help-17

[wary mantle]

This isn't really trig

[ruby vapor]

If you have two lines intersect then the opposite angles at the intersection points are of equal measure, it looks a bit more complicated because we have three angles, but it still applies.

[wary mantle]

I assume you are German, what Don describes are called "Scheitelwinkel"

[placid pier]

verstehe immer leider noch nicht wie ich das ausrechnen muss entschuldige

[placid pier]

aha habs

danke

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[fleet quiver]

if I have $\lambda_1 X_{n\cross m} A_{m\cross m} + \lambda_2 X_{n\cross m} I_{m\cross m} = 2X_{n\cross m} B_{m \cross m}$ with $n<m$, does that imply wlog $\alpha_1 A + \alpha_2 B = I$?

[twin meteor]

QuAnTuM

[hasty pulsar]

wlog?

[fleet quiver]

without loss of generality

im asking whether all the solutions for A and B to the second equation satisfy the first

[vocal sleet]

@fleet quiver Has your question been resolved?

[young blaze]

To critically engage with the problem, one should pay particular attention to the matrix equation (\mathbf{X} (\lambda_1 \mathbf{A} - 2 \mathbf{B} + \lambda_2 \mathbf{I}) = \mathbf{0}_{n \times m}) derived from the given equation. Consider the implications of (n < m) for the row-rank of (\mathbf{X}) and consequently for the matrix ( \lambda_1 \mathbf{A} - 2 \mathbf{B} + \lambda_2 \mathbf{I} ). This will allow you to characterize the properties of this matrix in relation to its singularity. Once the matrix's properties are delineated, you will be in a better position to consider whether or not such a matrix condition implies the existence of scalars ( \alpha_1 ) and ( \alpha_2 ) satisfying ( \alpha_1 \mathbf{A} + \alpha_2 \mathbf{B} = \mathbf{I} ).

[twin meteor]

adzetto

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[native kraken]

Find the length (in cm) of an arc of a circle with radius 6 cm if the arc subtends a central angle of
60°.

is it not 6.28?

[mild flower]

it is yes

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[waxen hinge]

I have a question about base changement with matrix

[waxen hinge]

i don't get why it's not A' = PAQ^-1

it seems more logical to first change the base with P then applying the function finally come back with the starting base (applying Q^-1)

[boreal remnant]

Yes

[round plover]

remember matrix multiplication is read right to left

Q^-1 A P is applying P first

[boreal remnant]

Matrix compositions are applied right to left

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@waxen hinge Has your question been resolved?

[waxen hinge]

ye i forgot it's right to left thx

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[full vessel]

I could provide a specific question if wanted but i just had a general quesiton abt composite functions: why do we have to find the domain of the inner(?) function when we're finding the domain of a composite function? wouldnt it makes sence to find all 3: the outer and the inner and the composite and then find the most restrictive for all 3?

[full vessel]

why does it make sense to only find the inner one and the total domain

[paper depot]

think of functions as machines that take numbers as input and return numbers as output

a composite function is made of two functions in which the output of one is fed directly as input into the other

the "inner" function is the one which sees the input from the outside world

so naturally said input must belong to the input function's domain

and the output of the inner function, which is the only thing that the outer function sees, must belong to the outer function's domain

[full vessel]

oh woww

[full vessel]

so just confirming; there will never be a time when the output from the inner function's domain will not "fit" in the outer function's domain?

[paper depot]

well... no

like if you compose g(x) = sqrt(x) with f(x) = 1 - x^2 with f as the inner

you get sqrt(1 - x^2)

and x=10 works fine with the inner f but you run into issues with the outer g because f(10) doesn't belong to its domain

[brisk moss]

f(1)?

[full vessel]

^^yeah i was confused abt that too

[paper depot]

typo sorry

had fixed it in one place but not the other

[full vessel]

bc its negative ?

and square roots n negatives dont work

[paper depot]

yes

the domain of the square root function is [0, +∞)

[full vessel]

so like in this case bc the outer funciton is more restrictive why wouldnt we find the domain of the outer function and the total domain to find the domain of the composite?

[paper depot]

truth be told idk what you mean by "total domain"

[full vessel]

oh the domain of the compoisite

yea i wored that confusingly

so like the domain of the inner and the domain of the composite and the actual domain of the composite is the overlap of those 2 right

[vocal sleet]

@full vessel Has your question been resolved?

[full vessel]

@paper depot ? (sorry to ping)

[paper depot]

8huedhgsjhj

sorry you managed to confuse me outright

[full vessel]

sadfa;sdlfjka;sd

OK

JUS

A:LWAYS WHEN FINDING DOMAIN OF COMPOSTIE FUNCION

WE FIND DOMAIN OF INNER FIRST THEN DOMAIN OF COMPOSITE THEN OVERLAP AND THATS OUR ANSWER RIGHT

WHO CARES ABT WHY

AS LONG AS THOSE RULES DONT CHANGE I DONT GOTTA KNOW WHY

RIGHT...??

[vast shale]

I think @full vessel

It's better for you to understand how mapping for a single function works

Like what does $\map f x$ even mean to you for example

[full vessel]

it means that when I enter a value x into a function f it gives me an output

[vast shale]

Right

So think of it this way

<tikz>
\begin{tikzpicture}
  \node (x) {$x$};
  \node[right=of x] (gx) {$g(x)$};
  \node[right=of gx] (f_gx) {$f(g(x))$};

  \draw[->] (x) -- (gx) node[midway, above] {$g$};
  \draw[->] (gx) -- (f_gx) node[midway, above] {$f$};
\end{tikzpicture}

[full vessel]

wthhh

thats crazy

where do u learn that stuff 😭

[full vessel]

yess i understand

[vast shale]

You read a 1.5k page manual

💀

[full vessel]

🥹

[vast shale]

But anyways

So you see how we are using g(x) for f(g(x)) the same way we had used x for g(x)

[full vessel]

yeaap

[vast shale]

So

Think of it this way

"You have an x value, and you map (or transform) that x value to g(x) under the transformation (or function) g"

Now the same exact process repeats

"You have a g(x) value, and you map (or transform) that g(x) value to f(g(x)) under the transformation (or function) f

[full vessel]

yes yes

i understanaaaanndd but how does that relate to why i have to find the domain of the inner functionnnnnnnn

[vast shale]

Because

Let's consider this example

Let's say we have

[
\map g x = -2x \tss{and} \map h x = \s x
]

[vast shale]

Can you tell me the domain of both of those

[full vessel]

domain of g is all real numbers and domain of h is x > or equal to 0

[vast shale]

Right

Now if we compose them such as h(g(x)) what do we have now

[full vessel]

sqrt(-2x)

[vast shale]

Right, that function on its own, what would it's domain be?

[full vessel]

x greater than or equal to 0

[vast shale]

You sure?

[full vessel]

eerrrrrrm

yee

[vast shale]

If you plug in x = 1 is that correct

if x is greater than 0 then theres negative in sqrt

[full vessel]

x less than or equal to 0

bc the sign flipped bc divide by negative 2 math makes me so sad

[vast shale]

Ye

[full vessel]

YE

[vast shale]

YEEE

What Ann was talking about earlier applied here

h(x) on its own was x greater or equal to 9, but under the composition with g(x), it changed to x Lesser or equal to 0

What you're doing here is making a totally new function

[full vessel]

yes

[vast shale]

f(g(x)) is different from both f(x) and g(x) obviously. But g(x), as the input of f, needs to satisfy the domain of f too

[full vessel]

yeees

when u say needs to

u mean the goal is for the answer to satisfy the domain of f too right

not that g(x) always satisfies f

[vast shale]

I mean

Yes I think I hope I didn't misread what you said

[full vessel]

bc a while back ann gave an example of how even though a number 10 worked for an inner function, it wouldnt work for an outer function ( a square root function(?) bc it turned the value negative

[vast shale]

Yeah I mean

Again

You are considering two totally different functions

[vast shale]

Here for example

[full vessel]

yess

[vast shale]

g(2) =-4, h(g(2) = undefined

[full vessel]

yeah

[vast shale]

Just take this as a takeaway:

Functions have inputs, the set of inputs is called the domain. Any input you have for a function must be in its domain, otherwise it is invalid

Composing a function with another requires the inner function to be in the domain of the outer function. Illustrated as g(x) needs to be in the domain of h(x) in the composition of h(g(x))

[full vessel]

alright :))

thank you smm@!

genuinely ty for being so patient w me

ygs should be paid for this its crazy how yall arnt

byee

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[vast shale]

Bai

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[vast shale]

Oh bruh

.close

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[vast shale]

Also the bot is showing my display name instead of my tag now

[river minnow]

$a$

[twin meteor]

A Lonely Bean

[river minnow]

I think it shows display name by default

[mild flower]

[vast shale]

no I am talking about

The help channel name

It was always lexqa which is my tag name

[river minnow]

Ah that you mean

[vast shale]

Now it changed to my display name

[mild flower]

i think that was just fixed today

[vast shale]

Darn

Thanks mniip u a real one fr 😎👍

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[trail shoal]

In the isosceles trapezoid ABCD where the diagonals cross the acute base angles given: AB = c, ADC = a
Given that the perimeter of the trapezoid is 5C find the angle a

[trail shoal]

I proved that AB = c

[merry python]

AB=c or AD=c?

[trail shoal]

both

[merry python]

Okay, so you'll have AD+AB+BC+OC+LD+LO=5C

and you have all of them in terms of c

[trail shoal]

Ya

[merry python]

So, just solve for cos(alpha)

[trail shoal]

wdym

[merry python]

You have,
C+C+C+C+Ccos(a)+Ccos(a)=5C

[trail shoal]

yes so c = 2cos(a)

[merry python]

NO

2Ccos(a)=C

[trail shoal]

whats the difference?

[merry python]

2C*cos(a)=C

not 2cos(a)=C

[trail shoal]

ok

but how do I find a?

I dont know what c is

[merry python]

just cancel it

divide both sides by C

cos(a)=1/2

[trail shoal]

Ok got it

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[elfin moon]

any short trick

[chrome raptor]

Use the double-angle formula

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@elfin moon Has your question been resolved?

[viral copper]

Shortest trick is letting a = b = c = 60 and see which option matches

[elfin moon]

Let me try it

I was trying with 90°

[viral copper]

nah 90, might cause a problem because it makes cos go 0

[elfin moon]

True

It works fine

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[fallow cloak]

can someone help me with this

i know i need to use one of the equations

i just dont know which one

[vocal sleet]

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.

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[simple mason]

Just look at this

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[simple mason]

Oops

.close

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[finite hatch]

hi!

[finite hatch]

can someone help me interpret this question

why is it 720 degrees

when in a circle theres only 360

[frigid root]

that's the point

[finite hatch]

what

[frigid root]

one circle have 360 degree, when the question goes beyond that, it's like you continuing doing laps around it

[finite hatch]

oh

so is the answer

4

[frigid root]

ye

[finite hatch]

oh...

ok thanks

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[gleaming geyser]

How do I solve further?!

[bright yew]

you've reached till here right $\frac{a\cdot\sqrt{x}}{\sqrt{x} + \sqrt{x - a}}$

[twin meteor]

nebula40

[bright yew]

try dividing the numerator and denominator by sqrt(x). I can't read the last part clearly, but if that's what you've done there you're on the right track

[gleaming geyser]

Ok wait I will try more clearly

I got the above one can u give me a hint on 7 no.

[bright yew]

try drawing/graphing it

[gleaming geyser]

Yeah but we usually don't draw graphs for such questions in our cls

[bright yew]

you don't *need * to graph it to get the answer, but it makes it very obvious for this case. and making a small sketch for a small range like (-5, 5) doesn't take any effort at all

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[dapper patio]

how is the formula simplified? what are the steps?

[dapper patio]

from step 2 to 3 basically

[soft walrus]

$\frac{\frac{(x+h)^2}{10}-\frac{x^2}{10}}{h}=\frac{\frac{x^2+2xh+h^2}{10}-\frac{x^2}{10}}{h}=\frac{\frac{\cancel{x^2}+2xh+h^2-\cancel{x^2}}{10}}{h}=\frac{\frac{2x\cancel{h}+\cancel{h}^2}{10}}{\cancel{h}}=\frac{2x+h}{10}=\frac{h}{10}+\frac{x}{5}$ :)

uh oh

[twin meteor]

MrFancy

[dapper patio]

thank you!

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[soft walrus]

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[high bison]

yo i just need someone to check if my answer is right

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@high bison Has your question been resolved?

[high bison]

<@&286206848099549185>

[viral pivot]

Not quite. Try looking at the third line (where you put in 2x into the most inner g(x))

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[mild narwhal]

I’m confused about how to draw the graph does it start at 2160 then decrease at 32 students per year for 8 years or is 2160 the currently enrolled number

[frigid root]

I think that 2160 is the current number

[mild narwhal]

Ok thanks would you be able to check another question?

[frigid root]

sure

[mild narwhal]

I got one of these questions wrong

Would number 3 be D

[frigid root]

i'll see, one moment

what the difference between the step D of 1 and the answer D of 3?

[mild narwhal]

On D of 1 it is subtracting the 6 when it is supposed to be adding it

Since it moved the -6 over

[frigid root]

only that changed?

[mild narwhal]

Ya

[frigid root]

sure?

check again

[mild narwhal]

I’m sure they are the exact same except for the symbol

3 C not d my bad

I said D but meant C

[frigid root]

hihi

your right

C is the answer

[mild narwhal]

Sorry for the confusion

I’ll try that

[frigid root]

np

if need, just send a mensage

or ping

[mild narwhal]

Thank you it worked have a nice day 😄

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[tame smelt]

Hello, I solved till I got the 2 roots and I got the line of symmetry but I am stuck now

x = -43 or x = 34

Line of symmetry = -4.5

How can i find b and c

[flat whale]

Show your work?

[tame smelt]

Any help?

[mild flower]

if it has the same roots it has the same line of symmetry

that should be enough to find b

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[tame smelt]

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[abstract sedge]

given 4 variables can someone calculate the other 4 ?

if yes please help me

names have no relation to the type of variable

[mild flower]

Here's one of them

X = (Left / 100) + (Width / 2)
108 = (Left / 100) + (75 / 2)
108 = (Left / 100) + 37.5
70.5 = Left / 100
7050 = Left

[abstract sedge]

yup i got it now

its kinda easy

lol

[mild flower]

it's algebra yeah

[abstract sedge]

got it

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[dim sail]

How do I do these kinds of problems

[dim sail]

Or set it up

[dull maple]

Do you know how to calculate slope of a line given two points on it?

[dim sail]

Yea but I struggle

[dull maple]

Umm... where exactly do you struggle?

[dim sail]

Like you do the math and get fractions right

[dull maple]

Well, not always, but yes.

[dim sail]

Yea I don’t know how to graph them

[dull maple]

Why do you have to graph it?

You are asking the ones highlighted in yellow. Right?

[dim sail]

Yea

[dull maple]

Yeah. You don't need to graph them to answer this question.

You sure could but you don't need to.

[dim sail]

To find out either the line are parallel I thought we’d have to graph it

[dull maple]

Umm... wait. You haven't been taught about gradient relationship of the parallel lines?

[dim sail]

Maybe we have im not sure I’m not good at math

[dull maple]

Well, let me tell you.

Two lines are parallel if and only if their gradient is same.

[mild flower]

(gradient = slope)

[dim sail]

Ohh

[dull maple]

I think their assignment said gradient so i went with that.

[dull maple]

Yes.

[dim sail]

But we don’t have to graph it to find the slope

[dull maple]

So, you just calculate gradient for both of the given lines. If they are same, they are parallel, or else, they aren't.

[dull maple]

Yes. I believe you already know how to calculate gradient/slope of a line if you are given two points on it.

[dim sail]

So like if both equations come out equal

[dull maple]

Where are you getting equations from?

[dim sail]

For like number 18 I though I’d have to do
3 - 1
-4 +5

[dull maple]

What? No.

[dim sail]

Oh

[dull maple]

I am not sure anymore if you know the formula. Wait.

[dim sail]

[twin meteor]

Enemagneto

[dull maple]

To apply the formula, you should be taking both the points which are on the line.

Like, you have for line 1: (3, -1) and (6, -4)

[twin meteor]

Enemagneto

[dull maple]

Now, calculate for line 2.

If that comes out to be -1 as well, they are parallel, else they aren't.

@dim sail

[dim sail]

Where did the -1-(-4) come from

[dull maple]

Read from here until the end.

[dull maple]

Read the formula

[dim sail]

OOO

How do I know if they’re perpendicular

[dull maple]

For perpendicular, we have this: Two lines are perpendicular if and only if product of their slopes is -1.

That is if a line has slope $m_{1}$ and another line has slope $m_{2}$, then they are perpendicular iff $m_{1}\cdot m_{2} = -1$.

[twin meteor]

Enemagneto

[dim sail]

One more question about this why did it change to a plus and where did the 4 come from

[dull maple]

Because negative multiplied with negative is positive.

You have -(-4), which is +4.

[dim sail]

Wouldn’t 4+(-6) be -2

[dull maple]

Yes

[dull maple]

Where are you getting those though?

[dim sail]

Wait nvm I get it lol

Omg it just clicked in not subtracting down I’m subtracting straight across

Ty for ur help I appreciate it:❤️:

[vocal sleet]

@dim sail Has your question been resolved?

[vocal sleet]

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[marble anchor]

need assistance

Its 0/0 obv

How would I go about factoring this?

[paper depot]

might want to add the fractions on the top first

[marble anchor]

gives 0

Adding the fractions (plugging in -6) makes it 0

[mild flower]

add them before plugging it in, make it all one big fraction instead of this mess of complex fraactions

[radiant spear]

it should give
(6+t)/6t(6+t)

first simplify algebraically

do you know the next step?

[marble anchor]

Sorry just looking at this

Let me look

[marble anchor]

How would I add something before plugging it in?

Would it become

1/t + 1t/6t?

[radiant spear]

don't add it yet, just treat it as an unkown

[marble anchor]

Treat the variable as unknown?

[radiant spear]

forget lim t—>-6 for a while and simplify the expression, so that you won't have t+6 in the denominator.

the problem is that t+6 in the denominator causes a problem (by turning into 0) so you have to get rid of this problem

[marble anchor]

mhm

so we need to make tht top fraction have the same denom right?

[radiant spear]

just something else. this fraction can be simplified to 1/6t

[marble anchor]

could i see the steps,

or rather ill take it out

talk*

[radiant spear]

ok, let me get my book so I can show you something at least

[marble anchor]

common denom would be 6t

Which would i multiply by 6t?

Rather why is the numberator still 1?

[mild flower]

common denominator would be 6t yes

[marble anchor]

shouldnt it be 6t/6t

[mild flower]

so if we convert each of those fractions to /6t, starting with the one on the left

$\frac1t = \frac{?}{6t}$

[twin meteor]

hayley!

[marble anchor]

should be 1t right there

Oh

hold on hold on

Why did you flip here

also how did you combine

[radiant spear]

the 6t goes below because it was being divided in the numerator. anything being divided in the numeratoe can shift to the denominator

[marble anchor]

oh ok

Ok i get it

thank you

Never knew everything could just go to the dom like that

[radiant spear]

i think changing t+6 to 6+t in step 2 to 3 caused the confusion

[marble anchor]

it did lol

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[glacial meadow]

how do i make an area model for this?

[vocal sleet]

@glacial meadow Has your question been resolved?

[zinc quail]

You make a 1x1 square

and divide it into two parts I presume

each are is equal to its probability

so splitting in half is already enough

@glacial meadow

[glacial meadow]

uh

ok so

how would that look??

[glacial meadow]

ive never seen an area model so small

[zinc quail]

I thought you always make a 1x1 square

such that the entire area is 1

isn't the purpose of it to let area and probability be equal

in this case each option has a probability of 0.5

so just split it into left & right

0.5x1 and 0.5x1

one half blue, the other red

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[blissful lark]

need help factoring a polynomial. the question is "Factor the following polynomial using the negative of the greatest common factor.
-27x^2 + 63"

[lime gorge]

What’s the gcf here

[blissful lark]

3

[lime gorge]

Nope

Check more values

[blissful lark]

oh its 9

[lime gorge]

Yep

So what’s the negative of 9

[blissful lark]

-9

[lime gorge]

Good

Now factor that out

[blissful lark]

i got 3x^2-7

[lime gorge]

Yep

-9(3x^2 - 7)

[blissful lark]

thats it

?

[lime gorge]

Ye

[blissful lark]

ok thanks!

[lime gorge]

Np

[blissful lark]

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[vital lark]

What does it mean to rationalize the denominator?

[round plover]

manipulate the fraction such that the denominator is a rational number

for instance, the denominator of the first fraction is sqrt(2), which is not rational

[vital lark]

How can I change that?

[round plover]

multiplying the numerator and denominator by the same thing doesn't change the value of the fraction

find something to multiply sqrt(2) by such that you get a rational number

[vital lark]

So the answer of 49 is rad2 over 2

[round plover]

yes

[vital lark]

Ok sounds good

Thanks

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[agile bramble]

Prove that n^2<3^n for every n is part of {0,1,2,3,...} using mathematical induction

[oblique wolf]

f

[agile bramble]

Mmm?

[round plover]

what have you done so far

[agile bramble]

Did the base case, induction hypothesis and started the induction step with

We wish to show that (k+1)^2<3^(k+1)

[steel walrus]

what inequality do you already have ?

[agile bramble]

k^2<3^k

[steel walrus]

is it simpler to transform k^2 into (k+1)^2, or to transform 3^k into 3^(k+1)?

[agile bramble]

3^(k+1)

[steel walrus]

how do you transform 3^k into 3^(k+1) ?

[agile bramble]

Multiply by 3 so 3*3^k=3^(k+1)

[steel walrus]

now modify the inequality k^2 < 3^k so there's a 3^(k+1) !

[agile bramble]

3k^2<3^(k+1)

[steel walrus]

to prove the desired inequality (k+1)^2 < 3^(k+1), is it enough to prove that (k+1)^2 < 3k^2 for some k ?

[agile bramble]

Yes

[steel walrus]

(in mathematical induction, you can start the induction at any k that you want so long as the cases beforewards are verified)

what do you get when you solve for k in (k+1)^2 < 3k^2 ?

[agile bramble]

2k^2-2k-1>0

[steel walrus]

it's a quadratic inequality, so you can solve it

also you got the direction flipped around

[agile bramble]

Well from looking at it, k>or=3

[steel walrus]

well tbh k >= 2 works as well

it just doesnt work for k = 0 and 1

[agile bramble]

Oh I see

[steel walrus]

so you only need to check the base cases 0, 1, 2 (because you need the inequality to hold if you want to use the induction principle)

after you verify that the base case n = 2 holds, the inductive argument will fill in the rest of the cases

[agile bramble]

Wait I just want to make sure I get what's happening. So we get to 2k^2-2k-1>0 and we realize that it works for k>=2. Because it doesn't work for k=0 and k=1, we just have to plug in those cases normally to the original equation. We don't have to plug in the other cases because those are covered by the inductive arguement since 2k^2-2k-1>0 works for k>=2.

Am I understanding that correctly?

[steel walrus]

you have to check the base cases k = 0 and 1 because they cannot be a part of the inductive argument

but you also have to check the base case k = 2 because that one will be the actual basis of your induction

[agile bramble]

But don't we already check it with k>=2?

[steel walrus]

k >= 2 only checks that [if the claim holds for n = k, then it necessarily holds for n = k+1]

you have to kickstart the engine with k = 2

basically you need to check if really n^2 < 3^n for n = 2

[agile bramble]

Oh I see now

Ty

[vocal sleet]

@agile bramble Has your question been resolved?

[vocal sleet]

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[frank geode]

Hello

[frank geode]

I need help with this

I am wondering why we add 2x <@&286206848099549185>

[daring dragon]

bc there are 2 xs on each side

[frank geode]

it does it for this to tho

[daring dragon]

they just replaced the ? with x

to make it easy

[frank geode]

So for every area problem its just gonna be 2x added

[daring dragon]

if there are those extra lengths on each side and they are all identical

yes

[frank geode]

alright thanks

[daring dragon]

np

[frank geode]

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[finite surge]

hii im writing my notes and i have a question, how do i write (-2+-59) properly? since there's a plus sign and a negative in the middle thank you

[misty belfry]

do you want to do -2-59

[finite surge]

im asking how to write (-2+-59) properly hehe, it looks wrong thats why bc of the two signs in the middle

or is it just how it is? 😭

[misty belfry]

there should only be one sign

[finite surge]

how do i write it?

[misty belfry]

you probably want to write (-2-59)

[finite surge]

ohh but what about the add sign ? 😭

[misty belfry]

it's implied

you don't have to write it

[finite surge]

OHH

THANK UU

[misty belfry]

👍

[finite surge]

.close

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[vast shale]

Halp

[vast shale]

Help me to know: if I did em correctly, and if my steps to find if its R or IR were wrong or right, and help me solve the ones where I put a “?” And yea

Ping me

[tough valve]

dot means recurring?@vast shale

[vast shale]

What dot

@tough valve

[tough valve]

Above Nos.

In part i

[vast shale]

Yess

[spiral turtle]

@vast shale your answers for the first one look good to me. For the ? one you have 0.111111.... hint is there a fraction that equals this (further hint, what if you consider the multiples of it, 0.2222..., 0.3333...., and so on, do you get to 1? How many times do you need to add it to itself?)

For the second one, also looks good to me, for sqrt(2) + sqrt(3), it's a little tricky to actually prove, but it is irrational.

Consider a^2. If a is rational, then a^2 is also rational. So consider (√2 + √3)^2 = 5 + 2√6. We know that this number is irrational because it's the sum of a rational and an irrational. Therefore, √2 + √3 cannot be rational

(proof by contrapositive)

(If p then q implies if not q then not p.)

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

@spiral turtle so the i part or Q1 is “0.11” ? Or is it 0.111..? If its 0.111.. den its rational and, sqrt 2 + sqrt 3 is irrational as they cant be added + rational nos are simple so ye, and are my steps right for Q2, part c and f?

[spiral turtle]

consider log base 10. log(2) is irrational. log(5) is irrational. log(2) + log(5) = 1. It's not always true that adding two irrational numbers always results in another irrational number.

My bad, c is rational.

That was incorrect, for Q2

f is correct though.

@vast shale

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[dapper crater]

I'm not sure how to solve when ln or e is part of the function, and got stuck on these two questions

[dapper crater]

They look similar, so is there some trick to solving these both?

[lime gorge]

Did u find f(3) and f(1)?

It’s just plugging in and some log rules

No trick or Anything

[dapper crater]

No, I don't know what to do with the ln once I get to ln(3^2+1) or ln(1^2+1)

Yea I'm mainly stuck on the actual log stuff

[lime gorge]

Ok so f(3) = ln (3^2 +1)

Simplify that

[dapper crater]

f(3)=ln(10)

[lime gorge]

Great, now do the same for f(1)

[dapper crater]

f(1)=ln(2)

[lime gorge]

Good, now what do we have altogether?

[dapper crater]

ln(10)-ln(2)/3-(-1)

[lime gorge]

Good

Now simplify denominator

[dapper crater]

ln(10)-ln(2)/4

[lime gorge]

$\frac {\ln (10) - \ln (2)}{4}$

[twin meteor]

Stephen

[lime gorge]

Use these rules to simplify the numerator

One of these rules applies to the numerator

[dapper crater]

Ln(10/3)=ln(5)?

[lime gorge]

10/2 but yea

So now

$\frac 14 \cdot \ln 5$

[twin meteor]

Stephen

[lime gorge]

Use another log rule so that the 1/4 enters the argument of the ln function

[dapper crater]

Log(5^1/4)?

[lime gorge]

Yea

$\ln(5^{\frac 14})$

[twin meteor]

Stephen

[lime gorge]

7 shouldn’t be all that hard, it’s just plugging in

[dapper crater]

So ln a would be ln 5^1/4, alr gotcha

Let me attempt 7 rq

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@dapper crater Has your question been resolved?

[dapper crater]

Alright, would this be right for 7?

Wait no

.close

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[vast shale]

Let $p(x)$ be a polynomial of degree 2010. Suppose $p(n)= \frac{n}{1+n} \forall n∈ {0,1,2...,2010}$. Find p(2012)

My solution: Let $p(x)= k(x)(x-1)(x-2)...(x-2009) + \frac{x}{1+x}$
Is it correct? If not, why?

[twin meteor]

Feynmann

[untold surge]

Use Lagrange polynomial

[vast shale]

Idk what that is

[untold surge]

Search it, under the term “Lagrange polynomial” on Wikipedia

[vast shale]

Can you explain why my method is wrong?

[untold surge]

Checking

Your p is not a polynomial

[vast shale]

Okay

Thanks

[untold surge]

Here is what it

Goes like:

$P(x)=\sum_{0 \leq i \leq 2010}\frac{i}{i+1}\frac{\prod_{0 \leq i \leq 2010, j \neq i}(x-j)}{\prod_{0 \leq i \leq 2010, j \neq i}(i-j)}$, therefore what you are looking for is $\sum_{0 \leq i \leq 2010}\frac{i}{i+1}\frac{\prod_{0 \leq i \leq 2010, j \neq i}(2012-j)}{\prod_{0 \leq i \leq 2010, j \neq i}(i-j)}$

[twin meteor]

Cogwheels of the mind

[untold surge]

Must be some way to simplify that, I leave calculations to you.

No problem

[vast shale]

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[sage wind]

hello everyone

[sage wind]

to bases of which vector space do i need to extend the LI sets

to bases of $\mathbb{R}^4$?

[twin meteor]

calculus is fun

[untold surge]

No

They didn’t ask you to do anything like that

Oh sorry didn’t see the last line

[sage wind]

np

i didnt see it too at first when i solved the question and wanted to jump to another one i noticed this

XD

[untold surge]

Yes you need to do this for c

[sage wind]

for d also

the sets in d are LI

[untold surge]

If d is really independent it is a basis itself

No action needed

[sage wind]

how do ik that it is a basis of $\mathbb{R}^4$

[twin meteor]

calculus is fun

[untold surge]

If you checked it’s linear independent then it is a basis already

[sage wind]

wait nvm

yes it is

nb of elements is same as dimension of R^4

[untold surge]

Okay

[sage wind]

to extend the set in c i just need to add vector in which the set remains LI

[untold surge]

Yeah

It’s sufficient to find one orthogonal to all that three

A system of linear equations, 3 equations 4 variables

[sage wind]

so triple cross product

[untold surge]

I don’t think there is such a thing

[sage wind]

why is there not such a thing

cross product of 2 vectors is a vector

so we can have a cross product with the resulting vector and another vector

or wait a sec

i made something wrong

these elements are just called vectors but they arent vectors in fact

they are different from vectors that we deal with

so we cant do a cross product like that

but we can consider 2 vectors made by these 3 elements and then perform cross product

after that we solve for the coordinates of the new point to get the 4th element of the basis of R^4

[untold surge]

Actually if you want , you can define such a concept, just not very necessary:

n-1 vectors in F^n

You write them as column vectors, n-1 columns

Add one more column, whose elements are e_1,…,e_n

Its determinant

You can name it yourself, it is a vector orthogonal to all these n-1 vectors

Work just fine as the three dimensional case. I don’t know or have a name for it.

[sage wind]

i was trying to do something like that which is to reach a cross product which will be computing a determinant

[untold surge]

(Just a basic result by the property of determinant

[sage wind]

but the problem i got is that i should have 3 vectors not 2

[untold surge]

Yeah 3 vectors in R^4

So you can do this way if you want

[sage wind]

otherwise i wont obtain a square matrix

but can i consider these elements to be vectors

i mean they are called vectors but arent the same as vectors $\vec{v}$

[twin meteor]

calculus is fun

[sage wind]

they are called vectors just because they are elements of vector spaces

[untold surge]

It doesn’t matter. Any n dimensional vector space over F is isomorphic to F^n

[sage wind]

but in fact they are just points

but then we can just consider them vectors staring from origin and ending at that point

i could just do that

i am stupid

[untold surge]

Good. People who think themselves being smart are actually stupid

[sage wind]

yea smart people dont say hey i am smart because this sentence itself is stupid

[untold surge]

Yeah

[sage wind]

so people who say hey i am smart are stupid XD

[untold surge]

Yeah

So in short

[untold surge]

You can do this

[sage wind]

yes

[sage wind]

is what i said here correct

[untold surge]

Yeah, I think that’s how people normally view point as a vector or conversely

[sage wind]

this will lead to the same determinant that you said but the elements are flipped 90 deg counter clockwise

which will give same result

the det ill obtain will have each row containing all 4 components of the same vector

the first row will be unit vector

tysm for you help

[untold surge]

Np

[sage wind]

.close

[vocal sleet]

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[sage wind]

hello everyone

[sage wind]

what is meant by C{1...n}

[hard atlas]

how should we know

probably defined somewhere else in your ressource

[sage wind]

no it is not this is the first time this appears

so there should be a way to know what this means

[paper depot]

what's your book called again

[sage wind]

this might be it

[paper depot]

yes that is in fact definitely it.

[sage wind]

linear algebra by werner greub

ok so ill try this and type what i get

yea they must be natural because they are used as indecies of lambda_i

arent these equal just because $C{1,.,n}={\lambda _1,...,\lambda _n}$ where $\lambda _i \in \Gamma$ for $i=1,2,..n$ and $\Gamma ^n$ is the set ${\lambda _1,...,\lambda _n}$?

[paper depot]

\{ ... \} for braces

[sage wind]

yea i forgot that

[paper depot]

also no "Gamma^n is the n-tuple (lambda_1, ..., lambda_n)" is very stinky

[hard atlas]

C(set) consists of functions

[paper depot]

you confuse a set of things for a generic thing from it

[hard atlas]

not tuples

[paper depot]

this is like confusing the concepts of "Spain" and "a Spaniard"

[sage wind]

wait i meant the set not the tuple because each lamba_i is just f(i)

[hard atlas]

well what you meant is not what you wrotr

lets be precise here

[sage wind]

and same for Gamma^n it is a said

i just miswrote

[hard atlas]

no

[sage wind]

ok

[twin meteor]

calculus is fun

[sage wind]

why

[hard atlas]

btw description of stuff inside {} should also be written inside those {}

[sage wind]

wdym

[paper depot]

people who confuse a set for its generic element don't like that.

[sage wind]

ahhh this thing again i just wasnt focusing on writing thats all

[paper depot]

bad!

you should never be unfocused on writing

[sage wind]

yes you are right but this happened already so i will do that from now on i cant travel back in time

[sage wind]

so what's the wrong part

[hard atlas]

the things we said

[sage wind]

so only notations

[hard atlas]

"only"

[sage wind]

so is this correct now after i corrected notation mistakes

[hard atlas]

that's like me writing lahfncpuw and the only wrong thing being spelling

[paper depot]

you have to rewrite it from scratch and properly.

[sage wind]

ok i will eventhough i can just copy paste it

[hard atlas]

lets not copy paste

lets start new

[sage wind]

arent $C{1,..,n}$ and $\Gamma ^n$ equal because $C{1,..,n}={\lambda _1,..,\lambda _n}$ , $(\lambda_1,..,\lambda_n) \in \Gamma ^n$ and $\Gamma ^n$ is a vector space over $\Gamma$ with dimension $n$

[hard atlas]

....

[sage wind]

what

where is the probem

[hard atlas]

you didn't fix the notation at all

and writing lambda_i in Gamma^n is wrong

[sage wind]

no its not because f maps ${1,..,n} \to \Gamma$

[twin meteor]

calculus is fun

[hard atlas]

and you wrote Gamma^n instead of Gamma

[twin meteor]

calculus is fun

[sage wind]

is this better now

[sage wind]

i thought the problem was writing that $C{1,..,n}=(\lambda_1,..,\lambda_n)$ instead of ${\lambda _1,..,\lambda _n}$

[twin meteor]

calculus is fun

[hard atlas]

both of thse are wrong

[sage wind]

isnt $C{1,..,n}$ a set so its elements are in braces

[hard atlas]

yes

[twin meteor]

calculus is fun

[hard atlas]

but what you wrote isnt its elements

[sage wind]

ok it is the set of functions f(i) for i=1,2,..,n

and f(i)=$\lambda _i$ for i=1,2,..,n

[twin meteor]

calculus is fun

[sage wind]

so how arent $\lambda _i$ the elements of this vector space

[twin meteor]

calculus is fun

[sage wind]

bro this ' is annoying

[hard atlas]

if a set consists of functions

then its elements are not numbers

[sage wind]

i needed alot of time to remove it

XD

[sage wind]

its elements are the functions evaluated at different values

wait wait i just said stupid stuff

let me rethink for a bit

how are we considering all maps f and specifying a single map f at the same time

are we talking about all maps f that satisfy $f(i)=\alpha_i$ for i=1,2,..,n and for all tuples $(\alpha _1,..,\alpha_n)$ where one tuple is $(\lambda_1,..,\lambda _n)$

[twin meteor]

calculus is fun

[hard atlas]

we are talking about all maps that satisfy some criteria

yes

[vocal sleet]

@sage wind Has your question been resolved?

[vocal sleet]

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[silk helm]

need help on simple quadratic equation I'm stuck on 4x²+4x=24 (if I'm even on the right track)

[river minnow]

You made the problem even harder by expanding (2x + 1)^2

[hybrid flicker]

Yeah don’t expand

[silk helm]

ohhhh

[river minnow]

Instead move the 25 to the other side and take sqrt of both sides

[silk helm]

hol up

[crimson jetty]

Difference of 2 squares.

25 is a square. (it actually didnt need to be - you can still use difference of 2 squares, but have square roots)

[river minnow]

Ah or that yeah

[silk helm]

yeah so 25 becomes 5 right, but what abt the other side, (2x+1)^2 how does that work when we sqrt

[crimson jetty]

no, do not square root both sides of an equation unless that is a last resort

it is better to factor

,,a^2 - b^2 = ???

[silk helm]

okay so I sqrt 25 and not the other side?

[crimson jetty]

do you recognise this?

[silk helm]

no..

[crimson jetty]

,,a^2 - b^2 = (a+b)(a-b)

[crimson jetty]

This?

multiply out the right hand side to check its true

This is called "the difference of 2 squares"

[silk helm]

ahhh

[crimson jetty]

You can directly use it here

[silk helm]

so, I have to apply this in the very first original equation of (2x+1)^2-25=0?

[crimson jetty]

yes.

[silk helm]

ill brb to figure this out

[crimson jetty]

So as an extra comment - you can do what was suggested at first. "square root both sides", but you must remember the plus or minus

x^2 = 1, means x is +1 or -1.

So thats why I didnt recommend it.

[silk helm]

I still dont get it like how to apply the equation itself, could you do it then I'll prob be able to learn from that example

specifically since there is a ^2 outside 2x + 1 and in the graphic shown its only a and b squared

[crimson jetty]

,,(2x+1)^2 - 25 = 0

[crimson jetty]

So I claim that the left hand side of your equation is in this form

so you need to tell me what a and b are if that is true.

[silk helm]

2x and 1 ????

[crimson jetty]

no.

[river minnow]

(If you haven't done so far, note that 25 is a perfect square)

[crimson jetty]

,,(2x+1)^2 - 25 = a^2 - b^2

[crimson jetty]

That's what I'm claiming

for the right choice of a and b

if a = 2x, b = 1 like you suggest
You just get (2x)^2 - 1^2 which is not what you got

[silk helm]

yeah im lost, can you just do the steps on how to use the thing you suggested and apply my equation on it. I'll learn that way than guessing it for myself cause I genuinely don't got a clue what all this means

[crimson jetty]

Right ok

[silk helm]

thank you sm tho

[crimson jetty]

let me change the numbers then u can do the same thing on your originnal

,,(3x-1)^2 - 36 = 0

[crimson jetty]

ok?

,,(3x-1)^2 - 6^2 = 0

[silk helm]

sure thatd be even better actually

[twin meteor]

\begin{gather*}
(3x-1)^2 - 6^2 = 0
\end{gather*}

[silk helm]

so i can do mine on my own

[crimson jetty]

thats how its a^2 - b^2

,,0 = (3x-1)^2 - 6^2 = ((3x - 1) + 6)((3x - 1) - 6)

[crimson jetty]

let a = 3x - 1, b = 6

[crimson jetty]

then we directly use this expansion

[crimson jetty]

so we have this, and this can be simplified with a bit of algebra

[crimson jetty]

Do you now know how to solve this?

[silk helm]

hol up not yet but this is for sure better now I can learn from dis

I get the procedure now just a bit confuse on some areas but def doable, Thank you sm for your time, I'll study this thoroughly tomorrow I gotta sleep now. Thanks again

[crimson jetty]

Alright no worries. I did make it more complicated than needs to be in some ways

but the usual way that is taught can often easily result in errors

And so I prefer this way

[silk helm]

hmm okay okay thanks

[vocal sleet]

@silk helm Has your question been resolved?

[vocal sleet]

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[cyan jay]

knowing that cos(2a) = 2cos²(a) - 1, find the value of cos²(pi/5) and sin²(pi/5)

[scenic ravine]

shouldn't it be $2cos^2(a) -1 =cos(2a)$

[cyan jay]

yeah

[obtuse sierra]

no gaps

[scenic ravine]

oops , thanks]

[twin meteor]

physicsrocks

[paper depot]

[scenic ravine]

which angles trig ratios are you allowed to use?

If you can use the trig ratios for 18deg for instance, it isn;t too hard

[cyan jay]

no mention of that

only thing given earlier is omega = e^(2ipi/5)

alpha = 2cos(2pi/5)

beta = 2cos(4pi/5)

i could give you the whole thing but its in french

if it can be of any help, i also have the value of cos(2pi/5)

[obtuse sierra]

just use the formula then

a = pi/5

[cyan jay]

i did that, but must have messed somewhere because i end with cos²(pi/5)-(1-cos²(pi/5) = 2cos(pi/5)-1, and cant find where is the problem

because cos(2pi/5) = cos²(pi/5) - sin²(pi/5)

fuck i'm just dumb, i forgot i had the numerical value

.close

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[half saffron]

this is wrong?

it looks right to me but idk maybe not

ohhh the x

OHH

.close

[vocal sleet]

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[half saffron]

.reopen

[vocal sleet]

✅

[half saffron]

actually how do I find the domain

[zinc quail]

Is this how the website uses domain notation?

usually you'd use [] ()

[crimson jetty]

that aint interval notation

[zinc quail]

(-∞,∞)

E.g.

which I presume you meant to type?

[crimson jetty]

your set builder is rather questionable even if it wanted that btw

{x : -infty < x < infty} makes better sense.
But this is not interval notation

[zinc quail]

yeah if they wanted set notation they'd also have to provide symbols for subsets etc

[half saffron]

it notates like,, {x:} I thought

but it might just be {} I'm thinking

[torpid linden]

I agree with you. It says in interval notation and it is linear function.

[half saffron]

so stupid

that it has to be a specific format

but doesn't tell you what that looks like

lemme see

nothing works so far

OH

I SEE

okay okay

I got it

thank you guys

[torpid linden]

typo ==> |

[half saffron]

one more question

weird size but how do you make a function that starts with division?

[torpid linden]

Let me see what you did, plz

[half saffron]

nothing yet

this is the earlier problem though

the one I asked for help with was an example problem

[torpid linden]

It is exactly the same process. divide 8; add 5; and multiply the whole thing by 2. I am assuming that you may need to simplify before it accepts the answer.

[half saffron]

hmmm