help-39

[vivid pilot]

|a_n| >= a_n

but it has to be the other way around

since we have a_n diverges, than |a_n|

[wraith hare]

no

can you state the comparison test for a series a_n <= b_n for divergence

[vivid pilot]

if a_n diverges, then b_n diverges, given that b_n>=a_n

[wraith hare]

yup, and here b_n = |a_n| >= a_n right

[cinder flower]

are we still on this question

[vivid pilot]

oh yeah

[cinder flower]

dawg, “if |a_n| converges, than a_n converges” is exactly the same statement as the one in the question

it’s the contrapositive

this comparison test argument isn’t valid

[vivid pilot]

why not?

[wraith hare]

oh yh abs test works mb, but why doesn't comparison test work

[cinder flower]

comparison test has extra hypotheses about the terms, like they need to be nonnegative and such

the a_n’s don’t need to be nonnegative here

[wraith hare]

right so?

[vivid pilot]

because if they are negative

we cant use comparison

i think

[wraith hare]

oh I forgot about that

super sorry

[cinder flower]

my point was this is more general - it’s also true when the a_n’s are not nonnegative

but to use comparison test you’d need to add that assumption

[vivid pilot]

i see

[vivid pilot]

so can i just say this?

[cinder flower]

yes

well i would

[vivid pilot]

is there no other way to derive it from the absolute test

[cinder flower]

🤷🏻‍♀️

it’s really saying the same thing as the abs value test

[vivid pilot]

yeah

thanks for the help!!

.close

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[high parrot]

how to simpliefy this cos^2(20)-1/4 (angle is in degrees) I keep getting 2cos(50)cos(10) but the answer is just cos(50)cos(10)

[mellow idol]

can i see ur work

[high parrot]

ahh I dont have discord on my phone just a sec

[pearl pond]

@high parrot Has your question been resolved?

[high parrot]

Here is my work

[high parrot]

.

[upper igloo]

(1+2cos(40°))/4 is not 1/2 - cos(40°) lol

[high parrot]

oh

any tips how to do this?

[upper igloo]

just divide everything by 2 after that

its just a factor of 2 off from that step

[high parrot]

Yes thank you had a brain fart

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[dusk otter]

I am confused about V_3 for part b) of this question
Where did the solutions get the V_1 + 1 from?

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[cosmic scroll]

hey

[cosmic scroll]

in this f(x) graph there inflection points on 2 and -1.

[silk sail]

I need help with this

[shrewd mist]

!occupied

[pearl pond]

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).

[cosmic scroll]

i can say that f''(x) has interception points with x axis on 2,0 and -1,0

but now they ask me to find the interception point with x axis in function f'(x).

i dont see no maximum or minimum point in the graph tho

i can only see that it has a domain on x not = 0.5

and its also known that f(x) is tangent to x axis if that helps

my question is how can i get the points in which the graph f'(x) is interepting the x axis.

[shrewd mist]

This is a graph of f(x) or f’’(x)?

[cosmic scroll]

this is a graph of f(x)

[shrewd mist]

So to be clear, both f(x) AND f’’(x) equal 0 when x = -1 and 2?

[cosmic scroll]

based on the graph, yes

the inflection happens to be on the x axis

thats why its tangent to it i think

[shrewd mist]

Alright. We know that when f’(x) = 0, f(x) is at a critical point. A critical point is defined as either a maximum, minimum or saddle point

[cosmic scroll]

saddle point?

[shrewd mist]

Yeah, I think you forgot that case

[cosmic scroll]

this point is a little new to me

tbh

[shrewd mist]

https://www.statisticshowto.com/saddle-point/

A saddle point is when the derivative is equal to zero (ie, function becomes flat), but it doesn’t hit a local max or min

Like a function might be increasing, flatten for an instant (saddle point) and then keep increasing

A critical point (f’(x)=0) doesn’t guarantee that the function has hit a local max or min, just that it might have

[cosmic scroll]

i thought it always has to change states of increasing and decreasing

thanks for the info

[shrewd mist]

Nope

[cosmic scroll]

how can i say if a point is a saddle point tho?

[shrewd mist]

How can you say if a point is a maximum?

[cosmic scroll]

i would plug values to x + something that sits in domain (x being the suspected maximum) and plus values x - something that sits in domain. and then see wheater the function is negative or positive or going up or down

for maximum it would be up and down

and its derivitive positive and negative

[shrewd mist]

And for minimum…

[cosmic scroll]

for minimum the opposite

[shrewd mist]

And for a saddle point, it would be up and up

Or down and down

That’s all

[cosmic scroll]

thats just an inflection point

[shrewd mist]

Do you know the second derivative test?

[cosmic scroll]

yes

thats to find the x of inflection point in f(x)

[shrewd mist]

One sec

What do you mean by inflection point, and what do you mean by critical point

Those are two seperate things

[cosmic scroll]

a critical point can be maximum or negative.
an inflection point is where the function continues to act in the same behavior before and after that point but just changes a curve.
before an inflection point the function could go down or up curved up or down and then it will continue going as it went "down or up" but it will change its curve.
They can happen when the f'(x) = 0 and f''(x)=0.

[shrewd mist]

So you agree that a critical point occurs when f’(x)=0 and an inflection point occurs when f’’(x)=0?

[cosmic scroll]

at Xa the function changes a curve while keeping the same behavior (going up)

[cosmic scroll]

yes

[shrewd mist]

Alright

[cosmic scroll]

but there are situations where an inflection point can happen at f'(x)

[shrewd mist]

So to simplify the definition, a saddle point is a critical point that is also an inflection point

[cosmic scroll]

thats why we check with values and see the behavior of the function

[cosmic scroll]

alright

makes sense

[shrewd mist]

Logically, if we hit a critical point, we check the curvature “behaviour” of the function at that point. If the second derivative is zero, that means there’s no change in the behaviour of the function. So an increasing function will continue to increase and vice versa

[cosmic scroll]

Yes

what about the question tho

[shrewd mist]

Try to reorder the logic

[cosmic scroll]

how can i get the points in which the graph f'(x) is interepting the x axis.

i only know where f''(x) is 0 tho

[shrewd mist]

Saddle points occur when f’(x) = 0, remember?

[cosmic scroll]

so you would say?

-1,0 and 2,0?

[shrewd mist]

When f’(x) = 0, we either have a maximum, minimum, or saddle. Looking at the graph of f(x), we see two saddles at -1 and 2. This means that for us to have a saddle, f’(x) must equal 0.

Yes, does the reasoning make sense?

A saddle is like a pseudo-min/max

[cosmic scroll]

it does

but the answer key doesnt

[shrewd mist]

What does the answer key say

[cosmic scroll]

so when a saddle occures i can claim 100% that at that x, f'(x) =0?

[cosmic scroll]

it only says 2,0

i understand and thats what i thought could be the asnwer but it kinda conused me when it was only 2,0

thats why i asked here

[shrewd mist]

Yes

[cosmic scroll]

yea makes sense

[shrewd mist]

Look at the picture though, there’s no saddle at -1

[shrewd mist]

This is a saddle

[cosmic scroll]

they said it

[shrewd mist]

Just cause there’s an inflection point doesn’t mean there’s also a saddle there

[shrewd mist]

Huh

[cosmic scroll]

inflection point is a saddle point no?

[shrewd mist]

No

Big difference

A saddle direction is when f’(x) = 0, and the concavity doesn’t change

Inflection point is when f’’(x) = 0

A saddle is visually defined as a flat instance,

An inflection point is visually a sort of “necking” of the function

See at 2, there’s an obvious flatness, but at -2, the function doesn’t become “flat”

It’s easier to visually identify saddles, not so much for inflection points

[cosmic scroll]

what would u say the domain of function where its curving up then?

0.5<x<2

and?

x<0.5 ?

bc by this logic

it didnt change the concavity

thats also when in the function f''(x) the function is changing between positive and negative

[shrewd mist]

Yeah

[shrewd mist]

You mean >

[cosmic scroll]

no

[shrewd mist]

Wait

[cosmic scroll]

the function curves up before 0.5

[shrewd mist]

You’re right

But still wrong

[cosmic scroll]

?

[shrewd mist]

It curves down until -1 no?

[cosmic scroll]

thats what i thought

but u said it doesnt change its concavity

[shrewd mist]

I said it doesn’t change direction, it can still change concavity

[shrewd mist]

What do you mean by this

[cosmic scroll]

here

oh u meant f'(x)?

this is kinda confusing me becuase they indeed say in the question there are inflection points in x=2 and x= -1

[shrewd mist]

Yes, my bad

That statement is incorrect, but -1 isn’t a saddle, done forget

Dont*

[cosmic scroll]

then every saddle is inflecton but not every inflection is a saddle?

im conused about this

[shrewd mist]

No, the fact that every saddle is an inflection in this question is pure coincidence

[cosmic scroll]

alright

then why is it only 2,0 who is a interecption with x axis with f'(x)?

[shrewd mist]

Actually

[cosmic scroll]

they both are inflections

they both are points in whic f'(x) =0

they both intercept x axis of that function

this is completely condradicting each other

if the only point is 2,0

[shrewd mist]

So -1 and 2 are inflections yes

[cosmic scroll]

yes

[shrewd mist]

Meaning f’’(0) in both points

[cosmic scroll]

Yes

f''(-1) and 2 = 0

idk about f''(0) =0

[shrewd mist]

An inflection point is defined* as a point where f”(x)=0

[shrewd mist]

I meant f”x

[cosmic scroll]

yea

yeah i see

[shrewd mist]

• there’s a particularity here, but I’ll get to that later. Not relevant for this q

A saddle point is defined as a point where f’(x) = 0

And the function doesn’t change direction

[cosmic scroll]

let me just ask you a question does it seem normal to you that only 2,0 is intercepting x axis if f'(x)?

[shrewd mist]

Yes

[cosmic scroll]

alright

[shrewd mist]

Because think of it numerically

At -1, we know that f’’(x) = 0

But this says nothing at all about f’(x)

[cosmic scroll]

why not?

[shrewd mist]

Consider the 2x at x=0

[cosmic scroll]

at f'(1) the function would also be 0

[shrewd mist]

f’’(0)=0, right?

[cosmic scroll]

why dont we take x^3 which has an inflection

[shrewd mist]

But f’(0)=2

[shrewd mist]

In x^3, the inflection and saddle point coincide

This means that both f’(x) and f’’(x) = 0 at zero

[cosmic scroll]

yea

[shrewd mist]

But this is coincidence, now law

f’’(x) = 0 just means that something has happened to f’(x)

[cosmic scroll]

yes

at that x f'(x) is also 0

[shrewd mist]

Just like how f’(x) = 0 means that f(x) is maximized, minimized or neither

Hold on

Do you agree:

The relationship between f’(x) and f”(x) is the same as f(x) and f’(x)?

Because it’s a differential relation

You differentiate one to get the other

[cosmic scroll]

yea

i think im out of time rn tho it would be nice to continue this later but for now i ahve to go

sorry and thank u for ur time

.close

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[grizzled iris]

for the top base I got 10cm finding hypotenuse if the small triangle 8^2 + 6^2 = a^2

then 20cm for the bottom base by doing 16^2 + 12^2 = b^2

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@grizzled iris Has your question been resolved?

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[elder pawn]

meant to send an image

[glad falcon]

Deal with it one part at a time

What's the formula for the area of a rectangle?

[elder pawn]

length x width.

[glad falcon]

Ye, and what is the length and width of the rectangle in that picture?

[elder pawn]

length x width

[glad falcon]

but what are the actual values of them

You have the numerical values of them labelled in the picture

[elder pawn]

im not sure im understanding this properly

[glad falcon]

Well you need numbers for the length and the width of the rectangle so you can type it into your calculator right

One of them is 3.9cm

What is the other one?

[elder pawn]

5.2

[glad falcon]

Yeah good

So length x width = 3.9 x 5.2

So plug that into your calculator and that's the area of the rectangle bit

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[elder pawn]

alr hold on

[elder pawn]

20.28

[glad falcon]

Ye

So that's the area of the rectangle

What's the other shape in the picture we need to work out the area of?

[pearl pond]

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[edgy wren]

Hey, how do I know when to use Binomcdf or Binompdf on the TI-84?

[marble sigil]

pdf is an exact number cdf is up to a certain number

so like "n successes" vs "at most n successes"

since the c stands for cumulative

[edgy wren]

So how do I know when a problem is asking for me to do one or the other? Is there any kind of indication? I currently have the sample size and the probability, but I keep screwing up x. Do I use cdf when x or more of something occurs?

Like I’m looking for some kind of indication for where I can clearly classify each one distinctively when solving a problem

[marble sigil]

you have to check if what you're trying to find allows for multiple possibilities

like "at least n cards" instead of "exactly n cards"

if a bunch of different scenarios would work you'd do cdf

[edgy wren]

Ohhh, okay. So pdf is an exact number, and cdf indicates an equal to or greater/less than equation

[marble sigil]

👍

[edgy wren]

Cdf also applies in “more than” and “less than” equations, right?

Not just the “equal to or greater/lesser” ones?

[sharp vigil]

cdf will always return the probability that a variable is less than or equal to n. but the other scenarios can be written in terms of that probability

e.g. less than 5 means less than or equal to 4, and greater than 7 means not less than or equal to 7

[edgy wren]

So in an equation where I’d be calculating something greater than 7, I’d use pdf to do that, I’m assuming

[sharp vigil]

pdf is only good for calculating single values (equal to 7, for example). but you can use the fact that the probability of a value being greater than 7 is the probability of the value not being less than or equal to 7 (and the probability of something not happening is 1 - the probability of something happening)

[edgy wren]

Ohhh, so I’d basically use binomcdf to calculate n,p,7, then do 1 - whatever the result was

Or is that wrong

[pearl pond]

@edgy wren Has your question been resolved?

[pearl pond]

@edgy wren Has your question been resolved?

[pearl pond]

@edgy wren Has your question been resolved?

[sharp vigil]

yes

[edgy wren]

Alright, thanks

[pearl pond]

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[fossil rose]

how does this make sense

i used the formula for averages of grouped datasets

and i end up with a mean of 12,36

which makes no sense

[acoustic path]

the sum of the right column is wrong

to be precise, it does not count the numbers who have a fractional component

[fossil rose]

wait what

can you explain

cus there is something i did wrong

i can feel it in my bones

[acoustic path]

i already explained everything relevant

309 is the number you get from adding 29 + 62 + 66 + 111 + 41

the real sum should be 309 + 52.5 + 19.5 + 64.5

[fossil rose]

oh my god

thank you

sorry you had to see that lmfao

i make so many stupid mistakes its incredible

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[loud dome]

Ok rq question

[loud dome]

What do I do here since I’m not sure how to simplify this

[versed remnant]

that question is very similar to
$$\sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta$$

[jolly parrot]

JustToPro

[loud dome]

So would it just be sin(pi/12-5pi/12) ?

[versed remnant]

yeah

[loud dome]

Oh

Thanks 🙏

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[gaunt crag]

hello

[gaunt crag]

for this problem

sorry

wrong one

I would use the half angle formula for this correct? so cos a/2 = +/- (sqrt1+cosA)/2 but I would square them correct?

[west sapphire]

there's a formula for cos^2 - sin^2

[gaunt crag]

oh

uh

[dry perch]

cos (2r) = cos2 (r) - sin2 (r)?

[gaunt crag]

well thats the double angle formula right?

[dry perch]

yea

[gaunt crag]

shouldnt i be using the half angle formula?

[west sapphire]

you could but that would be a very long way to do it

[dry perch]

yeah

[west sapphire]

using the double angle will be much quicker

[dry perch]

exactly

[gaunt crag]

ok

but i dont get it how would i since its not written as cos 2x

[west sapphire]

cos(2r) = cos^2(r) - sin^2(r)

let r = theta/2

[gaunt crag]

ok

so would i apply the same logic for sin and rewrite it using the double angle formula?

[west sapphire]

wha

[gaunt crag]

ok im still confused sorry lol

[west sapphire]

no, you already have cos^2 - sin^2

[gaunt crag]

alr so thats literally it?

[west sapphire]

well depends on what you mean by "that"

what's the answer?

[gaunt crag]

yea, is that the answer?

ok im sorry

im trying 💀

ok well i figured it out

its cos(theta)

[dry perch]

cos^2(θ/2) - sin^2(θ/2) = [1 + cos(θ)]/2 - [1 - cos(θ)]/2

so what do u think is the answer?

[gaunt crag]

i said it already

well atleast thats what the problem took

[dry perch]

oh my bad

[gaunt crag]

idk

this stuff is confusing and my professor does not explain anything

[dry perch]

for real bro

[gaunt crag]

but it took cos(theta) as the answer

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[quasi robin]

would someone mind checking this work

i am asked to find x and define it in terms of common log

my calculations get me 1.46773

but a calculator tells me that its -1.46773

im not sure where id be missing a negative

[gritty moat]

Lemme get some paper and do the question

[quasi robin]

thank you, i appreciate it

[gritty moat]

Easier to spot inaccuracies

Shoot

I have to go for like 30 monites

If no one else answers it I’ll do it

Sorry bro

[quasi robin]

no worries man

[tall flint]

valiant effort

I take issue with your final line

how does the thing on the left of the "or" equal the thing on the right?

[quasi robin]

i just mean that 1/5(144) is the same as 1/720

all i did was simply my answer for x a little more

[tall flint]

well that's not what you wrote

you said it's the same as 720

[quasi robin]

because its an equivalent value

[tall flint]

1/720 and 720 are not equivalent at all

[quasi robin]

oh i see

still though

[tall flint]

this discrepancy is precisely where the missing negative comes from

[quasi robin]

you’re absolutely correct

nice catch

and thanks

[tall flint]

[pearl pond]

@quasi robin Has your question been resolved?

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[raw axle]

i got that its not pairwise independent is this correct?

[compact ridge]

What's the reasoning for your answer?

The thought process is much more important than just the answer

[pearl pond]

@raw axle Has your question been resolved?

[raw axle]

because b and s arent independent

i wasnt too sure

[compact ridge]

Yeah, you're on the right lines

[compact ridge]

What formula can you use to show two events are independent (or not)?

[raw axle]

p(a n b) = p(a) * p(b)

[compact ridge]

Yep

So if we try calculating P(B and S) first, we have P(the sum of the red and blue dice is even, and the blue die is 3 or less)

So we could have the blue die is 3 and the red die is 1, 3, 5
Blue is 2 and the red die is 2, 4, 6
Or blue is 1 and the red die is 1, 3, 5

Each event has probability 1/6 * 3/6, so 1/6 * 3/6 * 3 = 1/4 in total

Now can you find P(B) * P(S)

[raw axle]

oh and p(b) = 1/2 * p(s) = 1/2

also 1/4

[compact ridge]

Yeah so it turns out that B and S are independent

So all you have to do is to do the same thing for R and S

[raw axle]

ahh i see

[compact ridge]

Like you can't guess they're independent or not

When it's not obvious

[raw axle]

yeah lol

would p(r n s)

also be 1/4?

same calculation as p(b n s)

[compact ridge]

Yeah it's a similar process

[compact ridge]

Yep, that's correct

[raw axle]

thnx

lastly how can i tell if r, b, and s are mutually independent?

[compact ridge]

Mutually independent is pairwise independent and P(A and B and C) = P(A) * P(B) * P(C)

[raw axle]

oh

so i need to find P(A and B and C)

[compact ridge]

Yes

[raw axle]

red die we have 2,4,6

blue die we have 1,2,3

[pearl pond]

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[compact ridge]

Ah shit

[compact ridge]

Ask your question in a new help channel

This one timed out

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[unkempt yacht]

How many positive integer values of a $(a \le 2023)$, such that there exists a real value of x satisfying $x(\ln a+e^x)\le e^x(1+\ln(x\ln a))$

[jolly parrot]

lemonsaurus aficionado

[unkempt yacht]

so i tried transform each hand side, but came up to dead end

$\ln a^x + xe^x \le e^x+e^x\ln(\ln a^x))$

[sharp smelt]

well, a has to be less than 0 if x is positive

[jolly parrot]

lemonsaurus aficionado

[sharp smelt]

*less than 1

[unkempt yacht]

how do you know?

[sharp smelt]

Less than 1

Because $xe^x>e^x$

[jolly parrot]

ƒ(Why am. I here)=I don't know

[sharp smelt]

If x>1

[unkempt yacht]

right

so how would that help us?

[sharp smelt]

Nvm, misread the question

Sorry

[unkempt yacht]

is ok

[unkempt yacht]

i dont know about this, but this looks really similar?

[pearl pond]

@unkempt yacht Has your question been resolved?

[pearl pond]

@unkempt yacht Has your question been resolved?

[pearl pond]

@unkempt yacht Has your question been resolved?

[pearl pond]

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[pearl pond]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight haven]

whats the mistake here ?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[versed void]

We have a,b,c are real numbers, a,b,c>=0 and a+b+c=3
Find the minimum value of P = 1/(a+1) + 1/(ab+1) +1/(abc+3)
Sorry I dont know how to use latex

[hardy hollow]

<@&286206848099549185>

i have a question reagarding square root

what is the square root of 8281

[hoary merlin]

91

[hardy hollow]

how

[halcyon plank]

HELP

<@&286206848099549185>

[hardy hollow]

through which method

[halcyon plank]

<@&286206848099549185>

[hardy hollow]

yes?

[lavish glen]

Is it complex no. Or quadratic equations question?

[hoary merlin]

well you can calculate 91*91=8281 or you use your calculator

[hardy hollow]

bro

[halcyon plank]

It's a quadratic

PLS HELP PLS

[tulip mason]

Lmao 3 people want this help channel

[hardy hollow]

seriously

i mean how

[tulip mason]

What have you tried?

[mental verge]

hold on

[halcyon plank]

NOTHING. I NEED HELP PLS TIME IS RUNNING OUT

[mental verge]

!status

[tulip mason]

Is it for a test?

[mental verge]

u cant and shouldnt be posting exams here

no tests

[halcyon plank]

Yes

[mental verge]

no

[halcyon plank]

IT'S NOT A TEST

[mental verge]

we shouldnt be helping

[tulip mason]

#rules

[halcyon plank]

IT'S A BLUEBOOK PREVIEW TEST

JUST HELP PLS

[hoary merlin]

dm me

[halcyon plank]

IT'S FOR PRACITXE

YOU CAN HELP

[tulip mason]

Come back after the test has closed and we can help

Or just look up the quadratic formula lol

[halcyon plank]

IT'S FOR EVERYONE OUT THERE WHO WANTS TO TAKE SAT PRATICE

[night monolith]

help the fella out

[tulip mason]

😂

So dramatic, feel it might be a troll

[night monolith]

perchance

[halcyon plank]

IT'S NOT YOU-

[mental verge]

check ur dm

[mental verge]

no we cant help

we shouldnt be

i did for this time but wont be doing it again

[lavish glen]

How can I find or prove injection of any function

[mental verge]

huh

[lavish glen]

<@&286206848099549185>

[mental verge]

okay so

wait

something is wrong bc i cant find available servers as well

[tulip mason]

What is the definition of injection?

[mental verge]

this channel is occupied

[lavish glen]

I know the algebra method

[mental verge]

anyways

[mental verge]

am i the only one with this problem=

?

[lavish glen]

Well any function f(a)=f(b) iff a=b

[hoary merlin]

f(x) = f(y) => x=y

[tulip mason]

<@&268886789983436800> the help channels are broken, there are like 5 different people in here wanting a channel

[opal lantern]

#changelog

we know

[tulip mason]

My bad

[tulip mason]

You need to show that then

[lavish glen]

I want to know the graphical method

[tulip mason]

Graphically it is one to one, ie. Each y has one x and each x has one y

[lavish glen]

Oh

[mental verge]

yes

[clear remnant]

(Z-12)(z+2)=0

Z is 12 or -2

[grave stream]

I need help please

[clear remnant]

With what

[grave stream]

6.5

Whether I multiply or square the equation I end up with a cubic which I cannot find the factors of

[grave stream]

I’m puzzled

[spiral sierra]

what did you end up with?

[grave stream]

X^3 -5x^2 -1=0

[grave stream]

Also in this module we are not allowed to use a calculator

[spiral sierra]

i dont think you can solve this by factoring tho

please don't post solutions.

[grave stream]

Same I went up until 10 and couldn’t find a factor

I’m guessing the solution had to do with complex numbers

[spiral sierra]

you can see that its around 5 but nah, factoring obv wont work

<@&268886789983436800> 🫡

[grave stream]

Agreed

[midnight haven]

is this channel available

[sullen leaf]

is anyone else havin g troubles understanding this problem?

[weak surge]

can anyone help me with this? where should I begin
can RREF be used?

[manic sparrow]

$\int\frac{\dd{x}}{3+x^2}$

[jolly parrot]

Jash

[tulip mason]

Multiply the two matrices and equate the items to find x and y

[tulip mason]

use the sub $x=\sqrt{3}\tan\theta$

[jolly parrot]

Max

[weak surge]

why is this channel not occupied

bug?

[manic sparrow]

why θ

[manic sparrow]

free for all channel lol

[tulip mason]

Which part

[manic sparrow]

ye #changelog

[tulip mason]

You can use whatever letter you want, I just use theta for angles

[manic sparrow]

isn’t it an arctan integral tho

[tulip mason]

Yeah exactly why I told you to use that sub

[manic sparrow]

isn’t it typically u=something of x

not x=something of another letter

[tulip mason]

In this example we do it this way round because we want the identity $1+tan^2\theta$ to appear

[jolly parrot]

Max

[tulip mason]

Recall $1+\tan^2\theta=\sec^2\theta$

[jolly parrot]

Max

[manic sparrow]

ohhh if u divide both sides by √3 you get x/√3=tan(θ)

[manic sparrow]

and you can factor √3 here

[tulip mason]

Yeah

To be good at maths (although it's not required) but you should learn where formulas come from as it will help in later areas

[manic sparrow]

$\frac{1}{3\sqrt{3}}\int\frac{\dd{x}}{1+\tan^2\theta}$

[tulip mason]

Well not quite

It's an x^2

So the bottom becomes $3+3\tan^2\theta$

[jolly parrot]

Max

Jash

[weak surge]

you mean like this?

[manic sparrow]

wait the integral is in terms of dx

[tulip mason]

Yup, now you have 4 equations if you equate the cells

[manic sparrow]

ohh i got it ty

[tulip mason]

Still not quite right

Do it by paper

[tulip mason]

You need to change the dx too

[manic sparrow]

ye u get 1/√3 arctan(x/√3)+C

[tulip mason]

Can't integrate something in $\theta$ wrt $x$

[jolly parrot]

Max

[tulip mason]

That is right but you should learn how to get that without using a formula...

[manic sparrow]

yea i will try to derive it in a bit

[weak surge]

how far should i go

here x is different

[tulip mason]

Then you made a mistake somewhere

[weak surge]

I'm just taking first entry of AB and equating it to first entry of BA

and second entry of AB, equating it to second entry of BA

maybe this question cannot be solved?

due to the values of A and B to begin with?

[wraith hare]

you made a mistake when rearranging for 3y from 9x = 3y-3

[weak surge]

oops

should be 1/9 not 1/3

ty

wait nvm

I don't see it

[wraith hare]

i dont think that's the one im talking about

on the rhs, 2nd line you showed that

[weak surge]

oh i see it now

[wraith hare]

9x = 3y-3

[weak surge]

should be -3y = ...

[wraith hare]

yes

[weak surge]

tyty

[wraith hare]

np

[weak surge]

so here I have x = 3 and x = x

[wraith hare]

ok so that showed your work was correct, and you can work out y

[weak surge]

i thought i did already

for equating second entry of AB with BA

[wraith hare]

I mean you still need to work out the number

[weak surge]

i swap x with 3?

y = 10?

x = 3

any need to keep going? that's what i mean not sure when to stop with this

i still need to equate 3rd and 4th entries?

i'm also not sure what they are asking for exactly

i need to provide a list of 2x2 matrices?

or did i already solve it

B =
1 3 -5 10

[hoary merlin]

Lemme see

[hoary merlin]

this is right

[hoary merlin]

this too

[weak surge]

OK

so that's the answer? i'm not sure how to phrase it

[hoary merlin]

so you made your calculation and you saw that there is only 1 possibility where the equation holds. So you say: For the Matrix A = (matrix A) and B = (matrix B with your calculated values) the equation AB = BA holds

[weak surge]

so I could have chosen any two entries to solve this?

i chose first and second

[hoary merlin]

yeah

[weak surge]

but i could have chosen third and fourth instead? same result

or second fourth

or 1st and fourth

etc

just need 2 to solve

[hoary merlin]

you can take everything as long as you get equations with which you can calculate your values

[hoary merlin]

exact

you can take any two, look what values they need and then you check up if the equation holds. If for one entry-pair there is exact one solution you are done

[jolly parrot]

Akti

[hoary merlin]

yes and no

but this is true

wait a sec????

there is something called unique factorisation theorem for Z (and also N)

and any number can be decomposed into a product of prime numbers, so if you take 2 distinct prime numbers, obvsly there product also devides c because they are from the decomposition

but you have to make sure that your if a prime number p only comes 1 time in your decomposition of c, you cannot set a = b = p

in your "theorem" you didn't exclude that case

[strong ether]

Heyooo

I need some help

It's seriously a small task left.

How do I do assignment c

[hoary merlin]

understood what I am saying?

[meager trellis]

counterexample: 3 | 3 and 3 | 3 but 9 does not divide 3
but yeah if a and b are different then this is true

[strong ether]

ehmm

I have this?

I have written here first

[meager trellis]

in fact i think you can replace the hypothesis that they're both prime with just that they're coprime

[hoary merlin]

so the thing is, exponential equations dont have tge value 0 (only if it is the 0 function itself), so we have to ask ourself when do we consider someone drug free? if there is a value (not 0) then calc that point. otherwise idk

[meager trellis]

which also deals with the case where they're the same, because a number isn't coprime to itself (unless like it's 1 but that's fine)

[strong ether]

True, but we have a function. Se here.

c(t) = 0,07

[frigid crest]

sorry to bother you lots conversation but have you lot heard of the math website sparx

[strong ether]

we have to isolate t

@hoary merlin

[hoary merlin]

ah yeah

[strong ether]

yes but how

that's all I'm missing

After I'm done

can u help plss @hoary merlin

[hoary merlin]

what is your exponential equation

[strong ether]

What u mean

[hoary merlin]

nice

[strong ether]

everything is there

[hoary merlin]

what is your solution for a)

[strong ether]

mhm not sure

but a must be 0.07?

[strong ether]

?

[hoary merlin]

let me think a bit

you guys are so inpatient

[strong ether]

ahah soryyy

[strong ether]

okay

[hoary merlin]

since I do this on paper now I cannot make sure my solutions are 100% correct but let us try that:

[strong ether]

yes is okay

[hoary merlin]

we can approximate the function to B(t) = 9*(⅚)^t
the question is, when does it have the value 0.07 so we make the following

[strong ether]

the answer to task b

9?

[hoary merlin]

well we start at a value of 9 and 8.9777 is nearly 9

[hoary merlin]

ok but lets work with that

[strong ether]

task b is where we have to determine the regulation, which you can see there

[strong ether]

Yesss

[hoary merlin]

when is c(t) = 0.07? we get

0.07 = 8.9777*0.8411^t

<=> 0.07/8.9777 = 0.8411^t

<=> log_(0.8411) (0.07/8.9777) = t

put that in your calc and that should be your answer

[rose marten]

Polygon JKLM is drawn with vertices J(−2, −5), K(−4, 0), L(−1, 2), M (0, −1). Determine the image coordinates of M′ if the preimage is translated 4 units up.

[strong ether]

don't we get 28.1 there?

[hoary merlin]

as a solution? idk i dont have a calculator with me

[strong ether]

don't you have a calculator on you?

[hoary merlin]

no bc I am outside

will be home in a couple of hours

[strong ether]

mhm

wait a minut

I'll check the answer again

[strong ether]

can you show me the bottom answer or explain what it looks like?

It could be that I'm typing wrong

[hoary merlin]

wait

[strong ether]

Okay

[hoary merlin]

log_(0.8411)(0,0077970973)

[signal basin]

i need help

[strong ether]

Yes, but can you show a picture or explain what should be written

[hoary merlin]

this is what you have to type in your calculator or what do you mean?

[signal basin]

can someone help me please

[hoary merlin]

y = -1

[serene kelp]

what is the spearman correlation for children and their mass

[signal basin]

can you tell me how you did it

[hoary merlin]

if you look at the blue line you see that it is always on the height -1, in school you describe the height as "y", that is the reason why we have "y=-1"

[signal basin]

thank you

[hoary merlin]

remember one thing: Horizontal lines are always "y=...", vertical lines are always "x=..."

[strong ether]

That's better, thanks - I'll do the math

[strong ether]

It is correct, thank you! 🙂

[hoary merlin]

np

[dense rover]

Need help with graph

$f' > 0$ $on$ $]-\infty, -3] U [0,3]$

$f' < 0$ $on$ $]-\infty, -2] U [3, \infty[$

$f'" < 0$ $on$ $]-\infty,-2] U [2, \infty[$

$f'' > 0$ $on$ $[-2,2]$

[jolly parrot]

Chrysanthemum(

[dense rover]

f(-5)=f(0)=f(5)=0 and f(-3) =f(3) =5

.

[minor ruin]

I have 8 vectors of 2 3D rectangles and I already use SAT to find if they are colliding but I need to find their contact points. Does anyone know how I might find them given the vectors of the cubes?

[prime kite]

Didn't I gave you the way to solve it yesterday?

[dense rover]

yes it's not working

try to graph it

[prime kite]

No, it's working, you might just be doing it wrong. Show me what you got

[dense rover]

whats the original function?

[prime kite]

Doesn't matter in this problem

[prime kite]

The graph you got

[dense rover]

I have a single line going up from -inf, -3

thats all

I don't think I understand the notation

[prime kite]

Why is there 2 interval going to $-\infty$ ? That's the problem

[jolly parrot]

Tittom_123

[dense rover]

I just posted what you wrote

: (

[prime kite]

Oh. Give me back your previous intervals for f' < 0, I might have just done a typo

@dense rover

[dense rover]

I might have worded it wrong

[pearl pond]

Please don't occupy multiple help channels.

[prime kite]

Give it back to me

[dense rover]

the original is
f'>0 on -inf, -3 and 0,3
f'<0 on -3, 0 and 3, inf
f" <0 on -inf <-3 and 2 inf
f">0 on -2, 2
f(-5)=f(0)=f(5)=0 and f(-3) =f(3) =5

[prime kite]

$f' > 0$ $on$ $]-\infty, -3] U [0,3]$

$f' < 0$ $on$ $]-3, 0] U [3, \infty[$

$f'" < 0$ $on$ $]-\infty,-2] U [2, \infty[$

$f'' > 0$ $on$ $[-2,2]$

[jolly parrot]

Tittom_123

[prime kite]

Then

[dense rover]

do you even need f(-5)=f(0)=f(5)=0 and f(-3) =f(3) =5?

[prime kite]

Yes

[dense rover]

I don't think the graph works, like draw a line from -inf,-3 up but it goes down at -3, 0'

f(-5)=f(0)=f(5)=0 and f(-3) =f(3) =5? I don't understand what this would look like on a graph either

[prime kite]

It should look like something like this

[dense rover]

so it's just x values

?

[prime kite]

Yes

[dense rover]

I thought you read it as x,y

I am stupid

[dense rover]

thanks man

[prime kite]

I strongly recommend that you do a table of value on intervals like this one

[dense rover]

I hope you go far in life

You really helped me out

.close

[prime kite]

I want to become a mathematician Iol. Also, np with the help

[pearl pond]

@versed void Has your question been resolved?

Channel closed

Closed by @versed void

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[indigo pewter]

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$. Find the unique number that is not in the range of $f$.

[jolly parrot]

Dork9399

[indigo pewter]

how would I find this?

I know that $\frac{a(\frac{ax+b}{cx+d}) + b}{c(\frac{ax+b}{cx+d}) + d} = x$

[jolly parrot]

Dork9399

[wraith hare]

think about how the graph of ax+b/cx+d looks like

[indigo pewter]

its a hyperbola

[wraith hare]

yes, more specificially a reciprocal graph

do you know about the asymptotes of a reciprocal graph?

[indigo pewter]

yea

we have to find the horizontal asy

[wraith hare]

that's right

and do you know how to find it?

[indigo pewter]

no

wait

its y=a/b right

[wraith hare]

no

[indigo pewter]

whoops

[wraith hare]

the horizontal asymptote of reciprocal f(x) is the limit of f(x) when x -> +- infinity

[indigo pewter]

yea ofc

but how do you find that

ohh

i got it

a/c right

[wraith hare]

yes

[indigo pewter]

so how do we find a/c

@wraith hare

[acoustic path]

in all honesty this is a perfectly fine and straightforward way of solving this problem

the hyperbola approach does provide an "alternative solution" but not one that i would call superior

[indigo pewter]

where do i go after that first step tho

[acoustic path]

fractions inside fractions is not pleasant

so getting rid of some of those fractions (how?) should be something you consider

then, it helps to think about under what condition the left side actually equals "x"

[indigo pewter]

so it simplifies to this

$\frac{a^2 x + a b + b c x + b d}{a c x + b c + c d x + d^2} = x$

[jolly parrot]

Dork9399

[indigo pewter]

wdym

[acoustic path]

which part of that statement is unclear?

i am saying that you must consider when that Left Hand Side equals "x"

[indigo pewter]

yea i understand

but how would i do that

[acoustic path]

it's a rational function

top and bottom are both polynomials in terms of x

[indigo pewter]

how would that help?

[acoustic path]

if u don't see then just get rid of that large fraction

then it will be fairly clear how to continue

[indigo pewter]

I got $cx^2 + dx - ax - b = 0$, assuming $a \neq -d$

now what

[acoustic path]

to be honest i have no idea what that is

[indigo pewter]

I cross multiplied and simplified

this is my work

$\frac{a^2 x + a b + b c x + b d}{a c x + b c + c d x + d^2} = x$ \
$a^2x + ab + bcx + bd = acx^2 + bcx + cdx^2 + d^2x$\
$acx^2 + cdx^2 + d^2x - a^2x - ab - bd = 0$\
$cx^2(a+d) + x(d-a)(a+d) - b(a+d) = 0$\
Assuming $d \neq -a$\
$cx^2 + dx - ax - b = 0$\

[acoustic path]

thanks

[jolly parrot]

Dork9399

Dork9399

[indigo pewter]

oops

[acoustic path]

so

why are you "Assuming d neq a"

[indigo pewter]

to divide by (a+d)

[acoustic path]

right ok

then in the equation cx^2 + dx - ax - b = 0

is it possible at all for equality to hold

[indigo pewter]

I would think so

[acoustic path]

no

a, b, c, and d are non-zero

therefore this equation is evidently just wrong

because cx^2 + dx - ax - b is not 0, the equality is false

[indigo pewter]

so is my work wrong?

[acoustic path]

also no

[indigo pewter]

or is my assumption wrong

[acoustic path]

what do you think?

[indigo pewter]

my assumption?

[acoustic path]

so what does that mean?

[indigo pewter]

a = -d

[acoustic path]

yes

using this argument, you now conclude that a + d = 0

[indigo pewter]

And how can we use that?

[acoustic path]

have you tried

[indigo pewter]

@acoustic path ?

have i tried what

[acoustic path]

tried anything

you have new information a + d = 0

i am asking if you have tried to do anything at all

[indigo pewter]

oh

[acoustic path]

well ig now i know the answer is no

anyway here i'd recommend seeing what you can do with that info

[indigo pewter]

38a + b = 361c

and 194a + b = 97^2c

156a = 116*78c

ohh

a/c = 116*78/156

right?

@acoustic path

[acoustic path]

oh nice you spotted the cancellation

and uhh

yes that is correct

[indigo pewter]

ayy

tysm!

.close

[pearl pond]

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[acoustic path]

🎉 🎉

[pearl pond]

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[stable iron]

How would I make an equation for this If I started with 24.34 oz of chicken (raw) and the nutrition value for 4 oz of raw chicken breast is 110 cal and 24 G protein and cooked it and it yielded 16 oz of cooked chicken how much protein and calories is in the cooked chicken per 4 oz

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@stable iron Has your question been resolved?

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[vivid pilot]

is there a nice proof for this?

[vivid pilot]

maybe mods?

[marble sigil]

it's always mods

[vivid pilot]

what are the remainders of powers of 5

lets see

[vivid pilot]

lol

[vivid pilot]

what modulus do we choose though

[vivid pilot]

wait wait. if the solutions have to be integers

could we try 0 and 1 for each

and you cant get 5

if you try 2, 2^5 is 32 which is already too large

maybe we dont need mods?

[worthy lance]

Integers

[pine moat]

5th powers grow fast, you can just enumerate the search space, this works perfectly

[worthy lance]

You have to prove for integers though

[pine moat]

ah i didnt see that im blind

[marble sigil]

mod 11 works for some awful reason

some of the variables will be negative to cancel it out so there's no way to brute force

I forget the theorem but you get mods of powers to be -1, 0, or 1 under certain cases, and 11 is the lowest prime that's big enough to make the argument of 5 being to far away

[pearl pond]

@vivid pilot Has your question been resolved?

[vivid pilot]

wait wdym

[marble sigil]

x^5 is -1, 0, or 1 mod 11

[vivid pilot]

oh

and with that you cant get 5 mod 11

with three of them

so you cant get 5

[marble sigil]

yea

[vivid pilot]

oh nice

thanks!

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[vivid pilot]

how would i do this?

[vivid pilot]

im not too good with percents

[obtuse bramble]

I would use a tree diagram to represent this. Put the values on the tree diagram as decimals like 90% as 0.9. And then multiply the branches that satisfy the answer. I hope that makes sense, lmk if u need a visual or more explanation

[vivid pilot]

do you have a visual?

that would be helpful

thanks

[obtuse bramble]

That should help

[pearl pond]

@vivid pilot Has your question been resolved?

[vivid pilot]

oh thanks

[pearl pond]

@vivid pilot Has your question been resolved?

[sharp tree]

7.2%?

[vivid pilot]

how did you find that?

[sharp tree]

Multiply them together?

[vivid pilot]

for the tasty, right?

[sharp tree]

Yeah

it asks for tasty

[vivid pilot]

oh yeah

thanks

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[unique crane]

can someone explain this

[pearl pond]

@unique crane Has your question been resolved?

[unique crane]

<@&286206848099549185>

[pearl pond]

@unique crane Has your question been resolved?

[pearl pond]

@unique crane Has your question been resolved?

[pearl pond]

@unique crane Has your question been resolved?

[narrow stirrup]

are you allowed to graph using a calculator for this question

if so, that would be the best approach for time

if not, you have a few options. You could plot your own graph, but depending on the function that may be difficult. Alternatively, you could plug in some values close to 1 from the right and left sides. The limit from the right side is 1+ and the left side is 1-. So if you plugged in like 2, 1.5, 1.2, etc and notice it is rapidly growing, then you might assume that it approaches infinity

then you can take that same approach and apply it to the left side

you're basically looking to see what does the function approach as we get closer to the asymptote at x=1 in both 4 and 5

[dapper kraken]

hey im not sure if this is correct, but for the limit as x->1+, x-1 would be slightly more than 0 no? so it would be approaching 0+, and im pretty sure the limit of something over 0+ is infinite, the same goes for lim x->1-, its slightly less than 0 so its negative, and that the limit of something over 0- is -infinite

please keep in mind im not that experienced in limits

[pearl pond]

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[drifting spire]

Hello

[drifting spire]

So I struggle understanding this topic

Due to extreme heat we were forced to be modular so there weren't any Discussion regarding the topic

[unkempt yacht]

do you know what a 45-45-90 triangle is?

[drifting spire]

Not quite

All I know is that it's an isosceles right triangle

For instance, it's an acute angle because its less than 90 degrees..

Base on my comprehension