help-26

[neon iron]

💀💀 im so sorry

[void crow]

don't worry

[neon iron]

thakb u 🙏

[void crow]

it's all good

yesyes

leave a 5 star rating

pls

[neon iron]

how

[void crow]

js send

5 star emojis

fr

[neon iron]

⭐️⭐️⭐️⭐️⭐️

[void crow]

YIPPIE

[neon iron]

YAY!!!

[void crow]

alright u can close now

get out

NO IM KIDDING

BYE BYE

[neon iron]

ok

bYE

[void crow]

😭

[neon iron]

.close

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[pale barn]

The number of kilograms of food they have spent at an animal shelter during a particular week can be calculated using the function f(t) = 10(-t3/8 +3t2/2 -9t/2 +10) , where t is the time in days and goes from day t = 1 (Monday) to day t = 8 (Monday of the following week).

[pale barn]

b) Determine the days of the week when spending on food was greater and the days when it was less. How many kilograms of food were used these days? [1.5 points]

i can only find it by drawing a graphic or there is another wway?

[maiden pelican]

By using the derivative of the function

[pale barn]

how to find it?

[maiden pelican]

Is this a linear function

[pale barn]

okk

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[neon iron]

idk if its right or not

[vernal matrix]

You're multiplying rather than applying it as a function

Remove the * and add

[neon iron]

where should i add brackets?

[vernal matrix]

ln(abs(t))

Oh also it should be $+t^{-6}$, you'd have had $6 \frac{t^{-6}}{-6}$ and that cancels out with the subtraction

[thorny flame]

@vernal matrix

[twin quest]

can i get help on 3 math problems?

im not fully understanding or knowing how to do it

[neon iron]

and thanks so much , i didnt catch that

[vernal matrix]

!occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[twin quest]

how do i make my own channel to ask for help?

[vernal matrix]

You just did: #help-15 is yoirs

[twin quest]

oh got it

[vernal matrix]

And unless I've missed something yet again, I think so

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[glad sierra]

they should have added a vertex with no edges aswell right?

here V has 6 elements

i dont get what they mean by H with 5 vertices and 8 edges

[rigid ivy]

Yes. They should had added the isolated vertex 6

At least that's how I would do it

[glad sierra]

makes sense

maybe some typo

[rigid ivy]

(but it can be omitted if you don't care about the isolated point)

[glad sierra]

If a graph G contains vertex set and edge set

how can we represent parallel edges in an edge set?

our familiar notion of sets only have notion of membership not repetition

<@&286206848099549185>

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[spark moss]

help with differentiation

[spark moss]

direct substitution gives 0/0 which means we have to go further right

like factoring or something

i have no idea what to do

[sweet shard]

Yes factor both polynomials in the numerator and denominator

[spark moss]

how the balls do i even factor the denominator

the numerator is just (t - 1)(t + 1) right

[sweet shard]

This tells you one of the factors of the bottom

[sweet shard]

Because you're getting 0/0

[neon iron]

L'hopital rule

[spark moss]

can i do -3(t^2 + (10/3)t + (7/3))

[neon iron]

Factor the denominator:-(x+1)(3x+7)

Cancel the term (x+1)

[spark moss]

can you explain how you factored

i dont understand this

[neon iron]

Ok, one sec

[spark moss]

oh

is it

numbers that are equal to the product of the outer terms

and numbers that are equal to the middle term

-7 * -3 = 21

and -10

so -7 and 3

[neon iron]

To factor ax²+bx+c, if X1 and X2 are its roots, then its factored form is: a(X-X1)(X-X2)

[spark moss]

whatdo you mean its roots

[neon iron]

The zeros of the equation, that is, the number X that makes the equation zero

In this equation the roots are X1=-7/3 and X2=-1

$-3t^2-10t-7 =0\
If \ t=-7/3 \ so \
-3(-7/3)^2-10(-3/7)-7=0 \
If t=-1\
-3(-1)^2-10(-1)-7=0$

You understood?

[thorny flame]

Danty ⛧

[grand solar]

You can also do factor by grouping

[spark moss]

haha yea im doing that right now

[spark moss]

yes i do understand

i looked back at the practice problems and she didnt solve it like this she used grouping

[peak delta]

Hello can anyone help me with calculus

[spark moss]

so she would take points off if i do it like htis

[neon iron]

What is your doubt?

[peak delta]

How do you do this

[grand solar]

multiply the leading coefficient by the c, or 7 in this case and so you get -3*-7 = 21 and do you’ll find the numbers that multiply to give you 21 and add to get -10 and that would be -3, -7 so you’d make the new equation like
-3x^2 -3x - 7x - 7 and then you can factor from there and you’ll end out with a (x+1) I think lol

[peak delta]

Is that for me?

[grand solar]

no the original posted

[peak delta]

Ok that’s def not for me

Oh I gotchu

[neon iron]

F(x) is constant for x<2 and f(x) is a straight line for x>=2, separate the integral in these limits and calculate separately

[peak delta]

Can you show me on paper I’m so lost 😭

[neon iron]

Create another post for this

[peak delta]

Uh how

[spark moss]

#❓how-to-get-help

[neon iron]

I I don't know >< I'm new to this server

[grand solar]

go to the math help available channels, right now there is help-20, 37 and 40

and just type in it

and it’ll give you the help channel

when your done getting help you can use “.close” to free the channel back up for someone else

[spark moss]

dawg

im still wrong

HOW DO I DO THIS

UGH

[neon iron]

Create a new POST and tag me

[grand solar]

what’d you get when you factored the equation

[spark moss]

i factored it and then re-substitutted the -1 into the equation

[spark moss]

(t-1) / (-3t - 7)

o wait sorry

[spark moss]

(-3t - 7)(t + 1)

was for the denominator

and (t + 1)(t - 1) was the numerator

[neon iron]

Good

[spark moss]

it said -1/4 is wrong

WAITTT

im so stupid

nvm yea im still wrong

((-1) - 1) = -2

(-3(-1)) - 7 = -4

-2 / -4 is 2

omg

its not

its 1/2 which is right thank you guys for all your help

im so tired im going to sleep

thank you fr

[grand solar]

np

[neon iron]

Anytime

[spark moss]

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[gilded ermine]

How is the change not 20 or 25? Is 2010 100 ?

[topaz sinew]

Please don't occupy multiple help channels.

[rigid ivy]

Interesting all your other channels seem closed

[gilded ermine]

yea u think its broken?

[rigid ivy]

I think so. You're fine here.

@gilded ermine

[gilded ermine]

thanks

[gilded ermine]

But year idk how its not 25 or 20 I estimated that 2010 was 135 or 130

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[mossy garnet]

,tex $$\lim_{{x \to +\infty}} \frac{2x^2}{x^2 + 1}$$

[thorny flame]

レナト (renato , ping if reply)

[worthy storm]

!status

[topaz sinew]

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

[mossy garnet]

can someone give me a hint

[worthy storm]

sure

[mossy garnet]

1

[worthy storm]

try dividing num and denom by x^2

[mossy garnet]

,tex \begin{align*}
\lim_{{x \to +\infty}} \frac{2x^2}{x^2 + 1} &= \lim_{{x \to +\infty}} \frac{\frac{2x^2}{x^2}}{\frac{x^2 + 1}{x^2}} \
&= \lim_{{x \to +\infty}} \frac{2}{1 + \frac{1}{x^2}}
\end{align*}

[thorny flame]

レナト (renato , ping if reply)

[mossy garnet]

something like this, kind sir? @worthy storm

[worthy storm]

yep that looks good

can you see now what happens as x->infinity?

[mossy garnet]

,tex \begin{align*}
\lim_{{x \to +\infty}} \frac{2x^2}{x^2 + 1} &= \lim_{{x \to +\infty}} \frac{\frac{2x^2}{x^2}}{\frac{x^2 + 1}{x^2}} \
&= \lim_{{x \to +\infty}} \frac{2}{1 + \frac{1}{x^2}} \
&= \frac{2}{1 + 0} \
&= 2
\end{align*}

[thorny flame]

レナト (renato , ping if reply)

[worthy storm]

yep perfect!

[mossy garnet]

tysm kind sir, cheers

big thanks

[worthy storm]

yw, enjoy

[mossy garnet]

.close

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[vapid notch]

15.00 is wrong

.close

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[vernal vale]

why do you keep doing this? 👀

[sonic ibex]

💀

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[stuck iron]

i got A and B and im not sure why they are wrong

[carmine marten]

You can't have A and B (presumably you mean the top two answers) because if the limits in A are not equal, the limit in B does not exist

[stuck iron]

oh okay where should i go from here im really bad at calculus

but that makes sense

[carmine marten]

did you find these limits?

[stuck iron]

i got 0, 0, and 1

[carmine marten]

ok so those are wrong (and I'm confused by why you said A and B because if you're listing them in the same order, that's not what A and B say? like you found the limits from above and below equal, and the overall limit equal to 1 which is the same as f(4)?)

the nice way to find the limits is to factorise the numerator and denominator

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[blazing marsh]

Wondering if someone can help me

im so lost

[keen tundra]

do you understand the basic idea of how to find solids of revolution

[blazing marsh]

not really

[keen tundra]

well

[blazing marsh]

im trying to figure out using subsituion but i just get stuck after putting it into u and du

[keen tundra]

your teacher probably explained it similar to how riemann sums work, right
Except instead of thinking about a sum of infinitesimally small rectangles, when you rotate a function around the x-axis it's a sum of cylinders

[keen tundra]

can i see the function you're trying to integrate

so i can verify that youre not integrating the wrong thing

if you have the correct function it's not super hard to get to the right u-substitution

[blazing marsh]

whats throwing me off is the x=0, x=1

[keen tundra]

thats the right integral

You dont actually need to do a trigonometric substitution here by the way

[blazing marsh]

oh really

[keen tundra]

well its possible to solve with a trig substitution but if you are familiar with the derivatives of inverse trig functions you don't have to

what i would do here is recognize that the function inside the integral looks a lot like some sort of inverse trig antiderivative thing, so i would try and turn the 36 in the denominator into a 1

[blazing marsh]

oh ok

[keen tundra]

like just try to cancel out all the 36s

[blazing marsh]

they probably never taught me that yet

[keen tundra]

oh

its actually a really clean substitution

you can use x = 6u

that way you remove all the 36s

[blazing marsh]

i been trying to use the washer method

but the way u make it sounds easier

[keen tundra]

i mean youre still doing like a solid of revolution thing here
everythings right so far you just have to solve the integral

[blazing marsh]

all i got is 36/6 pi so [theta]1/0

[keen tundra]

how did you get that

if thats your final answer its not correct btw 😔

[blazing marsh]

its not my final answer

mid step

damn i suck

[keen tundra]

ok

[blazing marsh]

i canceled out the sec^2

[keen tundra]

you did ?

[blazing marsh]

that left me with 36/6 pi integral 1, 0 d(theta)

[keen tundra]

hmmm

ok i can see where you got there from

i only have one issue with it and it's the bounds 1 and 0

[blazing marsh]

oh

[keen tundra]

When you do a substitution do you know how to change the bounds

?

[blazing marsh]

probably not

i been a good year since i did calc

im trying to recall as i go

[keen tundra]

Cause at this point you can't just do 36/6pi (1 - 0) because that gives you the wrong answer

okay

[blazing marsh]

yea so do i make the theta?

[keen tundra]

you can make the theta but if you want to resolve the integral with the original bounds 1 and 0 you have to put it back in terms of the original variable x

so you gotta reverse the substitutions

[blazing marsh]

reverse the subsitution?

[keen tundra]

Yeah

Like if you used the substitution tan(theta) = x, and your final integral resolved to just theta, then you would have to reverse the substitution to get the resolved expression in terms of x so you can plug in the bounds

so if your substitution were tan(theta) = x you would change all thetas to x by reversing the equation, theta = arctan(x)

[blazing marsh]

so i would get 36/6 pi[arctan(1/6x)]6 theta?

[keen tundra]

thats not exactly what i have its very close though

you actually get the right answer if you had mistyped the theta at the end (if you instead mean 36/6 pi[arctan(1/6x)]6x )

and you plug in the bounds 1 and 0 of course

[blazing marsh]

so it would something like 36/6 pi [arctan (??)-arctan(??)]?

[keen tundra]

actually yeah its just this with the bounds 1 and 0

[blazing marsh]

1/6, 0?

[keen tundra]

oh

yeah

[blazing marsh]

is it 3pi^2/2

nvm it was a decimal

thank you so much

i have one more question tho

not math related

[keen tundra]

oh ok

what

[blazing marsh]

what makes u want to help people for free

i always get curious about it

[keen tundra]

hmmmm today its just because i got bored of other things

sometimes its because i just want to do math

[blazing marsh]

ohh

[keen tundra]

sometimes people ask interesting questions !

[blazing marsh]

wahts the highest math course u taken

actually hardest

[keen tundra]

right now im in a matrix algebra course and a course about integration (though ive already taken this level of calc already so it feels easier)

the hardest math course for me personally that i took was actually a high school statistics course though lol

[blazing marsh]

DAMN

didnt expect that one

[keen tundra]

i do a little bit of math as a hobby so im mostly familiar with things like linear algebra and multivariable calculus

statistics was hard though cause there were so many statistical tests and sampling rules and things i had to remember

[blazing marsh]

yeah i can see that

u can probably breeze through them now

[keen tundra]

nah its still hard for me lol

i still remember a bunch of the important things from that class though

[blazing marsh]

i only took precalc in hs

[keen tundra]

that's fine

[blazing marsh]

it aint if u forget the whole thing

[keen tundra]

ehhh

its not too hard to revisit precalc concepts, i remember it being mostly about like . limits and elementary functions like logarithmic properties and whatever

[blazing marsh]

yea appreciate it tho bro u teach better than my professor

[keen tundra]

no problem :]

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@blazing marsh Has your question been resolved?

[keen tundra]

.close

oh

!done

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[lethal vessel]

What did I do wronggg 😭

Guys please

Help me

<@&286206848099549185>

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[charred hearth]

For c

F = Ma - kx - 3mnv

However this gets F = left hand side of what you need

Where is f + 3nft coming from???

<@&286206848099549185>

Anyone?

[acoustic pecan]

dy/dt=ft-dx/dt
d^2y/dt^2=f-d^x/dt^2

x''=f-y''
x'=ft-y'
x=0.5ft^2-y

x''+3nx'+2n^2x=(f-y'')+3n(ft-y')+2n^2 (0.5ft^2-y)
=f-y''+3nft-3ny'+fn^2t^2-2n^2y
=(f+3nft)+(-y''-3ny'-(2n^2)y+(fn^2t^2))
ig you need to show the stuff in the right bracket is 0

if i didnt mess anything up anyway

[topaz sinew]

@charred hearth Has your question been resolved?

[acoustic pecan]

my''=(2mn^2 x)-3mny
y''=2n^2 (0.5ft^2-y) -3ny
y''=fn^2t^2-2n^2y-3ny

@charred hearth yeah, thatll do it

sub that in and all goes away

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[glacial remnant]

how do I find the general solution to this

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@glacial remnant Has your question been resolved?

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[glacial rock]

I don’t understand how he got 3pi/4, 5pi/4, 7pi/4

[boreal mulch]

what are you asked to find

y?

[glacial rock]

No it’s just critical numbers

[neon iron]

arccosine 0 is pi/2

[glacial rock]

Huh

[neon iron]

he has cos2x= 0

to get 2x alone you need to bring cos to the other side so you do arccos

arccos0 is pi/2

maybe its called cos^-1 in english

nvm sry not arccos

cos0 ist pi/2

whatever i suck sry forget it

[glacial rock]

<@&286206848099549185>

[neon iron]

Theres multiple points within the interval that satisfy whats being asked

[glacial rock]

How did he get those numbers

[neon iron]

So hes working out the points where cos2x = 0, if u look at a graph of cos2x

u can see theres multiple points where the values 0

He just pulled those values from the graph

[glacial rock]

Im confused

why doesn’t he use the unit circle

And wdym

[neon iron]

u can use either

or u can work them out in ur head

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[tidal fern]

Question: Find a line in parametric form that lies in the plane x + y + z = 1 .

To solve this question, is it possible to use the dot product of the normal of the plane and another vector v that lies in the plane and equal it to 0, to get e.g. v = (-1,-1,-1), and can we then just apply a t(-1,-1,-1) where t is a real number?

[worthy storm]

the idea is fine

but is (-1,-1,-1) parallel to the plane?

[tidal fern]

I would assume so, as it is orthogonal to the normal vector of the plane (since the dot product is 0)?

[worthy storm]

check that dot product again...

you are computing (1,1,1) dot (-1,-1,-1)?

[tidal fern]

yes

[worthy storm]

that's not zero 😁

those two vectors are in fact parallel, not perpendicular

[tidal fern]

Okay, but if we find a vector v that does make the dot product 0, will the idea work?

[worthy storm]

it will give you a vector parallel to the plane

say v

but tv won't necessarily lie in the plane

[tidal fern]

do we need to add a starting vector that lies on the plane

[worthy storm]

(unless the plane contains the origin)

[tidal fern]

for e.g (0,0,1)?

[worthy storm]

you need a point in the plane yep

[tidal fern]

okay

[dim star]

I need help

[worthy storm]

#❓how-to-get-help

[tidal fern]

so, in short, we could write a line in parametric form that lies on the plane as (0,0,1) + t(2,-1,-1)?

as (2,-1,-1) dot the normal of the plane should equal 0.

[worthy storm]

yep that works

[tidal fern]

yeah I can see that now... Not sure what I was thinking before

[tidal fern]

thanks, that was all!

[worthy storm]

(there are infinitely many correct answers of course)

[tidal fern]

yes ofc

thanks!

[worthy storm]

cheers

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[past falcon]

James Cooke, an undercover agent, has received a tip from an anonymous entity. It’s come in the form of a code, and with enemies around every corner, Agent Cook must decipher the hidden number before it’s too late. Unfortunately, Agent Koek is, uhm… Not the tallest tree in the forest, so he’s going to need your help!

[To: Agent Cake]

{ 3 X 2ABCDE = ABCDE2 }

What is the product of A + B and D + E?

[Each letter represents a digit, making a complete six-digit number

[topaz sinew]

Please don't occupy multiple help channels.

[past falcon]

help i don't know what to do

[woven creek]

i like how the guys name changes 3 times. in 3 instances

[past falcon]

my school gave me this to do

[woven creek]

hiint try 1/7 * smtg

[past falcon]

?

i am in grade 8 and they gave me this

[woven creek]

what is 1/7

in decimas

[woven creek]

when u get 6 numbers multiplied by a constant giving the rearranged form of the same no. its usually smtg to with 1/7

[past falcon]

do you know what the answer is so I can work backwords

Please

.close

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[tacit panther]

How did they get 9.08?

Was it F(a) - F(b)?

[neon iron]

calculator

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[neon iron]

No idea how to do this

[neon iron]

I’m using

MU1 + MU2 = MV1 + MV2

And because they are both at rest mu1 and 2 are both 0

Therefore

[barren salmon]

yes

momentum is conserved

[neon iron]

0 = 0.3(m+0.5) + (0.5)(0.25)

[barren salmon]

yeah

[neon iron]

So

[neon iron]

Wait no

[barren salmon]

sign you can

change later

[neon iron]

0 = 0.3(m+0.5) - (0.5)(0.25)

0.3(m+0.5) = (0.5)(0.25)

(m+0.5) = .125/0.3

M = -0.083

Which is not correct

Anything wrong with my method @barren salmon ?

[barren salmon]

hold on

ahh

i dont get

why is it incorrect

[neon iron]

Negative mass?

Doesn’t make sense

[barren salmon]

exactly

[neon iron]

WHY IS THE MASS NEGATIVEEEEE 😭😭😭😭😭

[barren salmon]

shouldnt the trolley with extra mass have lower speed?

[neon iron]

True atuaclly

Actually

Lemme recalculate

PHEW

[barren salmon]

hahah

[neon iron]

0.1kg

I feel so relaxed now

That’s was scary

[barren salmon]

:)

[neon iron]

Tysm

[barren salmon]

have a nice day

[neon iron]

You too!

Wait

I don’t mean to be a pain

But can I add you? Becuase I don’t know many people on here that do physics stuff

[barren salmon]

sure!

[neon iron]

Tysm!

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[desert dagger]

for part b im getting x = 2 is that correct?
Image

[royal glacier]

Is this an exam?

[desert dagger]

this is a practice exam

its before my actual exam

[royal glacier]

ok

I think x=2 is right

[desert dagger]

okay thanks needed one more thing

can you give me a hint for this

im not really understanding the question

[royal glacier]

So first there is inital part of the circle AOB

[desert dagger]

yeah so i did theta/2pi * 2pir = ab

[royal glacier]

And you set B_1 as foot of perpendicular from A to OB

Then make arc with radius of OB_1 and center as O

And we repeat this process

right?

[desert dagger]

hmm i think i got it but are we treating radius value as 2?

[royal glacier]

OB is 2

but not for OB_1, OB_2, ...

[desert dagger]

so basically for value of ab we have to do theta/2pi2pir and we replace r by r-1, r-2 everytime

[royal glacier]

no

you need to use trig

[desert dagger]

im sorry i dint get it then

[royal glacier]

to get OB_1

point is OB_n is same in length with OA_n

[desert dagger]

yeah okay

[royal glacier]

you can use trig in this problem right?

[desert dagger]

sohcatoah?

that trig ratios thing?

[royal glacier]

yes

especially cosine

[desert dagger]

cos=base/hypo

cos theta = b1/a

[royal glacier]

so as OA=2, can you express lenght of OB_1 with theta?

and a1=b1

what for a1 and b2?

[desert dagger]

cos theta = cos theta = b1/2

right?

[royal glacier]

yes

[desert dagger]

then cos theta = ob2/0a1

what will we take as a1

if a2 = 2 and a1 = 1?

[royal glacier]

0a1=0b1

and you calculated the length of 0b1

[desert dagger]

0a1=2*cos theta

[royal glacier]

then 0a1=0a* cos theta

you get this?

[desert dagger]

yeah because 0a=0b1

which is equal to 0a

[royal glacier]

0a=0b

Then can you show that $OA_{n+1}=OA_n\cos{\theta}$?

[thorny flame]

Dri111

[desert dagger]

latex nice.

yeah got it

[royal glacier]

ok then

[desert dagger]

ty

[royal glacier]

first OA=1

no 2

[desert dagger]

so we just then replace n with 1,2,3

[royal glacier]

yes

[desert dagger]

and get a sequence in terms of theta

[royal glacier]

yes

[desert dagger]

okay got it thanks

[royal glacier]

good

[desert dagger]

.close

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[sudden trail]

Two triangles - ABC and DEF lie on the line ADBF, such that they intersect.

The height of ABC is 37.8
The height of DEF is 9.

A rectangle of arbitrary height is drawn with its vertical axis of symmetry intersecting with the intersection point of BC and DE, with width 0.61.

Both triangles have a base of 3.6, and are separated horizontally by a distance of 2.

[sudden trail]

(apologies for the bad diagram.)

I need to find the % area of the two triangles that is enclosed by the rectangle.

[torpid sparrow]

how do they lie on the line ADBF when B is up there?

[sudden trail]

... apparently my labelling is terrible. I meant ABCF, my bad

Already calculated the areas of the triangles to be 68.04 and 16.2.

Angle ACB = 1.52rad
Angle EDF = 0.048rad.
Have attempted to use some form of similar triangles to find the area of the small area either side of the lines, then subtracting.
This yielded a value of 13.6% of each triangle, which I have a feeling is incorrect - I tried the same problem with different values, but after a point resulted in negative area.

Oh, and just a note. The triangle points aren't on the vertical axis of symmetry.

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[latent solar]

Does anyone know how to do cumulative frequency? I don't understand how I can find q or r

[covert current]

Can you show the full question

Because it says question 4 (continued)

so there is information from the previous part that will help you solve

[latent solar]

I don't think this will help me necessarily though, because my tutor showed me how to do 4e previously, but I forgot his explanation now and can't understand my answers

[covert current]

Yeah that is really weird…

I don’t know exactly what upper class boundaries mean but hmmm

[latent solar]

OMG I FIGURED IT OUT sorry I just understood:

To find q, the upper class boundary is 30 < T < 40, meaning you add 173 + 21 = 194, which gives you q

And same for r; add 6 + 194 = 200

Sorry for wasting your time omigosh

[sudden trail]

No.

[latent solar]

?

[sudden trail]

Sorry, Discord did a screwy and didn't send my message earlier until now.

[latent solar]

Lmaoo hate when that happens

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[tight kiln]

i need help

[tight kiln]

hello'

anybody here

[haughty mural]

yeah, just write your question

[tight kiln]

can i screen shot it

the one on the right is answer choices

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[neon iron]

How would I approach this? I need to prove that this limit either exists or not

[outer portal]

try approaching this for when x = y or when x = -y, which should cause different/same results depending on whether n is even or odd

[neon iron]

Why should i pick x=y or x=-y?

[neon iron]

?

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[strong oak]

Because picking x=y and x=-y gives convenient cancellations to help prove/disprove the limit. I'd recommend you just try plugging them in and looking for cancellations

[neon iron]

Hey so I worked on it and what I have done is approach the limit from either axis. I first did it with the limit from (x,0) -> (0,0) and found that limit to be 1.

Likewise with the y-axis, I actually found that the limit is equal to (-1)^n.

Thus it alternates between 1 and -1

[neon iron]

you know what. This actually seems not a bad idea i will try it now

[strong oak]

What you've done is valid too, but it doesn't disprove the limit for even values of n (since they'll both be 1 in that case)

[neon iron]

You're right

that was exactly the issue

[neon iron]

I found multiple cancellations but they are all equal to 1 it seems

I have tried approaching from (0,-y), (-x,0), (y,y), (x,x), (-x,x) but they all seem to approach 1 or alternate

i mean (y,y) method is exactly like (x,x) method

n=2

zoomed closer to the origin

Oh wait

[strong oak]

Really? When y=x, you should get 0 in the numerator, and when y=-x, you get either an undefined number or 2^(n-1), so either way they're not the same

[neon iron]

Let me re-evaluate (btw thanks for doing the it with me)

[strong oak]

No problem

[neon iron]

When y=x you indeed get zero in the numerator and 2x^(n) as the demoninator

Since you a can only obtain 1/2 from the numerator you get 0/x^(n)

You cant really do much else from there am i right?

It is always challenging to determine when you cant go any further and just say that you have an undefined number as your limit

OH WAIT I GOT IT

[neon iron]

The limit for y=-x is indeed different because it gives you 2^(n-1) so for every natural n greater than 1 you will never get 1 but if n = 1 then in the case of y=-x you get 2^(0)=1 which is not the same limit in the case of approaching (0,0) via the y-axis, because then you get an alternating limit like i mentioned before of (-1)^n. If n equals 1 then the limit is -1 for any greater natural n it alternates between 1 and -1 which is DEFINITELY not equal to any greater power of 2

[strong oak]

Yep, I think you got it. My reason for the conclusion was a bit different but yours looks good as well.

What I said is that when y=x, the limit is always 0 and that when y=-x, the limit is either 2^(n-1) or undefined depending on n. Either way, they're never equal.

[neon iron]

Thank you so much

I have finished the question and written my conclusion

I look forward on how it is graded haha

I immediately knew this limit did not exists because the section of which this question was already part of mentioned that ''one of the following limits'' exists and the other don't.

The other limit i had to evaluate was

which i quickly determined to be zero.

So if that limit does exists the other didn't per implication of the question haha

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[limpid pulsar]

So I know that the denominator must equal to 0.

Would that mean that it's something similar to (x+1)(x-2) for the denominator?

[pastel salmon]

very well

[keen matrix]

that's precisley what it would be, yes :)

[limpid pulsar]

So in that case, Im struggling to write the exact answer. Would my answer be (x+2) and (x-1)?

Thank u guys for the responses too! Much appreciated

[pastel salmon]

you can write your example as:

[limpid pulsar]

Okay thank you so much!

@keen matrix you too!

(sorry for pings)

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[keen matrix]

no worries, yw!

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[grim ice]

I am struggling to understand how my teacher got this as an answer (this is a photo of the answer sheet). Because the negative 4 is in brackets, it would be interpreted as a horizontal translation of 1/4 correct?

[jade wharf]

you mean a horiztonal compression

[grim ice]

compression, yes.

[jade wharf]

right

[pale python]

I can give you the anser

[grim ice]

I miss typed that

[pale python]

The answer is z

[jade wharf]

so whats the problem ur supposed to be doing

[keen matrix]

write it as a piecewise

then preform each calcuation on the piecewise

[pale python]

Cos(f(x) = g(x - 4cy) / sin(pi)

This shows that the answer is

[grim ice]

I understand how to read the equation but plotting the points is where I am messing up

cause I would plot them differently

[jade wharf]

oh ok

so the +12 there is a translation by 12 to the left do u understand that

[grim ice]

yes

but it becomes -3 once you factor

[jade wharf]

do u also know that a negative coeficent means that theres a reflection

[jade wharf]

why are u factoring

dont factor

[grim ice]

well, I have been taught that since the -4 is with the x (-4x), then I have to isolate it and divide that 12 by -4

[jade wharf]

what?

this is transformations

[grim ice]

thats how he got (-4(x-3))

yep

[keen matrix]

[
g(x)=
\begin{cases}
-3x-15 & \text{for} & -6\leq x\leq -4\
\frac{3}{4}x & \text{for} & -4\leq x\leq0\
-x & \text{for} & 0\leq x\leq4
\end{cases}
]

From there preforming the calculations is easy then you can simply replot the piecewise :)

damn still cant use the cases package

[jade wharf]

so he factored out the -4

[grim ice]

yep

[jade wharf]

but why did he do that?

or she whatever

cuz theres aboslutely no reason why you should be doing that if ur graphing transformations

[grim ice]

-4 would be a compression about the y-axis, with a stretch of 4

and then x-3 would mean 3 to the right

[jade wharf]

compress

[grim ice]

compress

yes

its late mb

[jade wharf]

yeah that would be right except thats not the function

maybe im just wrong then

i dont get what ur teacher did

[grim ice]

and the negative symbol in front is what makes it a reflection, but the 4 is the value for the compression

so there is a reflection going on

[jade wharf]

yes

oh shit nvm i got it

u have to distrubte the -4 still

once u do x-3

[grim ice]

point (-6,3) ends up at (4.5,3)

[jade wharf]

u still have to multiply that by -4 and thats ur final input

to get y

so yeah

[grim ice]

its like, ive been doing these questions for a while, and my unit test is tomorrow.
yet, I cant understand how he plotted the points that way

cause I would read it differently

and put them somewhere else

[jade wharf]

did he translate 3 to the right

or 12

to the left

[grim ice]

thats a damn good question

[jade wharf]

which do u think is correct

[grim ice]

he must have traslated 3 to the right

otherwise, the factoring crap he taught us is no good

[jade wharf]

which is weird as hell becuase thats not what the original function does

[jade wharf]

yeah exactly because the factoring does nothing

R: reflect over y axis (negative sign in front of the 4x)
S: horizontal compress by 4 (4 in front of x)
T: translate to the left by 12 (+12)

well i said my piece and it doesnt seem like thats the way ur teacher taught u

[grim ice]

yeah, its fine if we cant figure it out, I can ask him before the test tomorrow

it could be an odd ball in the answer sheet thats wrong

today, I actually corrected him on a question in an assignment that had a typo

so he could have messed something up, idk.

Thanks anyways

.close

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[limpid pulsar]

I need help with a; not sure what the question is asking.

[acoustic pecan]

what is f

[sonic escarp]

for b: 9 = 3^2, 27 = 3^3, (a^b)^c=a^(bc)

[limpid pulsar]

y?

Not sure exaactly

[acoustic pecan]

you sure it wasnt defined somewhere prior?

[sonic escarp]

its your example. if you dont know what is given how should someone else do?

[limpid pulsar]

Oh i think theres a miscommunication

I am like HEAVILY confused by this problem and it wasnt defined anywhere else. This is the beginning of a problem

Does the question not make sense?

(I have no clue if it makes sense or not thats why Im asking)

[sonic escarp]

there is no equation in a)

so no that does not make any sense.

[limpid pulsar]

Okay

[sonic escarp]

you can do b.

[limpid pulsar]

Ok, got it

I have 3^(3x*x)=3^3(x+1)

[sonic escarp]

no.

wait

[limpid pulsar]

no.

[sonic escarp]

$3 \cdot 9^x=3 \cdot (3^2)^x=3\cdot 3^{2x}=3^{(2x+1)}$

[thorny flame]

ThM

[sonic escarp]

do the right side in an analoguos way.

[limpid pulsar]

why +1?

[sonic escarp]

$a \cdot a ^n = a ^{n+1}$

[thorny flame]

ThM

[sonic escarp]

ore more general: $a^m \cdot a ^n = a ^{m+n}$

[thorny flame]

ThM

[limpid pulsar]

Okay, that makes so much sense

I do not know what analogous means

[sonic escarp]

using the same rules.

[limpid pulsar]

Ah alright

[sonic escarp]

at the end you should have 3^something.

[limpid pulsar]

Yes

I got

3^3x+1

[sonic escarp]

use brackets, maybe you mean the right thing, but what you have written is wrong.

[limpid pulsar]

oh

3^(3x+1)?

[sonic escarp]

betrer, but still not right.

[limpid pulsar]

💀

Haha

Okay

Ill try again

27x+1 turns into (3^3)^(x+1)

(a^b)^c=a^(bc)

[sonic escarp]

use brackets.

[limpid pulsar]

oh

[sonic escarp]

now you are on the right way.

[limpid pulsar]

Noted

a=3

b=3

c=(x+1)

3^((3)(x+1))

:/

[sonic escarp]

yes, now you can remove some brackets.

[limpid pulsar]

oh

3^(3)(x+1)

ah

[sonic escarp]

no, not this one. 3(x+1)=3x+3

[limpid pulsar]

Confused.. so 3^(3x+3)?

[sonic escarp]

well, thats it.

[limpid pulsar]

Sorry... I'm not very intelligent

[sonic escarp]

thats just a practice thing.

[limpid pulsar]

Okay so currently

3^(2x+1)=3^(3x+3)

[sonic escarp]

well, if a^b=a^c -> b=c.

[limpid pulsar]

2x+1=3x+3

Got it

I should be able to solve from here (Id hope)

Thank you a lot for your help, ThM. I appreciate it a lot

[sonic escarp]

youre welcome.

[limpid pulsar]

Im gonna close the ticket now

.close

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[quasi peak]

hi

[quasi peak]

i believe the support is S = (-θ, θ) because it's given that x can only take values between -θ and θ

but how do i work out the total sample space?

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[quasi peak]

<@&286206848099549185>

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[true hamlet]

This question is weird.
The support is indeed (-theta, theta), but we have no sufficient information on the probability space on which X lives (in fact we have no information at all on it) to identify a sample space, assuming this is what Omega means here. It could be anything.

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[lusty goblet]

Yes u r right

[true hamlet]

@quasi peak ^

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[thorn viper]

3x + 2 = 3x 2

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[neon iron]

Hi just wondering for a simple math conversion question of litres per km to miles per gallon

[neon iron]

Question 10, am I able to change 6.9 L/100km to 14.5km/L or no

[copper zinc]

yes, you want to begin by converting liters per 100km to km per liter, from there you can convert to miles per gallon

[neon iron]

Okok thank you

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[proud kettle]

Hi

[proud kettle]

#13

This is polygons interior angles

[neon iron]

ok so the sum of the interior angles of a quadrilateral is always 360 degrees

so the equation you need to solve is the (sum of the interior angles) = 360

which is 110 + 130 + (x - 3) + x = 360

[proud kettle]

I wrote out 110 + 130 + x + x -3 = 360

[neon iron]

so now solve that for x

[proud kettle]

X = 61.5

because you combine like terms which is 237 +2x = 360

[neon iron]

so what’s your question?

[proud kettle]

Subtract 237 both sides so then 2x = 123

I was seeing if I did it right

[neon iron]

oh then say that next time

[proud kettle]

oh sorry

Did I do it right tho

[neon iron]

yes

[proud kettle]

The angle is a decimal, do I round it to the nearest whole?

(61.5)

[neon iron]

actually maybe not i don’t know a lot about this server

personally i wouldn’t unless the question or problem sheet says to because if you don’t you can generally convince the teacher that you didn’t know but if you round it that’s harder

[proud kettle]

it just said solve for x

[neon iron]

yeah don’t round it

[proud kettle]

ok

thanks

[neon iron]

np srry for the confusion

[proud kettle]

No no it’s my fault

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[hollow dew]

Can someone show me the simplest way to solve this problem

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<@&286206848099549185>

plzzzz some 😢 I have test tmrw and I need fastest way to solve this

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[buoyant rose]

yo, Trigo isn't working

[buoyant rose]

so basically: there's a problem.
triangle looks like this:

gotta find a

I tried using tan
so I wrote down: Tan55=a/6
and then switched 6 and turned it into multiplication
6*tan55=a

[barren salmon]

yes

[buoyant rose]

I get 7.025

In the answers, It says .8.57

[keen venture]

That's a b(22), not a 6

[buoyant rose]

oh- Sorry copied it wrong

but B is still 6

[keen venture]

Answer keys are wrong a lot

What's with the 22?

[buoyant rose]

there

[barren salmon]

oh

[keen venture]

But yeah I agree with your work

[barren salmon]

1.428 * 6

[buoyant rose]

so then what's da problem?

[keen venture]

,calc 6tan(55 degrees)

[thorny flame]

Result:

8.5688880404527

[buoyant rose]

eh?

how did you`

[keen venture]

Is your calc in radians?

[buoyant rose]

radians?

[keen venture]

,calc 6tan(55)

[thorny flame]

Result:

-271.09852746313

[keen venture]

No your calc is not in radians

[buoyant rose]

uh...

I just click on 6Xtan55

and =

[keen venture]

So I'm not sure! Something is going wrong when you punch your numbers in.

[buoyant rose]

do I need to press Shift?

[keen venture]

,calc 6tan(55 gradians)

[thorny flame]

Result:

7.0250973966752

[keen venture]

Your calculator is in gradians

That's a rare one

[buoyant rose]

gradians? eh

I got the casio fx-92MS

[keen venture]

So there's three common angle measurements, and your calculator can work in any of them.

You might see a menu that says "deg, rad, grad"

[buoyant rose]

this guy

How do I get the correct answer on this?

,calc 6tan(55 degrees)

[thorny flame]

Result:

8.5688880404527

[buoyant rose]

,calc 6tan(55)

[thorny flame]

Result:

-271.09852746313

[buoyant rose]

,calc tan(55 degrees)

[thorny flame]

Result:

1.4281480067421

[buoyant rose]

'

I'm really confused guys...

,calc 6tan(55 degrees)

[thorny flame]

Result:

8.5688880404527

[keen venture]

What's up?

It's that "mode" button

[buoyant rose]

calculator is going crazy

oh.. mode

what mode do I need?

got it working! Just had to put it on degree mode. thanks guys (Didn't really help, but I'll be polite lmao)

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[tawdry pebble]

Hey this is Calc AB and we are currently using U-Substitution. Can anyone help me solve this problem?

https://media.discordapp.net/attachments/775178173192863757/1207552044664094740/20240215_000006.jpg?ex=65e00f7a&is=65cd9a7a&hm=2913354c664792786a0d18e2e4b595d7b37febbdd0b0d86430488a4acd2666af&=&format=webp&width=1440&height=614

[versed cairn]

convert into the integral first and then solve

[tawdry pebble]

Typed out it is: Find the area between the graph of f(x) and the x-axis from x=0 to x=2

f(x) = 4x/(x^2+4)

and the answers are (MCQ):

A. ln4 B. ln8 C. ln16 D. 32

[tawdry pebble]

tried and ended up getting ln16, but not sure if its right or not

[versed cairn]

oh u were asking if ur answer is correct

can u show ur working?

[tawdry pebble]

https://media.discordapp.net/attachments/775178173192863757/1207553466403131403/20240215_000529.jpg?ex=65e010cd&is=65cd9bcd&hm=0b4313d5a56619b29d669f9eb8f99438a3bf613f5c2fe391ab231a5a03fe823c&=&format=webp&width=1141&height=671

[versed cairn]

u didnt change the integrals bounds

[tawdry pebble]

ohh yeah 1 sec

[versed cairn]

also ln0 is not 0 but undefined

[tawdry pebble]

ok just changed it

i changed the bounds so it's from 4 to 8, and then 2 * (ln|8|-ln|4|) =

(ln|8|-ln|4|) = ln|2|

2 * ln|2| = ln4

[versed cairn]

yea

[tawdry pebble]

is that correct?

[versed cairn]

yes

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[clear knot]

can yal explain this like its to a 5 year old

[carmine marten]

the observed average of a random sample is itself a random variable (because each sample you take the sample will be... random)

the expected value of this is the same as the expected value of the variable you're sampling

[clear knot]

Ya cuz the separate parts all follow the N(u, o^2) right

[carmine marten]

yes

the variance is proportional to the variance of the variable you're sampling, but also the more samples you take the less variance there is of the observed average (inversely proportional)

[clear knot]

but then why the o^2 is o^2/n

[carmine marten]

so e.g. if you roll a (normal) dice having an observed average of 6 means every roll was a 6 which is much much much more unlikely than getting a 6 ones, and in general the average of 100 rolls will be much closer to 3.5

[clear knot]

but then in this case if n gets larger than it will always return to 0 refuting the claim that it should be 3.5

[carmine marten]

the variance goes to 0

not the expected value

so it's still got the same expected value

but the distribution is less spread out around it

[clear knot]

so basically it get more precise the more n's

[carmine marten]

yes

it's also important that the observed average is also a normal random variable

[carmine marten]

e.g. here I was trying to illustrate the spread of observed averages as you repeat those sets of 100 dice rolls will cluster around 3.5 a lot more than the result of a single roll (because anything away from that becomes unlikely and 6 was an extreme example), and will do so more if you start doing sets of 200 or whatever

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[proud idol]

I want to understand how we will do the 3rd question and 4th part of 1 b)

[carmine marten]

the 4th part of 1b should be straightforward: just do the usual check(s) and remember that matrix multiplication distributes over addition

For 3, you should be aware (judging by 2) that you can uniquely define a linear transformation by defining it on a basis. So to find all linear transformations it is enough to find all the possible things it can do to a basis.

[proud idol]

How will i show it though?
I don't get what is BAB-

Like I know what it is
But how do I show it

[proud idol]

... uh, that went over my head

[carmine marten]

In Q2, they define T by defining it only on 3 vectors that form a basis

You can define any linear transformation this way, the resulting transformation is unique, and any such definition does give a linear map

[proud idol]

Yes this is clear to me

[carmine marten]

So given a basis v1, ..., vn of R^n, to find all linear transformations it's enough to describe all such definitions

[proud idol]

But linear transformations can be endless no?

[carmine marten]

well, yes

but you can still write all of them down using variables

[proud idol]

can you show how?

[carmine marten]

using matrices is the nicest way to write them all down

[proud idol]

matrices?
how will I get the entries?

[carmine marten]

any n x 1 matrix gives a linear map, all that's left is really to show that each linear map can be defined by a matrix like this

any linear map has a matrix associated to it

also Q4 gives you a really nice clue

[proud idol]

ughhh
all of this is so confusing

[proud idol]

by map,do you mean transformation?