help-10

[left mesa]

Ehy

[limber ocean]

you get 3^(y-1) equals to 10x

[left mesa]

Why

[limber ocean]

because on one side you're dividing by 5

you know that if you multiply each side of the equation by the same number, you keep the equation intact

so by getting rid of the /5 you just multiply by 5 on each side

[left mesa]

This one is a new given

[limber ocean]

they didn't teach you that?

[left mesa]

No they taught us differently

[limber ocean]

oh alright

[left mesa]

Sum different

[limber ocean]

well let's say you have 5 bananas and 10 apricots on each side of the equation

the ratio is the same if you had 1 banana and 2 apricots

it's all about ratio

if you have something on the left side, and something else on the right side of the equation

[left mesa]

Ion get it 😭

[limber ocean]

hmmm alright

let me explain it another way

[left mesa]

Can you just give me ur solution? And I'll tell you what I don't get?

For this one

[limber ocean]

2x equals 3 so x equals 1.5 right?

[limber ocean]

you're completely wrong and i'll give you the solution dw

[left mesa]

Thank you!

[limber ocean]

just understand this first cause it's a pillar of mathematics lol

[left mesa]

Aight

I feel a bit guilty for withholding you from ur breakfast tho 😭

You can eat first if u want

[limber ocean]

don't worry about it I'm happy to help

[left mesa]

Oh alright thanks!

[limber ocean]

I was making breakfast for my partner so dw 😉 I just have coffee in the morning

anyways do you see it written

[left mesa]

That's unhealthy ):

[limber ocean]

don't worry I'm very healthy I just don't get hungry in the morning unlike my partner hahah

[left mesa]

Oh that's good to hear

[limber ocean]

I'll repeat

2x equals 3 so x equals 1.5 right?

[left mesa]

Ye

[limber ocean]

you divided the equation by 2 to get that answer, correct?

same goes for your equation rn

you have 3x something

3x(exp(y-1)/5

equals to 2x

[left mesa]

No Not that given

The new given is

^3√3^y-1 = 5x+1

[limber ocean]

ah got you

so you have the right answer for the previous one correct?

[left mesa]

Ye

[left mesa]

This is my solution for the new given

The one with a square root

[limber ocean]

got u it's loading for me one sec

[left mesa]

Okii

[limber ocean]

got it

alright you're solving for y

you're wrong but let me see how to explain it

cause you crossed some work on your paper so idk haha

[left mesa]

Oki

[limber ocean]

did they teach you about the exponent rules already?

[left mesa]

Yeah

[limber ocean]

as in (x^a)^b is x^(ab)

[left mesa]

Yeah

They did

Way in junior high

Like 3^3/2 is the same as √3^3

[limber ocean]

yes so what would be ^3√ in exponent form?

wouldn't that just be 1/3?

[left mesa]

No since it's square root so it's supposed to be 2.. is what our teacher said

[limber ocean]

it's not a square root

it's a cube root

[left mesa]

Wait

[limber ocean]

just know that any square root is equivalent to exponent 1/2

[left mesa]

😭

[limber ocean]

do you have a calculator?

take square root of 16, we know it's 4

[left mesa]

Yeah

[limber ocean]

then do 16^(1/2)

isn't it 4 too?

[left mesa]

Oh

Ye

[limber ocean]

because square roots are "opposite" (much more complex than that lol) to exponents

so if you're taking something to the power of 1/2 it's like taking it's square root

if you're taking something to the power of 1/3 it's like taking it's square cubed

and on and on and on

[left mesa]

Then what should be the expo form of ^3√3^y-1 ?

[limber ocean]

it's a square cubed so..

how would you write it in an exponential form?

(3^y-1 ?)^(1/3) right?

let me use my pad one sec

[left mesa]

Oki

[limber ocean]

that's how you'd write it

[left mesa]

O

But our teacher taught us this instead wait lemme show u

[left mesa]

The second row

[limber ocean]

yep

he showed you by giving you the answer

what i'm telling you is what he wants you to do hahah

do you understand now?

[left mesa]

Oh

No I do not 😭

[limber ocean]

alright just one thing to remember for this question

step by step

if you're taking the square root of a number or an expression

doesn't matter if it's square root of 16 or square root of (3xflajfljfkezfezfnzp)

what you need to remember is that you can switch it up and change the expression in an exponential form instead

so square root of 4 is 2^(1/2)

square root of 16 is 4^(1/2)

understood?

[left mesa]

Ye

[left mesa]

So this one

[limber ocean]

so that's what we're doing here with the square cube

[left mesa]

What next?

[limber ocean]

you still have your five on the left

what would you do with it?

divide

on both sides

what are you left with?

(you gotta do it on your notebook to understand)

so now what's the solution in your opinion?

[left mesa]

O

Wait

Idk 😭

[limber ocean]

?

you just multiplied by 5 on both sides

[left mesa]

One sec

Give me about 5 minutes I just have to go out

[obtuse pebble]

@left mesa Has your question been resolved?

[obtuse pebble]

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[swift carbon]

i need to map:

1 >   1
2 >   2
3 >   1, 3
4 >   2
5 >   1, 3
6 >   2
7 >   1, 3 
.
.
.
11 >   3

how can i write this as a mathematical equation?
this is for the backrpop of a 1D cross-correlation (convolution with flipped kernel) and a stride of 2

[swift carbon]

because of the stride of 2, as u slide the kernel over the input, the input gets multiplied by 2 different weights at 2 different times, on the index 3, 5, 7, 9

i am solving for part iii) here

i think i figured it out with a reference from here:
https://medium.com/@mayank.utexas/backpropagation-for-convolution-with-strides-8137e4fc2710

things twisted

still not concrete on it tho so some help would be appreciated if possible

tbh the way its described here is also a bit difficult to write mathematically lol

[swift carbon]

would be ideal if i could come up with an expression for this

[obtuse pebble]

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[stiff stump]

When i solve this i get +1/4 not -1/4 where does it go wrong?

[marsh geyser]

Where does it says -1/4?

[stiff stump]

Here

the below expression is the solution

given

mb for not being clear bout that

[marsh geyser]

That is from the book?

[stiff stump]

It is an old exam question

given like 1 year ago

[marsh geyser]

,w lim x->0 (xe^x^2-x-x^3)/(2ln((e+x)^x^5))

[warm shale]

Results provided by WolframAlpha

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

Requested by _arkimond_

[stiff stump]

so it would most probably have been discovered

wtf

wait so the solutions are wrong

lol

[marsh geyser]

Unless i don’t see the original problem correctly yes, the solution is wrong

[stiff stump]

it says log here but that means ln right?

and even if it was would it change anything?

[marsh geyser]

.

Wolfram takes log as ln

[stiff stump]

oh ok

[marsh geyser]

Have this in mind

[stiff stump]

then this is wrong, ill email my prof, this is the 2nd thing i find "wrong" in that exam

unless he shows otherwise

[marsh geyser]

Do not solve the limit bow

[stiff stump]

bow?

[marsh geyser]

Just think about this

Now

When u use log properties

And put x^5 in front

When x<0, numerator is negative and denominator too

So the limit has to be positive

Although to be honest

U dont even need to do that

When x<0 u can see that the outcome is positive

[lone echo]

graph the function so its clearer

[stiff stump]

im trying to grasp what you're trying to communicate

[marsh geyser]

I made a bad wording

Anyways

[stiff stump]

The exam is done by hand with normal/no graphing calculator so im practicing without

[marsh geyser]

Ur solution is correct

[stiff stump]

okok is it an ok solution btw?

or would you suggest

a diff one in future?

something simpler maybe

[lone echo]

and that is perfectly fine, until you find an error. then you use everything at your disposal to check where the error comes from

it might be something subtle that you might not find otherwise

[stiff stump]

This is a fair point, i dont think ive graphed a lim of a function before though, is that just something you can chuck in desmos or something?

.close

[obtuse pebble]

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[glacial verge]

don't close this yet
i wanna explore that function

nvm

I wonder if there's a fixed point for that func

alr, I'll

infinity?

I like infinity more

[obtuse pebble]

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[slim ibex]

can someone help me

[lone echo]

,rotate

[slim ibex]

on

c

idk how to do it

[young sonnet]

@slim ibex hey what was the answer for a

[restive gorge]

g'(x) <= 0 means for which values of x is g(x) decreasing or neutral

[slim ibex]

(4,-3)

[slim ibex]

idk what that means

[restive gorge]

imagine you are on the curve

[slim ibex]

yes

[restive gorge]

for which values of x do you walk down?

from left to right

[slim ibex]

below the x axis

[restive gorge]

not quiet

if I am at x = 5 i am below the x-axis, but I am walking upwards, along the curve

[slim ibex]

between (0,5) and (4,-1.5)?

[restive gorge]

4.1 i would still be going upwards

4.2 too

[slim ibex]

but not 4.0???

[restive gorge]

exactly

imagine it's a hill

at x = 4 you are neutral

[slim ibex]

yes

[restive gorge]

not decreasing anyore

and if you continue

[slim ibex]

<4?

[restive gorge]

you are increasing meaning goinf upwards

[restive gorge]

yes exactly

also including 4

x <= 4

[slim ibex]

yes

is that it?

[restive gorge]

that's it

[slim ibex]

I got a question

follow up question

u there?

@restive gorge

[restive gorge]

bruh just ask your question what you waiting for

[slim ibex]

x <= 4 <= 0

why wont this be the final answer this “<=0”part doesn’t make sense to me

[restive gorge]

first of all 4 <= 0 is weird thing to say

[slim ibex]

ah

.solved

[obtuse pebble]

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[untold python]

need explanation on why I cant do this: if you have (2xlnx-x)/(lnx)^2. why cant I go (2xlnx-x)/lnx*lnx and cancel out the lnx in the denominator and get (2x-x)/lnx? (edited)

[restive gorge]

ln(x) is not a common factor in your numerator

[hoary flicker]

It is in (a-b)/c form but if you can get (a/c) - (b/c) form then you can do

[jovial flame]

x doesnt have lnx being multiplied to it

both need to have em

yes

exactly

[warm shale]

bacc the sigma😔🤞

[untold python]

thanks! .close

.close

[obtuse pebble]

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[warm shale]

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

[obtuse pebble]

@distant rampart Has your question been resolved?

[obtuse pebble]

@distant rampart Has your question been resolved?

[obtuse pebble]

@distant rampart Has your question been resolved?

[obtuse pebble]

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[tough flax]

Can someone pls help me solve this question pls

[tough flax]

Ik you gotta use trig

But don’t know where to begin

[obtuse pebble]

@tough flax Has your question been resolved?

[tough flax]

<@&286206848099549185>

<@&286206848099549185>

[finite storm]

Hi. Just min.

[finite storm]

And your age?

[tough flax]

i'm 14

[finite storm]

20km

[tough flax]

?

ik the final answer

i'ts 20sqr3 km

but idk how to do the working

[timid silo]

only use the ping once

[tough flax]

oh sry mb

can someone help me pls

[obtuse pebble]

@tough flax Has your question been resolved?

[obtuse pebble]

@tough flax Has your question been resolved?

[obtuse pebble]

@tough flax Has your question been resolved?

[obtuse pebble]

@tough flax Has your question been resolved?

[obtuse pebble]

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[lusty shore]

How does one solve question 24?

<@&286206848099549185>

[zenith tide]

easy

you don't.

[lusty shore]

Can you actually help or nah?

[timid silo]

@lusty shore what math is this

[strong pike]

do we assume lines are infinite?

[lusty shore]

Senior maths challenge

[lusty shore]

Yes

[obtuse pebble]

@lusty shore Has your question been resolved?

[obtuse pebble]

@lusty shore Has your question been resolved?

[lusty shore]

<@&286206848099549185>

[red ice]

http://www.alaricstephen.com/main-featured/2017/11/7/intersecting-lines

I found this online

11 is possible with a situation like this
all lines intersect at a single point and no two lines are collinear

[obtuse pebble]

@lusty shore Has your question been resolved?

[wild glade]

How is your SMC today

or is it tommorow

[shadow dagger]

is this from today's SMC?

[wild glade]

its from 2017

[shadow dagger]

ah k that's fine then

[wild glade]

didyou do SMC as well

[shadow dagger]

nah i'm too old now unfortunately

[wild glade]

I see

[shadow dagger]

i'm still involved with UKMT stuff tho

like i mark the olympiads etc.

[wild glade]

that's nice

[obtuse pebble]

@lusty shore Has your question been resolved?

[obtuse pebble]

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[atomic compass]

Number theory and logic

Why cant implication with something true lead to something false

[bright whale]

[frigid burrow]

Example?

Wait if you had an example why would you ask lol

[atomic compass]

A: "it starts to rain"
B: "I dont use my umbrella"

true A implies false B

Why cant this work

[frigid burrow]

What about proof by contradiction

[atomic compass]

Huh?

[frigid burrow]

Like in some questions we say let x = 5

And then while solving we see that x can never be 5

[fossil crag]

why can't something true lead to something false?

[atomic compass]

Yeah

[fossil crag]

first of all A => B is defined logically as (notA or B)

so if A is true

then notA is false

and if A => B is true

[frigid burrow]

Example is the proof that root 2 is irrational

[fossil crag]

the only way is for B to be true

[frigid burrow]

We assume it is rational then land at a contradiction so we conclude that it must be irrational

[obtuse pebble]

@atomic compass Has your question been resolved?

[obtuse pebble]

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[warped hearth]

Hey, I need help w/ 2 questions please i dont know how to do them both

[warped hearth]

ty :)

[trim portal]

So let's start with this one

do we want f to be large or small, if we want f/(mp) to be large?

[warped hearth]

we want f to be big

[trim portal]

correct

[oblique grotto]

Plz help me too after I beg

Bdkfkmfkf

[warped hearth]

so the hcf of 14 and 28, its 14

[trim portal]

correct

so f will be 14

[warped hearth]

please use another channel paste ur quetsion in #help-8 #help-14

[trim portal]

do we want m to be small or large?

[warped hearth]

small?

[trim portal]

correct, the smaller m is the larger will f/(mp) be

so we are looking for the least common multiple of 10 and 25

[warped hearth]

So f big, and m small

[trim portal]

yes, we want f to be big and m to be small

[warped hearth]

f = 14 m = 5

[trim portal]

because we want a large numerator and small denominator

[trim portal]

M has to be a common multiple of 10 and 25 though

[warped hearth]

isnt that a common multilpe tho

[trim portal]

5 isnt a multiple of 10, nor is it a multiple of 25

[warped hearth]

oh

[trim portal]

no, multiples of 10 are: 10, 20, 30, 40...

multiples of 25 are: 25, 50, 75, 100..

[warped hearth]

wait so what is 5 then am i thinking of lowest common factor

[trim portal]

5 is actually largest common factor

lowest common factor would be 1

[warped hearth]

oh

[trim portal]

so what is the least common multiple of 10 and 25?

[warped hearth]

uh 50?

[trim portal]

correct

so m = 50

now do we want p to be small or big?

[warped hearth]

small

[trim portal]

yes

and what is the smallest prime?

[warped hearth]

2?

[trim portal]

yep

so p = 2

[oblique grotto]

Sorryo

[warped hearth]

alright!

[oblique grotto]

Dd

Xd

[trim portal]

f = 14
m = 50
p = 2

What is f / (mp)?

[warped hearth]

this was a test question from earlier on today we had a quiz (test) and i got this wrong :(

14/100 = 0.14

[oblique grotto]

Put those number in it

And u get it

[warped hearth]

.. 14/100 = 0.14

[trim portal]

@alpine bison are you still here?

[warm shale]

wdg :)

[warped hearth]

could we go through second question please?

[alpine bison]

I am just lurking

as I am doing other things

[trim portal]

Oh, i see

[warped hearth]

its ok if u cant help w/ the 2nd question

[trim portal]

I'll have to go, i'll try asking someone in #「helpers-lounge」 if they can take over

[warped hearth]

No access?

its fine dont worry ill look for someone else, thanks a lot :)

[trim portal]

Sorry about that, bye

[warped hearth]

If anyone could help w/ these 3 questions that would b appreciated, theyre kinda quick questions, ty :))

[fossil crag]

let's try $m^{-\frac 14} = 27m^{-1}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

if it were m^(...) = constant, how would you solve for m?

[warped hearth]

I would times m by how many times the power says

like

,,m^(3) = mmm

[fossil crag]

uh ok

[warm shale]

wdg :)

[fossil crag]

but suppose you're already given the value of m^3

how do you get m itself

[warped hearth]

3rd root

[fossil crag]

cube root yes

so if you have m^a given

how do you get m

[warped hearth]

what do you mean by " m^a given "?

[fossil crag]

well just like we did

suppose you have the value of m^a

like m^a = 2

and a is some known rational number

how do we find m?

[warped hearth]

im not so sure

is it 2^1?

so m is 2

[fossil crag]

not really

if a = 4 for example then 2^4 is definitely not 2

[warped hearth]

oh

[fossil crag]

think about it this way

[warped hearth]

from m^a = 2
how do i find out the values?

like isnt there not enough info

[fossil crag]

oh there should be plenty of info

[warped hearth]

o

[fossil crag]

but the best way to see it is

a is a rational number

so we can write a = p/q

now, if we had an integer (q = 1)

you know how to solve that one

you just take the p-th root

[warped hearth]

yes

[fossil crag]

$m^p = b \Longrightarrow m = \sqrt[p]{b}$ (sometimes, provided $m$ is positive)

[warped hearth]

ok

[warm shale]

rafilou is not not born in 2003

[fossil crag]

well

if $m^{\frac pq} = b$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

then just raise everything to the power of q

$m^p = b^q$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

and just as before, we can now just take the p-th root

[warped hearth]

,, m = sqrt(b^p)

[fossil crag]

so when m^a = b with a,b known values (a rational)

we have plenty of info to solve for m

[fossil crag]

do you see how this equation

can be rewritten in our favor?

[warped hearth]

hm]

we do the 4th rot on the other side?

[fossil crag]

we shouldn't really do that

[fossil crag]

we need to obtain this form first

so powers of m all on one side

[warped hearth]

ohk

[fossil crag]

constants on the other

[warped hearth]

how do i do that w this question?

[fossil crag]

$m^{-\frac 14} = 27m^{-1}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

just divide/multiply

so that all the m's end up on one side

[warped hearth]

hm

how do i do that

like what happens to the powers if i divide 27

to bring that to the left side? if u get me

[fossil crag]

there's no need to move 27 around

the constant being on the right hand side is fine enough

what we need to do

is have ONLY constants on one side

ONLY m on the other

[warped hearth]

ok

so divide by m ?

[fossil crag]

divide by m?

[warped hearth]

likd bring the m on the lhs to the rhs?

[fossil crag]

ok so

1. you're doing it the worst way possible, because you're bringing MORE m with the constants

[warped hearth]

oh

[fossil crag]

2. it's still not clear what operation we're doing to "bring m on the rhs"

so

[fossil crag]

let's focus on this first

what is the easiest thing to do

to completely separate m from constants

[warped hearth]

😭 sorry bear with me im not v good at this im only 13, so what i understand is that I need to get it in the form
m^a = b and we need to know what a and b is but how do i do that?

[warped hearth]

like how do i do this

[fossil crag]

you can multiply/divide by anything you need

[warped hearth]

hm ok

[fossil crag]

ok so first of all

the only constant being 27

it's easy enough leaving it where it already is

so if we wanna separate 27 from the powers of m

what do we need to move to the other side?

[warped hearth]

m^-1 to the left side?

[fossil crag]

yes

so what operation does that

[warped hearth]

divide?

[fossil crag]

dividing by?

[warped hearth]

(m^-1/4) / (m^-1)

[fossil crag]

...

[warped hearth]

oops

[fossil crag]

yes

[warped hearth]

okay

[fossil crag]

$\frac{m^{-\frac 14}}{m^{-1}} = 27$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

now, you know how to simplify the LHS right?

you've done exercises like this before

[warped hearth]

(-1/4) - (-1)

[fossil crag]

yes

[warped hearth]

-1.25

I'm really sorry i gtg sleep as its late here, i will continue tmrw thankyou so much i think i get most of it now

.close

[obtuse pebble]

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[fossil crag]

Alright just be careful this isn't correct

[obtuse pebble]

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[main nexus]

task is to simplify the expression - when combining whats in the parentheses do you add or multiply the 4 and 5??

[brazen viper]

The 4 and 5 become exponents

[main nexus]

ohhhhhhh

thanks i forgot that part lol

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[paper sequoia]

i have had an had time learning Logarithmic equations an i was wondring if i could get some help on this equation:

[timid silo]

whats the base of the log

[boreal ether]

usually 10 if not written

[paper sequoia]

its 10

[timid silo]

yes i thought so
but was confirming from him maybe he forgot

[boreal ether]

👍

[timid silo]

can you do it now

@paper sequoia

[paper sequoia]

let me try

[timid silo]

b > 0

always

[paper sequoia]

i need help on 10 ^ -1

i dont understand what to do there

[timid silo]

10^-1 is 1/10

[paper sequoia]

wait so i have to 2x = 1/10 -1

2x = -9/10

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[obtuse pebble]

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[dawn falcon]

need help w b i

[north cradle]

f . g (x) is 4 -ln^3(x) right

[dawn falcon]

is it not 4 - ln(x^3) ?

[north cradle]

no

[dawn falcon]

what

or no it’s

4- (ln(x))^3 ?

why would it be only ln^3 if x is ^3

it’s 4 - (ln x)^3

[obtuse pebble]

@dawn falcon Has your question been resolved?

[obtuse pebble]

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[signal yacht]

Hello, I have a sneaky suspicion i did something illegal that will have me put in integration prison but I cannot find it ... Pls help

[signal yacht]

oop ignore the 2 infront of the answer

[unreal musk]

(As a side note, from the start, you could recognise you’re almost in the form v’/v)

[signal yacht]

oh...

true

how would you do it in that way tho? like how do you deal with the constant?

[last pilot]

constant doesn't matter

disappears upon differentiation

[unreal musk]

(Other than that, yours seems fine and is equivalent to what you’d get from the spotting v’/v form)

[signal yacht]

so ln(sinx) + C is correct??

no way

[signal yacht]

damn soptting that is so smart ngl

[unreal musk]

2ln|sin(y)| + C, but yea

[last pilot]

you get used to it

the fact that (ln v)' = v' / v is very very useful

[signal yacht]

wait but where does the 2 come from? I had the 2 initially but found a mistake in my work that made it cancel...

[signal yacht]

yeah..

[unreal musk]

Oh, I missed that, I’m blind

[unreal musk]

Yea it’s what you said here-

[signal yacht]

huh okay

this was the answer in book, not sure how to check equivilance...

omg wait i think i just saw it

[unreal musk]

(If you did it through the v’/v way, then you’d have it integrating to (1/2)ln|1 - cos(2y)| + C, and as 1 - cos(2y) is 2sin^2(y), that’s the same as (1/2)ln|2sin^2(y)| = ln|sin(y)| + ln(2)/2 + C)

[unreal musk]

Damn, just as I finished typing

[signal yacht]

wait no i did not actually see it

[unreal musk]

All the best then, so my typing didn’t go to waste

[signal yacht]

where is the ln2/2 coming from ahh

oh wait thats a constant

bruhhhhh

[unreal musk]

Yep and splitting the log up, as long as your answer differs from theirs by a constant, you’re fine

(“Absorb” any additional constants into your )

[signal yacht]

ok

thanks a lot!

.close

[obtuse pebble]

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[obtuse pebble]

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[foggy egret]

i know how to approach this question once i know what the sequences are however im struggling to find the sequences

[visual kernel]

Is the sequence not just $a_n = 18n-9$?

[warm shale]

NeptuneRises

[foggy egret]

yes this one is

but we need other ones

[vale pelican]

you can do it the stupid way and brute force it with code at least lmao

[foggy egret]

could do

take too long though

[vale pelican]

there's only 5 common ratios that work

sorry, 4

[foggy egret]

how do you know

[vale pelican]

because I brute forced it lol

[foggy egret]

oh ok

did you write a code?

[vale pelican]

yeah

[foggy egret]

can you add a part which states the starting value and increasing amount for each on that works

[vale pelican]

you could, yeah

the increasing amount is what I incorrectly called the common ratio

there's only four candidates for these common differences

[foggy egret]

what are they?

[vale pelican]

you could probably figure it out yourself with some exploration

and idk what the purpose of this exercise is

you could always code it yourself

[foggy egret]

+close

.close

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[wanton dock]

i dont even know how id begin, im so bad a mechanics😭

[obtuse pebble]

@wanton dock Has your question been resolved?

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[wanton dock]

.reopen

[obtuse pebble]

✅

[wanton dock]

no it hasnt😭

okay nvm ill js kms

.close

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[lost timber]

Hello, I have a few tough problems that I do not know how to get through! Could someone help me out through voice? It's one thing to post it on here and get an answer but I don't believe I will fully understand the concepts that way. I understand if you cannot, but if you can please DM me and I will invite you to my server.

[idle atlas]

You should probably post your questions

[tardy epoch]

just post one problem at a time with your attempt

[lost timber]

I understand how to factor the first problem

i believe it comes out to (X+2)(X-2)(X^2+4)

that being said the second problem (b.) I'm having trouble with because I'm not sure how to divide it properly

[tardy epoch]

,tex .exp rules

[warm shale]

riemann

[lost timber]

mmm

That definitely helps a lot'

now it just takes brute forcing it i imagine

is 1/8 x^-5 y^2 the same as (1/8)(1/x^5)(y^2)?

or does it have to be (1/8) (1/x^5 y^2)

[tardy epoch]

yes

[lost timber]

first or second?

is xy one unit?

[tardy epoch]

i only replied to one

[lost timber]

ahh

thank you

[obtuse pebble]

@lost timber Has your question been resolved?

[lost timber]

no

well yes but i'm still struggling on the problem

i dont know where to go for the next step

[tardy epoch]

why

[tardy epoch]

this is all you need so try it out

[lost timber]

im slowly attempting to

oh sorry

[obtuse pebble]

Channel closed

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[lost timber]

.reopen

[obtuse pebble]

✅

[lost timber]

i'm about to try prime factorizing the problem. I understand that the first side of the equation 1/8(x^-5)(y^2) looks like (1/2)(1/2)(1/2)(1/x)(1/x)(1/x)(1/x)(1/x)(y)(y) BUT

does the second part - (1/2)(x^3)(1/y^-2) look like (-1)(1/2)(x)(x)(x)(1/y)(1/y)?

i'm struggling to see how to apply the negative here

[wild glade]

the negative indicates reciprocal

so 1/y ^-2 will be the same as y^2

[lost timber]

wait so when i subtract a fraction its the same as adding the reciprocal?

[wild glade]

wait

[lost timber]

working on b

oh

can i not rewrite it to switrch the two sides arounbd?

that negative cant turn into a minus sign?

instead its plus (-1/2)

[wild glade]

yes

[lost timber]

mmm

so the full factored problem would look like

[wild glade]

wait as in 1/8 x^-5 y^2 - 1/2 x^3 y^-2?

[lost timber]

yes

[wild glade]

yes you can do that

[lost timber]

i tried to rewrite it to that

okay

so now how does the minus being there affect the right side of the equation

effect*

can i factor that to (-1)(1/2)(x)(x)(x)(1/y)(1/y)?

[wild glade]

yes

[lost timber]

okay

thank you

i'm still working on the problem though

so please keep this chat open

[wild glade]

yea alright

[lost timber]

is it possible to rewrite the right side as (2)(1/x)(1/x)(1/x)(y)(y) as well?

since i have to subtract it

[wild glade]

no because 1/2 x^3 is (x^3)/2 and the y bit it is to the power of -2, so it would become (1/y)(1/y)

sorry I might have confused you but the negative power -2 is different from the minus in -(1/2)

[lost timber]

not understanding what youre saying but that being said i cannot rewrite it

so it has to be this

[lost timber]

^

[wild glade]

yes

[obtuse pebble]

@lost timber Has your question been resolved?

[lost timber]

no

I came up with the answer (1/2 x^-5 y^-2) (1/4 y^4 - x^8) is that correct?

[wild glade]

where did the y^4 come from

[lost timber]

if you multiply exponents do you not get the addition of them?

[wild glade]

but where do you multiply the exponents

[lost timber]

so i figured (y^4)(y^-2) would give me y^2)

[wild glade]

yes but its addition here in the question

[lost timber]

yeah but im factoring no?

so here i went from (x^-5)(1/8y^2 - 1/2x^8 y^-2)

to

[wild glade]

oh yea I see it now

now you just need to group it up

[wild glade]

btw you can still factor out the last brackets
notice both the powers in the last bracket are even

[lost timber]

does that look correct?

or like

[wild glade]

yes but you will need to completely factor it

[lost timber]

mmm

so 1/4y^4-x^8 factors out to (1/2y^2 - x^6)^2

?

[wild glade]

its a difference of squares

[lost timber]

oh

aight

sec

so (1/2y^2 +x^4)(1/2h^2-x^4)?

[wild glade]

yes

[lost timber]

aaaah

i see

[wild glade]

yea and I think this is optional but you could combine the x^-5 and 1 together in (1/2 x^-5 y^-2) cuz its just multiplying by 1

[lost timber]

mmm

idk

[wild glade]

but I don't think that is needed
at least not when I did it

[lost timber]

rn what im getting is (1/2x^-5 y^-2) (1/2y^2 + x^4) (1/2y^2 - x^4)

[wild glade]

yea that's correct

[lost timber]

god bless it only took 2 hours

[wild glade]

: ) at least it's now done

[lost timber]

true

[wild glade]

👍

[lost timber]

i now have 4 more questions to answer

problesm

[wild glade]

that's rough

[lost timber]

it happens

i had like a week or two toi do it

proscrastination

procrastination*

[wild glade]

lol

I used to do that back in secondary

I still do it now but the amount of stuff to do is like triple the amount so I'm more screwed when that happens

whenever I see work I just can't be bothered

[lost timber]

absolutely

i feel that

speaking of which the next problem i have to figure out is this

i figured out I. a, II. a, and II. b, but now i'm working on I. b

i'm in cal 1 and the last time i took a math class was 4 years ago

so i'm struggling to remember all the basics of math

[wild glade]

when a number N is a root of an equation, you subsitute N into the unknown and it should be equal to 0

in this case N = -2 and -1

[lost timber]

so i just put n as x?

[wild glade]

yes

no wait

sorry Im tweaking out

if it has a root on [x,y] I think that means the solution of the equation is between the two values when you put the numbers in

so if you put -2 as x and then -1 as x one of your answers should have different signs as it passes through 0 ( the root)

[lost timber]

see i feel like that cant be the answer

or at least not what my teacher is looking for

if i plug in -2 i get -3 and if i plug in -1 i get 4. while in theory that makes sense what if it does something strange around 0

like curves in a wierd mannor?

[wild glade]

if you plug in zero you should get another change in sign as there is another solution that lies between -1 and 0

as the curve changes direction

[lost timber]

hmm?

if i plug in zero i get -4

-5

but how does that prove anything?

[wild glade]

if it change signs that shows that there is a zero between the two values as the line crosses through the x-axis

[lost timber]

aliright...

i can only hope thats what my teacher is asking for

[wild glade]

sorry Im not the best at explaning

[lost timber]

like

i get what you're saying

theoretically

if it passes from positive to negative then it has to pass zero

however i feel like my teacher is the type of guy to draw a graph like this

<-------o (nothing on zero) o-------------->

but yeahj

i might just be over thinking it

trying to process relearning the entirety of my previous math experience in a week is a little much

[wild glade]

I see

[lost timber]

alright now i just have to figure out III. a and III. b

can you show me the formula for average rate of change?

what is the difference between III. a.and III. b? are they not both equal to 4?

[lost timber]

just to be clear

[obtuse pebble]

@lost timber Has your question been resolved?

[obtuse pebble]

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[fallow tusk]

so for part b

I saw another problem where t was positive, and they said it passes through the equilibrium at that t value

so is the rule just that t has to be positive to pass the equilibrium?

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[modern bison]

Hello

[modern bison]

Want to learn about matrix from basic to advance but I don't have book

Can you recommend me good YouTube channel to learn from?

[lime wolf]

I like the Bright side of Math

there's always three blue one brown

he goes more into the intuition on the subjects

[lime wolf]

while this guy really gives you all the formulas you'll need to do stuff

[modern bison]

Does it teach from basic to advance properly?

[modern bison]

What kind of stuff?

[lime wolf]

well depending on how advance you wanna get

it gives you a solid base to learn more stuff

[void holly]

Khan academy, the organic chemistry tutor works

[lime wolf]

Oh yeah those too

Khan academy also offers free practice courses

[modern bison]

Okay thank you very much guys

!close

[lime wolf]

it's with a .

[modern bison]

.close

[obtuse pebble]

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[lime wolf]

there you go

Good luck!

[obtuse pebble]

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[void phoenix]

Can someone explain Limits, Derivatives and Integrals

[noble needle]

Oh yea id like to learn as well

[tardy epoch]

https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity

[tardy epoch]

https://www.khanacademy.org/math/calculus-1/cs1-derivatives-definition-and-basic-rules/cs1-average-vs-instantaneous-rate-of-change/e/secants-and-average-rate-of-change

[tardy epoch]

https://www.khanacademy.org/math/calculus-1/cs1-integrals/cs1-accumulations-of-change-introduction/e/area-under-a-rate-function-equals-net-change

[obtuse pebble]

@void phoenix Has your question been resolved?

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[teal turret]

show your work @high pawn

[obtuse pebble]

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[fathom kindle]

i think im tired and sleepy

but

why does it equal

=
0.56x

(im going crazy)

oh

HEKPL

AHAH

NVM

WHAT

BYE

[wary badger]

how about 💀

[fathom kindle]

1 - 0.44

[wary badger]

😹😹😹

[fathom kindle]

ya

i told u

[wary badger]

yes

[fathom kindle]

im going crazy.

too much studying

[wary badger]

sleep

[fathom kindle]

[wary badger]

🤝🏻

[fathom kindle]

ughhh

thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[steel goblet]

This is the formula im given to calculate the z score, where δ is the standard deviation of the mean. But Isn't this a t-test, not a z score?

[steel goblet]

because I am using this formula on 2 different datasets that appear to be consistent with each other, but this z score from the formula is very high (larger than 11), which makes me think that something is wrong with the formula

please @ me if anyone responds

[obtuse pebble]

@steel goblet Has your question been resolved?

[steel goblet]

<@&286206848099549185>

[gaunt thicket]

guys can you guys help with 3rd grade math

[tardy epoch]

#❓how-to-get-help

[obtuse pebble]

@steel goblet Has your question been resolved?

[obtuse pebble]

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[fathom schooner]

it’s ok

[still swift]

hi

[fathom schooner]

Hi

I only need help with aii

[still swift]

nvm its not secondary math

i’ll call the experts

[fathom schooner]

I’ve done the induction

Thanks

[still swift]

<@&286206848099549185>

np

[steep moth]

induction is wilding 😭

[still swift]

what the heck is an induction😭😭

[obtuse pebble]

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

[still swift]

ok

[fathom schooner]

The induction is fine it’s just part ii

[ionic oar]

would you not just do the sum form of r^n where a = r + 1

or does that not work

[fathom schooner]

Does not work @ionic oar

This is the answer that I do not understand

[ionic oar]

right mb

[fathom schooner]

But I don’t get the first line 😭

[ionic oar]

for the first line i think they meant to put T_r

we're just removing the first 2 terms since we can express T_r when r >=3

[ionic oar]

a stove

[still swift]

good to know

[ionic oar]

nah but fr induction is basically assuming a mathematical statement is true, then testing it for the next integer and proving its true

[fathom schooner]

@ionic oar does this answer make sense to u?

Obviously it not finished u jsut sub in from i) and get the final answer

[ionic oar]

yes

[ionic oar]

after this just substitute T_1 and T_2

and T_(r-1) from what you found

[obtuse pebble]

@fathom schooner Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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

[silk willow]

is there another formula i could use here?

.7

[fathom light]

No

You need to find all roots

Then sum = 20/3

It is straightforward, where are you getting confused?

[silk willow]

blocks of text kinda mess with my head

but i get it enough to repeat js wanted to know if there was a way to solve this without that version of the quadratic formula

.close

[obtuse pebble]

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[vapid tiger]

need help with this

for S1 i set z = k and my solution set is
(3i - (1-i)k, i-1+k, k)

for S2 i got y=j , [ (4-(2-8i)k)/2i, k]

S1 intersect S2 idk

[chrome compass]

oh you're basically there mate

[vapid tiger]

idk if the calculations are right

[chrome compass]

ah XD

[vapid tiger]

hopefully they are

[chrome compass]

i didn't check that but lms

[vapid tiger]

ight

[chrome compass]

might want to simplify that S2

the y-coord

get that i out of denom

[vapid tiger]

2i ?

[chrome compass]

yeye multiply by the conjugate

[vapid tiger]

x = 4-(2-8i)j/2i

so i multiply -2i both sides ?

[chrome compass]

nah, complex number arithmetic. multiply numerator and denominator by -2i

[vapid tiger]

can i just do that on the right side ?

ignoring x

[chrome compass]

if x = a/b, does x=ac/bc ?

[vapid tiger]

ig cx would be equal to ac/bc

oh wait nvrmind

-2i/-2i is basically 1

so we are multiplying by 1

[chrome compass]

mmm yeah okay these sets are a bit off

[vapid tiger]

shouldnt change anything

dang

[chrome compass]

right

um but okay let's work this

For $S_1$, if $z=k$ then $ix=3-(1+i)k$. Multiply both sides by -i to get $x=-3i+(i-1)k$.

[warm shale]

ENStucky

[chrome compass]

Ah okay, that wasn't quite as bad as i thought, just a sign error

[vapid tiger]

is S2 = [(i+4)j -2i, j]

[chrome compass]

No, S2 needs two variables

[vapid tiger]

y=j

and x is the first one

[chrome compass]

there's still a z component