help-10

[fossil crag]

So except for constants, everything else is higher than order 2 with kt²

[minor furnace]

right

final answer I obtained is $1+\frac{\Phi^2 x^2}{2} + \frac{\omega ^2 t^2}{2} - \Phi \omega tx + \omega t + \Phi x - kt^2$

[warm shale]

Triaxyz

[fossil crag]

Perfect, just regroup terms alike together

[minor furnace]

alright awesome

I appreciate your help

.close

[obtuse pebble]

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[obtuse pebble]

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[shut comet]

how do i determine intervals for a function from a table?

[tardy epoch]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[shut comet]

i just wanna know how they got the intervals

[tardy epoch]

this is just the answer. where's the question?

[shut comet]

theres no question, i am just confused on how they put the intervals in the function

[daring sorrel]

Just think of it as a sequence of events starting at t=0, and once you finish the first you move onto the next. After the swimming part of the race, 0.5 h have passed, so you begin biking at t=0.5h

[shut comet]

OHH thank you sm

that makes sm sense

.close

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[last hatch]

For part 2 of this question, what function should I be looking towards to model a function after? (linear, exponential) or am I on the wrong track?

[obtuse pebble]

@last hatch Has your question been resolved?

[last hatch]

<@&286206848099549185>

[brazen viper]

@last hatch you're wanting to solve this function for t. It's going to be logarithmic.

[last hatch]

Would I need to rearrange the function so it’s t = (rest of function)

[brazen viper]

Yup

@last hatch ^

[last hatch]

Sorry, I’m a bit stuck for what would go here while I rearrange the function

[brazen viper]

That's going to be L

Which is now your independent variable, and t is t(L)

But you can just use L and t

@last hatch

[last hatch]

L-infinity right?

[brazen viper]

Just L

[last hatch]

Alright I’ll give it a shot ty

[last hatch]

Does this look right? If not where did it go wrong

[brazen viper]

It looks wrong to me

You're violating log rules

Move the L_inf outside the parens to the left before taking the log

[last hatch]

Like this?

If it’s too much to explain is there any video on khan that would be able to

[brazen viper]

\begin{align*}
L &= L_\infty - (L_\infty - L_0) \mathrm{e}^{-kt} \
L - L_\infty &= - (L_\infty - L_0) \mathrm{e}^{-kt} \
L_\infty - L &= (L_\infty - L_0) \mathrm{e}^{-kt} \
\frac{L_\infty - L}{L_\infty - L_0} &= \mathrm{e}^{-kt} \
\ln\qty(\frac{L_\infty - L}{L_\infty - L_0}) &= -kt \
t &= - \frac1{k} \ln\qty(\frac{L_\infty - L}{L_\infty - L_0})
\end{align*}

[warm shale]

OmnipotentEntity

[brazen viper]

@last hatch

[last hatch]

Oh I see. Thank you very much for the help

.close

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[brisk arrow]

tf are the question is talking about

it is badly described

what the y-axis and x-axis stands for?

ridiculous

How do i solve it

[brazen viper]

The y-axis is the probability density and the x-axis is minutes

Hint: the area under the curve is 1.

Hint 2: ||what does the height of the triangle need to be for its area to be exactly 1, given its base as shown on the graph?||

[cold hazel]

is that bell curve

[brazen viper]

Triangular distribution.

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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[nocturne hinge]

help

[nocturne hinge]

Question 7 d)

<@&286206848099549185>

[surreal mason]

Do you know what an axis of symmetry is?

@nocturne hinge

[nocturne hinge]

yes

the midpoint of the two x-intercepts?

[surreal mason]

x intercepts?

no

[nocturne hinge]

wdym

[surreal mason]

It's a line you reflect your point across

in this cay x=6

so the point is how many units away from x=6?

[nocturne hinge]

hold up ill rewrite it

[surreal mason]

so you want (0,7) to be flipped 6 unit onto the other side of the ilne

so it's 6 away on the left

it needs to be 6 away on the right

[nocturne hinge]

12?

the other point

is it (12,7)?

[obtuse pebble]

@nocturne hinge Has your question been resolved?

[obtuse pebble]

@nocturne hinge Has your question been resolved?

[nocturne hinge]

.close

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[cunning burrow]

Yes that is correct. This comes directly from the definition

[obtuse pebble]

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[topaz echo]

Isn't this just the magnitude of the u vector???

ie:

sqrt(3^2 + 1^2)

Like what am I missing???

[tardy epoch]

,calc sqrt(10)

[warm shale]

Result:

3.1622776601684

[tardy epoch]

try just 2 decimal places

[topaz echo]

That didn't work

I tried putting in 3.16

wait figured it out

thanks

.close

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[warm flame]

I think A

Am I right

[royal basin]

how did you get A?

show us so we can tell you where you made a mistake.

[wheat kindle]

for this type of question it usually helps to start with a simple model

[warm flame]

5!/2!

?

[royal basin]

why divide by 2! ?

[warm flame]

That's what I usually see in YouTube videos

💀

So B?

I used 2 as R in permutation and combination formula though

[wheat kindle]

if u wanted to work through this logically u could first consider the same question for a word with 2 letters

[warm flame]

And I got 20 and 10

[royal basin]

oh yes, proof by somebody else did it.

[zenith raft]

proof by somebody else did it on a different problem

[obtuse pebble]

@warm flame Has your question been resolved?

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[exotic walrus]

reference angles

[exotic walrus]

what is the best way to find this reference angle

i did this but it's not a reference angle

so do i just subtract it from 3pi/3 now?

[high lily]

it may help to draw the terminal side on the unit circle
the reference angle will be the acute angle formed between that and the x-axis

[exotic walrus]

Yeah, I understand this

But what is the best way to find it algebraically?

[high lily]

visualise the location first

that'll give you an indication of whether you should
subtract pi
subtract from pi
subtract from 2pi
etc

[exotic walrus]

When would you need to subtract pi?

Quadrant I?

[high lily]

no

please draw this out

a unit circle and 4 terminal sides, one in each quadrant

and mark the acute angle formed each one and the x-axis

[obtuse pebble]

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[crimson tide]

Hello I asked for help for this question this morning. And now I'm back. I don't understand what I need to do to solve it. I beg can someone solve this for me and show me the work so i can have even the slightest idea of what I'm meant to do.

[crimson tide]

a little bit of context my math teacher is a man on a tiny screen in the corner of the room

iv had to go to the other math teachers to teach me so I can pass but recently they have been removed so Im close to screwed this upcoming final.

[obtuse pebble]

@crimson tide Has your question been resolved?

[static furnace]

can i ask what math class this is for?

[crimson tide]

calculus

[static furnace]

okay so you're given the equation for position, s=1.86t^2, do you know how to find the equation for velocity from this?

[crimson tide]

from that no

[static furnace]

so from the equation for position, you can derive it with respect to time to find the equation for velocity

[crimson tide]

which would make s=3.72t?

[static furnace]

yeah so we'd change it to v=3.72t for velocity

and you're given velocity, but need to find the time it takes to reach that velocity

[crimson tide]

but dont we need the displacement too?

I'm following the formula t=d/v

[static furnace]

is that the equation you're supposed to use?

[crimson tide]

im not sure really the last person to try and help me said something about velocity equation

[static furnace]

yeah so the easiest way to do this is by finding the equation for velocity from that of position

in my opinion

i think attempting to find displacement would require the time, where all your given is velocity

because now that you have the equation for velocity, you just need to plug in the values and find t

[crimson tide]

im a little confused sorry. so id have t=d/3.73t?

[static furnace]

it's alright. i just don't think this is the equation to use, and here's why:
In order to find the displacement, you need to know how long the object is moving for, which is exactly what you're missing. you can't find displacement because you need time, t, for that

so, we don't use that equation

[obtuse pebble]

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[static furnace]

we use the equation for velocity, which you quite literally derive from the position equation given to you

[crimson tide]

ohh okok

[static furnace]

.reopen

[obtuse pebble]

✅

[static furnace]

so for Mars, s=1.86t^2, and if we derive that with respect to time we get v=3.72t

and since you have v, you can now find t

s being position and v being velocity

[crimson tide]

ok I get that part its just how do I find t using v?

is there a formula im missing? TT

[static furnace]

the equation is v=3.72t. you are given v=27.8, now find t.

[crimson tide]

I plugged in the v into the equation

[static furnace]

the formula v=3.72t is just the derivative of s=1.86t^2

[crimson tide]

and then I devided the 3.72 on both sides

and I ended up with t=7.47

[static furnace]

perfect

and that's for Mars

[crimson tide]

REALLY AYYAAY finally something TT

[static furnace]

is there anything confusing you still?

yea!

[crimson tide]

no not anymore thank you TT

[static furnace]

okay so now let's do Jupiter yea?

[crimson tide]

yeah 👍

[static furnace]

alright so s=11.44t^2

what's the equation for velocity here

[crimson tide]

same thing right? 27.8=2.88t 22.88t being the derivative of 11.44t^2

[static furnace]

yes perfect. and the derivative of s is v, so v=22.88t

and yea 27.8=22.88t

and there's your time for Jupiter

[crimson tide]

ooooo okok i get it nowwww

[static furnace]

yayaya

[crimson tide]

now i js need to do this with the next 3 missing home i got ayayyaya 🥲

[static furnace]

well if you need help the discord is always here

and feel free to pm me if you want. i'm chronically online so

[crimson tide]

oh my days ty all my love to my last helper but I didnt really understand them TT

by pm you mean?

[static furnace]

dm

[crimson tide]

okok

[static furnace]

yea

[crimson tide]

anywho tysm you js ended a solid 2 hours of misery for me

[static furnace]

anytime!

[crimson tide]

okiii byee byeee ty againnn

idk how to close this thing : , D

[static furnace]

.close

[obtuse pebble]

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[static furnace]

(:

[obtuse pebble]

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[proper kelp]

for 3 odd's case, did 4C3 times 3! times 5!, logic being 4C3 ways to choose the 3 odds which can be arranged 3! ways, then 5! since its in a circle so (6-1)!

but i got the wrong answer

[timid silo]

for which part first of all

[proper kelp]

oh sorry

d)

I did it by cases, and the working out i wrote was for the case where there are 3 odd's together

[livid idol]

whats the formula of summation

[proper kelp]

of what now

[livid idol]

of 2

[timid silo]

well i think it should be like

$\binom 43 \cd 3! \cd 4! \cd 3$

[timid silo]

its like uh

you choose 3 out of the 4 odds together (3C4)

then you permute the the three odds (3!)

[proper kelp]

u mean 4C3

[timid silo]

yes

then the circular permutation of the odds and evens (4!)

and then finally

3 to consider all the ways to include the last odd number between the even numbers

[proper kelp]

what does this one mean

[timid silo]

like its the (n-1)! thing

[proper kelp]

oh

why is it not 5! though

since the three odds we treat as single term, and there are 5 other terms so (6-1)!

btw ur answer is right

[timid silo]

yeah but i considered like the last odd number seperately via the last multiplication of 3

so its really just permuting the group of odds we selected and the evens

and then accounting for the extra odd number

[proper kelp]

i see, so 3 spots since it has to be between even number

[timid silo]

yes

[proper kelp]

but I don't see why 5! wouldn't work

[timid silo]

like uh how should i explain this

[proper kelp]

its fine don't try too hard to make me understand, I'll just try follow whatever you explain it as

[timid silo]

its like the process is like, you have the restriction that three odd numbers are together in any arrangement. So what you can do is choose 3 of the 4 possible choices, then permute the 3 inside of their own group to account for all the possible arrangements

After doing so you can treat the whole group as a block. So you now have 6 positions, but because we want exactly 3 odd numbers together we cant overcount by simply permuting 5!. So we separate the extra odd case to be after

[proper kelp]

OH i get it 5! counts for the cases where the remaining odd is next to the other odds, meaning its 4 odd together

[timid silo]

yes

exactly

[proper kelp]

thanks

I was so confused

ty!

[timid silo]

yeah dw it took me a bit too

[proper kelp]

.close

[obtuse pebble]

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[proper kelp]

Isn't this question basically asking whats the probability that the men and women alternate, or is it saying that no more than 2 men can be together

[timid silo]

it is saying that no more than 2 men can be togheter

like uh

you could have:
MWMWMWW or
MWWMWMW

if you let the amount of men be 3

It says "no two men", meaning no bromance whatsoever can happen here

[proper kelp]

Oh so just all alternating

[timid silo]

Yep

There's so many jokes I could make but it seems like I should refrain

the only way the situation can occur is if two women sit next to each and then all the rest are alternating pretty much

[proper kelp]

oh what

i thought they could all just be alternating

[timid silo]

With n men and n women it would be the case

[proper kelp]

nvm

yeah i realised its different to a line since the ends of the line meet too

[timid silo]

i can provide my approach but can u verify my answer in case im spewing gibberish

im thinking its like uh

[
\f{n!(n+1)!}{(2n)!}
]

[timid silo]

possible ways

[proper kelp]

yep thats right

[timid silo]

ok thats great

so like

[timid silo]

no giving asnwrews!!1-1! BAN Qylo!!!

annyeong is a common helpee here i doubt they are solution sniping

[proper kelp]

I already have the answers though

[timid silo]

anyways yeah so we can go from this

you can consider the men to be fixed in their positions

so the females have (n+1)! ways to move around

now do the opposite and fix the women

you get n! for the men

so like it should be n!(n+1)! normally

but because its a circle we have to minus the extra shit

so like with n +1 +n = 2n+1 people you apply (n-1)!

so you get (2n+1-1)! = (2n)!

and ta da

[proper kelp]

yep i get the denominator bit

[timid silo]

what about the numerator

[proper kelp]

hmm

[timid silo]

Shouldn't that be done by dividing by the number of possible rotations?

[proper kelp]

there was this other question that seemed the exact same except with numbers, can I show you the working for this

[timid silo]

Which would be 2n+1...?

yeah

but like in a circle youre considering only (n-1)! permutations

instead of the normal n! permutations

Oh wait hmmm brain starting to brain again

[timid silo]

ye sure

[proper kelp]

This is for alternating

actually no

nvm

[timid silo]

[proper kelp]

the question was for 5 men 5 women

[timid silo]

whats ur confusion

[proper kelp]

this working is just wrong

[timid silo]

[proper kelp]

well if we fix 1 women, men have n! ways, then the rest of women have (n+1-1)!

and that one fixed women can be (n+1) different so yeah i think this checks out

[timid silo]

yes

[proper kelp]

would n+2 women and n men make it n!(n+2)!

[timid silo]

yes

uh wait

whats the restriction

no two men sit together?

[proper kelp]

yep

maybe there are more since there are 2 places where women are together

[timid silo]

n = 2

WMWMWW
MWMWWW

i dont think it matters it still will be alternating

[proper kelp]

is it n!(n+1)! x (2n+1)

since its like adding 1 women within the n men n+1 women scenario

and there are 2n+1 places for the women to be

I don't really know just a guess

[timid silo]

i think using stars and bars for the general case might be easier

[proper kelp]

haven't heard of stars and bars before but yeah it wasn't an actual question I was doing, just one which I thought would help solidify understanding

[timid silo]

im p sure it should remain the same but

[proper kelp]

pika cutie

anyways yep thanks for the help

.close

[obtuse pebble]

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[timid silo]

nw good luck

[obtuse pebble]

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[placid crescent]

In the diagram, ABC is a triangle, DE is a straight line which cuts the line at the points F and G, and DE is parallel to BC. Find the sizes of the following angles and giving reasons for your answers.

a) angle AGF
b)angle EGC
c) angle ACB
d) angle ABC
e) angle DFA

[latent walrus]

progress?

[placid crescent]

Hello

I’ll send it but idk if I’m right

@latent walrus

[latent walrus]

a seems fine
im not sure whats going on at b, your work process and notation isnt very clear, nor is the way you present your answers

[placid crescent]

That’s what I’m talking about I’m confused and shit pls help

[latent walrus]

so you understand how you got a right

[placid crescent]

Yes

[latent walrus]

for b, you can go directly to opposite angles being equal so AGF=EGC
or you can do it a longer way ie
AGE=180-AGF and then EGC=180-AGE, which shows the same thing

which seems like what you were going for

[placid crescent]

Alright 👍

Ima correct it real quick and the answer is 76

How about c?

@latent walrus

[latent walrus]

alternate angles

or corresponding angles, either rule works

[placid crescent]

Hm ok wait

[placid crescent]

So it’s the same with AGF?

[latent walrus]

indeed

[placid crescent]

Alr so it’ll be:

B = 50
A = 54
C = 76

How about d @latent walrus , I’m sure I can do e so just tell me d only if I don’t bother you please

[latent walrus]

what do you mean, d

you already did it

d) ABC, is the angle youve denoted as B

[placid crescent]

Oh ok

Ty

!close

.close

[obtuse pebble]

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[obtuse pebble]

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[cosmic cliff]

Another analytic geometry problem i need help with

We have a circle with an equation (x-3)^2+(y-2)^2=5

Thus centre has coordinates (3;2) and radius=sqrt5

The circle has a tangential line, to which belongs the point A(2;0)

What is the equation of said tangent?

[obtuse pebble]

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[wispy minnow]

(b)(i) had me working for days now. I've tried natural logarithm, e^ln() and expanding, but none of them worked out. Do anyone know other possible approaches? (The prove must make use of the first principle, idk why.)

This question is partially modified, and the given limit fact may not be useful.

[obtuse pebble]

@wispy minnow Has your question been resolved?

[wispy minnow]

<@&286206848099549185>

[obtuse pebble]

@wispy minnow Has your question been resolved?

[obtuse pebble]

@wispy minnow Has your question been resolved?

[wispy minnow]

<@&286206848099549185>

[calm junco]

im sorry i wont be able to help with this.

[signal halo]

I'll probably won't be able too, but I'll try just gimme some minutes

[signal halo]

@wispy minnow

What part you have problem with

Nvm

[wispy minnow]

I've written down the limit

and stuck on simplifying it

[remote skiff]

Here is a hint. You know from taking the first derivative using the usual methods that it has the form $x^{ln(x)}$ multiplied by some other junk, and that $x^{ln(x)} = e^{\ln(x)\ln(x)}$. Furthermore, notice that, for example, $e^{\ln(x+h)\ln(x+h)} = e^{(\ln(x)+ \ln(1+h/x))^{2}} = x^{\ln(x)} e^{2ln(x)\ln(1+h/x) + \ln(1+h/x)\ln(1+h/x)}$ giving you a way to factor out the term you need to start working grinding down the limit. Furthermore, notice that limit identity you are given is a lot more useful when taken inside a logarithm than outside.

[warm shale]

JessicaK

[wispy minnow]

good, I'll try it

[wispy minnow]

.close

[obtuse pebble]

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[timid silo]

Is it just a) 1000:1
b) 1:0.2= 5:1
c) 250:1

The reason I’m confused is the units

They are different quantities

Do I just say it’s 250mg:1capsule

1000units:1mL

Etc?

But we are not supposed to have different units in ratios

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[soft bluff]

If you think about something like car mileage, it's a ratio of miles:gallon or km:L (depending on where you are)
Ratios are very useful for determining the relationship between different units.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

yes but i've referred to different websites and they say that the physical quanties you are comparing in a ratio need to be in the same unit. im confused as to how to do that here?

[last pilot]

okay

i think they just mean one capsule, or one tablet

for example, if there's 150g in one packet of, say, corn, here you would write that as 150g : 1 packet

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

dont undersatnnd

[last pilot]

okay

so you asked before whether they meant one capsule contained 250 mg

i believe so

i gtg soon, it's 00:15 for me here so i'm hoping we can wrap this up

basically there they said represent the ratios in the respective units

what it means basically is that you need to represent the ratio--the relation in numbers, or how many of "x" there are in/for 1 "y"--in the respective "x" and "y"

so for example, in Q3, how many grams are there in one packet?

if it was "300 bags containing 1500 kg of rice", it would be how many kg of rice are there in one packet?

i'm sorry, i really gtg now
i hope i helped in solving ur question

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[languid magnet]

help

[latent walrus]

again?

if youre going to ask for help, actually stick around when people try to give it to you

[wary badger]

lol

[spare gale]

really

[surreal forge]

yep

[latent walrus]

same question 4 times in two hours

always timeout

[wary badger]

must be trolling

[obtuse pebble]

@languid magnet Has your question been resolved?

[obtuse pebble]

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[vagrant elbow]

Can someone help with this?

[timid silo]

do you know the product rule and the chain rulr

[vagrant elbow]

I think so but I get confused with the square root in there

I can find the derivative of everything there but I get confused when thats thrown in

[timid silo]

ok try doing it first

and then we can check if you messed up

[vagrant elbow]

hm okay

[timid silo]

can someone help me

[obtuse pebble]

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).

[vagrant elbow]

Okay so I think my first problem is not seeing how the derivative of a sqrt goes into a fraction

(sorry for the bad handwriting)

would it be something like 1 / 9sqrt of x?

[vagrant elbow]

.

[open sorrel]

,calc can do formal derivatives. that's helpful for self-checking

,calc derivative('sin(x^2)', 'x')

[warm shale]

Result:

2 * x * cos(x ^ 2)

[vagrant elbow]

hm okay ill try to use it, ive never done it before lol

[open sorrel]

you can also use CAS like
https://www.derivative-calculator.net/
it can generate step-by-step soln
so that you dun hav 2 w8 4 human response

[vagrant elbow]

,calc derivative('9xsqrtx')

[warm shale]

The following error occured while calculating:
Error: Too few arguments in function derivative (expected: SymbolNode or identifier, index: 1)

[vagrant elbow]

damn

[open sorrel]

,calc derivative('9 * x * sqrt(x)', 'x')

[warm shale]

Result:

(x * 9 / 2 + 9 * sqrt(x) ^ 2) / sqrt(x)

[open sorrel]

@warm shale won't recognise 9x. you hav 2 use * to separate stuff

[vagrant elbow]

ah okay

[open sorrel]

you can change the numbers a bit, then input it into CAS tool, and observe the structure of the solution.
it's like searching for the solution of another similar problem

[vagrant elbow]

hm okay ty

do you know how to keep the power inside the sqrt?

on the website you recommended

[open sorrel]

my pleasure

[open sorrel]

use parentheis (...) after the root sign

to group stuff

[vagrant elbow]

ah gotcha

tysm

alr i got it ty :)

.close

[obtuse pebble]

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[burnt thunder]

how is there no x-intercept

[leaden ginkgo]

because f(x) is never equal to 0

[burnt thunder]

?

i dont get what ur sayin

why not

[leaden ginkgo]

do you know what an x-intercept is?

[high lily]

try solving f(x) = 0

[burnt thunder]

asymptote

ouh

.close

[obtuse pebble]

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[queen lark]

i keep getting stuck at 6=(n-2)! and i dont know how to solve that without trial and error

[wary badger]

6=3!

[queen lark]

right but how do i show my work that i got n=5

[wary badger]

n-2=3

u just have to rewrite 6 as a factorial

because the RHS is also a factorial

[queen lark]

you mean write 6!=n-2?

[wary badger]

no like

6=3!

3!=(n-2)!

3=n-2

n=5

[queen lark]

ahaa ic, so if there is a factorial on either side i can cancel them both out?

[wary badger]

yes

because ur doing the same thing

[queen lark]

okok tysm!

.close

[obtuse pebble]

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[minor tide]

can someone just help me start this? i’m lost

i don’t think it can be partial fractions

[wary badger]

u=x^2+1

[minor tide]

but there’s the x + 4

[shut lagoon]

I'd split it into two integrals first.

[wary badger]

maybe no

yea

[minor tide]

oh wait

[shut lagoon]

Then one of them use u sub as knief* suggested

The other is a known integral

[minor tide]

ohh

[wary badger]

x/x^2+1

and 4/x^2+1

[minor tide]

thank you

[obtuse pebble]

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[nimble iris]

can someone explain to me the (n-i+1) part?

[gilded needle]

the i to n loop skips the first i-1 numbers and starts at i

so instead of looping n times it loops n - (i - 1) times, which is n - i + 1

[nimble iris]

why does it skip the first i-1 # s?

[gilded needle]

that's how it is written:
for j = i to n

[obtuse pebble]

@nimble iris Has your question been resolved?

[nimble iris]

uhhhhh

ill try plugging in numbers as example

ohhhhhh

ok i see it now, like if i is 2 and n=3 j would also start at 2 which means it skipped 2-1=1 numbers, thus it executes only 2 times (3-2+1=2)

[gilded needle]

yes that's right

[nimble iris]

ty!

.close

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[burnt thunder]

i did this wrong

u cant bring down

the denominator

why

theyre all under the same den

[cerulean ibex]

What do you mean you can't bring down the denominator? Could you clarify please?

[burnt thunder]

like

i took it out

bec theyre all under the same one

,rotate 270

[burnt thunder]

didnt they do that here

like they took out the denominator

i multiplied the rest with x

and took it out

bec theyre all under it

💀💀😭😭

[glacial obsidian]

This is the problem

You can't

[cerulean ibex]

^

[burnt thunder]

o

why

[glacial obsidian]

If you multiply by x, x could be a negative number

And replace < with >

[burnt thunder]

ddint they do that

here

[glacial obsidian]

No

[burnt thunder]

what they do

[glacial obsidian]

The x is still in the denominator

[glacial obsidian]

They factor the numerator, and then see when x-4, x+2 and x are positive or negative

[glacial obsidian]

One thing you can do is multiply by x^2

Because x^2 is always positive

[obtuse pebble]

@burnt thunder Has your question been resolved?

[burnt thunder]

.close

[obtuse pebble]

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[shell creek]

https://cdn.discordapp.com/attachments/839950625269088276/1198440640397189190/20240121_023336.jpg?ex=65bee9d2&is=65ac74d2&hm=e1cb7700a0cce03bea9e915e3877c77c63bf70da2e4e9eddd8f1983f1a63ad4c&

[shell creek]

do yall see whats on the picture good enough to read it?

[spare gale]

no

[shell creek]

god dmanit

[gilded needle]

haha, relax, just try to take it with different lighting

also if you can show the original problem statement that you're trying to solve, that would be helpful

[spare gale]

yes

[shell creek]

https://cdn.discordapp.com/attachments/870049820117721158/1198436019280822282/Screenshot_201.png?ex=65bee584&is=65ac7084&hm=6169ea2177a173b990d952e4d8d5472034560c4bb09e94257182ae8fde039a16&

from michael spivak calculus

ch 1 ex 21

i know there are much shorter proofs , but i really wanted to do it myself

and i made some proof that i fact checked a bunch times and seems right , but it's a bit long

i just wanna make sure that it's right

[gilded needle]

this looks like spivak's calculus?

oh wait, yea you said that

[shell creek]

it's chapter 1

[gilded needle]

i just recognized the font haha

[shell creek]

,rotate

[shell creek]

yes

i can see on my screen now

so hopefully you should too in theory

,rotate

[shell creek]

,rotate

[shell creek]

,rotate

[shell creek]

https://cdn.discordapp.com/attachments/839950625269088276/1198455719243022476/20240121_033459.jpg?ex=65bef7dd&is=65ac82dd&hm=a51c8c271518e756885cb433f5269bddd686d2b6fb14a6ad94223a07b32dd334&

thats all

[shell creek]

pls take a look at the proof

i can't find any error in it so seems right

[shell creek]

hmm let me see

[tough bolt]

It's here

[shell creek]

oh

you mean <= ?

as in

$|a| \geq 0 \implies |a| |x - x_0| \leq |a|$

[warm shale]

Bishop

[shell creek]

instead of

$|a| \geq 0 \implies |a| |x - x_0| < |a|$

[warm shale]

Bishop

[tough bolt]

Yes

[shell creek]

alright thanks i will add that in

alr i guess i will leave this open for like 30 minutes or 1 hour then close it.

[obtuse pebble]

@shell creek Has your question been resolved?

[obtuse pebble]

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[torn meadow]

√36a^12b^48

[torn meadow]

ik that its 6a^6b^24 but idk what to do with the remaining √

do i leave it there or erase

[exotic marsh]

do you mean $\sqrt{36a^{12b^{48}}}$?

[warm shale]

blahaquil

[torn meadow]

Ye

No

a^12 b^48

[exotic marsh]

or do you mean $\sqrt{36a^{12}b^{48}}$?

[warm shale]

blahaquil

[torn meadow]

That

[exotic marsh]

ok

[polar fossil]

$\sqrt{\green{36}\blue{a^{12}}\purple{b^{48}}} = \sqrt{36} \cdot \sqrt{a^{12}} \cdot \sqrt{b^{48}}$

[warm shale]

ヘイリー

[torn meadow]

Oh i just realized 48 is root of 7

[exotic marsh]

it is not

[torn meadow]

eh?

7x7=48?

OH wait

Im so sorry

[polar fossil]

it is just 6a^6b^12, what "remaining √" are you talking about?

err b^24

[torn meadow]

Oh i was taught to put the √ on the other side

[polar fossil]

other side

[torn meadow]

like 6^a6 b^12 √

[polar fossil]

oh. yeah i mean. there's a leftover 1 underneath i guess

[torn meadow]

so i just erase it?

not include i mean*

[polar fossil]

yeah you've taken the square root, there's nothing left underneath the square root

but again it's b^24 not b^12

[torn meadow]

ty

wait what

Oh

my math isnt mathing from the last 3hours

but ty

.close

[obtuse pebble]

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[modern heart]

Can someone help me prove this?

[honest forum]

Induction might work

[modern heart]

my goal is to show this initially for all natural numbers n, but my first screen shot comes from the induction step (k+1)

[honest forum]

Oh mb

[modern heart]

base case n=1:: 0<1, eventually

then i need it to hold true for k+1, which is the first ss

any ideas?

[tardy epoch]

Use the k case

[modern heart]

...

[tardy epoch]

Standard induction

If you don't know it do an easier problem first

[modern heart]

well, ik standard induction lol

i have the base case n=1, which holds true

then establish n=k

and show how the inequality holds true for n=(k+1)

[modern heart]

im at this step here

i just need to show how this holds true

[tardy epoch]

.

[tardy epoch]

doesn't look like you've used that here yet

[modern heart]

base case n=1

inductive hypothesis for all natural numbers k

prove for (k+1)

[tardy epoch]

yea use this

[modern heart]

wym

[tardy epoch]

find somewhere to insert the inductive hypothesis

[modern heart]

oh

[obtuse pebble]

@modern heart Has your question been resolved?

[obtuse pebble]

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[unkempt sparrow]

I know the correct answer is just 1, but I'm not sure what process was violated. My instinct is to say the constant rule but we learned about that in Calc 1 and this is Calc 2

Is my instinct correct, or is it something else?

[ruby path]

you're right

the derivative of a constant is 0

[unkempt sparrow]

Sweet, thanks

had me confused with that is not for a second lol

.close

[obtuse pebble]

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[sweet edge]

For this theorem, we have that $\langle S\rangle$ is the set of elements of G that can be written as words in S. So does that mean that $\langle S\rangle$ is a superset of G?

[warm shale]

Juke | ping me if no response

[sweet edge]

since there can be many more ways to express a single element from G?

[sage geode]

No, <S> is always a subset of G

Otherwise you are saying that G has no closure

[sweet edge]

Well

hmm

the thing is though, like for any element in G that we represent using S, we can just append an identity element onto it and get the same result

over and over, no?

[sage geode]

Well, surely

[sweet edge]

okay so the sizes aren't actually that important when saying that one is a subset of another?

[sage geode]

Yeah

[sweet edge]

since the "net action" of elements in <S> and G would be equal?

I see I see

interseting

.close

[obtuse pebble]

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[sweet edge]

.reopen

[obtuse pebble]

✅

[sweet edge]

okay soooo I have this written so far to prove this theorem

is that accruate?

the \implies should be \iff

@sage geode

[sage geode]

Not sure what you mean by <S>: S x S -> G

[sweet edge]

like if we think of it as a fucntion then <S> maps two elements from S to G

[sage geode]

Is that supposed to be * but restricted to S?

[sweet edge]

* as in multiplication?

no

the operation is not defined

[sage geode]

Doesn't it also map more than 2?

[sweet edge]

okay, yes

[sage geode]

• It could involve inverses of S which could be outside of S

[sweet edge]

why

S includes inverses too

[sage geode]

It's only defined as a subset of G

[sweet edge]

but wouldn't that include inverses though? bc G does

[sage geode]

No, it's not necessarily closed under inversion

[sweet edge]

huh

well in any case, <S> does include inverses

so hmm

[sage geode]

Right

Okay so what you can do is recall the following characterization of subgroups: A subset is a subgroup if and only if it is nonempty and is closed under division

[sweet edge]

division?

[sage geode]

Where by division I mean the mapping (a, b) -> ab^{-1}

[sweet edge]

ahh okay

[sage geode]

Start with why <S> is always nonempty

[sweet edge]

well groups cannot be empty and so G is not empty and <S> is a generating set for G so it must have at least 1 element that generates elements in G

right?

[sage geode]

I don't see where the assumption that <S> is a generating set of G comes from

[sweet edge]

that's how it was defined in our definition

[sweet edge]

here

oh wait

[sage geode]

That's <S> being defined as the subgroup generated by S

[sweet edge]

I'm getting ahead

if <S>=G then it's a generating set of G

but we don't know that yet

[sage geode]

Right

[sweet edge]

hmm

Okay so to show that it is a group, I need to show it's nonempty, that it's associative under * and that there is an identity element as well as inverse elements?

but how would I know if that is actually true or not?

[sage geode]

And closure

[sage geode]

Or just this

So, even if S was empty, what would <S> contain?

[sweet edge]

uhhh

lemme thinkg

it'd contain the empty word ,no?

so a word but no letters

[sage geode]

Right

[sweet edge]

okay I see

[sage geode]

So it's nonempty

[sweet edge]

why do we need to do this

since G is closed under * and S is a subset of G, wouldn't S also be closed?

or is that not how that works?

[sage geode]

A subgroup needs to be closed not under inversion but also under *

[sage geode]

No

[sweet edge]

can u show me an example?

I can't imagine where that doesn't work

hmm

[sage geode]

E.g., let G = {1, r, r^2} be a cyclic group of oder 3 and S = {1, r} so that S is a subset of G

[sweet edge]

idk what a cyclic group is

[sage geode]

S is not closed under * as r*r = r^2

[sweet edge]

we haven't gotten there yet

I see though

I get the idea now

hmm

the hard part for me though is that showing these properties to be true without actually having anything to work with

i.e. numbers or something

[sage geode]

Well you at least know that you are working with words so you could use that

[sweet edge]

good point

lemme just try it a bit more

[sage geode]

So take two words and consider their product

[sweet edge]

it's just a larger word that may or may not correspond to another element in G, right?

[sage geode]

Right, so <S> is closed under *

[sweet edge]

Ahh okay

[sage geode]

Now show closure under inversion

[sweet edge]

inversion would be something like I have a word but then add letter(s) to make it nothing?

[sage geode]

You may need to use the fact that
[ (x_1\cdots x_n)^{-1} = x_n^{-1}\cdots x_1^{-1} ]
In case you haven't proven this so far, you can do it by induction

[warm shale]

孤独な豆

[sweet edge]

That was one of my hw problems

it hasn't been checked though, wanna take a look at it?

[sage geode]

Good, so you can use this here

[sweet edge]

I think I got the idea

[sage geode]

Yeah show what you wrote

[sweet edge]

wait

wrong one

this one

wait I didn't prove what u wrote

I simply used it

I didn't prove anything actually

[sage geode]

Yeah that's different

[sweet edge]

this seems so familiar though

lemme check my notes

yeah we haven't done that

I coulda sworn we did

but I can't find it ig

[sage geode]

Yeah anyway it shouldn't be difficult to show

[sweet edge]

Alright

OH

okay so when we defined what an inverse in G was

we used the fact that the inverse of an element is just another element such that g*g' = e

where g is an element in G and g' is the inverse

and e is the identity

we later proved that g' is unique

so can I just do that for <S>?

Since we already have that the empty word is the identity from G?

[sage geode]

If for each word you provide another word which is the inverse of the first, sure

[sweet edge]

How would we gaurentee that though?

like say for example <S> is JUST the empty word, the inverse of that would be... well nothing?

[sage geode]

The inverse of identity is itself

[sweet edge]

oh right, ofc

hmm

[sage geode]

Once you can use this, you are basically done, because the inverse of a word is clearly also a word

[sweet edge]

in that case, what if <S> has a word but not the inverse of that word?

like it has a but not -a

-a just being the inverse

[sage geode]

No, <S> is closed under inversion

[sweet edge]

but why

[sage geode]

Because the inverse of a word $s_1\cdots s_n$ is $s_n^{-1}\cdots s_1^{-1}$ which is also a word

[warm shale]

孤独な豆

[sweet edge]

wait so inversion is not really a characteristic but rather something we do?

like an inverse isn't a thing but an action?

[sage geode]

By inversion I mean mapping each element to its inverse

[sweet edge]

the issue I'm having is introducing the notion of an inverse in a space where we aren't actually told how or why it's possible to do an inverse of something

does that make sense?

like why can we perform an inverse operation in the first place?

[sage geode]

<S> consists of some element of G, right?

[sweet edge]

ik that groups need that property but <S> isn't a group yet,

yeah it does

[sage geode]

And you can invert anything in G

[sweet edge]

yesh

oh

wait

DUH

[sage geode]

Since it's a group

[sweet edge]

I SEE

lol

okay yeah this makes sense

does this look good so far?

[sage geode]

Yeah no need to mention the case when S is empty though, just say "<S> always contains the empty word"

And you didn't really mention an argument for closure under inversion

[obtuse pebble]

@sweet edge Has your question been resolved?

[sweet edge]

mb i got busy

[sweet edge]

uhhh, well do I need to? I kinda just referenced that it's the case with G and so it should be with <S> too?

[sweet edge]

I'll have to revisit this another time

.close

[obtuse pebble]

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[obtuse pebble]

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[warm shale]

kyo

what is the radical form of $27(2xy)^\frac{1/2}$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.52 ...s the radical form of $27(2xy)^\frac{1/2}$
                                                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[obtuse pebble]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[normal warren]

.close

[warm shale]

kyo

[obtuse pebble]

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[lone sky]

help with hw

[lone sky]

<@&286206848099549185> Pls help bro 🙂

[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.

[timid silo]

what have you tried

[obtuse pebble]

@lone sky Has your question been resolved?

[lone sky]

i did 3c1 * 4c1 * 2c1

but its wrong

[timid silo]

Did your teacher told you about the multiplication principle

[lone sky]

yeah thats wat i did

but i got it wrong

[timid silo]

wait

The total number of selections of at least one fruit which can be made from 3 bananas, 4 apples and 2 oranges is 511 .

This can be calculated by adding the number of selections of one fruit, two fruits, three fruits, and so on, up to nine fruits. The formula for this is:

[lone sky]

why isnt mine correct?

any idea?

i used multiplication theorem

[timid silo]

Wait give me a sec

you can use the multiplication theorem of probability to solve this problem. According to the theorem, if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m × n ways .

Therefore, the total number of selections of fruit which can be made from 3 bananas, 4 apples, and 2 oranges is:

3 bananas can be selected in 3 ways. 4 apples can be selected in 4 ways. 2 oranges can be selected in 2 ways.

Using the multiplication theorem, the total number of selections of fruit which can be made is:

3 × 4 × 2 = 24

However, we need to consider the fact that we can select at least one fruit from the given fruits. Therefore, we need to subtract the number of selections in which no fruit is selected from the total number of selections.

The number of selections in which no fruit is selected is:

0 bananas can be selected in 1 way. 0 apples can be selected in 1 way. 0 oranges can be selected in 1 way.

Using the multiplication theorem, the number of selections in which no fruit is selected is:

1 × 1 × 1 = 1

Therefore, the total number of selections of fruit which can be made from 3 bananas, 4 apples, and 2 oranges is:

24 - 1 = 23

i hope you understand i gtg i have church

[lone sky]

wait

its contradicting ur previous answer

is it 23 or 511

[surreal forge]

not particularly surprising

[pure arrow]

foul ☠

[obtuse pebble]

@lone sky Has your question been resolved?

[timid silo]

what is this called

[surreal forge]

gptzero

[robust bloom]

Do not copy paste from gpt

[obtuse pebble]

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[south lantern]

Sorry- I know I just asked but for part C, I got one root but the answer says it has two roots

[south lantern]

I did x^2 + 6x +9 as my equaation and then put (6)^2 - 4(1)(9) in my calculator and got a 0

[wispy minnow]

0 means the two solutions are identical

[crystal geyser]

repeated root

[south lantern]

Ohh

Ohhhh so that's what

2 equal roots mean?

Whats the difference between two equal and two distinct roots?

(Respectfully idk why that sounded so agressive)

[pure arrow]

the roots are the same

like (x-4)²

[wispy minnow]

distinct means the two roots are not equal

(?

[south lantern]

Okay! Got it

[south lantern]

Makes sense

TYSMMM

.close

[obtuse pebble]

Channel closed

Closed by @south lantern

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[pure arrow]

one of the choices here cant be solved by factoring

but i cant tell u

[south lantern]

I did almost all by quadractic equation

I cant SPELL

[lone sky]

wow can u give me the website link for this

[surreal forge]

it's called gptzero; fun little tool for educators or anyone interested that checks text for signs of ai generation (it's free but has paid subscriptions)

https://gptzero.me/

[obtuse pebble]

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[lone sky]

use Discriminant b^2-4ac
if D > 0 non equal real roots
D = 0 equal roots roots are real
D < 0 complex solutions in the form of a-ib and a+ib, where a and b arbitrary constant

[south lantern]

Thanks!

[obtuse pebble]

@lone sky Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@opaque oxide Has your question been resolved?

[obtuse pebble]

@opaque oxide Has your question been resolved?

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[glass spindle]

i am beyond clueless