help-0

[shrewd otter]

Me neither.

Look at the wikipedia page for trigonometric functions.

You might find a formula or something.

[robust phoenix]

I watched the video it didn't involve any formula, he only said that cos and sin should be negative hence b is the ideal answer

[shrewd otter]

Aha.

I see.

You have to find which answer is always negative for m > 0.

Since square root is always positive, you have to find which numerator is always negative.

Since m > 0, 1 + 2m > 0, so A can't be it.

[robust phoenix]

why did you put the 1+2m

[shrewd otter]

That's the numerator in A.

[robust phoenix]

oh yeah you are right

[shrewd otter]

For C, if you choose m < 1/2, then 1 - 2m > 0.

So it can't be C either.

Similarly for D.

So by elimination, it must be B.

[true anchor]

how do i factor this with only positive exponents?

[arctic mango]

Can you see why we can write it as (20 - 45x²)/(x⁹)?

[true anchor]

hmm

why is the exponents positive though?

like i see can the other factors

[arctic mango]

x^(-a) = 1/(x^a)

That's the basic idea.

[true anchor]

ohh

like the flipping

[arctic mango]

Yes, that's how you define exponentiation for negative numbers.

[true anchor]

but dont you find the gcf of the 45 and 20 though?

[arctic mango]

And it makes sense, doesn't it? You want x^a * x^(-a) = x^(a-a) = x^0

That's the most important rule you need to know regarding exponents, actually!

x^a * x^b = x^(a+b). Most results follow from just this one rule.

[arctic mango]

Oh yeah, you can do that as well.

[arctic mango]

Just factorise this even more. Now it's straightforward, right?

[true anchor]

yes

[arctic mango]

Leave x⁹ in the denomination as it is.

[arctic mango]

This is all you need to know regarding exponents!

[true anchor]

the assignment said to factor it completely

[arctic mango]

Yes, then you must do it. Take out gcd(45,20).

[true anchor]

5

[arctic mango]

Yes. Outside the bracket it goes, then.

[arctic mango]

An interesting exercise regarding this. What is x²? It's basically x^2 = x^1 * x^1 = x * x. This rule is very powerful!

[true anchor]

i got 5x^-7(4x^-9 9)

hmm

its doesnt seem correct to me

[arctic mango]

Factor this

[true anchor]

oh ok

[arctic mango]

First write it in this form using the negative exponent rule.

[true anchor]

i got 5x^-9(4x^-7 9)

[arctic mango]

Can you send a photo of your work instead?

I think that'll give a better idea

[true anchor]

oh kk

[arctic mango]

First things first, bring the x^(-9) and x^(-7) down in the denominator using the negative exponent rule.

In the denominator, they will have positive exponents, and that is what you need

[arctic mango]

Wait, what happened to the minus sign?

[true anchor]

ohh oops

[arctic mango]

You don't need to deal with negative exponents, just bring them down right away.

[arctic mango]

No, this isn't right. The first term will be 20 x^(-16) in this case

Negative exponents adding up

[true anchor]

but isnt the 20 going to be factored?

[arctic mango]

See, first you do this:
20 x^(-9) - 45 x^(-7)
= (20/x⁹) - (45/x⁷)

[true anchor]

yes but what about the 5?

[arctic mango]

Later

At the end

First factor out (1/x⁹)

[true anchor]

oh ok

[arctic mango]

Rewrite it as
20/(x⁹) - (45 x²)/(x⁹)

Now you can factor out the x⁹ in the denominator

[true anchor]

oh

that puts a double fraction in the numerator

[arctic mango]

I think you might be have trouble making sense of it since I'm typing it

Give me a minute

[true anchor]

ok

[arctic mango]

Basic idea is that negative exponents in the numerator become positive exponents in the denominator

You get that, right?

[true anchor]

yes

[arctic mango]

Okay, I'll write it down

[true anchor]

kk

[arctic mango]

Make sure you realise how I got to the third step from the second. I multiplied (45/x⁷) with x² in the numerator as well as the denominator

[true anchor]

in the second step why is two in numoartor and 9 in denomintor

when it was first as 7?

[arctic mango]

I multiplied it by x² in the numerator and denominator

Numerator becomes 45 * x² = 45x²
Denominator becomes x⁷ * x² = x⁹

[true anchor]

okay

[arctic mango]

You can always multiply stuff by the same thing in both the numerator and denominator right?

[true anchor]

yes

[arctic mango]

Because it's the same as multiplying by x²/x² = 1

[true anchor]

but why is this step nessary?

[arctic mango]

Think about it

What does have x⁹ in the denominator accomplish?

The answer lies in the next step itself

[true anchor]

hmm it becomes a fraction

[arctic mango]

Both the terms have the same denominator

Like when you do 1/3 - 1/5
You turn it into (5/15) - (3/15)

[true anchor]

ohhh

[arctic mango]

Exactly the same idea

[true anchor]

so then you can subtract

[arctic mango]

What do you do before that?

[alpine sable]

Guys

[arctic mango]

You factor out (1/15)

Then you subtract without any trouble

That's actually what you are doing. Here you have variables so you can't actually subtract

[true anchor]

you put the numberators in the pertencases

[arctic mango]

So you just leave it like that

[true anchor]

oh

[arctic mango]

Like (4 - 9x²) in the end, right?

Did you get the steps till now?

[arctic mango]

Taken

[true anchor]

yes

[arctic mango]

Now there's one final step

You know a² - b² = (a+b)(a-b)?

[true anchor]

yeah

[arctic mango]

Use it

Write (4 - 9x)² as [2² - (3x)²]

Now you have a² - b²

[true anchor]

ohh

[arctic mango]

Yeah! Even I saw it just now 😂

You said simplest form

So we have to keep going

[true anchor]

really wow

[arctic mango]

Yeah, this is good enough now

[arctic mango]

Isn't it kinda amazing how much it factorises?

The reason is because someone has designed the question to be like that

Usually it won't be so "friendly"

At the end of the day, whenever you have A - B somewhere, always check if A and B exist as squares of some simple thing

Maybe you can factorise it into (a+b)(a-b) then

[true anchor]

thank you

[sweet kelp]

HELP

indicate the GCF of the polynomials
M(x) = x² - 25
N(x) = x² + 7x + 10

[arctic mango]

You're welcome!

[ionic jewel]

try factoring both of them

then find common factors

then combine all the mutual ones for the greatest one

[arctic mango]

One using a² - b², another using middle term splitting

[lofty abyss]

When the polynomial 3x^3+ax^2+bx-9 is divided by x - 2, the remainder is 29. When it is divided by x + 1, the remainder is -7. Determine the values of a and b.

[light sparrow]

"When the polynomial 3x^3+ax^2+bx-9 is divided by x - 2, the remainder is 29" this is the same as saying 3x^3 + ax^2 + bx-9 = (x-2)p(x)+29 for some quadratic p(x)

so if we sub in x=2, then you get 24 + 4a +2b -9 = 29, so thats one equation

can you do the same trick for the other division?

this trick is called the remainder theorem if you know that

[lofty abyss]

so you mean it's the same as sub in x=-1

after we find out two equation, simplify as simple as we can?

[light sparrow]

yeah so you'd get -3+a-b-9=-7

[light sparrow]

and then you got a pair of simultaneous equations for a and b which you can solve

[lofty abyss]

ok

how bout this one? I don't know how to do it
Solve the inequality (x^2-x-6)(2x-1) ≤ 0. State your answer using interval notation

[light sparrow]

usually you'd want to sketch a graph of the function

[lofty abyss]

what should I do after that

[light sparrow]

so the inequality corresponds to the values of x for which the graph is negative?

[lofty abyss]

yes

[round magnet]

Hello!
I wanted to ask if anyone has any material on sensor fusion with a Kalman filter

[silk coral]

Can someone explain trig to me

Cosine sine and tangent

I’m so lost on this for some reason

[sleek elbow]

did you try khanacademy? @silk coral

[keen owl]

can you post links here?

[sleek elbow]

if they are about the conversation I think yes

[keen owl]

https://www.youtube.com/watch?v=llA6iW043XU
some random video about it, should be alright I guess.
Just a little advice: Try figuring it out yourself first. That way you learn how to research and so on.
Also better because you can ask specific questions

[alpine sable]

Hello Luxury! Trigonometry deals with things on a triangle that remain unchanged when you shrink or enlarge the triangle. One obvious thing that is unchanged is the angles of the triangle. Try drawing a triangle and enlarge it and shrink it; you will see that the angles don't change. Another thing that is unchanged is ratios of the triangle sides; this sides themselves become larger or smaller of course, but their ratios do not. So there should be some connection between the unchanging angles and the unchanging ratios. These connections are sine, cosine and tangens.

[silk coral]

Ok thank you

How would you solve a problem like this

[keen owl]

You need to figure out which side is the hypotenuse, the adjacent and the opposite of the angle at first

[silk coral]

Hypotenuse is y adjacent is 75 and opposite is x

Of 50°

I just don’t know what to do past that

[keen owl]

Yeah, you have to remember some properties. Wait a second

[idle sapphire]

is this correct?

[alpine sable]

The cosine of the angle $50^\circ$ is equal to the ratio $75/y$. From a calculator you can obtain that $\cos 50^\circ$ is approximately $0.64$, so the ratio $75/y$ should be approximately $0.64$ too. This information gives you the length ($y$) of the hypotenuse.

[ocean seal]

AndersM

[silk coral]

Ok so that would be around 117 for hypotenuse

How do we get the opposite

[alpine sable]

The tangens of $50^\circ$ is equal to the ratio $x/75$, and a calculator tells you that $\tan 50^\circ$ is approximately equal to $1.19$; the ratio $x/75$ should be approximately equal to $1.19$ too, which allows you to compute the side length $x$.

[ocean seal]

AndersM

[silk coral]

Ok it’s starting to make sense

Thank you

What about this type

Ignore the 22/50

And ik the answer is there but I’m not sure how to solve

[glass lichen]

what about it?

in use.

[silk coral]

How do you solve it

[glass lichen]

well write out the ratio, and since you're solving for an angle you need inverse trig

[lilac pollen]

good moring

[silk coral]

How do you find the ratio

Is it inverse sine (22/50)

[alpine sable]

In this picture you know the ratio $22/50=0.44$. This ratio is also equal to the sine of angle $v$, so you can write equation $\sin v = 22/50.$ A calculator tells you that $\sin 26^\circ$ has a sine-value approximately equal to $0.44$, so the angle $v$ should be approximately equal to $26^\circ$.

[ocean seal]

AndersM

[silk coral]

Makes sense

[keen owl]

like just use what I wrote if you can read it

It's always the same method, find the sides (hypotenuse, adjacent, opposite) and use the relations that make sense, sine, cosine or tangens

[silk coral]

Ok

I’ll use that

[orchid dock]

I am really struggling with a question I tried to solve earlier...

Question: In a basketball league play a total of 10 teams. They play a first and a second round.
"How many games are played in the league in a season?"

---

possible solutions are
A: 45
B: 65
C: 80
D: 90

[whole prairie]

Is this channel in use? I have a quick question

[woeful pulsar]

is it two matches per matchup in a round robin tournament?>

[orchid dock]

I literally just have the information given there

I guess every match= 2 rounds?

or every matchup = 2 games

if thats what you mean

[woeful pulsar]

I think my interpretation seems reasonable @orchid dock how about try calculating that

[orchid dock]

I mean what is your interpretation?

I have no idea how to solve that problem as I dont know how im supposed to know how to solve a tournament related calculation without knowing how the tournament works

[woeful pulsar]

mine is just interpret as twice the number of matches as round robin

@orchid dock

[orchid dock]

so 20 is the solution?

or am i missing something

[woeful pulsar]

do you know how to count the number of matches in a round robin tournament?

[orchid dock]

no I have no idea what a robin tournament is, and to be honest, I cant imagine them to expect people to know this 👀

[woeful pulsar]

https://en.wikipedia.org/wiki/Round-robin_tournament

[orchid dock]

so do you think its impossible to solve without knowing what a robin tournament is?

[woeful pulsar]

it's kinda standard terminology

where every team plays every other team

there's also single elimination and double elimination and all sorts of formats

[orchid dock]

oh wow

[orchid dock]

so you say that with 2 rounds its N-2 instead of N-1 ?

[woeful pulsar]

when we multiply by 2, we multiply the entire expression by 2, not the -1 by 2

[orchid dock]

divide or multiply ?

[woeful pulsar]

there are two games for each pairing

[old swan]

Find the unknown angle measure.

40°
220°
180°
50°
0 / 25

I have no clues how to do this

[glass lichen]

angle sum of a triangle is...?

[old swan]

idk

[ionic jewel]

that's like the first rule of triangles

almost as important as the triangle rule

[ionic jewel]

interior angles of a triangle add up to 180

the square in one corner means "right angle" or 90°

[old swan]

ik that but i also need an explanation

[dark granite]

@orchid dock It's easier to see that $\binom{N}{2}=\frac{N(N-1)}{2}$ for any integer $N\geq2$.

[ionic jewel]

then why did you say you didn't know

[ocean seal]

logician_pdx

[ionic jewel]

and what do you need an explanation for

[woeful pulsar]

observe

[old swan]

how to get the anser because the assignment also had quadrilaterals

A quadrilateral has three angles that measure 74°, 84°, and 100°. Find the unknown angle measure.

thats one of the other questions

[ionic jewel]

okay so let's go through this one

[old swan]

k

[ionic jewel]

a triangle has interior angles that add up to 180

you know two of them

what are they?

[old swan]

50 and 90

[ionic jewel]

great so what's the last one to make it add up right?

[old swan]

thatll take me a second one sec

40

[ionic jewel]

yep

good job

[old swan]

thanks by the way I had a brain fart and completely forgot how to do this

[nimble meteor]

hey could someone pls help

not jk

[dark granite]

Sure I can help with this @nimble meteor

[nimble meteor]

lol

[nimble meteor]

thanks!

[dark granite]

You're welcome

kk so

We want to prove that

[ionic jewel]

ain't this just the guass sum but 6 you pull a 6 out first

i feel like nikh has done an almost identical problem iirc

[dark granite]

$\sum_{i=1}^n6i=3n(n+1)$ for every integer $n>0$ @nimble meteor

[ocean seal]

logician_pdx

[dark granite]

do you agree @nimble meteor ?

and I got this @ionic jewel

[nimble meteor]

nope have not

[nimble meteor]

yeah dyou mind seeing what I have so far?

I don't get what to do from here

[dark granite]

um so first off

we shouldn't assume for $n=k+1$, we should prove for $n=k+1$

[ocean seal]

logician_pdx

[dark granite]

You've also skipped a lot of steps

[nimble meteor]

ooh right

[dark granite]

and in proofs, we need to show our work

For instance, for the base case really you should write $6(1)=6=3(2)=3(1)(1+1)$

[ocean seal]

logician_pdx

[nimble meteor]

oh ok

[dark granite]

And then you assume that equation holds for some natural number $k$.

[ocean seal]

logician_pdx

[dark granite]

So then you look at $\sum_{i=1}^{k+1}6i$

[ocean seal]

logician_pdx

[dark granite]

and see that $\sum_{i=1}^{k+1}6i=\sum_{i=1}^{k}6i+6(k+1)$

[ocean seal]

logician_pdx

[nimble meteor]

ok I'm lost just about now

lol

also I was told to just do it with terms vs. sigma

[dark granite]

okay, for some reason that didn't typeset right

okay,

so for the first k terms we know what the formula is

so then for the first k+1 terms, we have this:

3k(k+1)+6(k+1), right?

[ocean seal]

Couldn't find a role matching ght?!

Jul 13, 2021, 11:43 PM

[nimble meteor]

on the right handside?

[dark granite]

yes

because we knew what the formula was for the first k terms

[nimble meteor]

ohhh

yeah I see what you mean now

[dark granite]

kk cool

so we currently have 3k(k+1)+6(k+1)

so then we can manipulate this

and get 3k(k+1)+6(k+1)=3k(k+1)+(2)(3)(k+1)=3(k+1)(k+2), which is what we needed

[nimble meteor]

and do we have this on the right hand side?

[dark granite]

no!

[nimble meteor]

oh

[dark granite]

look at this^

so we just showed that the sum of the first k+1 terms is 6(1)+6(2)+...+6(k+1)=3k(k+1)+6(k+1)=3k(k+1)+(2)(3)(k+1)=3(k+1)(k+2)

make sense @nimble meteor ?

[nimble meteor]

uh

where did you get the substitution tho

[dark granite]

from our assumption

we know what the sum of the first k terms is (by assumption)

[nimble meteor]

so like I substituted it and you told me what I was doing wrong...I forgot to substitute based on the assumption.

but like um I'm not sure what the right is supposed to look like?

[dark granite]

well you substituted it wrong

[dark granite]

look at this^

Why did you replace 3k(k+1) with 3(k+1)(k+1+1)? Those aren't equal!

Once you get 3k(k+1)+6(k+1), it's just algebra from there

[nimble meteor]

well like the original right hand side was 3n(n+1) right?

so I just substituted n for k+1

and got 3(k+1)(k+1+1)

and then I have to add the 6(k+1) on both sides of the eq

[dark granite]

look at this equation. This equation doesn't hold

you shouldn't replace variables with other things that aren't equal! You replaced k with k+1, midway through the equation...we can't do that since k doesn't equal k+1

This is what the sum of the first k+1 terms looks like:

$6(1)+6(2)+6(3)+\dots+6k+6(k+1)=3k(k+1)+6(k+1)=3k(k+1)+3(2)(k+1)=3(k+1)(k+2)$

[ocean seal]

logician_pdx

[late parcel]

Can I post my math question as a picture here?

[alpine sable]

ofc

[dark granite]

$6(1)+6(2)+6(3)+\dots+6k=3k(k+1)$

[ocean seal]

logician_pdx

[late parcel]

https://cdn.discordapp.com/attachments/862407974236061727/864653610046849044/image0.jpg

[dark granite]

This channel is busy

[dark granite]

This channel is busy

[alpine sable]

oop

[late parcel]

#2 and 3 are not making sense to me

[nimble meteor]

bruh

[dark granite]

Channel is still busy already lol

[nimble meteor]

in the middle of a problem lol

[late parcel]

my appoligize, should i come back later or go to a different channel

[nimble meteor]

no worries

either works

just make sure the channel you ask in isn't occupied first

ok master logician back to our discussion?

[dark granite]

@nimble meteor did my equation make sense?

[nimble meteor]

well you trying to help me not be retarded I mean

lol

[dark granite]

this one^

@nimble meteor

[nimble meteor]

well sort of

like where did the 2 go?

shouldn't it be (k+1)(3k+6)

cause you factored out the like k+1

right?

sorry honestly you should prob just give up

I'm a lost case lol

[dark granite]

the 2 came from noticing 6=3(2)

[nimble meteor]

oh ok I see

yeah

[dark granite]

yes now factor out a 3

from 3k+6

[nimble meteor]

so now (k+1)3(k+2)?

[dark granite]

yep

more precisely 3(k+1)(k+2)

which is what we wanted

[nimble meteor]

ohh right because multiplication is commutative right?}

[dark granite]

yes

[nimble meteor]

I see

[dark granite]

and then you say something like "By P.M.I, "$6(1)+6(2)+6(3)+\dots+6n=3n(n+1)$ holds for all integers $n>0$."

and you're finished with the proof!

[ocean seal]

logician_pdx

[nimble meteor]

wait so what was the last step we got?

[dark granite]

well we just showed that the sum of the first k+1 terms is given by 3(k+1)(k+2).

So we're done

[nimble meteor]

oh rightttt

thank you SO SOO SO SOOOOOO MUCH

you're amazing

I was about to give up lol

I'm self studying precalc to learn calc next year

[dark granite]

You're welcome! @nimble meteor

@nimble meteor The way I'd prove the statement you and I just proved would actually be to prove that the sum of the first n natural numbers is given by $n(n+1)/2$ and then multiply both sides of the equation by 6 to get this theorem (that you and I just proved).

[ocean seal]

logician_pdx

[nimble meteor]

oh I see

that's pretty smart

wait what theorem tho

[dark granite]

Bot theorems are equivalent because you can just divide 6 from both sides of the equation we just proved and get the theorem for the sum of the first n natural numbers

[dark granite]

the statement we just proved

by induction

We just proved $\sum_{i=1}^n6i=3n(n+1)$ for any integer $n>0$.

[ocean seal]

logician_pdx

[nimble meteor]

Wait but I have one question why did we not add (6k+1) on both sides...we just did it with one?

[dark granite]

The theorem for the sum of the first n natural numbers is: $\sum_{i=1}^ni=n(n+1)/2$ for any integer $n>0$.

[ocean seal]

logician_pdx

[dark granite]

huh?

what are you talking about

[nimble meteor]

if we added 6(k+1)

to the left

don't we have to do it to the right in order to satisfy algebra basically?

[dark granite]

because we are trying to prove what the sum of the first k+1 terms is

[nimble meteor]

oh

[dark granite]

you can see this more clearly since 3k(k+1) doesn't equal 3(k+1)(k+1+1)

[nimble meteor]

ohhh that's why its just the sum of all of the ones preceding that

plus it

I get it

[glacial hedge]

telescoping series?

[nimble meteor]

what is that?

sounds cool

[dark granite]

@nimble meteor , you got it from here?

[nimble meteor]

yes captain

[dark granite]

lol kk

[glacial hedge]

Something cool in calculus

[nimble meteor]

I appreciate the help and your time

[glacial hedge]

for calculating infite series

[nimble meteor]

bet..hopefully I'll get to that in calc ab?

[glacial hedge]

sadly I think its only a part of calc bc :(\

Is it possible for a linear transformation to be a translation (like move every point up 2 points)?

[nimble meteor]

oh ok

[nimble meteor]

like in geometry?

[glacial hedge]

nvm

[nimble meteor]

ok

[glacial hedge]

its something in la but i just realzied it obviously false

[nimble meteor]

what's la/

[alpine sable]

is there a name for the center point of a sphere?

[ionic jewel]

center

[silk coral]

<@&286206848099549185>

Can I get some help on this please

It’s none of these right

[upper pebble]

you're not supposed to ping the Helpers role unless your question has been unanswered for 15 minutes #❓how-to-get-help

[silk coral]

Oh wait nvm I got it it’s 8.9

Oh my bad

Sorry

[languid vessel]

hello, can i apply gauss elimination on nonsquare matrics ?

[silk coral]

Can I get some help on these

[pearl marlin]

Can you draw the diagram?

[silk coral]

I think it’s something like this

[warped phoenix]

if x is 2 and c is 1, then how is 2 < -1

[magic idol]

2 > 1

[warped phoenix]

yeah but it says x < -c

if x is 2 and c is 1, it's saying 2 < -1

which is false

[fathom violet]

It says or

one of them is true

[warped phoenix]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[magic idol]

x < -c or > c

[warped phoenix]

lol right

ty

[devout bronze]

Can someone explain why for a continuous random variable with the pdf p(x), the probability of it being a single value a is 0 instead of p(a)?

i get that $\int^a_ap(x)dx =0$, but why dont we just use p(a) for the probability since the integral is essentially a sum of probabilities?

[ocean seal]

hjebsjdhebdjd

[sinful monolith]

need some help

[nimble meteor]

its d 80

would you like me to explain? @sinful monolith

[sinful monolith]

yeah

pls

[nimble meteor]

gotcha

So basically the key thing to keep in mind is that it says ABC is EQUILATERAL (wish I could underline lol)

so do you know what that menas?

*means

@sinful monolith

[sinful monolith]

all sides is the same?

[thorn tapir]

Ah well, I'm not well versed here so take my words with a grain of salt-

let's say you assign p(a) = c which is not zero. Now, as p(x) is a continuous distrubution, there should be inifite points like a. If now every value of p(x) at all points is something greater than zero - if you add all of them discretely, you're bound to reach infinity.

but you do know that probability can at most be 1. So, only way of assigning values of p(x) is to make all the values infinitesimaly less, which you can regard as almost surely equal to 0.

[nimble meteor]

yes correct but more specifically what do we know about the angles?

[sinful monolith]

40?

[nimble meteor]

so are you aware of the rule of how all angles in an equilateral triangle is 60 degrees?

Well ok sorry wrong way to put it

BAISICALLY

each angle in an equilateral triangle is equal to 60 degrees

@sinful monolith do you follow?

[fathom violet]

@devout bronze a continuous random variable can realize an infinite amount of real numbers values, so we have an infinitude of values whose sum of probabilities must equal one, therefore these sum must each be infinitesimal aka almost sure equal zero

[sinful monolith]

uh why 60?

@nimble meteor

[nimble meteor]

https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-congruence-theorems/v/equilateral-triangle-sides-and-angles-congruent

@sinful monolith

sorry I had to do something real quick

basically since all sides are equal you can use the isoceles angle theorem

and then transitive I think

so they are all equal

and so x + x + x = 360

3x=180

x=60

[warped phoenix]

y = 16x^2

y/16 = x^2

sqrt(y/16) = x

sqrt(x/16) = y

x = t, y = d, so sqrt(t/16) = d

i chose 4, i am confused why it shows t = something instead of d = something, and also confused why it's wrong when i followed the steps it gives me

[jagged raptor]

bc its asking for the inverse

d=16t^2, t>=0
d/16 = t^2
t = sqrt(d)/4

[jagged raptor]

you shouldnt be switching these back around

[dull olive]

You have to apply the sqrt to both the numerator and denominator, not just the numerator

$\sqrt{d/16} = \sqrt{d}/4$

[ocean seal]

JohnL

[warped phoenix]

hm

[warped phoenix]

@jagged raptor @dull olive ohhhhhhhhhhhhhh

i see

thank you!!! i got it now

[sinful monolith]

need help

[gray isle]

which part are you struggling with?

[fresh bolt]

c

[nimble meteor]

which part do you need help with

[sinful monolith]

how did u get c?

[gray isle]

are you familiar with function notation?

[dull olive]

$f(.5) = 9(.5)+9^{.5}
=4.5+\sqrt{9}
=4.5 + 3
=7.5$

[ocean seal]

JohnL

[dull olive]

He's right, it's C

[eternal stag]

Im confused on how to do this question.

[sinful monolith]

where is the square root of 9?

[gray isle]

are you familiar with the link between fractional powers and radicals?

$a^{(1/n)} = \sqrt[n]{a}$

[ocean seal]

ℝamonov

[dull olive]

yeah

[sinful monolith]

i dont honestly

[tacit mortar]

I got [0, infinity) for domain am I right?

[dull olive]

"Power over root"

$x^{power/root}$

[ocean seal]

JohnL

[dull olive]

$=\sqrt[root]{x^{power}}$

[ocean seal]

JohnL

[dull olive]

You see? @sinful monolith

[sinful monolith]

i see

and also if 4more than 3 times the number is it 3x+4?

@dull olive

[dull olive]

Correct

[sinful monolith]

ty

[alpine sable]

someone help meh please

this is due in like 10 mins

[dull olive]

Area of picture = (12*16)/2=96

b = a * 12/(12+16)
(a * b) = (12 * 16)/2

a = 14.967
b = 14.967 * 12/(12+16) = 6.4

Dimensions of the picture: 6.4 X 14.967

[alpine sable]

uh anyone

[sinful monolith]

5x+8=88?

[alpine sable]

yes

thanks

[sinful monolith]

=16

[alpine sable]

ur a walking W bruh

[sinful monolith]

last question i need help

confused

[alpine sable]

hol up anyone know the second one

[fresh bolt]

just sub in x = 1/2 into the expression 9x + 9^x

[alpine sable]

someone help meh please

uh

[fresh bolt]

E

[alpine sable]

help me before I become

idk

jst

s0meone

help

please

[fresh bolt]

let x be length of rectangle

let y be width of rectangle

[dull olive]

E

[alpine sable]

ok

uh

[fresh bolt]

x = 4y - 5 is the first equation

2x + 2y = 136 is second equation

then you solve the simultaneous equation from there

sub x into second equation and solve

[alpine sable]

how u get 2x and 2y

[sinful monolith]

why E

[alpine sable]

wait

[fresh bolt]

its a rectangle

[alpine sable]

oh

I knew that

[mossy swift]

Everyone here is like 10 times smarter then me lmao

[dull olive]

$cupcakeCost = c/8 \
gingerbreadCost = g/6 \
(2c/8)+(g/6) = (3c+2g)/12$

[ocean seal]

JohnL

[fresh bolt]

@alpine sable the final ans should be length 53.4m x width 14.6m

[opal wyvern]

Can anybody help me understand this?

Why does A = xy turn to 2x + y = A? Makes no sense

[tough hatch]

it doesn't
A = xy gives area
2x+y = 2400 gives perimeter of the fencing

2400 is not A

[opal wyvern]

Oh wow I am not all there right now

[fervent valley]

please

[left citrus]

Hey I just wanted to double check and see that my proof is correct. Thnx

[onyx swallow]

eh

its 4cosx not 4

change it and ull find u dont actually need cosx=1 @left citrus

[left citrus]

Ha there was my problem…

Got it thank you

[onyx swallow]

imagine getting everything perfect but forgot to put one thing

[onyx swallow]

[left citrus]

Hey so I’m having trouble understanding why root(2i) = 1+i

[onyx swallow]

(1+i)(1+i)=2i?

[left citrus]

Alright today is just me being really stupid😂. Thx for taking your time out.

[onyx swallow]

[late vapor]

can someone help me with my homework please

[left citrus]

Of course it depends how hard the question is

[onyx swallow]

for cantor sets, if i change every close interval to open interval, is it still not countable? why or why not?

[late vapor]

umm ill just type an example in and see if you can teach me how to do it

[left citrus]

Lmao qwertytrewq what in the world is that. I’m scared now 😂

[onyx swallow]

i need help too

[late vapor]

7x + 7 = 2x + 2

[onyx swallow]

-1

[thorn tapir]

an identity

[late vapor]

how tf do i do this

[thorn tapir]

wait no

[onyx swallow]

[thorn tapir]

lmao

my brain ded

-1 yeah

[late vapor]

trying to find x im pretty sure

[left citrus]

Alright to do this you want all your Xs on one side of the equation and your numbers on one side

[late vapor]

ummm oki

actually i think i get that

[left citrus]

So if you follow the step

You should get 5X = -5 right?

[late vapor]

yuh

[left citrus]

And then you find X by diving -5 by 5…

[late vapor]

is -1

[left citrus]

Perfect

[late vapor]

but with the first bit, do i just right it out wit all thee Xs on one side and the numbers on the other or?

[plucky pier]

7x + 7 = 2x + 2
7x - 2x = 2 - 7
5x = -5
x= -5/5 = -1

right?

[left citrus]

Mmhm yeah just write out the Xs and numbers on different sides

You can never go wrong with that

[late vapor]

and with the operations how do i know when to do what

[plucky pier]

sorry, im new here. im in grade 8, will be going to grade 9 in a few months. does anyone know any websites or books that i can read/learn from for grade 9?

[left citrus]

Alright lemme show you

[late vapor]

im in year 9

14 yrs

[left citrus]

Sorry I’m from New Zealand so I wouldn’t really know the curriculum in the states

[onyx swallow]

aops maybe

[plucky pier]

ah, its ok. any curriculum is fine

[left citrus]

Alright so how old are you

[late vapor]

maths is still the same isnt it

[plucky pier]

im 14 this year

[left citrus]

Ofc haha

[late vapor]

ok cool

[left citrus]

So are you confident in things like quadratic equations?

[plucky pier]

yes

i learned that a few weeks ago

[left citrus]

Alright so that’s about the hardest stuff you can do?

[plucky pier]

i just want a few problems that i could solve

[late vapor]

guess ill just have to figure my work out from here thanks

[plucky pier]

harder than that is also fine. i would like to learn early

[plucky pier]

im sorry to interrupt u

[late vapor]

its alright

[left citrus]

Hey soz I can get back to you real quick haha

[left citrus]

I could search for a book of mine from last year and post a couple of problems and topics from there, how about that?

[plucky pier]

ooh, please

and thank you

[left citrus]

No problems

[late vapor]

so about this. for the first step how do i know whether to put a plus and a minus in between

[left citrus]

Alright so basically you want to get rid of the x from the right side and the number from the left side right?

[late vapor]

yes

[left citrus]

I found a bunch of stuff I’ll just put them up on here

[plucky pier]

thank you

[fervent valley]

thanks a lot!

[left citrus]

So you want to minus 2x from the right side to get rid of the x right?

[teal epoch]

Use the Subtraction Property of Equality to solve your problem

[late vapor]

ok idk wtf that means but im listening to mr red profile picture rn ok?

[onyx swallow]

<@&286206848099549185>

[left citrus]

So if yo want to get rid of 2x from the right side and minus it from the right side you must also minus it from the left side

[late vapor]

so i need to 7x minus 2x

[left citrus]

Yes

[plucky pier]

what you do on one side of the " = " you do the same to the other side. thats what my teacher told me

[late vapor]

and then 7 plus 2 on the other side. but how do we know to use plus?

right

[left citrus]

How do you reckon you’d fare with this stuff

[left citrus]

No not quite

Because on the left side you want to get rid of 7 right?

[late vapor]

so minus on both sides?

[left citrus]

Yes

[plucky pier]

oooh, ive seen some these before. ill try to solve it. thank you so much for your help.

[late vapor]

right ok

[left citrus]

That’s right

[left citrus]

Alright if you need help you can ask me and I can give you other stuff if you want them.

[left citrus]

So now you get -5 on the right, correct?

[late vapor]

9 - 6y = 10 - 2y. so for this one i plus on both sides after putting numbers on one side and Xs on the other

[left citrus]

Right so which side do you want to take out the Ys?

[late vapor]

so then it would go. 10 + 9 = 6y +2y

is that right?

[teal epoch]

Can I join?

[left citrus]

No that wouldn’t be

Of course

[late vapor]

oh fck

[left citrus]

I’ll tell you why

[late vapor]

i though i could do it but i cant

[plucky pier]

its 9-10=-2y +6y ?

[left citrus]

Correct

[late vapor]

oh because there already on there sides

[left citrus]

I do feel like this would be alot easier to explain on call

Do u wanna get on to one of the server voices

[late vapor]

wish i could but im unable to atm

[left citrus]

Alright that’s no worries

But you minus 10 on the left side because you get rid of the 10 on the right side

[warm granite]

when you're done could i get some help

[left citrus]

Of course

[plucky pier]

no matter which number you want to move, when a negative number go over the = , it will change to positive. and vice versa

[warm granite]

okay thank you

[plucky pier]

right?

[left citrus]

As long as I’m able to haha

[left citrus]

Yep that’s right

[left citrus]

You could just ask your question we could just have 2 going at once

[late vapor]

ok i give up bye

[left citrus]

Cmon you are so close to getting it

[teal epoch]

Quack-

[late vapor]

meow

[late vapor]

really?

[left citrus]

Yeah you are

[late vapor]

go through the steps of that last qeustion

[left citrus]

Alright I’ll do it on paper and give an explanation for why as I do it. You’ve gotta wait a minute though

[thorn tapir]

You're welcome - I did my best to write what I did, if you have any problems in it you can ask me

[late vapor]

ok cool thanks

[left citrus]

Ask me if you don’t get something

[left citrus]

Would you be interested in something like this?

[thorn tapir]

nice explanation 👍

[left citrus]

Thank you. I tried.

[thorn tapir]

[dawn crown]

anyone know why U isnt U^-1 in the second line

[thorn tapir]

is this matrix

[dawn crown]

yes

[thorn tapir]

damn

[late vapor]

ok thanks for trying to help me. but i just cant get my head around this. im going to have to surrender

[thorn tapir]

uhm afaik (AB)^T = B^TA^T

but not sure about inverses

[dawn crown]

oh i see

good point ill take a look online

thanks mate!

[thorn tapir]

[dawn crown]

huh

this is weird

[thorn tapir]

wat

I probably haven't studied this yet

[left citrus]

Alright! Hopefully tomorrow’s better.

[tacit mortar]

I got [0, infinity) for domain am I right?

[tough hatch]

no

[tacit mortar]

u sure?

[tough hatch]

why can't u put negative values in for t?

[tacit mortar]

it cant be negative

[tough hatch]

t^2 + 2 is never negative

[tacit mortar]

Time can't be negative

[tough hatch]

then yes, that is the domain

[tacit mortar]

ty and would the grap look like this

[thorn tapir]

the graph migt be that but Desmos doesn't know it is tim

time*

just take the positive part

[left citrus]

Yeah [0,infinite] is the correct domain.

And yeah that’s how the graph would look like

[late vapor]

PEOPLE I FIGURED IT OUT

I CAN DO MY HOMEWORK NOW

thx for the help red profile pic

[thorn tapir]

ayyy nice

[left citrus]

Algs haha. That’s a W.

[late vapor]

if i am correct. 5w + 3 = 2w + 21. w is 6???

[topaz scaffold]

Yep

[thorn tapir]

good 👍

[topaz scaffold]

You can always plug it in to check

[late vapor]

no waaaaaaay

i can actually do it

[topaz scaffold]

Ayyy

[late vapor]

i am so smart 😆

[topaz scaffold]

Indeed

[late vapor]

i joined this server just to learn how to do my homework

it worked. yall are so smart thankyou heaps

[topaz scaffold]

All I know from joining this server is that know nothing oof

[late vapor]

same ay

ok now i have to do all the questions 💀

see ya

[topaz scaffold]

Good luck

[late vapor]

thanks

[thorn tapir]

have fun and good luck

😎👍

[late vapor]

👍

[outer ocean]

pls help me

pls help

[thorn tapir]

uh yes

Your question?

[left citrus]

Lol what’s the question

[ionic jewel]

what is the mean?

[left citrus]

?

[ionic jewel]

just reminds me someone who asked that everytime someone posts questions like this

[left citrus]

😂

[obtuse fox]

help

someone

Hello

[plucky pier]

these are fun to do, so im very interested.

[plucky pier]

hi

[obtuse fox]

Yes

Help

[plucky pier]

sorry, i wont be much of a help. im just a newbie at math

[obtuse fox]

I suck more

[plucky pier]

oh

which one?

[obtuse fox]

the first 3

Actually wait

Lemme see

[topaz scaffold]

For 8 think about using exponents of 10

[obtuse fox]

12 1 9

Thx

[shut tartan]

45.000=30000(1.11)^n

[left citrus]

Alr one second…

[shut tartan]

how do i solve?

[topaz scaffold]

For 9 you can you divide the top the divide the bottom

[obtuse fox]

4578.307

raised to 1000

[topaz scaffold]

Channel occupied at the moment

[obtuse fox]

4/6?

2/3

[plucky pier]

flip the second fraction

[obtuse fox]

I did just that

[topaz scaffold]

I think what they want you to do something like 321 expressed as 3* 10^2 + 2* 10 + 1

[obtuse fox]

eh what?

[left citrus]

Hey using the example b4, you should try and solve these and if you need help, you can always ask me…

[topaz scaffold]

It works if you don't flip it and do division, just gets a bit messy sometimes

[plucky pier]

yep

[plucky pier]

eee omg thanks so much

[topaz scaffold]

Sorry discord formatting

[obtuse fox]

not that I didnt get it lol

[left citrus]

Np 🙂

[topaz scaffold]

Calculate it out and you'll get 321

[obtuse fox]

Are you saying about the question 10?

[topaz scaffold]

8

Its an example

[obtuse fox]

oh ok

thx

I suck at Maths

Do you know I can become good at it?

Can you help me do 11 and 12 too

[topaz scaffold]

For me it's about finding the right explanation that makes sense for me

[obtuse fox]

ok

[obtuse fox]

how to do 10?

Can you explain it pls?

[topaz scaffold]

So do you know what a negative exponent means?

[obtuse fox]

Yeh

[obtuse fox]

yes

[topaz scaffold]

Actually nvm that doesn't really matter here I read the question wrong

[obtuse fox]

Lol

can you give me the just the answer

I have online classes in a bit

[topaz scaffold]

Here a hint: you can get the base 3 by raising to a power it to get 81

[topaz scaffold]

Well you gotta learn somehow

[plucky pier]

(3^4)^-3 or 3^-12

i think

[obtuse fox]

Do you mean raising 3 so it becomes 81?

[topaz scaffold]

Yea to some power

[obtuse fox]

3^27?

[obtuse fox]

3^27

[plucky pier]

no

[obtuse fox]

What

[obtuse fox]

But you get 81 by doing that

[plucky pier]

o-o

[obtuse fox]

3 raised to 27 = 81

How is it wrong?

[plucky pier]

its not 3*27

[obtuse fox]

^

[topaz scaffold]

3^something

[obtuse fox]

that means raised to

[plucky pier]

yes

[obtuse fox]

This means raised to the power

I did 3 raised to the power 27

is that wrong?

[plucky pier]

but that means 3 * 3 * 3 * 3 * 3 ... ans so on till there are 27 3's

[obtuse fox]

Yes

[plucky pier]

but the answer wont be 81

[obtuse fox]

It said

[plucky pier]

it would be more than 81

[obtuse fox]

You need the base to be 3

[plucky pier]

yes

for example, 3^2 = 9 = 3 * 3

[obtuse fox]

But 3 * 3 * 3 *3 *............. 27 times = 81

[plucky pier]

not 3 * 2

[obtuse fox]

yeh

OH WAIT

I am dumb

So sorry

[plucky pier]

then try 3 * 3 * 3 27 times on ur calculator

[obtuse fox]

I just realised

[plucky pier]

lol

ik

[obtuse fox]

oh god

[plucky pier]

its fine, most of us got confused about it too

so now that u realized, do u know the answer now?

i mean, i already told u the answer a while ago

but i dont think u saw it

[obtuse fox]

Yes

3^4

[plucky pier]

mhm, but what r u gonna do about the ^-3 ?

[obtuse fox]

what

Oh

[plucky pier]

ur question

[obtuse fox]

I forgot about that

What do I do about that

[plucky pier]

so its (3^4)^-3

[obtuse fox]

Oh ok

In which grade are you btw

[plucky pier]

when theres an exponent inside and outside, u can multiply them together too

grade 8

[obtuse fox]

same

[plucky pier]

aye

[obtuse fox]

And you are so much better than me

[plucky pier]

the answer could also be 3^-12

[obtuse fox]

[obtuse fox]

ok

[plucky pier]

dont beat urself about it, its just math is my favorite subject

[obtuse fox]

My favorite is Science

[plucky pier]

i came here to find more stuff about math

[topaz scaffold]

Oh wow exponents in grade 8

[obtuse fox]

Rational Numbers too

[plucky pier]

those r fun

[topaz scaffold]

That's early I think idk

[obtuse fox]

more like suffering but ok

what is 11

[obtuse fox]

11

This is kinda fun ngl

[plucky pier]

u draw the fractions in a number line

[topaz scaffold]

Science has math later on...

[plucky pier]

just a number line and u pinpoint where the fraction is

[obtuse fox]

Oh no

[obtuse fox]

This is pain

[plucky pier]

lol, its just formulas with math

[topaz scaffold]

It's not really

[topaz scaffold]

Kinda ig

[obtuse fox]

At least I remember how to do this

Rational numbers are kinda good

[plucky pier]

to make it a bit easier, find the same denominators

[obtuse fox]

what

Oh that

Ok

I dont get it

[plucky pier]

o

[obtuse fox]

I remember watching a vid about that yesterday

So I remember how to pinpoint the numbers

[topaz scaffold]

Ok good

[plucky pier]

alright

im glad i get to help someone here :DD

[topaz scaffold]

Same lol

[obtuse fox]

I really dont like Math, but getting someone to help me understand it is fun

[topaz scaffold]

A bunch of the math here gets too advanced for me lol

[obtuse fox]

Lol

Dont worry

I am here

I am dumb

[topaz scaffold]

Oh yea having someone explain helps a ton

[obtuse fox]

So basic questions go right by my head

So I will need your help

and Nisha's help too

[plucky pier]

is there a channel here that let us talk freely about anything? because this is a channel for asking questions

[obtuse fox]

There is a Math - general

And a chess - go

cats

scroll

scroll down you will find htem

them*

Finally

The Last question

[obtuse fox]

found it

https://discord.gg/tw9Q8WqR

[left citrus]

There’s a chill channel

[obtuse fox]

ello

[plucky pier]

oh ok

[obtuse fox]

Last question

[plucky pier]

i actually dont know how to put it in solution form

[obtuse fox]

oh

Ok

This is pain

[plucky pier]

i mean u can mess around with it

[obtuse fox]

what?

[plucky pier]

use any theories or multiply it around

[obtuse fox]

How do I find x tho

[topaz scaffold]

Oh just think about the exponent rules

If you multiply exponents with the same base you can add the powers they're being raised by

[plucky pier]

?

its ok, u r bz with other channels

[topaz scaffold]

For example 3^2 * 3^3 = 3^(2+3) = 3^5

[plucky pier]

yes, i know that

[austere elk]

Does anyone have chegg please??

[topaz scaffold]

You can use that to solve question 11

Oof I meant 12

[alpine sable]

stop

[plucky pier]

oo okk

[alpine sable]

never take the easy way out

[topaz scaffold]

Isn't that a physics thing?

[alpine sable]

https://tenor.com/view/brave-girls-rollin-eunji-minyoung-yujeong-gif-21179011

[topaz scaffold]

Ehhh?

I don't get the reference

[left citrus]

Did you solve it?

[plucky pier]

yep i did

[left citrus]

Alright what did you get