help-13

[rigid pelican]

So correct me if I'm wrong

[hazy trench]

Go on

[rigid pelican]

But ^3 means number X 3 right?

Power of 3

[hazy trench]

Power of 3

[rigid pelican]

That means that number X 3 right?

[hazy trench]

No, hold up

Give me a sec

[rigid pelican]

Ohhhh

So it's essentially sqrt

But instead of multiplying the number by itself you multiply the number by itself three times

So for example 3x3x3

[hazy trench]

Yeah, but in this case, we're not multiplying the number but the unit which is cm

Exactly

[rigid pelican]

I'm confused

We need to multiply a number

[hazy trench]

Take a look at this: https://www.youtube.com/watch?v=M2g3KQ_Uaag

Got it?

@rigid pelican ?

[rigid pelican]

Not quite

We already did that

We do the length X width X height

[hazy trench]

Yes

[rigid pelican]

Which is 0.94 X 0.61 X 0.21

[hazy trench]

Yes

That's also right

[rigid pelican]

That gives us 0.120414

[hazy trench]

Yeah, 0.120414 m^3

[rigid pelican]

Yep

[hazy trench]

Since we're doing meter x meter x meter

Also because the cuboid is a 3-D shape

[rigid pelican]

But we need this in the nearest cm

How do we measure to the nearest cm?

[hazy trench]

Re-read your question,

[rigid pelican]

Oh

So what do we do now?

[hazy trench]

We don't measure to the nearest cm, the values are measured to the nearest cm

That's the answer

0.120414 m^3

[rigid pelican]

According to the book the answer is 0.12

Why do we leave the 0414?

[hazy trench]

That's because they rounded the answer

https://www.gauthmath.com/solution/6-A-cuboid-measures-0-94-m-by-0-61-m-by-0-21-m-each-measured-to-the-nearest-cm-F-1714650467703830

That's the verification

No clue why the book suggests that 0.12 is the answer

High-five

[rigid pelican]

🙌

Ok that's all well and good. Maybe we were meant to leave out any numbers after and including 0?

[hazy trench]

Maybe

But why is the question

They never told us to round anything

[rigid pelican]

The book also suggests a weird method here in the answers

[hazy trench]

Throw in a pic

[hazy trench]

Wh...what?

[rigid pelican]

I'm confused too

[hazy trench]

I'm stumped

Try this: https://www.google.com/search?q=Lower+bound+and+upper+bound+for+volumes&oq=Lower+bound+and+upper+bound+for+volumes&aqs=chrome..69i57j33i22i29i30.10989j0j7&sourceid=chrome&ie=UTF-8#:~:text=The lower bound is the smallest value that would round up to the estimated value. The upper bound is the smallest value that would round up to the next estimated value.

I haven't been exposed to the concept of lower and upper bounds

[rigid pelican]

What would the next estimated value be though?

Just looked back in the book and found this

[hazy trench]

Ah, there we go

That cleared things up

[rigid pelican]

Anyways we pretty much got the right answer, I'm gonna need to go now but thanks for your help!

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[void pecan]

Im trying to prove that for all prime p and for all a {0,...,p-1} there exists a unique x {0,...,p-1} such that ax is congruant to 1 mod p

[void pecan]

Ive managed to prove that the x exists but im unsure how to prove that its unique

I did it by using bezouts identity by the way

[terse roost]

Suppose that you have $y$ in $[![0,p-1]!]$ such that $ay\equiv 1 [p]$. Then $ay\equiv ax [p]$ and $a(x-y)\equiv 0 [p]$. That means that $p$ is a factor of $a(x-y)$, so the only possibility is $a(x-y)=0$ (because $p$ is prime) which means $x=y$. Hence $x$ is unique.

[wraith dagger]

Silfer

[void pecan]

thank you, clearly wasnt thinking that was pretty obvious lol thanks

[terse roost]

np

its ofter a good way to prove something is unique

suppose y verifies the property and prove y = x

[void pecan]

while im here im using this as part of a proof of wilsons theorem and im kinda stuck as to why when i pair each term in (p-1)! mod p with its inverse im left with (p-1)*(1), like i can see why thatd be true but im not sure how to prove it

[terse roost]

hum

I have to think a bit

this theorem isnt trivial to prove

well, when you have (p-1)*(1) you know it's congruent to p-1 which is to -1

but I forgot one of the steps to get there

Oh yeah I remember now

[void pecan]

Yeah i know thats why i can see that to be true but i just dont know how to show that youre left with those 2 terms

[terse roost]

i'm writing it

[void pecan]

because im actually not trying to prove wilsons theorem so i cant state like this gives -1 as desired

[terse roost]

You have $(p-1)!$ which is the product of every $k$ between 1 and $p-1$, separate the product into 2 products $P_1$ and $P_2$, $P_1$ contains the integers which are their own inverse.
\\
Suppose $a$ is its own inverse, thus $a^2 \equiv 1[p]$ which bring $a\equiv 1 [p]$ or $a\equiv -1 [p]$. To prove this, you just say that $a^2 - 1 = (a-1)(a+1)$ and use the fact that $p$ has to be a factor of it. Therefore, $P_1 = 1\times (p-1)=p-1$
\\
Then, about $P_2$, every number in there also finds its inverse in there, thus $P_2\equiv 1 [p]$.
\\
Conclusion : $(p-1)!=P_1\times P_2\equiv p-1 [p]$, so $$(p-1)!\equiv -1 [p]$$

[wraith dagger]

Silfer

[void pecan]

Ahh that makes sense thank you! I could see from just trying some examples i shouldve been left with p-1 but i couldnt think of a way to actually show that, that makes sense though so thank you

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[toxic flame]

when we square root an inequality, do we have to swich the direction of the inequality sign?

[toxic flame]

for this question, this must be true

but the bit in red

doesn't make sense

[dapper cargo]

it doesn't

(something)^2 greater than or equal to 0 means something can be any integer or fraction(negative or positive)

[toxic flame]

yup

so this doesn't tell us anything?

[toxic flame]

would it be correct to say that this tells us nothing

[dapper cargo]

well u can only find intervals that satisfies that inequality

[toxic flame]

would that not be all real values?

[gloomy minnow]

both the expressions in the roots can be expressed as a square of a polynomial

[dapper cargo]

yessir

[gloomy minnow]

x^2-6x+9=(x-3)^2

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

[toxic flame]

yep

[dapper cargo]

was that ur actual question or was it the inequality

[toxic flame]

just the inequality

so just solve this?

[gloomy minnow]

yes

[dapper cargo]

yes

is any domain given in the question btw

[gloomy minnow]

you need to handle the cases for when x-3 or x+1 is negative cuz square root is always positive

[dusty hazel]

Exactly.

[dapper cargo]

yea that's why i asked him

[dusty hazel]

$$\sqrt{x^2-6x+9} \neq (x-3)$$
It's
$$|x-3|$$

[toxic flame]

no domain give

[dapper cargo]

solve for both cases then

[gloomy minnow]

you need to solve the cases x<-1, -1<x<3 and x>3

[wraith dagger]

What the hell am I doing here?

[toxic flame]

because we also have (-x+3)^2 ?

[dusty hazel]

Right, but that is done to ensure the sqrt is positive

[toxic flame]

so when x < 3 then we consider this case

[dusty hazel]

Uhm.

For x<3 it becomes 3-x

[toxic flame]

yep

[dusty hazel]

For x≥3 it's x-3.

So yeah, solve like that.

[toxic flame]

cool, thanks everyone

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[stiff kraken]

why is it 0 to pi/2

and not 0 to 1

[glad kestrel]

why would it be 0 to 1

[stiff kraken]

ik on a unit circle its pi/2 but this isn't really a unit cirlce

cause thats point is 1

(0,1)

[glad kestrel]

?

[stiff kraken]

the point

on the graph is

(0,1)

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[toxic flame]

did you find out why?

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[toxic flame]

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[ember phoenix]

this is the question

and im struggling to understand how to draw the x^2+y^2+4z^2 ≤ 1

could someone help me with this

[cedar kiln]

@ember phoenix Has your question been resolved?

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@ember phoenix Has your question been resolved?

[dire geode]

https://mathinsight.org/ellipsoid

[ember phoenix]

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[proper mesa]

im not sure how to approach this problem

is this a matrix function?

and input is a matrix? im not sure what to do here

[torpid python]

@proper mesa product of 2 matrices

[proper mesa]

so A * 3B?

[torpid python]

Yup

[proper mesa]

ahh okok

ty!

[torpid python]

For matrix multiplication use row x column

[proper mesa]

ok

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[rustic ridge]

im having a hard time understanding the FTC part 1

[rustic ridge]

im not rly grasping the idea as to why when we differentiate the integral we get back the function

[azure horizon]

https://www.youtube.com/watch?v=rfG8ce4nNh0

[cedar kiln]

@rustic ridge Has your question been resolved?

[rustic ridge]

thank u very much! ill watch this

[upper abyss]

Let A(x) be the area under f(t) between (a,x)

For a small enough interval, this is approximately a rectangle. That is,
A(x + h) - A(x) = hf(x)

But that's the definition of the derivative in disguise!
A'(x) = f(x)

[rustic ridge]

but why hf(x)

thats what im not understanding

[upper abyss]

Length of rectangle × height of rectangle

[rustic ridge]

Ohh

so the height is f(x)

[upper abyss]

h × f(x)

[rustic ridge]

and h is uhm

the length

[upper abyss]

Another good intuition, if a function is really tall, the area under it is increasing fast.

[rustic ridge]

but the hell do i go from the integral to hf(x)

[upper abyss]

A(x + h) - A(x) = hf(x)

[A(x + h) - A(x)] / h = f(x)

For an infinitesimal h, left is the definition of the derivative.
A'(x) = f(x)

Or if you will, ∫ f(x) dx gives an area function

[rustic ridge]

i see that

i mean

how do u go from the integral from a to x of f(t)dt

to hf(x)

[upper abyss]

I didn't kek

[rustic ridge]

i mean this

i dont know how to explain my question

like how did u end up with A(x+h)-A(x)=hf(x)

[upper abyss]

Left is the area under f between (x, x+h). This is a small interval and is approximately rectangular

So we embrace that. The area is length×width = h×f(x)

[rustic ridge]

oh

[upper abyss]

This is made perfect with the limit

h → 0

[rustic ridge]

still not understanding the definition with the constant going up to x of f(t) dt

i just dont see how it goes from differentiation to integration

[azure horizon]

try the video

[rustic ridge]

kay thank u both, if i have more questions ull probably find me back !

[upper abyss]

So I defined A(x) as "the area between a to x under f(t)"

[rustic ridge]

yes

[upper abyss]

Using the correct notation, that's just
∫ f(t) dt between a and x

[rustic ridge]

yes

[upper abyss]

The argument established that taking the derivative of this, gives back f(x)

Ergo, first FToC

[rustic ridge]

o

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[hazy sleet]

Does this seem right? I’m pretty dumb

[hazy sleet]

There’s also a part B saying (How many times faster is the speed of light than the speed of sound

For part A I just subtracted

This is part B

[upper abyss]

So that's:
300000000

•            340
  

The speed of light is so large that subtracting the speed of sound from it does nothing.

[hazy sleet]

Ohh ok

[upper abyss]

You basically need to convert back from scientific notation to add/subtract

[hazy sleet]

Alright

[upper abyss]

https://www.sparknotes.com/math/algebra1/scientificnotation/section2/#:~:text=Since all number in scientific,coefficients and subtract their exponents.

[hazy sleet]

Ok ima check it out

I also have notes on scientific notation

Thanks for the help

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[forest dust]

Consider the sequence 60,71,82,93,…,11n+5.
How many terms are there in the sequence? Your answer will be in terms of n.
What is the second-to-last term?
Find the sum of all the terms in the sequence, in terms of n.

[forest dust]

i dont get it why would this sequence ever have to end?

[wicked brook]

Let's say n=100

Where does the sequence end?

How long is the sequence?

What is the second to last term

What is the sum

[forest dust]

why are you setting n = 100

[wicked brook]

To show an example

For any n the last term is 11n+5

So if n is 10, the last term is 115

If n is 11, the last term is 126

Etc.

The problem revolves around the fact that n is finite

So eventually we will reach an end of the sequence

[forest dust]

ok but then how would i answer a problem like this in terms of n

[wicked brook]

The second to last will be with n-1 right

[forest dust]

yes

[wicked brook]

So plug that in and you get 11(n-1)+5

To get how many you need to ask what n gives 60

[forest dust]

well n = 5 gives 60

[wicked brook]

So it ranges from 5 to n, with n>5

How many terms are there

It goes 5,6,7,....,n

[forest dust]

honestly kind of lost me there

[wicked brook]

ok

lets try setting n to 10

5,6,7,8,9,10

how many numbers are there?

[forest dust]

6

ok cool so (n-5)+1

[wicked brook]

yea

finally we do the sum

its [11(5) + 5] + [11(6) + 5] + ... + [11(n) + 5]

so the first thing to notice is that there are (n-4) fives

so we have 11(5) + 11(6) + ... + 11(n) + 5(n-4)

what do you think the next step is?

[forest dust]

11(n-5)+5(n-4)

?

[wicked brook]

well let's check

if we factor out 11

11(5 + 6 + ... + n) + 5(n-4)

does that help you?

[forest dust]

not really, sorry

[wicked brook]

do you know how to evaluate 1 + 2 + 3 + ... + n? @forest dust

[forest dust]

$\sum_{n=1}^{k}$

oh wth lol

[wicked brook]

\sum

[wraith dagger]

M00NLIG7

[forest dust]

yeah my bad im new to TeXit

[wicked brook]

ok so you know $$\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$$

So do you see that: $$\sum_{k=1}^{n}k = \sum_{k=1}^{4} k + \sum_{k=5}^{n} k$$

[wraith dagger]

ohNoiAmHere

ohNoiAmHere

[wicked brook]

Also, do you see why this is useful to the question we're trying to answer?

[forest dust]

Well it kind of makes sense but its useful to the question were tryign to answer because our sum is 11(5 + 6 + ... + n) + 5(n-4) and this is basically the part inside of the 11(...)

[wicked brook]

yea

can you get an answer now?

[forest dust]

I think im getting closer im just having trouble with the summation format. I only know $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$ because we went over it in class

[wraith dagger]

M00NLIG7

[wicked brook]

so what i did here is I said: $$1 + 2 + 3 + 4 + 5 + .... + n = (1 + 2 + 3 + 4) + (5 + 6 + 7 + ... + n)$$

[wraith dagger]

ohNoiAmHere

[wicked brook]

and I can evaluate each seperately

bring one side over and (5 + 6 + 7 +....+ n) = (1 + 2 + ... + n) - (1 + 2 + 3 + 4)

[forest dust]

so $$11((n(n+1))/2 -10)+5(n-4)$$

[wraith dagger]

M00NLIG7

[wicked brook]

yea

[forest dust]

so ${\sum_{k=1}^{4} }k$ is just saying at the numbers starting at 1 (k) until it reaches 4 (n). So n in that case is essentially the limit?

[wraith dagger]

M00NLIG7

[wicked brook]

yes

[forest dust]

ok thank you

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[dense wing]

I have a collected set of data (82,25,4,2) and from that I found a maximum likelyhood estimator and got expected frequency of the data to be (in same order) (78.6,28.5,5.2,0.7)

How would I explain if the initial modelling was reasonable/accurate/whatever by comparing the list of numbers

[dense wing]

Question for reference, though mainly asking about the general idea, specifics of the question aren't as important to me

[cedar kiln]

@dense wing Has your question been resolved?

[astral crown]

Maybe a chi squared test?

[dense wing]

What's a chi squared test?

[astral crown]

It basically gives you how good some distribution fits onto a set of values

[dense wing]

Is there a way to do it just comparing values?

Rather not go above the content we have access to.

[astral crown]

Yea fair

[astral crown]

Not sure

I don't know of any

[dense wing]

Ok I'll ask my prof

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[fleet kernel]

idk how to do this

[fleet kernel]

do i just plug the bounds in

[bold vine]

IBP

Integration by Parts

[fleet kernel]

i dont understand how to do that though

[bold vine]

You can look up the IBP formula

You'll also need to move the e^(2y) to the numerator

[bright surge]

rewrite integrand

[bold vine]

^^

[fleet kernel]

9ye^-2y

[bright surge]

yeah, then do ibp

[fleet kernel]

i've only seen the way to do it with tabular notation

but im not very sure how that works either

[bold vine]

You can use tabular too

First learn IBP before you try applying it

[fleet kernel]

u= 9y & dv= e^-2y

i differentiate 9y till it gets to 0 right

[bold vine]

ye

[fleet kernel]

so would it be -9/2ye^-2y-9/4e^-2y+C

[fleet kernel]

?

[bold vine]

Idk I haven't done the integration myself

You can always check integrals online though

[fleet kernel]

i've done it correctly according to this calculator but how do i plug in the bounds?

[bold vine]

FTOC Part 2

Just evaluate the antiderivative at the upper bound - subtract it evaluated at the lower bound

If you've done any definite integrals that's what you'd do

[fleet kernel]

so you just do it how you normally would

i've done that and it says that the answer is incorrect

[cedar kiln]

@fleet kernel Has your question been resolved?

[fleet kernel]

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[shut summit]

Help bruh how can it be π, it should be 2π clearly since it has power 1 ie. Odd?

[crimson sedge]

I don't get what you mean, could you explain?

anyway sin(x) has period of 2pi

so sin(2x) should have period of pi

[balmy ivy]

the coefficient (2 here) changes the period. what you see with the odd even powers only apply if theres coefficient is 1

[shut summit]

Thanks for explaining guys

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[cedar kiln]

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[cedar kiln]

@sinful bear Has your question been resolved?

[cedar kiln]

@sinful bear Has your question been resolved?

[bold hinge]

What exactly do you mean by this

Is there more context

[cedar kiln]

@sinful bear Has your question been resolved?

[dire geode]

i and ii look unrelated

C stands for continuous function

What is the original question

[cedar kiln]

@sinful bear Has your question been resolved?

[dense wing]

Check if this is a subspace

If you think it is, just check it meets the requirements

[cedar kiln]

@sinful bear Has your question been resolved?

[cold oasis]

Can anyone help me solve this?

[vernal palm]

@cold oasis #❓how-to-get-help

[crimson sedge]

the function in the root in denominator has only odd powers of x

maybe a sub?

(it seems one-to-one)

actually the numerator has d/dx of that expression

hm

[dire geode]

take it to a new channel. this one's occupied

[crimson sedge]

you mean to respond to @vernal palm

or actually @cold oasis

[dire geode]

i don't see how anything you said answers
#help-13 message

[crimson sedge]

@cold oasis Change channel to an available one and ping me

[cedar kiln]

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[carmine pilot]

Anyone have idea about how to proof this? Please help me to find it out

[celest ledge]

That is the coefficient of x^(n-1) in (1+x)^2n

I mean, you compare the coefficient of x^(n-1) both sides of the equation (1+x)^n (1+x)^n=(1+x)^2n

[cedar kiln]

@carmine pilot Has your question been resolved?

[cedar kiln]

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[carmine pilot]

Thx for helping I am still working on it

[cedar kiln]

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[astral moon]

I'm unsure how the second and third terms of line two combine into the second term of line 3

[dire geode]

did you do the suggestion?

"change summation variable n to n+2"

[astral moon]

I understand what happened to the first term of line 3@dire geode but how did the second term of line 3 come about

[dire geode]

plug in n=0 and n=1 to see

[astral moon]

I see thank you

*S̴̤̃e̶̪͗e̵͉̋

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Please don't occupy multiple help channels.

[glad kestrel]

i don't even know whether or not to bother the mods

[cedar kiln]

Channel closed

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[wicked brook]

<@&268886789983436800>

[calm sierra]

ty

[cedar kiln]

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[crimson sedge]

<@&286206848099549185> yo help ITS URGENT

[glad kestrel]

• After 15 minutes, feel free to ping @Helpers.

[crimson sedge]

Hi

YO

[glad kestrel]

stop pinging helpers

[crimson sedge]

...

15 MINUTES <@&286206848099549185>

[glad kestrel]

still hasn't been 15 minutes

• you already pinged helpers

stop pinging helpers

<@&268886789983436800>

[calm sierra]

ty

[wicked brook]

@crimson sedge whats this problem for?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[frosty oxide]

just a quick question, do I have to use integration by parts to prove every reduction formula?

[frosty oxide]

Or is it possible to prove it without integration by parts

Maybe it's possible for some but not others?

[glad kestrel]

if there is another way to prove them i'm not aware of it

[frosty oxide]

Oh I think I know where I saw something

it was tan^n of x

they just used an identity

[wicked brook]

I can tell you that by parts will always work, but you can might a different reduction formula if you dont integrate by parts for trig functions. Namely, if you have $\int cos^nx\ dx$, you can split it up as $\int cos^2x \cdot cos^{n-2}x\ dx$ And use $$cos^2x = \frac{1+cos2x}{2}$$ I don't know if this leads to anything, but you can try it.

[wraith dagger]

ohNoiAmHere

[frosty oxide]

oh interesting

[glad kestrel]

ah yeah i guess an identity does work there

[frosty oxide]

ok so if an identity works then thats fine, if not then integration by parts

I see now

thank you guys!

[glad kestrel]

are you only asking about trig reduction btw

[frosty oxide]

well I guess in general

[glad kestrel]

cause i'm sure other reduction formulas could be derived otherwise

[frosty oxide]

we just learned reduction formula today

[wicked brook]

there can be identities with exponents that you can use too if you wanted to integrate log^n(x)

[frosty oxide]

I'm still trying to understand it properly

[glad kestrel]

some of those reduction formulas look gross

[frosty oxide]

we'll have to prove it before using so I guess that's one less thing to have to memorize

which is nice

[glad kestrel]

[frosty oxide]

I'm gonna do a bunch of problems today and try doing some basic ones on my own

[frosty oxide]

oh god

that is nasty

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[willow hill]

Here's a really simple one. I want to solve for x in y = (x)(6 - x). I expect there to be a +/- and a sqrt somewhere but I cannot for the life of me find a way to phrase this as x = ...

[willow hill]

I tried solving for the vertex and i got something using that and point testing, but it was a bit wonky and I want something i can do without a second thought on the exam (plus I'm not sure how reliable my strategy is with other polynomials)

My ultimate goal is to rotate (x)(6 - x) around the y axis though but I couldn't find a more direct way to do it so I'm just trying to solve for x and I remember nothing from Grade X when we learned this I guess

[cedar kiln]

@willow hill Has your question been resolved?

[willow hill]

uhhh nope

thanks for the concern bot

[mighty drift]

you forgot about the quadratic formula ?

just complete the square so x is only present once, and work your way to isolating it. The result you get is the quadratic formula

[umbral jackal]

can I help a bit

[frosty oxide]

I think they want the inverse

But I'm not sure

[mighty drift]

that's still the quadratic formula, except now c is a variable

[frosty oxide]

That's fair

[willow hill]

hmm

it wasn't super intuitive as to how i could apply QF until you mentioned completing the square

so im imagining something like

y = 9 - (x - 3)^2

(x - 3)^2 = 9 - y => x - 3 = +/- sqrt(9 - y) => x = 3 +/- sqrt(9 - y)

alright, a bit painful but sounds good

completing the square is always scary

as you know

T_T

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[cedar kiln]

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[cedar kiln]

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[silent imp]

can someone help

[silent imp]

I dont know how to do the next problem ill write it

It says "State one solution to the system" y<2x-1 & y>=10-x

<@&286206848099549185> pls help

[umbral jackal]

pls make that latex

$y<2x-1$

[wraith dagger]

sunflame

[umbral jackal]

$y /geq 10-x$

[wraith dagger]

sunflame

[umbral jackal]

wait

$y \geq 10-x$

[wraith dagger]

sunflame

[umbral jackal]

solve both

[livid hound]

consider graphing and shading in each region

[umbral jackal]

any point in the overlapped region would work

[silent imp]

I don't get what exactly I should do

It tells me to state on solution so am I supposed to put an order pair from the double shaded area

[umbral jackal]

yes

[silent imp]

Ok thankyou

that was very helpful

.close

[cedar kiln]

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[gusty lotus]

Hi

[gusty lotus]

How would I do the chain rule for cos^5 (x)

it's not even a composite function

or is it?

[steel canopy]

(cos(x))^5

[gusty lotus]

they mean the same thing?

/have the same value?

[steel canopy]

Ye they mean same thing

And same value depend on what you put in x

[twin coral]

set u to cosx

y=u^5

then multiply y' and u'

[gusty lotus]

oh alright

got it

thank you guys

have a good day/night

.close

[cedar kiln]

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[cedar kiln]

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[verbal grove]

Hello can someone help me with this please? I need to f and e

[verbal grove]

Oh wait i think i found it

Can someone help me with this one, i need to find b and a

[cedar kiln]

@verbal grove Has your question been resolved?

[cedar kiln]

@verbal grove Has your question been resolved?

[dreamy plume]

I think it's congruent

So b is 90 and a is 12

Not 100 sure tho

[verbal grove]

oh right!!! Thank you so much!

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[still hawk]

how would i figure this out

[still hawk]

find x y and z in terms of a and b

[umbral jackal]

x+y+z=360º-a-b

[still hawk]

uhhh

so

i dont know what a or b are

tho

x + y + z is equal to 360 - 144 (because 360 / 5 is 72)

?

im confused

so x = 72, y = 72, z = 72?

[umbral jackal]

did it say a=b

[glad kestrel]

post the question

[still hawk]

i did

find the size of the angles marked x, y, and z in terms of a and b

[glad kestrel]

you posted a drawing on microsoft paint

post the original question

[still hawk]

,rotate

[still hawk]

so

[cedar kiln]

@still hawk Has your question been resolved?

[cedar kiln]

@still hawk Has your question been resolved?

[cedar kiln]

Channel closed

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[midnight owl]

@flint plinth Just wanted to make sure I did it right. I got 1 additional officer.

[midnight owl]

(9 total)

[flint plinth]

lemme check.. i didn't work out a solution earlier

we had what

[midnight owl]

np

[flint plinth]

t(p) = 56 - 4p

[midnight owl]

I'll send a pic sec

[flint plinth]

where p = total # of officers

t(p) = average # of tickets per officer

[flint plinth]

so then the total number of tickets, let's call that N(p)?

N(p) = p t(p)

[midnight owl]

here's my work

[flint plinth]

one second, I'll go through it with my notation and then compare with yours

[midnight owl]

ya np 😄

[flint plinth]

so N(p) = p * (56 - 4p) = 56p - 4p^2

if I did that algebra right

[midnight owl]

oh wait what x should be -1

not 1

[flint plinth]

and we want to maximize N(p)

so naively taking the derivative (even though p is an integer variable, not continuous...)

N'(p) = 56 - 8p

set that equal to zero

56 - 8p = 0

8p = 56

p = 7

uh oh

[midnight owl]

hmmm

what did I do wrong

[flint plinth]

we're in that gray area where we were wondering, does this formula even apply if p < 8?

[midnight owl]

I'm assuming you're right because you can't add a negative person lol

What did I do wrong though in mine?

I took the foiled binomial from online and didn't check if it's right

could be why

[flint plinth]

btw here's a plot of N(p) vs p:

lemme check yours now

[midnight owl]

hmm the binomial looks right

[midnight owl]

is that pymatlab?

[flint plinth]

yours is right except (as I think you mentioned) that final answer should be x = -1 not 1

and your x = -1 is the same as # of officers = 7

since your x is the number of additional officers

[midnight owl]

oh sorry I misread yours

I thought it was saying 7 additional officers lol

[flint plinth]

no, mine is just # of officers total

[midnight owl]

so -1 is definitely right then

[flint plinth]

so we agree about 7

but I don't know if it's valid

[midnight owl]

yea

then just plug 7 back in

or wait no

[flint plinth]

it's only valid if the rule still applies when you remove officers

[midnight owl]

it has to be right

[flint plinth]

i.e. if you remove officers does their average number of tickets go UP by 4?

[midnight owl]

good thing is this professor doesn't really check if the answer is right just wants to make sure if you do all of them lol

[flint plinth]

haha

you may want to check with the professor or TA or whatever

[midnight owl]

There were 10 sections due last week and I purposely left out like 10 problems total throughout the sections just because I couldn't figure them out lol

still gave me a 20/20 though

[flint plinth]

the math is right but only assuming that the formula can still be applied for # of officers less than 8

[midnight owl]

eh too much work if the math is right I trust it lol

[flint plinth]

ha ok

btw, if the rule doesn't allow you to go less than 8

[midnight owl]

Yea just keep it the same

[flint plinth]

then calculus doesn't actually give you the solution, the answer is just 8

[midnight owl]

^^

[flint plinth]

which you can see from the graph

[midnight owl]

is that pymatlab?

the graph

[flint plinth]

just plain matlab

[midnight owl]

ah

I've done a little within python using matlab lol

I have one more problem

[flint plinth]

i've used matplotlib in python, and what's the other one... bokeh

[midnight owl]

If you're willing to check my answer when I'm done?

[flint plinth]

sure

you can ping me

[midnight owl]

I'm definitely a beginner when it comes to python but still learning 😄

kk

[flint plinth]

@midnight owl I'm gonna tune out for a while to watch tv so maybe DM me when you're done (I think the channel will go away eventually)

[midnight owl]

sure np 😄

I'll dm if the channel gets closed if not I'll just ping here

[cedar kiln]

@midnight owl Has your question been resolved?

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[cedar kiln]

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[mental mesa]

Hi @harsh relic you are very helpful is this correct

[harsh relic]

correct

[mental mesa]

Is this correct

[harsh relic]

wrong

2-5 is not 3

[mental mesa]

Oh -3

[harsh relic]

yes

[tired charm]

heres a simpler approach

(a-b)(a+b) = a^2-b^2

hope this helps

nevermind

im just blind

💀

[dusty hazel]

That's exactly what they did.

💀

[tired charm]

lmao tf is wrong with me

wrong channel

[harsh relic]

put this in #help-11

[mental mesa]

Is this correct @harsh relic

[harsh relic]

wrong

[dusty hazel]

Is your question,
$$(\sqrt{x+2})(\sqrt{2x+3})$$
If it is, then your work is wrong.

[harsh relic]

in step 2, you should multiply (x+2) and (2x+3) instead of adding up

[wraith dagger]

What the hell am I doing here?

[harsh relic]

or it is adding up instead of multiplying?

[mental mesa]

ThT is my question

It’s adding

Not multiplying adding

[harsh relic]

adding is also wrong btw

shift sqrt(2x+3) to the right first

[dusty hazel]

Right.

[harsh relic]

then square both sides

[dusty hazel]

$$\sqrt{x+2} + \sqrt{2x+3}=0$$

[wraith dagger]

What the hell am I doing here?

[tired charm]

oh

[dusty hazel]

There are no solutions to this.

Real

[tired charm]

x+2=-2x-3?

[dusty hazel]

No.

[tired charm]

oh wait

bro i need to stop being dum b

[dusty hazel]

Agreed.

Mwahahahaha.

It alright alright.

[harsh relic]

why shift to right and squaring both sides cannot work?

[dusty hazel]

No. It won't.

We have the domain

x+2≥0
2x+3 ≥0

So yea.

[mental mesa]

It doesn’t work?

[dusty hazel]

Think of it as, sqrt of something is always non negative.
Sum of two non negative numbers can only be 0, when both of them are zero.
And x+2 and 2x+3 can't be simultaneously zero.

[tired charm]

oh yea

[dusty hazel]

That's your q right?

@mental mesa

[mental mesa]

Yes

That is

[dusty hazel]

Then, there is no solution, read whatever I wrote.
Ask if anything does not make sense.

[mental mesa]

This is question

[dusty hazel]

Exactly then, you should know.

x+2 must be non neg.

2x+3 as well

[mental mesa]

So there is no solutions

[dusty hazel]

Right.

[mental mesa]

Oh

Okay

[uneven steeple]

Wait

[dusty hazel]

...

[uneven steeple]

I don't get why we can't shift one of the square roots to the right and then square both sides though

[dusty hazel]

Because

[mental mesa]

That’s. what I was thinking

[dusty hazel]

Square root of anything is positive number.

Always.

If we take it to the left

[dusty hazel]

It does not work mate.

[uneven steeple]

Yeah so wouldn't it be like x+2 = 2x+3 afterwards?

[dusty hazel]

That doesn't matter.

Because

[mental mesa]

Ok so I will put down no solution

[dusty hazel]

Answer is not important, realising why there's no solution is.

Mate.

√x+2 is positive.
√2x+3 is also positive.
So if you take it to the other side
√(x+2) = -√(2x+3)

[uneven steeple]

Yeah

[dusty hazel]

So like

Positive= negative is what you did here.

Which is wrong.

[uneven steeple]

Wouldnt it be like -(√(2x+3))

[dusty hazel]

Right.

It would be.

[uneven steeple]

So something like -(√4)

-(2)

Isn't it the same thing

[dusty hazel]

2≠-2
But 2²=(-2)²

It's not.

[uneven steeple]

oh

[dusty hazel]

That is why you get answers.

It's incorrect though.

Since the first one is our original question.

[mental mesa]

https://youtu.be/aKZlwb2vDUI

[dusty hazel]

Squaring is dangerous.
As you can see, horrors of squaring.

[uneven steeple]

OH I GET WHAT U MEAN NOW

Bruhhh I brain farted

[dusty hazel]

Basically long short story

√4 ≠ (-2)

[uneven steeple]

Yeah no i took another look at the equation and it hit me

Lmao

[dusty hazel]

Right.

[uneven steeple]

Isnt there something where like if u square root a number you can get a negative number somehow though?

Like an imaginary number

[dusty hazel]

That's what my pfp says.

[uneven steeple]

Oh

[dusty hazel]

Mwahahahaha

[mental mesa]

Guys how do solve this can you teach me

I’m new to graphing concept

[dusty hazel]

Domain.

Do you know what that is?

[mental mesa]

Across. The graph

And range up or down the graph

[dusty hazel]

Ah. I mean define domain.

Kind of.

Domain of any function.

[mental mesa]

I don’t know to be honest

[harsh relic]

firstly, you will need to know the domain for sqrt x

[cedar kiln]

@mental mesa Has your question been resolved?

[dusty hazel]

That is all they need to know.

[dusty hazel]

Domain means, the accepted value for the input. The ones which make sense.

For example the domain of y=1/x would be all real values, except x=0.
Since the denominator can't be zero.

For square roots, like sqrt{x} x here can NOT be negative.

Right?

[harsh relic]

yes

[dusty hazel]

Oh that is for them.

Ahura that is.

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[junior kestrel]

for this, we could use a limit comparison test but couldnt we also use a direct comparison

[junior kestrel]

say, this

which is clearly bigger, and simplifies to 1/n^2 which converges

so you dont even have to compare their limits right?

[split mirage]

The problem is that it's not clear why that is bigger.

[junior kestrel]

Is it not?

you're dividing by something smaller

[split mirage]

No, because we know that n/(n³ + 6n + 3) < n/n³, but when we add 1 to the numerator, we aren't sure any more.

[junior kestrel]

Isn't it a thing where you can disregard the +1 since its just a small constant

pretty sure we do that sometimes for direct comparison but i might be wrong

[split mirage]

Not really. In this case though, you can still use direct comparison as (n + 1)/(n³ + 6n + 3) < (n + 1)/n³ = 1/n² + 1/n³ < 2/n².

But you can't compare to 1/n².

[junior kestrel]

oh yea just thought of that

i see

yea looking through my direct comparisons rn and youre right

ty

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[crimson sedge]

I need help with logarithmic equations

[crimson sedge]

so for example

I get normal logarithms

like log2(4) is obviously 2

but then there are things like log3(sqrt27) that I just don't get

[dense wing]

Why is log_2(4)=2?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

because 2^2=4

I don't know whether I'm writing it correctly or not

but a^x=b translates into loga(b)=x

[dense wing]

This is correct

Apply that idea to log_3(sqrt(27))

[crimson sedge]

yea so 2^2=4 translates to log2(4)

[crimson sedge]

3 to the power of 3 is 27

but we need the square root of 27

[crimson sedge]

I mean, the correct answer is 3^3/2

but I don't think I understand why

[dense wing]

log_3(sqrt(27))=x iff 3^x=sqrt(27)

From what you said

[crimson sedge]

yup

log3(sqrt27)
3^x=sqrt27

but how does that get me any further

<@&286206848099549185>

[dense wing]

What's sqrt(27) as 3^something...?

[crimson sedge]

not sure

that's why I'm asking

I've got no idea

[jaunty mural]

sqrt 2 = 2^0.5

[crimson sedge]

oh

so 3^1/2 is sqrt3

but we want sqrt 27

so we simply do 1/2*3

which is 3/2

so 3^3/2=sqrt27

[dense wing]

Yes

So what's x?

[crimson sedge]

3/2

[crimson sedge]

what happens when we use a negative number

[dense wing]

Exponent laws

[crimson sedge]

how do they work

[dense wing]

Google

[crimson sedge]

ok, thanks!

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[gilded nest]

So, I’m in school right now, I’m doing an continuation of an algebra 2 test in a few periods, I need to find compound interest with contributions for a word problem and it’s worth more than a quarter of the grade, I do not know how to find compound interest when accounting for contributions

[gilded nest]

The question is something like “max wants to buy a car when he turns 17, he starts saving at 14. He uses an account that adds 3.6% interest monthly, and he puts $100 into it each month, how much does he have after 36 months?”

I know how to compound interest with P(1+r^t=A

But my teacher never taught how to add contributions

[cedar kiln]

@gilded nest Has your question been resolved?

[gilded nest]

<@&286206848099549185>

[upper abyss]

That's an annuity. You sure you don't have a formula for that?

[gilded nest]

I don’t believe I was given one

[gilded nest]

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✅

[junior kestrel]

why do we naerly have the same name

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[night pumice]

I need a bit of help with this question (sorry but the image is way more blurry)

How do I find 'x' in

1⅔:x=1/24

[night pumice]

if image is required i will send it

[dusty hazel]

Your question
$$1\frac{2}{3} : x=\frac{1}{24}$$

[night pumice]

no hold on

[dusty hazel]

Alright.

[night pumice]

part (a) (ii)

[dusty hazel]

Got it.

[wraith dagger]

What the hell am I doing here?

[night pumice]

yea this question

[dusty hazel]

Convert into improper fraction first.

[night pumice]

so 5/3 :x = 1/24?

[dusty hazel]

Correct.

Now this also means

5/3 ÷ x = 1/24

Right?

[night pumice]

mhm

[dusty hazel]

You know how to divide fraction?

For example

4/3 ÷ 2?

[night pumice]

yea

change it into like

4/3 * 1/2?

[dusty hazel]

Correct.

Do it for your question too now.

Well?

[night pumice]

is it 40?

[dusty hazel]

What, the answer?

[night pumice]

yea

[dusty hazel]

Right.

It is.

Well done.

[night pumice]

thank you

but how does ':' mean divide?

[dusty hazel]

Haha.

Have you been taught ratio and stuff?

[night pumice]

yea just ratio i guess

[dusty hazel]

So you should know, it represents same thing.

[night pumice]

i see, im pretty sure i have forgotten about it

[dusty hazel]

That's alright, not anymore!

[night pumice]

Thank you so much!

[dusty hazel]

You're welcome!

[night pumice]

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[carmine fern]

aren't both examples of f discontinuous because of [x]?

[cedar kiln]

@carmine fern Has your question been resolved?

[carmine fern]

<@&286206848099549185>

it was probably a typo in the book unless I'm missing something

[dire geode]

What are you trying to show

[carmine fern]

I'm trying to understand how the f they give is continuous, it seems discontinuous to me

[dire geode]

Depends on A

It can be continuous on an A

[carmine fern]

since it says "a continuous and bounded function" but the function they give isn't continuous (it has a jump at every nonzero integer)

i guess

they said A was arbitrary though

[dire geode]

Where

[cedar kiln]

@carmine fern Has your question been resolved?

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[crimson sedge]

Ηελπ

[crimson sedge]

Help

So

Im trying to solve this problem here

that has to do with venn diagrams

E= whole numbers from 1 to 100

A= multiplies of 5

B= multiplies of 7

a) n(A)= 20

n(B) =14

Also there is another set C which is AUC=nothing empty basically and

B U C= B so c Is in B

C=AUB. Explain why n(C) is 34??/

hello?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

no

it's not solved

[crimson sedge]

:(?

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@crimson sedge Has your question been resolved?

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[valid nexus]

What is the difference between these two equations?

[glad kestrel]

they're rearranged differently

[valid nexus]

In second x is defined as function.

Am I right?

@glad kestrel

[glad kestrel]

yes, a function of x in terms of y

[valid nexus]

Cool

Thank you 👍

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