help-23

[lean otter]

its basically 1/2 *4

[marsh walrus]

id think 2/3

like make 3 (Pieces) of 2

i guess pieces doing some heavy lifting there

[lean otter]

mhm 3 parts of 2

[sudden nova]

This doesnt really matter, in the end its just a matter of convention. With division and subtraction just being another version of addition and multiplication.

[lean otter]

well to compute something like 92-[18+16/4{26-(10)}] it does matter, one could confuse of with multiplication and end up multiplying after resolving the brackets and then the whole answers turns out to be wrong

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[thorny dock]

Why do they do this step?

$T_{\epsilon,\mathcal{B}}$ is the transition matrix from the basis $\mathcal{B}$ to the basis $\epsilon$

[flat frigate]

normalAtmosphericPa=101,325

[thorny dock]

this is for a "change of basis" in linear algebra

[winter pivot]

This just ensures that the basis vectors of e get sent to the basis vectors of Β

[lean otter]

Bro I need help on basic triginometry

Can u help

[winter pivot]

#❓how-to-get-help

[lean otter]

I pinged u see

[thorny dock]

ah you see

I am having trouble understanding the signifiance of this theorem

because all vector spaces have a unique basis right?

i.e. all bases are isomorphic

so what is the point in "sending" vectors of $\mathcal{B}$ to $\epsilon$?

[flat frigate]

normalAtmosphericPa=101,325

[winter pivot]

It allows you to describe transformations

[thorny dock]

you mean the transition matrix described above?

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[strange plaza]

How can I calculate b) without having to list everything out

is there a shorter way than just going 1^2×1/20 + 2^2×1/20...

[lean otter]

factor out the 1/20, and use series of squares formula for n = 20

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[strange plaza]

Thanks

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[wise venture]

Why was the interval split at 1/2? (7th line)

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@wise venture Has your question been resolved?

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@wise venture Has your question been resolved?

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@wise venture Has your question been resolved?

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[obtuse jackal]

That's simply where the actual values change (since [x^2 +- 1/2] changes) so it makes sense to split there and just evaluate by hand in each case

[wise venture]

Right.

Had misread it as x instead of x^2 thus was wondering why the limit had been split at an arbitrary number. Thank you for your time and assistance.

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[karmic garden]

can someone please explain this

this makes NO sense

im unable to understand where the numbers are coming from in the answer either

[marsh holly]

what don't you understand?

[marsh holly]

the answer is only using numbers that are allready in the question

replace the -7w with x, then you have that 17+27 = x now you know that x = 44

then you have to do 10(-21w+1) now if -7w is x then -21w is 3 x which is 3x44 = 132 that plus 1 = 133 times 10 = 1330

I think that the -7w part confused you, did this help @karmic garden ?

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[lean otter]

I know I’m supposed to make the derivative equal 0 but I don’t know how to do that

[iron flare]

do you think a tangent is a line with gradient 0?

[lean otter]

I’ve got no idea what gradient means

[iron flare]

slope

do you think a tangent is a line with slope 0

[lean otter]

Is the equation the tangent line?

[iron flare]

you're finding the tangent at (4,4) of 16x-xy+y^2=64

i'm asking what you think a tangent actually is

[lean otter]

I’ve got no idea what a tangent is

[iron flare]

a line that touches only one given point

[lean otter]

Ohhh okay

[iron flare]

if that red dot was say, (3,4)

that would be the tangent at (3,4)

the important idea to note is the slope of a tangent is the slope of the curve at that exact point

[lean otter]

So the given equation is the black line, the coordinate is what the red line passes through and we are looking for the equation that gives us the red line?

[iron flare]

Yes

You have the means (through calculus) to find the slope of the red line

You also have one pair of points, (4,4), the coordinate in which the line intersects with the equation

you should be able to find the equation of a straight line given a point and the gradient

[lean otter]

I find the derivative of the first equation first right

[iron flare]

yeah

the derivative is essentially a slope function

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[haughty sorrel]

Anyone know what on earth this question is asking? Thank you in advance

[haughty sorrel]

is it saying [[0,1,2],[3,4,5],[6,7,8]]?

[lean otter]

If a 3 digit number is divisible by 10

Then the last digit has to be 0

So c f and i are 0

[haughty sorrel]

ohhh

and the rest can be any numbers shown aboive

[lean otter]

Yea

[haughty sorrel]

gotcha

sweet thanks

[lean otter]

Yep

[haughty sorrel]

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[main heart]

Can y'all please correct my wrong ones? Also I don't really know what definition this one is.
Reason:
1.) Given
2.) Definition of ???
3.) Reflexive Property of Congruence
4.) Angle Angle Side Congruence Postulate

Statements:
1.) ∠G ≈ ∠I ; F̅H̅ bisects ∠GFI
2.) ∠GFH = ∠IFH
3.) FH ≈ F̅H̅
4.) △GFH ≈ △IFH

[main heart]

I think reason number 2 is probably Definition of Bisect? I'm not sure.

And another one.
Reasons:
1.) Given
2.) Reflexive Property of Congruence.
3.) Side Side Side Congruence Postulate

Statements:

1. JK = LK ; JM = LM
2. KM = FM
3. △KJM = △KLM

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[ebon mesa]

if i have 2 whales, and one of them is certainly a male, what's the probability that they're both male?

[ebon mesa]

this problem is the reason why i don't understand probability questions

[lean otter]

There is a 1/3 chance that they are both male

there are 3 possible combinations for the pair: (male, female), (female, male), and (male, male)

[safe radish]

@ebon mesa Has your question been resolved?

[ebon mesa]

but if there's two, and i know that one of them is male, there's 1/2 chance that the other one is going to be a male

i cant spot what's wrong with my answer

[earnest rapids]

You don't know which one of them is_certainly male_ either 1st or 2nd, hence that arises a new possibility so total=3

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[narrow vault]

can someone explain to me how this is the squeeze theorm? because I know that if we take the limits of whatever is written before the <= and its the same as the limit of whatever is written after the <= on the right side, it means that the inner limit is also that value

[narrow vault]

but here im seing only a limit in the middle and a value "2" on the left and right

[quasi bison]

because I know that if we take the limits of whatever is written before the <= and its the same as the limit of whatever is written after the <= on the right side, it means that the inner limit is also that value
yes, that is the squeeze theorem

[quasi bison]

the limit of 2 - |xy|/6 as (x,y) approaches the origin is 2

[narrow vault]

right

thats the left side

on the right?

we're okay with the value 2?

so as the limit of 2 - |xy|/6 as (x,y) approaches the origin is 2, and the value on the far left is 2, we can say the inner limit is 2 as well?

@quasi bison ?

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[green fable]

help me with this please

i derived that y=35 but after that I dont know what should i do

this is my working so far

I dont know what should I do futher

further*

[pastel verge]

How is this done?

[green fable]

see i assumed that they are written in ascending order

so there are a total of 9 factors odd number

which means it is a perfect square

and a pair of factors also give that same number when multiplied

so just equated that

and got y=35

but what should i do after that

[green fable]

please help bruh

[pastel verge]

I'm not even sure about what you did, idk how to do this

[green fable]

oh nice

<@&286206848099549185>

Please help

<@&286206848099549185>

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[green fable]

<@&286206848099549185>

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@green fable Has your question been resolved?

[green fable]

<@&286206848099549185>

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@green fable Has your question been resolved?

[vestal walrus]

Answer: D

Z+20 is the maximum number among all of given 9 factors.
So Z+20 itself is equal to X+Y+Z.
Thus
Z+20 = X+Y+Z
20 = X+Y
X+Y = 20

and X-7, Y-10, Z-12 are positive numbers, and X, Y, Z are positive numbers too.
So X>7, Y>10, Z>12

Because only 9 factors are there,
X+Y+Z can be denoted as p^2q^2 or p^8, either you can cofirm that X+Y+Z is perfect square.
Therefore statement 3 is true.
This also means that X+Y+Z only contains one or two primes

And if Y is prime, it should be 11 because Y is 11, 13, 17, or 19 from conditions Y>10 and X+Y=20 (means Y is less than 20), and the condition that X should be more than 7.

Now assign Y=11 to X+Y=20, we get X=9.
And let's assign variables X and Y to factors X-7 and X-6, we get 2 and 3. Remember that X+Y+Z only contains one or two primes . Let's assume it only has two primes and they are 2 and 3. then X+Y+Z is 2^2*3^3 =36 so Z=36-20=16

Therefore X=9, Y=11, Z=16, and by assigning these variables to given factors, you can cofirm that all of them are actually being factors of X+Z+Y(=36). X is perfect square and Y is prime. So statements 1 and 2 are true too.

[green fable]

Oh man thnx a lor

I will give this a read

[green fable]

Wonderful explanation man

Got it

Thnx a lot

.close

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[regal hatch]

Hey so i have a question

[regal hatch]

Like for any random series

Can we write the general term formula ??

For example

2 ,9 ,28 , 65 ,126

Here 5 finite numbers are written

[split ether]

Finite or infinite?

[regal hatch]

And we know general term will be n^3 +1

[regal hatch]

Finite

[split ether]

Then you can always come up with some polynomial function

[regal hatch]

Now this was a pretty basic thing so
What if i write 2, 4 , 6 , 17 , 30 , 91

This was random

So can i write its genral term

Finite series

[ebon mesa]

there must be something that satisfies it

as long as its finite

[regal hatch]

So i need to form a polynomial with these as its roots?

[split ether]

Consider looking into Lagrange polynomials

[regal hatch]

Oh thanks , what about infinte series? , is their a way for it too?

[split ether]

Not really sure

[regal hatch]

.close

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[spring herald]

Dont know where to start, Wanted to know how to do 2a and 3a

[lean otter]

[
\parens{\f ab}^x = \f{a^x}{b^x}
]

[spring herald]

so is the answer is just x^27 2/3 / 243

?

[lean otter]

No

[spring herald]

i just times it?

[lean otter]

Also make use of [
\parens{a^b}^c = a^{b \cdot c}
]

[spring herald]

then whats the thing that like adding

but smth similar to this

[lean otter]

[spring herald]

...

x^2 /5 X x^7 / 6?

is it those ones?

[spring herald]

Nvm, so is the answer of 2a is x^18 / 243 or x^18 / 9???

;-;;;

@woven thistle ;-;

@lean otter ;-;;;;;;;

[tall bough]

,w 27^(2/3)

[flat frigate]

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[tall bough]

Second

[spring herald]

My calculator is kinda weird, so how do i work that out?

through paper?

[tall bough]

Huh

[spring herald]

it said 243

[tall bough]

It's wrong

,w ((x^(27)/27))^(2/3)

[spring herald]

;-;

[flat frigate]

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[spring herald]

srry i dont understand that, but ok, nvm
can you help me with question 3a? @tall bough

[tall bough]

Do same thing

[spring herald]

3a?

[tall bough]

Take log

[spring herald]

srry im not 1st language english, i dont understand]

[tall bough]

What language are you

[spring herald]

im thai

[tall bough]

$log(x)$ do u know this?

[spring herald]

take log means like, use new channel?

[flat frigate]

Fucktalogist

[spring herald]

for that question?

[spring herald]

no

my teacher never teach me that

[tall bough]

🤣🤣🤣 no

[spring herald]

srry

[tall bough]

Bruh

[spring herald]

Are there other way of doing it without log?

[tall bough]

Idk

[spring herald]

srry

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[plain crown]

The car runs the distance between 2 cities in 5 hours. If he passes every kilometre 3/7 minutes faster, then he will take 3 hours to travel the entire distance. Find vehicle speed

[velvet nebula]

what have you tried

[plain crown]

Well we need variables right

S for distance t for time and v for speed

[velvet nebula]

what equations can we make for them

[plain crown]

first of all we will get first speed = S/5

every kilometre 3/7 minutes faster confuses me

can u tell how do I write it as a variable

[velvet nebula]

be consistent with your units

if you want to use hours, use hours consistently

basically look at the amount of time the car takes for a kilometer and decrease it by 3/7 of a minute

[plain crown]

how do I do that

[fallen stirrup]

I think 1 indication that will help in the future is this:don’t think about numbers too much, just think about how can you obtain the relationship between them, and then use the numbers
Basically, leave the numbers at the and; solve the question and then substitute the symbols with the values you had, always in the same units

[plain crown]

Thanks, thats really good advice I will try to use

in this case the final equation would be first time x speed = second time x new speed

so

[fallen stirrup]

Yes

[plain crown]

5v=3(?)

[fallen stirrup]

Yes, something like that

[plain crown]

well I knew that much honestly, I just dont get how do I use this

[fallen stirrup]

Igtg back to my class for now, if you want you can dm me later

If you still don’t figure it out

[plain crown]

alr ty and good luck in class

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[snow sigil]

any number at the power of 0 is equal to 1. that means n is 0, so that you do 32 at the power of 0 and that is 1.

[snow sigil]

oh I didn't mean to open channel, sorry

.close

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[spring herald]

my bad ;-;

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[coarse kelp]

anything wrong here?

[quasi bison]

seems ok to me

[coarse kelp]

AYYYO

IM GETTING BETTER

LETS GOO

i went from nothing

to something

lets foajflajfldajfaf

go

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[coarse kelp]

guys give me a clue on how to proceed, I don't know how to deal with e^-n/n

[quasi bison]

if the series was just $\sum_{n=1}^{\infty} e^{-n}$ could you say whether it converges or diverges

[flat frigate]

Ann

[coarse kelp]

(1/e)^n

| r | < 1

converges

yeah

[quasi bison]

take this as your hint.

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@coarse kelp Has your question been resolved?

[coarse kelp]

does this make sense to you?

because an <= cn <= bn

so I compared an to bn

with an <= bn

is that allowed

?

[obsidian oracle]

Yes it is allowed, it's transitivity

[coarse kelp]

so my solution and answer is right?

also that is a cool pfp

[obsidian oracle]

Thanks, and yes your solution is correct

[coarse kelp]

it moves when i put my mouse on top of it

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[tawdry yacht]

i need to get from the left side of the equal sign to the right side

[tawdry yacht]

Ive been tryin for like 30 minutes to think of how can i factor these terms so i would get to those 2 paranthesis

but i just cant seem to figure it out

[obsidian oracle]

You want to start from the left hand side or you have to?

[tawdry yacht]

i have to

[peak estuary]

do you know polynomial division

[tawdry yacht]

sorta we just got into it today

[peak estuary]

well then try doing that with the LHS and the factor (a+1)

[obsidian oracle]

Well, one of the important results of polynomial division is that : if P(a) = 0, then (x-a)|P(x)

[peak estuary]

why you try it with a+1 we'll discuss after that

[tawdry yacht]

Okay ill get to it rn thanks

okay i understood honestly

but yeah i dont get why we use a+1

like that wouldnt have crossed my mind lol

[peak estuary]

do you know the rational root theorem

[tawdry yacht]

nop

[peak estuary]

if a polynomial with integer coefficients has a rational root p/q, then p divides the last coefficient and q divides the first coefficient. so here p divides -2 and q divides +1

that only gives very few options for what a rational root could be

[tawdry yacht]

oh

[peak estuary]

p could be 1, -1, 2, -2 and q could be 1 or -1, so p/q could only be 1, -1, 2, -2

[peak estuary]

now we combine it with this

so we only have to test which of these are roots, then we know what factors to check

here -1 is a root which you can check by hand

so (a-(-1))=(a+1) is a factor

[tawdry yacht]

I see

[peak estuary]

also 2 is a root so (a-2) is a factor

so we could have also done the polynomial division with that

[tawdry yacht]

Yeah i was jus about to ask that

Fair enough it makes sense i just didn't know about it

thanks!

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[hallow condor]

I’m stuck in This problem. Not sure what to do with loops as well.

[quasi bison]

an euler path is a path that goes through every edge exactly once.

a loop is not hard to take care of -- just go through it once at any time you enter the vertex it's on

[safe radish]

@hallow condor Has your question been resolved?

[hallow condor]

@quasi bison so what would the path be, or what would I start on? I guess I have trouble figuring out where it should start to create the Euler path.

[quasi bison]

are there any odd-degree vertices?

if so, there should be at most 2 (otherwise there is no euler path at all), and one of them should be your starting point

[hallow condor]

Soo I should start with an odd degree vertices? @quasi bison

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[rapid lantern]

Hello

[rapid lantern]

So we have to find the distance between those two points in polar coordinates

And im confused why we do that on line 2

Why do you use cosine on line two?

[calm girder]

you shud probably ping the helpers

it's been 15 mins

also sorry idk how to help you solve this

[rapid lantern]

All i know is this

For distance formula

[rapid lantern]

But on line 2, they used d^2 = something

And i don’t know where that something comes from

It doesn’t seem like the distance formula

Anyone?

<@&286206848099549185>

[obsidian oracle]

Do you know how to get cartesian coordinates from polar coordinates?

[rapid lantern]

Yes

You use the pythagorean theorem

To find the radius(hypote)

And then also find the angle

So we can write it as (radius, angle)

But im still confused on line 2

d^2=x^2+y^2-2xy*cos(theta2-theta1)

This is what im not understanding

Does such formula exist?

Oh nvm

I think i know why

I keep reading it as rectangular coordinates

Nonono i keep making the same mistake

Thanks for your help though

.close

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[graceful gull]

Shouldn't this answer be correct?

I'm given the diameter of a sphere and I want the radius. So 76/2 = 38
r = 38

I found the derivative of 4/3 * pi * r^2 * h which is 4 *pi * r^2

Then I plugged 38 into the derivative
4 pi (38)^2 = 18145.9392 (rounded to 4th decimal placed as asked)

[graceful gull]

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[strong crown]

Given y(0)=0 and y’(0)=0 , what are the values of c1 and c2? I got them both as 0 but it’s saying that is wrong

[strong crown]

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[strong crown]

Nevermind I sorted it

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[frosty granite]

For part c u don’t need to consider the domain of f(x) or g(x)?

[lean otter]

ok

[frosty granite]

I don’t understand why u wouldn’t consider domains if ur combining those functions

[neat kiln]

Who said that?

[frosty granite]

so i do consider them?

ms said x can be plus or minus 6

probably wrong knowing my spec tho😭

[safe radish]

@frosty granite Has your question been resolved?

[safe radish]

@frosty granite Has your question been resolved?

[frosty granite]

.close

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[slim urchin]

trying to calculate the volume of the red shape. is the length of the blue segment equal to 1/2?

[hybrid thunder]

Not exactly

How did you get there?

[slim urchin]

just because the length of the cube is 1

[slim urchin]

but you're right, it doesnt really make sense

[hybrid thunder]

Try messing around with a sketch on paper

[slim urchin]

how about the dark blue segment? and yeah okay ill try that

[hybrid thunder]

All the sides of an octahedron are equal

[slim urchin]

oh yeah good point... umm

[slim urchin]

is the length and height equal to 1 btw? because the width seems like it is equal to less than that, not to 1.

[hybrid thunder]

Are you talking about the long diagonals of the octahedron?

[slim urchin]

nah the cube, sorry should have mentioned

[slim urchin]

cause for my last cube it was like a proper cube, but this one is just a rectangular prism it looks like, so all the lengths arent equal to 1

[hybrid thunder]

Yeah all sides are 1(for the cube)

Since it was given that it was a cube and volume is 1

[slim urchin]

thats so weird though because this side doesn't look like its equal to 1 - ill show u apic

the blue side seems a shorter length than the red sides

[hybrid thunder]

Yeah that's how 3d perspective works

[slim urchin]

but it doesnt look that way here? ok my spatial intelligence is really poor, so i'm just gonna accept that every side in the other cube is also equal to 1 esp since he mentioned explicitly that it is a cube

[hybrid thunder]

Just assume all 1 and proceed

[slim urchin]

is one of the diagonals of a square face of the cube also a diagonal of one of the equilateral triangles of the octahedron?

[hybrid thunder]

Is this for the octahedron or tetrahedron?

[slim urchin]

octa, i only need to figure out for the octa

[slim urchin]

i've done the green shape already

[hybrid thunder]

Can you rephrase your question, it was a bit unclear?

[slim urchin]

yeah sure. actually now that i look at it

it looks like half the length of the diagonal would give me half the length of the height? or would half the diagonal length give me the length of one of the equilateral sides of the triangle faces making up the octahedron?

[slim urchin]

like, half the blue length is equal to the yellow length?

[hybrid thunder]

Yes

Try using visual guessing less, and more of algebraic work

Pythagorean theorem should be used a lot here

[slim urchin]

or maybe i just calculate the volume of 8 pyramids with height of 1/2 and the length of the equlateral triangles being equal to sqrt 2, which will help me calc the base

and subtract it from 1?

but the height wouldn't be 1/2 i don't think..

[slim urchin]

is the length of the green segment here 1/2?

[hybrid thunder]

I think finding the volume of one of the 8 pyramids of the octahedron and then multiplying by 8 is the easiest way to find the volume

Sorry, there are 2 pyramids

(Top and bottom)

8 of the triangle-like sectors

[slim urchin]

the length of one of the equilateral sides is sqrt 2 right?

sorry, sqrt2/2

[slim urchin]

is this right so far, the labelling?

[hybrid thunder]

Yes

[slim urchin]

ok so then i just have to find the volume of a square based pyramid and times it by 2

[slim urchin]

the height of the triangular face is also equal to sqrt2/2 right?

i'm just not sure how to get the height

[hybrid thunder]

Do you know how the formula for volume of a square pyramid?

[slim urchin]

base edge length squared times height over 3, right?

[slim urchin]

so base edge must be sqrt2/2 ?

[hybrid thunder]

Yes

[slim urchin]

bit confused abt the height tbh

[hybrid thunder]

What is the height of the entire octahedron(both prisms)?

[slim urchin]

i mean i thought it was 1 originally

since the cube is height 1

[hybrid thunder]

Right

[slim urchin]

which would make the height of the pyramid 1/2

[hybrid thunder]

Exactly

You got it

[slim urchin]

oh yeah, i think thats what i was getting at before when i asked this but i confused the height of the octa with the length of one of the sides

[hybrid thunder]

Not your fault, it isn't exactly the easiest process to explain to someone else online

Anything else you need help with?

[slim urchin]

thanks a lot for all the help. if its okay with u i might just confirm i've calculated the vol correctly? i just wanna make sure i dont make any careless mistakes

im just gonna input all the vals now

[hybrid thunder]

Go ahead

[slim urchin]

ok i got 1/12 for the vol?

[slim urchin]

i can send the working out? i tend to make weird/careless mistakes

wait i need to multiply it by 2

vol = 2/12 for the entire octa?

[safe radish]

@slim urchin Has your question been resolved?

[safe radish]

@slim urchin Has your question been resolved?

[safe radish]

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[peak crane]

How do u do 1-12

Written exercises

[hybrid thunder]

Do you remember the rational root theorem?

[peak crane]

nope

is it 6/1

for number 1

nvm idk what it is

[hybrid thunder]

Here

https://www.youtube.com/watch?v=GTx-rXTQg1o&ab_channel=patrickJMT

First watch the video, learn the theorem, then see if you have further questions

Best of luck

[peak crane]

alr thx

[hybrid thunder]

np

[peak crane]

@hybrid thunder

i watched the video and it made sense

but my only concern is

when you find the rational roots

how do you represent your answer

[hybrid thunder]

wdym

[peak crane]

like

for example

you have 2 and 3 as your answer

[hybrid thunder]

since the polynomial is = 0, then the zeroes of the polynomial are the solutions of the equation

[peak crane]

so

how would u show that

[hybrid thunder]

Has your book gone over synthetic division yet?

[peak crane]

yea

lol

light work

[hybrid thunder]

K, so if say 2 is a root of your polynomial

That means the polynomial = 0 at 2

[peak crane]

yea

[hybrid thunder]

So (x-2) must be a factor of the polynomial

Since at 2, it would drive the entire expression to 0

[peak crane]

why x-2

-2

ohhh

because

its a positive 2

and so it would have to = to 0

[hybrid thunder]

Right, you got it

[peak crane]

bett

can i do number 1 and u check it

[hybrid thunder]

so you divide the entire polynomial by x-2 using synthetic division

[peak crane]

yea

[hybrid thunder]

and see if the remaining polynomial has any remaining roots, etc..

[peak crane]

yea

[hybrid thunder]

Sure go ahead

[peak crane]

k

[hybrid thunder]

$x^{3}-7x-6=0$

[flat frigate]

BetterThanYou

[peak crane]

yep

[hybrid thunder]

yeah im just practicing using TeXit

[peak crane]

oh nice

wait

how would u do 1-7-6

💀

which one would go first

@hybrid thunder

[hybrid thunder]

for synthetic division or rational root theorem?

[peak crane]

no just simplyfng

[hybrid thunder]

You are asking for what is 1 - 7 - 6?

[peak crane]

yea

[hybrid thunder]

-12

[peak crane]

7-6 would go fisrt right?

[hybrid thunder]

No, order of operations

right to left

[peak crane]

yea 7-6

1-1=0

[hybrid thunder]

mb left to right

[peak crane]

yea

alr

also i have another question

[hybrid thunder]

ask away

[peak crane]

u know when you find the roots

[hybrid thunder]

yeah

[peak crane]

and its + or -

do you find both values

of that number

[hybrid thunder]

See, when you test the rational root theorem, it gives all POSSIBLE rational roots of the expression

[peak crane]

like do you plug in the + and - of that number

[hybrid thunder]

You have to individually test each one

[peak crane]

bruh

that takes a long time

[hybrid thunder]

Yeah no shit, but if you get good at synthetic division it shouldn't be too bad

[peak crane]

no

im not using synthetic

im pluging them into the polynomial

[hybrid thunder]

That is a shortcut, but consider...say... a 7th degree polynomial

[peak crane]

wdym

[hybrid thunder]

With possible roots of like ±1, ±3

then plugging in wouldn't be easy to calculate

[peak crane]

wait

using synthetic division

u would also have to plug in those seperatly

[hybrid thunder]

I don't think you would

[peak crane]

do u mind explaining

[hybrid thunder]

you would just divide by all (x-r), the r's being all possible roots

until you get no remainder at the end of synthetic

[peak crane]

okokokokok

can we do number 1 together

[hybrid thunder]

Yeah

$x^{3}-7x-6=0$

[flat frigate]

BetterThanYou

[hybrid thunder]

Go ahead

[peak crane]

ok so

rational roots are

• or - (2,3,6,1)

[hybrid thunder]

good

now test each with synthetic

[peak crane]

so 8 values using synthetic?

[hybrid thunder]

Well, the polynomial prob had like 3 rational roots

So you have a 3/8 chance of being right first time

and then once you are successful once

You are left with a quadratic

Which tbh isn't worth using synthetic division on

[hybrid thunder]

but yes

[peak crane]

bruh

what is this

thats to much work 😭

[peak crane]

how do you know that

[hybrid thunder]

its a third degree polynomial

[peak crane]

o

[hybrid thunder]

so it will have 3 roots by the fundamental theorem of algebra

they CAN be complex

[peak crane]

yea

[hybrid thunder]

but i don't think that would be the case in the first example

[peak crane]

ok so

[hybrid thunder]

and at the end of the day synthetic is just some multiplication and addition

[peak crane]

i didnt use syntehtic for this

[hybrid thunder]

what did you use?

[peak crane]

substitution

i substituted the roots into x

[hybrid thunder]

yeah, that works until you get to higher order polynomials like i mentioned

then there are hardly any shortcuts

[peak crane]

up to what polynimal do u think

[hybrid thunder]

It also depends on the coefficients, because if you have possible roots like 12 or smth

even 3rd degree isn't easy

and don't even get me started on when the roots are fractions

[peak crane]

so u would say syntheric is easier?

[hybrid thunder]

yes

[peak crane]

im still confused on the synthetic way

so we have + or - (2,3,6,1)

[hybrid thunder]

k, what next?

[peak crane]

uhhh

no clue

[hybrid thunder]

we test each root with synthetic

start with test 1

by dividing the polynomial by (x-1)

using synthetic

you know how to do synthetic right?

[peak crane]

yea

i only know how to use synthetic using 1 value

like 3 or 4

[hybrid thunder]

oh see, if you do it once

and you get a remainder

you basically scrap the attempt

and start from the polynomial before attempted division

[peak crane]

??

[hybrid thunder]

because if you have a remainder after synthetic, the term couldn't be a root of the polynomial

[peak crane]

yea

[hybrid thunder]

so you test something else

[peak crane]

cuz its not 0

[hybrid thunder]

its a trial and error process basically

[peak crane]

but what do you plug in

using synthetic

thats my question

[hybrid thunder]

the possible root

[peak crane]

so 6

in this case

or -6

[hybrid thunder]

yeah, any of the possible roots

[peak crane]

wait

so u gotta use synthetic division 8 times

just to be clear

[hybrid thunder]

no

once you get a remainder of 0

you found a root

[peak crane]

mhm

and you do it untill u find 3

right?

[hybrid thunder]

for this problem yes

[peak crane]

ohh ok

i get it

[hybrid thunder]

wait

[peak crane]

bruh

[hybrid thunder]

so once you divide once, you will have a remainder

which will become the new polynomial you have to test

[peak crane]

???????

[hybrid thunder]

let me write out the solution for a random example just so that you can see

[peak crane]

ok

[hybrid thunder]

sayyyyy...$x^{4}-2x^{3}-7x^{2}+20x-12$

[peak crane]

mhm

[flat frigate]

BetterThanYou

[hybrid thunder]

one moment as a write it

it should take no more than 5 mins

[peak crane]

yeye take your time

[hybrid thunder]

This is part one of the solution

ran out of space, might take a bit more time

[peak crane]

all good

thanks for the effort

wait

why did u choose 1 over 2

or 3

or-4

[hybrid thunder]

im just going in order

[peak crane]

og

bruh

thats a lot of work for one problem

[hybrid thunder]

well..sometimes that just how math is

k im gonna have to leave for like 10-15 mins

pick apart the example for now and try to do the first few problems

Did you make any progress?

@peak crane

[peak crane]

Yoo

Not really

I was planning on doing it tmr

@hybrid thunder

So after 2

[hybrid thunder]

Alr, as long as you understood the material

[peak crane]

U gotta do all of those roots

3 4 6 12

Oh wait

Nvm

Ngl I’m still a bit confised

What if 2 was not a root

And u got a different remainder

[hybrid thunder]

Then I would have kept testing

[peak crane]

With the new polynomial correct?

[hybrid thunder]

Yeah

[peak crane]

Bruh

Wait

In the beginning

Why didn’t u use the polynomial of -1

Since that also didn’t work

@hybrid thunder

[hybrid thunder]

Well, once I saw that 1 was a factor, testing any further would have been unnecessary

[peak crane]

U said 1 was not a factor

[hybrid thunder]

We are talking about the example right?

[peak crane]

Yea

[hybrid thunder]

Yeah so

1 was a factor of the original expression

After the first test was successful I had a new polynomial to factor

And when I tested 1 again I found out that it wasn’t a factor of the new polynomial

Is that clear?

[peak crane]

How was it successful

[hybrid thunder]

The remainder when dividing by (x-1), in other words testing if 1 is a root, was zero

[peak crane]

Butttt

U said it didn’t work

[hybrid thunder]

It worked once

There was no guarantee it would be a factor again

[peak crane]

Wdym

So 1

When u put it through synthetic

Was it = to 0

[hybrid thunder]

The original polynomial: yes, when I tried synthetic again with the remainder: no

So (x-1) is only a factor of the whole polynomial ONCE

[peak crane]

Wdym when I tried again with the remainder

@hybrid thunder

[hybrid thunder]

When I divided by (x-1) once, I got a remainder of 0, and the other coefficients left from synthetic division were the polynomial left after I divided by x-1

[peak crane]

Ye

Sooo?

[hybrid thunder]

When I divided this new polynomial by (x-1) again, I found that it wasn’t a factor

So I moved on to other possible roots

[peak crane]

Why would u divide it again

[hybrid thunder]

Because it can be a repeated root

Let’s pretend we are back in the arithmetic days

And your teacher asked you to factor 28 into its prime factors

You see that it end in an even digit, so you divide it by 2 and get 14

You see that it ends in an even digit again so you divide by 2 again, so the prime factorization of 28 = 2 * 2 * 7

Meaning 2 is a repeated factor here

So just because it was a factor once, doesn’t mean it can’t be a factor again

That is why I tested (x-1) twice

[peak crane]

Ye

I got it

Thanks!

Thanks for the help I appreciate it

Good bye

.close

[safe radish]

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[hybrid thunder]

Anytime

[safe radish]

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[zenith remnant]

Can someone explain how this is no solution

[elfin yew]

erm

could you post the original problem

?

[marsh walrus]

no solution? do you mean its undefined?

[zenith remnant]

yes undefined

[marsh walrus]

so square roots have to be of positive numbers

,w graph 9-x^2-16

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jan Niku#2119

[marsh walrus]

this is never positive

[zenith remnant]

Ohhh ok

Ty

.close

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[dreamy finch]

how would you find the remainder of 10^115 when divided by 7? This is what i did, but i feel like i just got super lucky

[dreamy finch]

first i found 10 mod 7 which is 3 and not really helpful. Then i found 100 mod 7 which is 2 which is better, but still not so useful because of how high the power is. Then i found 1000 mod 7 which is saw is 6, which is congruent to -1. Then I rewrote 10^115 as 1000^38 x 10

and replaced 1000 with -1, -1^38 is even, so now i have 1x10 which is 10 and 10 mod 7 is 3

that's how i got the answer, but my concern i guess is what if 1000 mod 7 was not 6 and thus not congruent to minus 1?

[lean otter]

So your final answer was 3 right

[dreamy finch]

yes

[lean otter]

I have an idea on how to approach it but I’m not 100% sure

oh cool

[dreamy finch]

well i guess what im asking is if there's a general thing i should be looking for

like should i be trying to figure out how i can somehow get a -1(mod m) congruence?

because i feel like doing that can be pretty difficult

[lean otter]

the way i was thinking probably doesn’t work

[dreamy finch]

oh okay. Also my professor find 6^27 mod 7 by doing this:

and im just confused because that looks so complicated

cuz 6 mod 7 is congruent to -1 mod 7, and then its really easy to do

[lean otter]

ah ok

[dreamy finch]

because -1^27 is -1

which is congruent to 6

isn't that the simplest way of doing it?

[lean otter]

I think so

[dreamy finch]

okay awesome

so in general, am i just trying to look for a congruence to -1

or can that not always be the case

[lean otter]

Idk much about this stuff, but I found this

https://en.m.wikipedia.org/wiki/Fermat's_little_theorem

[dreamy finch]

Thank you! I studied that theorem a bit and i hate it lol. But thank you for your help nonetheless!

[lean otter]

np

[dreamy finch]

.close

[safe radish]

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[languid lintel]

and how this can be simplified

[kind fractal]

$A\oplus B = A\bar B + \bar A B$

[flat frigate]

ELeonardo

[kind fractal]

Use DeMorgan rules

[languid lintel]

does this look right as the simplified version?

[kind fractal]

Mmm I don't think so

[kind fractal]

Could you show your work please?

[languid lintel]

^

[kind fractal]

If you consider C = 1, then the whole expression is 1

[languid lintel]

that makes no sense tho

it can't all be 1

[kind fractal]

Yes

That's why I think it's wrong

[languid lintel]

crap

I don't know how to make it right

this was what I thought the correct schematic was

[kind fractal]

The schematic for the original expression? Or the simplified form?

[languid lintel]

simplified

unless maybe it isnt simplifyable?

[kind fractal]

Every boolean expressions can be simplified to a sum of products (SoP)

Try using DeMorgan first, considering 2 terms:
(A xor B)! and C! D

In that way you get rid of the complement in A xor B

Use DeMorgan again for (C! D)!

Finally, just distribute

[languid lintel]

I made what I think is the unsimplified version

can you check if its correct?

if so

I can match up the TT

and see if the simplified version is correct myself

this is what I think the unsimplified one is

When I set D=1 the unsimplified turns off but the simplified stays on

thats not a good sign

[kind fractal]

Yeah, the simplified version you got is not correct, but you can try the steps I mentioned

Also this diagram is not correct. The expression has the complement of A xor B, but this diagram shows A! xor B!

(A! XOR B!) ≠ (A XOR B)!

[safe radish]

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[bold ferry]

can someone explain what they did at step 4 and after?

how does theta = 0 work?

and why replace r with -r?

[marsh walrus]

my guess is

if you are somewhere in polar coordinates right

then its like

picture standing at the origin

facing up, pi/2

if you walk out 1 radius from there

its the same as if you first turn 180 degrees around

and then walked 1 radius backwards

make sense?

any point $(r,\theta)$ is just the same as $(-r, \theta + \pi)$

[flat frigate]

jan Niku

[bold ferry]

ohhh

yea

wait so they negated the second one

-1 + cos(theta)

but then why set it equal to 1 - cos(theta)

[marsh walrus]

im not sure what you mean, since you want them equal

[bold ferry]

I guess I'm confused where the theta + pi comes in here

[marsh walrus]

lets think hrm

$\cos(\theta+\pi)$

[flat frigate]

jan Niku

[marsh walrus]

so were a half period ahead

then i think

$-\cos \theta = \cos(\theta + \pi)$

[flat frigate]

jan Niku

[marsh walrus]

wait

yea, that makes sense to me

[bold ferry]

[bold ferry]

okay that kinda makes sense

cuz adding pi is just the other side

[marsh walrus]

hmm

how do you flip that 1

[bold ferry]

oh so they started with 1 - cos(theta) = 1 + cos(theta)
then flipped the left side cuz -r, and flipped just the cos in the RHS

[marsh walrus]

oh right,

[bold ferry]

couldn't you do it any order tho?

[marsh walrus]

1 is an r

[bold ferry]

maybe i can test it

[marsh walrus]

so you flip 1

and the minus becomes a plus

makes sense

[bold ferry]

okay but

what if we started with 1 + cos(theta) = 1 - cos(theta)

flip the left side

and flip the cos in the right side

-1 - cos(theta) = 1 + cos(theta)

[marsh walrus]

you cant flip just the cos

at least not this way

the r comes along for the ride

[bold ferry]

oh

?

[marsh walrus]

flipping the cosine like

cosx into -cosx you mean?

[bold ferry]

yea

[marsh walrus]

that depends on r -> -r

remember

you have to both turn around 180 degrees

and walk backwards

if you only do one you end up at a different place

the flipping the cosine is part of this turning and walking backward

(it comes because were adding 180 to the angle)

[bold ferry]

okay like im confused

like all i get is

[marsh walrus]

i mean in general -cosx doesnt equal cosx right

[bold ferry]

we wanna flip r and add 180 degrees

yea

[marsh walrus]

but it can

if you replace say

y=x+180

and some p = -r

give them different letters if it helps

[marsh walrus]

okay lets start here

which equation do you wanna work on?

and which side

or still the original

[bold ferry]

1 + cos(theta) i guess

[marsh walrus]

also at some point

my sleeping pills are gonna kick in

[bold ferry]

[marsh walrus]

okay so

[bold ferry]

it's too late for me to process anything too

[marsh walrus]

$r \to -r$ and $x \to x + \pi$

[flat frigate]

jan Niku

[marsh walrus]

maybe changing letters is not fair because this transformation is like

the coordinates are all the same

because were doing both of these things at once

seems fine to use the same letters

okay so 1 is an r

it gets flipped

and the x inside gets replaced

[bold ferry]

okay

[marsh walrus]

$1+\cos x = -1 + \cos(x + \pi)$

[flat frigate]

jan Niku

[bold ferry]

so -1 - cos(x)

[marsh walrus]

,w graph cos(x+pi) and cos(x)

[flat frigate]

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[marsh walrus]

yes

[bold ferry]

oh wait

[marsh walrus]

[bold ferry]

OHH

i get it now