark otter]

Yes

[undone veldt]

ah and then thats the same as the aswer in question d

so is that done?

[spark otter]

Yes

We asked you to prove that this was true

And we did it

Saying "y(curve) = y(line)"

[undone veldt]

ok thank you very much

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[flat wigeon]

can someone tell me why does this have two solutions? I understand the pi/3 but not the -pi/3.

[lone heart]

Unit circle

[versed lava]

cosine is positive in the first and the fourth quadrant

[lone heart]

-pi/3 = 5pi/3

[versed lava]

a value in the fourth quadrant can be defined as negative of the value in the first quadrant

i.e, if you have a value of 330 degree, you can write it as -30 degree

[flat wigeon]

oooooohhhhh

[flat wigeon]

ohh because you just basically do 360-θ, and you only write -θ

[versed lava]

yes

to sum it up

in the first quadrant, your value is θ
in the second, it is π - θ
in the third, its θ - π
in the fourth, its -θ

[flat wigeon]

thank you very much!

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[versed lava]

no worries

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[lusty basalt]

can anyone tell me why it is not 1/2(x)^2

i.e 1/2(sint^2[t]) and the same for the cost value

[desert dirge]

?

its integrating trig functions

[lusty basalt]

yes?

[desert dirge]

why would there be a 1/2 (sin^2(t))

[lusty basalt]

i don't understand why there wouldn't be

[desert dirge]

because youre just integrating cos(t)

it goes to sin(t)

[lusty basalt]

when integrating the power ^1 then divide

[desert dirge]

cos(t) isnt a variable, its a function of a variable, power rule doesnt just apply

[lusty basalt]

i know

i understand\

sorry

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[calm cloud]

Does someone know how I would get rid of the cube roots on top. I cant just multiply by the conjugate so I am stuck.

[wise jackal]

do you know the limit defintion of the derivative?

[calm cloud]

no

[stoic saddle]

I cant just multiply by the conjugate
you can actually, it'll just look different

the conjugate you're looking for is (x+h)^(2/3) + (x(x+h))^(1/3) + x^(2/3)

[calm cloud]

how did you get that

[marsh citrus]

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[frail goblet]

how to do these?

[paper raptor]

for the first one maybe consider using a u-sub

then for the second looks like a partial fraction

[frail goblet]

u = 8x + 16?

du = 8dx

[paper raptor]

$\sqrt{8x+16}=\sqrt{8}\sqrt{x+2}$ :)

[elfin berry]

PajamaMamaLlama

[frail goblet]

ahhh

[paper raptor]

wow that first one simplifies really nicely actually

[frail goblet]

then 1/sqrt8 (int(e^u/sqrtu)du)?

idk how to use bot sry

[paper raptor]

$\frac{1}{sqrt{8}}\int\frac{e^{u}}{\sqrt{u}}\dd u$

[elfin berry]

PajamaMamaLlama

[paper raptor]

be careful when ur subbing in the differential

$\frac{\dd u}{\dd x}=\frac{1}{2\sqrt{x+2}}=\frac{1}{2\sqrt{u}}$

[elfin berry]

PajamaMamaLlama

[frail goblet]

du = 1/(2(sqrt(x+2)))

if u = sqrt(x+2) why is it e^u/sqrt(u) again?

should bottom be not sqrt

[paper raptor]

$\frac{du}{dx}=\frac{1}{2\sqrt{u}}\implies du=\frac{dx}{2\sqrt{u}}$, can you isolate dx?

[elfin berry]

PajamaMamaLlama

[frail goblet]

2sqrt(u) du = dx

[paper raptor]

now we have $\int \frac{e^u}{\sqrt{u}}\dd x=\int \frac{e^u}{\sqrt{u}}(2\sqrt{u}\dd u)$ :)

[elfin berry]

PajamaMamaLlama

[frail goblet]

ohhh okok

[paper raptor]

and don't forget about that 1/sqrt8 on the outside

very important

[frail goblet]

so the sqrtu cancel out?

[paper raptor]

[frail goblet]

so

(e^(srt(x+2)))/sqrt(2)?

[paper raptor]

🎉

[frail goblet]

ty

how about the second one

partial fraction decomp?

[paper raptor]

that's a simple partial fraction decomp, you get some nice integrals out of it :)

[frail goblet]

wait

you can't do partial fraction on it though

[paper raptor]

oh wait

you can just split the fraction

[frail goblet]

uhhh wdym?

[paper raptor]

$\int\frac{x+2}{x^2+4}\dd x=\int\left(\frac{x}{x^2+4}+\frac{2}{x^2+4}\right)\dd x=\int\frac{x}{x^2+4}\dd x+\int\frac{2}{x^2+4}\dd x$

[frail goblet]

OHHH

ok

i split into two integrals

[paper raptor]

you will do that anyways tho

[frail goblet]

huh

[elfin berry]

PajamaMamaLlama

[frail goblet]

o ye

for first one i got 1/2 int(1/u du)

[marsh citrus]

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[carmine ice]

Confused

[wise jackal]

The solution to a system of equations is the point where the lines intersect

I think they give the graph below so I’d just use that

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[carmine ice]

Oh

[marsh citrus]

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[ashen crown]

I am confused with this question

[still temple]

Sorry mb

[oblique geyser]

do you know what the law of sines is?

[ashen crown]

Can I not just go 33sin(120)/45 = sin theta

[oblique geyser]

yes

[ashen crown]

But the answer I get is wrong

[oblique geyser]

sin theta / c = sin 120 / 45

sin theta = c sin 120 / 45

theta = arcsin(c sin 120 / 45)

so arcsin(33sin120/45)

[ashen crown]

Ohhh

arcsin

Why is it arcsin

[oblique geyser]

arcsin = sin^-1

[ashen crown]

And not sin

[oblique geyser]

thats to find the degrees

sin is the ratio between sides

and arcsin is the degree found by that ratio

try sin 60

[ashen crown]

GOTCHA

[oblique geyser]

and then do arcsin(sin 60)

[ashen crown]

That makes a lot of sense

[oblique geyser]

its the inverse of sin

[ashen crown]

Yup yup this is clicking

Thank you

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[sage grotto]

How does the limit when b is infinity on 2arctanx = pie

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[violet prairie]

When I plug this into the calculator i get the crazy number 97656250

does anyone know why

[cobalt sentinel]

Maybe cuz that exponent of X is at the moon

Maybe it’s some weird formatting error

[violet prairie]

its just me pressing the ^ button after the closed bracket

just defaults to the moon idk why its so big

[cobalt sentinel]

Hmm

Quite the oddity

[native frigate]

,w sum from x = 1 to 5 of (-1)^x*x

[elfin berry]

Results provided by WolframAlpha

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

Requested by tuzis

[native frigate]

okay yeah calculator quirk not some really cursed math

[violet prairie]

how unforunate

thanks for looking into it

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[violet prairie]

if i keep it all within the brackets it works for some reason so w/e

[native frigate]

oh it's doing the sum then multiplying by X

[native frigate]

try calculating just X by itself, I expect a big negative number

[violet prairie]

.reopen

[marsh citrus]

✅

[violet prairie]

like

5 Sigma x = 1 (x) or

idk how to do the cool formating so the best i have is the cursed text format lol

or just x itself in the calculator? if so thats 10

[native frigate]

it's using the value of 10 somehow

but even sum from 1-5 of 10 is way lower than that

[violet prairie]

i imagine it sees (-1)^10 * 10

[native frigate]

sure, but that doesn't make it a high enough number even if true

[violet prairie]

thats true...

[native frigate]

interesting, just use parenthesis in the future

it was worth a shot

[violet prairie]

alrighty, thanks for the help again

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[autumn forum]

I know I have to use cosine law

[autumn forum]

but how do I know which one of these sides I have to subtract?

Do I do it for all three angles one by one?

but the question says only calculate the largest angle

[dry prawn]

The largest angle is always opposite the longest side

[autumn forum]

OH

so

DK/ Side E

right?

is the biggest

[dry prawn]

The angle at E is the biggest yes

[autumn forum]

so Angle E is the biggest therefore, equation would be:
cos-1(35^2+48^2-52^2/ 2x35x48)

right

[dry prawn]

Seems right

[autumn forum]

thank u so much

[dry prawn]

[marsh citrus]

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[dawn socket]

what the heck

[main idol]

?

[dawn socket]

Problem 7

[marsh citrus]

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[dawn socket]

,rotate

[marsh citrus]

@dawn socket Has your question been resolved?

[lofty fulcrum]

I gotchu.

[dawn socket]

Omfg

[lofty fulcrum]

Np. Ima write it down on paper as on computer its getting messy.

[dawn socket]

Omg

Thank you

[marsh citrus]

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[dawn socket]

.reopen

[marsh citrus]

✅

[lofty fulcrum]

There yah go. If they want exact answers jus do the algebra then. Max volume would be 96.8 or 96.77 depending on how many decimal figures they want.

Have you done Calculus leah?

Or was this an algebraic problem?

Because it can be solved algebraically with the use of a graphing calculator, but this is how to obtain the dimensions and volume via calculus.

[marsh citrus]

@dawn socket Has your question been resolved?

[dawn socket]

This pre calc

Sooo

Maybe

[dawn socket]

I think In pre calc we use the graphing calculator

[lofty fulcrum]

Ohhhhh okay. Then let me think about this. You must be on Remainder Theorem and Polynomials I believe?

[lofty fulcrum]

Hmmmmm let me see if there is a way we can analytically solve this, cause I don't beleive we can use the polynomial remainder theorem as that is just for factoring.

[dawn socket]

yaaaa polynomials yes

remainder probablt idrk

[lofty fulcrum]

Okay, so assuming your up to date with Remainder Theorem and Polynomials. We know that for an odd leading degree its going to behave like a linear function. In our instance the degree is odd and the coefficient associted is positive, hence its going to behave like a positive linear function. Now according to the rules of multiplicity for Polynomials we can determine whether the function will bounce or cross the axis according to the degere of the multiplicity. Now its also apparant that we can't have negative values hence we disregard 5-6 and (-infinity, 0]. [6, infinity) Those values would not make physical sense if we were to plug them in, therefore we can deduce that the maximum volume will be between [0,5]

[marsh citrus]

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[marsh citrus]

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[buoyant turret]

does it mean to apply the theorum to ax^2y+bxy^2?

because when i do that i still cant determine a and b

actually is the point (0,0) in R^2?

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[onyx edge]

I just started limits and have no idea how to prove this, any help to just walk me through the process?

[stark trail]

What’s your definition of a limit of a function converging?

[marsh citrus]

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[lethal bridge]

How do I approach this? $\int_{ }^{ }\frac{x+1}{x-1}dx$

[elfin berry]

water beam

[lethal bridge]

I thought about splitting it up into $\int_{ }^{ }\frac{x}{x-1}dx+\int_{ }^{ }\frac{1}{x-1}dx$

[elfin berry]

water beam

[lethal bridge]

but that wouldnt help

and a u-sub of the denominator wouldnt help

because theres nothing to cancel out the x's with

[devout mauve]

if you wanna go overkill: polynomial division

[proper zodiac]

You can do the "add 0" trick in the numerator, try adding -1+1

[lethal bridge]

I dont know polynomial division

[devout mauve]

then add 0

[lethal bridge]

i havent learnt that yet

add 0?

[devout mauve]

.

[proper zodiac]

x+1 = (x-1) + 2

[lethal bridge]

$\int_{ }^{ }\frac{x+1-1+1}{x-1}dx$

[elfin berry]

water beam

[lethal bridge]

like this?

[proper zodiac]

Yeah but now group it so you have an x-1 in the numerator

So that it's nice when you split it

[lethal bridge]

how do you group x + 1 - 1 + 1

isnt that just

x + 1

[proper zodiac]

The point isn't to simplify it as that just gets you back to where you started

Make it look like x-1 + something

[lethal bridge]

im sorry i dont understand what you mean

[proper zodiac]

.

[vernal forge]

x+1 = x+1-1+1 = x-1+2

[devout mauve]

and then you can separate the fractions as (x-1)/(x-1) +2/(x-1)

and now the first fraction simplifies nicely. which is the reason why we did that

[lethal bridge]

$\int_{ }^{ }\frac{x+1-1+1}{x-1}dx=\int_{ }^{ }\frac{x-1+2}{x-1}dx$

[elfin berry]

water beam

[lethal bridge]

is this what youre talking about

[proper zodiac]

Yes

[lethal bridge]

so what happens next

[proper zodiac]

.

[lethal bridge]

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

[elfin berry]

water beam

[lethal bridge]

?

[proper zodiac]

Yes

[lethal bridge]

this feels like a glitch

lol

[proper zodiac]

Common trick that's good to know

[lethal bridge]

how do people even come up with this stuff

[proper zodiac]

"Can we add zero in a clever way somewhere"

I think if you stared at this integral for long enough it's something you might try

[devout mauve]

you can motivate it slightly differently

[lethal bridge]

honestly the only thing that crossed my mind was multiplying the top and bottom by x-1 but that sounds illegal so i didnt try it

[devout mauve]

you start with x+1

you would want a x-1 there

so lets write it for fun

x-1 +1

but now we changed the value

so we have to add 1 again

x-1+1+1

[proper zodiac]

Not a terrible idea, probably makes it more complicated though

[lethal bridge]

ah i see

[devout mauve]

remember how you did for example completing the square. you start with x^2+bx+c. you would want to have x^2+bx+(b/2)^2 there to have a square. so you add (b/2)^2 and then you subtract it again

[lethal bridge]

seems like integration requires a lot of outside of the box and rather creative thinking for some problems

[devout mauve]

definitely

integration is a lot harder than differentiation

but addign zero is a very common trick also outside of integration

[lethal bridge]

yeah ive been beginning to see that when i initially started integration like a month ago i think

[lethal bridge]

where would you use this?

[devout mauve]

well I just gave you another example with completing the square

also just generally pops up all over the place

[lethal bridge]

oh yeah

[devout mauve]

similarly is multiplying by 1 for example. which is what you would have done if you multiplied top and bottom by x-1

[lethal bridge]

right

[devout mauve]

I dont think it would have helped here but it is also useful often

[lethal bridge]

yeah i was thinking about like trig identities and how sometimes its useful to do that to simplify things a bit

anyways thanks guys for the help

ill take note of this nifty trick

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[keen gulch]

Can someone check if my math is correct and help me to better understand what's happening on this exercise (i mean that i do not know how to interpreter it)

It's about limits.

[keen gulch]

(it's written in italian but it just says what i need to do to verify the limit)

I would interpreter it like that this limit is not verified, so it doesn't exist, as there is no I+(x0)

[hidden plaza]

Clearly the limit exists so something has gone wrong with your proof

Also what do you mean by -2+?

[keen gulch]

How can you say that the limit exists?
Because its domain is R?

[hidden plaza]

The function is a parabola and it’s continuous everywhere

[keen gulch]

(-2)^+

[keen gulch]

Ok, makes sense

[hidden plaza]

Why would you put a plus there

[keen gulch]

Because the limit is the right of -2

[hidden plaza]

Weird but okay

Firstly your proof set up is not very good

[keen gulch]

Weird?
It means that the function, for x -> 3, has values above -2 (so like -1,99998) (not less than -2 (ex: -2,00001or -2 itself)

[keen gulch]

???

[hidden plaza]

One would normally say the limit is -2

[keen gulch]

No, because this exercise forces me to use this formula

[hidden plaza]

Yes I understand

[hidden plaza]

But do you know what the formula means

Because that’s the thing you’re trying to prove

You can’t just start off by stating it

[keen gulch]

Yes, that i have to verify the limit normally and check if the I(x0) has his 'right-part' (Sorry for translation), so I+(x0) exists

Am i right?

That's the definition

@hidden plaza ?

[elfin berry]

Quantum

[keen gulch]

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

@keen gulch Has your question been resolved?

[keen gulch]

.

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@keen gulch Has your question been resolved?

[keen gulch]

.close

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[olive dragon]

hello, potentially super quick question, I have a "series of odd numbers" is there a convinient way to rewrite it as "normal series" ? Picture for reference

[olive dragon]

My initial thought, use $\Sigma^{n/2}_{l=0}$ and $(1+2l)$ within expression, however possible non integers in the sum make me uncomfortable

[elfin berry]

NZzska

[whole hazel]

floor function?

[olive dragon]

I think I found solution, without doing much danggling with the series, however if I needed to, the notation would be $floor(n/2)$ ?

[elfin berry]

NZzska

[olive dragon]

anyway thanks for suggestion Im closing

/close

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[random pasture]

guys i am stuck in this problem (e^i^i)
this is basically a cimplex number and we have to find its polar form
so can we do it?

[rocky sapphire]

we can do it

[marsh citrus]

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[shut stone]

Can you integrate this function using substitution or integration by parts?

[grizzled bobcat]

linear sub

u = x^2 + 1

wait

nvm

[wide grail]

Partial fractions

[shut stone]

I've tried this but I didnt reach an answer

dont know them

[wide grail]

Time to learn them

[shut stone]

we have not done polynomials

[grizzled bobcat]

you need a good hour to learn partial fractions confidently

[wide grail]

You cant use u sub here to get anything less complex

[shut stone]

this execrise was given to me by a private tutor so maybe he gave me this exercise by mistake

I was just wondering if it could be done in another way than partial fractions

[wide grail]

Also integration by parts doesnt help out no matter what you take as u or v

[wide grail]

Maybe but not in an easy way

[grizzled bobcat]

lemme try with u sub

you can’t via u sub

[shut stone]

thanks anyway 🙂

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[pale sparrow]

Hello! I need help with this problem

[marsh citrus]

@pale sparrow Has your question been resolved?

[pale sparrow]

<@&286206848099549185>

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.close

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[marsh citrus]

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[inner summit]

<@&286206848099549185>

I analyzed it and got the answer 7. Is this true?

[mystic heath]

no... I don't believe so

[inner summit]

what is your answer

[mystic heath]

sorry wrong channel

uhh

7 looks right

[inner summit]

oh ok

[inner summit]

<@&286206848099549185>

[marsh citrus]

@inner summit Has your question been resolved?

[inner summit]

!15 min

!15min

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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.

[inner summit]

<@&286206848099549185>

@gleaming oracle

[inner summit]

.close

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[sturdy lagoon]

Is this correct?

[vernal forge]

yes

[sturdy lagoon]

thank you ~ how about these two?

[vernal forge]

1. wrong

2. also wrong

[vernal forge]

first

let's find what's the inverse of y=x

[sturdy lagoon]

ok

[vernal forge]

what is it

it's very simple

[sturdy lagoon]

x=y

[vernal forge]

yep

it is it's own inverse

[sturdy lagoon]

Alright

[vernal forge]

so the function you have there is clearly not the inverse

[vernal forge]

as for the second one

[sturdy lagoon]

I understand

[vernal forge]

why would they not be inverses?

[sturdy lagoon]

thats what I thought to but they hadn't switched up the question yet so I wanted to be sure that if I thought it was actually inverse I'd know to pick it

as for this question though how would I know to pick this graph based on the red line

[vernal forge]

huh

i mean

your new slope will be 1/(original slope)

[marsh citrus]

@sturdy lagoon Has your question been resolved?