help-39

[lean orbit]

it says 451?

[drowsy adder]

the name rahul and alok in your worksheet shows that youre probably an indian

[lean orbit]

makes sense

[drowsy adder]

im indian too anyways how did yall get the answer im stilkl stuck

[lean orbit]

i can't even start

[stone portal]

Sum of page numbers on the last leaf is 63, that means 2n-1=63
n=32 which is the number on the last page of the book. Probably 16 leafs in the unripped book.
That means both sides had an even 8 pages.
To get the sum of page numbers to a maximum while still having an even-odd split we can remove the first page of the book.
The sum of the page numbers will be 64•65/2-3

If I read it correctly

[drowsy adder]

WOAH WOAH WOAH

wowww

[stone portal]

That's way too large

[onyx kernel]

Let us check again first case odd so 2n-1
Second case even so 2n
Totally = 2n-1+2n=4n-1
4n-1 is of the form??

[drowsy adder]

how did u get the last linetho

[midnight haven]

you need to remove pages from the second half

there are 32 pages so 16 leaves

i suppose remove two from the second half and one from the first?

[stone portal]

Didn't do that

So first one from the first half and first 2 ones from the second half

[stone portal]

The sum on all numbers from 1 to n is n(n+1)/2

[slate tulip]

451 is the answer?

[midnight haven]

i think that means you miss 1, 2, 17, 18, 19, and 20?

[drowsy adder]

im getting 390 😦

[lean orbit]

yep im failing

[midnight haven]

32*33/2 = 528 so subtracting 77 would give… 451?

[midnight haven]

oh cool we agree

and look at that it’s even one of the options

[drowsy adder]

oh mb im getting 441 too

451

[lean orbit]

the correct answer is 451

[stone portal]

Oh my algebra's totally wrong

[lean orbit]

(book has answers too)

[midnight haven]

ok it seems everyone agrees

[slate tulip]

That’s what I did

[drowsy adder]

yeeah!

[lean orbit]

so

why

is

it

451

[drowsy adder]

no it isnt the 1st 3 lines are correct

[midnight haven]

• the last leaf adds to 63
• so the last two pages are 31 and 32
• there are 16 leaves, 8 in each half
• we need to remove leaves from both halves so that the first half has an odd number of leaves and the second half has an even number of leaves

[drowsy adder]

those are the main steps needed to solve it

[midnight haven]

• we remove 1 leaves from first half and 2 from second half
• minimise the numbers on the removed leaves: 1, 2 for the first half; 17, 18, 19, 20 for the second half

[drowsy adder]

dayum ur all smartiess

[stone portal]

Convinced as well

[midnight haven]

• sum of all pages in book is 32*33/2 = 528
• sum of removed pages is 1+2+17+18+19+20 = 77
• 528-77 = 451

this is my solution

[drowsy adder]

the formula for a sw

[lean orbit]

holy

[drowsy adder]

series

the formula to calculate the sum of a series till the nth term is the (nth term(nth term+1))/2

but the series should be in arithmetic progression

+1

eg 3,4,5,6,7,8,9,10

like that

[slate tulip]

Okay that was a great question!! @lean orbit

[lean orbit]

thank you all

[drowsy adder]

.

[stone portal]

And

[lean orbit]

for helping me

[drowsy adder]

i didnt have to type it was alrerady given by mebasically

yw

[lean orbit]

ill be back with more questions soon as i don't wanna

[drowsy adder]

even tho i didnt contribute

[lean orbit]

fail

[drowsy adder]

dw just practice math

u will love it

[lean orbit]

hopefully

well thanks yall again

[drowsy adder]

byee

[lean orbit]

may i close this?

[drowsy adder]

sure

[lean orbit]

.close

[pearl pond]

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[stone portal]

Everything I wrote had wrong numbers plugged in oh yh except 16,32

[drowsy adder]

ooh

imma go now

byee!!!

smart people

[pearl pond]

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[uncut coral]

i get (sinAcosB + cosAsinB) (sinAcosB - cosAsinB)
diff of 2 squares so
(sinAcosB)^2 - (cosAsinB)^2

What do I do from there

[brisk steeple]

you do what you have to do

[uncut coral]

wow

[autumn fossil]

Hmm maybe try using pythagorean identity to express cos(B)^2 and cos(A)^2 in terms of sin

then it should all simplify nicely

[cosmic garnet]

remember (sinx)^2 + (cosx)^2 = 1?

try using that

[uncut coral]

ill try astars method first it seems simplier

[autumn fossil]

Spoiler: ||They are the same method lol||

pythagorean identity is (sinx)^2 + (cosx)^2 = 1?

[uncut coral]

oh lol

idk i just couldn't think of sin^2 + cos^2 = 1 when you said what you said

[autumn fossil]

yeah, sorry. I didn't know you don't know the name "pythagorean identity"

[uncut coral]

im kinda tired rn

do I just rewrite cosa^2 as sinx^2-1 ?

[cosmic garnet]

1-sinx^^2*

[uncut coral]

sinx^2 + cosx^2 = 1
sinx^2 - 1 = -cosx^2

sorry should've been more specific

[cosmic garnet]

yeah that works too

[uncut coral]

thank you it works out

btw for b) do I just turn it back into sin(45+30)sin(45-30)

[cosmic garnet]

yes

[uncut coral]

alright thank you

[cosmic garnet]

then apply the formulas that u did before (of sin(a+b) and sin(a-b))

and solve it

[uncut coral]

im so confused

[autumn fossil]

you sure?

[cosmic garnet]

yes

[uncut coral]

when i eval sin75 square - sin15 squared i get root 3 on 2

[autumn fossil]

why not just let A = 75 and B = 15

[cosmic garnet]

that works too

thats simpler too

[uncut coral]

if i eval sin(45+30)sin(45-30) aka sin75sin15 i get 1/4

1/4

[cosmic garnet]

yeah thats right

[autumn fossil]

I dont see how that helps?

[cosmic garnet]

wot

[autumn fossil]

So now we know that sin(45+30)sin(45-30) = 1/4

how is it helpful?

How is sin75sin15 = 1/4 helpful?

[cosmic garnet]

i didnt say its 1/4, they said its 1/4 which i assumed they got by calculating

[autumn fossil]

it is correct

but how is it helpful?

[cosmic garnet]

how is answer helpful

[autumn fossil]

we are finding sin^2(75) - sin^2(15)

meanwhile we found sin75sin15

[cosmic garnet]

but thats after using the identity that we just proved above

[uncut coral]

hmm when i put it into verif mode its false

[autumn fossil]

yeah, it is false

because you found sin75sin15

[autumn fossil]

but you are supposed to find this

[cosmic garnet]

nvm im dumb

[autumn fossil]

Try setting A = 75 and B = 15

it will give sin^2(75) - sin^2(15) = .......

[cosmic garnet]

^

[uncut coral]

alright

ill try it tommorow

its 11 rn

thanks for the help

[pearl pond]

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[cunning stump]

how do you solve this?

[tropic saddle]

you start by guessing a solution using the rational root theorem

then you use polynomial division and then quadratic formula

[lethal palm]

u could start by splitting -3x to -4x + x and see if u are able to factorise anything

[pearl pond]

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[crisp walrus]

hey ppl, um anyone maybe help me with that? https://discord.com/channels/268882317391429632/1271438594296385556 ( im sorry i know i should not post it here, but i really need it :< don't even know where to start solving )

[pearl pond]

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[naive hemlock]

@crisp walrus

!noadvert

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[nova kestrel]

i need help

[nova kestrel]

nvm

[uneven smelt]

help

[golden fjord]

ask in an available help channel

[pearl pond]

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[sharp smelt]

what am I doing wrong here?

[glass meadow]

You did R_2 - 2R_1 -> R_2

1 - 2(-1) = 3

not -3

[sharp smelt]

Ah

Is the rest of it right?

[glass meadow]

Looks ok to me

[sharp smelt]

Thanks!

.cose

.close

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[sterile turtle]

can i get confirmation this is the right way to do this

i said coeff, but its just the term whoops

[cursive wraith]

constant coeff = constant term

because multiplying by x^0 = 1 changes nothing

this is the right way

[sterile turtle]

thanks

.close

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[white coral]

Hello, I'm studyin projective geometry, I don't have any idea of how to approach this exercise, the book never give an example with numbers, only theory, but, I don't know how to use that theory to solve the exercise, I know that P^2 means that the vectorial space is V^3, but, I don't sure if just use geogebra graph the two points and draw the line between them it's the solution, in advance thank you <3

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[sterile turtle]

help with 9 pls

[sterile turtle]

my expansion for the terms

idk where to go from here

<@&286206848099549185>

[pearl pond]

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[inland ivy]

$$\begin{vmatrix}
bc-a^2&ca-b^2&ab-c^2\
ca-b^2&ab-c^2&bc-a^2\
ab-c^2&bc-a^2&ca-b^2\end{vmatrix}$$

[jolly parrot]

kheerii

[inland ivy]

how do I simplify this

[sharp smelt]

is that supposed to be a det or a matrix

[inland ivy]

determinant

[sharp smelt]

Not too sure, sorry.

[inland ivy]

all good

[jolly parrot]

kheerii

[inland ivy]

I have this till now

oh wait shit

just R3 -> R3 - R2 now

I think

no that doesn't help nvm

[last basin]

i don't know if that would help but i just noticed (1 , 1 , 1) is an eigenvector of this matrix xD

[inland ivy]

how did you manage to find that out lmao

[last basin]

I just added all columns 😂

[inland ivy]

ah right

I need to simplify the determinant though

[last basin]

first thing I do in any matrix problem is add all columns before even reading what i want

well the matrix is symmetric too which means the eigenvectors are orthogonal

and we landed one of them so we know one of the eigenvalues

if we can somehow find just one more eigenvalue

we can use the trace to get the third

then multiply them to get the determinant

[inland ivy]

I don't know how that works

can you explain what that means?

[last basin]

you know the product of eigenvalues is the determinant of the matrix right

[inland ivy]

no

we haven't even been taught eigenvalues yet

though I know what they are/ how to calculate them

the exercise just expected us to use properties of determinants

[last basin]

oh nevermind then, i guess i went too far xD

[inland ivy]

no it's all good your way is probably faster

but not the way required of me unfortunately

[last basin]

my way relies more on guessing, so it's not systematic. probably why it isn't prefered. i didn't see a way to simplify it yet to be honest so I will look into it

[inland ivy]

alright thanks!

[pearl pond]

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[vestal venture]

how do i start

[pearl pond]

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[quiet goblet]

Hi

[quiet goblet]

I have a question: integrate sin^(x) wrt sin^(x). On solving this by 2 methods, first by replacing d(sin^(x)) with (cosx)dx and other method is direct substitution of sinx by t.

[light helm]

wdym by integrate
$\sin^x$

[jolly parrot]

ℝαμΩℕωⅤ

[quiet goblet]

I think I spelled sin^2 by mistake and then forgot to erase ^

it was just sinx

wrt sinx

[quiet goblet]

is*

[light helm]

ahk, so
$$\int\sin(x)\dd(\sin(x))$$

[jolly parrot]

ℝαμΩℕωⅤ

[quiet goblet]

Yeah

I want to know why there is 2 seperate answers for 2 methods I tried

[versed mica]

udu?

[verbal whale]

This is equivalent to integral of u du which is u²/2 + c

[quiet goblet]

Yeah

[verbal whale]

In your case u = sinx

[quiet goblet]

Then I converted it to cos

using 1-Cos2x = 2 Sin^2 (x)

[verbal whale]

No, don't do that

[versed mica]

why?

[verbal whale]

Not that you can't, but there's no need to do it

[versed mica]

it’s the integral of udu

[quiet goblet]

actually I wanted to verify the answer of this method by first method

thats why I converted

[verbal whale]

Ahn ok

[quiet goblet]

But there was some slight difference

1/4 was left out

[verbal whale]

All fine, that's just a constant, that's why you have to write + C when finding indefinite integrals

Or you mean 1/4 multiplied?

[quiet goblet]

so whenever I get a numerical constant in integral I just assume it a additional whole of C?

[verbal whale]

If it's added/substracted to the primitive you found, yes

[quiet goblet]

No it was 1/4 added

to 1/4 cos (2x)

[verbal whale]

C is just a name for "some unknown number"

[quiet goblet]

I see, then that answers to my doubt

[verbal whale]

You're welcome

[quiet goblet]

Thank you

.close

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[white coral]

¿Two unit vectors in 2D that are independent are necessarily orthogonal?

[verbal whale]

Nope

Take (1, 0) and (1/√2, 1/√2) as a counterexample

[white coral]

1/sqrt(2) is different of zero

thank you, another quick question ¿rotate with respect to the origin and translate produce the same effect on a 2D point, regardless of the order in which they are performed?

[verbal whale]

Nope, the order matters

[white coral]

thank you so much

[verbal whale]

You're welcome 🤗

[white coral]

.close

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[pearl pond]

@surreal magnet Has your question been resolved?

[pearl pond]

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[surreal magnet]

I figured out a periodic function that when integrated from 0 to 2pi gets you the area of a circle, and if the circle has a hole, you can simply treat that hole (as long as the hole is also a circle lol) as the same function and subtract that from the first function and still get the area in the end
(-cos(x) + 1) * (r^2 /2)
I like -cos just because it starts at 0 and goes up lol
so for an annulus the formula would be (-cos(x) + 1) * (r^2 /2) - (-cos(x) + 1) * (r^2 /2)
the idea is to take any 2d closed surface, and turn it into a sum of 1d continuous periodic functions
where if you take the integral from 0 to 2pi you get the area

My question is, is this already a thing, and how do I find the thing? (I have no education)

[summer imp]

Integrating cos or -cos from 0 to 2pi just gives 0

The +1 bit just yields x which gives 2pi at the end of integration

So all in all it’s just pi r^2

[surreal magnet]

you still have to multiply by r^2/2 in the formula (-cos(x) + 1) * (r^2 /2)

I want to do this for more complex shapes than a circle in the end

[summer imp]

That’s what I mean. You’ve just rewritten it but the cos part is insignificant

What you’re doing is equivalent to integrating r^2/2 from 0 to 2pi

That gives 2pi * r^2/2, which is just pi r^2

[surreal magnet]

Maybe if you slice a 2d surface in half along the x axis, you can get a piece wise function for the shape in 1d. Then you could assume it to be periodic, and use fourier.

Then once you have a non piecewise function in terms of sins and stuff, you could use that formula.

Thats the maybe dumb idea

[surreal magnet]

That is a neat insite into why its working in the first place though!

[summer imp]

But the point is that cos(x) is insignificant. You could’ve chosen sin(x), it’s just adding 0

[surreal magnet]

True, I guess, I just wanted to start with a circle as its simple, but I want to do more complex shapes.

[summer imp]

And as it is, r^2/2 is a periodic function that integrates to the area of a circle when integrated from 0 to 2pi

[surreal magnet]

Oh thats neat, I didnt know that!

What about the idea of slicing a more complex shape along the x and making a piece wise function where the top half is the first part, and the bottom half is the second part. Then assuming that to oscillate on an infinite domain so you can use a fourier transform to get a non piecewise function back in terms of the sine's?

[pearl pond]

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[west vault]

I'm looking to prove that (provided the necessary conditions are satisfied), if $(a_n) \to a$, then $f(a_n) \to f(a)$

[west vault]

Also this would be the first time I'm working with the definition of a limit using epsilon-delta

[jolly parrot]

jewels!

[lime river]

Isn't that obvious

[west vault]

Probably is, I just don't know how to word the proof

It is probably exactly epsilon-delta

[lime river]

So in other words u have to proove
Lim an --> a f(an) = a

[west vault]

yeah

I suppose

[lime river]

Use the Delta Epsilon definition to do it

[west vault]

I know 😭

So I suppose I start off as normal

Let $\epsilon > 0$, and choose $N \in \mathbb{N}$ such that for all $n \ge N$, we have:
[|a_n - a| < \epsilon]

[jolly parrot]

jewels!

[west vault]

But I need to show that this implies that |f(x) - f(a)| < delta

If I'm not wrong

which is where I'm stuck

[summer imp]

This requires continuity of f, so you'll need to use that :p

[lime river]

Yeah exactly was thinking that

[west vault]

f is continuous at a and defined for all a_n yeah

[lime river]

The above statement is wrong since if the function isn't continous at a then the limit won't exist so functuon won't tend to f(a)

[west vault]

I did say the necessary conditions are satisfied

I didn't want to list them out

[lime river]

then what's the damn proof
If function is continous it means
Limit is defined and value of limiti equalw the value of function at that point

Limit *
Equals*

[west vault]

But is there not a need to show that the sequence defined by f(a_n) converges to f(a)?

[summer imp]

When inputs of f get really close, outputs get close too. That's continuity.
You've written down that the points of the sequence are arbitrarily close to a, so you should be able to conclude that the outputs are close as well.

[west vault]

I'm sorry I'm kind of slow I need some time to think about this

[summer imp]

But I think you started from the wrong place. You will use the fact that an converges to a, but I recommend you start with the definition of continuity of f first.
Use that to find some delta such that whenever |x-y| < delta, |f(x) - f(y)| < epsilon.
Then you can use the convergence of a_n to find a suitable N that fits this delta.

[west vault]

So I can just say that

Since $f$ is continuous, for some $x, y \in D$, and $\epsilon, \delta > 0$:
[|x - y| < \delta \implies |f(x) - f(y)| < \epsilon]
Let $\epsilon' > 0$ and choose $N \in \mathbb{N}$ such that for all $n \geq N$:
[|a_n - a| < \epsilon']
From the fact that $f$ is continuous, there exists $\delta' > 0$ such that:
[|f(a_n) - f(a)| < \delta']
Thus, $(a_n) \to a \implies f(a_n) \to f(a)$.

[jolly parrot]

jewels!

[west vault]

D is the domain of the function i'll write it out better

[summer imp]

If f is continuous on D, the first line hold for every x,y in D.

[west vault]

wait

its only continuous around f(a) im pretty sure

its defined at all a_n

[summer imp]

You also don't need to invoke epsilon', just find some N such that |a_n - a| < delta.
Then, whenever n >= N, we have |a_n - a| < delta, and as such |f(a_n) - f(a)| < epsilon.
You're still proving that f(a_n) converges to f(a). It's just that you're finding your N through continuity.

[west vault]

Ohhh

That makes a lot more sense actually

[summer imp]

Yeah that's fine, you only need it to be continuous at a.

[west vault]

Since $f$ is continuous, for some $x, y \in D$, and $\epsilon, \delta > 0$:
[|x - y| < \delta \implies |f(x) - f(y)| < \epsilon]
Now choose $N \in \mathbb{N}$ such that:
[|a_n - a| < \delta]
From the fact that $f$ is continuous:
[|f(a_n) - f(a)| < \epsilon]
Thus, $(a_n) \to a \implies f(a_n) \to f(a)$.

[jolly parrot]

jewels!

[west vault]

But doesn't that make my first statement wrong

I'd need to somehow rephrase it

[summer imp]

Yeah I think I should've written the definition as :
$f$ is continuous at $x=a$ $\iff$ given $\epsilon > 0$ there is some $\delta > 0$ such that
$$|x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$$

Sorry about that

[jolly parrot]

Azyrashacorki

[summer imp]

That should be better

[west vault]

Yeah alright

Thanks for having the patience and sticking with me btw it means a lot

[summer imp]

No worries! Happy to have helped a bit

[west vault]

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[cloud yarrow]

Not sure how ii) is related to the first part

[cloud yarrow]

For i) i got +- [sqrt(10)/2 + isqrt(2)/2]

using

(x+iy)^2 = 2+isqrt(5)

[spare lark]

And what did you try for the (ii) ?

[cloud yarrow]

Nothing

Idk how I’m supposed to use the last answer

[spare lark]

Let Z = z^2

[cloud yarrow]

Alright lemme see

[cloud yarrow]

Oh I see now u end up with z^2 = (first part)

thanks

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[fervent galleon]

asking...

[fervent galleon]

I'm trying to isolate x: 3x - 8 = 0. I tried to move x to the right side by diving by x, but I get bad answer:

3-8 = 0/x

I know that I can first move the 8 by adding 8 to both sides, and then divide by 3.

but what's the correct way to just move the x?

[elfin cloud]

3x=8 (add 8 to both sides)

[elfin cloud]

So this yes that is the correct way

Afterwards you could divide on both sides to isolate x

[fervent galleon]

But, my question is, why do I get a nonsense answer if I divide both sides by x?

[elfin cloud]

Because you did not divide both sides by x

You forgot to divide the 8 🙂

[fervent galleon]

oh

[elfin cloud]

You would need to do: $\frac{3x-8}{x} = 0/x$

[jolly parrot]

thijs2725

[elfin cloud]

Where 0/x would just be 0 BUT

There is a big problem with this

When you divide by x you assume x does not equal 0

Because you cant divide by 0

If x=0 would have been an answer you would no longer get this result and thus it is generally not used to divide by a variable unless we know it does not equal 0 or something

Do you understand this last part because this is a important part? 🙂

[fervent galleon]

yes

But, if I did go that route anyway, would that simply to:

$\frac{3-8}{x} = 0/x$

[jolly parrot]

Vulkanoid

[fervent galleon]

sorry

$3 \cdot \frac{8}{x} = 0/x$

[elfin cloud]

You can edit your msg the math text will update

[jolly parrot]

Vulkanoid

[elfin cloud]

It would still be a minus sign between the 3 and the 8/x not a multiplier

And 0/x would just simplify to 0 because 0/anything is 0 (assuming x does absolutely not equal to 0)

[fervent galleon]

$\frac{-5}{x} = 0/x$

[jolly parrot]

Vulkanoid

[elfin cloud]

No

The 3 would be outside of the fraction

[fervent galleon]

my point is, what want to follow through on simplifying the equation, if I decide to initially move the x to the right.

[elfin cloud]

$3 - 8/x = 0 $ is the correct way

[elfin cloud]

Yeah that wont work because there is a 0

[fervent galleon]

ah, so it's like the x never even moved.

[elfin cloud]

Yes

And again it is best to avoid dividing by 0 so it wont become a bad habit

Could become a real problem in equations such as $x^2-x = 0$

[jolly parrot]

thijs2725

[fervent galleon]

I'm more trying to understand the rules, rather than finding a specific answer. When I first tried to move the x, I got that 0 answer, and I wanted to understand where I was going wrong.

I think I get it now.

Thank you

[elfin cloud]

Good luck

[fervent galleon]

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[lilac fern]

How does this work?

[west sapphire]

which line is the first one you don't understand?

[lilac fern]

How does (k(k+1))/2 + (k+1) go to the next line?

[west sapphire]

notice that both terms in that line contain k+1 as a common factor

[lilac fern]

Hmm

[glass meadow]

$\frac{k(k+1)}{2} + (k+1) = \frac{k}{2}(k+1) + 1(k+1)$

[jolly parrot]

Nel

[lilac fern]

$\frac{k}{2}(k+1) + 1(k+1)$

[jolly parrot]

🐚🐳 𝕄σ𝑜Ｓ𝓗ⓡ𝐎Ｏ𝕞 ๓ỖόＮ ☯💋

[lilac fern]

I don’t understand

[glass meadow]

What's 3/2 cm + 1 cm?

[lilac fern]

2.5 cm

2 1/2 cm

5/2 cm

[glass meadow]

What's 2a kg + 5b kg?

[lilac fern]

2a kg + 5b kg = 2a kg + 5b kg

[glass meadow]

Can't you rewrite it using "kg" one single time?

[lilac fern]

2a + 5b kg

[glass meadow]

Right

What's k/2 days + 1 day?

[lilac fern]

k/2 + 1 day

[glass meadow]

Same with that, except it's (k+1) instead of days

[lilac fern]

ah I see so

[lilac fern]

Now I see why this is right

[pearl pond]

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[warm burrow]

How do I calculate this limit

[warm burrow]

I have been applying taylor series and stuff but I get nowhere

[cursive wraith]

I suggest putting everything on the same denominator

$\frac{(x-1)^2-x(\log x)^2}{x(\log x)^2(x-1)^2}$

[jolly parrot]

rafilou2003

[worthy lance]

Honestly i am not sure what is exactly the denominator notation of the first fraction

[cursive wraith]

then use equivalence on the denominator

$\frac{(x-1)^2-x(\log x)^2}{(x-1)^4}$

[jolly parrot]

rafilou2003

[cursive wraith]

you're gonna need 4th order taylor approximation of the numerator

[warm burrow]

Ah yes so basically I have to apply the log(x+1) taylor series and manipulate that right?

[cursive wraith]

well you can use taylor series of log(t+1)

to find taylor series of (log(t+1))^2

[warm burrow]

and why is the denominator (x-1)^4

[cursive wraith]

x ~ 1, log(x) ~ x-1

[warm burrow]

aight i gotchu

Thanks!!

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[stuck mural]

I think I forgot something

how to let the 9 write in it

[light helm]

consider simplifying log(9)

and your work for Q74 is wrong

[stuck mural]

Did I need to divide the log9 alone?

[light helm]

no

[stuck mural]

Emmm

Maybe I have some thoughts

[gaunt crest]

what formula is this?

[stuck mural]

Like that?

[light helm]

that doesn't really help you

[stuck mural]

I already corrected Q74 thanks

[stuck mural]

Emmm

[light helm]

consider simplifying log(9)

74 is still wrong

mistake is in that first step

[light helm]

that's ok now

[pearl pond]

@stuck mural Has your question been resolved?

[stuck mural]

Did I need to turn to3^2?

[light helm]

that will help

[stuck mural]

Is that right?

[light helm]

no

9/2 isn't 3

[stuck mural]

So just turn in to 3

But the ans will have log3

[light helm]

consider simplifying log(9)

[stuck mural]

?

[light helm]

no

are you ignoring

consider simplifying log(9)

if you were asked to do just that, (forget about the 150)
what would you do?

[stuck mural]

450?

I don't know

Change into 3^2

But if 2log3/log3 it will be 2

Then it cannot be log 150

[light helm]

(forget about the 150)

just simplify log(9) and what do you get

[stuck mural]

2log3

[light helm]

yes

from that you'll have
2log(3) = b
and can get an expression for log(3) in terms of b

[stuck mural]

Oh

🫠

Can I do it like that

[light helm]

yes

[stuck mural]

I apologize for my mistake

Thx

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[brave helm]

what is the relationship between:
(a/b + c/d + e/f) * (1/3)
and
(a+c+e)/(b+d+f)

[brave helm]

as in: which is larger? when does equality hold

[midnight haven]

Well, are there any values associated with each variable, or is this for universal situations?

[brave helm]

positive real

[midnight haven]

Ok

[pearl pond]

@brave helm Has your question been resolved?

[pearl pond]

@brave helm Has your question been resolved?

[tight cairn]

Gotta have the same denominators to compare

So the question is how to get the same denominator

[pearl pond]

@brave helm Has your question been resolved?

[prime pond]

Damnnnn

Look at that pfp

[pearl pond]

@brave helm Has your question been resolved?

[midnight haven]

(1/2 + 1/3 + 1/4)/3 > 1/3 = (1+1+1)/(2+3+4)

but (1000000/2000000+ 1/3 + 1/4)/3 = 1/4+1/9 < 3/8 < 1000001/2000007

i do not think either side is, as a rule, larger than the other

[brave helm]

aight

ggs

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[west vault]

Is this negation correct?

[west vault]

To keep it more concise I guess I should replace "at least one sequence" with just "a sequence"

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[random ermine]

what's x0

what you wrote doesn't really make sense

[pearl pond]

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[jolly parrot]

z

[light helm]

is that the original question

seems to be a mistake in the question

sum of complex conjugates will be real

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[frozen wharf]

qst 5

plz provide me a solution

plz

ALSO this qst

plz

[frozen wharf]

<@&286206848099549185>

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[spark mist]

Looking for help with question 7.c please

[spark mist]

this is the solved table

[vital estuary]

a mass greater than 104g has a z score greater than what?

[spark mist]

2

[vital estuary]

whats the probability of having z score greater than 2?

[spark mist]

uh

2.5%

i think

oh

2.5% of 2500

ty

[vital estuary]

youre slightly off

i looked up a lookup table and it says 2.3%

[spark mist]

oh i think yours might me different to mine

2.5% of 2500 is 62.5

[vital estuary]

oh weird

[spark mist]

i gotta check my text book hold up

yeah 62.5

idk why they are different

[vital estuary]

go with the numbers in your book

[spark mist]

alright cool, thanks for the help 👍

[vital estuary]

youre welcome

[spark mist]

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[solemn pollen]

Greetings seniors, may I ask how they got A+2B=0 and 2A-B=1. Where did Ax and 2Bx go??

[verbal whale]

What's the coefficient of x on the RHS?

[verbal whale]

I'm referring to this equality

[solemn pollen]

Oh wait for a min

It's unknown

[tulip matrix]

@solemn pollen

did they give the denominators in the question itself?

[verbal whale]

Ok, but what's the "value"?

[tulip matrix]

on the rhs

[solemn pollen]

Should I send the whole question?? Because I don't really get it myself, I'm sorry

[tulip matrix]

sure

[verbal whale]

No, this is sufficient for answering your question

[solemn pollen]

My teacher graded my friend's answer as 10 as in 100% correct

But I simply could not understand where the hell did the ax and bx go on the answer at the bottom

[verbal whale]

I was explaining that 😅

[solemn pollen]

I'm sorry my English is bad hehe

Might've misunderstood ur point

[verbal whale]

RHS means right-hand side

[verbal whale]

Here

[solemn pollen]

I don't know

[solemn pollen]

That's the whole question

[verbal whale]

You just have to look at what's written there

[solemn pollen]

It wasn't mentioned

[verbal whale]

Here in the green box

[solemn pollen]

I don't understand

[verbal whale]

Do you know what a coefficient is?

[solemn pollen]

The number

[verbal whale]

In the polynomial 3x - 1 what's the coefficient of x?

[solemn pollen]

3

[verbal whale]

Nice

And in the polynomial 4 - 6x + 7x?

[solemn pollen]

6 and 7?

[verbal whale]

The coefficient is just one number

[solemn pollen]

Oh 1

[verbal whale]

You have to simplify the polynomial, first of all

[verbal whale]

Exactly

[verbal whale]

Now, can you tell me what's the coefficient of x in the right hand side of the equation in green?

[solemn pollen]

Umm wait

I'm sorry I may have brain damage but Im guessing 2?? Because there's 2x??

[verbal whale]

Ok, let's start from scratch

[tulip matrix]

why dont u teach him from the start?

like from the splitting part

[verbal whale]

Because I think they got that part, as far as I understood from their question

[tulip matrix]

ye but he needs to understand

[verbal whale]

I believe he already knows that

[tulip matrix]

o

[solemn pollen]

I kinda understand the other processes but I just don't get where ax and bx go

[verbal whale]

What's the coefficient of x in the polynomial 4x² + 7 ?

[solemn pollen]

2

What

Wait

Uhh should I square root it? Since it's asking me to give the cooficient of x

[verbal whale]

No no

The coefficient of x is the number in front of the x

[solemn pollen]

So 4?

[verbal whale]

If you don't have any x, what's the only option for the coefficient?

[solemn pollen]

Eh I'm a dumb dumb

[verbal whale]

Is 4 multiplied by x?

[solemn pollen]

7

[verbal whale]

Is 7 multiplied by x?

[solemn pollen]

😭

[verbal whale]

If you can answer these two simple questions you'll see why they're not the correct coefficient 😅

[solemn pollen]

If I don't have any x, how are there any options for the coefficient?

From. 4x²-7

So none?

[solemn pollen]

Yes?

Sir I'm sorry

Yeah Im giving up on a levels

[tulip matrix]

bruh

[solemn pollen]

Is it really that obvious???

[tulip matrix]

can u vc??

[solemn pollen]

Sure

[tulip matrix]

im with a friend rn so do u mind a gc?

[solemn pollen]

Ok ok

[tulip matrix]

just wanna use whiteboard

[verbal whale]

If you don't have any x, it means you have how many??

[solemn pollen]

Do I VC my answer sheets?

0??

[verbal whale]

There you go!!

[tulip matrix]

vc means voice call

we resolved it

👍

[solemn pollen]

Sir Albert I apologize for my stupidity that has caused you immense pain and stress🥲

[verbal whale]

Don't worry 🤗 we're here to help you all

I just wanted not to give you the answer directly, but instead taking it out from your reasoning, which is more productive

[solemn pollen]

Thank you for your patience

🫡🫡

.close

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[sharp smelt]

I was thinking they could be either odd or even for both

[twin hedge]

even means that f(-x)=f(x) and g(-x)=g(x) right?

[sharp smelt]

ys

*&es

*yes

[twin hedge]

so check whether this holds when you plug in -x for f+g

[sharp smelt]

I was thinking that f(x)=x^2, (x)=-(x^2+1) for f+g

That gives an even function

[twin hedge]

yes, for this example but we need to know whether this is for all functions

[sharp smelt]

however, f(x)=x^2, g(x)=-x^2 gives 0, a function that's both odd or even

so odd(say)

so f+g can either be odd or even

[midnight haven]

there is an easier and more general way of doing this

i will work f+g and let fg for you

[sharp smelt]

Let me think then

If that's so

Thanks

[twin hedge]

so if we plug in -x for f+g, we get (f+g)(-x)=f(-x)+g(-x) which is equal to f(x)+g(x) which is just (f+g)(x), so (f+g)(-x)=(f+g)(x) which is even

[midnight haven]

@sharp smelt that is what i was going to say

try doing the same with fg

[sharp smelt]

I see.

Thanks

I feel that even f(x)g(x) should be even

Y/N

[twin hedge]

yes

that's correct

[sharp smelt]

Cool, thanks

I suspect that's not true for odd functions though

consider f(x)g(x) when f(x)=x^3, g(x)=x^5

[twin hedge]

then f*g is even

and that's for all odd f and g

do you think you can prove why?

[spare lark]

And if f is even, and g odd, then product is odd

[twin hedge]

also true

[spare lark]

That is useful for integration sometimes

[sharp smelt]

I know, I've done integration, just revising functions for my uni's calculus course

We'll be focusing on proofs now, so probably for the better

[sharp smelt]

Hmm.Yes

f(-x)=-f(x)

g(-x)=-g(x)

so f(-x)g(-x)=f(x)g(x)

[twin hedge]

very good

[sharp smelt]

Now to deal with the sum

[tropic saddle]

just because its also odd doesnt mean its not even. that would be the case for numbers but not here

[sharp smelt]

Got it. Thank you.

[tropic saddle]

the claim doesnt say anything about f+g or fg being odd. it only cares about them being even

[sharp smelt]

That's actually the next part of the question. By conicidence, @twin hedge suggested it as an exercise.

I haven't posted that as I needed help with this question first.

Was hoping to apply my learnings from this to that

I suspect f(x)+g(x) will be odd, if f(x) and g(x) are odd

[twin hedge]

yes, very good

[sharp smelt]

Thanks

I think the answer is true here

That is ,Yes, h is always an even function

[twin hedge]

yes

[sharp smelt]

Awesome!

Here it would depend on ƒ's nature, right?

[twin hedge]

yeah, if f is odd then h is odd, if f is even then h is even

[sharp smelt]

Thanks a lot!

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[midnight haven]

hallo I didnt pay attention at school and I dont know how to solve this

[midnight haven]

how many degrees is a straight line

0

no

Oh

I dont know

[fluid dagger]

how many degrees in a right angle?

[midnight haven]

90¿

90 ti what

to

my question or his question

his

correct

now if a right angle is 90 degrees

how many degrees is a straight line

I thought a null angle was 0°, like a straight line

[midnight haven]

null angle != straight line

Ummmmm

ok so

for starters

a straight line is double a right angle

so a straight line has 180 degrees

you have to learn this like english

now

the thingy in red is a straight line

how much angle is x

20°?

no

180-126

goo

d

vety goo

what is that equal to

54°

ok

[hushed heath]

Do uk vertically opposite angles?

[warm sonnet]

you should start with the basics

if you really want to learn this

[midnight haven]

yeah you are right.........

[warm sonnet]

it doesn't take a long time

[fluid dagger]

just watch a quick vid

[warm sonnet]

you should buy a geometry book from a book store

• use youtube as extra resource

[midnight haven]

tbh we can't really help you if you don't know the basics

like we are not asking to be a math olympic

but the basics...

I think I just don't know english very well either

[midnight haven]

why

[warm sonnet]

you can get a book in your native language

a lot of resources on the internet are in english though

[midnight haven]

I understand

20° is also an angle inside?

[warm sonnet]

you can apply this

[midnight haven]

Ohh I think I understand

So

54+20+x=180°

?

[karmic sluice]

Its quite simple. If you consider the size of an angle in a whole circle, its 360 degrees. Then when you cut a circle in half, the angle at the centre splits to 180 degrees. So therefore angles on a straight line would add to 180 degrees. So the angle x is 180-126 (half a circle) which is 54 degrees.

[karmic sluice]

yeah

[warm sonnet]

yea, u have to know those little properties to solve those problems

vertical angles, sum of angles in a triangle, etc

[midnight haven]

This image was really helpful

[warm sonnet]

just knowing that a straight line is 180 degrees is useful

[midnight haven]

Thank you

[karmic sluice]

best to learn all the circle theorems, search up 'circle theorems and basic angles explained in 1 video'

maximum 30 mins and ur a pro

[midnight haven]

Ill be at the nasa next

Thank you

!!!

[karmic sluice]

nws

[midnight haven]

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[brisk steeple]

They don't know you lil bro

[brisk steeple]

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[uneven smelt]

lil bro

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[sturdy hamlet]

why can a = 0

[spice snow]

on the first step, we multiplied by 1/sqrt(a)
thus, we lost the solution a=0, since 1/sqrt(0) is 1/0 which is an indeterminate form (we would be dividing by 0!!)
so the only solution that we'll get after that is 1/4. that doesn't mean that a=0 doesn't satisfy the first equation: we can clearly see that it does without doing any work, so we add the solution a=0 at the end

[sturdy hamlet]

ok thanks

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[sharp smelt]

I was thinking of using the general form of the equation of a circle to obtain this, upon comparison

that is $x^2+y^2+2gx+2fy+c=0$

[jolly parrot]

f(why am i here )= I don't know

[calm wing]

hint: ||when a=0, can you see why it's not a circle?||

[tropic saddle]

well how would you wanna do it with that form

[sharp smelt]

Yeah, just realised that won't help much

oops

[calm wing]

check my hint

[sharp smelt]

I have one more idea, I;ll check it after that ?

[calm wing]

oki

[sharp smelt]

My other idea is to use the defn of a circle

that is

the locus of points , equidistant from a fixed point.

so the point in this case, would be $\frac{x_1+x_2+x_3}{3} , \frac{y_1+y_2+y_3}{3}$

Now to prove sucha point doesn't exist if all 3 points are co-linear

[cursive wraith]

so (0,1), (1,0) and (-1,0) would be on a circle of center?

[sharp smelt]

(0,0)

[tropic saddle]

and your formula gives you what?

[sharp smelt]

(0,1/3)

[tropic saddle]

well your formula also adds +3 for the y coords for some reason

[cursive wraith]

sounds like that's not much of a valid formula

[tropic saddle]

but ignoring that, yes

[cursive wraith]

y_3 instead of y+3

[jolly parrot]

f(why am i here )= I don't know

[sharp smelt]

I supposed I could solve a system of equations to find this point

[sharp smelt]

Yes

[tropic saddle]

which system of equations

from what book is this question

[sharp smelt]

Linear algebra and its applications by Keith nicholson

[tropic saddle]

well and what chapter is it in?

[calm wing]

great, do you see the connection between what i said and your problem?

if three points are not collinear, can they all satisfy a linear equation

[sharp smelt]

Systems of linear equations

[sharp smelt]

Not a single one, no.

[tropic saddle]

so then an approach using a system of equations is a good idea

[sharp smelt]

I was thinking of the following system

Let the centre be $(a,b)$
$$\sqrt{(a-x_1)^2 + (b-y_1)^2} = \sqrt{a-x_2)^2 + (b-y_2)^2}$$
\
and
\
$$\sqrt{(a-x_2)^2 + (b-y_2)^2} = \sqrt{a-x_3)^2 + (b-y_3)^2}$$
\
and
\
$$\sqrt{(a-x_1)^2 + (b-y_1)^2} = \sqrt{a-x_3)^2 + (b-y_3)^2}$$
\

[jolly parrot]

f(why am i here )= I don't know

[sharp smelt]

Now we can square both sides of all equations

[tropic saddle]

you are so much overthinking

traumatised from jee etc

use the form for the circle you are given

imagine the exercise used the word oazgn instead of circle

[cursive wraith]

I suggest you keep it with this formula for circle given

and you can "pick a = 1" when a !=0

[sharp smelt]

Okay. Thanks

[cursive wraith]

if three points lie on a circle of this equation, maybe the best thing to do is find the values of b,c,d in that case

[sharp smelt]

Let me think. Thanks

[cursive wraith]

alr

[sharp smelt]

hmm

I have to make it a linear eq. froma. quadratic

[tropic saddle]

book exercises are usually not trying to trick you. often they really only require the knowledge that the book just taught you

[tropic saddle]

what are the variables you are searching for

[sharp smelt]

b,c,d.

Oh

right

so $(x_1^2+y_1^2)+bx_1+cy_1+d=0$

[jolly parrot]

f(why am i here )= I don't know

[sharp smelt]

and similarly for (x_2,y_2)

and (x_3,y_3)

[cursive wraith]

great

so that's 3 equations for b,c,d

[sharp smelt]

Now, I just solve them

so $$(x_1^2+y_1^2)+bx_1+cy_1+d=0$$
\
$$(x_2^2+y_2^2)+bx_2+cy_2+d=0$$
\
$$(x_3^2+y_3^2)+bx_3+cy_3+d=0$$

[jolly parrot]

f(why am i here )= I don't know

[sharp smelt]

now $(x_i^2+x_j^2)$ is a constant

[jolly parrot]

f(why am i here )= I don't know

[sharp smelt]

so I can club that with d too

to obtain

[tropic saddle]

d is not a constant

[sharp smelt]

oh, rigt

*right

Okay, so tha challenge now is to solve this mess of a system

[tropic saddle]

not quite

its to show the system is solvable

[sharp smelt]

Hmm

No two planes are parallel.

But there's no gurantee that they'll all intersect at one point either

[tropic saddle]

well presumably your book went through some theory to check whether a system is solvable

(solving it of course shows its solvable but is a bit more messy)

[sharp smelt]

I could check that, but idts

[tropic saddle]

I doubt that

[sharp smelt]

Ah

one thing

we did say that if the number of variables is more than the rank, solutions exist

[tropic saddle]

not quite

[sharp smelt]

No of variables is more than the number of equations

[tropic saddle]

no

[sharp smelt]

I'm not too sure, sorry

[tropic saddle]

this is for homogenous systems. we actually had that at the start, but decided to set a=1. we could apply that result but then we would have to argue that a!=0.

thankfully you got a hint earlier on how to do that

[sharp smelt]

OOh

right

if a=0

there AREN'T more equations than variables

so only trivial solns exist

Thanks!

[tropic saddle]

uhh

if we let a be a variable, how many equations and variables do we have

[sharp smelt]

we take a=1,no?

[tropic saddle]

no

[sharp smelt]

we can divid the eqn by a , assuming a!=0

[tropic saddle]

we dont know that yet

[sharp smelt]

And when a=0, it's automatically a line

[tropic saddle]

we want to show that a,b,c,d exist in the first place

[sharp smelt]

So we have 4 equations, 3 vars

[tropic saddle]

no

[sharp smelt]

Uh