help-19

[cursive musk]

but what dose that get me

[olive needle]

And now you have your ratio.

[cursive musk]

dose 2a go under 2x+30?

[digital kindle]

BFE~BAC

[olive needle]

2 is the ratio for the third side lengths.

[cursive musk]

whats the third side lengths 2x+30 or 2x+27

2x+30 right

kuje tgus

like dis

[digital kindle]

There you go kid

[cursive musk]

thx

[digital kindle]

No worries

[odd edge]

@cursive musk Has your question been resolved?

[odd edge]

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[random maple]

The Future Value of $1500 saved each year for 20 years at 13%, compounded monthly

[random maple]

im stuck on how to compound it properly

[odd edge]

@random maple Has your question been resolved?

[odd edge]

@random maple Has your question been resolved?

[snow steeple]

P = principal ($1500), r = 0.13, n = 12, t = 20

[random maple]

The formula is a bit off afaik

this is the base formula without adding compounding intervals

[random maple]

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[cold tapir]

this is what I get at the end when I do the ratio test

[leaden widget]

what is your question?

[cold tapir]

how do I solve it

[lone elbow]

are you trying to see if the series converges?

[cold tapir]

yes

[lone elbow]

next to the limit < 1

also, show your work

[cold tapir]

this is as far as I got

[lone elbow]

how did you cancel sqrt(n) with sqrt(n+1)

everything else is perfect

[cold tapir]

well i though that when they have the same degree you grab the ratio of there coeffcients

so their both to the 1/2

right?

[lone elbow]

yeah, is just that you skipped steps, but is fine

[cold tapir]

soo how do I solve for n

[lone elbow]

the limit 1/(n+1) approaches 0

why are you trying to solve for n?

[cold tapir]

would about interval of convergence

[lone elbow]

this is not radius of convergence

[cold tapir]

oh there are specific type of series that allow you to solve for the interval of convergence

?

[lone elbow]

the ones that have x inside

[cold tapir]

ohhhhhh

[lone elbow]

but n is the variable for the limit, there is no x here

[cold tapir]

oh no wonder this one looked odd

ok i see thanks

.close

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[drifting pine]

can someone help me with graphing polar functions
specifically limacons and rose graphs

[drifting pine]

like ts makes no sense

[odd edge]

@drifting pine Has your question been resolved?

[drifting pine]

<@&286206848099549185> plz

[stone rain]

Hey there

I have this amazing yt video on Calc II graphing polar functions

[drifting pine]

hey

[stone rain]

https://youtu.be/xx4JN8x2E24?si=MFzcFPwxV4Qoy4i2

It’s an amazing lecture with practice problems too

[neon knot]

whys that third one looking at me funny

[stone rain]

?

[neon knot]

nvm if u dont see it

[drifting pine]

ill give it a watch

tho something i dont understand still is where exactly do the petals point when n is odd

like this for example

[stone rain]

Watch the video

It goes in detail how transformations affect the function

[drifting pine]

what timestamp

[stone rain]

Calc II isn’t fresh to me but trust that video will give you answers

Uhhh lemme see

It should have them in the video

Yeah the video has time stamps

Just scroll around it explains graphing every type of equations

Organic chem teacher is also fire

[drifting pine]

oh, so i just have to evenly distribute the petals so long as they stay withing the r = a?

[drifting pine]

.close

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[stiff kettle]

how do i do d) for the second part

the image point (2,1)

im kinda confused

this is log functions btw

[brittle beacon]

What did you get for the first part?

[stiff kettle]

hiii

vertical stretch baf of 3 and shift 1 unit up

cant i put it in some kind of table and do it

chat gpt isnt helping

[brittle beacon]

Basically you have the point (2, 1) on the "original" y = log_2(x), and using those transformations, you wanna figure out where that point ends up getting "sent to" under these

[stiff kettle]

mhm

but for other functions for the transformed version u could do the table and then draw out the points cant u do the same for this

do u kinda get what i mean

[brittle beacon]

I'm not super sure if I do, do you have an example?

[stiff kettle]

like this typa thing

[brittle beacon]

I mean, I guess you could, but then it's like only one point you're really wondering about, if you get me?

[stiff kettle]

okok

[brittle beacon]

So like you know that for the point (2, 1), the y component gets "vertically stretched" by a factor of 3, and then shifted 3 units up

[stiff kettle]

its that easy

ok so and if there was a HA translation and horizontal shifts then that would affect x?

[brittle beacon]

Yep just like that

[stiff kettle]

ah i see

and

how would i find the x intercept

do i just go like

0 = 3log_2x + 1
-1 = 3log_2x

-1/3 = log_2x

how do i do this without a calculator omg

oh wait

2^(-1/3) = x

so 3sqrt-2?

where do i put this negativeuhmmm

[brittle beacon]

Remember x^{-a} is 1/x^a

[stiff kettle]

oh oops

oh

1/2

AH

[brittle beacon]

(and cube root )

[stiff kettle]

3sqrt0.5

ya

thats what i meant LOLL

i dont think i can type the small 3 on my computer

[brittle beacon]

I'll let you have that one

[stiff kettle]

yayy

tysmm

[brittle beacon]

It's quite difficult to, I usually just use TeX for it $\sqrt[3]{2}$

[clever fjord]

@brittle beacon

[stiff kettle]

olala

damn

thats cool

[brittle beacon]

[stiff kettle]

thank uu

💗

.close

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[mystic saffron]

screwing up some order of operations on what should be an easy question

[mystic saffron]

have to plug in 0.97a0 for r

the a0 is 0.529E-10

i literally put the whole thing in my calculator and it came out wrong

Correct answer is 1.644E14 I believe

[royal herald]

r = 0.97 * a_0 ?

[mystic saffron]

yes

the a0 cancels in 2 places if i did that part right

[royal herald]

u said a_0 = 0.529

but in calculator its 0.0529

[mystic saffron]

holy crap

thank u

i have been staring at this for 20 min wondering if i should be in this class even

[royal herald]

$\frac 1{4 \sqrt {2\pi}} (0.529^{-1} \cdot 10^{10})^{1.5} (2 - 0.97) \cdot e^{-\frac {0.97}2}$

[mystic saffron]

yea that looks correct

i got the right answer now

thank u so much

.close

[royal herald]

np

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[glacial tapir]

What does [aij] part mean?

[glacial tapir]

Defined for every element in the 3x3 matrix?

or every elemment cubed

This is the subspace and the LT

[low locust]

its just notation for a 3x3 matrix

it has entries a_ij where both i and j are between 1 and 3

[glacial tapir]

Oh thanks, never seen this notation in class before and was a lil confused

.close

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[wanton bison]

Hi guys, I have a question.
$\$ I know that [ \langle 1,2,3 \rangle \circ \langle 1,3,2 \rangle = \iota \delta ]
but is it also correct to write
[ \langle 1,2,3 \rangle \circ \langle 1,3,2 \rangle = \langle 1 \rangle \circ \langle 2 \rangle \circ \langle 2 \rangle ?]
Is that also essentially the identity?

[clever fjord]

𝔸dωn𝓲²s

[frosty mango]

is not wrong, but why do you need that notation?

[wanton bison]

So I can understand what I am doing, and what the id actually represents

Actually there is a typ

mid one is <3>

[frosty mango]

right, no problem with that

[wanton bison]

ok thanks mary

.solved

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[mystic saffron]

Yo

[mystic saffron]

Is the casio 991 cw a good calculator to get

Rn I have the 300ms

And it's terrible ngl

<@&286206848099549185>

[lost crane]

It's very good but it can't do graphing

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@mystic saffron Has your question been resolved?

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[wary harbor]

If you have an image and a frame, where the ratio between the image's width and height (width / height) is r_i, and the ratio between the frame's width and height is r_f, then what should the ratio between the image width and frame width be so that the portion of the image that is outside the frame is as large as the amount of free space inside the frame?

[wary harbor]

I had no idea how to solve this outright, but I started with a similar problem, where I have an image that's w_i wide and h_i tall, and a frame that's w_f by h_f.

Then we could choose a resizing ratio r for the image such that it would fit the problem statement. Logically, this should eventually factor out somewhere.

If the amount of image that is outside the frame is just as large as the amount of free space inside the frame, then clearly we should be able to "cut off" the part outside the frame and fit it inside the frame, filling the frame exactly. Thus the image and the frame must be the same size.
So w_i * h_i * r^2 = w_f * h_f

And we're looking for the ratio between the image width and frame width, so x = w_i * r / w_f

Now, this is kind of where I got stuck, because I couldn't solve this set of algebraic equations.

But I plugged it into Wolfram Alpha, and it gave me this, which indeed has no r in it.

And I can simplify this into just sqrt(w_i / h_i * h_f / w_f), which is sqrt(r_i / r_f), a very beautiful answer.

But... I don't understand where it came from.
Can anyone assist me in actually doing the intermediate steps for this, or to help me understand on a conceptual level, why this answer is correct?

[mystic saffron]

Is calculator Casio fx911ex good?

[wary harbor]

This channel is occupied.

[mystic saffron]

?

?

[odd edge]

@wary harbor Has your question been resolved?

[odd edge]

@wary harbor Has your question been resolved?

[odd edge]

@wary harbor Has your question been resolved?

[fresh badge]

yap yap yap

[odd edge]

@wary harbor Has your question been resolved?

[noble cypress]

put it in chatgpt

🙂

who reacted x

why not

use it to clarify concepts

its acc sick

[wary harbor]

Unless it's a well-known problem, AI will hallucinate a random answer that looks coherent.

[noble cypress]

oh ok then

[wary harbor]

Yeah, it just gives me completely wrong answers that keep getting more wrong as I try to tell it where it makes a mistake.

Language models are okay for short, imaginative stories, and maybe as a sort of search engine, not for solving mathematical problems.

[odd edge]

@wary harbor Has your question been resolved?

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@coarse swallow Has your question been resolved?

[quasi sparrow]

Use slope = rise / run for any two points

One point would be (0,0)

What point did you use

[quasi sparrow]

Two points on the line if that wasn't clear

[odd edge]

@coarse swallow Has your question been resolved?

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[empty tinsel]

hi

[empty tinsel]

pls help

[shy ore]

I really want this.

[empty tinsel]

yo

[amber schooner]

hello

[empty tinsel]

uh how i do this

[amber schooner]

^

[empty tinsel]

but like ion know

[amber schooner]

🤔

you haven’t tried anything

[empty tinsel]

area under curve??

[amber schooner]

differentiate implicitly

[amber schooner]

lol

[empty tinsel]

bru ion know

[amber schooner]

can you at least tell me what the derivative of the right side is?

[empty tinsel]

my brain so fried rn i been doin this homework for 6 hrs shits like 150 questions

[empty tinsel]

hm ok

uh

idk

wdym right side

[amber schooner]

of the equation

3y

[shy ore]

it helps if you give detail on what part or what exactly you don't know. like is it a conceptual thing, do you understand derivatives but don't know how to approach this particular problem?

[empty tinsel]

o

its dy/dx

[amber schooner]

are you sure

[empty tinsel]

wait

dy /dy???

[amber schooner]

nope

what happened to the 3?

[empty tinsel]

idk aaaaaaaaaaaa

you deriv the 3 right

[amber schooner]

3 is a constant multiple

[empty tinsel]

wa

so its 3 dy/dx?

[amber schooner]

$\frac{d}{dx} cf(x) = cf’(x)$

[clever fjord]

knief

[amber schooner]

yes

[empty tinsel]

ok

[amber schooner]

now for the left side we use power rule and chain rule

can you try the left

[empty tinsel]

and ight side is 5(x+y^2) (1+2ydy/dx)

[amber schooner]

hmm what’s 5-1

[empty tinsel]

5 - 1?

[amber schooner]

close though

[shy ore]

@empty tinsel maybe slow down a little and try to figure out what's going on conceptually if/when you have the time.

[amber schooner]

you wrote (x+y^2)^1

[empty tinsel]

oh yea 4 im slow

[empty tinsel]

i dont its 1 am rn im just tryina sleep

[amber schooner]

$5(x+y^2)^4 (1+2y\frac{dy}{dx}) = 3\frac{dy}{dx}$

[clever fjord]

knief

[empty tinsel]

oh i forgor 4 yea

[amber schooner]

now we need to isolate dy/dx

[empty tinsel]

how

[amber schooner]

distribute the 5(x+y^2)^4 on the left side then move all terms with dy/dx on one side

[empty tinsel]

-5(x+y^2)^4)/10y(x+y^2)^4 -3

[mystic saffron]

You can replace the whole $5(x + y^2)^4$ with some variable $a$ to avoid confusion

[clever fjord]

Edmund Cloudsley

[amber schooner]

$(10y(x+y^2)^4 -3)\frac{dy}{dx} = -5(x+y^2)^4$

[clever fjord]

knief

[amber schooner]

hence

$\frac{dy}{dx} = \frac{-5(x+y^2)^4}{10y(x+y^2)^4-3}$

[clever fjord]

knief

[mystic saffron]

Yup and just move it over to the other side

[amber schooner]

yep

all good

sleep

[empty tinsel]

no....

we did one

there are more

[amber schooner]

how many

[empty tinsel]

like 30 more

[empty tinsel]

gg

na im playing like prolly 5-10

[amber schooner]

did you procrastinate for the entire semester?

[empty tinsel]

no she gave it use like last week bruh

[amber schooner]

caffeine

👍🏻

[empty tinsel]

guys i entered the answer wrong

now i have to do 2 problems like it

🙂

🙂

🙂

🙂

🙂

FUCK

[amber schooner]

💀💀

[amber schooner]

this is the same bullshit almost

[empty tinsel]

-sin(2x^3 +5y^2 )*6x^2 / 2y +10ysin(2x^3 + 5y^2)

[amber schooner]

what

where did that came from

[empty tinsel]

my friend sent

idk if its right

[amber schooner]

do you see sine there?

[empty tinsel]

no

😭

bruh

[amber schooner]

bruh

||derivativecalculator.com||

[empty tinsel]

4x-50x(5x^2+y)^4 / 5(5x^2 + y )^4

[empty tinsel]

:0

[amber schooner]

are you using that site

[empty tinsel]

no

gpt

[amber schooner]

you just copied and pasted

oh lol

[empty tinsel]

from gpt

look bruh how many questions there r

been doin this shit for 6 hours

gave up half way tried to gpt but its kinda buns

[amber schooner]

$5(5x^2+y)^4(10x+y’) = 4x \implies y’ = \frac{4x-50x(5x^2+y)^4}{5(5x^2+y)^4}$

[clever fjord]

knief

[amber schooner]

yep

[empty tinsel]

4 more

we got this

[empty tinsel]

how old is u btw

[amber schooner]

18

[empty tinsel]

o

senior?

[amber schooner]

freshman in uni

[empty tinsel]

o tf im 17 junior bruh

[amber schooner]

are you going to try this

as in copy and paste a gpt answer

[empty tinsel]

gpt

[amber schooner]

$6(x+y)^5(1+y’) = 8x \implies y’ = \frac{4x-3(x+y)^5}{3(x+y)^5}$

[clever fjord]

knief

[amber schooner]

you can cancel a factor of 2

but it’s correct yes

[empty tinsel]

alr done next topic

[amber schooner]

chain rule

$h(x) = f(g(x)) \implies h’(x) = f’(g(x)) \cdot g’(x)$

[clever fjord]

knief

[amber schooner]

would you like to try

or are you still relying on gpt

[empty tinsel]

i would if i still didnt have...

[amber schooner]

calc bc

[empty tinsel]

didnt even start on the other shi bruh

[amber schooner]

🔥🔥

[empty tinsel]

mhm

[amber schooner]

those are optional

[empty tinsel]

we skipped ab

[empty tinsel]

they extra credit

i need that shit im go from 98 to 100 for Q2

so yea go ahead and leak answer

[amber schooner]

!nosols

[odd edge]

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.

[empty tinsel]

o

[amber schooner]

^

let x = 6

what’s g(6)

[empty tinsel]

1

[amber schooner]

yep now what’s f’(1)

slope formula

[empty tinsel]

1/2

[amber schooner]

yep and now what’s g’(6)

[empty tinsel]

idk its sharp turn

[amber schooner]

oh wait

it said -6

😭

[empty tinsel]

1/3

[amber schooner]

ok what’s g(-6)

[empty tinsel]

-

[empty tinsel]

<

-2

[amber schooner]

yes g’(-6) = -1/3

g(-6) = -2

what’s f’(-2)

[empty tinsel]

2

[amber schooner]

mhm so tell me what h’(-6) is

[empty tinsel]

0

huh

[amber schooner]

it’s multiplication

not addition

(2)(-1/3)

-2/3

[empty tinsel]

yea it was 2/3 we got it wrong

[amber schooner]

oh bro

[amber schooner]

g(-4) = ?

[empty tinsel]

-1 * -3

[amber schooner]

yea

3

did you gpt

[empty tinsel]

1/3 * 1/2

[amber schooner]

what

[empty tinsel]

no i would if it worked

[amber schooner]

lol

[empty tinsel]

its not that hard my brain lowky fried

[amber schooner]

h’(3) = f’(3)g(3) + g’(3)f(3)

[empty tinsel]

tell me if im right bru

[amber schooner]

ok

[empty tinsel]

<

[amber schooner]

how did you get that

[amber schooner]

you didn’t use this

that was just the first part of the product rule

[empty tinsel]

what product rule

bro i used chain rule

i didnt see questions

[amber schooner]

f’(3) = 1/2, g(3) = 3

[empty tinsel]

1.5 + -1/3

[amber schooner]

it’s a product of two functions

[empty tinsel]

this?

[amber schooner]

then g’(3) = 1/3 and f(3) =-1

yes

3/2 - 1/3 = 7/6

[empty tinsel]

u sure

[amber schooner]

🤔

sure about what

[empty tinsel]

perhaps you misclicked and gave me the answer

[amber schooner]

perhaps

what does the feature part want

[empty tinsel]

if its positive neg etc//

[amber schooner]

increasing/decreasing?

what does it say though

[empty tinsel]

value part

[amber schooner]

ok

oh well i mean you can do the value one i hope

[empty tinsel]

oh

its too hard

yes i can do it

i think

[amber schooner]

oh this is the graph of f’

[empty tinsel]

nvm i g i cant

[amber schooner]

lol

what’s f’(-3)

you can do that

[empty tinsel]

0

[amber schooner]

yep

can we deduce anything about the value of f at -3 given the graph of f’

[empty tinsel]

idk

uh

[amber schooner]

🤔

don’t gpt bruh

[empty tinsel]

0?

[amber schooner]

no

we can’t know

[empty tinsel]

bru i wouldnt be here if it worked

[amber schooner]

ok now for the features

f’ changes from - to 0 to +

[empty tinsel]

inc

[amber schooner]

what does this mean for f at -3

by first derivative test

[empty tinsel]

mi

n

[amber schooner]

yep

now what did it want for f’?

[empty tinsel]

sign change

[amber schooner]

no but the feature box

[empty tinsel]

o incr

[amber schooner]

same thing?

[empty tinsel]

ye

[amber schooner]

i think you can do the sign change

[empty tinsel]

idk

oops got it wrong i gotta redo all 5 questions again

🙂

[empty tinsel]

na jokes

i got it irhgt

nxt one

omfg my head hurts like shit man

been getting 2hrs sleep consitantly

speed run

[amber schooner]

hello

[amber schooner]

nice

[amber schooner]

happens

i averaged 2 all nighters over the last 2 weeks

my sleep schedule is cooked

i go to bed at 5-7 am

need to fix that

i have my last finals tomorrow

so i’m chilling

[empty tinsel]

me too lately

i been doin some freaky shit lately

i shud prolly stop like all night like im cooked

[amber schooner]

what

[empty tinsel]

yea dw bout it

[amber schooner]

lol

[empty tinsel]

wait u think im girl?

[amber schooner]

🤔

you’re a mouse

[empty tinsel]

mhm

il tell u after we finish 🙂

[amber schooner]

🤔

[empty tinsel]

so les go

[amber schooner]

bro

😭😭

[empty tinsel]

bro im a guy chil

[amber schooner]

we know

[empty tinsel]

continue with our convo on hand

[empty tinsel]

aint no girl be using my vocab ofc u know

[amber schooner]

no shit

"ion"

[empty tinsel]

😭

continue

gogo

[amber schooner]

you have to attempt the problem

[empty tinsel]

i did

[amber schooner]

and

[empty tinsel]

no idea how to do it

🙂

[amber schooner]

any important things

lol

speed increasing means x’ > 0 and x’’ > 0 (velocity and acceleration in same direction) or x’ < 0 and x’’ < 0

[empty tinsel]

ok

wat i do with that info

[amber schooner]

look at the graph, find intervals where it’s both concave up and increasing or intervals where it’s both concave down and decreasing

[empty tinsel]

5 6 8 11

13 18

[amber schooner]

what

write it in interval notation

[empty tinsel]

nooooo

5 to 6

8 to 11

13 to 18

[amber schooner]

yes

[empty tinsel]

A particle moves along the x-axis so that its velocity at time 0, is less than, t, is less than, 180<t<18 is given by v, left parenthesis, t, right parenthesisv(t), whose graph is shown in the figure below. The inflection points of the graph are marked with green circles. Find all values of t when the particle is at rest.

[amber schooner]

at rest -> v(t) = 0

[empty tinsel]

o k

so

@amber schooner

wat i do

[amber schooner]

brother

[amber schooner]

^

[empty tinsel]

oh its vt graph

[amber schooner]

do you fr need to check this

[empty tinsel]

yes

I FIna doze off bru

[amber schooner]

it’s a semi circle

[empty tinsel]

16PI?

8pi

[errant gust]

yes

[amber schooner]

lol no

[errant gust]

8pi is the semicircle

[empty tinsel]

WHY BRO I HAVE TO DO 10 MORE PROBLEMS NOW

bruh

[amber schooner]

yea but that’s not the answer

[amber schooner]

^

[empty tinsel]

gg

[amber schooner]

the circle went from -5 to 3

[errant gust]

add the triangle from 3 to 5

[empty tinsel]

im blind i didnt c

[amber schooner]

same shit

[empty tinsel]

12.5 pi

[amber schooner]

just add the triangle on the right

[amber schooner]

no

[empty tinsel]

yes

[amber schooner]

🤔

[empty tinsel]

what is it

-4

-12.5pi

[amber schooner]

radius 5 btw, -7 -> 3 is 10 so r = 5

[empty tinsel]

-1

+1

20

[amber schooner]

-12.5pi - 4

[empty tinsel]

12.5pi

[amber schooner]

lol what

[errant gust]

why 25pi

its a semi circle

[empty tinsel]

3pi

[amber schooner]

oh you’re right

semi circle

[empty tinsel]

• 4

[amber schooner]

-12.5pi - 4

[empty tinsel]

-10

6.25pi -10

-6.25*

alr gg

gn twin

[amber schooner]

you’re welcome

[empty tinsel]

.close

[odd edge]

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[steady tide]

i don't really understand the first and second inequality? what does it mean to integrate wrt a vector? how does the second equality come to be?

[signal crown]

there is a dot product here so it makes sense

work is generally defined as :
$\delta W= \vec{F} \cdot \vec{dr}$

[clever fjord]

Herels

[signal crown]

dr is a small displacement

[steady tide]

small changes in work equals to dot product of force vector and small changes in displacement vector?

[signal crown]

yes

[steady tide]

okay that makes sense, but what did the second equality mean?

[signal crown]

the one with the kx i ?

[steady tide]

yeah

[signal crown]

they just rewrote F and dr with their expression

[elfin zodiac]

$d\vec{W} = \vec{F} \cdot d\vec{r}$

[clever fjord]

Samuel

[signal crown]

no, we use delta for work

[elfin zodiac]

Oh ok

[signal crown]

but anyway, it means the same thing

[signal crown]

and its a dot product here, W isnt a vector

[steady tide]

ahh i see

[elfin zodiac]

Lol that was a typo sorr

[steady tide]

alright thanks

.close

[signal crown]

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[worn swallow]

.solved

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[ruby narwhal]

Hi this is a quick question, I remember in math class when I had to find the value of say, sin( 2pi / 5 ) without a calculator, I remember taking 2pi / 5 and increasing the values in the fraction (value stays the same but bigger numerators and denominators), eg: 4pi / 10, or 10pi / 25

[ruby narwhal]

Then from there I was able to put that in terms of some of the values on the sine table

(pi/2, pi/3, pi/4, pi/6 etc)

meaning I was able to solve the thing, but I completely forgot how to do it and I don't have access to my notes, does anyone know how to do this?

[frigid grove]

i dont think you can do that on sin(2pi/5)

[ruby narwhal]

ah alright, by any chance do you know of any where it would work?

[odd edge]

@ruby narwhal Has your question been resolved?

[ruby narwhal]

.close

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[late pine]

Hello, im trying to compine this expression

[late pine]

This is my result

but the correct is:

why is this??

[narrow crypt]

• log (x-1)^2

No bracket

Don’t add imaginary bracket

[late pine]

so everytime is a logarithm addition it goes in the numerator

just if it has a bracket in the substraction that is in the denominator?

[signal oar]

The 2log(x-1) is positive

It needs to go in the numerator

[late pine]

understood🫡

[signal oar]

Since $\log(a) + \log(b) = \log(ab)$, only for $\log(a) - \log(b)$ you get that this is $\log(a) + \log(\frac 1b) = \log(\frac ab)$

[late pine]

thanks

yeah, I though that if there is a substraction and then an addition, you should multiply it to the denominator

thanks :))

[clever fjord]

Kepe

[late pine]

understood

thank you very much

.close

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[signal oar]

np

[odd edge]

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[eager night]

help

[loud stone]

cross multiply

[bitter folio]

Get quadratic equation and solve for x

[loud stone]

(x+2)(x+1)=(x-1)(x-4)

[eager night]

okey now

[loud stone]

try it

[eager night]

It doesn't give me the result

I mean I did all the steps

[loud stone]

no but it's easy from there

show me what you did

[eager night]

wait for me

I will do it again

I looked for equivalent denominators

[loud stone]

no

did you cross multiply

[eager night]

okok

I already did

and now

[loud stone]

show me what you have

[eager night]

go

[loud stone]

what?

yes thats correct

[eager night]

I don't know how to continue

[loud stone]

you can cancel out the x ^ 2 because it's on both sides

[eager night]

okok

[loud stone]

and you can rearrange to have all the x terms on one side

[eager night]

okok

Five is still negative

?

that is, if I pass it, it is still negative.

[loud stone]

no, you have to add positive 5x to both sides

[eager night]

okokok

look

[loud stone]

correct

[eager night]

okok

[loud stone]

and then combine the like terms

[eager night]

yes

8x=2

Ready, now I know how to follow it

Thank you, you are the best

[loud stone]

nice

youre welcome

[eager night]

I love you

let's be friends

[loud stone]

ok lol

[eager night]

.close

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[candid tangle]

i do not understand how does the answer obtain the straight line. i tried subbing 2pi into 2x/pi but it doesnt work

[sweet goblet]

Can someone explain how to get the limit of this function? I would have actually gotten 0, but the correct result should be infinity.

[candid tangle]

hi this is already in use

[pale jay]

for the straight line you can treat 2/pi as the slope for the linear function 2x/pi

[sweet goblet]

?

ahh sorry

[candid tangle]

it's fine

[pale jay]

so if x = 2pi, the result of the linear function in the right side is 2*(2pi)/pi or simply 4

so you just draw a straight line from (0,0) to (2pi, 4) to get that specific straight line

and since solutions of an equation can just be thought of as the points of intersection of the graphs of both left and right sides of the equal sign

[candid tangle]

but if this is what im doing, wouldn't it not match with the drawing of the graph. my graph shows 3 intersections but according the the answer it's 4

[pale jay]

i think you have to factor the +1 y intercept here

because the problem said "hence using the same axes"

not sure tho

because if you look at the graph in the textbook you can definitely see that the function used is 3/2 cos2x + 1

instead of the 3 cos2x = 2x/pi

[candid tangle]

this is the graph, the red line is me factoring in the 1 but this still doesnt make sense to me

[pale jay]

ohh

you're not supposed to add the +1 intercept to the straight line

try graphing it as just y = x/pi

[pale jay]

because this is for the cosine

[candid tangle]

oh wait, how did we get to x/pi ? i did the +1 because i tried subbing in 3cos(2x) =2x/pi into y=3/2 cos2x +1

[candid tangle]

it's correct if it's graphed as x/pi , but i dont know how did we get here

[pale jay]

because 3cos2x = 2x/pi can be written as (3/2)cos2x = x/pi

[candid tangle]

so i do not need to subsitute that into y=3/2 cos2x + 1

but why can i just use x/pi to draw the line? it feels a bit random because i still dont understand

[odd edge]

@candid tangle Has your question been resolved?

[odd edge]

@candid tangle Has your question been resolved?

[odd edge]

@candid tangle Has your question been resolved?

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[wary harbor]

Is there any way I can systematically find numbers in the vicinity of 1 that are neither repeating decimals themselves, nor is their reciprocal a repeating decimal?

[wary harbor]

e.g. 1.25 and 0.8

Do we get a repeating decimal when dividing by every number that isn't 2, 5, or something's that's entirely divisible into 2s and 5s? 🤔

So I'd need a fraction where both the numerator and denominator only have 2s and 5s as factors?

1/2, 2/2, 1/4, 2/4, 4/4, 1/5, 2/5, 4/5, 5/5, 1/8, 2/8, 4/8, 5/8, 8/8, 1/10, 2/10, 4/10, 5/10, 8/10, 10/10, and so on...
Some of those are duplicates, and the reciprocals of many of those are not near 1, but otherwise they seem to fit...

[fickle silo]

if you cancel terms, then you are looking for a way to approximate 1 with 2^n / 5^m

[wary harbor]

Oh yeah, 'cause there's no point in putting the same factor in both sides of the fraction.

So if I'm over 1, I increase 5 by 1. If I'm under 1, I increase n by 1

Mm, well, might actually miss something, because for every time I raise m, I need to raise n a bit more than twice.

I think

I suppose it doesn't matter that much. If I'm already over 1, increasing n would already give me a number that's not close to 1 at all.

Yeah, I think that would do it.

Thanks @fickle silo

Heh, orbit

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[fickle silo]

what is the integral of |sin(t)| from 0 to pi?

[heady plover]

Ooh shoot

Thx

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[granite hazel]

Under what topic does Finite State Machine designing topic comes?

I have many doubts:

1. How to choose the number of states? I know how to choose when it comes to simple Sequence Detector problems. It is states = number of digits to detect. But in some other twisted questions like this one, I can't get it.

2. How to know my state machine is in reduced form? If not, how to reduce it?

[odd edge]

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[granite hazel]

.reopen

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✅

@granite hazel Has your question been resolved?

[granite hazel]

No

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[pure barn]

What’s the 48 dgree for?

[pure barn]

And how do I find po and the others

My teacher is using theorem 4 n 5 (central angel and arc and inscribed angles)

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[mystic saffron]

I'm trying to create a formula which follows the following rule

For each 10x multiplication of an X value, Y doubles.

I got close with x^0.30 but its not exactly it, 10 million is at 126 instead of 128.

both X and Y values should be integers but i believe i can handle that via some rounding.

[quasi sparrow]

can you write a table of (x, y) values you want

[mystic saffron]

1 1
10 2
100 4
1000 8
10000 16
100000 32
1000000 64
10000000 128
100000000 256

and so on

i tried using websites which fit curves but i didnt get any perfect fits

which is wild considering it is a pretty simple formula in my opinion. But perhaps its not so

[quasi sparrow]

does $y = \log_{10}(x) + c$ not work for you

[clever fjord]

riemann

[quasi sparrow]

oh did you give (y, x) values instead of (x, y)

[mystic saffron]

they should be x,y

if one travels a million units in the x direction, they should be at y=128

[quasi sparrow]

,calc log(10^6, 10)

[clever fjord]

Result:

6

[quasi sparrow]

,calc 2^(6+1)

[clever fjord]

Result:

128

[quasi sparrow]

oh i forgot 2^y

[mystic saffron]

doesnt seem like a match to me

[quasi sparrow]

$2^y = \log_{10}(x) + c$. find c with a couple points

[clever fjord]

riemann

[mystic saffron]

what do you mean by that exactly?

i'm uncertain what C is supposed to be here

[quasi sparrow]

plug a couple datapoints (x, y) into here and solve for c

the c should be the same for all of them

oh oops maybe the +c should be on y

$2^{y+c} = \log_{10}(x)$

[clever fjord]

riemann

[quasi sparrow]

fixed my earlier mistake

you should be able to use algebra and solve for y or x now

[mystic saffron]

so forexample

2^128+c = log10(10,000,000)?

[quasi sparrow]

errr no my bad

[mystic saffron]

This is getting really confusing

I'm just a programmer trying to make this work

[forest sky]

i think it should be something like y = 2^(log_10(x))

[quasi sparrow]

another way to write this is
$10^n, 2^{n}$ for $n =0, 1, 2, ...$

[clever fjord]

riemann

[quasi sparrow]

,calc log(10, 2)

[mystic saffron]

oh wow these match

let me try to plot the numbers

[clever fjord]

Result:

3.3219280948874

[quasi sparrow]

there we go

$y = x^{log_2(10)}$

[mystic saffron]

looks good....

[quasi sparrow]

thanks for fixing my nonsense

[clever fjord]

riemann

[quasi sparrow]

this should get you far enough 3.3219280948874

but log_2(10) is exact

[mystic saffron]

this works beautifully!

i knew ethere was gonna be a clean expression

[mystic saffron]

it's all gonna be done by machine so i'M not sure how intensive a log10(x) operation is

but it shouldnt be so bad

[quasi sparrow]

Yea just get like 20 decimals and you should be good

,w log(10, 2)

[mystic saffron]

what equation are you proposing as an approximation exactly?

i'M a little confused

[forest sky]

this is equivalent to the expression i posted earlier. since log_2(10) is a constant you can use the same approximation for all inputs

[mystic saffron]

oh i see

[forest sky]

actually it should be log_10(2)

[mystic saffron]

yeah

[quasi sparrow]

oops

[mystic saffron]

they did not match lol

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[mystic saffron]

oh wow

my original x^0.30 was damn close

[quasi sparrow]

yea that's why you had to go way far out before you found discrepencies

[mystic saffron]

I only really need it to work around the 10s of millions

30-60 million max

[quasi sparrow]

yea 20 digit decimal expansion should be enough for you then

just put in comment # technically log_10(2) is exact

[mystic saffron]

zoomed in as much as i could

can tell no difference at 10 mill

especially since everything will be rounded to the nearest integer

difference between the 2 expressions is.... 0 lets just say that

this will work perfectly

thank you for your help

SOLVED!

idk how to end a help channel

[quasi sparrow]

do .close

[mystic saffron]

.close

[odd edge]

Channel closed

Closed by @quaint pollen

Use .reopen if this was a mistake.

[odd edge]

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[pallid reef]

$\angle PQA = \angle RQB = 60^{\circ}$ and $\angle RQP = 90^{\circ}$

[clever fjord]

如月あやみ　Kisaragi Ayami

[pallid reef]

Firstly, what does that tell you about $\angle AQB$?

@mystic saffron

[clever fjord]

如月あやみ　Kisaragi Ayami

[pallid reef]

How is it 180?

no, 180 is what you get when you add all three angles of a triangle together

i'm only asking you about angle AQB

what do you get when you add all the angles around the same point

yeah

can you point to the angle AQB for me on the picture?

and take a picture of that

i wanna know if you understand what i mean when I say "angle AQB"

[odd edge]

@mystic saffron Has your question been resolved?

[high sentinel]

what is angle b

what is angle abq

what is the measure of angle abq

take a picture of the whole problem

[high sentinel]

aqb = 150

360 - (90 + 60 + 60) = 150

equilateral triangles are equiangular

60 degrees each

huh

you must name angles with the middle on being vertex

measure of angle AQB = 150 degrees

oh shoot youre trying to find abq

ok so its isosceles

180-150 = 30

30/2 = 15

measure of ABQ = 15 degrees

aqb

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

Channel closed

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[low cradle]

Why are the base cases 5 and 6 for part B

[ember oak]

I answer your question with another question: What do you think it should be instead?

[low cradle]

Can’t it be anything lower

Like

3 and 4

Why would nothing else work

[brittle beacon]

What would $F_0$ and $F_{-1}$ be, given that your sequence is defined as $F_1, F_2, \ldots$?

[clever fjord]

@brittle beacon

[low cradle]

Omg yea

What if you take 4 and 5 tho?

Isn’t F(0) 1

No

Nvm

Just saw

Sequence starts at F1

Tysm u the goats

.close

[odd edge]

Channel closed

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Use .reopen if this was a mistake.

[ember oak]

classic chartbit helping style

[odd edge]

Available help channel!

Send your question here to claim the channel.

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

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

[steady tide]

what have you tried?

[versed depot]

so

i put 3x-7 for y

and i did product rule

but idk how to set it up

do i set it euqal to 0

[hallow pelican]

What is the domain of x

Is it all reals > 0

[versed depot]

sry i solved it by myself

its optimization

[hallow pelican]

O

[versed depot]

i solve for absolute minimum by second deriative test