help-17

[night oriole]

i see

[odd dawn]

yeah

i don't think you can brute force your way

[night oriole]

hmm

well

we could take away the cube on each of them

by doing 2017/10
instead of 2017^3/10

[odd dawn]

can you actually do that?

[night oriole]

no idea :D

i think i have failed

and i have no clue how to solve that

[odd dawn]

cuz like for example
3^3 / 10^3 = 27/1000
and
3/10 = 3/10

[night oriole]

without making a function whichll take an hour if i try to do it

[odd dawn]

it's not the same

[odd dawn]

wait how old are you

[night oriole]

four

[odd dawn]

nah fr

[night oriole]

...

[odd dawn]

[night oriole]

fine ive been caught

five

[odd dawn]

[night oriole]

🗿

[odd dawn]

ain't no way

[night oriole]

y

[odd dawn]

ain't no wayyy

[night oriole]

why

are you being agiest

[odd dawn]

nah

[night oriole]

i can determine my own age....

[odd dawn]

br

i'll just open a new channel 🗿

.close

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[trim acorn]

Hi I have no idea how to figure any of the ones past E out. I know I got E right Cus my teacher said so but idk the rest

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[dense inlet]

In a rising geometric sequence with n indices, the sum of the initial n-2 indices is 9 times smaller than the last n-2 indices. find the ratio of the sequence.

[dense inlet]

I have tried everything, I just can't seem to get it right.

[flat whale]

do you have the exact problem

[dense inlet]

yes but I've translated it, in my own words.

let me try to be more exact,
in a geometric sequence a1,a2,a3,....an, the sum of the first n-2 indices are 9 times smaller than the sum of the last n-2 indices. find the ratio of the sequence

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@dense inlet Has your question been resolved?

[dense inlet]

<@&286206848099549185>

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@dense inlet Has your question been resolved?

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[wild oar]

I have to calculate the colored area by means of the algebraic sum of definite integrals, but I don't know how to set it

[violet sonnet]

guys I need help with a question. I have a test tomorrow

the guy whose video I was watching just concluded from here that h^2=12*6

how can we conclude that h^2=12*6

[silk hollow]

#❓how-to-get-help

[silk hollow]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
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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

[rugged vortex]

Are you familiar with the area between two curves?

[wild oar]

I know I have to calculate the difference between the areas under the curves

[silk hollow]

you basically do integration with upper curve - lower curve

,w Integrate[1-x^2-x^2+3,{x,-2^0.5,2^0.5}]

[twin meteor]

Results provided by WolframAlpha

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Requested by shik#9660

[silk hollow]

this is random but it doesnt have to be upper curve - lower curve, just minus one equation from another and if you get a result with negative, change to positive

[wild oar]

And what about the intervals [- square root 3 ; -1 ] and [ 1 ; square root 3]? Do they cancel each other out?

[silk hollow]

oh, are you talking abt when y = x^2 - 3 is at x-axis?

[wild oar]

My teacher didn’t explain this, thank you for it

[wild oar]

Yes

[silk hollow]

it depends on the area you are finding, but the bounds that matter are where the two curves meet

[wild oar]

Ok

[silk hollow]

so in this case -sqrt2 and sqrt2

<@&268886789983436800>

It seems that @void relic is spamming "None" in every channel

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[peak shard]

can I also ask here if calculated something correct?

[hard atlas]

just ask

although depending on what it is you could also just check an online calc

like wolframalpha

[peak shard]

$\sum_{n=1}^\infty \frac{n^2}{n!}x^n = e^x(x+x^2)$

[twin meteor]

tobi.

[peak shard]

I proved that the Series is absolutely convergent for all x before that so I am allowed to split up like I did

I had the idea by myself so I would really like for this to be correct

[boreal remnant]

Uh let's see

f(x) = e^x - 1 = Σ x^n/n!
f'(x)= e^x = Σ n/n! x^(n-1)
xf'(x) = xe^x = Σ n/n! x^(n)
(xf'(x))' = e^x + xe^x = Σ n^2/n! x^(n-1)
x(xf'(x))' = xe^x + x²e^x = Σ n^2/n! x^(n)

Looks good!

All sums are from 1 to infinity

@peak shard

[peak shard]

thats cool!

how did you come up with that idea

[boreal remnant]

Yeah, in general for sums like these you want to start with a known power series

Then just multiply by x and take derivatives and such

Notice how we get that pattern?

x(xf'(x))'

We're basically just manipulating the series on both sides so that we arrive at the desired result

[peak shard]

but that only works when I know what the series is in the first place right?

or what I want

[boreal remnant]

No

This works whenever you have a known series that looks similar enough to the desired series

Lets say you wanted to find:

Σ n^99/n! x^n

You could use that exact same process

All the factors of n just come from taking the derivatives and multiplying by x

[peak shard]

yeah thanks a lot doing it like that is a lot cleaner than what I did

.close

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[rugged vortex]

Well this one just looks like total ass

Alright well let's break it into parts

The arccos(1 - {x})

I haven't seen {x} in a while lol

Do you know the domain of {x}?

[vast shale]

yes probably 0 less than equal to fractional part of x greater than 1

[rugged vortex]

That doesn't answer my question

What's the domain of {x}?

[vast shale]

in this question ?

or in general

[rugged vortex]

Answer this

[vast shale]

between 0 to 1

excluding 1

all real numbers - integers

[rugged vortex]

That's its range...

I want the domain

[vast shale]

all real numbers - integers

[rugged vortex]

That's not the domain

[vast shale]

🥹

[rugged vortex]

Didn't I just tell you what a domain was

[vast shale]

yes

[rugged vortex]

What's the set of all possible inputs of {x}

I want to know what you can and cannot put into {x}, not what it does

[vast shale]

we can put all the decimal values in {x} and can't put integers

or simply integers with decimal values

[rugged vortex]

Are you sure we cannot put in integers?

You're telling me that 5 - floor(5) doesn't exist?

Hm I'd review your functions then

[vast shale]

yes if we put {3.2} our teacher told 3 get out side the fractional part

[rugged vortex]

3.2 - floor(3.2) = 0.2

But if you put in an integer

5 - floor(5) = 5 - 5 = 0

You can input any real number into {x}

I'd review your functions

[vast shale]

ok

[rugged vortex]

Then analyze arccos(1 - {x})

Do you know how to find the domain of arccos(1-{x})?

[vast shale]

yes

we would put it like 1-x greater than equal to -1 to less than equal to 1

right ?

[rugged vortex]

Yes but you can get away with analyses

I would just do an analysis rather than algebra for this

Well I guess you can do algebra

But I'd rather do analysis

Do you know the domain of arccos(x)?

[vast shale]

-1 to 1

[rugged vortex]

And you already gave the range of {x}, which is [0,1)

[vast shale]

Yes

[rugged vortex]

That means that the range of 1 - {x} is (0, 1] right

[vast shale]

Yes

[rugged vortex]

That means that 1 - {x} stays within [-1, 1]

So what's the domain of arccos(1 - {x})

[vast shale]

[-1,1]

[rugged vortex]

No

[vast shale]

Oh yeah sorry

The domain will be between (0,1]

[rugged vortex]

No

[vast shale]

As (0,1] lies between [-1,1]

[vast shale]

ok

Pls give me time to solve and think

[rugged vortex]

Sure

[vast shale]

is the answer 0 ?

@rugged vortex

[rugged vortex]

For the domain of arccos(1 - {x})?

No

[rugged vortex]

Think about this

Recall that the domain of {x} is all real values of x

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

i dont think i can solve this

pls help

[rugged vortex]

The domain of $\arccos(a)$ is $a \in [-1, 1]$. The range of $1 - \setb{x}$ is $x \in (0, 1]$.
$$\text{}$$

This means for all values of $x$, $0 < 1 - \setb{x} < 1$ which is a subset of $[-1,1]$.

[twin meteor]

Umbraleviathan

[vast shale]

so the final answer is -1 to 1 ?

@rugged vortex ?

but i had also given u the answer -1 to 1

[rugged vortex]

no

...

My guy the range of 1 - {x} only kicks out values between (0,1], which is a subset of [-1,1]

Which means FOR ALL VALUES OF X, it will NEVER make arccos(1-{x}) undefined

Have you ever done composite functions before?

[vast shale]

nope

so the final answer was phie

[rugged vortex]

#help-7｜zen1thxyz message read this. Get used to it

Shit

Wrong link

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:composite

There

[vast shale]

ok sir thanks !!!

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[cold glen]

can someone help with this pls lol

[vocal sleet]

Please don't occupy multiple help channels.

[cold glen]

someone pls help with tihs question

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@cold glen Has your question been resolved?

[dull bear]

.close

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[dull bear]

.reopen

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✅

[dull bear]

For that I assume that they want you to use the statement here

And then note of course the gradients of perpendicular lines and what the tangent's gradient would be

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[rugged orchid]

im having a bit of a hard time understanding what it means by "large" or "small" values of \tau

[rugged orchid]

suppose xbar = µ (H_0 is true), then we'd expect \tau to be close to 0 right?

so when it says smaller does it mean close to 0 or close to -inf

[pallid zenith]

im guessing its a poor way of saying "if xbar and mu are more distant"

im thinking like radius on a real number line

[rugged orchid]

hmm

and if n was small (not many samples), then you'd get $\frac{(\bar{x}-\mu)\sqrt{n}}{\sigma}$

[twin meteor]

Frosst

[rugged orchid]

which means less samples exaggerate the top more

suppose xbar = µ again, small or large sample doesn't matter

suppose xbar ≠ µ, small samples exaggerate it less

so if they were different you'd expect \tau to increase as n increases?

whereas if they were the same you'd expect \tau to not really change? (or change less?)

[vocal sleet]

@rugged orchid Has your question been resolved?

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@rugged orchid Has your question been resolved?

[vocal sleet]

@rugged orchid Has your question been resolved?

[cobalt crypt]

what's still troubling you

[rugged orchid]

i wonder how big is "big" for our test statistic here

i assume close to 0 is good but how far is bad

[cobalt crypt]

for tau?

[rugged orchid]

yeah

[cobalt crypt]

that's up to the significance level you decide on ig

you calculate the p value associated with your test statistic and decide based on whether it's smaller or larger than the significance level

[rugged orchid]

right, and the pvalue i calculate for my test statistic depends on the distribution of my test statistic

[cobalt crypt]

yes

i assume it's normal here

[rugged orchid]

so here we take p value from a normal dist because sample mean is distributed normally

[cobalt crypt]

yeah

[rugged orchid]

but if it were something else then p value will be of another dist

[cobalt crypt]

yep

[rugged orchid]

whatever the test statistic's sampling dist is

right, that is pretty cool

[cobalt crypt]

possibly t-dist or chi^2 or F etc

[rugged orchid]

right right

[cobalt crypt]

but in this case since you standardise the statistic you can also check how many standard deviations it is away from 0

well it's the same thing but like

it allows you to eyeball it

[rugged orchid]

wdym same thing

[cobalt crypt]

as in like

calculating a p value for a normally distributed statistic is basically just looking at how many standard deviations it is away from the mean

but there are tables for these things

which you can look up to kinda eyeball the p-value

[vocal sleet]

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[hushed atlas]

any hints for part b? found f(X) and factorised but not sure what to do w that

[hushed atlas]

nvm

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[keen sequoia]

i dont how how to use the semicircle and the washer formula at the same time

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[lost geyser]

How can this be solved?

[lost geyser]

firstly multiplying the parenthesis by x and then looking at the method of the photo?

[hushed pewter]

Multiply by $\frac{1/x^2}{1/x^2}$ then lhoptial

[fervent heart]

Hello

[twin meteor]

SWR

[lost geyser]

the exercise means that you have to prove that f(x) is an infinite when x->inf

[wraith venture]

You learned the linear approximation for sqrt(1+x) no ?

[fervent heart]

Nope

[wraith venture]

You're not the helpee

[lost geyser]

can u remember me?

[wraith venture]

For now

[lost geyser]

by the name i dont know what is it

[dull bear]

Infinity as in either ±♾️ right?

[wraith venture]

#help-32 message
makes it possible to find the one for sqrt

[dull bear]

You could also multiply by $\frac{\sqrt{x^{2} - x + 1} + x }{\sqrt{x^{2} - x + 1} + x }$ then do algebra

[twin meteor]

@dull bear

[wraith venture]

Why do that when Taylor
Is how I always think

[lost geyser]

but here x tends to inf, so i can't apply this

[wraith venture]

1/x goes to 0 though

Factor x out of the complicated expression

[vocal sleet]

@lost geyser Has your question been resolved?

[lost geyser]

is this well done?

[wraith venture]

surely a mistake somewhere

you end up with an equivalent to -x^3

[wraith venture]

this is equivalent to -x/2 I think

[lost geyser]

why?

[wraith venture]

Taylor

[lost geyser]

well i don't know that much

[wraith venture]

x²(sqrt(1-1/x + 1/x²) - 1) = x²(1/2 (-1/x+1/x²) + o(1/x)) ~ -x/2

[wraith venture]

hopefully this makes my tastes clear

[lost geyser]

and what i have done is not correct?

[wraith venture]

I need to go right now, but the end result feels wrong

can't proofread all that to tell where the mistake is

[lost geyser]

ok, i will ask to my teacher how to do it then

[wraith venture]

your method probably works, just don't make a mistake

[wraith venture]

also, Taylor > algebra

[lost geyser]

ok, thanks for the help

.close

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[ocean marlin]

No clue how to start

[rugged orchid]

so you have that curve

you have the power to change k (move the curve up or down)

such that the point of y-intercept is k

you can move it to the 4 following spots

which would would give you 3 real solutions?

(imagine sliding the curve up and down)

[ocean marlin]

2?

because it y is at 2 it would curved down and back up into the x axis along with the other solution at ~2,0 so it would be 3 solutions?

crrect

[rugged orchid]

kinda

but do you think at k = 2 that it will go down far enough on the positive side?

[ocean marlin]

oh

no

cuz its up 2 but the curve only goes down 1

zero?

oh

wait

[burnt locust]

look at the graph

[ocean marlin]

y=k is just a striaght line

i just need a line horizonalitty that hits 3 real soltion

so -3

[burnt locust]

yes

[ocean marlin]

oh ok

thx

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[ocean marlin]

need walkthrough

[ocean marlin]

pls ping when help arrvies

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[ocean marlin]

.reopen

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✅

[ocean marlin]

help?

assign needed

<@&286206848099549185> Know u guys r busy but i kinda need the help rn

[vast shale]

use trig

[lone quest]

draw a line from A to B so that it forms 90° angle
then you have the lengh of the base of the trianle formed which is square root of 3
and height which is 1

[vocal sleet]

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

[lone quest]

which trig function can you use in this case?

[vocal sleet]

@ocean marlin Has your question been resolved?

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[gaunt panther]

need some help on part c plz

[vocal sleet]

@gaunt panther Has your question been resolved?

[gaunt panther]

<@&286206848099549185>

[upper iron]

Because sinh²(x) is an even function

Only the coefficients of even powers should be present in the Taylor series expansion of sinh²(x)

[gaunt panther]

ok kl

and what about part d

[vast shale]

use the cosh double angle formula then do an inductive argument

[gaunt panther]

cool cool

btw any chance i could show u a quick matrix invariance question

if ur available to see

part ii idk what to do now

[vast shale]

matrix invariants = eigenvectors?

[gaunt panther]

the MS says to do this

idk how they find m and c tho

ive just done what the MS says to do

[vast shale]

it says to "verify"

which means substituting in m=2 and c=-3 is sufficient

[gaunt panther]

oh ok

[vast shale]

if you had to show it from scratch, equate coefficients

group all the x terms to get -2m^2+m+6=0

and all the constant terms (-2mc+4c)=0

[gaunt panther]

that gives m = 2

but if u sub that into the constant terms

youd get that the c's cancel

but shouldn't c = -3

since its y = 2x- 3

[jolly shadow]

@gaunt panther you still freaking doing this Hahaha

[gaunt panther]

HARPER

NO WAY

HOW R U

HOLY SHITE

i havent seen u in so long

[jolly shadow]

Because comp off spec was so nasty i removed myself from the server to preserve my dignity and wellbeing :D

[gaunt panther]

ah ok

anyway how u been

[jolly shadow]

im good

[gaunt panther]

its been so long even in this server

[jolly shadow]

chilling still

[gaunt panther]

havent seen u in a while

good good

[jolly shadow]

yea i was off discord a while too

needed that

[gaunt panther]

we all need a break from it tbh

[jolly shadow]

much better now

[gaunt panther]

good glad ur well

👍

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[hasty pulsar]

!status

[vocal sleet]

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

[civic olive]

Between 1 and 2. I know what substitution is and with easy functions, I know how it works, but I suck with sin & cos

[echo badger]

$u = cos(x)$

[twin meteor]

Rub05

[civic olive]

Doesn't seem right to me...

[hasty pulsar]

no

whatsthe derivative of cos(x)

[civic olive]

• sin(x)

[hasty pulsar]

so sin(x) gets cancelled by -sin(x)

you are left with - u² du

[civic olive]

Oh, damn. Forgot the derivative with dx = ...
right? that was my mistake?

[hasty pulsar]

your dx is just wrong

[civic olive]

yeah, okay thanks a lot!

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[hollow moss]

Is this correct?

[gray yarrow]

the graph looks correct

[hollow moss]

But it's only intersection only?

Or it doesn't matter

They included the whole A alongside the intersection

[gray yarrow]

well, we have an intersection of two sets. and both sets contain A
thus the intersection will contain A

[hollow moss]

Oh

[hollow moss]

But is it optional to include it or not

Or a must

[bronze lily]

hi can you help me with a sets question?

[gray yarrow]

wdym, optional
we can't just decide to ignore A

[bronze lily]

it’s y z and the other alphabet inside it

[hollow moss]

Oh I see

[bronze lily]

i don’t think it contains a tho there’s an intersection

[gray yarrow]

intersection of two sets
and both contain A

[bronze lily]

oh shit mb i saw it as b union c

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[desert pier]

why is it false?

i think it's a textbook error

[lapis marten]

I think the limit could be 1 as well

So it's not necessarily true

[median harbor]

? why

it said f(x) > 1

[lapis marten]

Take the function y = {x^2 + 1 for all x != 0 and 2 for x = 0}

for example

[lapis marten]

Yes

[desert pier]

The limit seems to be greater than one

[lapis marten]

What is the limit of x^2 + 1 as x approaches 0

[desert pier]

One

[lapis marten]

Yeah and the limit of this function? With f(0) = 2

[desert pier]

https://cdn.discordapp.com/attachments/903463353417072641/1104419224660738160/Screenshot_2023-05-07_004227.png

It seems to make sense with the textbook

[lapis marten]

Yes, false is correct

[median harbor]

oh damn

that is so tricky

wait

[median harbor]

what

[lapis marten]

It's 1 since the limit doesn't depend on the value of f(0)

[median harbor]

and 1 isnt >1

[lapis marten]

Yes

[median harbor]

AHA

WOW

that is really tricky

[vocal sleet]

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[scarlet bough]

How would I solve this question?

[viral copper]

Integrate

And plug in that point to solve for the mystery constant

[scarlet bough]

So I got 10x^3/2 - 80x^0.5 then substituted 4 into x and I got -80, is that correct?

[viral copper]

,w integrate 15*sqrt(x^3) - 40/sqrt(x)

[twin meteor]

Results provided by WolframAlpha

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Requested by NEONPerseus#2128

[viral copper]

Your first term seems to be incorrect

[scarlet bough]

Forgot about the x next the 15

[scarlet bough]

Got 15x^5/2 - 80x^0.5 which got me 320

[viral copper]

Should be 6 not 15

15*2/5

[scarlet bough]

Ye so now I got 32

[viral copper]

For your value of c?

[scarlet bough]

When I submitted 4 in

[viral copper]

,calc 32*6 - 160

[twin meteor]

Result:

32

[viral copper]

Alright

19 = 32 + c

You need to find the value of the constant to get the equation for h(x)

Remember this whole thing was equal to h(x)

[scarlet bough]

-13?

[viral copper]

Yes

So whats the final equation for h(x)

[scarlet bough]

19=32-13?

[viral copper]

Uh

h(x)

What is it

h(x) is a function

[scarlet bough]

6x^5/2 -80x^1/2?

[viral copper]

-13

We calculated the constant

[scarlet bough]

Ye forgot

[viral copper]

Yup so h(x) = 6x^5/2 - 80x^1/2 - 13

[scarlet bough]

Thanks

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[violet adder]

i dont know how to calculate this plz help

[violet adder]

Two loads of 1.5 kW with cos φ = 0.78 and 1.8 kW with cos φ = 0.72 respectively must be connected to a separate single-phase 50 Hz, 230 V switchboard. The central unit is now phase compensated with a common capacitor (parallel) which has a capacitance of 50 mF.

a. Calculate the main line current.

b. Determine the power factor.

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[solemn kelp]

I'm confused as to where the 2 in the numerator went on the second line

Wouldn't it be multiplied by ln(2x-5)?

wait nvm im an idiot

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[worthy citrus]

what happens if you differentiate ln(2x-5)?

[vocal sleet]

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[cunning saddle]

?

[cunning saddle]

how i can solve that

[runic marten]

chain rule

[cunning saddle]

how ? log(u) , u =5inx?

[runic marten]

yeah

[cunning saddle]

I don't understand, log u becomes 1/in10

[hushed pewter]

ln, not in

$\ln{x} :=\log_e{x}$

[twin meteor]

SWR

[cunning saddle]

how you can derivade logu

[hushed pewter]

Do you know derivative of ln x?

[cunning saddle]

1/x

[hushed pewter]

yup

Then use this property: $\log_a{x}=\frac{\ln{x}}{\ln{a}}$

[twin meteor]

SWR

[cunning saddle]

f'(x)/f(x)

then is 1/uin10?

[hushed pewter]

uin10?

[cunning saddle]

uln10

[hushed pewter]

u?

[cunning saddle]

yes i use chain rule

[hushed pewter]

ah

Then yes

[cunning saddle]

i have this expression

[hushed pewter]

show your work

[hushed pewter]

You had a 7 here, not a 6

[cunning saddle]

is diff excersie

you cancel 5?

[cunning saddle]

i need cancelm 5? sooo i can cancel?

I don't understand why they cancel? It is my first exercise that cancels that way, I mean, you always cancel with a common factor and all that.

[hushed pewter]

5/5=1

5 on top cancels 5 on bottom

[cunning saddle]

thx

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[rare swift]

Can somebody quickly see if this is correct? The Row operations are written, and the picture in the last is the result. idk why the last two matrix wont come out as they should in latex, but the last picture is the result i get

[rare swift]

And the third row operation is -1/a

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@rare swift Has your question been resolved?

[rare swift]

mo

no

[rare swift]

so what should i change

[full hatch]

The things I wrote into it

[rare swift]

in the last one, its −2(1 − a)r2 + r3 -> r3

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@rare swift Has your question been resolved?

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@rare swift Has your question been resolved?

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[rotund escarp]

The answer should be B

[vocal sleet]

@normal plaza Has your question been resolved?

[normal plaza]

what is this

[rotund escarp]

Thats just the way I solve these type of questions

[half imp]

The definition of integral involves f(x_k) and Δx

So you have to find out what x_k and Δx are, both in terms of k

Then just fit them together to get the sum

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[normal plaza]

.close

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[slate sapphire]

find the value of each of the following algebraic expression when a=-2 and b= 1 over 2. i) 3a+b

[median harbor]

where are you stuck on?

[slate sapphire]

I can't solve that question

[median harbor]

whats trickign you?

[fluid obsidian]

My favorite question

[slate sapphire]

3a+b

I stuck on b

[median harbor]

a is -1

b is 1/2

sooooooo maybe putem in?

[slate sapphire]

a is -2 and b is 1/2

so how to solve b

[median harbor]

wdym how to solve b

b = 1/2

[slate sapphire]

OH I got answer thx for helping little bit

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[slate sapphire]

if this is add what is the answer ? 2+3/2

[slate sapphire]

.close

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[slate sapphire]

.close

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[paper depot]

could probably get more attention by posting your problem

"integer roots" can mean multiple different things in different contexts

also weren't you like, going to leave this server a couple hours ago

[grizzled igloo]

When you say integer roots do you mean as in say the roots of some function. Or the number theory concept of integer square roots

[paper depot]

please don't call me bro.

fuck you.

[grizzled igloo]

The roots of a function are the places where the function takes a value of 0. It is possible but not necessary that those domain values are integeres.

[flat whale]

jesus christ why are you still here

[winter hawk]

.close

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[empty frigate]

,rccw

[clever parcel]

Help this isn't even my homework it's for kids two years uounger than me and idk how to teach it 💀

Idek how to solve it tbh

[grizzled igloo]

Have you tried some algebraic manipulation?

[clever parcel]

In what way?

[thin vale]

try expanding out the product of the two binomials

[clever parcel]

Yup

[grizzled igloo]

First multiply our the left side, or FOIL it

[clever parcel]

So I get x +3 rt x + rt 2x

[thin vale]

typing sqrt(and using parenthesis) makes it easier for us to understand what you mean

[clever parcel]

Aight and thank you

[hushed pewter]

Do you know how to solve quadratic equations?

[clever parcel]

Yes

[hushed pewter]

Try substituting u=sqrt(x) and solve the quadratic

[clever parcel]

x + 3sqrt(x) + sqrt(2x)

[grizzled igloo]

Ok good.

Now can you make this new equation into a quadratic?

I would move something over from this new left side to the right side

[clever parcel]

Alright thank you and gimme a min

Nvm what do I do with the u sqrt (2) at the end of the equation

Subbing u in for sqrt(x)

[grizzled igloo]

I wouldn't take that approach of subbing in.

[clever parcel]

I see

[clever parcel]

Then idk how to get a quadratic from this

[grizzled igloo]

can you show what you have so far?

[clever parcel]

I'm still only up to x + 3 sqrt(x) + sqrt(2x)

= the right side

[grizzled igloo]

sure great

ok what if you moved just the x to the other side.

And squared to get rid of the sqrts

[clever parcel]

I feel like I did smth wrong

As I got x^2 -11x +344

[silk comet]

how did you get 3sqrt(x)

[clever parcel]

Because I accidentally did 2 times rt x instead of the other way around 😭

[grizzled igloo]

Lol I didn't even spot that

[vocal sleet]

@clever parcel Has your question been resolved?

[clever parcel]

Yes

[vocal sleet]

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[keen jetty]

I'm trying to solve this probability question and I know I need to use probability with combinations but I don't know how to calculate getting at least two questions correct

[paper depot]

have you heard of the binomial distribution before

[keen jetty]

not yet

[paper depot]

ok thats not a problem

in how many ways could the student fill in their answers for the test?

[keen jetty]

2^9 right

[paper depot]

yes

now

instead of finding the number of ways to answer at least two questions correctly, let's count the ways to NOT do that.

[keen jetty]

to got 0 or 1

[paper depot]

yes

[keen jetty]

get*

[paper depot]

so how many ways are there to get 0 questions right?

[keen jetty]

1

[paper depot]

yes

and exactly 1 question right?

[keen jetty]

9?

[paper depot]

yes, 9

do you see how to proceed?

[keen jetty]

not really. I know all the facts but I don't know how to put it together

[paper depot]

how many ways are there to not get at least 2 questions right

[keen jetty]

10

[paper depot]

so how many ways are there to get at least 2 questions right

[keen jetty]

2^9-10?

so its something/502

[paper depot]

yes

502 ways out of 512 possible

[keen jetty]

oh shoot i get it now

there ten ways not to get at least two so you minus that from all different ways of answering and you get 502/512 which simplifies to 251/256

right?

so then this question also has to how probabiltiy of combinations but using a different strategy

but im a bit stuck

she launches 6 fireworks but there are 14 left and then im stuck find the chance exactly 4 of them are bule

<@&286206848099549185>

could someone hint me in the right direction please

.close

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[low obsidian]

Hello

[low obsidian]

Find the gradient of the scalar field $w =(\vec{a} \cross \vec{b}) \cdot \vec{r}$ if $\vec{a} = (2,3,6)$ and $\vec{b}=(-1,4,0)$ where $\vec{r}$ is the radius vector of the point $M(x,y,z)$.

How do I solve this problem

[twin meteor]

RickyDicky

[low obsidian]

what exactly is r equal to

i dont get it

I found the cross product of a x b

but i dont know what r is equal to.

is it just r = x*i + y*j + z*k?

anyone...?

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@low obsidian Has your question been resolved?

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[maiden iron]

what have you tried

[vocal sleet]

@vast shale Has your question been resolved?

[maiden iron]

did you use the formula they gave u

[vocal sleet]

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[vestal shuttle]

how would you long divide this? would you write out the denominator?

[vestal shuttle]

found it, but why do theyy do this>?

oh im stupid

you integrate it

lmao

[gentle swift]

they have intergrated that already and moved it outside

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[keen inlet]

[x]=[x/2]+[(x+1)/2] prove this. [.] denotes the greatest integer function. I cant understand what the beginning step will be and what approach to take to prove it

[keen inlet]

I dont need the full explanation, just would like to know the approach to take

[vast shale]

there are two cases, either the fractional part of x/2 ≤ 1/2, or > 1/2

this tells you whether [x/2] and [(x+1)/2] are different or the same

[keen inlet]

okay...lemme try to solve

and if i cant i'll ask for more help

[vocal sleet]

@keen inlet Has your question been resolved?

[keen inlet]

.close

[vocal sleet]

Channel closed

Closed by @keen inlet

Use .reopen if this was a mistake.

Available help channel!

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• 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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[scarlet bough]

How would I solve this question?

[river minnow]

Generally the area between curves y = f(x) and y = g(x) is the integral of |f(x) - g(x)|

So, here, the integral that needs to be set up is $\int_0^6(6x - x^2 - 2x)\dd{x}$

[twin meteor]

A Lonely Bean

[scarlet bough]

Where did you get -2x from?

[median harbor]

y=2x

[river minnow]

-g(x)

[median harbor]

were subtracting L from C

[scarlet bough]

So 2x= 6x-x^2

[median harbor]

woah what

[median harbor]

here

6x-x^2 is curve C

2x is curve L

we need to do C - L

[river minnow]

We are not setting them equal

[median harbor]

soooo 6x - x^2 - 2x

[scarlet bough]

Can't I integrate each one separately and then take away?

[median harbor]

that's also valid

its the same thing we just combined them up

[scarlet bough]

So would both integrations be from 0 to 6?

[median harbor]

where did you get 6?

[soft walrus]

it's given

[median harbor]

wait what

[soft walrus]

the curves intersect and 0 and 6

[scarlet bough]

^

[soft walrus]

^

[median harbor]

didnt it say that

L intersects C at 4, 8

[soft walrus]

oh wait

intersect the x-axis

[blissful ledge]

Hello

[soft walrus]

mb mb 💀

[median harbor]

oopy

[blissful ledge]

I'm new

[median harbor]

hi new im candies

[soft walrus]

hello new

[median harbor]

welcome

[median harbor]

same braincell

[blissful ledge]

No, I just joined

[soft walrus]

😂

[blissful ledge]

My name is not new

[soft walrus]

dad joke

[blissful ledge]

Alright

I see

[soft walrus]

but if you need help #❓how-to-get-help and if you want to chat #discussion @blissful ledge

[scarlet bough]

So integrate from 4 to 8 for both?

[median harbor]

no 4, 8 is the coordinate where they intersect

not 4 to 8

integrate from where the shade starts, to where the shade ends

[soft walrus]

You have two coordinates at which the functions intersect

$(0,0)$ and $(4,8)$

[twin meteor]

XxMrFancyu2xX

[soft walrus]

what are the two x values?

[scarlet bough]

0 4

[soft walrus]

yes

so we have $\int_{0}^{4}{6x-x^2-2x}\mathrm{d}x$

[twin meteor]

XxMrFancyu2xX

[soft walrus]

does that make sense?

[scarlet bough]

So you always integrate the x values?

[median harbor]

you always integrate what the dx lil thingy is

if its dy then y it is

[soft walrus]

when your variable of integration is dx yea I'd suggest it

[scarlet bough]

So if I want to integrate the 2 lines seperatly I still do 0 to 4?

[median harbor]

yep

[median harbor]

but again this does look a bit more simple to me

considering 6x-2x is kinda tasty

[soft walrus]

yes if you use $\int f(x)+g(x)\mathrm{d}x=\int f(x)\mathrm{d}x+\int g(x)\mathrm{d}x$ you must keep your bounds the same

[twin meteor]

XxMrFancyu2xX

[soft walrus]

👀 "tasty"

[scarlet bough]

So is this the answer?

[soft walrus]

how did you get a decimal for 2x?

[scarlet bough]

I got 16

[median harbor]

just consider all my positive adjectives the same

all my positive adjectives are interchangable in my definition

if i dont know what the word is i'll just resort to "hot" "tasty" "yum"

[scarlet bough]

That dosnt seem like alot of work for 6 marks

[median harbor]

well then you're in luck

[soft walrus]

lmao I love using the word "delicious"

[median harbor]

yum

[soft walrus]

"That simplification is looking quite delicious"

omg green name!!

[median harbor]

ooo

any colour roles?

[soft walrus]

i've been promoted 🫡

[soft walrus]

its based on activity I think

[median harbor]

i want a purple name

aw

[soft walrus]

Unfortunately i think the closest you can to that in this server in the donate rank 😅

[median harbor]

oops 🤮

[soft walrus]

also yes that answer is completely correct

you can close this channel if that's your question

have a good day

[scarlet bough]

.close

[vocal sleet]

Channel closed

Closed by @scarlet bough

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[native bloom]

.reopen

[elfin moon]

I have to find out the value of arg(-z)- argz =? Where z <0 so here I did this

[soft walrus]

for z<0 looks right to me

[viral copper]

$\arg(ab) = \arg(a) + \arg(b)$

[twin meteor]

NEONPerseus

[viral copper]

Use this property

Yes they work like logarithms

No it's not a coincidence

Oh you're asking for verification

Yes it's correct

[elfin moon]

My doubt is i thought z and -z will be opposite side

Is it correct?

[soft walrus]

wait z less than 0

[viral copper]

z < 0 doesn't make sense

You can't compare complex numbers

[soft walrus]

does that mean $\arg\left(\frac{a}{b}\right)=\arg(a)-\arg(b)$?

[twin meteor]

XxMrFancyu2xX

[viral copper]

Yes

[soft walrus]

oh nice that's cool

[elfin moon]

How to use it....it will be arg (-z/z)

[soft walrus]

$\arg_{b}(a)=\frac{\arg_{c}(a)}{\arg_{c}(b)}$

[twin meteor]

XxMrFancyu2xX

[soft walrus]

[viral copper]

$$\arg(-z) - \arg(z)$$
$$\arg(-1) + \arg(z) - \arg(z)$$
$$\pi$$

[twin meteor]

NEONPerseus

[elfin moon]

I didn't understand this step

[viral copper]

Which one?

[elfin moon]

Arg(-1)

And u added +

[soft walrus]

$\arg(-z)=\arg(-1\cdot z)$

[twin meteor]

XxMrFancyu2xX

[viral copper]

Do you understand that^

[elfin moon]

Perfect

[viral copper]

a = -1, b = z

[elfin moon]

And arg(-1)

Why is it pi

[viral copper]

Is pi

Isn't it obvious

Look at -1 on the complex plane

[elfin moon]

Ohh but is it not -pi

[viral copper]

Do you understand what the argument is?

Nope

I mean

[elfin moon]

It lies on -x axis

[viral copper]

You can write $-1$ as $e^{-\pi i}$ but the amplitude is $\pi$

[twin meteor]

NEONPerseus

[elfin moon]

My concepts regarding argz and angle is confusing

[viral copper]

$\arg(z) \in (-\pi , \pi]$

[twin meteor]

NEONPerseus

[elfin moon]

That's the confusion

[viral copper]

Do you understand the difference between ( and ]

[elfin moon]

Ohhh i didn't see it

Pi included

[viral copper]

Exactly

[elfin moon]

But why?

[viral copper]

That's just how the argument is defined

It's like asking why sin x is opposite by hypotenuse

[soft walrus]

because I said so

[viral copper]

Because Descartes said so

Or whoever did this

Argand

[soft walrus]

Aryabhata

or Brahmagupta

i forget

it was one of those guys

[elfin moon]

E^pi I

[viral copper]

Aryabhata defined the argument of a complex number?

[soft walrus]

no sine

lmao

[viral copper]

Yes

[viral copper]

Ah alright

[elfin moon]

Why you added -

[viral copper]

Both of them are valid representations for -1

Ykw just forget it

[elfin moon]

Let me check

[viral copper]

Dont add the -

[elfin moon]

Yes it works for

[viral copper]

It's just 1/e^(ipi) so yeah it works

[soft walrus]

$-1=e^{i\pi+2ki\pi}$ where $k\in\mathbb{Z}$

i didn't mean to reply to you

[viral copper]

It should be multiplied by i btw

2kπ

[elfin moon]

So?

[soft walrus]

oh yea thanks

[twin meteor]

XxMrFancyu2xX

[viral copper]

It's wrong otherwise

Because e^(2kpi) would be a real number

You can't apply Euler's formula

[elfin moon]

Yeah it works for general terms but here we need to find value of arg(-1)

[soft walrus]

still pi

-1 or e^(ipi) or e^(-ipi) or e^(-69pi)

[elfin moon]

Why pi

Wait a minute

[soft walrus]

cause that's -1 in polar form...

[elfin moon]

Can we say it is -1 +0.i

And it's in third quardent

[soft walrus]

what?

[elfin moon]

Can we not plot -1 and then see it's angle

Ohh it's pi

[soft walrus]

yea it's angle is pi

all the angles are coterminal to pi

[elfin moon]

I am thinking things in graphic mode

[soft walrus]

pi, -pi, -69pi

are all coterminal

[elfin moon]

Why do they not include -pi

[soft walrus]

because -pi+2pi=pi

they're all coterminal angles

meaning if rotate -pi radians with radius 1

you'll end up at -1

same if you rotate pi radians in the positive direction

[elfin moon]

Yes

[empty frigate]

yeah if you included both -pi and pi you'd have two ways to "represent" -1

[elfin moon]

Do I know you?@empty frigate

I meant there was someone like your name

[empty frigate]

well that might have been a different bee

[elfin moon]

And we have only one choice to represent

[soft walrus]

?

[empty frigate]

yeah well we want arg(-1) to be some number instead of "-pi and also pi"

[elfin moon]

I meant we can't use -pi and pi for -1 at the same time

We have to pick only?

[viral copper]

I guess you could think of it as a function from C to R

So you can't have two images for -1

[elfin moon]

So is there any other way to solve that question

I wanted to know all the possible ways to solve it

[viral copper]

Well

Sub in a value for z