help-33

[marsh citrus]

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[karmic shore]

.reopen

[marsh citrus]

✅

[marsh citrus]

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[noble ingot]

I don’t know how to solve this problem

I think you put the height on the other one like this

10/16

Which is 5/8

But here I’m kinda lost

[normal sky]

what's the reason you're taking the ratio of heights

[noble ingot]

To get the ratio

Because they’re similar figures

[normal sky]

right

so if we say that the unknown radius of the larger cylinder is r

can you set up an analgous ratio for the radii

[noble ingot]

Oh ok. How can I set up this ratio?

[normal sky]

well

[noble ingot]

That’s mostly what I don’t know how to rlly do

[normal sky]

so

how did you set up the ratio of the heights?

the smaller height was 10, the larger 16

you took 10/16

[noble ingot]

yes

[normal sky]

for the radii, the smaller radii is 2.5, and the larger is r

[noble ingot]

So I assume you set it equal to the smaller over the larger

5/8 = 2.5/r

[normal sky]

yep

do you know how to proceed from here

[noble ingot]

Okay so radius is 4

Then you just plug it in

Ty

[normal sky]

[noble ingot]

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[marsh citrus]

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[stone bolt]

how do i approach this excersice?

[marsh citrus]

@stone bolt Has your question been resolved?

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[azure flare]

Could anyone assist me with the annihilator method please?

[azure flare]

for differentials

[whole sleet]

Have an example we can look at?

[azure flare]

Okay

[azure flare]

for example here

and here

My question for the second one is, why can we use (D - 1)^3 for the two terms?

<@&286206848099549185>

[whole sleet]

It's not clear what the second picture is

If I had to guess:
D-1 is annihilating e^x? If that's true, then (D-1)³ definitely annihilates it

As that's just more derivatives after the annihilation

For the first, I suggest taking a derivative and seeing if both sides cancel nicely / what needs to be changed for them to do so

[azure flare]

Hmm

So if we had, for example, e^(-x) + xsin(x)

the annihilator e^(-x) be (D + 1)

how would you calculuate the annihilator for xsin(x)?

the annihlator for sin(x) is D^2 + B^2, B in this case is 1 so (D^2 + 1)

the annihlator for x would be D^2

so do you just multiply them all? D^2(D^2 + 1)(D + 1)?

<@&286206848099549185>

[whole sleet]

Note you can go from D + 1 to D² + 1 by multiplying D - 1

So D² + 1 already annihlates both

[azure flare]

wtf

where does D - 1 come from

[whole sleet]

Like, imagine applying D+1 to both sides. You've annihlated e^(-x)

Then, apply D-1. In total you've now applied D²+1 and have annihlated sin(x)

As (D+1)(D-1) = D²+1

[azure flare]

Shouldn't it be D^2 - 1?

[whole sleet]

Oh lol good for me. I just kept writing D² + 1 and running with it. Yeah that doesn't factor, does it?

Okay then yeah (D² + 1)(D + 1) will annihlate everything

[azure flare]

wait you omitted D^2

why did you do that?

does D^2 + 1 annihilate x?

[whole sleet]

I didn't see the x on your example. We need to cook up an annihlator for xsin(x)

[azure flare]

oh but i thought that D^2(D^2 + 1) would annihilate xsin(x)

i thought you had to like make seperate annihilations for all functions (so an annihiliation for x and another for sin(x)) and then combine them

[whole sleet]

I don't think they work through products like that, but I'm not certain. I'll look it up

[azure flare]

its what i found but i may be wrong idk

i mean i am doing that yes

[whole sleet]

(D² + 1)xsin(x) = 2cos(x)

So applying D² + 1 onto it again gives 0

So the annihlator we want here is (D² + 1)²

[azure flare]

What??

no

deriviate sin(x) twice

-sin(x) + sin(x) = 0

[whole sleet]

Sorry I meant xsin(x)

Edited

[azure flare]

ooh ok

[azure flare]

how do you know when 1 annihilator can be used for multiple functions?

[whole sleet]

Basically when one is a factor of another

If (D-1) annihlates a term, then (D-1)³ definitely will

[azure flare]

I see

[azure flare]

so the annihilator for this

it would be (D - 1)(D^2 + 2^2) right?

and the annihilator for this would be D(D - 1)

[marsh citrus]

@azure flare Has your question been resolved?

[azure flare]

<@&286206848099549185>

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@azure flare Has your question been resolved?

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[low iris]

Does 1. Look correct?

[cobalt sentinel]

U got the right answer but in the first line it should be -(x^2 + 4x + 1)

[glass silo]

Also in the simplification, you're missing what happens to that h^2 term

And they want you to find that when x = 0 too

[low iris]

So I thought that I had taken care of it when I factored it out

[glass silo]

That only accounts for the 4h, and not the h^2 you had

The numerator of the line above is effectively 2xh + h^2 + 4h, what did you do with the h^2? Where did it go?

(there are also some notation comments, the $\lim_{h\to 0}$ should remain until the very last line, or you can do the steps without it, and the last line, state that you approach that, so like $\to 2x + 4$ as $h\to 0$)

[elfin berry]

@glass silo

[low iris]

@glass silo would this work?

[glass silo]

That's fine (though as before, notation)

[low iris]

HAHA sorry about notation my Prof isn’t picky about it but I suppose it’s better to create good habits when it comes to that

[marsh citrus]

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✅

[low iris]

Better?

[cobalt sentinel]

u put the parentheses on line 1, and u canceled out in line 2, but u didnt change the signs appropriately

and the lim notation stuff is still a little bit icky but u have the right idea

[low iris]

Oh I forgot to do that so it should be -x^2 -4-1

Because when we distributed that negative

[marsh citrus]

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[oblique veldt]

value for f(1) = 79,5
value for f(4) = 33,4

I have that f(x) = C(a^x) = 79,5 * 0,75^x (by moving x by 1 step)

i just dont know how to integrate correctly or if what I am doing is correct at all

[lyric bay]

?

Do you know power rule?

[oblique veldt]

the power rule?

nope

[outer lodge]

Power rule doesn’t apply here

[oblique veldt]

i might know it just not what its called in english tho

[red nimbus]

they mean the rule how you would integrate polynomials, which doesnt apply here

[oblique veldt]

oh ok

[lyric bay]

Why not

Oh wait

[oblique veldt]

because there is no polynomial

[red nimbus]

it's an exponential function

[lyric bay]

I misread

My bad

[oblique veldt]

👍

The question reads as follows btw, idk if this will help.

In a city, the density of the population is described as an exponential function. 1 km away from the city, the density is 79,5 citizens/ha (idk why they use ha, but i think its 1/100 km). 4km away the density is 33,4 citizens/ha. How many citizens live within 1km and 4km of the city

[red nimbus]

You can integrate $a^x$ by rewriting it as $a^x = e^{\ln(a^x)} = e^{\ln(a)x}$

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

Also I think you would need first to find f(x) and then integrate

[oblique veldt]

f(x) = 79,5 * 0,75^x (i think)

$f(x) = 79.5 * 0.75^x$ I think

[elfin berry]

Hassan

[oblique veldt]

the integral that ive done to this gives 160, the answer is 229

[red nimbus]

The numbers are really ugly as I just tried

$f(x) = C \cdot a^x \implies \begin{cases} f(1) = 79.5 \ f(4) = 33.4 \end{cases}$

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I think it's kinda off

[oblique veldt]

idk how its off tho

[oblique veldt]

it seems right, just that its shot backwards by 1 x value

[red nimbus]

not really you prob did some mistake

[oblique veldt]

the integrals for 1 -> 4 and 0 -> 3 are the same however

for both methods

[red nimbus]

what you did is f(0) = 79.5

[oblique veldt]

yeah

[red nimbus]

text however says 1km

not 0km

or beginning

[oblique veldt]

yeah but if we look at it relatively

[red nimbus]

i mean it's up to you what you want to do

[red nimbus]

But yeah in order to integrate f(x) you would use the trick from above

[oblique veldt]

i did but the answer still isnt correct

[red nimbus]

hmm

can you show what you did?

[oblique veldt]

ok 1 sec

[red nimbus]

where is the factor

75.9

[oblique veldt]

a i wrote 0.75 instead of 79.5 mb

[red nimbus]

also a = 0.75

[oblique veldt]

but thats just cuz i doodled on paper

on computer i did it right

[red nimbus]

,,\int_0^3 75.9 \cdot 0.75^x : \dd x = 75.9 \int_0^3 e^{\ln(0.75)x} : \dd x

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

and what's that

value wise

also

you did 1 to 4

no wonder man

[oblique veldt]

120

[red nimbus]

if you decide to use your wrong function

at least use the right bounds

[oblique veldt]

oh right im kinda tired sorry

but i still get 160

[red nimbus]

no sorry

yea that seems about right

[oblique veldt]

but the book says 229

i dont know if its some hectar to km confusion

or because its a radius

[red nimbus]

oh ofc

the input in terms of km

output gives us ha

we need to convert this shit

that was bad from my part

[oblique veldt]

the difference isnt that big tho, 1km^2 is 100ha

[red nimbus]

hmm ok lemme see

but then we should differ by a factor of 100

[oblique veldt]

yes but the answer is 229 * 1000

[red nimbus]

1 person per ha is 100 persons per km²

[oblique veldt]

the area of which the people live in is 4^2 * pi - 1^2 * pi which is 15 pi km^2

[marsh citrus]

@oblique veldt Has your question been resolved?

[oblique veldt]

.close

[marsh citrus]

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[red nimbus]

I found the problem

[oblique veldt]

.reopen

[marsh citrus]

✅

[red nimbus]

You draw a circle

[oblique veldt]

yes

[red nimbus]

So the region is actually bigger

So

,,2\pi \int_a^b x \cdot f(x) : \dd x

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

we had to change units

mb

but yea

so

https://teachingcalculus.com/2017/01/10/density-functions/

we then get 229000 something

so it was these two mistakes

[oblique veldt]

yooo

ily man

[red nimbus]

conversion and integrating what actually was needed

np itachi

it said density function and then i looked it up

[oblique veldt]

i see that the answer is right but i kinda dont get it

[red nimbus]

somehow it clicked with the circle you made so yeah

[oblique veldt]

it also only works with 1->4

so thats another mistake i made

thx man

.close

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[red nimbus]

I think this form reminds me of the shell method

[oblique veldt]

shell method?

[red nimbus]

So I think we took that piece of area and integrated it around the y-axis

hold on

basically we calculated a volume

[oblique veldt]

oh

[red nimbus]

which we kinda wanted

radius

around the city

if we think of it in 3D

[oblique veldt]

kinda understand

anyways i have to go sleep now

thank you so much

[marsh citrus]

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[bitter kettle]

i need help with part C

[inland lark]

It's 2

[bitter kettle]

can u explain how please?

like all parts of C

[inland lark]

1 +1

[bitter kettle]

like where are u getting that tho

<@&286206848099549185>

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[still temple]

Can someone guide me with 7?

[still temple]

Idk where to start

What does it mean to find the equation of the tangent line there

Uh I was absent that day so I’ve been trying to catch up

https://tenor.com/view/rebecca-mel-john-trending-gif-4994843

[frail orbit]

do you know the point-slope equation of a line?

@still temple

[still temple]

yes

[frail orbit]

you know the derivative represents the slope of th tangent line at a particular x value?

[still temple]

yes

but nothing other than that fact

[frail orbit]

so you can replace m with something

what would be a good thing to replace m with?

[still temple]

uh

square root 3 (?)

i dont kno

https://media.discordapp.net/attachments/744073200651993109/1199061015703343214/ED6EFBC5-A68E-48C6-9103-B123B1ECC22D.gif

how would i find m

[frail orbit]

well the derivative is the SLOPE OF THE TANGENT LINE at a particular x1 value

[still temple]

but how do i find that

with my

derivitive thingy

[marsh citrus]

@still temple Has your question been resolved?

[gilded roost]

Point-Slope Form of a Line:

[
(y - y_1) = m(x - x_1)
]

Components:

• (m): The slope of the line. It represents the change in (y) for a unit change in (x).
• ((x_1, y_1)): A point on the line. Any point on the line can be used for ((x_1, y_1)).

Explanation:

• Slope ((m)): The ratio of the vertical change ((\Delta y)) to the horizontal change ((\Delta x)) between any two points on the line. It is calculated as:
  [
  m = \frac{y_2 - y_1}{x_2 - x_1}
  ]
• Point (((x_1, y_1))): Any specific point through which the line passes. This form is particularly useful when you know the slope of the line and one point on it.

Example:

Suppose you have a line with a slope (m = 2) that passes through the point ((3, 4)).

Using the point-slope form:
[
(y - 4) = 2(x - 3)
]

You can then simplify or rearrange this equation to other forms, such as the slope-intercept form ((y = mx + b)) or the standard form ((Ax + By = C)).

[marsh citrus]

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[ripe python]

Hello again. I have no idea what am i doing. Does this makes any sense at all? The question is: 20 barrels with a volume of 220 liters were used to empty the tank. How many barrels with a volume of 1200 would be needed to empty three such tanks? The key: 110 barrels

[dawn niche]

ok as an example

lets say x = volume of a singular tank

and y represents the number of barrels (specifically for the ones with a volume of 1200)

so we have x = 20*220

[ripe python]

1200?

Isnt 120?

[dawn niche]

you said the volume is 1200

in your questions

[ripe python]

Oh translator

Its 120

I apologise

[dawn niche]

oh ok

its fine

so with x = 20*220 we get x = 4400

so we have 3x = 120y

3*4400 = 120y

13200 = 120y

[ripe python]

Aaa

[dawn niche]

110 = y

[ripe python]

Thank you

.close

[marsh citrus]

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[clever bronze]

the smaller one is the newer one. I dont get what its asking. Do I have to translate it or do scale factor

the smaller shape has different size and points, so it wasnt translated? so confused

[marsh citrus]

@clever bronze Has your question been resolved?

[clever bronze]

no

<@&286206848099549185>

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[prime otter]

guys i may get the solution but the answer in the book is diffrent

[prime otter]

thw answer in the book is

x^4y^4(x^2+y^2)(x-y)(x+y)(x^2+y^2-root2xy)(x^2+y^2+root2xy)

how do i get this plj

help

or my teacher whoop my ass

<@&286206848099549185> \

bruh

who pinged me

<@&286206848099549185>

[remote grove]

whoop your teachers ass instead

you got this!

[quaint elm]

so like

[prime otter]

bruh

[quaint elm]

basically $x^4 + y^4 = ($x^2 + y^2 + \sqrt2xy)(x^2+y^2 - \sqrt2xy)$ is like

[elfin berry]

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

[quaint elm]

a form that you can prove with weird factoring tricks

or just prove by expanding the right hand side

but also it's stupid because it involves square roots

which aren't really how factoring normally works

to me i'd say your answer is perfectly fine

[latent coral]

yeah i was going to respond originally but this is some really nitpicky factoring stuff

almost every math teacher i know would consider your answer fully factored

and if they saw someone factoring x^4 + y^4 they would think they're insane

[prime otter]

oh

then thank you

but may i just know those tricks

maybe it will be useful

[quaint elm]

tbh just memorize that one

if you really want to derive it, rewrite it as

$x^4 + y^4 = x^4 - i^2y^4$

[elfin berry]

hayley

[quaint elm]

and apply difference of squares

[latent coral]

yucky i

[prime otter]

bro idk wht is i i am in 9th grade

ahah

[latent coral]

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

[quaint elm]

then you really don't need to be worried about that

[quaint elm]

[latent coral]

then you notice that (x^2)^2 + (y^2)^2 + 2(x^2)(y^2) is a trinomial that you can factor into (x^2 + y^2)^2

and you'll get (x^2 + y^2)^2 - (sqrt2 * x * y)^2

and you factor this as a difference of squares

no imaginary numbers needed

[prime otter]

bro i am in india

matsh here too hard

for no fucking reason

[latent coral]

india's putting out some really strong stem students though

embrace the grind

[quaint elm]

if you want to know, i = √-1

[prime otter]

ahahhahhaha

[prime otter]

k

k then thank you il close the

channel

.close

[marsh citrus]

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[prime otter]

bruh

[marsh citrus]

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[still temple]

For [
\int_\SS\pdv[\vv B]t \vd \dd \vv s = \dv t \int_\SS\vv B\vd \dd \vv s
]
why does the partial derivative turn into a normal derivative? $\vv B$ is a 4 dimensional vector field with space coordinates and time as parameters

[lucid zenith]

I guess after integrating the function becomes a pure function of t

[marsh citrus]

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[steep ruin]

I dont get what the difference between qestion a) and b) is

[pastel granite]

i believe a asks for the derivative while b just asks you to evaluate the function

[marsh citrus]

@steep ruin Has your question been resolved?

[amber birch]

Yes a) is $f'(4)$ whereas b) is just $f(4)$

[elfin berry]

south

[amber birch]

terrible typography from the IB

[marsh citrus]

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[lyric relic]

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lyric relic]

I know my answer is wrong

It's suppose to be a positive 42.25

What did I do wrong

[dusky viper]

u distributed the negative wrong

-(a)(b) = (-a)(b) not (-a)(-b)

[pastel granite]

try to conisder seeing the negative sign before a parentheses as a (-1)

[lyric relic]

yes

but a negative times a negative is a positive

-1 * -2 = a positive 2

[dusky viper]

yeah but u distributed the negative wrong

[pastel granite]

you're right! but the negative one only distributes once

[dusky viper]

2(2)(3) = (4)(3) not (4)(6)

[lyric relic]

this is so stupid

khan academy is stupid

yes i understand

but theres other stupid things

like they say -2^2 is 4

but when you do it on the calculator

it says its -4

this is absolutely ridiculous

[pastel granite]

perhaps -(2^2)

[lyric relic]

this is so frusturating lol

ok thx

im gonna lose my mind

[marsh citrus]

@lyric relic Has your question been resolved?

[marsh citrus]

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[quick sage]

Hi can you help me solve this grade 12 homework problem I don’t know what formulas to use and what to input into them

[marsh citrus]

@quick sage Has your question been resolved?

[quick sage]

No

[buoyant summit]

i can try to help with a since I recently learned that but I don't know what a z-score is

for mean, you just take all the values and divide them by the number of values, the arithmetic average

for variance, you take every value one by one, and from each value, you subtract the average, and take the square of the result, then add the results together, and divide the result by the number of values given

for standard deviation, take the square root of that

in case I worded that badly, here's a simple example (I'm not gonna do latex or I'll be here for an eternity figuring that out):
let's say you have the values 1, 3, 4, 3, 2, 4
there's 6 values
The mean is (1+3+4+3+2+4)/6 = 2.8(3) = 2 + (5/6)
variance is ((1-2.8(3))^2 + (3-2.8(3))^2 + (4-2.8(3))^2 + (3-2.8(3))^2 + (2-2.8(3))^2 + (4-2.8(3))^2)/6

[quick sage]

Ok thanks

[buoyant summit]

np

i hope i didnt get anything wrong

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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Please don't occupy multiple help channels.

[lucid zenith]

delete this

!nosols

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[cunning fiber]

tbf I checked off everything lol

[lucid zenith]

oh

nvm

my bad

[cunning fiber]

Here are the answers

[red nimbus]

I would add q.e.d. so thats -1 point

[cunning fiber]

Your teacher is wrong

All answers are correct

[lucid zenith]

I think all of you should create a group chat and discuss your paper over there, insteda of creating 6 different channels asking for just the answers of the same test...

[weary ibex]

Don't troll, even in response to low-quality questions.

[cunning fiber]

I am not troll

[weary ibex]

.close

[marsh citrus]

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[dreamy hull]

Hello, can you help me with this combinatorics exercise pls?
Ex.15) To integrate a commission, 4 people must be chosen from a group of 8 engineers and 5 programmers.
a. How many ways can the election be made?
b. How many if we impose the condition that at least 2 of the members must be programmers?

[obsidian kayak]

For part a
We have to consider the following cases to form a group of 4

1. 1 from eng and 3 from prog
2. 2 from eng and 2 from prog
3. 3 from eng and 1 from prog

Now I think you can solve it from here

And for the part b you can consider the first two cases only

@dreamy hull

[marsh citrus]

@dreamy hull Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[fathom mountain]

What method can I use?

[marsh citrus]

@fathom mountain Has your question been resolved?

[fathom mountain]

<@&286206848099549185>

[red nimbus]

Method to achieve what?

[fathom mountain]

If it goes inf or not

[red nimbus]

Maybe try ratio test or some

Think it wont work out

[fathom mountain]

It didnt tell me much r u sure?

[red nimbus]

comparison test?

[fathom mountain]

To what

[fathom mountain]

<@&286206848099549185>

[marsh citrus]

@fathom mountain Has your question been resolved?

[fathom mountain]

<@&286206848099549185>

[fathom mountain]

<@&286206848099549185>

[surreal flax]

What have you tried?

[fathom mountain]

Idk what to try tbh because I cant see solutions

The most promising was ^nsqrt

But idk if thats the right choice

What do u think

I just dont see anything specific

Maybe asymptotes? But idk to what

[fathom mountain]

<@&286206848099549185>

[fathom mountain]

<@&286206848099549185>

[minor vortex]

what u need help@with

@fathom mountain

[fathom mountain]

i need to say if it goes to infinite or not

idk what method to use

i cant see any asymptotes

[fathom mountain]

<@&286206848099549185>

[thorn copper]

In which grade are you @fathom mountain

[fathom mountain]

uni

[thorn copper]

Okay

Wait let me try

[fathom mountain]

wait i must have just figured out

isnt it asymtote to 1-1+e^(sqrt1/n)?

[thorn copper]

Use Limit Comparison Test Method

Or

Ratio Test Method

[fathom mountain]

i get 1/n^(3/2)

is it right

[thorn copper]

Yes, that's correct! The function ( f(x) = \frac{1}{\sqrt{n}} ) satisfies the condition of being continuous everywhere except at ( x = 0 ). Well done!

[elfin berry]

Mehul Das

[fathom mountain]

3/2 not 1/2

[thorn copper]

My bad

Wait

[fathom mountain]

let me ask you this tho

same exercise but there is + instead of -

ok?

dont you get the same result?

[thorn copper]

You can use convergence test to determine if the series converges or diverges

[fathom mountain]

im using comparison test

isnt that ok too

using this

wait i did one calculation wrong

[thorn copper]

Using the comparison test is a good approach. To use the comparison test, you'll need to find another series that you can compare to, which you know converges or diverges.

[fathom mountain]

I get this do u agree

[thorn copper]

If you're comparing the original series to the series ( \sum_{n=1}^{\infty} \frac{\sqrt{1/\kappa}}{\kappa} ), where ( \kappa ) is a positive constant, then you're essentially comparing it to a constant times the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{\kappa} n^{3/2}} ).This new series looks like the convergent p-series with ( p = \frac{3}{2} ). Since ( \frac{3}{2} > 1 ), this series converges.

[fathom mountain]

Sry bro but I cant read latex

[thorn copper]

Wait

If you're comparing the original series to the series ( \sum_{n=1}^{\infty} \frac{\sqrt{1/\kappa}}{\kappa} ), where ( \kappa ) is a positive constant, then you're essentially comparing it to a constant times the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{\kappa} n^{3/2}} ).This new series looks like the convergent p-series with ( p = \frac{3}{2} ). Since ( \frac{3}{2} > 1 ), this series converges.

It should have automatically converted

[fathom mountain]

Maybe 1 per time

[thorn copper]

Maybe

If you're comparing series to

( \sum_{n=1}^{\infty} \frac{\sqrt{1/\kappa}}{\kappa} ),

Why isn't it converting

[fathom mountain]

Idk maybe ask mods?

( \sum_{n=1}^{\infty} \frac{\sqrt{1/\kappa}}{\kappa} )

[fathom mountain]

Well can u tell me some other way meanwhile please?

[thorn copper]

Use Integration

Express the series as a function of ( n ).

Rewrite the series in integral form, which involves converting the series into an integral using the properties of Riemann sums.

Integrate the function over the appropriate interval.Evaluate the definite integral to find the sum of the series.

[fathom mountain]

Can I not just use this for comparison

[fathom mountain]

.

[thorn copper]

You can

[fathom mountain]

And I obtain this .

Do u agree

[thorn copper]

Yes

[fathom mountain]

Ok but now lets say u have the same exercise ok? But instead of 1-e theres 1+e

For n->inf u obtain the same right?

[thorn copper]

I don't know about that

[fathom mountain]

I mean + here @thorn copper

[thorn copper]

Myself a little lost here

[fathom mountain]

Well you see theres a - here

[thorn copper]

Hmmm

[fathom mountain]

Here bro theres a minus

[thorn copper]

Got it. If the series expression is ( \sum_{n=1}^{\infty} \frac{1 + e^{\sqrt{1/n}}}{n} ), we can still attempt to use integration to find its sum.

[fathom mountain]

Yes and what i get is (2+sqrt(1/n))/n

Do u agree

[thorn copper]

By the way in which uni are you?

[fathom mountain]

Engineering

[thorn copper]

No the university name

[fathom mountain]

Politecnico

[thorn copper]

Agree

[fathom mountain]

Ok but theres an issue

[thorn copper]

What?

[thorn copper]

Must be nice in Milan right?

[fathom mountain]

That solution we are saying is the same as before with n->+infinite

But this second solution must be infinite

Diverge

[fathom mountain]

Nah

[thorn copper]

If we're comparing the series ( \sum_{n=1}^{\infty} \frac{2+\sqrt{\frac{1}{n}}}{n} ) to ( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} ), and we acknowledge that ( \frac{2+\sqrt{\frac{1}{n}}}{n} ) grows larger than ( \frac{1}{n^{3/2}} ) as ( n ) approaches infinity, then we conclude that the series ( \sum_{n=1}^{\infty} \frac{2+\sqrt{\frac{1}{n}}}{n} ) also diverges.

[thorn copper]

Thanks for the clarification

[thorn copper]

Your uni has a world ranking of 123 that's better than the top uni in my country

Yours acceptance rate is 28% still has a ranking of 123 mine has acceptance rate of 0.5-1% still is at 150

[keen reef]

mine uni is 16th

in the world

uft

[thorn copper]

Name?

[thorn copper]

University Of Toronto?

[keen reef]

yes

[fathom mountain]

Can u help with the exercise then lol

[keen reef]

i graduated in business

[thorn copper]

It's 18th

[thorn copper]

MBA?

[fathom mountain]

Oh

[keen reef]

12th

its 12th

[thorn copper]

Then what are you doing in a Math discord?

[keen reef]

i did accounting also

in business

[thorn copper]

No it's 21

[fathom mountain]

here bro i got the pic

[keen reef]

i can help u

but no time

[fathom mountain]

hows it different than the other

i dont get it

[keen reef]

yes phd

[fathom mountain]

.close

[marsh citrus]

Channel closed

Closed by @fathom mountain

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[thorn copper]

This in in specific Business

I was talking overall

[twilit grove]

take non-math stuff to #discussion

[thorn copper]

I would love to help but it's 2 Am here

[marsh citrus]

Available help channel!

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[thorn copper]

It's like 10:30 Pm in Milan

[keen reef]

yes i was taking business

[thorn copper]

Okay

Good

[keen reef]

top 12

only 11 schools beter

[thorn copper]

Then what are you doing in a Math discord?

[thorn copper]

MIT 💀

Top 1

[keen reef]

mit?

[thorn copper]

Massuute Institute Of Technology?

[thorn copper]

Didn't tell me this

[keen reef]

stanford

is # 1

[thorn copper]

In Business?

[boreal rose]

what

are you guys talking about

PLEASE stay on task

[sand fable]

on task abt what bro

[keen reef]

yes

[sand fable]

have you read the channel before saying that

[boreal rose]

oh they're just yapping since the start

[sand fable]

lmao

[thorn copper]

There is no task

. close

[boreal rose]

.close

[thorn copper]

Isn't working

[olive sky]

no

space

between

.close

[thorn copper]

. close

[boreal rose]

thanks honey

[keen reef]

im in love with coco

[thorn copper]

💀

[olive sky]

ninja

. close ❌ .close 😀

[keen reef]

baking soda i got baking soda

[marsh citrus]

@thorn copper Has your question been resolved?

[marsh citrus]

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[queen echo]

Can someone help with this?

[boreal rose]

depends

if uoft is university of toronto, then no

[marsh citrus]

@queen echo Has your question been resolved?

[trim quest]

OP left the server

.close

[marsh citrus]

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[potent token]

so i just realized today that i have zero clue as to how the generalized binomial theorem (for all real numbers) works. almost all of the proofs that i've seen use taylor series, which would be nice if taylor series didnt need the derivative of that function, which needs the power rule, which uses the generalized binomial theorem to be proven. if i could get an independent set of proofs for all of them ,that'd be nice

[potent token]

or a proof of the generalized binomial theorem that doesnt depend on taylor series

[devout mauve]

you can prove the power rule by writing x^k = e^(k ln(x)) and then chain rule

[potent token]

oh shit that does work

but wouldn't that only make sense for x>0?

[devout mauve]

well x^k for x negative and k not a natural number is very questionable anyway

but you should probably also be able to do something for x^1/k with k odd using the chain rule

as its the inverse of x^k

[potent token]

ic

is there a place i can find a proof for the generalized binomial theorem? when i searched it up i only found like 2 and the rest were for the regular one (which i understand at an okay level)

is there like another name for it?

[devout mauve]

binomial series

but thats basically the same name so you probably found that along the way

[potent token]

i just realized
why do i care about generalized binomial theorem if i have taylor series

[devout mauve]

always good to know what the series for a specific function is

[potent token]

aight im gonna close

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[north obsidian]

while integrating a definite partial derivative in terms of x where x is independent and y is some function of x, so $$f(x, y) = (1 - x)y$$ and $$\int_0^n\frac{\delta f}{\delta x}dx$$

[elfin berry]

Avo

[north obsidian]

would the answer include just $y(x)$ or would it have $y(n)$?

[elfin berry]

Avo

[stray hound]

well you're integrating with respect to x, so all of the x's will be gone

so it will only have n and y

oh but y is some function of x I see, so it'll only have n

[north obsidian]

oh wait

I was doing it wrong

when treating y as a constant, I took it outside of the integral but it has to multiply the partial derivative of (1 - x) right?

Does this look right?

[stray hound]

y is a function of x, so you can't treat it as a constant. imagine if y was x^2, then you couldn't just ignore it

[north obsidian]

you would treat it as a constant while taking the partial of x no?

[stray hound]

no, since y is a function of x. if y and x are two separate variables that don't depend on each other at all, you could

[north obsidian]

so the partial of f in terms of x will just be -y in this case

[marsh citrus]

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[tulip marsh]

for question c i do not udnerstand why we can do this

[tulip marsh]

yes this is FTC but the integral has a f'

[quaint elm]

i mean it's exactly FTC

,tex .FTC1

[elfin berry]

hayley

[quaint elm]

f is the antiderivative of f '

[tulip marsh]

but there is a prime

[quaint elm]

yes

[tulip marsh]

this has no prime

[quaint elm]

true...

do u agree that f is the antiderivative of f ' ?

[tulip marsh]

yes

[quaint elm]

being that f ' is the derivative of f

okay

[tulip marsh]

yes

[quaint elm]

so... if we look at FTC again

,tex .FTC1

[elfin berry]

hayley

[quaint elm]

this is just a pattern match

Let $\rsq(x)$ be the antiderivative of $\gsq(x)$:
$$\int_a^b \gsq(x)\dd{x} = \rsq(b) - \rsq(a)$$

[elfin berry]

hayley

[quaint elm]

in this case, our red function is f, and our green function is f '

[tulip marsh]

um, im not sure i understand this way

[quaint elm]

do you see how what I wrote is the same both times

or is there some difference between them

[tulip marsh]

the derivative of fx is FX
yes?

[quaint elm]

F and f are just symbols

we usually use F to mean the antiderivative of f

[tulip marsh]

okay

[quaint elm]

so I restated the theorem but instead of using symbols I used colore

[tulip marsh]

hm

[quaint elm]

all I did was replace F with a red square and replace f with a green square

[tulip marsh]

alright

[quaint elm]

I could put anything in there

[tulip marsh]

okok

(repost for reference)

[quaint elm]

as long as the red thing is the antiderivative of the green thing, that will work

Let $\red f(x)$ be the antiderivative of $\green{f'}(x)$:
$$\int_a^b \green{f'}(x)\dd{x} = \red f(b) - \red f(a)$$

[elfin berry]

hayley

[tulip marsh]

ok

[quaint elm]

does this make sense to you now?

[tulip marsh]

well maybe when we do the problem i can see it better. i do know that notation tho yes

[quaint elm]

there isn't that much left to do

just replace x with t

[tulip marsh]

then that means

hx = f'x

yes?

[quaint elm]

no...

$h(x) = \int_{-6}^x f'(t)\dd{t}$

[elfin berry]

hayley

[quaint elm]

and you need to find h(6), so replace x with 6

lucky for you, we just got done talking about how to evaluate that thing on the right

[tulip marsh]

wait hold on

i am meant to find the area under the curve for -6 to 6

[quaint elm]

are you?

what curve

[tulip marsh]

well this curve

[quaint elm]

why?

that's the graph of f

if you wanted to find $\int_{-6}^6 f(t)\dd{t}$ then by all means yes you'd want to find the area under that curve

[elfin berry]

hayley

[quaint elm]

is that your goal?

[wary kite]

this is from calc bc?

[tulip marsh]

AB

[tulip marsh]

no

goal is to find f'

[quaint elm]

no

goal is to calculate that integral

the one from -6 to 6, of f '

[quaint elm]

ye

[quaint elm]

use this

[tulip marsh]

i see what we are doing

but

i dont like it

[quaint elm]

uhhh why not

[tulip marsh]

maybe becasue i dont understand it

well

understand it well enough

becasue like

[wary kite]

it’s just FTC

[tulip marsh]

understood

[quaint elm]

i didn't really understand calc until much later

like on a fundamental level

[tulip marsh]

no cuz like we are setting an integral equal to ints anti derivarives limits subctracted

[quaint elm]

i just learned pattern matching

[quaint elm]

uh... yes 🧐

[tulip marsh]

which is what does not make sense to me as to y that can be done. but that is probably a solo researching thinge for later

also

what about h''6

how can we do that

[wary kite]

did you find h’

[tulip marsh]

i found h6

[quaint elm]

yeah I mean that's the fundamental theorem of calculus, it's a beautiful result and you should research it yes

you will need the more advanced form of FTC

,tex .FTC2

[elfin berry]

hayley

[quaint elm]

(and I will need to drive so I can't keep going here sorry!)

[tulip marsh]

thanks

.close

[marsh citrus]

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[left charm]

i dont get this problem

[left charm]

what is center in the interior of the square

[sand kindle]

that isn't what the question says

[rich bough]

😬

[sand kindle]

I believe it's lookig for a generic solution

a good starting point is finding the intersections with a given vertical/horizontal line

[left charm]

is it possible to have 3 points

[marsh citrus]

@left charm Has your question been resolved?

[marsh citrus]

@left charm Has your question been resolved?

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[stuck folio]

Can anyone help me understand how to get b?

[marsh citrus]

@stuck folio Has your question been resolved?

[stuck folio]

Need help with this too

<@&286206848099549185>

[half kettle]

Okay for b what have you tried and how did you get that answer?

[stuck folio]

i don't even know how to get the answer. i just guessed 😭

[half kettle]

😭

Okay so to find Q3 with this one

It's asking us for the mean of the 49 years, so we have to use the properties of the normal distribution we have

So it tells us that the mean of the snowfall is normally distributed with a mean of 94 inches and a variance of 196 inches, the standard deviation 𝜎 can be calculated as the square root of the variance

[stuck folio]

what's after that?

[half kettle]

Okay so what we need to do is σ = sqrroot196

And that will give us 14 inches

Now we need to look at the mean of 49 years, and for that we'll use the standard of error formula

[stuck folio]

where in the question does it say to use standard of error formula 0-0

[half kettle]

Oh yea u already have all info necessary to solve it

[half kettle]

U have to assume but in this case it's already done

14/7 = 2 would basically be the standard of error

[stuck folio]

basically standard deviation?

[half kettle]

In a way

But u already can solve it

[stuck folio]

how 😭

[half kettle]

X = μ + (z x Standard of Error)

This is the formula you'll basically use

μ is your mean of 94 inches and z will be your z-score of Q3 and you already have your standard of error which is 2

[stuck folio]

i forgot rearranging the z score formula was a thing

[stuck folio]

can you help with this now?

[half kettle]

Lmao yea but hey u did get ur answer in the end!

[half kettle]

Let's see that one is a bit tricky but it's doable

With ur standard deviation of 8.4 and the mean of 83 u will need to find the sum of the test scores but this is kinda to give context for now u can just find the mean of ur test scores

What would the mean of ur test score be?

[stuck folio]

2075?

[half kettle]

Yep that's correct

So we know our mean of our test scores is 2075

Now we need to find the standard deviation of our test scores

What would that be now?

[stuck folio]

42

I got it

[half kettle]

Yea that's the answer good job

U need an explanation for b too or u think u can do it?

[stuck folio]

i'm gonna try to do it myself first

[half kettle]

Okay then if u get confused or anything @ me

[stuck folio]

It would’ve been nice if I figured out how to do this for my actual quiz instead of after the fact 🫠

[half kettle]

Oh yea correct congrats

And yea I get u

That moment when all the info fluxes to ur brain after the quiz/test lol

[stuck folio]

btw how do i tell whether the mean is mu or x bar

[half kettle]

μ represents population mean, whilst x bar represents a sample mean

Need more context or?

[stuck folio]

so scores on an english test is a sample yeah?

[half kettle]

Well that would kind of depend on context, if you have the scores of all students who took the English test, then these scores represent the entire population

but

If you have the scores of only a subset of students who took the English test, then these scores represent a sample

[half kettle]

For this one u need to 1st find the percentile u are looking for

What would the percentile you'll use be?

[stuck folio]

0.59 and 0.41 right?

[half kettle]

Half correct

For this question you'll only use one percentile

Which of those 2 is it?

[stuck folio]

idk 😢

uhh the larger one ig

0.59?

[half kettle]

Nop

We want the score that separates the top 59% from the bottom 41%, so with that in mind it means we are looking for the 41st percentile of the distribution

This means we need to find the score below which 41% of the observations falls

[stuck folio]

sorry how do i do that?

i gtg soon 😢

[half kettle]

Okay so basically our percentile we're looking for is 41% or 0.41

So this means we need to find the z score of the 41st percentile which is?

[half kettle]

Rip

[stuck folio]

-0.23?

[half kettle]

Yep

And with that in mind next thing u need to do is use the z score formula

But in this case we'll need to make x the subject of the formula

What would the formula for this be?

[stuck folio]

x=mu+(z)(standard dev)

[half kettle]

Yep and with that plug in ur values, calculate, and u should get ur answer

That's it

[stuck folio]

tysm for your help!!

[half kettle]

Np but hey we're finally done now

And wait didn't it say to round to the nearest tenth?

[stuck folio]

oh yeah it did

34.6

[half kettle]

Yea it's not written there but anyways done finally

Cyaa

[stuck folio]

byee!

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[marsh citrus]

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[marsh citrus]

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[bold valve]

if i graph 2-f(x) do i just shift two up

[stray fern]

from f(x)?

you'll have to reflect it over x-axis first

[marsh citrus]

@bold valve Has your question been resolved?

[marsh citrus]

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[keen hemlock]

$( f(x, y) = x^2 y + y^3 ):$

[elfin berry]

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

[keen hemlock]

Sweet, lets say we have this function

This would be its differential

[
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = (2xy) dx + (x^2 + 3y^2) dy
]

[elfin berry]

TheKingPin

[keen hemlock]

This is its mixed partial:

[
\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial y} (2xy) = 2x
]
[
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} (x^2 + 3y^2) = 2x
]

[elfin berry]

TheKingPin

[keen hemlock]

what is the difference between mixed partial and differential?

because even though the formulas are different, I cant distinguish them

they both look at slope for multiple variables

[marsh citrus]

@keen hemlock Has your question been resolved?

[marsh citrus]

@keen hemlock Has your question been resolved?

[keen hemlock]

.close

[marsh citrus]

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[marsh citrus]

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[hybrid nova]

so i did a math test, and there was a question with an image similar to this one, and it basically said that a company makes cardboard boxes (without one of the faces of the cube) and they want to waste the least amount of material when cutting out the shape to make the boxes, so the question asks which one of the example cutouts waste the least amount of material (wasted material is painted black), knowing, of course, that both have the same measurements

[hybrid nova]

i answered that both waste the same amount of material, since the measurements are the same, but my math teacher disagrees, am i wrong? if so, why?

[mighty gyro]

in first one u have 4 squares

in second one

all corners become 1 square

the edge ones give 2

so only 3

ig like tht

[still temple]

2 corner give 1 sq no?

[still temple]

andthe edge gives 1

[mighty gyro]

yea nvm i took the wrong square

[hybrid nova]

but its the same area

[mighty gyro]

they give same waste

yes

[hybrid nova]

oh alr, thanks

.close

[marsh citrus]

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[still temple]

wait, consider they trisect

[hybrid nova]

what does that mean?

[mighty gyro]

then what i said would be right

[still temple]

then the above one gives 4a^2, and the lower one, assuming that it too trisects, then, we get, 3a^2

and thus, the square one wastes more area?

[still temple]

assume that, the sides of the original square are 3a, and then, all these lengths that are created by placing that shape as a, of course this is an ideal case

[still temple]

yep

[hybrid nova]

but if i call the total area X, and the area of the shape Y, then the waste would be X - Y, since the area on both examples remain the same, shouldnt the waste be the same as well?

[still temple]

yea that makes sense

that ideal case isn't possible

you're correct

[marsh citrus]

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[bold abyss]

Alright i got a couple of questions so ill take a lil while with sending them in

[bold abyss]

1

I have to solve f(x)=g(x) only with the picture as info and idk how

2

[twilit grove]

post one at a time

[bold abyss]

Oh alr