help-13

[uncut steeple]

Yes

This is a central angle right

[lone ferry]

if the angle has the center O, then yes

[uncut steeple]

So 1052 and 360-(1052)

The multiplication signs disappeared

[lone ferry]

use x

[uncut steeple]

105x2 and 360-(105x2)

[lone ferry]

thats eight

right

[uncut steeple]

So final answer is 150

Ok thank his

[lone ferry]

wait

[uncut steeple]

Ye?

[lone ferry]

yes fhats right

i did the basic math wrong mb

[uncut steeple]

Ok two more

I think it’s either equilateral or

I was gonna say isosceles…..

[lone ferry]

draw a diagram

sorry i gtg to eat dinner

[uncut steeple]

Oof ok

[lone ferry]

u drew the diagram wrong

those are not chords

[uncut steeple]

Oh yea

Is it equilateral?

[lone ferry]

redraw rhe diagram

[uncut steeple]

I don’t know how to make it correct

[lone ferry]

whats a chord

[uncut steeple]

Do all the points touch the circle

[uncut steeple]

Isn’t that what a chord is

[lone ferry]

its basicallt a triangle inscribed inside a circle

[uncut steeple]

Oof I meant

It’s obtuse then or scalene

[lone ferry]

draw the diagram

[uncut steeple]

I’ve made two

[lone ferry]

send

[uncut steeple]

Idk how to mark its correct diagram

Make*

[lone ferry]

mark the center

okat i act gotta go 😭 byee

[uncut steeple]

Bye

The hanks

Thanks*

.close

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[unique isle]

im trying to find FKG, BKC, KDF, and BCK

and i kind of did

but now i need to check my work

and also like

solve

tan FKG= opposite adjacent=tan60=sin60/cos60

tan BKC= opposite/adjacent= tan45 = sin45/cos45

tan KDF= opposite over adjacent= tan30= sin30/cos30

tan BCK= opposite/adjacent= tan45 = sin45/cos45

this one is the same as BKC i think??

[cedar kiln]

@unique isle Has your question been resolved?

[cedar kiln]

@unique isle Has your question been resolved?

[hearty arch]

what exactly is your question

is it find the tangents of those angles?

[cedar kiln]

@unique isle Has your question been resolved?

[unique isle]

yep

like

im gonna look real stupid if this doesnt work let me try the wolfram bot

,w tan BCK= opposite/adjacent= tan45 = sin45/cos45

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by dryforest#0

[unique isle]

,w 1 = sin45/cos45

[wraith dagger]

Results provided by WolframAlpha

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

Requested by dryforest#0

[unique isle]

ok nevermind

that didnt work

[hearty arch]

can't you just put tan60

the angles are given, just find their tangential values

[unique isle]

liiike this

[hearty arch]

oh you have to do that?

[unique isle]

what is that

whats it

called

[hearty arch]

what is what called

[unique isle]

like

solving

😔

[hearty arch]

you're just solving for the tangent of the angle

[unique isle]

yes

that

[hearty arch]

yes

[unique isle]

i have most of it

why do they all involve square rootd

[hearty arch]

that's the nature of trig

that's just what they are at certain values

[unique isle]

ok

so

the last step, how

how did they get that

[unique isle]

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[crimson sedge]

How do I do this?

[tropic oxide]

part 1 or part 2?

part 1 is some mildly annoying integration by parts

[crimson sedge]

Yeah part 1

[tropic oxide]

here's some lubricant: let $J_n = \int_{0}^{\pi/2} \cos^n(x) \dd{x}$, then $I_n = J_n - J_{n+2}$ is readily seen

[wraith dagger]

Ann

[tropic oxide]

derive a recurrence for J_n and from it derive one for I_n

[crimson sedge]

Ohhh ok

But how do you know this is true?

[tropic oxide]

$\cos^n(x) \sin^2(x) = \cos^n(x) (1 - \cos^2(x)) = \cos^n(x) - \cos^{n+2}(x)$

[wraith dagger]

Ann

[crimson sedge]

Thank you

[tropic oxide]

does this suffice

[crimson sedge]

Just one more thing

From this point

How would I get the expression in part i?

If I sub in the bounds

I get 0

Wait

I see my mistake

[tropic oxide]

multiple counts of bad notation there

[cedar kiln]

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[final bronze]

23. Find the value of (Sin 300 +Cos 300) – (Sin600 + Cos 600)

[final bronze]

this will be 0 right?

[cedar kiln]

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

[tiny socket]

,M (Sin 300 +Cos 300) – (Sin600 + Cos 600)

,m

I swear I saw someone using wolfram alpha inside the server

[livid hound]

wolfram starts with a w

[tiny socket]

,W (Sin 300 +Cos 300) – (Sin600 + Cos 600)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by _shachar#0

[tiny socket]

Lol just remembered that he used M but that’s make sense

[cedar kiln]

@final bronze Has your question been resolved?

[final bronze]

how

also its 30degrees and 60degres

not 300

and 600

this will be 0 right

?

[dusk finch]

Yes, this is 0

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[crimson sedge]

do i just sub 4 into the first equation and then draw the parabola

[magic solar]

Why do you draw a parabola

[crimson sedge]

i meant a graph

they intersect when x =2?

im a bit confused tbh

[hearty arch]

yes just sub 4 in, but you don't have to graph it

[magic solar]

So go find the intersection of two

[crimson sedge]

so 4=2^2

x=2

thats it?

[hearty arch]

and y=4

yes

[crimson sedge]

so they intersect at 2,4

[hearty arch]

your answer should be a coordinate (x,y)

[crimson sedge]

(2,4)

[hearty arch]

noice

[crimson sedge]

easy!

[hearty arch]

sure is!

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[spare carbon]

so im doing my revision, and i keep seeing this formula show up. what actually IS this?

google has not been helpful

[solemn torrent]

you workin on classical mechanics?

harmonic oscillator?

[spare carbon]

kinda

trigometric functions

im aware that they are oscillations, im mostly just curious on WHAT that formula is

im aware it dampens as time increases

but... what is w?

angle?

.close

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[unique isle]

The 10 inch diagonal of a rectangle forms a 30 degree angle with the base. What is the area
of the rectangle to the nearest tenth?

[unique isle]

 Determine whether you will use the special right triangles or a trig function and then set up
the problem.

i understand what trig functions are but whats a special right triangle??

can someone walk me through this first problem

[dusk finch]

Special right triangles are used to calculate sin and cos of some specific angles

[timid pawn]

Special right triangles I presume are triangles with angles as multiples of 3 and 5 in degrees.

[unique isle]

ohh

[timid pawn]

Yep. So the triangle formed here is a special right triangle.

[unique isle]

how would i set up that as a problem?

[timid pawn]

I am not sure. It's best if you refer notes or ask your teacher.

I think what they mean is to literally "set up" the problem, i.e. draw a figure and set the things you need to achieve in order to solve the problem.

[sonic thistle]

yeah just draw the triangle

then use trig

[sonic thistle]

or this

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[viral mural]

is this valid?

[crimson delta]

not without further restrictions on y or x

[viral mural]

one second

@crimson delta

https://cdn.discordapp.com/attachments/903430334681608232/1132659445877440652/image.png discord tends to downscale images

[crimson delta]

well the problem is that you could also have y=-x-1

I am not seeing right now where that is excluded

[viral mural]

oh wait obviously this is right since the values diverge

idk why I didn't just think about it for like 3 seconds, sorry

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[old falcon]

Is it possible to make this equation continuous? $x\prod_{n=1}^{x-1}\frac{100-n}{100}$
I know it can be simplified, but I don't want to accidentally do an XY problem.

[wraith dagger]

defiantburger

[gritty galleon]

expand it out by letting x-1 = 100 for now

u will observe a pattern

then u can decrease x to say 50

also the denominators pretty obvious

u need help with numerator right?

[old falcon]

yeah

[gritty galleon]

u need to simplify it, or solve it?

for solving there should be an =

[old falcon]

I want to make it continuous, like get rid of the product

[gritty galleon]

that would be expanding it imo

are u able to see what i was talking about? the pattern?

[old falcon]

I mean I can expand it and I get a result that sort of looks like combinatorics, but the only way I can think of generalizing requires a sigma which makes it not continuous all over again

Simpifies to: $\frac{x}{100^{\left(x-1\right)}}\prod_{n=1}^{x-1}\left(100-n\right)$
So for x=4
$\frac{4}{100^{\left(4-1\right)}}\cdot\left(100-1\right)\left(100-2\right)\left(100-3\right)$
Then I just need to do the right part: $\left(100-1\right)\left(100-2\right)\left(100-3\right)$
$100^{4-1}+\left(\left(1\right)+\left(2\right)+\left(3\right)\right)\left(-1\right)^{1}\left(100\right)^{2}+\left(\left(3\right)\left(1\right)+\left(3\right)\left(2\right)+\left(1\right)\left(2\right)\right)\left(-1\right)^{2}\left(100\right)+\left(-1\right)^{4-1}\left(4-1\right)!$
And that can be represented by combinations of digits I think: https://www.geeksforgeeks.org/print-all-possible-combinations-of-r-elements-in-a-given-array-of-size-n/
But then I think I would just need another sigma making it not continuous

[wraith dagger]

defiantburger

[old falcon]

WolframAlpha gets me to a point where I have to make this continous:
$-100^{-x}\left(x-101\right)x\prod_{n=102-x}^{100}n \$ But I'm still not sure what to do here

[wraith dagger]

defiantburger

[old falcon]

Get rid of the product symbol so that I can get values between integers

make it not do that

make it do that ^ (btw this is just lines drawn between points at integer values)

[wraith dagger]

kebesque

[old falcon]

Yes, and I have simplified it to this: $-100^{-x}\left(x-101\right)x\prod_{n=102-x}^{100}n$

But i'm not sure where to go now

[wraith dagger]

defiantburger

[old falcon]

I thought it might be possible like $\sum_{n=1}^{x}2=2x$

[wraith dagger]

defiantburger

kebesque

[old falcon]

Maybe I can make it after solving the Collatz Conjecture 😂

yeah

thanks for helping me though

If I do I'll let you know

!close

wait huh, close didn't work

oh whoops

.close

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[cedar kiln]

Please don't occupy multiple help channels.

[prisma mauve]

I don’t know how to draw the diagram

[regal oak]

,rotate

[lunar lynx]

What does from 070° mean ?

[upper garnet]

Pretty sure they meant 70°

[prisma mauve]

70 degrees angle

[crystal raptor]

It'll probably be because they're bearings, always written with 3 digits

The angle clockwise from the north

[lunar lynx]

Okay. Then it should be easy. Right?

[cedar kiln]

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[livid tundra]

can someone the part (b) I missed my class and can't understand anything anymore

[livid tundra]

if someone can send a video or can explain it to me

[livid tundra]

youtube video

<@&286206848099549185>

<@&286206848099549185>

[cedar kiln]

@livid tundra Has your question been resolved?

[lunar lynx]

Alright.

Do you know what is a plane's equation ?

[livid tundra]

not really

but I know how to solve part a

(x-x0)+(y-y0)+(z-z0)?

[cedar kiln]

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[fleet terrace]

Where is the mistake here?

[crimson sedge]

the derivative of a constant is 0

so whenever we integrate, its only unique up to a constant (+c)

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[zinc hamlet]

help

[cedar kiln]

Please don't occupy multiple help channels.

[ancient valley]

keep to your own channel.

[zinc hamlet]

this is

[quartz frost]

!status

[cedar kiln]

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

[zinc hamlet]

2

[ancient valley]

there was no need to open another

[zinc hamlet]

no one was answering

so i technically had no choice

[quartz frost]

if you have begun and got stuck, then pls post what you tried

[zinc hamlet]

:/

[ancient valley]

i was answering lol i just got sidetracked

[zinc hamlet]

aw

[ancient valley]

you should use a(1+r)^t. however in your formula in the book im not sure why it has the negative one

[zinc hamlet]

ok

[ancient valley]

first lets fine a

the initial ammount

do you know what that is?

[zinc hamlet]

yea

10000

[ancient valley]

yep

so lets find r

[zinc hamlet]

its

1.05

[ancient valley]

1+r is 1.05

r itself is .05

but correct nonetheless

now lets find time

how many years has it been since 2005?

[zinc hamlet]

10

[ancient valley]

yep

so now put all of those values into the formula and what would you get?

[zinc hamlet]

16200

round to 100

soo

16300

right?

[ancient valley]

yep

[zinc hamlet]

thx

[ancient valley]

np :)

[zinc hamlet]

.close

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[ancient valley]

next time you should wait in your own channel though

[zinc hamlet]

dam this server is strict

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[vagrant temple]

Anybody know how to solve e^-2 02 without a calculator?

[regal oak]

What do you mean?

Like e^(-2)*2=0?

[vagrant temple]

e^-2.02

without a calculator

[dusk finch]

by e you mean the number 2.718... ?

[vagrant temple]

yes

[dire geode]

Why can't you use a calculator

[vagrant temple]

The MCAT will not allow it

I have an example of what I am doing

I can do the rest till I get to lnK then from there I need help

[dire geode]

[violet night]

does it require the exact evaluation

[vagrant temple]

no

[violet night]

then leave it

[vagrant temple]

aproximation will do

[dire geode]

Guess it's Taylor series then

Oh yea

exp(-2.02) is probably fine

[violet night]

e^(-2.02) and then just realising its small is enough for the next part

[dire geode]

Does the question say you need to give a decimal approximation

[vagrant temple]

I was curious if anybody knew a method of how to find a decimal approx

[dire geode]

.

Did you learn that in your calculus class?

[vagrant temple]

I do not recall it has been a while since I took calculus

[dire geode]

,tex .maclaurin

[dusk finch]

You would need to calculate a lot of terms to get 0.132

[wraith dagger]

rie.mann

[dire geode]

First row

[vagrant temple]

ya that might take too long

[dire geode]

.

[dusk finch]

8 terms arent enough

[violet night]

should get a good approx shouldn't you? its a very small x value

if you require extra reasoning taylor is best bet if saying e^(-2.02) is obviously very small is not enough

[vagrant temple]

problem it is an timed exam honistly it is fucking stupid they do not want us to use a calculator

even the PCAT allows a basic computer calculator

[dire geode]

Then stop trying to get the decimal approximation since no one's asking

[vagrant temple]

ya thank you then

[cedar kiln]

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[wild belfry]

https://i.imgur.com/xQtdvh6.png

[wild belfry]

I checked the answer it says that 0 is not a critical point, how come?

e^x - e = (e)(e^(x-1) - 1)
So the critical points are 0 and 1, no?

[crystal raptor]

what are you meaning by critical points

because thats usually when the derivative is 0

[wild belfry]

Ah ok I got mixed up

The places where the area changes from above the x axis to below the x axis

I guess it is the roots

[crystal raptor]

okay then yes x=1 makes the function 0

but i dont know why you think x=0 does too

[wild belfry]

Because of the (e), no?

Since theres no x it just becomes zero

[crystal raptor]

e is just a number

[wild belfry]

Ah ok

I think I see what you mean

[crystal raptor]

you wouldn't think 0 was a root of 2(x-1) right

[wild belfry]

Yeye I see your point

Okok, that was all I wanted to ask

Thank you!

❤️

.close

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[stuck galleon]

How do I write the equation of the line In Fully simplified slope intercept form

[stuck galleon]

It’s so hard and idk why I must of js forgot 😭

How do I do it

.close

[cedar kiln]

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[stuck galleon]

.reopen

[cedar kiln]

✅

[stuck galleon]

Man I’m so CONFUSED how do you know the right points to pick?

[ancient valley]

slope intercept is written in the form y=mx+b where m is the slope and b is the y-intercept

do you know how to find the slope?

[stuck galleon]

I DONT know how to do that

[ancient valley]

ok np, are you familiar with rise/run?

[stuck galleon]

Nope

It’s hard for me to understand math

Sorry

[ancient valley]

its all good dont worry

[stuck galleon]

Like how do I know where to start on this

[ancient valley]

if we take a look at 2 points like (2,0) and (3,-1), we can take a look at how much they change in their y-axis and how much they change on their x axis. then divide the change in y by the change in x

first lets take a look at the y parts of these two points

what are they?

[stuck galleon]

It’s (x,y) Right?

[ancient valley]

yeah

[stuck galleon]

0 and -1

[ancient valley]

im gonna actually label the points rq

[stuck galleon]

Okay

[ancient valley]

point 1 would be (2,0) and point 2 would be (3,-1)

good

now we need to see how much the y values change from point 1 to point 2

[stuck galleon]

Okay

By 1?

[ancient valley]

not quite by positive 1

we start at 0, how much do we go down by to get to -1?

[stuck galleon]

1

[ancient valley]

and going down by one is the same as subtracting by 1

so our rise would be -1

[stuck galleon]

Okay

[ancient valley]

now we neet to see the change in x values

[stuck galleon]

By positive 1

[ancient valley]

what must we do to get from 2 to 3?

perfect :)

[stuck galleon]

Add 1

[ancient valley]

so our rise would be -1 and our run would be 1

now remember rise/run?

[stuck galleon]

yes

-1/1

Rise over run

[ancient valley]

great

so our slope would be -1/1, which is just the same as -1

so now we got y=-1x+b

the y-intercept should be no problem to identify

[stuck galleon]

1

[ancient valley]

where does our line intercet the y-axis?

[stuck galleon]

OH

it intercepts at 2

[ancient valley]

perfect

so b=2

[stuck galleon]

y=1x+2?

[ancient valley]

-1x*

[stuck galleon]

Oh I missed that

How do you find the points to do the math on the line tho? And how do you find point 1 and point 2? Is it a random pick?

[ancient valley]

its a random pick. you can choose any point as long as its on the line

[stuck galleon]

Okay hold on

So like on the picture I sent u

I can pick

(-2,4)

That was on the line right?

[ancient valley]

ye

[stuck galleon]

And then The other one be (-5,7)

[ancient valley]

yeah

also, i want to point out that the method we used was done using point slope formula

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

[wraith dagger]

judgementalsnail

[stuck galleon]

I havnt done that since 8th grade

[ancient valley]

with the two points you provided just now we can use this formula to find the slope

[stuck galleon]

This is so embarrassing bcuz I’m literally in 12th rn

UH

[ancient valley]

its all good :)

[stuck galleon]

7 and 4

Y2 and y1

And -5,-2

X2 and x1

[ancient valley]

yep

[stuck galleon]

Arnt u supposed to multiply or divide them across

Like ↖️↗️

[ancient valley]

wdym

we would do (7-4)/(-5+2)

[stuck galleon]

Maybe it was js a math trick my teacher taught me

Ooooh

So subtract on top and add on bottom?

or since they are both positives u subtract

[ancient valley]

nono, subtract on both top and bottom

the bottom should be (7-4)/(-5-(-2))

but the negatives cancel out to make a positive

[stuck galleon]

Okay okay okay

So it would be

3/ -3

[ancient valley]

yep

[stuck galleon]

y=-3x+b right?

[ancient valley]

what is 3/-3?

[stuck galleon]

1

Or negative 1 I think

[ancient valley]

pos/neg=neg

[stuck galleon]

oooooh

[ancient valley]

yep so our slope would be -1

[stuck galleon]

So it’s y=-1+b

x*

forgot to add that

[ancient valley]

see how we got the same answer even though we used completely different points?

[stuck galleon]

yoooooooooo

What

that’s cool

Now I NEEDA figure out B

[ancient valley]

b is y-intercept

[stuck galleon]

3

Y=-1x+3

[ancient valley]

where does the line intercect with the y-axis?

just like last time

[stuck galleon]

2

[ancient valley]

yep

so b=2 not 3

[stuck galleon]

So to simplify a slope form it’s always gonna be y=-1x+2?

[ancient valley]

it depends on your line, its not always gonne be y=-1x+2

for example we can have y=2x+3

[stuck galleon]

So if I js use the math we js learned I should always be able to figure it out?

[ancient valley]

yep

be sure to remember point slope formula

its pretty important even after highschool

[stuck galleon]

Thank you so much big bro 🙏

[ancient valley]

np :)

[stuck galleon]

Do I do the close thing now?

[ancient valley]

ye, i think i can do it actually

.close

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[pallid pond]

Please check my steps. My calculus is very rusty. I know the final answer is correct, but I don’t trust all my steps.

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@pallid pond Has your question been resolved?

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[smoky stump]

How would we do this question?
I'm thinking it is 6

but i feel like you can get 4

<@&286206848099549185>

[cedar kiln]

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[smoky stump]

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[smoky stump]

@cold briar

<@&286206848099549185>

[slate lintel]

well there are only a handful of connected graphs with 4 vertices

i found one with 5 but i haven't with 4 but i haven't checked all of them

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[ivory finch]

aside from remainder theorem, is there other ways I can solve this a lot easier?

[slate lintel]

i don't know any

it mayyyy be faster to newton's method your way to a solution

probably not though

[ancient valley]

wouldnt newtons method be a bit too extreme in this case?

[slate lintel]

probably

i just see 108 with its like 20 factors

ok 12 factors

[short blade]

if they give you a question like this you just try the smallest ones

[short blade]

24 if you include the negatives

[slate lintel]

[short blade]

you're here to factor, not to guess the roots

so chances are they give you an easy root

and make you use remainder theorem for the rest

[cunning forum]

2 is a root

it doesn’t need you to prime factorise 108

Lol

[muted bear]

You can make a rough estimate since 48x and 108 are the biggest

-48(2)+108 is almost 0, so you see if it works with the rest of the equation

[ivory finch]

it takes a lot of time to solve using remainder theorem, I'll have to try many
(+-) factors

[short blade]

testing 2 is "justified" by seeing 2 is a factor of 108

obviously you don't need to know all 12 positive divisors

[ivory finch]

maybe theres' and easier method

[short blade]

but if you're going by the p/q method you do

[muted bear]

This is the "easiest" methos

[short blade]

start with the smallest

how many did you really try

[ivory finch]

it already works for -2

[short blade]

yeah it's not that many

[ancient valley]

rational root theorem seems to be the best way so far...

[ivory finch]

Yes its almost similar with remainder theorem

thanks a lot everyone : ))

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[brave pike]

Hi, I keep getting this problem wrong and I have no idea where I am messing up. Its just a tripple intigral already given, no steps beforehand:

[cedar kiln]

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[nova snow]

How can I evaluate log base 5 of 5^2

[nova snow]

I know the answer is 2 but I don’t know how to get there

[short blade]

do you know any log rules?

[nova snow]

My only thought is to get 2log base 5 of 5

[short blade]

yup

[nova snow]

But I don’t know what to do from there

[short blade]

log_b(b) = 1

for positive b

[nova snow]

Ouhhh

then it’s just 2 x 1 = 2

Alright cool

How abut this one

Given log 7 base of 2 = 0.36 and log base 7 of 5 = 0.83

Find log base 7 of 10

I’m not sure how to start

[short blade]

factor 10

[nova snow]

Wait nvm

I figured it out

[cedar kiln]

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[cedar kiln]

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[fleet terrace]

How to solve the integral of 1/x?

[short blade]

can you think of a function that has a derivative of 1/x?

[fleet terrace]

Yes

Ln(x)

But, how to integrate 1/x??

I know that the answer is ln|x|+c

But how?

[dire geode]

like you want the proof?

[fleet terrace]

Yeah

[dire geode]

depends on your definition of log

[fleet terrace]

Wdym

[dire geode]

natural log

log is the same as natural log in math

[fleet terrace]

Whaaa???

their bases are different

[dire geode]

no

[fleet terrace]

They’re not the same

[dire geode]

log is log base 10 on a calculator

[fleet terrace]

Yes

Ln is log base e

[dire geode]

in a majority of math books, log is natural log

[fleet terrace]

Alr let’s say they’re the same, what now?

[dire geode]

what's your definition of log

[fleet terrace]

log base 10

[short blade]

what is "log"

[dire geode]

log(x) = ?

finish the right side

[fleet terrace]

Gives the exponent

[fleet terrace]

I don’t understand

[short blade]

let me give you a "proof" skittle

[
\dv{x}\int_a^xf(t), dt
]

[wraith dagger]

maximo

[short blade]

what does this evaluate to?

i am also asking you, do you know the fundamental theorems of calculus?

[fleet terrace]

Yes

[gritty viper]

Are you also asking why the derivative of log(x) is 1/x

[fleet terrace]

Not really

[fleet terrace]

-f(t)

[short blade]

not quite

[gritty viper]

maximo is saying the integral of 1/x is log|x| because the derivative of log(x) is 1/x

[short blade]

by the FTC that's f(x)

[gritty viper]

so

can we start by assuming the derivative of e^x is e^x

[fleet terrace]

f(x)

[gritty viper]

is that one good

[fleet terrace]

Yea

[gritty viper]

ok

e^(log(x)) = x

right

[fleet terrace]

Yes

[gritty viper]

So now I'm going to derive both sides

The derivative of the right side is 1

[fleet terrace]

Mhm

[gritty viper]

The derivative of the left side, do using chain rule

[fleet terrace]

Chain rule on the left side

[gritty viper]

when you get to deriving log(x), just write it as log(x)'

[fleet terrace]

log(x)’e^log(x)=1

[gritty viper]

exactly correct

now

simplify e^log(x)

[fleet terrace]

1*log(x)’=1

[gritty viper]

nah, e^log(x) isn't 1

[fleet terrace]

oh ary

sry

[gritty viper]

ur good

[fleet terrace]

X*log(x)’=1

[gritty viper]

right

[fleet terrace]

And now

log(x)’=1/x

[gritty viper]

exactly

[fleet terrace]

Got it

But how to do the antiderivativeeee

[gritty viper]

the other way is you could do it looking at the graph of log(x) which is a reflection of e^x

[fleet terrace]

Mhm

[gritty viper]

antiderivative is derivative in reverse

[fleet terrace]

But let’s say we’re not allowed to do the anti derivative using the derivative

Just

Integrate

[short blade]

"integrate" is "antiderive"

[fleet terrace]

.

[short blade]

im telling you they are the same thing

so how are you supposed to do one without the other

[fleet terrace]

I understand

[gritty viper]

i don't see how to do it without FTC

[short blade]

what you are asking for is a primitive of 1/x

which is exactly a function y such that y' = 1/x

[dire geode]

https://math.stackexchange.com/a/4266989/142904

[gritty viper]

the best way to do it has gotta be using FTC, even if that requires proving FTC

[cedar kiln]

@fleet terrace Has your question been resolved?

[cedar kiln]

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[molten canyon]

<@&286206848099549185>

[molten canyon]

l need some help

[slate lintel]

that is the worst possible way you could have gone about this

!15m

[cedar kiln]

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

[molten canyon]

Need help with this

<@&286206848099549185>

Hello

<@&286206848099549185>

[flint sapphire]

Formula for the volume of a cone

[molten canyon]

Ok

[charred briar]

I can Help

So you know the formula for a cone’s volume?

[molten canyon]

Yes

The volume right

[crimson sedge]

4/3πr³

[molten canyon]

Yeah

[crimson sedge]

ig that was it

[molten canyon]

Ok

[charred briar]

That’s a sober

Sphere

1/3πr^2 is the formula for a coke

[crimson sedge]

Lol, I am messed up

[charred briar]

Cone

[molten canyon]

Oh

Even l messed up

[crimson sedge]

Yeah 1/3 of the volume of cylinder they say

[charred briar]

Okay so now the small cone has no radius so you need to find the scale factor

[crimson sedge]

Look up, they have given it

Two radius

Lemme solve that

[charred briar]

oh yes my bad 😭😭

[molten canyon]

Oh yeah

[charred briar]

1/3πr^2 x height is the formula , i forgot height sorry 😭😭

Right so you do the (volume of the whole cone) - (volume of the small cone)

[molten canyon]

Yes

[crimson sedge]

That's it

[charred briar]

So (1/3 x pi x 8^2 x 4) - (1/3 x pi x 4^2 x 2)

[crimson sedge]

Owo I forgot the height

ig, I need to revise my 10th grade

lol

[charred briar]

Calculate that and you have your final answer @molten canyon

[molten canyon]

Ok thank you

[charred briar]

The shape that is formed then is called a frustrum

So it’s a cone which has its top taken off

[molten canyon]

Ok

[charred briar]

you’re welcome

[molten canyon]

Good to know

[molten canyon]

Thank you

[charred briar]

Anytime

[cedar kiln]

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[south tundra]

Try rewriting the second equation as 2x - 3(x^2 + 1) = -4 since we know that y = x^2 + 1

[cedar kiln]

@worthy notch Has your question been resolved?

[royal loom]

How is that "what you got" if you don't even know what you wrote down is?

[cedar kiln]

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[royal loom]

!status

[cedar kiln]

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

[royal loom]

what have you tried

[ivory finch]

what is the formula for sum of coefficient

[royal loom]

coefficient of the expansion

have you expanded it yet?

[ivory finch]

yes I got...

[royal loom]

my next question is going to be, do you know what coefficients are?

[short blade]

there's a good trick to this one

[ivory finch]

16x^4 - 32x^3 + 24x^2 -8x+1

[short blade]

would you like to know the trick eleazar

[ivory finch]

yes

is it the nCr?

[short blade]

let's say our polynoimal after expanding is
ax^4 + bx^3 + cx^2 + dx + e

we want

a + b + c + d + e

agreed?

[short blade]

no

[ivory finch]

oh I should only simply add the coefficients

[short blade]

well the trick is

a + b + c + d + e
= a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e

which is exactly just plugging in x = 1

so we can just do

(2(1) - 1)^4

[ivory finch]

if x = 1 my answer to this is

1

[short blade]

yes

[ivory finch]

oh it's so simple, so I can just substitute x=1 to the binomial

thanks so much

.close

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[slow apex]

How do you test this series for convergence?

[crimson delta]

you can try splitting it into two series and check those. if they both converge then the original converges aswell

[slow apex]

ok I got converges for both

what happens if only one of them converge?

[cedar kiln]

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[spiral marsh]

oh wait I get it now

.close

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[ivory spear]

Find the next number in this sequence: 2, -1, 6, 7, 18, 23, x?

[cedar kiln]

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

[ivory spear]

1

[tropic oxide]

19

malicious compliance with underspecification

[sonic thistle]

-90

[ivory spear]

Why?

[tropic oxide]

literally any number you could put as the next term, you could figure out a rule that gives it

[sonic thistle]

yeah it's unsolvable

[tropic oxide]

so you need more info

[vague rapids]

are there any options given for this question?

[ivory spear]

no

[vague rapids]

then you need more info

[cedar kiln]

@ivory spear Has your question been resolved?

[cedar kiln]

@ivory spear Has your question been resolved?

[slate lintel]

54 is the best answer you're going to get I think

bad question

[ivory spear]

ok I figured it out

[cedar kiln]

@ivory spear Has your question been resolved?

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[placid flower]

I need help with this integral from my complex analysis class

It's in german but it basically just wants the value of the integral along the curve

[cedar kiln]

@placid flower Has your question been resolved?

[placid flower]

I have figured it out myself

.close

[cedar kiln]

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[crimson sedge]

rank of matrix

[crimson sedge]

I don't see any linear dependence here

so then first I need to bring 1 in position a11

than by multiplying the second column by -1 and adding 1

Did I start good?

by dimensions we know that the highest rank of the matrix can be 3

[tropic oxide]

a10?

[crimson sedge]

sry a11

[tropic oxide]

than by multiplying the second column by -1 and adding 1
adding 1?

[tropic oxide]

oh so you mean adding it into column 1

i mean sure that works

there are also other things that work

i'd probably switch cols 1 and 2 and then zero out either the first row or the first column or both

gtg sorry

[grand kernel]

You can also swap column 2 ans column 3 from here

[grand kernel]

How did u get this ?

[crimson sedge]

by multiplying the second column by -1 and adding 1 column

[grand kernel]

Then why do u have -4 and not 4 ?

And 6 and not 5?

[crimson sedge]

oh

[grand kernel]

Maybe i misunderstood what you've done ?

[crimson sedge]

I didn't calculate well so I got confused apparently

probably

[grand kernel]

Yeah you should have 1 4 5 as first column

[tropic oxide]

coming back to this i think you messed something up here

or did sth that doesnt add up

[grand kernel]

Yeah that's what i found too

[tropic oxide]

honestly the lazy thing at this point would be to just compute det(A) and see if we end up with 0 or not

if not we know rk(A) = 3 for sure

,w det {{2,1,3},{4,0,5},{7,2,5}}

[wraith dagger]

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Requested by ann_dec#0

[tropic oxide]

well there you have it

[grand kernel]

Yeah or to just check if the 3 columns verify a linear relation

[crimson sedge]

I still haven't reached the part where we talk about the determinants of what this should mean

[grand kernel]

If this family if free then rank(A) = 3

[tropic oxide]

$\bmqty{2&1&3 \ 4&0&5 \ 7&2&5} \xrightarrow{C_1 \leftrightarrow C_2} \bmqty{1&2&3 \ 0&4&5 \ 2&7&5} \xrightarrow{R_3 \gets R_3 - 2R_1} \bmqty{1&2&3 \ 0&4&5 \ 0&3&-1}$

[wraith dagger]

Ann

[tropic oxide]

this is the route i would go

and then fuck with the last two rows

[crimson sedge]

thank you anyways

.close

[cedar kiln]

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[crimson sedge]

How do I calculate the taylor polynomial of order 3 at x = 0? Or more generally speaking: What is a systematic way to solve this type of question?

[weary vessel]

Use the well-known Taylor Series formula

$$\hat{f}(x)=\sum_{n=0}^{N}\frac{f^{(n)}(x_0)}{n!} \cdot (x- x_0 )^{n}$$

[floral obsidian]

Hat

🤠

[wraith dagger]

luna7427

[cedar kiln]

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[cursive inlet]

how do i find the other two sides if the hypotenuse is 10

[cursive inlet]

like which angle is the adjacent

and what do i do with sin cos tan

oh

can you use sin cos tan to find angles

[livid hound]

you'd use their (function) inverses to find angles

[slate lintel]

those are known as arcsin, arccos, and arctan and are often notated as sin^-1, cos^-1, and tan^-1

[cursive inlet]

oh

okay if you know the hypotenuse is 10 do you multiply 10 by cos

[slate lintel]

that would get you the Adjacent side length

[livid hound]

you'd need the angle

cos is not a number

by itself has no value

[cedar kiln]

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[mental crag]

Can someone explain how to do the weighted-average for this particular question? I tried doing it but I feel like I’m doing it wrong and my professor won’t even show the correct way

[cedar kiln]

@mental crag Has your question been resolved?

[mental crag]

<@&286206848099549185>

[fallen moat]

whats your formula for total$$

[mental crag]

Let me show

So basically, (Total $ above on the far right side + Total $ in the same row but far left side - Total $ in the same row but in the middle)

[mental crag]

Hopefully, that makes sense😭

<@&286206848099549185>

[fallen moat]

back, now reading

wouldn't sales be gaining money?, if so

the total $ should be
right-left+middle

[mental crag]

I’m just copying how my prof does it

[fallen moat]

i see

[mental crag]

But the screenshot I just recently shared with you, is that correct for weighted-average?

Because I double checked my answer and I got the same answer when I did units available - units sold and GAFS - COGS

[fallen moat]

doesn't look good, since from the info given, it looks like it's gaining money, so it wouldn't even do those sales and purchases if it's not gaining

[mental crag]

How should I do it

[fallen moat]

to be fair, I don't understand the abbreviations.

[mental crag]

I was thinking I was wrong because I was getting negative numbers lol

[fallen moat]

this

[mental crag]

So on the right far column it says Total $, right

[fallen moat]

yea

K238 should be
=K237-E238+H238

[mental crag]

So basically, you just do Total $ (far right column side) + Total $ (far left column side) - Total $ (middle)

Yes LOL

I made it easier bc I thought Excel would confuse you more

Wait no, K238+E238-H215

[mental crag]

K238+E238-H215 is what I did

[fallen moat]

i mean this is my attempt

[mental crag]

Oh, I see

[fallen moat]

and then you can use the excel sliding to autofill those remaining ones

[mental crag]

Yea

So what I did is wrong since I got closing inventory as negative, right

[fallen moat]

that's around the idea, yea

[mental crag]

I don’t know how to solve this question because my prof. would only show us questions when sale at cost units are given for sale

But this question, he gave us # units and $/unit

[fallen moat]

i think they made this HW intentionally the other way round to see if you get which is positive and which is negative

[mental crag]

OMG I think I know what you mean

I hate my prof., he’s a complete asshole

[fallen moat]

lol

[mental crag]

Wait so, how should I do it to make it positive

The complete opposite?

[fallen moat]

nah, I'll explain a bit

[mental crag]

Take your time