help-49

[drowsy briar]

ok

[stable owl]

Np.

[drowsy briar]

i got it now

[stable owl]

What is the logical thing to do then?

Nice.

[drowsy briar]

flip it

thanks man

[stable owl]

Yes.

[drowsy briar]

.close

[midnight plank]

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[midnight jasper]

so I have a function f(x) = (3x)/(x-2)

[midnight jasper]

how would I solve for the inequality f(|x|)<3/2

I know what to do if it is |f(x)|

but when modulus is inside I'm unsure

[golden cloak]

If modulus is inside

Then

Everything of X>0 will be copied to X<0

[midnight jasper]

right

[golden cloak]

With y-axis as symmetry

[midnight jasper]

yeah

[golden cloak]

So do you know f(X) graph?

[midnight jasper]

yeah

[golden cloak]

Then can you proceed from now?

[midnight jasper]

I'll try

thanks

[midnight jasper]

I'm confused with what I'd do algebraically though

[golden cloak]

Then find the values for which the condition satisfy for f(x)

[tribal temple]

(You may find it easier to first set u = |x|, solve the inequality f(u) < 3/2, then from there solve the resulting inequalities you find in terms of u = |x| for x)

[midnight jasper]

so when I set (3u)/(u-2)=3/2 I get u=-2

so the inequality is satisfied when u>-2

but the answer is -2<x<2

is it because the graph is the same for x>0

wait nvm I got it

because of the asymptote

ok

thank you

.clos3

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[hexed elbow]

need help with this

[cobalt swan]

I saw this question somewhere

hm

[hexed elbow]

cengage

textbook

[cobalt swan]

ohh ok

so what is ur doubt here

[hexed elbow]

i dont know how to solve it

my own approach gives me the wrong answer so

[cobalt swan]

what did u try to do?

[hexed elbow]

it might seem stupid but

i divided it into

2 cases

1st case the two profs that want to be in roorkee end up in the same centre in roorkee

second case the two profs are in sep centres in roorkee

[cobalt swan]

no first separate ones who want to go to other, who only want roorkee, anything fine dudes

[hexed elbow]

waiiit

[hexed elbow]

i made a mistake in this

i think i am getting the answer

could you tell me if my approach is right though?

[hexed elbow]

uh i

kiiinda did that

[hexed elbow]

in this method

[cobalt swan]

ohh

[hexed elbow]

could u verify my logic tho

[cobalt swan]

separating will be easier to choose tho

[hexed elbow]

when i considered this case

for the second roorkee centre

[cobalt swan]

ok

[hexed elbow]

i selected 2 out 5 profs

[cobalt swan]

mhm

[hexed elbow]

excluding the 3 that wanted outside centres

[cobalt swan]

mhm

[hexed elbow]

so is that considered as separating?

[cobalt swan]

yes

[hexed elbow]

how would you do it?

[cobalt swan]

ok so my approach would be

2 roorkee
3 outside
5 any

[hexed elbow]

yes

[cobalt swan]

2 centres in roorkee right

[hexed elbow]

mhm

3 outside

[cobalt swan]

i can select any 2 from 5

2C2*5C2

now shuffle

*4!

now exclude siilar cases

/2!2!

so this would be

------------
2!2!```

this

and then

i can shuffle the other left guys

[hexed elbow]

im a lil slow why 2C2 ?

2 centres out of 2 centres selection?

[cobalt swan]

u r selecting 2 roorkee people from group of 2 roorkee

[cobalt swan]

no

[hexed elbow]

ohhhh

okay got it

[cobalt swan]

yep

now 6 are left

choosing 6 from 6 is 6C6

shuffle them in 6!

exclude similar cases in /2!2!2!

-----
2!2!2!```

[hexed elbow]

what would those be

oh wait nvm

distribution no?

[cobalt swan]

basically we r considering a,b and b,a while shuffling

but since both are same we are dividing by 2

[cobalt swan]

ye

[hexed elbow]

ohhh right i had a doubt about this

[cobalt swan]

oh

[hexed elbow]

like if the groups are distinct right

and wehave 1 case as a,b and c,d

2nd as c,d and a,b

[cobalt swan]

both are diff

[hexed elbow]

since the groups are distinct wouldnt this be two different cases?

[cobalt swan]

a,b and c,d
b,a and d,c
these kinds

these we are excluding

[hexed elbow]

oh right

[cobalt swan]

yep

u prep for mains or adv?

[hexed elbow]

adv

[cobalt swan]

ohh same here

[hexed elbow]

im gone bro

1 month wat prep i can do

[cobalt swan]

wat 1 month

[hexed elbow]

how was mains?

[cobalt swan]

10 days left lol

[cobalt swan]

was good wbu?

[hexed elbow]

nah i started right after session 2 only

[hexed elbow]

good enough

[cobalt swan]

oh

[hexed elbow]

anyways

doubt cleared

thanks man

[cobalt swan]

no worries

[hexed elbow]

.close

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[oak epoch]

my chem professor is known for getting questions incorrect. Myself, my friend, and chat gpt got the correct answer as C. My professor listed the answer as A and i was wondering if she is correct.

Consider the following reaction: PCl3 + Cl2 → PCl5. If the percent yield is 66.3%, what mass of PCl5 is actually
formed when 61.3 g of Cl2 reacts with excess PCl3? Molar mass Cl2 = 70.90 g/mol, PCl5 = 208.22 g/mol
a. 40.6 g
b. 92.5 g
c. 119. g
d. 145. g
e. 79.2 g

[cobalt swan]

Im also getting C

[ivory crescent]

Also got C

[oak epoch]

hmm, okay. Its not uncommon as on the first test there was a question asking "what is the element that has the sign Ni" and she listed the correct answer as nitrogen instead of nickel

thanks for the confirmation!

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[normal plover]

A pilot is steering N20E with an airspeed of 400km/h. The plane encounters a wind from the south @ 60km/h. find groundspeed + track

HOW DO I FIND G ??????

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[midnight plank]

@bleak pier Has your question been resolved?

[wanton shore]

@bleak pier is there any value given?

[bleak pier]

No sir

[wanton shore]

Are options available?

@bleak pier

[bleak pier]

Yes options

@wanton shore

[cobalt sigil]

Hey the soln is available online

[bleak pier]

Yes it is available but

[cobalt sigil]

?

[bleak pier]

I do not understand it

[cobalt sigil]

umm

ar = 4^r + 5a(r-1) - 6a(r-2)

4a(r-1) = 4^r + 20a(r-2) - 24a(r-3)

ar = 9a(r-1) - 26a(r-2) +24a(r-3)

@bleak pier https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/08%3A_Recursion_and_Recurrence_Relations/8.03%3A_Recurrence_Relations#:~:text=Find all roots of the,bnank.

Here is a detailed explanation , hope it helps !

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[minor sierra]

When we add vectors we connect them for example
A. B.
--->---->

A and b are two vectors so am confused what is the connection between these vector and why are we allowed to connect them despite them not being related at the given instance

[minor sierra]

I k we can't change their direction but can position and when I tried to solve it I did get an answer but still I don't get how

How does this work exactly

[topaz geyser]

vectors represent a change in quantities

a vector is only defined by a change in quantities

so you can add them or subtract them

[minor sierra]

I see, that makes sense since when we use force on an object it moves forward and changes quantity am I correct?

[topaz geyser]

yes

it can also represent a change in position

or change in velocity

[minor sierra]

But can't we just like take 1 vector from 100 metre away and another at 200 metre and try to add or subtract em since they are both vectors

How would that work

That's why I can't understand on what basis are we doing this

There isn't a connection or use ;-;

Ping me when someone is here !

[topaz geyser]

vectors do not have to be touching in order to add or subtract them

they are not defined by position in space

only by a change in a direction

[minor sierra]

That's interesting now.. nevr thought of it that way

[minor sierra]

Where are vector additions used?

Or subtraction

[topaz geyser]

In statics

for example you may have 2 force vectors <3,6> and <x,2> that act on an object

you want to find what value of x causes the object to not accelerate along the x axis

you would have to add the vectors to do this

<3+x, 6+2> = <0, some constant>

3+x = 0

can also be used to add or subtract torques

anything that requires a magnitude and direction

[minor sierra]

: o

Thanks!

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[nimble matrix]

If there is a cylindrical rod of mass M(uniformly distributed mass) and length L, I need to find radius.
If L length has mass M
Unit length has mass M/L
If pi * r^2 * L has M
Unit volume has mass M/pi * r^2 * L
I equated the mass of unit volume to mass of unit length and I'm getting radius = r=√(1/pi)
Is this correct?

[nimble matrix]

I guess not, never mind

.close

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[last slate]

What's the difference between these 2?

[twin stump]

the 2 eqns are different ways of representing the same thing

[frail burrow]

I think the first one is written using a normalized vector, meaning the r at the front there has magnitude 1.

In the second line, it's written with the vector already having the length r.

[last slate]

Can u just construct a simple problem and solve it with these 2

I would understand better ig

[frail burrow]

Do you know what a vector is?

[last slate]

Yea

[frail burrow]

In this formula, the constant r represents the distance between the two objects.

[last slate]

Yea

[frail burrow]

The r with the little ^ at the top, is a vector with magnitude 1 that represents the direction between the objects

Meaning a vector from one of the objects to the other.

(But magnitude 1)

And in the second line, they replace the r^ with:

$\frac{1}{r} \vec{r}$

[grand pond]

RedstonePlayz09

[frail burrow]

Where this time, this vector r-> represents the direction vector again,

But it has magnitude r this time

[last slate]

Can u just construct a question

[frail burrow]

I don't really have a question to construct

Just try to understand whats the difference

They just changed the vector they multiply it by

[last slate]

Take 2 charges on any points on Cartesian plane
And can u solve it using this

Let's say a charge is at 0,0 and another at 10,0

[frail burrow]

Sorry maybe someone else can. The difference is not so complicated that you'd need a question for it

It's just written with a vector of different magnitude

So there's an extra r constant at the denominator to fix the length

[frail burrow]

Replace r^ in the first formula with this

[last slate]

Oh cool

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

What do you need help with?

13b?

[fossil wagon]

to find the value of a,b and c

[last slate]

Oh

[fossil wagon]

yes

[last slate]

Write it out

[fossil wagon]

wdym

[last slate]

Write out the HCF and LCM tables

[fossil wagon]

you mean HCF and LCM of 8000?

[last slate]

Ya

Write it out

[fossil wagon]

how

[last slate]

Just write it down

[fossil wagon]

can you give me an example for the HCF and I’ll do the same for LCM

[midnight plank]

@fossil wagon Has your question been resolved?

[fathom atlas]

Did you find the value of X?

[fossil wagon]

I think its 8000

[fathom atlas]

No, how did you get to 8000? It should be 11250

[fossil wagon]

how you get that?

[fathom atlas]

HCF (a,b) * LCM (a,b) = a * b

[fossil wagon]

I’m kinda confused

[fathom atlas]

This is a formula

[fossil wagon]

ok?

[last slate]

It’s ok, I’ll just show him

Let me grab my pen

[fossil wagon]

sorry its just that I’m a pretty slow learner. I’m not that smart either 😅

[last slate]

It’s fine

When I first started, I was TERRIBLE

So look at this

LCM means the power of EACH prime must be the highest comparing the 2

So since ONLY X can have 3^something, 3^b=3^6

@fossil wagon understand so far?

[fossil wagon]

yeah

[last slate]

OK

So5^3 and 5^c

The bigger number must end up as 5^4

So what’s C?

[fossil wagon]

4

[last slate]

There you go

Now for a, do HCF

[fossil wagon]

why can’t we do it for LCM as well?

[last slate]

We don’t know how big a is

It’s smaller than or equal to 6

So 2^a is SMALLER than 2^6

And the HCF is 2^1 x 5^3

What’s a?

[fossil wagon]

1

[last slate]

There you go

But make sure to show working or your teachers will be like, “where tf did you get this?”

[fossil wagon]

Ok but can I ask a doubt?

[last slate]

Sure

[fossil wagon]

how do we know whether to use HCF or LCM to compare the ^

[last slate]

LCM is when the power of the unknown must be greater than the power of a number

HCF is when the power is less than or could be equal to

[fossil wagon]

hmm let me have a min to understand this if you don’t mind

[last slate]

Ok

[fossil wagon]

ok I understand better now

[midnight plank]

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[last slate]

Help

Pls

[delicate gale]

intuitively, 2^n + n goes to 2^n if n goes to infinity, from that it should be simple.

[pastel spoke]

As a former Expert of high school calculus, I will answer your question

That’s beyond my level

[last slate]

👍

[delicate gale]

what happens if you change 2^n + n to 2^n?

[carmine void]

You can use 2^n •n >= 2^n+n >= 2^n and apply sandwich theorem

[queen ermine]

Desmos says the answer is 1/2

I mean the top part will always be 1 since nth root of 1 is always 1 (I guess)

so you just gotta find a way to reduce the bottom part

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

.close

[last slate]

this is gonna look bad

[last slate]

but

for preperation for my igcse tomorrow

cos if that the case isnt P and Q also touching the circle

nvm

i just clocked

.close

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[iron stone]

3.14 Find the length of the chord of a circle of radius 20 cm subtended by a central angle of 150.

can someone help me with this? I'm sharing my work with you.

[iron stone]

I've asked this to channel 34 a while ago got no answer and chat got closed

answer is 39

This is the explanation
But i dont get it why dod we use sin
Or what is that C there I havent worked with ‘chords’ yet

I asked this question at 3:35 PM so I can tag helpers. Do not write "We got 15 min rule." I know this rule, I am doing nothing against this rule as the channel got closed without me getting an answer.

<@&286206848099549185>

[cosmic pollen]

what is the question

[slender walrus]

firstly, your diagram is wrong

[iron stone]

this is the question

[iron stone]

its the first thing written

[iron stone]

idk how or what to do tbh

i drew something random

[cosmic pollen]

chord means basically a line that cuts a slice off the edge of the circle

you need to draw that towards the edge not the centre

[slender walrus]

are you familiar with circle geo terms like
subtends,
central angle
etc

[iron stone]

i know central angle but idk subtends

i know how to find the sector area, length of arc. etc.

[cosmic pollen]

you need to disregard the circle really in this situation it is just trigonometry

we use sine because the chord length you are calculating is "opposite" the 75 degree angle

[iron stone]

how am i supposed to use central anle here

angle

[iron stone]

maybe its cos how do we know

[slender walrus]

start with drawing a central angle of 150°

[cosmic pollen]

yeah because the hypotenuse is given, you seek to find the "opposite" edge to the angle, you use sine because it is "opposite" / hypotenuse

[iron stone]

done

[slender walrus]

extend those two lines to the circumference

[iron stone]

ok

[slender walrus]

connect the two intersection points

that will be the desired chord

[iron stone]

Like this?

[cosmic pollen]

the angle 150 is much wider than that but yes

[iron stone]

oh yeah hold on lemme adjust it a bit

maybe thats why i was confused, because the shapes dont matter i always draw like this lol

[cosmic pollen]

the next step is to recall that the two edges coming from the centre are 20cm because they are the radii

[iron stone]

done

[cosmic pollen]

split the angle 150 down the middle so you get two right-angled triangles

[iron stone]

done

[cosmic pollen]

you can now isolate the triangle and notice it has hypotenuse of length 20 and you seek to find the side opposite to angle 75 degrees

[iron stone]

Chord?

Is this red part

[cosmic pollen]

yes

[slender walrus]

if you know cosine rule, you could just apply that directly

[iron stone]

all of them or only one side?

[cosmic pollen]

yeah true

the whole red line is the chord

[iron stone]

alr so im gonna do 20 sin 75

and double it?

[cosmic pollen]

yes

[iron stone]

alright

thank youu

[cosmic pollen]

or you could use the cosine rule c^2 = a^2 + b^2 - 2ab cosx

[iron stone]

i got the answer

ohh

[cosmic pollen]

great

[iron stone]

isnt hypo. cos x = opposite better to use

[cosmic pollen]

that would equal the adjacent side

[iron stone]

yeah i know this rule but i tend not to use it very often with these simple exercises

oh

[cosmic pollen]

cos is adjacent sin is opposite

[iron stone]

sorry i meant sin

yeah

[cosmic pollen]

yeah so it is simpler that way yes

[iron stone]

thanks a lot to both of you

im closing the chat have a great day

.close

[midnight plank]

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[stray raven]

Set theory/set building:

How does one define a set in relation to some of it's sub sets relating to each other?

E = (u, v) ∈ V : (any item in u) not in v

[stray raven]

.close

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[queen ermine]

cursed fr

[hard pewter]

arctanx+arctany=arctan((x+y)/(1-xy)), given that x and y are >0 and xy<1

[cobalt sigil]

arctanx+arctany arctanz=arctan((x+y+z -3xyz)/(1-xy-yz-zx))

[tidal token]

ew

[scenic umbra]

ty, wasnt familiar with these formulas

[cobalt sigil]

welc.

[midnight plank]

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[quasi flame]

Intergrate this

So I trued doing u sub

I did u=1-x^2

[queen ermine]

Whenever you see a root like this

[quasi flame]

Didn't work

[queen ermine]

The answer is usually a trig sub

[quasi flame]

Oh ?

I don't do trig

Is there any other way or is trig the best

[queen ermine]

Trig would probably insta solve this

Other than that..... hmmm...

Like, if x=sin(t) then the bottom part just becomes cos(t), the top part becomes sin(t)^3 and dx becomes cos(t) dt and then..

[quasi flame]

Oh okay
I haven't done trig in a few years but that's useful

Thank you

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[dawn tundra]

I'm stuck on a math question I found online.
It's like this
"Consider a hypothetical mathematical structure, X, devise an algorithm that efficiently computes the maximum prime gap, where a prime gap is the difference between consecutive prime numbers. The twist: your algorithm must be capable of handling an infinite set of prime numbers within X, while also accounting for the potential existence of infinitely many twin primes, i.e., prime pairs with a gap of 2, within X. Furthermore, prove or disprove the existence of a polynomial-time algorithm for this task within the framework of X. But use non-Euclidian and algebraic solutions"
I went and thought it would be a good idea to tackle this in an non non-Euclidian aspect
Sure, let's dive a bit deeper.

To tackle the non-Euclidean aspect, i consider the mathematical structure X and its properties. If X involves non-standard or non-Euclidean geometries, it might affect the distribution of prime numbers and twin primes within that space. I would need to understand how the geometry of X influences the distribution of primes and prime gaps.
Now adding non-Euclidean factors into the formula for computing prime gaps within the structure X:

[ G_n = p_{n+1} - p_n \times f(X) ]

Here's an explanation:

1. ( p_n ) and ( p_{n+1} ): These represent consecutive prime numbers within the structure X, similar to the Euclidean case.

2. ( G_n ): This still denotes the maximum prime gap up to the ( n )-th prime number, calculated by subtracting the ( n )-th prime from the ( (n+1) )-th prime.

3. ( f(X) ): This term represents a function that encapsulates the non-Euclidean factors of the structure X. It could involve parameters or functions derived from the geometry, curvature, topology, or other properties of X that influence prime number distributions by incorporating ( f(X) ).

[grand pond]

ika

[dawn tundra]

But the thing is I don't understand how should I add algebra into this since it has no use. I tried adding it in Manny ways but it didn't work. The formulas I tried were:

1. ( G_n = p_{n+1} - p_n ): This formula represents the traditional Euclidean calculation of prime gaps and does not incorporate any non-Euclidean factors related to the structure X.

2. ( G_n = p_{n+1} \times p_n ): Multiplying consecutive prime numbers would not give a meaningful representation of prime gaps.

3. ( G_n = p_{n+1} - p_n \times e^{f(X)} ): Exponentiating ( f(X) ) introduces an exponential factor that is not typically associated with prime gap calculations.

4. ( G_n = p_{n+1} - p_n + f(X) ): Adding ( f(X) ) directly to the difference between consecutive primes does not accurately reflect the influence of non-Euclidean factors on prime gaps.

5. ( G_n = p_{n+1} \div p_n \times f(X) ): Dividing consecutive primes and then multiplying by ( f(X) ) does not align with traditional prime gap calculations and lacks a clear mathematical basis.

[grand pond]

ika

[midnight plank]

@dawn tundra Has your question been resolved?

[dawn tundra]

@help

<@&286206848099549185>

[chilly vale]

Ya

Can u give me details of sum pls

So I can solve

@dawn tundra

[dawn tundra]

G
n

=p
n+1

−p
n

+f(X)

Bruh

Gn=pn+1−pn+f(X)

Wait I'm just so confused rn

[chilly vale]

Ok

[dawn tundra]

The sum of the sequence up to the ( n )-th term ( G_n ) can be represented as:

[ G_n = \sum_{i=1}^{n} p_i ]

[grand pond]

ika

[chilly vale]

Ok

Ika

To represent the sum of a sequence up to the ( n )-th term ( G_n ), given the individual terms of the sequence are ( p_1, p_2, p_3, \ldots, p_n ), we use the summation notation:

[ G_n = \sum_{i=1}^{n} p_i ]

This notation signifies that ( G_n ) is the sum of the terms ( p_1 ) through ( p_n ) in the sequence. In other words, it is the cumulative sum of the first ( n ) terms.

To expand on this, if we write out the summation explicitly, it would look like this:

[ G_n = p_1 + p_2 + p_3 + \cdots + p_n ]

This formula captures the essence of summing up the terms from the first term ( p_1 ) to the ( n )-th term ( p_n ) of the sequence.

[grand pond]

Daksh_GamerYT

To represent the sum of a sequence up to the \( n \)-th term \( G_n \), given the individual terms of the sequence are \( p_1, p_2, p_3, \ldots, p_n \), we use the summation notation:

\[ G_n = \sum_{i=1}^{n} p_i \]

This notation signifies that \( G_n \) is the sum of the terms \( p_1 \) through \( p_n \) in the sequence. In other words, it is the cumulative sum of the first \( n \) terms. 

To expand on this, if we write out the summation explicitly, it would look like this:

\[ G_n = p_1 + p_2 + p_3 + \cdots + p_n \]

This formula captures the essence of summing up the terms from the first term \( p_1 \) to the \( n \)-th term \( p_n \) of the sequence.

[chilly vale]

To represent the sum of a sequence up to the ( n )-th term ( G_n ), given the individual terms of the sequence are ( p_1, p_2, p_3, \ldots, p_n ), we use the summation notation:

[ G_n = \sum_{i=1}^{n} p_i ]

This notation signifies that ( G_n ) is the sum of the terms ( p_1 ) through ( p_n ) in the sequence. In other words, it is the cumulative sum of the first ( n ) terms.

To expand on this, if we write out the summation explicitly, it would look like this:

[ G_n = p_1 + p_2 + p_3 + \cdots + p_n ]

This formula captures the essence of summing up the terms from the first term ( p_1 ) to the ( n )-th term ( p_n ) of the sequence.

[grand pond]

Daksh_GamerYT

[chilly vale]

This is good?

[dawn tundra]

THANK YOU!

[chilly vale]

Idk if iam right

But I followed math rules

[dawn tundra]

You did. I checked it. THANK YOU!!!

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[burnt vigil]

with the numers 1-7 you can get 5040 seven digit numbers. if every number shuold include 7 different digits. Think that you are writing every number in order of small to large. which numbers holds place 2161

[burnt vigil]

if u dont understand pls ask

i need to do this quickly

[polar mortar]

count, how many numbers start with 1?

[untold harness]

bruh its like 720

[burnt vigil]

is it 7 x 6 x 5 x 4 x 3 x 2 x 1

nvm

no 7

[polar mortar]

right, so there are 720 numbers that start with 1

[burnt vigil]

so 720

yes

[polar mortar]

but there's nothign special about 1. There's also 720 that start with 2 and 3

[untold harness]

yes

[burnt vigil]

yes

[untold harness]

but you should start with how many numbers total

wich is 7 x 6 x5 x4 x3 x2 x1

[polar mortar]

which is 5040, the total number they gave in the question.

[burnt vigil]

yes

[polar mortar]

since there's 720 per starting number, you know the small numbers that start with 1, 2 or 3 there's 720 + 720 + 720 = 2160

[burnt vigil]

what small numbers?

[polar mortar]

numbers that start with 1, numbers that start with 2, etc.

[burnt vigil]

yes

720 x 7

[untold harness]

so ther number 2161 would start with four

[burnt vigil]

ohhhh

[untold harness]

and would be the smallest number that starts with four which is 4123567

[burnt vigil]

thank you guys

i understand now

much better than how my teacher explains

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[carmine patrol]

is this true, that if r = 3^2 and if x = 2^2, y = SQRT 5

[formal blade]

Your original equation is x^2 + y^2 = r^2 ?

[carmine patrol]

yup

[queen ermine]

So if you have a circle of radius 3

[carmine patrol]

ye

[queen ermine]

and you intersect it wih line x = 2

[carmine patrol]

ye

[queen ermine]

you get points root 5

[formal blade]

I think you are mixing up x=2 and x=2^2 for ex

[last slate]

ye ur right

wait\

hol up

[formal blade]

So you have x=2 and r=3 right?

[queen ermine]

Why is there a plus minus

[carmine patrol]

cause its a relation

[last slate]

if 2^2 + y^2 = 3^2 then y = SQRT of 5

[queen ermine]

that root 5 is whack

it works but the way you reasoned it is whack

[carmine patrol]

how are you supposed to reason it? Cause i just figured you could rearrange it

[queen ermine]

yeah well just do 9 - 4

to be 5

and then take the root

[formal blade]

This seems wrong

Am I failing algebra today? Lol

[queen ermine]

don't rewrite 5 as sqrt5 squared

[last slate]

just write it down

[carmine patrol]

yea

[formal blade]

No. Think, what are the solutions to y^2 = 5?

One is sqrt(5) the other is -sqrt(5).

[carmine patrol]

but putting it into an equation would it be 2^2 + sqrt(5)^2 = 3^2? or would it just be five

[queen ermine]

you'd have 4 + 5 = 9

which is true

[carmine patrol]

ye

[queen ermine]

but don't keep the squares when solving

just compute 4 and 9

and then take roots

you'll confuse yourself

[carmine patrol]

yea okay, but on a test or anything would they expect me to write it out as 2^2 + sqrt(5)^2 = 3^2, or do whole numbers?

[formal blade]

You're solving for y right?

[carmine patrol]

yep

[formal blade]

Because on a test they'd want y=blahblah

[carmine patrol]

rather t?

[formal blade]

Rather than the original equation with the value of y plugged in

Sorry phone glitch

[carmine patrol]

so then y would just equal +-sqrt(5)?

[formal blade]

Yes

[carmine patrol]

okay sweet

awesome

[formal blade]

To be clear

The issue people are pointing out to you about your method is on the second to last line

You have y^2 = 5 essentially

[carmine patrol]

yea

[formal blade]

So rather than jumping to something weird like y^2= (+/-sqrt(5))^2 they're saying you can take the positive and negative root of both sides of y^2 = 5 to isolate y

[carmine patrol]

so ultimately
-use whole numbers
-solve
-take the roots?

[formal blade]

Well sometimes you might not have whole numbers but it's easier to try and keep things in terms of them I guess.

[carmine patrol]

mk

thank you sm

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[regal kelp]

Hi, you can solve this equation using the inverse matrix method (if possible with a step-by-step explanation)

[midnight plank]

Please don't occupy multiple help channels.

[surreal moon]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[surreal moon]

Stay in one channel please.

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[errant shale]

Need help, dont know how to do this

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[dense quarry]

hello

[dense quarry]

need help solving this, step by step

daily practice, this is not a test.

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[burnt flame]

try labeling the points at the stations and satellites. write down what you know, and the things you're trying to solve for

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[dense quarry]

cholo

[dense quarry]

a) The signal travels approximately 48,136 miles.
b) The technician would fly approximately 599 miles along the surface of the Earth.

Double checking, are my answers correct?

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[dense quarry]

<@&286206848099549185>

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[obsidian hedge]

how does a^d - 1 divides a^m - 1 imply d divides m?

[midnight plank]

@obsidian hedge Has your question been resolved?

[tidal token]

i cant prove it but i can give u examples: $$\frac{a^2-1}{a-1} = a+1$$ $$\frac{a^3-1}{a-1} = a^2+a+1$$ $$\frac{a^3-1}{a^2-1} = \frac{a^2+a+1}{a+1}$$ $$\frac{a^8-1}{a^4-1} = a^4+1$$

[grand pond]

Obotron

[tidal token]

you only get integer quotients when d divides m

[obsidian hedge]

yeah, id like a more rigorous proof tho

its easy to see that $(a^m - 1) \mid (a^{mk}-1)$ for all m,k. How would you prove the other direction tho? that $(a^m - 1) \mid (a^{n}-1) \Rightarrow n = mk$?

[grand pond]

SirGareth

[tidal token]

I'm not sure. i was just trying proof by contradiction and idk hwo to continue

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[dawn prism]

I have to find the total derivative of f: R^n to R with f(x) =||x||^2 , ||x|| is the euclidean norm

[dawn prism]

f(x) = f(xo) + A(x-xo) + o(x-x0)
A is linear and continuous. I get A(x) = 2<x,x0>.
What is the total derivative of f?

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[granite bluff]

Im so confused

[midnight plank]

Please don't occupy multiple help channels.

[granite bluff]

wait i cant see other help channels

oh nvm

.close

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[dawn prism]

I have to show that f: R^m → R^n is totally differentiable
K>0, a>1

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[inland patio]

I'm reading about discrete distributions and the probability mass function. Why is it that a probability mass function is non-zero for at most a countable number of values?

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[burnt flame]

contact modmail

@shadow scaffold

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[fallow rune]

I don’t know how to work with cords all that well, I just know that it’s usually trigonometry. I was thinking that, since AE is 2 greater than AB, you would subtract 2 from each side of BC to get DE. But since 14 isn’t an answer, it must be more complicated than I am making it out to be

[slender walrus]

this isn't trig

[fallow rune]

I’m not sure what it is lol

[slender walrus]

it's circle geomrtry

[fallow rune]

Unfortunately my Geo teacher didn’t actually teach us anything besides planes and converse/inverse

[slender walrus]

look up circle geometry theorems

[fallow rune]

I didn’t think there would be something like that so simply online because these types of questions are not good examples for a beginner like me

I looked that up and it did not help :/

[slender walrus]

what came up when you searched what I recommended

you should have a list of around 10 theorems

depending on the resource

[fallow rune]

I found a lot of intersecting lines and triangles theorems but nothing about two rays/chords starting at a single point outside the triangle

circle* lol

I was looking at the image charts for the theorems and one did look similar but it was lines that were touching the circumference instead of going through the circle

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[slender walrus]

this is missing a few

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[plush pebble]

yo can someone find the derivative of this

[wanton quest]

Use the chain rule three times

[slender walrus]

apply it one level at a time if it's too intimidating

[plush pebble]

u don't have to fully simplify btw

[plush pebble]

yeah thats what I figured

I just want a non simplified answer

[slender walrus]

substitute as well if you prefer that

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[sturdy cave]

i got 1/5 of these and im not sure what the issue was

[midnight plank]

Please don't occupy multiple help channels.

[sturdy cave]

oops

.close

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[half meteor]

Is my work correct?

Mathematical induction

[formal blade]

The line as "2+blahblah" is true but probably requires more explanation.

The rest seems fine also, but I would avoid saying things like P(n)=blahblah when you are also using equivalence rather than equality elsewhere.

The notation P(n) equiv blah blah might be a little more symbol heavy than some people prefer too.

[half meteor]

Is it okay if i just removed it

?

Then the rest stays the same

[formal blade]

I think you need it. Do you see why it is true?

[half meteor]

Yeah

But, idk how to elaborate it further

[formal blade]

The 2+[k(k-1)...1] < blahblah stuff is the line I'm talking about fwiw.

Well how do you know k is not 1 for example?

That is the ambiguity I think might be worth pointing out.

[half meteor]

oh i add k is an element of the set {4, 5, 6…}

[formal blade]

Think of your base case

[half meteor]

?

[formal blade]

Yeah, our base case starts from 4.

It's not a huge issue or anything, just easy for a tired brain to not realize at first why that line is true lol.

[half meteor]

i see i see

is that all i needed to add?

[formal blade]

Well that's all I see. I get what you're going for and the ideas seem correct. A good rule of thumb is to ask yourself "if I handed this to a classmate and walked away could they figure out the proof from what I have?"

[half meteor]

ooooo okay okay

[formal blade]

Ideally you want it to be something they could easily figure out too. Rather than sit for hours scratching their head to figure out.

[half meteor]

alright, got it

Thanks so much! @formal blade

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[bleak pier]

i am getting answer 0 is this correct?

[pine thicket]

how did you get your answer?

[bleak pier]

i expand the arctanx

then i put x=0

[pine thicket]

what do you mean by expand the arctan

[bleak pier]

x-x^3/3+x^5/5+...

[pine thicket]

the polynomial expansion keeps going, its infinite

[bleak pier]

yes

so when we derivate it 9 times

we will get same

and when we set x=0 all terms will be 0?

[pine thicket]

which term would not be 0 in the ninth derivative?

[bleak pier]

i guess no term

[bleak pier]

ohh 9th derivative

i was integrating it

[pine thicket]

yeah, the notation was signifying the ninth derivative

[bleak pier]

my mistake

there will be some terms

8!

[pine thicket]

yes

[bleak pier]

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[dawn trail]

Where have i miscalculated

srry if i skipped steps im on mouse and didnt want to write it all out

[cobalt elk]

But this is correct

[hybrid crane]

The missing factor of a 1/6 is one thing.
The second is that just have to find the limit

• 1/x as x approaches infinity
  x is very large compared to the 1
  It's "easy" to see 1/x tends to 0

[cobalt elk]

And -1/infty is 0

[dawn trail]

ohh I should mention I tried to submit it as Diverging

and that was wrong ^

I searched up 1/infinity and it said undefined

[hybrid crane]

It doesn't diverge

[dawn trail]

ohhh

[cobalt elk]

Well if the denominator becomes really big (e.g 1000000). Then the number will become very small

[dawn trail]

Ohhhh I see so we can represent it as 0?

[cobalt elk]

So for approaching Infinity it's 0

Ye

[dawn trail]

I see ty!!

StrangeQuark thank you for putting so much thought into a reply 😭

[hybrid crane]

BTW, subbing in infinity will lead to mistakes if you don't know enough about limits

[hybrid crane]

No problem

[dawn trail]

Ill review the infinity limit rules! ty

[cobalt elk]

Yeah that is very true. Mathematically you aren't even allowed to calculate with infinity

[dawn trail]

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[lunar dome]

Help

[topaz geyser]

,r ccw

,rccw

[bold beacon]

You don't seem to have made any attempt

[lunar dome]

I did but i rubbed it out

I didn’t understand

[bold beacon]

So the first thing is how things are defined

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[lunar dome]

Ok so you use dot product right?

[bold beacon]

no

[lunar dome]

Then I have no idea

[bold beacon]

Maybe I will have see full working out to determine

[lunar dome]

Ok

[bold beacon]

,rotate

[lunar dome]

This is all i know

[bold beacon]

Yes you want use dot product my apologies

[lunar dome]

Ohk

[bold beacon]

How can you <APB?

[lunar dome]

Cos rule?

[bold beacon]

dot product. You rearrange the dot product

[lunar dome]

Oh right

I did that in my working no?

[bold beacon]

yep. Now you make phi the subject

[lunar dome]

What’s phi

[bold beacon]

You used the symbol phi

[lunar dome]

Oh right

,rotate

[lunar dome]

Ok is this right?

[bold beacon]

,rotate

[bold beacon]

I am having trouble understanding your work

I take back what I said about making phi the subject. We will need to do this but not now

I do not know. What I would do is check a textbook like James Stewart and see if there similar examples

[midnight plank]

@lunar dome Has your question been resolved?

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[lunar dome]

Is a right?

[wary thorn]

, rotate

[cobalt swan]

3/root(10) is right idk how u got 3/root(2) from there

[midnight plank]

@lunar dome Has your question been resolved?

[lunar dome]

The answer is (1.5, 1.5)

@cobalt swan

[cobalt swan]

which one

oh ok he asked to find vector shit

[rain wasp]

ignore below

[midnight plank]

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.

[cobalt swan]

lol wot she asked

[rain wasp]

oh fuck sorry

[cobalt swan]

lol

[rain wasp]

a bit tired so misread

[rain wasp]

do you need to check your work or?

[cobalt swan]

so he asked a on b
so the final vector will be parallel to b
magnitude will be equal to the found 3/root(10)

[cobalt swan]

u did half right but thats not final ans @lunar dome

[lunar dome]

Ohk

[lunar dome]

Wait can u like fully tell me what it even means if a is projected on b

Our tutor taught us formulas but didn’t explain what it means

[cobalt swan]

https://www.cuemath.com/geometry/projection-vector/

check this out @lunar dome

[lunar dome]

Ty

[midnight plank]

@lunar dome Has your question been resolved?

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[iron stone]

hey this is going to sound weird but can someone teach me how "bearing" thing in trigonometry works? English is not my first language and I've came across this question and I don't know what bearing is. Other than bearing I got the whole question incorrect :/

[iron stone]

highlighted part is the question and there is the solution below but

1. I don't get how we got huge numbers like 529.1 or why did we use sec at all? Is that the first thing written in 3 lines a formula?

2. how does "bearing" work? "while B, due south of C, bears S25°10'E from A" what does this mean?

[summer remnant]

Bearing is basically the clockwise angle measured between them

So like starting from B, move clockwise till you reach A

[iron stone]

this is what is written online as well but how i draw it here?

[iron stone]

but that doesnt make B 25°10'

or the thing between A and the line

[summer remnant]

For the first part of your question, the question has provided us with one side and two angles(90⁰ too, indirectly), so we can use trig to find it out... I don't quite get why you find the answer huge

[iron stone]

my solution is less than the shown. the solution i provided belongs to the book

[iron stone]

this is what i did tbh

here let me share my work

[summer remnant]

Kk

[iron stone]

the parts written in black is my solution and red + blue is notes sort of thing

[summer remnant]

,calc 1/6

[grand pond]

Result:

0.16666666666667

[summer remnant]

You switched up the angles, in the fig, ABC is 25.2⁰

[iron stone]

oh yeah

okay so i still couldnt figure out bearing here tbh so starting from B how do i get to A

it doesnt make sense or create a specific circle

Is this it?

[summer remnant]

Yeah it's good

But note that in the ans they have mentioned W

[iron stone]

so b is 360 - the bearing angle?

[summer remnant]

You could do that too
In the ans they went the opposite way(anticlockwise) but mentioned W too

Either way, it should work

[summer remnant]

That's why the magnitude of the angle remains the same

[iron stone]

but wait alr i just chekced the solution again i think my answer is wrong tho

[iron stone]

if bearing from A to B is S25°10'E then why Angle B is 25°10'

shouldnt it be 360 - the angle ( 25.2 apprx. )

<@&286206848099549185>

[slender walrus]

the green angle isn't 25°10'

the red angle is 25°10'

the true bearing of A from B will be

360 - the angle ( 25.2 approx).
but they're not using true bearings here

[iron stone]

it says from A

how do i draw it then

its confusing

[slender walrus]

[North or South] [rotation towards] [East or West]

[iron stone]

North and west?

this is what i did with the green thing though

oh no

i did from B to A

[slender walrus]

should've mentioned that the rotation would be acute here

[iron stone]

yeah no idk what u mean by acute

[slender walrus]

between 0 and 90°

[iron stone]

oh

okay

[iron stone]

it says S25°10'E does it mean from south to east?

[slender walrus]

yes

[iron stone]

oh okay

[slender walrus]

here, you're missing the W to indicate the rotation towards the west direction

[iron stone]

but the rotation is towards east?

isn't it?

south to east we did

[slender walrus]

that's what you were given

the bearing of B from A is S25³10'E

[iron stone]

yeah

[slender walrus]

and you're trying to find bearing of A from B

which is the blue part i've marked

[iron stone]

oh uhm i thought the green part would be from A to B

because in the examples it shows like that

[slender walrus]

no

(don't get your to and from mixed up)
again as mentioned earlier that would be the true bearing of A from B

[slender walrus]

where the angle is measured clockwise from north

however here, they bearing notation they use (which is also accepted)
[North or South] [acute angle of rotation towards] [East or West]

[iron stone]

so we got 2 ways to solve it?

im new to this term sorry if im making you confused

[slender walrus]

i.e. here from the** North** of B,
you'd rotate 25°10' in the West direction to get to A
thus that bearing could be represented as
N 25°10' W

as a true bearing, that'd be what you mentioned earlier
360° - 25°10'

[iron stone]

oh okay if it was North of B to East that would be the true as well

[slender walrus]

yes

N 42° E
would be a true bearing of 42°

[iron stone]

alright

thank you so much

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[spark nebula]

why did they cancel the second line

[still zinc]

they operated on columns

[midnight plank]

@spark nebula Has your question been resolved?

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[bleak forge]

Question 17 (6 Marks)
The Rundle Mall Balls are a sculpture in Adelaide that are made up of two, identical, chrome-plated spheres stacked on top of each other. They are starting to lose their shine and are required to be re-plated with chrome. 1 L of chrome covers 7.5 m^2. 16 L was purchased, which was thought to be exactly enough to coat both balls. However, they accidentally used the diameter instead of the radius when calculating the surface area of the sculpture. To the nearest litre, how much extra chrome will they be left with?

[bleak forge]

I just dont know where to start

7.5m^2 x 16 = 4/3Pi d^2

Thats as far in the working as i can get

120m^2 = 4/3 Pi d^2

<@&286206848099549185>

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@bleak forge Has your question been resolved?

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[last slate]

So

[last slate]

This

Do we find AX

Or AN as a whole

[midnight plank]

@last slate Has your question been resolved?

[torn compass]

find AX and AN and then find the ratio between them , isnt that what question asks?

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[boreal geyser]

How do I set this up??

[slender walrus]

power chain rule

[boreal geyser]

so 4?

Move the 3 to the front

take deriv of x^2

and put the original function back

[still nacelle]

yh

[boreal geyser]

And what about this one?

How do I set this up

[still nacelle]

what have you tried?

[slender walrus]

first differentiate using chain rule

[boreal geyser]

product rule seems appealing here

[slender walrus]

no

there is no quotient/fraction here
why would you think quotient rule

[boreal geyser]

wait product

my bad

[slender walrus]

there is no product here either

this is a function composition

apply chain rule directly

lagrange form of the chain rule tells you directly what the derivative will be

[boreal geyser]

h'(x)=f'(g(x))g'(x)

[slender walrus]

yes

then plug in 3 and use the given values

[boreal geyser]

so we put in h'(x)=3(2(3)6

wait

no

[slender walrus]

also your question uses capital F, not h, so try sticking with that

first replace all x with 3

[boreal geyser]

F'(3)=f'(g(3))g'(3)