help-10

[tall bramble]

ohhhhh 2 and -2

[trail musk]

You actually have -2 and 2

Order matters here

[tall bramble]

ok lol

so -2 is assigned to a

and 2 is assigned to b

[trail musk]

(a component of CB) / (b component of CB) =
(a component of AB) / (b component of AB)

[tall bramble]

isnt it gonna be over a component of AB

(a component of CB) / (a component of AB) =
(b component of CB) / (b component of AB)

like this

[trail musk]

Those are both algebraically equivalent

[tall bramble]

oh so i can do it my way aswell

[trail musk]

You can rearrange one equation to get the other

[tall bramble]

cause thats how i undersntand it

so now ive got 2-9k/2 = -3k/-2

do u cross mutiply?

how do they get 3k/2 then

[trail musk]

Looks good

Simplify

[tall bramble]

so u cross multiply

then simplify

[trail musk]

You don't need to

Just multiply both sides by 2

[tall bramble]

huh

why do u multiply by 2

[trail musk]

Why wouldn't you?

don't you see 2 in both denominators

[tall bramble]

u mutliply each fraction by 2?

one denominator is 2 the other is -2

[trail musk]

Yep, so then the negative cancels in the right fraction

[tall bramble]

ohhhhh

[trail musk]

2 - 9k = 3k

[tall bramble]

k = 6

?

nvm

1/6

ok thank u

i undestand now

.close

[obtuse pebble]

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[trail cloak]

[zenith ibex]

How do I differentiate between these two

without solving it ? ?

[trail cloak]

You mean you wanna know if the equations have a common solution or not without solving?

[limber quartz]

What is the QUESTION

...that you are trying to answer

[zenith ibex]

wait 1 sec

[limber quartz]

En Español es bueno

I can just Google it if I can't interpret it

[zenith ibex]

I have this linear ecuation system, and I know that It has only one solution which is (x2,y1)

and that this is the graph that represents the system

[limber quartz]

(2,1)

Ok

[zenith ibex]

without solving it, I came to the realization that these 3 weren't parallels or coincidents lines, using this

so I know that they all intersect themselves

[limber quartz]

Good job

[zenith ibex]

and since the third equation is the sum of the first two, I know that it passes through the same point as the other two

[limber quartz]

Ok

So you are onto several things here

For a system to have a single point solution, all the slopes must be different

That doesn't on its own guarantee a single point solution though

You are right about eq3 being a linear combination of 1 and 2

[zenith ibex]

ok

so when I have systems like this one and I can't solve it, (because the prompt tells me not to) , How do i know that it doesn't have a solution?

[limber quartz]

Ah, this is a different question

Just look at it

There are 3 intersections

An independent system of linear equations (a system with ONE solution) has ONE intersection.

[zenith ibex]

sorry

it just comes like this

I can't graph and I can't solve

[limber quartz]

There is no solution to the system because no single point lies on all 3 lines.

[limber quartz]

The instructions say not to solve?

[zenith ibex]

yes

I just have to analize if it is compatible, incompatible and to say how it would be graphed

for example for this one I did this to find out that it is an incompatible system

where two lines are parallels, and there is one that intersects those two

[limber quartz]

It's easy to weed out parallels and coincident lines

[zenith ibex]

yes

[limber quartz]

This isn't though

[zenith ibex]

but since in this ecuation they all intersect, (but in different points) I don't know how to know that

[limber quartz]

I wonder if the inconsistent linear combination is all you need?

Since eq1+eq2 != eq3

Specifically, the LHS are the same but the RHS are not

The parts on the left match, but the part on the right does not match.

Implying there is NO linear combination of the first two that will equal the third

[zenith ibex]

I didn't get that

I if sum equation 1 and equation 2 and it doesn't give me equation 3 then I know that the system doesn't have a solution because they don't intersect in the same point?

[limber quartz]

Not quite what I was saying

I said that because the LEFT side of that result did match equation 3, but the RIGHT didn't, that there is not any linear combination of them that will equal equation 3

And that, I think, is the giveaway that this system has no solution

[zenith ibex]

What is the Left and the Right side

[limber quartz]

<@&286206848099549185> somebody please confirm this

[limber quartz]

Equations have left sides and right sides

When you add the first two equations together, they also have a left side and a right side

[zenith ibex]

oh, before the = and after the =

left= right

[limber quartz]

The left side of that result matched equation 3, but the right did not.

[limber quartz]

Yes

[obtuse pebble]

@zenith ibex Has your question been resolved?

[limber quartz]

you know @zenith ibex I don't think my idea about linear combinations of equations is correct because I know it doesn't hold for all systems of two linear equations which do have solutions.

[zenith ibex]

yea i tried to do the math and I can't do it

maybe this exercise just isn't meant to be done

[limber quartz]

Read your notes and the chapter, it will tell you what to look for and can't be more than 4-6 pages long

(The examples/ definitions)

[obtuse pebble]

@zenith ibex Has your question been resolved?

[zenith ibex]

yeah ill try that

[obtuse pebble]

@zenith ibex Has your question been resolved?

[obtuse pebble]

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[quaint crescent]

Does there exist a continuous map [0, 1] -> [0, 1] such that the pre-image of every point is infinite? I've been stuck thinking about this for a while...

[upbeat gazelle]

Take xsin(1/x)

I think it works

[quaint crescent]

Maybe just sin(1/x) though

Is that continuous at 0?

No it is not

Hmm

[upbeat gazelle]

xsin(1/x) is continuous

[quaint crescent]

Yeah but not every pre image is infinite

In fact it looks like all pre images are finite for xsin(1/x)

My feeling at the moment is that no such function exists but I can't prove it

[high pivot]

it is not continuous, it is only connected

we need (0,0) to be united to the region

[quaint crescent]

I think xsin(1/x) is continuous, but it doesn't work for my problem because the pre image of some (all?) points in [0, 1] are finite

[upbeat gazelle]

Do you want f to be surjective ?

Because otherwise 0 works

I know it's a trivial solution

[quaint crescent]

Empty pre images are still finite

[upbeat gazelle]

If you want to prove it does not exist if f takes more than one value, you can maybe build a sequence of values such that f(x_n) has multiple adherence points

[quaint crescent]

Eg with 0, the pre image of 1 is empty which is finite

[quaint crescent]

Yeah I mean f must take every value infinity often, so something like this might work, but I haven't been able to put it all together yet

[obtuse pebble]

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[obtuse pebble]

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[still oxide]

(not sure if I can ask about logic equivalency questions here)

Prove that (~p ^ q ^ r) v (p ^ ~q ^ ~r) and (p (+) q) ^ (q <-> r) are logically equivalent. Start from the left side.

[still oxide]

I'm dumbfounded because i can't seem to apply any of these rules here

[obtuse pebble]

@still oxide Has your question been resolved?

[obtuse pebble]

@still oxide Has your question been resolved?

[high pivot]

i assume we are doing analysis instead of a truth table

lets start with $(\neg p\land q\land r)\lor(p\land \neg q\land\neg r)$

[warm shale]

ThisGuy

[high pivot]

we can use distributivity of and through or

just like you would with multiplication

show me your work when you get there

[high pivot]

@ me when you get there if youre still on

[still oxide]

i have no clue

[high pivot]

i take it (+) is xor?

its not on the table but ~(a(+)b) <-> (a<->b)

that one is useful in this problem

can you show me what you have?

[obtuse pebble]

@still oxide Has your question been resolved?

[obtuse pebble]

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[worthy moss]

Guys how 2 * 2 = 5? My teacher give a project that we should give him the reason why 2 * 2 =5

[oak bear]

Ah

Yes

This one

Maybe ur teacher is asking u for a trick

[worthy moss]

Yep

[oak bear]

anything that is a x a = wrong answer there would be a 0/0 somewhere

Ur gonna divide something there by 0

[worthy moss]

I done this trick and he said no

I told this trick already 🫤

Is there any other trick that he could accept

?

[wanton hull]

mathematically it will have an exeption for some rule like dividing by zero

[oak bear]

I only know that one, have u tried making 2 = a variable and 5 into something

[worthy moss]

can you explain more?

[oak bear]

like a x a = (a+1) or something

[worthy moss]

Yes yes I done that but he stills says know

[oak bear]

or sometimes they do
a x b = b+1
and then make
a = b or something

[worthy moss]

Yes I done that but he is not accepting IDK why

I done any way in Google

[oak bear]

I found it

[worthy moss]

He gives A+ on our math exam if I find this

[oak bear]

This one

[worthy moss]

I will try this one

[oak bear]

The mistake there is b-a-c

sine it would be 5-4-1

which is 0

they divided it by 0 on both sides

But yeah, u can also look up for the proof of 2 + 2 =5

By proving that,
u also get 4 = 5

which is 2 x 2 =5

[worthy moss]

Ok I try it tomorrow

[oak bear]

yeah u can probably look for the proof of 2 + 2 first

u gotta prove that 4 = 5 then u can do 2 x 2

since 2 x 2 =4

U can probably use a better solution than the one I've sent

[worthy moss]

Ok👍

Thanks, see you later 👋

[obtuse pebble]

@worthy moss Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[sinful kraken]

What is the vector equation for the tangent line of the intersection of x2+y2=25 and y2+z2=20 at the point (3,4,2)?

[jolly ginkgo]

their gradient should be in same direction

[sinful kraken]

im supposed to use parametric equations

[trail musk]

Δ • f = k Δ • g

[jolly ginkgo]

what is the dot about?

[obtuse pebble]

@sinful kraken Has your question been resolved?

[sinful kraken]

i need vector equation

[jolly ginkgo]

say one thing

[sinful kraken]

???????

[jolly ginkgo]

find both gradients, then the plane normal to gradient passing through (3,4,2) and finally their intersection

[obtuse pebble]

@sinful kraken Has your question been resolved?

[obtuse pebble]

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[next reef]

@rain meadow, don't keep posting the image in other's channels, read #❓how-to-get-help

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[warm harbor]

2

is f(x) = dy/dx
and f'(x) = d²y/dx²

i am confused

[torpid forum]

no f'(x) = dy/dx

and f''(x) = d²y/dx²

[warm harbor]

ok

[hasty salmon]

f’(x) is Lagrange notation for the derivative of f(x).
df/dx or dy/dx is Leibniz notation for the derivative of f(x).

So f’(x) = df/dx = dy/dx (Derivative)
And f’’(x) = d²f/dx² = d²y/dx² (Second Derivative)

And derivative is just the function representing rate of change of a function with respect to some variable.

[warm harbor]

thanks so f(x) = y??

f'(x) = y'

f''(x) = y''

[hasty salmon]

if you want to define it that way

then yes

[warm harbor]

yeah

aight ty

[hasty salmon]

you can say a = f(x) and so a' = f'(x)

y is just a convention

[warm harbor]

right

[hasty salmon]

Also we like to do

$\frac{d}{dx}[f(x)]$

[warm shale]

Blissful

[obtuse pebble]

@warm harbor Has your question been resolved?

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[balmy crescent]

hello im having trouble finding n

i know that the value is 0.364 or something tried learning through photomath but they offered no explanation

[sage geode]

Tried dividing both sides by 1.08^n ?

[balmy crescent]

how would that help?

[royal basin]

do it and see

[balmy crescent]

ok

[viral knot]

hello

@viral knot

[balmy crescent]

how do i divide 1.08^2n with 1.08^n

[sage geode]

$\frac{a^b}{a^c}=a^{b-c}$ so $\frac{1.08^{2n}}{1.08}=1.08^{2n-n}$

[balmy crescent]

ty

wait that doesnt look right

[warm shale]

A Lonely Bean

[sage geode]

Had a typo there

[balmy crescent]

nvm

yes i got the answer right

thank you all so much you were very helpful

[obtuse pebble]

@balmy crescent Has your question been resolved?

[obtuse pebble]

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[spark vector]

Larger triangle with hypotenuse length 15

Smaller triangle with hypotenuse length 5

Am I right in saying that the other legs of the smaller triangle are 3 and 4?

[obtuse pebble]

@spark vector Has your question been resolved?

[spark vector]

<@&286206848099549185>

I just see that I could just double the area of the 3 squares?

Seems to fit in half of the large rectangle

<@&286206848099549185>

[obtuse pebble]

@spark vector Has your question been resolved?

[hasty salmon]

well you would have to prove that

[spark vector]

Yeah that I am not sure of

[hasty salmon]

you know the Pythagorean theorem:
a^2 + b^2 = 15^2 (for the large triangle)
a^2 + b^2 = 5^2 (for the little triangle)

solve for a and b

[spark vector]

Yeah a and b is 3 and 4

Small triangle

[hasty salmon]

and for large one?

[spark vector]

Not sure

[lusty cave]

bro what did i do wrong

[hasty salmon]

9 and 12

[spark vector]

Oh right ya

[lusty cave]

i drew a line through the 3 squares and got 15.811 as the length of the rectangle

[hasty salmon]

so the long side of the big box is 12+3
and the small side is 9+4

[lusty cave]

because sqr(5^2+15^2)

[hasty salmon]

15*13 is 195

hmmm

okay

so it wants big box area - small boxes area

[spark vector]

195 - 75

?

[hasty salmon]

well yeah

but that isnt an answer

in the question

[spark vector]

Right

[lusty cave]

it says of the white rectangle that dosent imply that it wants white minus grey

[hasty salmon]

well in the other case if would just be white

which is 195

195 isnt an answer either

[lusty cave]

i got 150

[hasty salmon]

how

[lusty cave]

draw a line through the 3 rectangles in the middle

solve

15.811

[hasty salmon]

why

[lusty cave]

that gives you the long side of the rectangle

[hasty salmon]

why

prove it

[lusty cave]

because it touches with the diagonals

[lusty cave]

i cant

[hasty salmon]

???

[lusty cave]

Oh

i ment through the diagonal sorry

which is a triangle with a=5 and b=15

[hasty salmon]

yeah

that works

[lusty cave]

yes

[hasty salmon]

why does my way not work tho

it should be the same result

[spark vector]

I agree

[lusty cave]

and then the triangl you make has an area of 37.5

times it by 4 to get 150

[hasty salmon]

how do you get the small side

of the box

[lusty cave]

area = 150

15.811 * x = 150

x = 150/15.811

[hasty salmon]

how do you get the small side

[lusty cave]

that is the small side

[hasty salmon]

without knowing about 150

[lusty cave]

idk

thats how i got it

[hasty salmon]

how did you get 150

[spark vector]

I am not sure if decimals should come into the question though

[lusty cave]

^

[lusty cave]

^

[hasty salmon]

why do you times it by 4

[lusty cave]

because it specifies in the question that the rectangle touches the midpoint

so the line bisects the rectangle

and a triangle is half the area of the rectangle iti is contained in

so times it by 2 to make it the same area of the rectangle

then 2 to be the same area of the entire white rectangle

[hasty salmon]

sqrt(15^2 + 5^2) = 5*sqrt(10)

big box has width 5*sqrt(10)

then you do times 8

[lusty cave]

thats what i got

[lusty cave]

why

[hasty salmon]

height of big box is 8

[lusty cave]

where did u get 4,3 and 12 from

[hasty salmon]

theorem of pythagoras

solve a^2 + b^2 = 15^2 and a^2 + b^2 = 5^2

[spark vector]

Should be 120 right?

[hasty salmon]

this makes no sense

[spark vector]

Agreed

Their dimensions are wrong

[hasty salmon]

ya

on one side the halflength is 9 and on the other its 4

like how???

[spark vector]

Right

[spark vector]

I’m thinking they just want us to prove that the small squares are half the size of the rectangle

[lusty cave]

is this correct?

[hasty salmon]

how did you get this

[lusty cave]

gimme a sec

[hasty salmon]

how is it 4.744

and not 4

we are dealing with right triangles

so a^2 + b^c = 5 so a = 3 and b = 4

[lusty cave]

the other side is 1.579

so the other side of the big triangle is 15.811-1.579

[spark vector]

I thought it was always given that if the hypotenuse is 5 then the other two legs must be 3 and 4?

[lusty cave]

14.232

[lusty cave]

only if the values have to be an integert

[spark vector]

Oh

I see

[lusty cave]

what grade are you in

[spark vector]

Got it

[lusty cave]

because this question stumped me for a while

i was sitting on the post for like 20 minutes

[spark vector]

Grade 7 and 8 math competition paper

[lusty cave]

alright

[spark vector]

But we calculators can’t be used

How can they expect us to know that lol

[lusty cave]

hmm

maybe the question is wrong

ohhhh

no its right

because you only need to find the area

which has no decimals

the area of the triangle is ((5x15)/2)x4

bro

i need to learn how to use the math bot

[spark vector]

Oh ok

[lusty cave]

,help

[warm shale]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[spark vector]

So we are not expected to use Pythagoras here

[lusty cave]

no

you just need to find the area of the triangle and times it by 4

[spark vector]

How to prove this is equal to the blue one

@lusty cave

[lusty cave]

hmm

idk how to prove it

but the equation to a right angle triangle is $bh/2$ and the equation for a rectangle is $bh$

[warm shale]

Xav

[spark vector]

Yeah

[hasty salmon]

ahhhhhhhhhh

yes

[spark vector]

?

[spark vector]

.close

[obtuse pebble]

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[limber sable]

how

[limber sable]

how 4- disappeared, can i get the extended explanation?

[whole dock]

$(4 - (2 - x) = 4 - 2 + x$

[warm shale]

ColdTee

[limber sable]

ik but where did the 4 go

[whole dock]

$4 - 2 = 2$

[warm shale]

ColdTee

[limber sable]

OH BrRUH

man i rly should get a sleep

tysm

.close

[obtuse pebble]

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[brazen coral]

hey

[brazen coral]

For the start of this solution, I'm a bit confused where the (1 0), (1 1), (1 -1), and (2 0) are coming from?

[alpine raven]

image of each matrix from the basis B

by S

[obtuse pebble]

@brazen coral Has your question been resolved?

[brazen coral]

sorry, but just to make sure i understand it correctly, how would that be done?

[alpine raven]

can u give the whole exercise ?

not the solution :/

[brazen coral]

in part a and b we found

[brazen coral]

This is a workshop already completed (not for marks)

[alpine raven]

so S is an application that takes 2x2 matrix as input and returns a vector

and they gave us how S works

brb rq

so $S\begin{pmatrix}r & s \ t & u \end{pmatrix} = \begin{pmatrix} r+u \s-t \end{pmatrix}$

[warm shale]

Ｈｅｒｅｌｓ

[brazen coral]

ohhh its found by using that given identity

[alpine raven]

Now they asked you to find a matrix representation of this application in the basis E

So you should find the image of each matrix in B by S

you have them all but its not done yet

[brazen coral]

so if that was not given, how would we find the image of each?

[alpine raven]

you just cant

and how would you find a matrix representation of an application without even giving how that application works

[brazen coral]

well im trying to learn this stuff, so I wasnt sure if there was a general rule

but thx 🙂

.close

[obtuse pebble]

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[slim copper]

So far what i've done is say we have φ([k]_n)=[k]_m

and because its a homomorphism we have φ([k+l]_n)=[k]_m+[l]_m

but not sure what do next

i think it's along the lines of noting φ([1]_n goes to [1]_m and that by adding [1]_m to itself several times you can get every value in Z/mZ

but not sure how to word it precisely

<@&286206848099549185>

[obtuse pebble]

@slim copper Has your question been resolved?

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@slim copper Has your question been resolved?

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[sage geode]

Tried simplifying f(x)?

[warm canopy]

what is it you are trying to prove isn't possible?

probably easier to look at the cubic instead

oh is this f(x) or f(x)-11?

[obtuse pebble]

@wintry fable Has your question been resolved?

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[void chasm]

Let a and b be 2 real numbers such that |a| > 0. if the equation ||x-a|-b| = b has 3 distinct roots, find value of b.

[void chasm]

tried doing this a couple of ways

unable to get an answer

[obtuse pebble]

@void chasm Has your question been resolved?

[obtuse pebble]

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[river basin]

The first files are my answers to the MTH311 practice exam. I need help writing some of the from the practice exam 2 or need help clarifying some of them.

[tardy epoch]

don't make ppl download files to help you. screenshot and upload images

[idle thunder]

also be specific

[river basin]

These are the first three from the 2nd practice exam question MTH311 Intro to higher mathematics and Proofs. I need help with questions 1, 3? and then some others but I need some clarification.

Questions 6 and 7, but mostly 7 because I make making multiple assumptions.

→ MY ANSWER PROOF: We let n be a part of the set containing all positive integers, or ℤ+, where n ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …}.
We are asked to prove that the square root(n) ∈ ℚ leads to the square root(n) ∈ ℤ, and vice versa. This means that there exists a square root of any chosen positive integer is a rational number, and that the square root of a chosen positive integer n is any integer. We should know that the set dealing with both positive and negative integers is a proper subset of the set dealing with rational numbers.
For example, √1 = ±1, which shows two rational integers since -1/1 = -1 and 1/1 = 1. (0 already classifies as a rational number and is an integer.)
According to Definition #14.3 from the textbook, -1/1 is described to be an infinite subset of Z x (Z – {0}) and is the equivalence class of (2, 5).
-1/1 = [(-1, 1)] = {(-1, 1), (-2, 2), (-3, 3),…., (1, -1), (2, -2), (3, -3),….}.
Equivalently, -1/5 = [(-1, 5)] = {(a, b) ∈ Z x (Z – {0}): -1 • b = 5 • a}.
√2 is an irrational number, which then leads to √2 not being an integer. This is especially true when we negate the statement "square root(n) ∈ ℚ <==> square root(n) ∈ ℤ" such that n ∈ ℤ+.
√4 = ±2, which shows 2 rational integers since 2/-1 = -2 and 2/1 = 1.
So, if n^(1/2) ∈ ℚ, then n^(1/2) ∈ ℤ such that n ∈ ℤ+. And, if n^(1/2) ∈ ℤ, then n^(1/2) ∈ ℚ such that n ∈ ℤ+.
∴ If n ∈ ℤ+, Then √(n) ∈ ℚ <==> √(n) ∈ ℤ. ☐ㅁ◻ or Q.E.D

I don't know if I'm right.

[tardy epoch]

pick ONE problem

[river basin]

Question 3 then

PROBLEM #3: Prove that 11│a^20 + a^11 + 9 for any 'a' such that gcd(a, 11) = 1. Answer Proof:
By Bezaut's Lemma, ꓱ x, y ∈ ℤ, such that 1 = a • x + 11 • y since gcd(a, 11) = 1. If gcd(a, 11) = 1, then we can say, by the definition of the greatest common divisor, that a + 11 are coprime (or relatively prime numbers). Meaning a and 11 are coprime (or each prime). For example, since a and 11 are coprime, let's see what happens when we choose a = 1. The gcd(a, 11) = 1, and 11 would be divisible by 11. I have trouble figuring out where to go from thereon.

[obtuse pebble]

@river basin Has your question been resolved?

[river basin]

No, still waiting for clarification.

[obtuse pebble]

@river basin Has your question been resolved?

[river basin]

No

[obtuse pebble]

@river basin Has your question been resolved?

[river basin]

→ PROBLEM #11: Let n be a positive integer.
Then square root(n) ∈ ℚ <==> square root(n) ∈ ℤ.
→ MY ANSWER PROOF: We let n be a part of the set containing all positive integers, or ℤ+, where n ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …}.
We are asked to prove that the square root(n) ∈ ℚ leads to the square root(n) ∈ ℤ, and vice versa. This means that there exists a square root of any chosen positive integer is a rational number, and that the square root of a chosen positive integer n is any integer. We should know that the set dealing with both positive and negative integers is a proper subset of the set dealing with rational numbers.
For example, √1 = ±1, which shows two rational integers since -1/1 = -1 and 1/1 = 1. (0 already classifies as a rational number and is an integer.)
According to Definition #14.3 from the textbook, -1/1 is described to be an infinite subset of Z x (Z – {0}) and is the equivalence class of (2, 5).
-1/1 = [(-1, 1)] = {(-1, 1), (-2, 2), (-3, 3),…., (1, -1), (2, -2), (3, -3),….}.
Equivalently, -1/5 = [(-1, 5)] = {(a, b) ∈ Z x (Z – {0}): -1 • b = 5 • a}.
√2 is an irrational number, which then leads to √2 not being an integer. This is especially true when we negate the statement "square root(n) ∈ ℚ <==> square root(n) ∈ ℤ" such that n ∈ ℤ+.
√4 = ±2, which shows 2 rational integers since 2/-1 = -2 and 2/1 = 1.
So, if n^(1/2) ∈ ℚ, then n^(1/2) ∈ ℤ such that n ∈ ℤ+. And, if n^(1/2) ∈ ℤ, then n^(1/2) ∈ ℚ such that n ∈ ℤ+.
∴ If n ∈ ℤ+, Then √(n) ∈ ℚ <==> √(n) ∈ ℤ. ☐ㅁ◻ or Q.E.D
Does this make sense?

[viral blade]

Is this just a very long proof that sqrt(n) in Z implies sqrt(n) in Q

[obtuse pebble]

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[serene rain]

Question 7

using the 30, 60, 90 degree theorem

(not in decimals please, just leave as radicals)

[oak plover]

Since it's looking to the 30 and the bottom is root 3 the left side should be 1.
Two 15s should divide the left side of the big triangle into two pieces.
So we have 0.5 and 0.5 on the left side.
Then you can find the hypotenuse using the Pythagoras Theorem.
Calling the bottom a, splitted left part b and the hypotenuse c.
3 + 0.25 = c^2
3.25 = c^2

a ≈ 1.73205080757
b = 0.5
c ≈ 1.80277563773

[serene rain]

u sure its 0.5? BD is just specified as an angle bisector. doesn't that mean that AB and BC are not necessarily equal to each other?

wouldn't you need to do AD/AB=DC/BC

since all we know is that its an angle bisector

[oak plover]

Yes that is what I said

[serene rain]

I thought that was the right way but I'm having trouble getting farther than that

[oak plover]

I called BC the b

CD the a and BD the c

[serene rain]

no but your saying AC is equal to 1 and that because its an angle bisector (two 15s) it should divide the left side of the triangle into two equivalent pieces, but those two sides might not be equal to one another

all we know is that it equally divides the angles not the segment

[oak plover]

Oh wait, I think you are right

One second

[oak plover]

The first thing that came to my mind was the trigonometric values of 15 degrees

[serene rain]

oh

[oak plover]

But they are kind of long

So I tried to do it with geometry. Let me send the trigonometric values of it

[serene rain]

can u substitute one piece for x and another for 1-x instead?

[oak plover]

Which piece?

[serene rain]

BC as x and AB as 1-x

since we know the total value

[oak plover]

$sin(15deg) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

$cos(15deg) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

$tan(15deg) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

[warm shale]

Oğuzhan

[oak plover]

But solving it with pure geometry still seems to be a better option

[oak plover]

Maybe we can go to somewhere from that

Not sure

[serene rain]

yeah cause that was the route I was trying to take and I kept getting confused

i would get to 2/1-x=radical 3/x

I think that simplifies to x=2sqrt3 minus 3

[oak plover]

It might be

[oak plover]

But how did you get here?

[serene rain]

and from there I plugged the known values into the Pythagorean theorem

[serene rain]

angle bisector theorem? or am I doing it wrong

[oak plover]

I didn't see a similarity

[oak plover]

Oh wait, right

We know both sides

That should work?

[serene rain]

but when I plugged it into the Pythagorean theorem I got like a weird radical in a radical when I tried to simplify

[oak plover]

They are tricky and fun to simplify

Let's try

[serene rain]

I got sqrt(24-12sqrt3)

[oak plover]

So at that point you should first get something like

$\sqrt{ab - 2\sqrt{a + b}}$

[warm shale]

Oğuzhan

[oak plover]

Get it in this format

$a > b, \sqrt{ab - 2\sqrt{a + b}} = \sqrt{a} - \sqrt{b}$

[warm shale]

Oğuzhan

[oak plover]

This works if the sign in the middle is plus too. You just have to change the output's sign

[oak plover]

But this seems like it's not gonna end well

[serene rain]

yea

well i plugged it into a calculator

and it came out to be 3sqrt2-squrt6

[oak plover]

It actually only wants the approximation I suppose

[oak plover]

Also my apologies this was the opposite*

not ab and then a + b

a + b then the ab

Just realised it

[serene rain]

yes, but I might have a similar problem on a future test only instead of decimals it asks me in simplest radical form

[oak plover]

$\sqrt{24 - 2\sqrt{108}}$

$a + b = 24, ab = 108, b = \frac{108}{a}$

$a + \frac{108}{a} = 24$

$a^2 - 24a + 108 = 0$

$a = 12 +- \sqrt{144 - 108} = 12 +- \sqrt{36} = 18$

[warm shale]

Oğuzhan

[oak plover]

I meant + or - by typing +-

[serene rain]

okay

[oak plover]

So it was simplifiable

[oak plover]

The hypotenuse is $3\sqrt{2} - \sqrt{6}$

[warm shale]

Oğuzhan

[oak plover]

Because a = 18 and b = 6*

[serene rain]

okay

thx

[oak plover]

You're welcome!

[serene rain]

is that correct tho?

[oak plover]

Let's try to calculate?

[serene rain]

alr

[oak plover]

This gives: 1.79315094434

[oak plover]

This gives: 1.79315094434

Great!

[serene rain]

but like does it work in the problem. hol up

[oak plover]

Also this might seem kind of long. But you could just seen that 18 * 6 = 108 and 18 + 6 = 24

[serene rain]

I gtg now, thanks for all the help

.close

[obtuse pebble]

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[golden current]

Hi everyone! I need verification if my solution for this problem is logical/correct

[golden current]

"Segment RP and segment TP are tangent to circle S and circle W. What is RP?"

First, I marked up the picture using what I know about tangent lines and radii.

Then, I looked at quadrilateral RPTS and quadrilateral QPUW.

I noticed how they are similar to each other by AA~ (both have 90 degree angles and both share angle P)

This means that UP/TP = QP/RP

So first I will solve for UP:

Looking at triangle WPU

(WP)^2 = (WU)^2 + (UP)^2

101^2 = 20^2 + (UP)^2

10,201 = 400 + (UP)^2

9,801 = (UP)^2

Therefore, UP = 99

Also, notice how triangle QPW and triangle UPW are congruent by HL congruency

So this means that QP is congruent to UP

So therefore, QP = 99

[golden current]

So now I can subsitute UP, TP, and QP into this proportion

[quick pine]

Is this help for geometry

[golden current]

Yes

I just need to know if my logic here is correct

for solving

[quick pine]

Hi I need help like understanding this

[golden current]

like what specificallt

specifically

[quick pine]

These problems

[golden current]

ohh this is my channel

if u need help, u have to make ur own channel

#❓how-to-get-help

[quick pine]

Ohhh

[golden current]

99/165 = 99/RP

RP = 99(165)/99

Therefore, RP = 165

So is this solution logical and correct? Please let me know!

Thanks in advance ❤️

<@&286206848099549185>

D:

[obtuse pebble]

@golden current Has your question been resolved?

[obtuse pebble]

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[mild cargo]

Write Viet's formulas for the polynomial x^3 + ax + b and determine its zeros, if one of them is of the second order.

[mild cargo]

i got system of eqs: $\ 2x1 + \ x3 = 0$

$\ x1^2 + \ 2x1x3 = a$

$\ x1x2x3 = -b$

[warm shale]

ghhgjhgjhjgjhjh

[mild cargo]

is it even possible to find roots just from that information?

[wispy wadi]

use your first equation in the second

[mild cargo]

x3 = -2x1?

[wispy wadi]

yes for example

and in last equation, you can sub x2 by x1 since you already assumed one root is of second order (x1 here)

eventually you'll be able to express both x1 and x3 in terms of a and b

and you're done

[obtuse pebble]

@mild cargo Has your question been resolved?

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[obtuse pebble]

@rain meadow Has your question been resolved?

[worldly helm]

what needs to happen for the product to be divisible by 4?

[rain meadow]

4, 8, 12, 16, 20 will be divisible by 4

[worldly helm]

right, but we're picking 3 numbers and multiplying them

[rain meadow]

20 of 5 selection

[worldly helm]

if i pick 2, 6, 7 as my three numbers is the product divisible by 4?

[rain meadow]

Unit place is fix

[rain meadow]

No

Because there is no factor of 4

[worldly helm]

actually it is, since 2.6.7=2.2.3.7=4.3.7

im using . for multiplication there

[rain meadow]

84

It is

Factor of 4

[worldly helm]

yeah

notice what happened, even though none of the numbers were divisible by 4, two of the numbers were divisible by 2

and so the product became divisible by 4

[rain meadow]

Why we are finding divisibility of 2

We want for 4

[worldly helm]

right

but since we are multipliying, two factors of 2 becomes one factor of 4

we want to figure out all the cases that the product might end up as a factor of 4, and then count how many ways there are for the different cases

[rain meadow]

Okay in this case if i want to find divisibility of 6 I can check out with two divisible of 3 will give divisible of 6

[worldly helm]

not quite, to check for divisibility by 6 we check for divisibilty by 2 and 3 since 2 * 3 = 6

[rain meadow]

Oh we needs to multiply

[worldly helm]

lets think about the problem you posted, we're taking 3 numbers from 1 to 20 and then multiplying them together

if one of the numbers we picked was divisible by 4, then the product definitely will be too

so like if we choose 8, 9, 13

then since 8 is divisible by 4, so is the product 8 times 9 times 13

[rain meadow]

936 is divisible

[worldly helm]

another way the product could be divisible by 4 is if two of the numbers we pick are divisible by 2

[rain meadow]

Yes when we took two numbers divisible by 2 is also divisible by 4

[worldly helm]

but lets think about the first case for now (where one of the numbers we picked is already divisible by 4), and try to figure out how many ways there are to choose 3 numbers that have a product divisible by 4 in that case

[rain meadow]

5

!

[worldly helm]

thats the first step, but we need to keep in mind that we are also picking two more numbers and theres lots of ways to do that

[rain meadow]

We have 20 fir first digit and 20 for second

[worldly helm]

remember that the numbers also have to be distinct

[rain meadow]

Ok

Then 20, 19

[worldly helm]

well we picked one number already, that is one of the 5 numbers divisible by 4

so actually it will 5 choices, then 19 choices, then 18 choices

[rain meadow]

15 , 14

When we already chose 5, we remain with 15

?

[worldly helm]

there are 20 numbers to begin, and we pick one of those 20 that is divisible by 4 (there are 5 of them), then pick two numbers from the 19 remaining (so 19 choices, and then 18 choices)

[rain meadow]

Ohhhh ok that's 5!, 19!, 18!

[mild cargo]

There are 20⋅19⋅18 / 3! =1140
combinations of numbers to choose from

[rain meadow]

This is for choosing three digit number

[mild cargo]

yes wait

[rain meadow]

We want divisible by 4 also when multiply

[worldly helm]

so for this first case it should 5.19.18/3!

dividing by 3! since order doesn't matter

[mild cargo]

The product P of the three numbers is not divisible by 4 if 2 ∤ P, or 2 ∣ P and 4 ∤ P

first case arises out of choosing all three numbers from the set {1,3,5,…,19}; there are (10, 3)=120 such selections.

second case arises out of choosing two numbers from the set {1,3,5,…,19} and one from the set {2,6,10,14,18}; there are (10, 2)⋅(5, 1)=225 selections.

so n.o divisible = total combinations - case 1 - case 2

[rain meadow]

What '∤' symbol means

[mild cargo]

not divisible

does not diivide*** this case

[rain meadow]

Okay

[rain meadow]

Why we choose 19 instead 20

[mild cargo]

even number

cannot be divisible by 4

sorry odd**

1,3,5..

[rain meadow]

Oh

Odd numbers are not divisible by 4

120 way

[mild cargo]

#include <stdio.h>

int main(){
int a, b, c, t = 0;
for(a = 1; a<=20; a++)
for(b = a + 1; b<=20; b++)
for(c = b + 1; c<=20; c++)
if(!((abc) % 4)) t++;
printf("%d\n", t);
} try this

[rain meadow]

We have to remove these odd

[mild cargo]

yep, total number is 20 * 19 * 18

/ 6

so, the product of three numbers a, b, c is not divisible by 4 if: it is not divisible by 2 (odd* numbers), or it is divisible by 2 BUT not by 4.

1. it can be {1,3,5,7...}

[rain meadow]

See there are 10 odd and 10 even

[mild cargo]

2. pick TWO odd from set {1,3,5,7...19} and ONE from even but not divisible by 4 {2, 6, 10, 14, 18}

that means their product is divisible by 2 but not 4

[rain meadow]

Then product will not divisible by 4

[mild cargo]

yes, we solve for total number of numbers that arent divisible by 4

and subtract

total combinations - ways that arent divisible by 4

[rain meadow]

Select 2 from 10 odd number

And 1 from 5 even number

[mild cargo]

yes

combination 10,2 * combination 5, 1

[rain meadow]

10×9/2

Plus 5

[mild cargo]

• 5

**

[rain meadow]

Multiply

[mild cargo]

cause for every TWO numbers you pick first choice

i mean every combination

there is also 5 different number you can choose for third

so *

[rain meadow]

We need to multiply all these or add

[mild cargo]

45 * 5

[rain meadow]

225

[mild cargo]

yes

and for first choice

[rain meadow]

225 is not divisible

[mild cargo]

yes yes

[rain meadow]

By 4

[mild cargo]

and if you ahve

3 odd numbers they wont be divisible aswell

so c(10,3)

10! / 3! 7!

120

[rain meadow]

Ok from 10 odd number we select 3

[mild cargo]

these are combinations that AREN'T divisible by 4

we get the total of combinations from 3 digits

[rain meadow]

225+120

[mild cargo]

20 * 19 * 18

yes

now, we SOLVE for cases that AREN'T divisible by 4

so then we can subtract the total with the cases we solved

[rain meadow]

325 way not divisible by 4

[mild cargo]

and we get the remaining, which is divisible by 4

yes, but we're solving for ways to arrange 3 numbers from a set of 20 numbers {1,2,...,20}

that means that 325 ways

we dont get divisilbe by 4

[rain meadow]

1140-325

[mild cargo]

and the other (total - 325)

is divisible by 4

good luck.

[rain meadow]

815

[mild cargo]

225 + 120 = 445

345

'

[rain meadow]

Oh sorry

[mild cargo]

np did you get it now?

[rain meadow]

Oh yes it's right

Thank

[mild cargo]

np

[rain meadow]

You

. close

.close

[obtuse pebble]

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[wind hedge]

I'm having trouble figuring out which equations to use for this optimization problem: "Given a square sheet of side a, you want to build a box with it without a lid, cutting equal square corners at its corners and conveniently folding the remaining part. Determine the side of the squares that must be cut so that the volume of the box is the greatest possible."

[frigid prism]

what equations do you think you could use

there are only so many

how do you optimize something

broo

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[frigid prism]

you do all that work but then dont read the server rules

<@&268886789983436800> chatgpt i think

they responded "I hope that helps" oml

[wind hedge]

lol.

wellz, i still got lost cuz he sent two different responses so.

[frigid prism]

read my original response

[wind hedge]

yes yes.

i'm using the V = x * (a-2x)^2

but i don't know which other to use

[frigid prism]

thats it

you dont need anything else

if you had another equation you could literally find the number value for x or a

[wind hedge]

but, don't i need two equations for these types of problems?

[frigid prism]

but obviously that shouldnt be possible

you are finding x in terms of a

you dont need to find x or a

you cant

[wind hedge]

and here i thought i needed to find the actual value of x

[frigid prism]

then you would break math

that would mean:

if I give you a square paper of size anything you would give me the same number to use to get the max volume

[wind hedge]

ahhhh. So that's why i just leave it in terms of x in terms of a.

it's about finding a general equation

[frigid prism]

yes

[wind hedge]

then i just need to find the derivative of V and get x in terms of a

thank you a lot^^

[tough ingot]

Hope this helps but I have a list of steps to solve optimization problems:

1. Draw a picture to represent the situation
2. Assign variables and be specific
3. Use the given information to create a function of a one variable whose output is what you're trying to optimize
4. Find the applied interval for the optimization function
5. Find the absolute maximum and minimum to determine the optimal situation including endpoints
6. State your conclusion in a full sentence with units

Use that for future reference

[obtuse pebble]

@wind hedge Has your question been resolved?

[obtuse pebble]

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[tough hawk]

Hello

[tough hawk]

Are these correct? The ranges and domains?

[obtuse pebble]

@tough hawk Has your question been resolved?

[tough hawk]

Toggaf

.close

[obtuse pebble]

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[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[whole yoke]

hi! i’m doing a worksheet on arithmetic and geometric sequences, and i’m stuck on this problem. i think it involves logs (?), but that’s not my strongest suit :/

[ancient crystal]

https://youtu.be/sULa9Lc4pck

[sage geode]

Firstly you need to spot a pattern, do you see it?

[ancient crystal]

Yeah, the pattern is there, x3. So, three to the what is 4000. (You're right, it's logs).

[sage geode]

Right, so you need to solve 3^n being equal to that divided by the initial term

[ancient crystal]

^smrt

(^smrt meaning the comment above is smart, to the power of smart, and is a Simpson's reference)

[obtuse pebble]

@whole yoke Has your question been resolved?

[whole yoke]

oh wait

[whole yoke]

right now i’m at log400,000 = (n-1)log(3) but i’m not sure what to do next

oh wait it’s obvious 😭

thank you guys so much for your help! i got it

[obtuse pebble]

@whole yoke Has your question been resolved?

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

how do i find the lenght of that curve

r is a vector function equation parametric

0 <= t <= 1

[jolly ginkgo]

Find r'(t) then integrate norm{r'(t)} wrt t

[sinful kraken]

i got this equation but idk how to do the rest

[wooden cipher]

Hint: try making a common denominator

[jolly ginkgo]

Are you sure that's what you got

[sinful kraken]

yes

thats r'(t)

[jolly ginkgo]

Uh?

Where did the 2 come from

[sinful kraken]

sqrt(2)^2=2

[jolly ginkgo]

Ok

[sinful kraken]

wdym ?

[jolly ginkgo]

e^(-2t) = 1/e^(2t)

So it's a perfect square

[sinful kraken]

?

[jolly ginkgo]

Perfect square buddie

[sinful kraken]

whats that

[jolly ginkgo]

(Something)²

[sinful kraken]

wtf

[jolly ginkgo]

What is (a+b)²

[sinful kraken]

a^2+2ab+b^2

[jolly ginkgo]

Oh wait the minus

[sinful kraken]

this ?

[jolly ginkgo]

If the - was + it would've been perfect

[sinful kraken]

how ?

anyways

how do i solve it

[jolly ginkgo]

Leave it we don't get the (a+b)² form

[sinful kraken]

answer is e-e^(-1)

btw

[jolly ginkgo]

Is this right?

[sinful kraken]

pretty sure

[jolly ginkgo]

Don't be

[sinful kraken]

derivateive of sqrt(2)t is sqrt(2)

then e^t is e^t

[jolly ginkgo]

,w norm(sqrt(2) , e^t , -e^(-t))

See

[sinful kraken]

?

[jolly ginkgo]

The signs

You did a sign mistake

[sinful kraken]

hm

how ?

derivative of e^-t is -e^-t

ooooooooooooooooooo

[warm shale]

Results provided by WolframAlpha

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

Requested by yes sam#5028

[sinful kraken]

i understand now

then what we do

[jolly ginkgo]

You know what (a+b)² is

[sinful kraken]

yea

[jolly ginkgo]

Then do it

[sinful kraken]

o_o

[jolly ginkgo]

Do it adam

[sinful kraken]

my name isnt adam but ok ill do it stay nearby

[jolly ginkgo]

If you know (a+b)² that means you know how to do maths, so you can solve it

@sinful kraken is it done?

[sinful kraken]

yea

but its weird

i got

[jolly ginkgo]

What?

[sinful kraken]

this inside the integral

[jolly ginkgo]

That's not what you get

[sinful kraken]

i meant a + instead of a *

[jolly ginkgo]

So why is that weird

[sinful kraken]

because do the intergal

you get

e^1+ e^-1 -2

[jolly ginkgo]

I get e - (1/e)

[sinful kraken]

wait nvm

im dumb

true

[jolly ginkgo]

Yeah

What did you do after square root

What is sqrt(x²) btw

[sinful kraken]

x

+- x

[jolly ginkgo]

Give me the right answer

[sinful kraken]

+- x

not x

[jolly ginkgo]

No

[sinful kraken]

how did u get that

the asnwer is not that btw

[jolly ginkgo]

After integration

[sinful kraken]

answer is e^1-e^-1

[jolly ginkgo]

Adam can you see they are equivalent

If yes then I want you to say yes

[sinful kraken]

yes

i have 3 brain cells remaining

[jolly ginkgo]

That question wasn't for you

[sinful kraken]

?

[jolly ginkgo]

I didn't ask that to you

[sinful kraken]

then to who ?

[jolly ginkgo]

Adam

[sinful kraken]

who is adam ?

[jolly ginkgo]

Idk, whoever it is

[sinful kraken]

wtf

[jolly ginkgo]

Any Adam works

Anyway

Show me your work

[sinful kraken]

ugh

[jolly ginkgo]

You want me to write that integration?

[sinful kraken]

no

ok il lsend pic but its annoying

wait im doing the intergation

ok works

good ?

[jolly ginkgo]

What did you do about sqrt()

[sinful kraken]

i removed ^2

thanks @jolly ginkgo

.close

[obtuse pebble]

Channel closed

Closed by @sinful kraken

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Available help channel!

Send your question here to claim the channel.

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

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

[timid silo]

not suer what to do

any help is needed

[jolly ginkgo]

Not that way

You have to keep mod

[timid silo]

?

@sinful krakenhelp?

[sinful kraken]

i was the previous channel owner faf is talking about my question

[timid silo]

@jolly ginkgocould you help me

[timid silo]

ok

[jolly ginkgo]

Not free rn

[sinful kraken]

do a graph money in y year in x

[timid silo]

:C

[sinful kraken]

then graph 2 lines

one if he smokes one if he doesnt

[timid silo]

could you do it and show me?

[sinful kraken]

no

[timid silo]

o

[sinful kraken]

sorry

[timid silo]

ok