help-0

[alpine sable]

-b?

[thorn sundial]

do y'all know this?

[dire whale]

the shift from B to A is the same as the shift from C to D. So you just take -b in the first coordinate and keep the second one the same

[alpine sable]

long division

nvm

[harsh nimbus]

Do you still need help with this one?

[alpine sable]

idunno

[harsh nimbus]

What’s the full question

[alpine sable]

synthetic

[dire whale]

why cant you just put x=i+1 and calculate

[leaden sundial]

how do i do this?

[gritty tulip]

So wait what letter does that make it

[dire whale]

shift all the coefficients into exponents on the arguments, then use the rule for combining logs

[dire whale]

G

[dire whale]

this is some 3rd grade stuff

[leaden sundial]

visuals? im stuck on the argument part

[alpine sable]

can you slove it?

[little whale]

Does anyone know principle of virtual work method for Trusses?

[fresh hatch]

@dire whale
I understand what is derivative on a point on function line, but what is antiderivative in the same point?
This is just breaking my mind. Derivative is just another function that describes a tangent of a function in this point, but what will be antiderivative? wot

[dire whale]

$2logx - 3log(x^2+8)+4log(x-6)$\
$log(x^2)-log(x^2+8)^2+log(x-6)^4$

[ocean seal]

Pepa

[gritty tulip]

Thanks bro @dire whale

[glass lichen]

Area under the curve

[fresh hatch]

point

not curve

[glass lichen]

Ok, so 0

Since the area is 0.

[fresh hatch]

I need answer to this question in order to understand how integral related to area, but no vice versa

[leaden sundial]

so what step am i from that, i gotta do the rule rn right?

[glass lichen]

I answered the question

[fresh hatch]

this a bit not what I need

[glass lichen]

You asked what the antiderivative of a point is

It's 0

[fresh hatch]

What on fundamental level connects integral and area?

[leaden sundial]

would it not be 8)^3?

[coral frigate]

Ive got it, thanks, just to clarify, solutions are 5, 65 ,125

[fresh hatch]

how antiderivative related to area at all

[dire whale]

$a-b = -\frac{5}{4}$, $8a+2b = 2$

[ocean seal]

Pepa

[latent ginkgo]

i need help

someone

helppp

pls

i need help

[harsh nimbus]

Question?

[latent ginkgo]

this

[woeful summit]

Can someone give me a video or tut about change SI i mean M change to m and else

[latent ginkgo]

idk why im not gettin it

[harsh nimbus]

What do you not get

[latent ginkgo]

im tryin

[alpine sable]

Yo

[latent ginkgo]

idk what to do

can ya help

[harsh nimbus]

What’s the question asking

Isn’t it just delta velocity over delta time = acceleration

[latent ginkgo]

how 6m-0ms/30 is eqyak to0.2

[latent ginkgo]

idk

can u solve and help

pls

ifi get one answer ill remember the whole process

i forgot the basics

?

u there

[harsh nimbus]

I am confused on what you don’t get it’s just division

[latent ginkgo]

lol

sry

idk

[harsh nimbus]

That s was supposed to be blue

[latent ginkgo]

how 6/30 = 0.2

im gettin somethin else

[harsh nimbus]

6/30=2/10=0.2

[latent ginkgo]

oh

how 6/30 = 2/10 ?

im confuse

d

[harsh nimbus]

6=2 * 3
30=10 * 3
Cancel 3 from both
You are left with 2/10

[latent ginkgo]

tnx a lot

idk how i didnt get this

i usually do this stuff

i forgot the basics

for some reason

[harsh nimbus]

Happens to the best of us as long you understand in the end

[latent ginkgo]

ig

tnx a bunch

[small gale]

alright guys help me

everyone knows about pythagorean triplet

[harsh nimbus]

Yeah

[small gale]

a² + b² = c²

I have a pythagorean triplet

7, 24, 25

[harsh nimbus]

Yes

[small gale]

7² + 24² = 25²

49 + 576 = 625

nice

this is a triplet

[harsh nimbus]

Yes

[small gale]

but in school we were taught

a formula

[tacit kernel]

Is it 12 or 18 or both??

[small gale]

2m, m²+1, m²-1

for finding triplets

now my question is

[alpine sable]

12 isnt it

[harsh nimbus]

12

[latent ginkgo]

12

[small gale]

what is the value of m in this triplet

[harsh nimbus]

Yeah it’s one formula

[tacit kernel]

Why not 18?

[alpine sable]

howd u get 18

[harsh nimbus]

It’s not always just one thing

[tacit kernel]

6*3 🥲 🥲 😭 😭

[latent ginkgo]

wow

[alpine sable]

oh nice

but i think it 12

but

ambigous qn

[small gale]

what do you mean

man come in a channel which isn't getting other questions

you there?

[alpine sable]

It's 12

[tacit kernel]

Why not 18?

[harsh nimbus]

So like the formula you have of 2m, m^2-1, m^2+1 does not work for 3,4,5
But it still is a triplet meaning that not necessarily it has to follow that pattern there are other

[fresh hatch]

Integral in point x is a coefficient of tangent to a integral of that function in that point x. Here in x=6 for x^2/8 tangent coefficient of tangent is 1.5 and for x/4 function argument equal to 1.5. And also integral of x/4 is equal to x^2/8. Thanks. If I am wrong then correct me. This message is related to my questions earlier

[small gale]

so basically i should trash this formula for finding third member if 2 are given

[harsh nimbus]

Yeah you can

[small gale]

given was 7 and 25

and i found 24 by other method

but i wanted to see if this formula works

and it fekcin doesn't

lol

thanks man

[harsh nimbus]

Wait also I think there is a similar version to that formula by dividing 2 all around
And get

[autumn plaza]

Can I get my help here?

[harsh nimbus]

Sure

[autumn plaza]

I need help with similar triangles.

[harsh nimbus]

Question?

[autumn plaza]

Okay. So. In a triangle ABC a line splits the 90 degrees angle in two 45 degree angles. The line cuts hypotenuse in the point D. Cathetus AC=12cm and BC=6cm. You need to find CD.
Hope you understand. I had to translate it from another language.

[crisp grove]

assuming B is the right angle..?

[alpine sable]

ok can someone help me with this quiz

please

[buoyant kayak]

no help on quizzes/tests

[autumn plaza]

So I found the solution in the textbook. And they added another point, E. So DE = CE = x And CD = x(squareroot)2.

But how did they get that point E..?

[native temple]

Does pythagorian therim give estimations or exact values?

[oak chasm]

Exact values if you have exact right angles.

And exact lengths.

[native temple]

there was a problem

where I had similar angles

one triangle had

16 and 25 as a and b

for c

that was x

the other triangle had the left leg as 12

and hypotenuse as 18

the answer was 24

but sqrt(16^2 + 25^2) is 25…

roughly

….

[oak chasm]

No, sqrt(0² + 25²) is 25.

[native temple]

roughly

closer to 26

[oak chasm]

,calc sqrt(16^2 + 25^2)

[ocean seal]

Result:

29.681644159312

[native temple]

what?

sorry

16 ^2+ 20^2

[oak chasm]

So, c is sqrt(16² + 20²) = sqrt(656) = 4 sqrt(41).

[native temple]

but the answer was 24….

[oak chasm]

The other triangle's hypotenuse is 18/(4 sqrt(41)) times as large in the lengths.

Are you sure they're similar triangles.

[native temple]

yep

[oak chasm]

Legs 16 and 20 have different angles than leg 12 and hypotenuse 18, I think. Let me check.

,w sqrt(18^2 - 12^2)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[oak chasm]

So, the shorter leg in both is 16 and 12.

[native temple]

…

[oak chasm]

,w sqrt(16² + 20²)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[oak chasm]

The hypotenuse in both is 4 sqrt(41) and 18.

So, the sine of the larger angle is 16/(4 sqrt(41)) and 12/18.

One is irrational, one is rational, so the sines are unequal.

So, the larger acute angles aren't equal, so the triangles aren't similar.

The measurements must be wrong if they're similar.

[native temple]

oh

[open girder]

can someone tell me how to do the type of questions boxed in red? i think they are all the same type, im getting stuck on them

[alpine sable]

seperate x in such a way that if you multiply the coefficient of x^2 and the constant term you get the product of the values of the seperated terms

for example ,
x^2-x-90
= x^2-10x+9x-90 [ notice that (-10)(9) = -90 which is also the product of coefficient of x^2 (which is 1 ) and the constant term (which is -90) ]
now ,
x^2-10x+9x-90
= x(x-10)+9(x-10)
=(x+9)(x-10)

you can do the same for the other questions

[open girder]

so i can split the next one as 2x+2x?

[alpine sable]

yup

[open girder]

so for (b)(iv) i multiply 10 and -2, so -20

i can split as -2x and 10x?

or 2x and -10x

[alpine sable]

yup

no no wait

whatever u choose its sum must be equal to (-x) in this qn

which would be -5x + 4x

[open girder]

ohh okay

thanks

[alpine sable]

and -5*4 = -20 too

[nocturne totem]

Could i have some help for this question:

The perimeter of a rectangle is 68cm. If the diagonal is 26cm long, what is its area?

I have seen the way that people do it but idk how they get there

[winter rock]

Wassap

[sweet ice]

This is a practice question. How would you go about answering it?

[winter rock]

I need help for an demonstration

[sleek elbow]

#❓how-to-get-help

[thin berry]

Can someone help me with this real quick ?

[alpine sable]

I don't have my copy but can tell you the process

[thin berry]

Ofc

[alpine sable]

Take the coor.of point A (x,0)

By distance formula, find the value of that x

[thin berry]

Yes

Wait imma do it

[alpine sable]

Now, as O is (0,0) measure the distance of it too.

Then apply Pythagorous Theorem, (take the y intercept 0,y) and find it.

[alpine sable]

Kind of?

[winter rock]

@alpine sable vectorials

Look,

[thin berry]

Wait so you mean (x,0)right?

[alpine sable]

Oh yeah

[thin berry]

Yes it is -20

[alpine sable]

Then follow the process above

[winter rock]

The 20th,

[thin berry]

Thank you so much @alpine sable

[alpine sable]

You're welcome. It's cuz of you as I think you're decent in calculating stuff :)

[alpine sable]

I can't even understand the language + I don't have a copy rip

[winter rock]

Sry,

I can translate

[thin berry]

Yes , i am not that much in graph question and stuff like this

[winter rock]

I just have to demonstrate that AB² - AM² =MB × DM

[alpine sable]

Hmm, I see

[winter rock]

And it's just said that M is a random point on BD

[winter rock]

Anyone can help me?

[strong zealot]

can someone help me with this

[hoary shell]

Sure, if for a, it is the limit as x approaches zero from the right, so you use x^2 for the limit because it is defined when x > 0

Then for the limit as x approaches zero, the right and left hand limits would need to agree for the limit to exist, so you would find the limit as x approaches zero for x and x^2

The left hand limit you would use x because that is how h(x) is defined when x < 0

One small thing I can’t remember is if the actual value of the function needs to be defined at that point for the limit to exist, which i think it doesn’t. h(0) is not defined but I think the limit does exist

[winter rock]

@hoary shell could u help me?

For the 20th, I have to make an demonstration of
AB² - AM² = MB × DM

[hoary shell]

I wish I could help but I am not very good at geometry problems 😦

[winter rock]

Could u try w me?

I already find out 2answer

[hoary shell]

sure

[winter rock]

The problem is that the 1st answer seems too short 4 me

But the 2nd answer seems too long 4 me

Also the 2nd answer is just the 1st answer but going farther

And I don't know if it's like that that I should do

But I did the demonstration in a reverse way

So I don't know when to stop

Here I got that

@hoary shell

[hoary shell]

hmm

[winter rock]

Yup

Like I told u

I'm doing the demonstration in a reverse way

Idk when to stop🤣

[uncut tapir]

Geometrically, there's probably two things you'd need to use. Congruent triangles and Pythagoras

[winter rock]

The 2nd answer is going from 2AB²=2AO²+(AO+OM)+(AB×AO)+(AB×OM)+(AO×OD)+(AO×OM)+(OM×OD)

@uncut tapir shit, I didn't used pythagoras at all

But it's vectorial

And dot product

With 2vectors

[hoary shell]

ohhhhh

[uncut tapir]

Ohhh

Ok

[hoary shell]

so that is magnitude squared still?

[winter rock]

I did everything in a reverse way btw cuz it was too hard for me to find it like that

Magnitude squared still?

Idk what it means english isn't my main language

It's 11th grade btw so if what u just said is from an uni level then no it's not that

It's vectorial like I said yh

[hoary shell]

magnitude is just length yeah

[uncut tapir]

Or when you square a vector is it that vector dot itself

[winter rock]

Yeah that's it

[hoary shell]

ohhh

[uncut tapir]

The notation is a bit unclear

[winter rock]

AB²=AB×AB

And because AB=AD

[uncut tapir]

That is the cross product

[winter rock]

AB²=AB×AD

[hoary shell]

AB*AB haha ABxAB is cross product 😛

[winter rock]

This is cross product?

It's equal to AB² right?

[bold herald]

i need help

[uncut tapir]

The notation is a bit confusing

[hoary shell]

i think in this problem AB^2 is dot product

[bold herald]

Simplify
11
b
3
+
5
c
5
+
23
+
a
2
−
6
c
5
−
3
b
3
−
6
a
2
−
16

[winter rock]

@bold herald pls go to another channel

@hoary shell wait what do u mean

[modern ginkgo]

anyone got an idea for this?

right triangle

[uncut tapir]

What does it mean to square a vector

[modern ginkgo]

need the ratio between black area and whole triangle

[uncut tapir]

There are 3 possible interpretations

[winter rock]

What are those?

[uncut tapir]

Dot, cross or magnitude squared

[winter rock]

I have absolutely no idea about what ure talking about

What is magnitude?

[hoary shell]

length

like AB length

[winter rock]

Maybe look at the problem it could help u

[tranquil parcel]

Abs(A)

[uncut tapir]

It can't be cross product because cross returns a vector

[winter rock]

There's no information

[tranquil parcel]

For an arbitrary vector A

[winter rock]

Wait

Are u telling me that AB² isn't equal to AB×AB?

I got false on everything I did from the first step😭?

[tranquil parcel]

Or the dot product with itself

[tranquil parcel]

No atleast you shouldn’t use x for the multiplication

Otherwise ur confusing yourself with the cross product

[hoary shell]

with vectors there is two types of multiplication

[winter rock]

OW THATS WHY YALL ARE SAYING THAT

OW DONT WORRY YALL

It's just cuz I didn't find the point

This ;

•

[hoary shell]

yeah

[winter rock]

AB•AB

Huh let's continue

😭

I was gonna kms for a second

I thought I had false on everything

Cuz it was the first thing I did lol

So AB²-AM² = MB • DM
I have to do the demonstration,

[hoary shell]

in other words its (AB * AB) - (AM * AM) = MB * DM ?

[winter rock]

Idk why are u asking me😭

Yeah I typed that but I'm here cuz I'm not sure of my answer 😭

[hoary shell]

not sure if this can help but I do know from my calc 3 class that since MB and DM are parallel the dot product MB * DM = |MB|x|DM|

[winter rock]

But

I never saw that in classes

So I can tell u 4 sure that this shouldn't be used

[hoary shell]

have you learned any properties about dot products of a vector with itself?

my instinct would be to try to use pythag but since it is all dot products I am not sure how they could be combined

huh I am dumb

a vector dotted to itself is the same as its magnitude squared XD

AB*AB = |AB|^2

[glass lichen]

yes, and this holds in general

[junior hazel]

guys i kinda need some help

which graph would it be

[hoary shell]

try plugging in 0?

[junior hazel]

B?

[glass lichen]

you know 2^0=1

so (0,1) needs to be on the graph

[rose fulcrum]

And 2^-1 = 1/2

The rest are fine

[junior hazel]

so instead of -2 i write 0.5?

[glass lichen]

yes

[junior hazel]

ty

[edgy cape]

In this question, I know how to solve for theta however I do not know how to express the answer as an inequality, theta= 30,150,222 and 318 as sintheta = 1/2 or -2/3

plz help

[oak chasm]

@edgy cape Are you allowed to use a graphing calculator?

If not, both sides are continuous except when tangent is undefined.

The solutions you got are where the sides are equal.

So on each side of a solution and on each side of a place where tangent is undefined, find out whether the left or right side of the inequality is higher.

That will tell you the regions where the inequality is true (the left side is higher).

[edgy cape]

I am

[oak chasm]

OK, subtract the right side from both sides.

You'll get something > 0.

Graph that something on the left.

See what regions it's over the x axis in.

You'll know the exact x axis crossing values because those are the solutions you got.

You'll know the infinite points because those are where tangent is undefined.

And you just say it's above the x axis from this known point (solution you got or tangent undefined point) to this other known point.

And then you make point1 < θ < point2 as one of the intervals that work.

[edgy cape]

So I graph: 6sin^2θ + sinθ -2 , and then say: 0<θ<30 , 30<θ<150, 150<θ<318 and 318<θ<360 ?

[oak chasm]

Well, 6 cos²(θ) - sin(θ) - 4.

And then you have those intervals, yes.

But which of those intervals is above the x axis?

Cross off the ones that aren't.

It's that > 0, so the y value has to be greater than zero, so that interval will be above the x axis.

[edgy cape]

i think they're all above the x axis

[oak chasm]

Oh, mistake.

There we go.

As you can see, it's above the x axis from 0 to 30, from 150 to 222, and from 318 to 360.

@edgy cape See how that's essentially equivalent to your yellow regions?

[edgy cape]

yea

[oak chasm]

OK, so it repeats forever.

[hoary shell]

so one solution would be 0 < theta < 30 ? and another 150 < theta < 222?

[oak chasm]

So, you get

-42° + 360n° < θ < 30° + 360n°
150° + 360n° < θ < 222° + 360n°

Where n is some integer.

[hoary shell]

I think it is asking only within 0 - 360 degrees

[oak chasm]

Oh!

Well, then:

0° < θ < 30°
150° < θ < 222°
318° < θ < 360°

Far left, middle, and far right rises above x axis.

[edgy cape]

Oh that makes sense

thanks

[oak chasm]

You're welcome.

[edgy cape]

but why does your graph look different

[hoary shell]

bc urs is two graphs

[oak chasm]

Mine does the subtraction.

[edgy cape]

oh right

[oak chasm]

Yours has the two sides graphed. Mine has the two sides subtracted graphed.

If you can get it < 0 or > 0 then you can look at where it goes below or above the x axis, which makes things easier to read.

And your solutions to the equality is when the subtraction will be zero (both sides are equal, so subtracting a number from an equal number is zero), so those will be the x axis crossings.

So, you know the x axis crossings and you know which crossings are the endpoints of what you want (interval above or below x axis).

And you can give exact answers.

[edgy cape]

i understand

thanks

[oak chasm]

No problem.

[wild olive]

@oak chasm question

what does the V looking symbol mean?

geeft means gives

[small bear]

I'm not them, but that is the logical operator or also called disjunction.
It means that the first equation gives a=0 or a=4 as solutions

[wild olive]

ooh ok ty

[edgy cape]

So I tried to solve (a) by first expressing the eqn as log2(x)=log2(kx-1)+log2(8) and then getting rid of logs but that leaves me with x^2-kx+7 which is wrong

where did I go wrong

[oak chasm]

@edgy cape

y log(x) = log(xʸ).

log(x) + log(y) = log(xy).

[edgy cape]

oof

yeah of course

thanks

[oak chasm]

No problem.

[quaint pine]

4x<2x+1<=3x+2
how do I get x on one side only?

[oak jewel]

hi @quaint pine

[quaint pine]

I tried subtracting but then I have to subtract on all sides and I'm stuck on a loop of never ending variables on all sides:(

[oak chasm]

@quaint pine When you have that, it means there are two inequalities that are both true:

4x < 2x + 1
2x + 1 ≤ 3x + 2

Now you can move the xs in each inequality to one side.

[quaint pine]

omg why didnt i think of that

[oak jewel]

looks like you got an answer, i was writing it down

[quaint pine]

thx both<3

[oak chasm]

No problem.

[oak jewel]

now that this one is open

f(x)=x and g(x)=sin(x)

how would you, with mean value theorem prove that

f(x)>g(x)

[oak chasm]

Well,

f(x) > g(x)
f(x) - g(x) > 0

[oak jewel]

i agree

[oak chasm]

What interval are you using?

[oak jewel]

from (0,->)

not 0

greater than 0

at 0 they are the same

are you still here chai?

[oak chasm]

@oak jewel Sorry, I'm out of ideas on how to use MVT for that.

[oak jewel]

same here

been trying to do it all day

[oak chasm]

I mean, x > 1 when x > 1. sin(x) ≤ 1 when x > 1, so that proves it on x ∈ (1, ∞).

Other than that, I don't see how the conditions of MVT can be met, since you need the endpoints x ∈ [a, b] to have f(a) = f(b).

[oak jewel]

well, you can just take any point

say 2

say 1

(0,1)

its pretty easy to prove otherwise though. if you take the definition of a triangle

its so that any angle will result in the hypo being greater than the two (english sucks) other sides

so thats proven...

[oak chasm]

The hypotenuse is always greater than either of the other two sides unless one angle of the triangle is 0.

[oak jewel]

yeah so proving it is not an issue

dead simple

but, mean value theorem

that was the task

i would think, that you could take the d/dx sinx

[oak chasm]

But I mean if x ∈ [a, b] is your interval, then f(a) = f(b) must be true to use MVT.

x ∈ [0, 1] means that 0 - sin(0) = 1 - sin(1) or you can't use MVT for that interval.

[oak jewel]

and its slope would be less than that of x

but you still have to resort to trigonometry to prove that

so, whats the point of doing it?

proofs suck

[oak chasm]

Well, the proof is easy without MVT.

[oak jewel]

yeah

it was a task given by the professor, not in calculus

i think he is making fake news tbh

[oak chasm]

In (1, ∞), you have 1 < x and sin(x) ≤ 1, so sin(x) < x.

And then the derivatives are higher on x than sin(x) for [0, 1].

[alpine sable]

Can someone help

[oak chasm]

@alpine sable Sorry, this channel is busy.

And the values are equal at x = 0.

[oak jewel]

yes but they arent = to 0 ever

[oak chasm]

No, I mean to prove without MVT.

[oak jewel]

this example made me super confused on mvt

i think i should solve some easier ones

[oak chasm]

You might ask on #calculus.

Maybe someone there will know.

[oak jewel]

will do, channel is open in the mean time

if you wanna help killua

[tardy plank]

confused on some algebra, where did the 4 next to t go ?

[oak chasm]

@tardy plank Factor out 4 from under the left square root.

sqrt(4(1 + t²))

Then take the 4 out of the square root.

2 sqrt(1 + t²)

[oak jewel]

sqrt*(4+4t)=sqrt(4(1+t))

[tardy plank]

ahh got it thanks all

[oak chasm]

No problem.

[pearl galleon]

i like your funny words magic man

[oak jewel]

?

nvm its just a bot

[errant wolf]

does anyone know how to do discrete math?

i need help with this

[oak chasm]

OK, so what's the contrapositive of if n² is a multiple of 3, then n is a multiple of 3?

[tardy plank]

does anyone know how this problem went from the 1st to the 2nd step

[oak chasm]

@tardy plank Sorry, this channel is busy.

[errant wolf]

is it like 3 is not a multiple of n^2?

[oak chasm]

No, it needs to be an if then statement.

a → b's contrapositive is ¬b → ¬ a.

So:

If n² is a multiple of 3, then n is a multiple of 3.

If n is not a multiple of 3, then n² is not a multiple of 3.

[errant wolf]

oh

[oak chasm]

See how the things change places and become their negation?

[errant wolf]

i get that

yea

[oak chasm]

So prove that contrapositive.

That will also prove the original statement since contrapositives are as true as each other.

[errant wolf]

dont know how tho

[oak chasm]

OK, do you know the form of a multiple of 3?

[errant wolf]

not really

[oak chasm]

3k is the form of a number that's a multiple of 3.

[errant wolf]

ok

[oak chasm]

3k + 1 and 3k + 2 are the forms of numbers that aren't a multiple of 3.

[errant wolf]

ok

[oak chasm]

So, we have if n is not a multiple of 3.

So, n = 3k + 1 or n = 3k + 2.

Does that make sense?

[errant wolf]

um kind of

[oak chasm]

Well, there are three kinds of numbers: 3k, 3k + 1, and 3k + 2.

The number you add to 3k is the remainder after you divide by 3.

Like 7.

7 is 3 · 2 + 1.

So, it's a number that has the 3k + 1 form.

If you divide 7 by 3, you get 2r1 (quotients and remainders from grade school).

[errant wolf]

yeah

[oak chasm]

So, 7 = 3 · 2 + 1.

Multiples of 3 are like 9.

9 divided by 3 is 3r0.

So, 9 = 3 · 3 + 0.

That's a number that's of the form 3k.

Does that make sense?

[errant wolf]

let me think about everything you just said for a second

[oak chasm]

OK.

[errant wolf]

how would i use 3k+1 to prove that 3 is not a multiple of n^2

[oak chasm]

Well, you have the contrapositive as if n is not a multiple of 3, then n² is not a multiple of 3.

So, we take the if as true.

n is not a multiple of 3.

So n = 3k + 1 or n = 3k + 2.

Then you find n² for each case.

n² = (3k + 1)² or n² = (3k + 2)².

[errant wolf]

ok

[oak chasm]

So expand each.

n² = 9k² + 6k + 1 or n² = 9k² + 12k + 4.

[errant wolf]

yes

[oak chasm]

Then, you need to get both in the form 3j + 1 or 3j + 2.

[alpine sable]

anyone know how to solve a cubic equation without calculator

[oak chasm]

@alpine sable Sorry, this channel is busy.

[alpine sable]

ok np

[oak chasm]

So, the first one has + 1.

[errant wolf]

yea

[oak chasm]

n² = 9k² + 6k + 1.

Factor out the 3 from the other parts.

[errant wolf]

ok

[oak chasm]

n² = 3(3k² + 2k) + 1.

So, n² is of the form 3j + 1.

Where j = 3k² + 2k.

[errant wolf]

yea

[oak chasm]

So, n² isn't a multiple of 3 in that case.

What about the other case?

We have + 4.

That's not + 0 or + 1 or + 2.

So, we get it down to that by separating out a multiple of 3.

• 4 = + 3 + 1

So,

n² = 9k² + 12k + 4
n² = 9k² + 12k + 3 + 1
n² = 3(3k² + 4k + 1) + 1

n² is of the form 3j + 1, which is not the form of a multiple of 3.

Does that make sense?

[errant wolf]

yeah

[oak chasm]

OK, so both cases where n isn't a multiple of 3 give n² isn't a multiple of 3.

So, if n isn't a multiple of 3, then n² isn't a multiple of 3 is true.

So, the contrapositive of the original statement is true.

So, the original statement is true.

Does that make sense?

[errant wolf]

yes

so does the n² = 3(3k² + 2k) + 1 prove that it is not a multiple of 3?

[oak chasm]

It proves that if n = 3k + 1, then n² = 3j + 1.

Any multiple of 3 plus 1 is not a multiple of 3.

3 + 1 = 4, not a multiple of 3.
6 + 1 = 7, not a multiple of 3.
And so on.

[winter rock]

@oak chasm wassap

[oak chasm]

To get from one multiple of 3 to another, you have to add a multiple of 3.

[winter rock]

Could u help me with one small small thing when u finished w that guy

[errant wolf]

you can help them now ill be back

[oak chasm]

@winter rock You can post your question since the channel is now open.

[winter rock]

O nice

Here,

Im trying to understood

How he went from AM×AC×COS O - AM²

to

AB²-AM²

@oak chasm

[oak chasm]

Well, by SOHCAHTOA, cos(θ) = (L sqrt(2)/2)/|AM|, right?

@winter rock

So, |AM| cos(θ) = L sqrt(2)/2.

And |AC| is L sqrt(2).

So, |AM| cos(θ) |AC| = L sqrt(2)/2 · sqrt(2) L.

Which is L².

And then I assume |AB| is length L.

[winter rock]

Yeah

[oak chasm]

So, L² = |AB|².

[winter rock]

Ok thx

[oak chasm]

No problem.

[winter rock]

Just a quickly question

Sorry but yh

Huh

Heu

Yeah

[rare lion]

hello

[winter rock]

In the square

[rare lion]

Is this sub occupied?

[winter rock]

They weren't any length mentioned

[oak chasm]

@rare lion Yes.

[winter rock]

The person who sent me the screen

He added the L by himself

[rare lion]

Thank you.

[winter rock]

And I don't know how he found out sqrt 2 too

[oak chasm]

Oh, OK. I assume B is either the top right or bottom left corner.

[winter rock]

The top right yup

[oak chasm]

Oh, a square has a diagonal that's sqrt(2) times its side length.

[winter rock]

Ow yeahhhh ure right

[hoary shell]

if u set AB=1 AC=sqrt(2)

[winter rock]

So

Ac= Lsqrt2/2

[hoary shell]

AC is Lsqrt2

[oak chasm]

Right, AC is the diagonal, and the side of the square is length L, so the diagonal is sqrt(2) L, so |AC| is sqrt(2) L.

[hoary shell]

then AO, the line from corner to origin, is Lsqrt(2)/2

[oak chasm]

And A to the midpoint of the square is half the diagonal or sqrt(2) L/2.

[winter rock]

Thanks everyone! Yall are really amazing in this serv, I have to say Im getting really good grades because of yall

[oak chasm]

You're welcome.

[upbeat heath]

what is -7a -10 = 2- -3a, I got -6/5 but it's not appearing as the answer to the problem

[oak chasm]

Can you show the problem?

[upbeat heath]

yeah

[winter rock]

2- -3a?@upbeat heath

[hoary shell]

2 - (-3)

[winter rock]

Ow okay

[upbeat heath]

I feel like my teacher is on steroids

who puts two negatives together

;-;

[winter rock]

He forgor

[hoary shell]

hey well he fooled you didnt he XD

[upbeat heath]

yeah

[oak chasm]

OK, so can you show the answers?

[upbeat heath]

yeah

[oak chasm]

I get -6/5 as well.

[errant wolf]

is a just -3 or something?

[hoary shell]

well I think this is the time you get to tell the teacher they are wrong XD

[upbeat heath]

lmaooo

I'll just skip it and talk to her tmrw ig

[hoary shell]

if its just -3a, then the answer -3

[oak chasm]

Yeah, that would be my guess.

[hoary shell]

otherwise its -6/5

[upbeat heath]

alright

[oak chasm]

Bring it up with the instructor.

[upbeat heath]

I will thank you

[oak chasm]

No problem.

[errant wolf]

can we go back to my problem?

[oak chasm]

Sure.

[upbeat heath]

yes

[errant wolf]

ok so how does n² = 3(3k² + 2k) + 1 prove that the original statement is true

[oak chasm]

Well, you have n² isn't a multiple of 3 in that case.

[errant wolf]

ok

[oak chasm]

So, if n isn't a multiple of 3, then n² isn't a multiple of 3 is true for n = 3k + 1.

It's also true for n = 3k + 2.

So, it's true for all cases where n isn't a multiple of 3.

So, the contrapositive is always true.

So, the original statement is always true.

Does that make sense?

[errant wolf]

so that proves that if n^2 is a multiple of 3 then n must also be a multiple of 3?

[oak chasm]

Yes, because we proved the contrapositive of that.

A statement and its contrapositive are either both true or both false.

[errant wolf]

ok

[oak chasm]

They have the same truth or falsehood.

The contrapositive is true, so the original must be true.

That's how you prove by contrapositive.

[errant wolf]

what about contradiction

[oak chasm]

OK, so contradiction means to assume the statement is false.

If n² is a multiple of 3, then n is a multiple of 3.

3 | n² → 3 | n

a | b means b is a multiple of a.

[errant wolf]

ok

[oak chasm]

It's said as "a divides b" or "b is a multiple of a".

So, now we convert → to ∨

3 | n² → 3 | n
¬(3 | n²) ∨ (3 | n)

So, n² isn't a multiple of 3 or n is a multiple of 3.

[errant wolf]

yeah

[oak chasm]

We need to show that that's never true.

For any n.

If we can find an n where that's true, then the negation is sometimes true.

So, the original statement is sometimes false.

Does that make sense?

Oh, wait.

That's wrong.

[errant wolf]

oh

[oak chasm]

This is equivalent to the original statement.

Forgot to do the negation.

A → B is equivalent to ¬A ∨ B.

¬(A → B) is equivalent to ¬(¬ A ∨ B).

And then by DeMorgan's laws, we have ¬(¬ A ∨ B) is equivalent to A ∧ ¬B.

So that's the negation.

[errant wolf]

ok

does that prove the original statement?

[oak chasm]

n² is a multiple of 3 and n is not a multiple of 3.

No, this is the negation of the original statement.

[errant wolf]

ok

[oak chasm]

We need to show that there are no ns that ever make n² a multiple of 3 and n not a multiple of 3.

If we can find one, then the negation is sometimes true.

Which means that the original statement is sometimes false.

That wouldn't be good, since we're trying to prove it, not show that it has some holes in it.

[errant wolf]

then how would we prove it

[oak chasm]

OK, we need to show that n² is a multiple of 3 and n is not a multiple of 3 can never both be true for any n.

We do that by finding a contradiction.

First, we start with n² is a multiple of 3 and we get to some statement.

Then we start with n is not a multiple of 3 and get to the negation of that statement.

That's a contradiction.

A ∧ ¬ A is a contradiction.

A and its negation can't both be true.

Does that make sense?

[errant wolf]

yes

[oak chasm]

OK, so we already showed that n is not a multiple of 3 leads to n² is not a multiple of 3 in the contrapositive proof, right?

[errant wolf]

yes

[oak chasm]

So, we have:

n² is a multiple of 3 and n is not a multiple of 3.
n² is a multiple of 3 and n² is not a multiple of 3.

[errant wolf]

ok

[plush gale]

how would i prove that angles 1 and 3 are = to 180

and how would i find out if angles 1 and 2 and =

[oak chasm]

@plush gale Sorry, this channel is busy.

@errant wolf So we have that n² is a multiple of 3 and n² isn't a multiple of 3.

That can't be true.

It's a contradiction.

[errant wolf]

yes

[oak chasm]

So, if the negation of the original statement leads to a contradiction, the original statement is true.

That's how proof by contradiction works.

[errant wolf]

ok

[oak chasm]

Contradictions are always false.

The negation is false.

And the only way to negate something and get false is to negate true.

[errant wolf]

ok

how would i write this on paper to prove the original statement

do i just explain everything thats happening?

[oak chasm]

Yes.

Proof by contradiction means that if the negation of a statement leads to a contradiction, then the statement is true.

The negation of the statement is ....

This leads to ..., which is a contradiction.

So, the statement is true.

[errant wolf]

yeah

[alpine sable]

anyone have experience wit binary that can join chat and help

me @ me

[oak chasm]

@alpine sable What is your question?

[alpine sable]

realted to binary

odd even

stuff

and checksum

[oak chasm]

Right, but what is the actual question?

[alpine sable]

this is the qeustion

[rocky cape]

\

Hey does anyone know a tutorial to solve this ?

[oak chasm]

@rocky cape Sorry, this channel is busy.

@alpine sable Well, the parity bit works by making the number of 1 bits even.

Both received bytes have an even number of 1 bits.

So the parity says they're OK.

Does that make sense?

[alpine sable]

so if the number of 1 are even in the received what does that signify

exactly

that they are okay?

there are no erros?

[oak chasm]

That means there might not be an error.

[alpine sable]

right

[oak chasm]

If it's odd, there's definitely an error.

[alpine sable]

ok

yes

so would the justification for why

there wouldnt be an error detected

that because the received parity bits have an even number of 1 bits, therefore no error

[oak chasm]

Well, all the 1 bits in a byte are an even number.

There's only one parity bit per byte.

The last bit in each byte according to your question.

If the parity bit had to be an even number of 1s, it would have to be a 0.

Instead, parity works by setting the parity bit so that the total count of 1 bits in a byte is even.

If you have 1011100 as the first 7 bits.

[rocky cape]

Alright.

[oak chasm]

That has an even number of 1s already.

So, you don't want to change the number of 1 bits.

10111000

So the parity bit is zero.

1001010 has an odd number of 1s.

So, we want to change it to even.

10010101

So we set the parity bit to 1.

Does that make sense?

The first seven bits are the data.

The last bit is a bit that's not data.

It's just set so that the number of 1 bits ends up as even.

And then if the receiver gets an odd number of 1 bits, it knows something went wrong.

Otherwise, it can't tell if it's good or not.

[alpine sable]

is this always the case

[oak chasm]

In this setup, yes.

You can choose to make the count of 1s odd in a scheme.

Or choose a different bit index as the parity bit.

All of those are parity bit schemes.

[alpine sable]

so how would i structure a response to this

since its 2 marks

[oak chasm]

The reason no error was detected was because the numbers of 1 bits in both bytes were even, so an error wasn't detected.

I'm not sure which checksum they're using.

[alpine sable]

however if theree were an odd number of 1 bits in either

would that mean there could be an error

[oak chasm]

No, that would mean that there definitely was an error.

Parity bit schemes have two things they can know:

· there is definitely an error
· there is not definitely an error

If there is definitely an error, we know an error happened.

If there is not definitely an error, maybe there's an error, maybe not.

That's the only thing it can tell us: one of those responses.

[runic timber]

hey

why is -4--1=3 but -1--4=3?

https://cdn.discordapp.com/emojis/747606584104058880.gif?v=1

[oak chasm]

@runic timber Sorry, this channel is busy.

[runic timber]

ok

[thick flame]

Go to #help-2

Make that 3

[alpine sable]

ok just to make sure i got it

What I have understood is that, If the recieved bytes have an even number of 1 bits that means that there is not an error detected, meaning that it is not corrupted data

and if the recived bytes have an odd number of 1 bits that means there is an error

[oak chasm]

Not in math means all other cases, not opposite.

Opposite is the other extreme. The opposite of there definitely being an error is there definitely being no error.

Negation is all other cases. There are three cases: definitely an error, we're not sure, definitely no error.

The negation of definitely an error is that one of the other cases left over must be true. So, either we're not sure or there's definitely no error.

[hoary shell]

so the only way u could have an error and it not be detected is if there was an even number of errors?

[oak chasm]

You can have an odd number of transposition errors.

One transposition error will swap two bits, leaving the number of 1 bits unchanged.

[hoary shell]

oh nice

[oak chasm]

@alpine sable Does this make sense?

[alpine sable]

ok

yea

[alpine sable]

is this right

[oak chasm]

No.

You did opposite instead of negation.

Here's how it works.

You have two situations:

Even number of 1 bits
Odd number of 1 bits

[radiant barn]

Hello can i ask something about arithmetic

[oak chasm]

There are three cases:

Definitely an error
Not sure
Definitely no error

@radiant barn Sorry, this channel is busy.

[hoary shell]

@radiant barn maybe i could help u in different channel

[oak chasm]

So, we have to handle all three cases with those two situations.

We can put all three cases on one situation.

We can put two cases on one situation and the other case on the other.

Three cases on one situation would be useless.

If we got that situation, we wouldn't know anything about the error status.

[oak chasm]

@alpine sable Sorry, this channel is busy.

[alpine sable]

Can someone help me with this

[radiant barn]

In what channel

[surreal sinew]

go to a diff channel

[alpine sable]

@oak chasm

all good

i think i got it

thanks'

[oak chasm]

No problem.

[errant wolf]

@oak chasm

[hoary shell]

XD

[oak chasm]

@errant wolf OK, even and odd are a multiple of 2 and something that's not a multiple of 2.

[errant wolf]

since n is even would i just do 2k^2

[oak chasm]

So, the forms are 2k and 2k + 1.

Yes, but (2k)².

[errant wolf]

yea

after i get (2k)² what do i do after that

[oak chasm]

Expand.

[errant wolf]

so 4k

[oak chasm]

Almost.

[errant wolf]

4k^2

[oak chasm]

Right.

Now factor out the 4, since they want the form 4s.

4(k²)

[errant wolf]

ok

[oak chasm]

Now it's of the form 4s.

[errant wolf]

thats it?

[oak chasm]

Yes, you can put s = k² if you want to be more detailed.

[errant wolf]

ok thanks

[oak chasm]

No problem.

[errant wolf]

im going to fail this class

[hoary shell]

what class is it?

[errant wolf]

discrete math

[oak chasm]

Discrete math is hard.

[hoary shell]

I like these problems

I guess I have not reached discrete math yet

[oak chasm]

It's usually done for computer science.

If you're doing a math degree, you'll see stuff like that in number theory.

[hoary shell]

yeah thats my major

[errant wolf]

how do you know so much math @oak chasm

[oak chasm]

I just know the lower level stuff.

I did part of a computer science degree.

And took a few extra math classes, like abstract algebra and linear programming.

@dry hare What's the area of a circle formula?

Right, but you need to know the area of a circle formula to do it.

Do you know it?

A = πr²

Area of a circle is pi times r squared.

r is radius of the circle.

So,

A = πr²
A = π(3 cm)²

See how I filled in r with the radius?

You're welcome.

[edgy sand]

anybody help me with this? ik im supposed to use the formula but im lost on calculatoions

[oak chasm]

What's the definition of a derivative?

@edgy sand

[astral dagger]

[edgy sand]

you dont know or you are just asking me? @oak chasm

[oak chasm]

Asking you since it's the first step.

[edgy sand]

i got 9+15t for the first part

[oak chasm]

OK, that's right.

[edgy sand]

alright

now the domains

[oak chasm]

Are there any t values that make the expressions undefined?

Like can you choose a t that makes 7.5t² + 9t undefined?

[edgy sand]

0?

[oak chasm]

OK, let's try it.

[edgy sand]

if not then i dont think so

[oak chasm]

7.5(0)² + 9(0)

What's that come out to?

[edgy sand]

0

[oak chasm]

Right, which is defined.

(as 0)

[edgy sand]

makes sense

so it would be all real numbers then?

[oak chasm]

Yes.

All real numbers will give you a real number after they're filled in.

[edgy sand]

but for botht the derivate and the function?

[oak chasm]

Yes.

Polynomials have the reals or the complexes or whatever as the domain.

And these are polynomials.

[edgy sand]

makes sense

so its asking for interval notation though

[oak chasm]

Well, what's the left bound?

[edgy sand]

-inf < x < inf

?

[oak chasm]

Yes, and the interval would be (-∞, ∞) if you write it that way.

Not sure which way they want you to write it.

[edgy sand]

ohh yea i guess you could do it both ways

interval notation

is the first way or second way?

[oak chasm]

The one with ( or [ on the left, a number, a comma, a number and ) or ] on the right.

There's an alternative for some countries, where ( is replaced with ] and ) is replaced with [.

So, ]-∞, ∞[.

[edgy sand]

ive only seen ( so im assuming its that haha

would you know how to do this one as well?

[oak chasm]

Well, is the function continuous at u = -1?

[edgy sand]

i think it is

because its not over 0

[oak chasm]

Well, square roots can be undefined with negative inputs.

Depending on how you define square root.

[edgy sand]

hm

i mean im assuming i need to get rid of the radical first

[oak chasm]

No.

Figure out whether the function is continuous at u = -1.

Polynomials are continuous everywhere.

[edgy sand]

by graphing?

[oak chasm]

And square root is continuous when the input is continuous and positive.

So, the polynomial under the square root is continuous.

Is the input to the square root positive?

[edgy sand]

1 s

[oak chasm]

Sorry?

[edgy sand]

calculating

[oak chasm]

Oh, OK.

[edgy sand]

its pos

[oak chasm]

OK, so the function is continuous at u = -1.

[edgy sand]

so sqrt 1

[oak chasm]

So, the limit of the function as u → -1 is f(-1).

Whenever your limit is going towards an input where the function is continuous, you can just fill in the input to the function, and that's the value of the limit.

That will save some time.

But with the limit laws.

They want you to use limit laws.

So, use the root law.

@granite yacht Sorry, this channel is busy.

If you're taking the nth root and n is a positive integer, you can use the root law.

[edgy sand]

for some reason, im getting 1 when calulcating

could you double check to see if its -1 and im messing up?

[oak chasm]

How can square root give a negative number?

[edgy sand]

you said its -1?

or just typo

[oak chasm]

Yes, u is.

The input to the function is -1.

[edgy sand]

right

and the limit is 1?

correct?

[oak chasm]

That doesn't mean the output is -1.

Yes.

[edgy sand]

oh okay now i get it

[oak chasm]

OK, now we do it with the limit laws like they say.

That's a shortcut to skip doing a lot of work getting the limit.

So, you have a square root on the outside of the function.

A square root is the 2nd root and 2 is a positive integer, so you can use the root law.

[ocean seal]

Chai T. Rex

[oak chasm]

Do you see how the limit moved under the root, so the part we have left to take the limit of is simpler?

[edgy sand]

yep

[oak chasm]

Then you use the sum law.